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Calculus Refresher, version 2008.4

©1997-2008, Paul Garrett, garrett@math.umn.edu

http.//www. math. umn.edu/garrett/

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Contents

(1) Introduction

(2) Inequalities

(3) Domain of functions

(4) Lines (and other items in Analytic Geometry) (5) Elementary limits

(6) Limits with cancellation

(7) Limits at infinity

(8) Limits of exponential functions at ifinity

(9) The idea of the derivative of a function

(10) Derivatives of polynomials

(11) More general power functions

(12) Quotient rule

(13) Product Rule

(14) Chain rule

(15) Tangent and Normal Lines

(16) Critical points, monotone increase and decrease

(17) Minimization and Maximization

(18) Local minima and maxima (First Derivative Test)

(19) An algebra trick

(20) Linear approximations: approximation by differentials (21) Implicit differentiation

(22) Related rates

(23) Intermediate Value Theorem, location of roots

(24) Newton's method

(25) Derivatives of transcendental functions

(26) L'Hospital's rule

(27) Exponential growth and decay: a differential equation (28) The second and higher derivatives

(29) Inflection points, concavity upward and downward

(30) Another differential equation: projectile motion

(31) Graphing rational functions, asymptotes

(32) Basic integration formulas

(33) The simplest substitutions

(34) Substitutions

(35) Area and definite integrals

(36) Lengths of Curves

(37) Numerical integration

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(38) Averages and Weighted Averages

(39) Centers of Mass (Centroids)

(40) Volumes by Cross Sections

(41) Solids of Revolution

(42) Surfaces of Revolution

(43) Integration by parts

(44) Partial Fractions

(45) Trigonometric Integrals

(46) Trigonometric Substitutions

(47) Historical and theoretical comments: Mean Value Theorem (48) Taylor polynomials: formulas

(49) Classic examples of Taylor polynomials

(50) Computational tricks regarding Taylor polynomials

(51) Prototypes: More serious questions about Taylor polynomials (52) Determining Tolerance/Error

(53) How large an interval with given tolerance?

(54) Achieving desired tolerance on desired interval

(55) Integrating Taylor polynomials: first example

(56) Integrating the error term: example

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1. Introduction

The usual trouble that people have with 'calculus' (not counting general math phobias) is with algebra, not to mention arithmetic and other more elementary things.

Calculus itself just involves two new processes, differentiation and integration, and applications of these new things to solution of problems that would have been impossible otherwise.

Some things which were very important when calculators and computers didn't exist are not so important now. Some things are just as important. Some things are more important. Some things are important but with a different emphasis.

At the same time, the essential ideas of much of calculus can be very well illustrated without using calcu- lators at all! (Some not, too).

Likewise, many essential ideas of calculus can be very well illustrated without getting embroiled in awful algebra or arithmetic, not to mention trigonometry.

At the same time, study of calculus makes clear how important it is to be able to do the necessary algebra and arithmetic, whether by calculator or by hand.

2. Inequalities

It is worth reviewing some elementary but important points:

First, a person must remember that the only way for a product of numbers to be zero is that one or more of the individual numbers be zero. As silly as this may seem, it is indispensable.

Next, there is the collection of slogans:

positive times positive is positive

negative times negative is positive

negative times positive is negative

positive times negative is negative

Or, more cutely: the product of two numbers of the same sign is positive, while the product of two num- bers of opposite signs is negative.

Extending this just a little: for a product of real numbers to be positive, the number of negative ones must be even. If the number of negative ones is odd then the product is negative. And, of course, if there are any zeros, then the product is zero.

Solving inequalities: This can be very hard in greatest generality, but there are some kinds of problems

that are very ζdo-able'. One important class contains problems like Solve:

5 (x-1)(x+4)(x-2)(x+3)<0

That is, we are asking where a polynomial is negative (or we could ask where it's positive, too). One im- portant point is that the polynomial is already factored: to solve this problem we need to have the polyno- mial factored, and if it isn't already factored this can be a lot of additional work. There are many ways to format the solution to such a problem, and we just choose one, which does have the merit of being more efficient than many.

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We put the roots of the polynomial

P(x)=5(x-1)(x+4)(x-2)(x+3)=5(x-1)(x-(-4))(x-2)(x-(-3))

in order: in this case, the roots are 1, -4,2, -3, which we put in order (from left to right)

. . . <-4<-3<1<2<...

The roots of the polynomial P break the numberline into the intervals

(-, -4), (-4, -3), (-3,1), (1,2), (2, +)

On each of these intervals the polynomial is either positive all the time, or negative all the time, since if it were positive at one point and negative at another then it would have to be zero at some intermediate point!

For input x to the right (larger than) all the roots, all the factors x+4, x+3, x-1, x-2 are positive, and the number 5 in front also happens to be positive. Therefore, on the interval (2, +) the polynomial P(x) is positive.

Next, moving across the root 2 to the interval (1, 2) , we see that the factor x-2 changes sign from posi- tive to negative, while all the other factors x-1, x+3, and x+4 do not change sign. (After all, if they would have done so, then they would have had to be 0 at some intermediate point, but they weren't, since we know where they are zero Of course the 5 in front stays the same sign. Therefore, since the func- tion was positive on (2, +) and just one factor changed sign in crossing over the point 2, the function is negative on (1, 2) .

Similarly, moving across the root 1 to the interval (-3,1) , we see that the factor x-1 changes sign from positive to negative, while all the other factors x-2, x+3, and x+4 do not change sign. (After all, if they would have done so, then they would have had to be 0 at some intermediate point). The 5 in front

stays the same sign. Therefore, since the function was negative on (1, 2) and just one factor changed sign in crossing over the point 1, the function is positive on (-3,1) .

Similarly, moving across the root -3 to the interval (-4, -3) , we see that the factor x+3=x-(-3)

changes sign from positive to negative, while all the other factors x-2, x-1, and x+4 do not change sign. (If they would have done so, then they would have had to be 0 at some intermediate point) . The

5 in front stays the same sign. Therefore, since the function was positive on (-3,1) and just one factor

changed sign in crossing over the point -3, the function is negative on (-4, -3) .

Last, moving across the root -4 to the interval (-, -4) , we see that the factor x+4=x-(-4) changes sign from positive to negative, while all the other factors x-2, x-1, and x+3 do not change sign. (If they would have done so, then they would have had to be 0 at some intermediate point). The 5 in front

stays the same sign. Therefore, since the function was negative on (-4, -3) and just one factor changed

sign in crossing over the point -4, the function is positive on (-, -4) .

In summary, we have

P(x)=5(x-1)(x+4)(x-2)(x+3)>0 on (2, +)

P(x)=5(x-1)(x+4)(x-2)(x+3)<0 on (1, 2)

P(x)=5(x-1)(x+4)(x-2)(x+3)>0 on (-3,1)

P(x)=5(x-1)(x+4)(x-2)(x+3)<0 on (-4, -3)

P(x)=5(x-1)(x+4)(x-2)(x+3)>0 on (-, -4)

In particular, P(x)<0 on the union

(1, 2)(-4, -3)

of the intervals (1, 2) and (-4, -3) . That's it.

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As another example, let's see on which intervals

P(x)=-3(1+x2)(x2-4)(x2-2x+1)

is positive and and on which it's negative. We have to factor it a bit more: recall that we have nice facts

x2-a2=(x-a)(x+a)=(x-a)(x-(-a))

x2-2ax+a2=(x-a)(x-a)

so that we get

P(x)=-3(1+x2)(x-2)(x+2)(x-1)(x-1)

It is important to note that the equation x2+1=0 has no real roots, since the square of any real number is non-negative. Thus, we can't factor any further than this over the real numbers. That is, the roots of P, in order, are

-2<<1 (twice!) <2

These numbers break the real line up into the intervals

(-, -2), (-2,1), (1,2), (2, +)

For x larger than all the roots (meaning x>2) all the factors x+2, x-1, x-1, x-2 are positive, while the factor of -3 in front is negative. Thus, on the interval (2, +)P(x) is negative.

Next, moving across the root 2 to the interval (1, 2) , we see that the factor x-2 changes sign from pos- itive to negative, while all the other factors 1+x2, (x-1)2, and x+2 do not change sign. (After all,

if they would have done so, then they would have be 0 at some intermediate point, but they aren't). The -3 in front stays the same sign. Therefore, since the function was negative on (2, +) and just one factor changed sign in crossing over the point 2, the function is positive on (1, 2) .

A new feature in this example is that the root 1 occurs twice in the factorization, so that crossing over the root 1 from the interval (1, 2) to the interval (-2,1) really means crossing over two roots. That is, two

changes of sign means no changes of sign, in effect. And the other factors (1+x2), x+2, x-2 do not change sign, and the -3 does not change sign, so since P(x) was positive on (1, 2) it is still positive on

(-2,1) . (The rest of this example is the same as the first example).

Again, the point is that each time a root of the polynomial is crossed over, the polynomial changes sign. So if two are crossed at once (if there is a double root) then there is really no change in sign. If three roots are crossed at once, then the effect is to change sign.

Generally, if an even number of roots are crossed-over, then there is no change in sign, while if an odd number of roots are crossed-over then there is a change in sign.

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#2.1 Find the intervals on which f(x)=x(x-1)(x+1) is positive, and the intervals on which it is negative.

#2.2 Find the intervals on which f(x)=(3x-2)(x-1)(x+1) is positive, and the intervals on which it is negative.

#2.3 Find the intervals on which f(x)=(3x-2)(3-x)(x+1) is positive, and the intervals on which it is negative.

3. Domain of functions

A function f is a procedure or process which converts input to output in some way. A traditional mathe- matics name for the input is argument, but this certainly is confusing when compared with ordinary En- glish usage.

The collection of all 'legal' 'reasonable' or 'sensible' inputs is called the domain of the function. The col- lection of all possible outputs is the range. (Contrary to the impression some books might give, it can be very difficult to figure out all possible outputs!)

The question 'What's the domain of this function?' is usually not what it appears to be. For one thing, if we are being formal, then a function hasn't even been described if its domain hasn't been described!

What is really meant, usually, is something far less mysterious. The question usually really is 'What

numbers can be used as inputs to this function without anything bad happening?'.

For our purposes, 'something bad happening' just refers to one of

trying to take the square root of a negative number

trying to take a logarithm of a negative number

trying to divide by zero

trying to find arc-cosine or arc-sine of a number bigger than 1 or less than -1

Of course, dividing by zero is the worst of these, but as long as we insist that everything be real numbers (rather than complex numbers) we can't do the other things either.

For example, what is the domain of the function

f(x)=x2-1?

Well, what could go wrong here? No division is indicated at all, so there is no risk of dividing by 0. But we are taking a square root, so we must insist that x2-10 to avoid having complex numbers come up. That is, a preliminary description of the 'domain' of this function is that it is the set of real numbers x so that x2-10.

But we can be clearer than this: we know how to solve such inequalities. Often it's simplest to see what to exclude rather than include: here we want to exclude from the domain any numbers x so that x2-1<0 from the domain.

We recognize that we can factor

x2-1=(x-1)(x+1)=(x-1)(x-(-1))

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This is negative exactly on the interval (-1,1) , so this is the interval we must prohibit in order to have just the domain of the function. That is, the domain is the union of two intervals:

(-, -1][1, +)

#3.4 Find the domain of the function

f(x)=x-2x2+x-2

That is, find the largest subset of the real line on which this formula can be evaluated meaningfully.

#3.5 Find the domain of the function

f(x)=x-2x2+x-2

#3.6 Find the domain of the function

f(x)=x(x-1)(x+1)

4. Lin es (and other items in Ana/yTic Geome try)

Let's review some basic analytic geometry: this is description of geometric objects by numbers and by algebra.

The first thing is that we have to pick a special point, the origin, from which we'll measure everything else. Then, implicitly, we need to choose a unit of measure for distances, but this is indeed usually only implicit, so we don't worry about it.

The second step is that points are described by ordered pairs of numbers: the first of the two numbers

tells how far to the right horizontally the point is from the origin (and negative means go left instead

of right), and the second of the two numbers tells how far up from the origin the point is (and negative

means go down instead of up). The first number is the horizontal coordinate and the second is the ver- tical coordinate. The old-fashioned names abscissa and ordinate also are used sometimes.

Often the horizontal coordinate is called the x-coordinate, and often the vertical coordinate is called the y- coordinate, but the letters x, y can be used for many other purposes as well, so don't rely on this labelling.'

The next idea is that an equation can describe a curve. It is important to be a little careful with use of

language here: for example, a correct assertion is

The set of points (x, y) so that x2+y2=1 is a circle.

It is not strictly correct to say that x2+y2=1 is a circle, mostly because an equation is not a circle, even though it may describe a circle. And conceivably the x, y might be being used for something other than horizontal and vertical coordinates. Still, very often the language is shortened so that the phrase ζThe set of points (x, y) so that' is omitted. Just be careful.

The simplest curves are lines. The main things to remember are:

Slope of a line is rise over run, meaning vertical change divided by horizontal change (moving from left to right in the usual coordinate system).

The equation of a line passing through a point (xo, y0) and having slope m can be written (in so-called

point-slope form)

y=m(x-xo)+y0 or y-y0=m(x-xo)

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The equation of the line passing through two points (x1, y1), (x2, y2) can be written (in so-called two- point form) as

y=y1-y2x1-x2(x-x1)+y1

...unless x=x2, in which case the two points are aligned vertically, and the line can't be written that way. Instead, the description of a vertical line through a point with horizontal coordinate x1 is just

x=x1

Of course, the two-point form can be derived from the point-slope form, since the slope m of a line through two points (x1, y1), (x2, y2) is that possibly irritating expression which occurs above:

m=y1-y2x1-x2

And now is maybe a good time to point out that there is nothing sacred about the horizontal coordinate being called 'x' and the vertical coordinate 'y'. Very often these do happen to be the names, but it can be otherwise, so just pay attention.

#4.7 Write the equation for the line passing through the two points (1, 2) and (3, 8) .

#4.8 Write the equation for the line passing through the two points (-1,2) and (3, 8) . 4.9 Write the equation for the line passing through the point (1, 2) with slope 3.

#4.10 Write the equation for the line passing through the point (11, -5) with slope -1.

5. Ele mentary limits

The idea of limit is intended to be merely a slight extension of our intuition. The so-called , δ-definition was invented after people had been doing calculus for hundreds of years, in response to certain relatively pathological technical difficulties. For quite a while, we will be entirely concerned with situations in which we can either 'directly' see the value of a limit by plugging the limit value in, or where we transform the expression into one where we can just plug in.

So long as we are dealing with functions no more complicated than polynomials, most limits are easy to understand: for example,

limx34x2+3x-7=4(3)2+3(3)-7=38

limx34x2+3x-72-x2=4(3)2+3(3)-72-(3)2=38-7

The point is that we just substituted the '3' in and nothing bad happened. This is the way people evalu- ated easy limits for hundreds of years, and should always be the first thing a person does, just to see what happens.

#5.11 Find limx52x2-3x+4.

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#5.12 Find limx2x+1x2+3. #5.13 Find limx1x+1.

6. Limits with cance//a tion

But sometimes things 'blow up' when the limit number is substituted:

limx3x2-9x-3=00?????

Ick. This is not good. However, in this example, as in many examples, doing a bit of simplifying algebra first gets rid of the factors in the numerator and denominator which cause them to vanish:

limx3x2-9x-3=limx3(x-3)(x+3)x-3=limx3(x+3)1=(3+3)1=6

Here at the very end we did just plug in, after all.

The lesson here is that some of those darn algebra tricks ('identities') are helpful, after all. If you have a 'bad' limit, always look for some cancellation of factors in the numerator and denominator.

In fact, for hundreds of years people only evaluated limits in this style! After all, human beings can't re- ally execute ifinite limiting processes, and so on.

#6.14 Find limx2x-2x2-4 6.15 Find limx3x2-9x-3 6.16 Find limx3x2x-3

7. Limits a T infinity

Next, let's consider

limx2x+35-x

The hazard here is that is not a number that we can do arithmetic with in the normal way. Don't even try it. So we can't really just 'plug in' to the expression to see what we get.

On the other hand, what we really mean anyway is not that x 'becomes ifinite' in some mystical sense, but rather that it just 'gets larger and larger'. In this context, the crucial observation is that, as x gets larger and larger, 1/x gets smaller and smaller (going to 0). Thus, just based on what we want this all to mean,

limx1x=0

limx1x2=0

limx1x3=0

and so on.

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This is the essential idea for evaluating simple kinds of limits as x: rearrange the whole thing so that everything is expressed in terms of 1/x instead of x, and then realize that

limx is the same as lim1x0

So, in the example above, divide numerator and denominator both by the largest power of x appearing

anywhere:

limx2x+35-x=limx2+3x5x-1=limy02+3y5y-1=2+3050-1=-2

The point is that we called 1/x by a new name, 'y', and rewrote the original limit as x as a limit as y0. Since 0 is a genuine number that we can do arithmetic with, this brought us back to ordinary everyday arithmetic. Of course, it was necessary to rewrite the thing we were taking the limit of in terms of 1/x (renamed 'y').

Notice that this is an example of a situation where we used the letter 'y' for something other than the name or value of the vertical coordinate.

#7.17 Find limxx+1x2+3.

#7.18 Find limxx2+3x+1.

#7.19 Find limxx2+33x2+x+1. 7.20 Find limx1-x25x2+x+1.

8. Limits of exponential functions a τ infinity

It is important to appreciate the behavior of exponential functions as the input to them becomes a large positive number, or a large negative number. This behavior is different from the behavior of polynomials or rational functions, which behave similarly for large inputs regardless of whether the input is large pos- itive or large negative. By contrast, for exponential functions, the behavior is radically different for large positive or large negative.

As a reminder and an explanation, let's remember that exponential notation started out simply as an ab-

breviation: for positive integer n,

2n=2×2×2×...×2 ( n factors)

10n=10×10×10×...×10 ( n factors)

(12)n=(12)×(12)×(12)×...×(12) ( n factors)

From this idea it's not hard to understand the fundamental properties of exponents (they're not laws at all):

( m+n factors)

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and also

at least for positive integers m, n. Even though we can only easily see that these properties are true when the exponents are positive integers, the extended notation is guaranteed (by its meaning, not by law) to follow the same rules.

Use of other numbers in the exponent is something that came later, and is also just an abbreviation, which happily was arranged to match the more intuitive simpler version. For example,

a-1=1a

and (as consequences)

a-n=an×(-1)=(an)-1=1an

(whether n is positive or not). Just to check one example of consistency with the properties above, notice that

a=a1=a(-1)×(-1)=1a-1=11/a=a

This is not supposed to be surprising, but rather reassuring that we won't reach false conclusions by such manipulations.

Also, fractional exponents fit into this scheme. For example

a1/2=a a1/3=γ[3]a

a1/4=γ[4]a a1/5=γ[5]a

This is consistent with earlier notation: the fundamental property of the nth root of a number is that its nth power is the original number. We can check:

a=a1=(a1/n)n=a

Again, this is not supposed to be a surprise, but rather a consistency check.

Then for arbitrary rational exponents m/n we can maintain the same properties: first, the definition is

just

am/n=(γ[n]a)m

One hazard is that, if we want to have only real numbers (as opposed to complex numbers) come up, then we should not try to take square roots, 4th roots, 6th roots, or any even order root of negative numbers.

For general real exponents x we likewise should not try to understand ax except for a>0 or we'll have to use complex numbers (which wouldn't be so terrible). But the value of ax can only be defined as a limit: let r1, r2, ... be a sequence of rational numbers approaching x, and define

ax=limiari

We would have to check that this definition does not accidentally depend upon the sequence approaching x (it doesn't), and that the same properties still work (they do).

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The number e is not something that would come up in really elementary mathematics, because its reason for existence is not really elementary. Anyway, it's approximately

e=2.71828182845905

but if this ever really mattered you'd have a calculator at your side, hopefully.

With the definitions in mind it is easier to make sense of questions about limits of exponential functions. The two companion issues are to evaluate

limx+ax

limx-ax

Since we are allowing the exponent x to be real, we'd better demand that a be a positive real number (if we want to avoid complex numbers, anyway). Then

limx+ax={+ if a>11 if a=10 if 0<a<1

limx-ax={0 if a>11 if a=1+ if 0<a<1

To remember which is which, it is sufficient to use 2 for a>1 and 12 for 0<a<1, and just let x run through positive integers as it goes to +. Likewise, it is sufficient to use 2 for a>1 and 12 for 0<a<1, and just let x run through negative integers as it goes to -.

9. Th e idea of th e deriva tive of a function

First we can tell what the idea of a derivative is. But the issue of computing derivatives is another thing entirely: a person can understand the idea without being able to effectively compute, and vice-versa.

Suppose that f is a function of interest for some reason. We can give f some sort of 'geometric life' by

thinking about the set of points (x, y) so that

f(x)=y

We would say that this describes a curve in the (x, y)-plane. (And sometimes we think of x as 'moving' from left to right, imparting further intuitive or physical content to the story).

For some particular number xo, let y0 be the value f(xo) obtained as output by plugging xo into f as in- put. Then the point (xo, y0) is a point on our curve. The tangent line to the curve at the point (xo, y0) is a line passing through (xo, y0) and 'flat against' the curve. (As opposed to crossing it at some definite angle).

The idea of the derivative f(xo) is that it is the slope of the tangent line at xo to the curve. But this isn't the way to compute these things

10. De riva tives of polynomials

There are just four simple facts which suffice to take the derivative of any polynomial, and actually of somewhat more general things.

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First, there is the rule for taking the derivative of a power function which takes the nth power of its in- put. That is, these functions are functions of the form f(x)=xn. The formula is

ddxxn=nxn-1

That is, the exponent comes down to become a coefficient in front of the thing, and the exponent is de- creased by 1.

The second rule, which is really a special case of this power-function rule, is that derivatives of constants are zero:

ddxc=0

for any constant c.

The third thing, which reflects the innocuous role of constants in calculus, is that for any functions f of x

ddxcf=cddxf

The fourth is that for any two functions f, g of x, the derivative of the sum is the sum of the derivatives:

ddx(f+g)=ddxf+ddxg

Putting these four things together, we can write general formulas like

ddx(axm+bxn+cxp)=amxm-1+bnxn-1+cpxp-1

and so on, with more summands than just the three, if so desired. And in any case here are some exam- ples with numbers instead of letters:

ddx5x3=53x3-1=15x2

ddx(3x7+5x3-11)=37x6+53x2-0=21x6+15x2

ddx(2-3x2-2x3)=0-32x-23x2=-6x-6x2

ddx(-x4+2x5+1)=-4x3+25x4+0=-4x3+10x4

Even if you do catch on to this idea right away, it is wise to practice the technique so that not only can you do it in principle, but also in practice.

#10.21 Find ddx(3x7+5x3-11) #10.22 Find ddx(x2+5x3+2)

#10.23 Find ddx(-x4+2x5+1) #10.24 Find ddx(-3x2-x3-11)

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11. More general power functions

It's important to remember some of the other possibilities for the exponential notation xn. For example

x12=x

-1 1

x =x¯

x-12=1x

and so on. The good news is that the rule given just above for taking the derivative of powers of x still is correct here, even for exponents which are negative or fractions or even real numbers:

ddxxr=rxr-1

Thus, in particular,

ddxx=ddxx12=12x-12

ddx1x=ddxx-1=-1x-2=-1x2

When combined with the sum rule and so on from above, we have the obvious possibilities:

ddx(3x2-7x+5x2=ddx(3x2-7x12+5x-2)=6x-72x-12-10x-3

The possibility of expressing square roots, cube roots, inverses, etc., in terms of exponents is a very impor- tant idea in algebra, and can't be overlooked.

#11.25 Find ddx(3x7+5x-11) #11.26 Find ddx(2x+5x3+3) #11.27 Find ddx(7-5x3+5x7)

12. Quotient rule

The quotient rule is one of the more irritating and goofy things in elementary calculus, but it just couldn't have been any other way. The general principle is

ddx(fg)=fg-gfg2

The main hazard is remembering that the numerator is as it is, rather than accidentally reversing the roles of f and g, and then being off by±, which could be fatal in real life.

ddx(1x-2)=ddx1(x-2)-1ddx(x-2)(x-2)2=0(x-2)-11(x-2)2=-1(x-2)2

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ddx(x-1x-2)=(x-1)(x-2)-(x-1)(x-2)(x-2)2=1(x-2)-(x-1)1(x-2)2

=(x-2)-(x-1)(x-2)2=-1(x-2)2

ddx(5x3+x2-x7)=(5x3+x)(2-x7)-(5x3+x)(2-x7)(2-x7)2

=(15x2+1)(2-x7)-(5x3+x)(-7x6)(2-x7)2

and there's hardly any point in simplifying the last expression, unless someone gives you a good reason. In general, it's not so easy to see how much may or may not be gained in 'simplifying', and we won't make ourselves crazy over it.

#12.28 Find ddx(x-1x-2) #12.29 Find ddx(1x-2) #12.30 Find ddx(x-1x2-5) #12.31 Find ddx(1-x32+x)

13. Product Rule

Not only will the product rule be of use in general and later on, but it's already helpful in perhaps unex- pected ways in dealing with polynomials. Anyway, the general rule is

ddx(fg)=fg+fg

While this is certainly not as awful as the quotient rule just above, it is not as simple as the rule for sums, which was the good-sounding slogan that the derivative of the sum is the sum of the derivatives. It is not true that the derivative of the product is the product of the derivatives. Too bad. Still, it's not as bad as the quotient rule.

One way that the product rule can be useful is in postponing or eliminating a lot of algebra. For example, to evaluate

ddx((x3+x2+x+1)(x4+x3+2x+1))

we could multiply out and then take the derivative term-by-term as we did with several polynomials

above. This would be at least mildly irritating because we'd have to do a bit of algebra. Rather, just ap- ply the product rule without feeling compelled first to do any algebra:

ddx((x3+x2+x+1)(x4+x3+2x+1))

=(x3+x2+x+1)(x4+x3+2x+1)+(x3+x2+x+1)(x4+x3+2x+1)

=(3x2+2x+1)(x4+x3+2x+1)+(x3+x2+x+1)(4x3+3x2+2)

16

Now if we were somehow still obliged to multiply out, then we'd still have to do some algebra. But we can take the derivative without multiplying out, if we want to, by using the product rule.

For that matter, once we see that there is a choice about doing algebra either before or after we take the derivative, it might be possible to make a choice which minimizes our computational labor. This could

matter.

#13.32 Find ddx(x3-1)(x6+x3+1) )

#13.33 Find ddx(x2+x+1)(x4-x2+1) .

#13.34 Find ddx(x3+x2+x+1)(x4+x2+1) ) #13.35 Find ddx(x3+x2+x+1)(2x+x) )

14. Ch a in rule

The chain rule is subtler than the previous rules, so if it seems trickier to you, then you're right. OK. But it is absolutely indispensable in general and later, and already is very helpful in dealing with polynomials. The general assertion may be a little hard to fathom because it is of a different nature than the previous ones. For one thing, now we will be talking about a composite function instead of just adding or multiply- ing functions in a more ordinary way. So, for two functions f and g,

ddx((f(g(x)))=f(g(x))g(x)

There is also the standard notation

(fg)(x)=f(g(x))

for this composite function, but using this notation doesn't accomplish so very much.

A problem in successful use of the chain rule is that often it requires a little thought to recognize that

some formula is (or can be looked at as) a composite function. And the very nature of the chain rule picks on weaknesses in our understanding of the notation. For example, the function

F(x)=(1+x2)100

is really obtained by first using x as input to the function which squares and adds 1 to its input. Then the result of that is used as input to the function which takes the 100 th power. It is necessary to think about it this way or we'll make a mistake. The derivative is evaluated as

ddx(1+x2)100=100(1+x2)992x

To see that this is a special case of the general formula, we need to see what corresponds to the f and g in the general formula. Specifically, let

f (input) = (input )

g(input)=1+(input)

The reason for writing 'input' and not 'x' for the moment is to avoid a certain kind of mistake. But we can compute that

f (input) =100 (input)99

g (input) =2 (input)

17

The hazard here is that the input to f is not x, but rather is g(x) . So the general formula gives

ddx(1+x2)100=f(g(x))g(x)=100g(x)992x=100(1+x2)992x

More examples:

ddx3x+2=ddx(3x+2)1/2=12(3x+2)-1/23

ddx(3x5-x+14)11=11(3x5-x+14)10(15x4-1)

It is very important to recognize situations like

ddx(ax+b)n=n(ax+b)n-1a

for any constants a, b, n. And, of course, this includes

ddxax+b=12(ax+b)-1/2a

ddx1ax+b=-(ax+b)-2a=-a(ax+b)2

Of course, this idea can be combined with polynomials, quotients, and products to give enormous and ex- cruciating things where we need to use the chain rule, the quotient rule, the product rule, etc., and pos- sibly several times each. But this is not hard, merely tedious, since the only things we really do come in small steps. For example:

ddx(1+x+2(1+7x)33)=(1+x+2)(1+7x)33-(1+x+2)((1+7x)33)((1+7x)33)2

by the quotient rule, which is then

(12(x+2)-1/2)(1+7x)33-(1+x+2)((1+7x)33)((1+7x)33)2

because our observations just above (chain rule tell us that

ddxx+2=12(x+2)-1/2(x+2)=12(x+2)-1/2

Then we use the chain rule again to take the derivative of that big power of 1+7x, so the whole thing becomes

(12(x+2)-1/2)(1+7x)33-(1+x+2)(33(1+7x)327)((1+7x)33)2

Although we could simplify a bit here, let's not. The point about having to do several things in a row to take a derivative is pretty clear without doing algebra just now.

#14.36 Find ddx((1-x2)100) #14.37 Find ddxx-3

18

#14.38 Find ddx(x2-x2-3)

#14.39 Find ddx(x2+x+1)

#14.40 Find ddx(x3+x2+x+13) #14.41 Find ddx((x3+x+1)10)

15. Tangent and Normal Lin es

One fundamental interpretation of the derivative of a function is that it is the slope of the tangent line to the graph of the function. (Still, it is important to realize that this is not the definition of the thing, and that there are other possible and important interpretations as well).

The precise statement of this fundamental idea is as follows. Let f be a function. For each fixed value xo of the input to f, the value f(xo) of the derivative f of f evaluated at xo is the slope of the tangent line to the graph of f at the particular point (xo, f(xo)) on the graph.

Recall the point-slope form of a line with slope m through a point (xo, y0) :

y-y0=m(x-xo)

In the present context, the slope is f(xo) and the point is (xo, f(xo)) , so the equation of the tangent line to the graph of f at (xo, f(xo)) is

y-f(xo)=f(xo)(x-xo)

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent. A person might remember from analytic geometry that the slope of any line perpendicular to a line with slope m is the negative reciprocal -1/m. Thus, just changing this aspect of the equation for the tangent line, we can say generally that the equation of the normal line to the graph of f at (xo, f(xo)) is

y-f(xo)=-1f(xo)(x-xo)

The main conceptual hazard is to mistakenly name the fixed point 'x', as well as naming the variable coor- dinate on the tangent line 'x'. This causes a person to write down some equation which, whatever it may be, is not the equation of a line at all.

Another popular boo-boo is to forget the subtraction -f(xo) on the left hand side. Don't do it.

So, as the simplest example: let's write the equation for the tangent line to the curve y=x2 at the point where x=3. The derivative of the function is y=2x, which has value 23=6 when x=3. And the value of the function is 33=9 when x=3. Thus, the tangent line at that point is

y-9=6(x-3)

The normal line at the point where x=3 is

y-9=-16(x-3)

So the question of finding the tangent and normal lines at various points of the graph of a function is just a combination of the two processes: computing the derivative at the point in question, and invoking the point-slope form of the equation for a straight line.

19

#15.42 Write the equation for both the tangent line and normal line to the curve y=3x2-x+1 at the point where x=1.

#15.43 Write the equation for both the tangent line and normal line to the curve y=(x-1)/(x+1) at the point where x=0.

16. Critical points, monotone increase and decrease

A function is called increasing if it increases as the input x moves from left to right, and is called de-

creasing if it decreases as x moves from left to right. Of course, a function can be increasing in some

places and decreasing in others: that's the complication.

We can notice that a function is increasing if the slope of its tangent is positive, and decreasing if the

slope of its tangent is negative. Continuing with the idea that the slope of the tangent is the derivative: a function is increasing where its derivative is positive, and is decreasing where its derivative is negative. This is a great principle, because we don't have to graph the function or otherwise list lots of values to

figure out where it's increasing and decreasing. If anything, it should be a big help in graphing to know in advance where the graph goes up and where it goes down.

And the points where the tangent line is horizontal, that is, where the derivative is zero, are critical

points. The points where the graph has a peak or a trough will certainly lie among the critical points, al- though there are other possibilities for critical points, as well.

Further, for the kind of functions we'll deal with here, there is a fairly systematic way to get all this infor- mation: to find the intervals of increase and decrease of a function f:

Compute the derivative f of f, and solve the equation f(x)=0 for x to find all the critical points, which we list in order as x1<x2<...<xn.

(If there are points of discontinuity or non-differentiability, these points should be added to the list! But points of discontinuity or non-differentiability are not called critical points.)

We need some auxiliary points: To the left of the leftmost critical point x1 pick any convenient point to, between each pair of consecutive critical points xi, xi+1 choose any convenient point ti, and to the right of the rightmost critical point xn choose a convenient point tn.

Evaluate the derivative f at all the auxiliary points ti.

Conclusion: if f(ti+1)>0, then f is increasing on (xi, xi+1) , while if f(ti+1)<0, then f is decreasing on that interval.

Conclusion: on the 'outside' interval (-, xo) , the function f is increasing if f(to)>0 and is decreasing if f(to)<0. Similarly, on (xn, ) , the function f is increasing if f(tn)>0 and is decreasing if f(tn)<0. It is certainly true that there are many possible shortcuts to this procedure, especially for polynomials of low degree or other rather special functions. However, if you are able to quickly compute values of (deriva- tives of!) functions on your calculator, you may as well use this procedure as any other.

Exactly which auxiliary points we choose does not matter, as long as they fall in the correct intervals,

since we just need a single sample on each interval to find out whether f is positive or negative there.

Usually we pick integers or some other kind of number to make computation of the derivative there as

easy as possible.

20

It's important to realize that even if a question does not directly ask for critical points, and maybe does

not ask about intervals either, still it is implicit that we have to find the critical points and see whether

the functions is increasing or decreasing on the intervals between critical points. Examples:

Find the critical points and intervals on which f(x)=x2+2x+9 is increasing and decreasing: Compute f(x)=2x+2. Solve 2x+2=0 to find only one critical point -1. To the left of -1 let's use the auxiliary point to=-2 and to the right use t1=0. Then f(-2)=-2<0, so f is decreasing on the interval

(-, -1) . And f(O)=2>0, so f is increasing on the interval (-1, ) .

Find the critical points and intervals on which f(x)=x3-12x+3 is increasing, decreasing. Compute

f(x)=3x2-12. Solve 3x2-12=0: this simplifies to x2-4=0, so the critical points are ±2. To the left of -2 choose auxiliary point to=-3, between -2 and =2 choose auxiliary point t1=0, and to the right of +2 choose t2=3. Plugging in the auxiliary points to the derivative, we find that f(-3)=27-12>0, so f is increasing on (-, -2) . Since f(O)=-12<0, f is decreasing on (-2, +2) , and since f(3)=

27-12>0, f is increasing on (2, ) .

Notice too that we don't really need to know the exact value of the derivative at the auxiliary points: all we care about is whether the derivative is positive or negative. The point is that sometimes some tedious computation can be avoided by stopping as soon as it becomes clear whether the derivative is positive or negative.

#16.44 Find the critical points and intervals on which f(x)=x2+2x+9 is increasing, decreasing. #16.45 Find the critical points and intervals on which f(x)=3x2-6x+7 is increasing, decreasing. #16.46 Find the critical points and intervals on which f(x)=x3-12x+3 is increasing, decreasing.

17. Minimization and Maximiza tion

The fundamental idea which makes calculus useful in understanding problems of maximizing and minimiz- ing things is that at a peak of the graph of a function, or at the bottom of a trough, the tangent is horizon- tal. That is, the derivative f(xo) is 0 at points xo at which f(xo) is a maximum or a minimum.

Well, a little sharpening of this is necessary: sometimes for either natural or artificial reasons the variable x is restricted to some interval [a, b]. In that case, we can say that the maximum and minimum values of f on the interval [a, b] occur among the list of critical points and endpoints of the interval.

And, if there are points where f is not differentiable, or is discontinuous, then these have to be added in, too. But let's stick with the basic idea, and just ignore some of these complications.

Let's describe a systematic procedure to find the minimum and maximum values of a function f on an

interval [a, b].

Solve f(x)=0 to find the list of critical points of f.

Exclude any critical points not inside the interval [a, b].

Add to the list the endpoints a, b of the interval (and any points of discontinuity or non-differentiability!)

At each point on the list, evaluate the function f: the biggest number that occurs is the maximum, and the littlest number that occurs is the minimum.

Find the minima and maxima of the function f(x)=x4-8x2+5 on the interval [-1, 3]. First, take the derivative and set it equal to zero to solve for critical points: this is

4x3-16x=0

21

or, more simply, dividing by 4, it is x3-4x=0. Luckily, we can see how to factor this: it is

x(x-2)(x+2)

So the critical points are -2,0, +2. Since the interval does not include -2, we drop it from our list.

And we add to the list the endpoints -1,3. So the list of numbers to consider as potential spots for

minima and maxima are -1,0,2,3. Plugging these numbers into the function, we get (in that order) -2,5, -11,14. Therefore, the maximum is 14, which occurs at x=3, and the minimum is -11, which occurs at x=2.

Notice that in the previous example the maximum did not occur at a critical point, but by coincidence did occur at an endpoint.

You have 200 feet of fencing with which you wish to enclose the largest possible rectangular garden. What is the largest garden you can have?

Let x be the length of the garden, and y the width. Then the area is simply xy. Since the perimeter is

200, we know that 2x+2y=200, which we can solve to express y as a function of x: we find that y= 100-x. Now we can rewrite the area as a function of x alone, which sets us up to execute our procedure:

area=xy=x(100-x)

The derivative of this function with respect to x is 100-2x. Setting this equal to 0 gives the equation

100-2x=0

to solve for critical points: we find just one, namely x=50.

Now what about endpoints? What is the interval? In this example we must look at 'physical' considera- tions to figure out what interval x is restricted to. Certainly a width must be a positive number, so x>0 and y>0. Since y=100-x, the inequality on y gives another inequality on x, namely that x<100. So x is in [0,100].

When we plug the values 0,50,100 into the function x(100-x) , we get 0, 2500, 0, in that order. Thus, the corresponding value of y is 100-50=50, and the maximal possible area is 5050=2500.

#17.47 Olivia has 200 feet of fencing with which she wishes to enclose the largest possible rectangular garden. What is the largest garden she can have?

#17.48 Find the minima and maxima of the function f(x)=3x4-4x3+5 on the interval [-2, 3].

#17.49 The cost per hour of fuel to run a locomotive is v2/25 dollars, where v is speed, and other costs are $100 per hour regardless of speed. What is the speed that minimizes cost per mile?

#17.50 The product of two numbers x, y is 16. We know x1 and y1. What is the greatest possible sum of the two numbers?

#17.51 Find both the minimum and the maximum of the function f(x)=x3+3x+1 on the interval [-2, 2].

18. Local minim a and maxima (First Deriva tive Test)

A function f has a local maximum or relative maximum at a point xo if the values f(x) of f for x 'near' xo are all less than f(xo) . Thus, the graph of f near xo has a peak at xo. A function f has a 10- cal minimum or relative minimum at a point xo if the values f(x) of f for x 'near' xo are all greater

22

than f(xo) . Thus, the graph of f near xo has a trough at xo. (To make the distinction clear, sometimes the 'plain' maximum and minimum are called absolute maximum and minimum.)

Yes, in both these definitions' we are tolerating ambiguity about what 'near' would mean, although the peak/trough requirement on the graph could be translated into a less ambiguous definition. But in any case we'll be able to execute the procedure given below to find local maxima and minima without worry- ing over a formal definition.

This procedure is just a variant of things we've already done to analyze the intervals of increase and de- crease of a function, or to find absolute maxima and minima. This procedure starts out the same way as does the analysis of intervals of increase/decrease, and also the procedure for finding ('absolute') maxima and minima of functions.

To find the local maxima and minima of a function f on an interval [a, b]:

Solve f(x)=0 to find critical points of f.

Drop from the list any critical points that aren't in the interval [a, b].

Add to the list the endpoints (and any points of discontinuity or non-differentiability): we have an or- dered list of special points in the interval:

a=xo<x1<...<xn=b

Between each pair xi<xi+1 of points in the list, choose an auxiliary point ti+1. Evaluate the derivative f at all the auxiliary points.

For each critical point xi, we have the auxiliary points to each side of it: ti<xi<ti+1. There are four cases best remembered by drawing a picture

if f(ti)>0 and f(ti+1)<0 (so f is increasing to the left of xi and decreasing to the right of xi, then f has a local maximum at xo.

if f(ti)<0 and f(ti+1)>0 (so f is decreasing to the left of xi and increasing to the right of xi, then f has a local minimum at xo.

if f(ti)<0 and f(ti+1)<0 (so f is decreasing to the left of xi and also decreasing to the right of xi, then f has neither a local maximum nor a local minimum at xo.

if f(ti)>0 and f(ti+1)>0 (so f is increasing to the left of xi and also increasing to the right of xi, then f has neither a local maximum nor a local minimum at xo.

The endpoints require separate treatment: There is the auxiliary point to just to the right of the left end- point a, and the auxiliary point tn just to the left of the right endpoint b:

At the left endpoint a, if f(to)<0 (so f is decreasing to the right of a) then a is a local maximum.

At the left endpoint a, if f(to)> (so f is increasing to the right of a) then a is a local minimum.

At the right endpoint b, if f(tn)<0 (so f is decreasing as b is approached from the left) then b is a l0- cal minimum.

At the right endpoint b, if f(tn)> (so f is increasing as b is approached from the left) then b is a local maximum.

The possibly bewildering list of possibilities really shouldn't be bewildering after you get used to them. We are already acquainted with evaluation of f at auxiliary points between critical points in order to see whether the function is increasing or decreasing, and now we're just applying that information to see whether the graph peaks, troughs, or does neither around each critical point and endpoints. That is, the geometric meaning of the derivative's being positive or negative is easily translated into conclusions about local maxima or minima.

Find all the local (=relative) minima and maxima of the function f(x)=2x3-9x2+1 on the interval [-2, 2]: To find critical points, solve f(x)=0: this is 6x2-18x=0 or x(x-3)=0, so there are two critical points, 0 and 3. Since 3 is not in the interval we care about, we drop it from our list. Adding the endpoints to the list, we have

-2<0<2

23

as our ordered list of special points. Let's use auxiliary points -1,1. At -1 the derivative is f(-1)= 24>0, so the function is increasing there. At +1 the derivative is f(1)=-12<0, so the function is decreasing. Thus, since it is increasing to the left and decreasing to the right of 0, it must be that 0 is a local maximum. Since f is increasing to the right of the left endpoint -2, that left endpoint must give a local minimum. Since it is decreasing to the left of the right endpoint +2, the right endpoint must be a local minimum.

Notice that although the processes of finding absolute maxima and minima and local maxima and minima have a lot in common, they have essential differences. In particular, the only relations between them are that critical points and endpoints (and points of discontinuity, etc.) play a big role in both, and that the absolute maximum is certainly a local maximum, and likewise the absolute minimum is certainly a local minimum.

For example, just plugging critical points into the function does not reliably indicate which points are l0- cal maxima and minima. And, on the other hand, knowing which of the critical points are local maxima and minima generally is only a small step toward figuring out which are absolute: values still have to be plugged into the funciton! So don't confuse the two procedures.'

(By the way: while it's fairly easy to make up story-problems where the issue is to find the maximum or minimum value of some function on some interval, it's harder to think of a simple application of local max- ima or minima).

#18.52 Find all the local (=relative) minima and maxima of the function f(x)=(x+1)3-3(x+1) on the interval [-2, 1].

#18.53 Find the local (=relative) minima and maxima on the interval [-3, 2] of the function f(x)=(x+ 1)3-3(x+1) .

#18.54 Find the local (relative) minima and maxima of the function f(x)=1-12x+x3 on the interval [-3, 3].

#18.55 Find the local (relative) minima and maxima of the function f(x)=3x4-8x3+6x2+17 on the interval [-3, 3].

19. An algebra trick

The algebra trick here goes back at least 350 years. This is worth looking at if only as an additional re- view of algebra, but is actually of considerable value in a variety of hand computations as well.

The algebraic identity we use here starts with a product of factors each of which may occur with a frac- tional or negative exponent. For example, with 3 such factors:

f(x)=(x-a)k(x-b)(x-c)m

The derivative can be computed by using the product rule twice:

f(x)=

=k(x-a)k-1(x-b)(x-c)m+(x-a)k(x-b)-1(x-c)m+(x-a)k(x-b)m(x-c)m-1

Now all three summands here have a common factor of

(x-a)k-1(x-b)-1(x-c)m-1

24

which we can take out, using the distributive law in reverse: we have

f(x)=

=(x-a)k-1(x-b)-1(x-c)m-1[k(x-b)(x-c)+(x-a)(x-c)+m(x-a)(x-b)]

The minor miracle is that the big expression inside the square brackets is a mere quadratic polynomial in x.

Then to determine critical points we have to figure out the roots of the equation f(x)=0: If k-1>0 then x=a is a critical point, if k-10 it isn't. If -1>0 then x=b is a critical point, if -10 it isn't. If m-1>0 then x=c is a critical point, if m-10 it isn't. And, last but not least, the two roots of the quadratic equation

k(x-b)(x-c)+(x-a)(x-c)+m(x-a)(x-b)=0

are critical points.

There is also another issue here, about not wanting to take square roots (and so on) of negative numbers. We would exclude from the domain of the function any values of x which would make us try to take a

square root of a negative number. But this might also force us to give up some critical points.' Still, this is not the main point here, so we will do examples which avoid this additional worry.

A very simple numerical example: suppose we are to find the critical points of the function

f(x)=x5/2(x-1)4/3

Implicitly, we have to find the critical points first. We compute the derivative by using the product rule, the power function rule, and a tiny bit of chain rule:

f(x)=52x3/2(x-1)4/3+x5/243(x-1)1/3

And now solve this for x? It's not at all a polynomial, and it is a little ugly.

But our algebra trick transforms this issue into something as simple as solving a linear equation: first fig- ure out the largest power of x that occurs in all the terms: it is x3/2, since x5/2 occurs in the first term

and x3/2 in the second. The largest power of x-1 that occurs in all the terms is (x-1)1/3, since (x-1)4/3 occurs in the first, and (x-1)1/3 in the second. Taking these common factors out (using the distributive law 'backward'), we rearrange to

f(x)=52x3/2(x-1)4/3+x5/243(x-1)1/3

=x3/2(x-1)1/3(52(x-1)+43x)

=x3/2(x-1)1/3(52x-52+43x)

=x3/2(x-1)1/3(236x-52)

Now to see when this is 0 is not so hard: first, since the power of x appearing in front is positive, x=0

make this expression 0. Second, since the power of x+1 appearing in front is positive, if x-1=0 then the whole expression is 0. Third, and perhaps unexpectedly, from the simplified form of the complicated factor, if 236x-52=0 then the whole expression is 0, as well. So, altogether, the critical points would appear to be

x=0, 1523,1

25

Many people would overlook the critical point 1523, which is visible only after the algebra we did.

#19.56 Find the critical points and intervals of increase and decrease of f(x)=x10(x-1)12.

#19.57 Find the critical points and intervals of increase and decrease of f(x)=x10(x-2)11(x+2)3.

#19.58 Find the critical points and intervals of increase and decrease of f(x)=x5/3(x+1)6/5.

#19.59 Find the critical points and intervals of increase and decrease of f(x)=x1/2(x+1)4/3(x-1)-11/3.

20. Lin ear approxim a Tions. approxim a Tion by differentials

The idea here in 'geometric' terms is that in some vague sense a curved line can be approximated by a

straight line tangent to it. Of course, this approximation is only good at all 'near' the point of tangency, and so on. So the only formula here is secretly the formula for the tangent line to the graph of a function. There is some hassle due to the fact that there are so many different choices of symbols to write it.

We can write some formulas: Let f be a function, and fix a point xo. The idea is that for xζnear'xo we have an ζapproximate ' equality

f(x)f(xo)+f(xo)(x-xo)

We do not attempt to clarify what either 'near' or 'approximate' mean in this context. What is really true here is that for a given value x, the quantity

f(xo)+f(xo)(x-xo)

is exactly the y-coordinate of the line tangent to the graph at xo

The aproximation statement has many paraphrases in varying choices of symbols, and a person needs to be able to recognize all of them. For example, one of the more traditional paraphrases, which introduces some slightly silly but oh-so-traditional notation, is the following one. We might also say that y is a func- tion of x given by y=f(x) . Let

x= small change in x

y= corresponding change in y=f(x+x)-f(x)

Then the assertion is that

yf(x)x

Sometimes some texts introduce the following questionable (but traditionally popular!) notation:

dy=f(x)dx= approximation to change in y

dx=x

and call the dx and dyζdifferentials '. And then this whole procedure is 'approximation by differen-

tials'. A not particularly enlightening paraphrase, using the previous notation, is

dyy

Even though you may see people writing this, don't do it.

More paraphrases, with varying symbols:

f(x+x)f(x)+f(x)x

26

f(x+δ)f(x)+f(x)δ

f(x+h)f(x)+f(x)h

f(x+x)-f(x)f(x)x

y+yf(x)+f(x)x

yf(x)x

A little history: Until just 20 or 30 years ago, calculators were not widely available, and especially not typ- ically able to evaluate trigonometric, exponential, and logarithm functions. In that context, the kind of

vague and unreliable 'approximation' furnished by 'differentials' was certainly worthwhile in many situa- tions.

By contrast, now that pretty sophisticated calculators are widely available, some things that once seemed sensible are no longer. For example, a very traditional type of question is to 'approximate 10 by differ- entials'. A reasonable contemporary response would be to simply punch in 1, 0, sqrt' on your calculator and get the answer immediately to 10 decimal places. But this was possible only relatively recently.

Example: For example let's approximate 17 by differentials. For this problem to make sense at all imag- ine that you have no calculator. We take f(x)=x=x1/2. The idea here is that we can easily evaluate ζby hand' both f and f at the point x=16 which is ζnear'17. (Here f(x)=12x-1/2) . Thus, here

x=17-16=1

and

17=f(17)f(16)+f(16)x=16+121161=4+18

Example: Similarly, if we wanted to approximate 18 'by differentials', we'd again take f(x)=x=x1/2. Still we imagine that we are doing this 'by hand', and then of course we can 'easily evaluate' the function f and its derivative f at the point x=16 which is 'near' 18. Thus, here

x=18-16=2

and

18=f(18)f(16)+f(16)x=16+121162=4+14

Why not use the 'good' point 25 as the 'nearby' point to find 18? Well, in broad terms, the further away your 'good' point is, the worse the approximation will be. Yes, it is true that we have little idea how good or bad the approximation is anyway.

It is somewhat more sensible to not use this idea for numerical work, but rather to say things like

x+1x+121x

and

x+hx+121xh

This kind of assertion is more than any particular numerical example would give, because it gives a rela- tionship, telling how much the output changes for given change in input, depending what regime (=inter- val) the input is generally in. In this example, we can make the qualitative observation that as x increases the difference x+1-x decreases.

Example: Another numerical example: Approximate sin31o 'by differentials'. Again, the point is not to hit 3, 1, sin on your calculator (after switching to degrees), but rather to imagine that you have no calcu- lator. And we are supposed to remember from pre-calculator days the 'special angles' and the values of

27

trig functions at them: sin30o=12 and cos30o=32. So we'd use the function f(x)=sinx, and we'd imagine that we can evaluate f and f easily by hand at 30o. Then

x=31o-30o=1o=1o2πradians360o=2π360 radians

We have to rewrite things in radians since we really only can compute derivatives of trig functions in radi- ans. Yes, this is a complication in our supposed 'computation by hand'. Anyway, we have

sin31o=f(31o)=f(30o)+f(30o)x=sin30o+cos30o2π360

=12+322π360

Evidently we are to also imagine that we know or can easily find 3 (by differentials?) as well as a value of π. Yes, this is a lot of trouble in comparison to just punching the buttons, and from a contemporary perspective may seem senseless.

Example: Approximate ln(x+2) 'by differentials', in terms of lnx and x: This non-numerical question is somewhat more sensible. Take f(x)=lnx, so that f(x)=1x. Then

x=(x+2)-x=2

and by the formulas above

ln(x+2)=f(x+2)f(x)+f(x)2=lnx+2x

Example: Approximate ln(e+2) in terms of differentials: Use f(x)=lnx again, so f(x)=1x. We prob- ably have to imagine that we can 'easily evaluate' both lnx and 1x at x=e. (Do we know a numerical approximation to e?). Now

x=(e+2)-e=2

so we have

ln(e+2)=f(e+2)f(e)+f(e)2=lne+2e=1+2e

since lne=1.

#20.60 Approximate 101 'by differentials' in terms of 100=10.

#20.61 Approximate x+1 'by differentials', in terms of x.

#20.62 Granting that ddxlnx=1x, approximate ln(x+1) 'by differentials', in terms of lnx and x. #20.63 Granting that ddxex=ex, approximate ex+1 in terms of ex.

#20.64 Granting that ddxcosx=-sinx, approximate cos(x+1) in terms of cosx and sinx.

21. Implicit differentia tion

There is nothing 'implicit' about the differentiation we do here, it is quite 'explicit'. The difference from earlier situations is that we have a function defined ζimplicitly '. What this means is that, instead of a clear-cut (if complicated) formula for the value of the function in terms of the input value, we only have a relation between the two. This is best illustrated by examples.

28

For example, suppose that y is a function of x and

y5-xy+x5=1

and we are to find some useful expression for dy/dx. Notice that it is not likely that we'd be able to solve this equation for y as a function of x (nor vice-versa, either), so our previous methods do not obviously do anything here! But both sides of that equality are functions of x, and are equal, so their derivatives are

equal, surely. That is,

5y4dydx-1y-xdydx+5x4=0

Here the trick is that we can 'take the derivative' without knowing exactly what y is as a function of x,

but just using the rules for differentiation.

Specifically, to take the derivative of the term y5, we view this as a composite function, obtained by ap- plying the take-the-fifth-power function after applying the (not clearly known!) function y. Then use the chain rule!

Likewise, to differentiate the term xy, we use the product rule

ddx(xy)=dxdxy+xdydx=y+xdydx

since, after all,

dxdx=1

And the term x5 is easy to differentiate, obtaining the 5x4. The other side of the equation, the function '1', is constant, so its derivative is 0. (The fact that this means that the left-hand side is also constant

should not be mis-used: we need to use the very non-trivial looking expression we have for that constant function, there on the left-hand side of that equation!).

Now the amazing part is that this equation can be solved for y, if we tolerate a formula involving not only x, but also y: first, regroup terms depending on whether they have a y or not:

y(5y4-x)+(-y+5x4)=0

Then move the non-y' terms to the other side

y(5y4-x)=y-5x4

and divide by the 'coefficient' of the y:

y=y-5x45y4-x

Yes, this is not as good as if there were a formula for y not needing the y. But, on the other hand, the

initial situation we had did not present us with a formula for y in terms of x, so it was necessary to lower our expectations.

Yes, if we are given a value of x and told to find the corresponding y, it would be impossible without luck or some additional information. For example, in the case we just looked at, if we were asked to find y

when x=1 and y=1, it's easy: just plug these values into the formula for y in terms of both x and y: when x=1 and y=1, the corresponding value of y is

y=1-514514-1=-4/4=-1

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If, instead, we were asked to find y and y when x=1, not knowing in advance that y=1 fits into the equation when x=1, we'd have to hope for some luck. First, we'd have to try to solve the original equa- tion for y with x replace by its value 1: solve

y5-y+1=1

By luck indeed, there is some cancellation, and the equation becomes

y5-y=0

By further luck, we can factor this 'by hand': it is

0=y(y4-1)=y(y2-1)(y2+1)=y(y-1)(y+1)(y2+1)

So there are actually three real numbers which work as y for x=1: the values -1,0, +1. There is no clear way to see which is 'best'. But in any case, any one of these three values could be used as y in substitut- ing into the formula

y=y-5x45y4-x

we obtained above.

Yes, there are really three solutions, three functions, etc.

Note that we could have used the Intermediate Value Theorem and/or Newton's Method to numerically solve the equation, even without too much luck. In 'real life' a person should be prepared to do such

things.

#21.65 Suppose that y is a function of x and

y5-xy+x5=1

Find dy/dx at the point x=1, y=0.

#21.66 Suppose that y is a function of x and

y3-xy2+x2y+x5=7

Find dy/dx at the point x=1, y=2. Find d2ydx2 at that point.

22. Re/aTedra tes

In this section, most functions will be functions of a parameter t which we will think of as time. There is a convention coming from physics to write the derivative of any function y of t as y˙=dy/dt, that is, with just a dot over the functions, rather than a prime.

The issues here are variants and continuations of the previous section's idea about implicit differentiation. Traditionally, there are other (non-calculus!) issues introduced at this point, involving both story-problem stuff as well as requirement to be able to deal with similar triangles, the Pythagorean Theorem, and to re- call formulas for volumes of cones and such.

Continuing with the idea of describing a function by a relation, we could have two unknown functions x and y of t, related by some formula such as

x2+y2=25

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A typical question of this genre is 'What is y˙ when x=4 and x˙=6?'

The fundamental rule of thumb in this kind of situation is differentiate the relation with respect to t: so we differentiate the relation x2+y2=25 with respect to t, even though we don't know any details about those two function x and y of t:

2xx˙+2yy˙=0

using the chain rule. We can solve this for y˙ :

y˙=-xx˙y

So at any particular moment, if we knew the values of x, x, y then we could find y˙ at that moment.

Here it's easy to solve the original relation to find y when x=4: we get y=±3. Substituting, we get

y˙=-46±3=±8

( The±notation means that we take +chosen if we take y=-3 and -if we take y=+3).

#22.67 Suppose that x, y are both functions of t, and that x2+y2=25. Express dxdt in terms of x, y, and dydt. When x=3 and y=4 and dydt=6, what is dxdt ?

#22.68 A 2-foot tall dog is walking away from a streetlight which is on a 10-foot pole. At a certain mo- ment, the tip of the dog's shadow is moving away from the streetlight at 5 feet per second. How fast is the dog walking at that moment?

#22.69 A ladder 13 feet long leans against a house, but is sliding down. How fast is the top of the ladder moving at a moment when the base of the ladder is 12 feet from the house and moving outward at 10 feet per second?

23. Intermediate Value Theorem, location ofroots

The assertion of the Intermediate Value Theorem is something which is probably 'intuitively obvious', and is also provably true: if a function f is continuous on an interval [a, b] and if f(a)<0 and f(b)>0 (or vice-versa), then there is some third point c with a<c<b so that f(c)=0. This result has many rel- atively 'theoretical' uses, but for our purposes can be used to give a crude but simple way to locate the

roots of functions. There is a lot of guessing, or trial-and-error, involved here, but that is fair. Again,

in this situation, it is to our advantage if we are reasonably proficient in using a calculator to do simple

tasks like evaluating polynomials! If this approach to estimating roots is taken to its logical conclusion,

it is called the method of interval bisection, for a reason we'll see below. We will not pursue this method very far, because there are better methods to use once we have invoked this just to get going.

For example, we probably don't know a formula to solve the cubic equation

x3-x+1=0

But the function f(x)=x3-x+1 is certainly continuous, so we can invoke the Intermediate Value The- orem as much as we'd like. For example, f(2)=7>0 and f(-2)=-5<0, so we know that there is a root in the interval [-2, 2]. We'd like to cut down the size of the interval, so we look at what happens at the midpoint, bisecting the interval [-2, 2]: we have f(0)=1>0. Therefore, since f(-2)=-5<0, we can conclude that there is a root in [-2, 0]. Since both f(O)>0 and f(2)>0, we can't say anything at this point about whether or not there are roots in [0,2]. Again bisecting the interval [-2, 0] where we

31

know there is a root, we compute f(-1)=1>0. Thus, since f(-2)<0, we know that there is a root in [-2, -1] (and have no information about [-1,0

If we continue with this method, we can obtain as good an approximation as we want! But there are faster ways to get a really good approximation, as we'll see.

Unless a person has an amazing intuition for polynomials (or whatever), there is really no way to antici- pate what guess is better than any other in getting started.

Invoke the Intermediate Value Theorem to find an interval of length 1 or less in which there is a root of

x3+x+3=0: Let f(x)=x3+x+3. Just, guessing, we compute f(O)=3>0. Realizing that the x3 term probably 'dominates' f when x is large positive or large negative, and since we want to find a point where f is negative, our next guess will be a 'large' negative number: how about -1? Well, f(-1)=1>0,

so evidently -1 is not negative enough. How about -2? Well, f(-2)=-7<0, so we have succeeded.

Further, the failed guess -1 actually was worthwhile, since now we know that f(-2)<0 and f(-1)>0. Then, invoking the Intermediate Value Theorem, there is a root in the interval [-2, -1].

Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. We can use the Intermediate Value Theorem to get an idea where all of them are.

Invoke the Intermediate Value Theorem to find three different intervals of length 1 or less in each of which there is a root of x3-4x+1=0: first, just starting anywhere, f(O)=1>0. Next, f(1)=-2<0. So, since f(0)>0 and f(1)<0, there is at least one root in [0,1], by the Intermediate Value Theorem. Next, f(2)=1>0. So, with some luck here, since f(1)<0 and f(2)>0, by the Intermediate Value Theorem there is a root in [1, 2]. Now if we somehow imagine that there is a negative root as well, then we try -1: f(-1)=4>0. So we know nothing about roots in [-1, 0]. But continue: f(-2)=1>0, and still no new conclusion. Continue: f(-3)=-14<0. Aha! So since f(-3)<0 and f(2)>0, by the Intermediate Value Theorem there is a third root in the interval [-3, -2].

Notice how even the 'bad' guesses were not entirely wasted.

24. Newton's method

This is a method which, once you get started, quickly gives a very good approximation to a root of poly- nomial (and other) equations. The idea is that, if xo is not a root of a polynomial equation, but is pretty close to a root, then sliding down the tangent line at xo to the graph of f gives a good approximation to the actual root. The point is that this process can be repeated as much as necessary to give as good an ap-

proximation as you want.

Let's derive the relevant formula: if our blind guess for a root of f is xo, then the tangent line is

y-f(xo)=f(xo)(x-xo)

'Sliding down' the tangent line to hit the x-axis means to find the intersection of this line with the x-axis: this is where y=0. Thus, we solve for x in

0-f(xo)=f(xo)(x-xo)

to find

x=xo-f(xo)f'(xo)

Well, let's call this first serious guess x1. Then, repeating this process, the second serious guess would be

x2=x1-f(x1)f'(x1)

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and generally, if we have the nth guess xn then the n+1th guess xn+1 is

xn+1=xn-f(xn)f'(xn)

OK, that's the formula for improving our guesses. How do we decide when to quit? Well, it depends upon to how many decimal places we want our approximation to be good. Basically, if we want, for example, 3 decimal places accuracy, then as soon as xn and xn+1 agree to three decimal places, we can presume that those are the true decimals of the true root of the equation. This will be illustrated in the examples below. It is important to realize that there is some uncertainty in Newton's method, both because it alone cannot assure that we have a root, and also because the idea just described for approximating roots to a given

accuracy is not foolproof. But to worry about what could go wrong here is counter-productive.

Approximate a root of x3-x+1=0 using the intermediate value theorem to get started, and then New- ton's method:

First let's see what happens if we are a little foolish here, in terms of the 'blind' guess we start with. If we ignore the advice about using the intermediate value theorem to guarantee a root in some known interval, we'll waste time. Let's see: The general formula

xn+1=xn-f(xn)f'(xn)

becomes

xn+1=xn-x3-x+13x2-1

If we take x1=1 as our 'blind' guess, then plugging into the formula gives

x2=0.5

x3=3

x4=2.0384615384615383249

This is discouraging, since the numbers are jumping around somewhat. But if we are stubborn and can compute quickly with a calculator (not by hand!), we'd see what happens:

x5 = 1.3902821472167361527

x6 = 0.9116118977179270555

x7 = 0.34502849674816926662

x8 = 1.4277507040272707783

x9 = 0.94241791250948314662

x10 = 0.40494935719938018881

x11 = 1.7069046451828553401

x12 = 1.1557563610748160521

x13 = 0.69419181332954971175

x14 = -0.74249429872066285974

x15 = -2.7812959406781381233

x16 = -1.9827252470441485421

x17 = -1.5369273797584126484

x18 = -1.3572624831877750928

x19 = -1.3256630944288703144

x20 = -1.324718788615257159

x21 = -1.3247179572453899876

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Well, after quite a few iterations of 'sliding down the tangent', the last two numbers we got, x20 and x21, agree to 5 decimal places. This would make us think that the true root is approximated to five decimal places by -1.32471.

The stupid aspect of this little scenario was that our initial 'blind' guess was too far from an actual root, so that there was some wacky jumping around of the numbers before things settled down. If we had been computing by hand this would have been hopeless.

Let's try this example again using the Intermediate Value Theorem to pin down a root with some degree of accuracy: First, f(1)=1>0. Then f(0)=+1>0 also, so we might doubt that there is a root in [0,1]. Continue: f(-1)=1>0 again, so we might doubt that there is a root in [-1, 0], either. Continue: at last f(-2)=-5<0, so since f(-1)>0 by the Intermediate Value Theorem we do indeed know that there is a root between -2 and -1. Now to start using Newton's Method, we would reasonably guess

xo=-1.5

since this is the midpoint of the interval on which we know there is a root. Then computing by Newton's method gives:

x1 = -1.3478260869565217295

x2 = -1.3252003989509069104

x3 = -1.324718173999053672

x4 = -1.3247179572447898011

so right away we have what appears to be 5 decimal places accuracy, in 4 steps rather than 21. Getting off to a good start is important.

Approximate all three roots of x3-3x+1=0 using the intermediate value theorem to get started, and then Newton's method. Here you have to take a little care in choice of beginning 'guess' for Newton's method: In this case, since we are told that there are three roots, then we should certainly be wary about where

we start: presumably we have to start in different places in order to successfully use Newton's method to find the different roots. So, starting thinking in terms of the intermediate value theorem: letting f(x)= x3-3x+1, we have f(2)=3>0. Next, f(1)=-1<0, so we by the Intermediate Value Theorem

we know there is a root in [1, 2]. Let's try to approximate it pretty well before looking for the other roots: The general formula for Newton's method becomes

xn+1=xn-x3-3x+13x2-3

Our initial 'blind' guess might reasonably be the midpoint of the interval in which we know there is a root: take

xo=1.5

Then we can compute

x1 = 1.533333333333333437

x2 = 1.5320906432748537807

x3 = 1.5320888862414665521

x4 = 1.5320888862379560269

x5 = 1.5320888862379560269

x6 = 1.5320888862379560269

So it appears that we have quickly approximated a root in that interval! To what looks like 19 decimal

places!

Continuing with this example: f(O)=1>0, so since f(1)<0 we know by the intermediate value theo- rem that there is a root in [0,1], since f(1)=-1<0. So as our blind gues let's use the midpoint of this

34

interval to start Newton's Method: that is, now take xo=0.5:

x1 = 0.33333333333333337034

x2 = 0.3472222222222222654

x3 = 0.34729635316386797683

x4 = 0.34729635533386071788

x5 = 0.34729635533386060686

x6 = 0.34729635533386071788

x7 = 0.34729635533386060686

x8 = 0.34729635533386071788

so we have a root evidently approximated to 3 decimal places after just 2 applications of Newton's method. After 8 applications, we have apparently 15 correct decimal places.

#24.70 Approximate a root of x3-x+1=0 using the intermediate value theorem to get started, and then Newton's method.

#24.71 Approximate a root of 3x4-16x3+18x2+1=0 using the intermediate value theorem to get started, and then Newton's method. You might have to be sure to get sufficiently close to a root to start so that things don't 'blow up'.

#24.72 Approximate all three roots of x3-3x+1=0 using the intermediate value theorem to get started, and then Newton's method. Here you have to take a little care in choice of beginning 'guess' for Newton's method.

#24.73 Approximate the unique positive root of cosx=x. #24.74 Approximate a root of ex=2x.

#24.75 Approximate a root of sinx=lnx. Watch out.

25. De riva tives of transcendental functions

The new material here is just a list of formulas for taking derivatives of exponential, logarithm, trigono- metric, and inverse trigonometric functions. Then any function made by composing these with polynomi- als or with each other can be differentiated by using the chain rule, product rule, etc. (These new formu- las are not easy to derive, but we don't have to worry about that).

The first two are the essentials for exponential and logarithms:

ddxex = ex

ddxlnx = 1x

The next three are essential for trig functions:

ddxsinx = cosx

ddxcosx = -sinx

ddxtanx = sec2x

The next three are essential for inverse trig functions

ddxarctanx =

ddxarcsinx = xx2-112,1-x11+x21

ddx arcsec x =

35

The previous formulas are the indispensable ones in practice, and are the only ones that I personally re- member (if I'm lucky). Other formulas one might like to have seen are (with a>0 in the first two):

ddxax = lnaax

ddxlogax =

1lnax

ddxsecx = tanxsecx

ddxcscx = -cotxcscx

ddxcotx = -csc2x

ddx arccot x =

ddxarccosx = xx2-1-12,1-x-11+x2-1

ddx arccsc x =

(There are always some difficulties in figuring out which of the infinitely-many possibilities to take for the values of the inverse trig functions, and this is especially bad with arccsc, for example. But we won't have time to worry about such things).

To be able to use the above formulas it is not necessary to know very many other properties of these func- tions. For example, it is not necessary to be able to graph these functions to take their derivatives.'

#25.76 Find ddx(ecosx)

#25.77 Find ddx(arctan(2-ex))

#25.78 Find ddx(ln(x-1))

#25.79 Find ddx(e2cosx+5)

#25.80 Find ddx(arctan(1+sin2x)) #25.81 Find ddxcos(ex-x2)

#25.82 Find ddxJ[3]1-ln2x

#25.83 Find ddxex-1ex+1

#25.84 Find ddx(ln(1x))

26. LEospiTa /s rule

L'Hospital's rule is the definitive way to simplify evaluation of limits. It does not directly evaluate limits, but only simplifies evaluation if used appropriately.

In effect, this rule is the ultimate version of 'cancellation tricks', applicable in situations where a more

down-to-earth genuine algebraic cancellation may be hidden or invisible.

Suppose we want to evaluate

limxaf(x)g(x)

where the limit a could also be + or - in addition to 'ordinary' numbers. Suppose that either

limxaf(x)=0 and limxag(x)=0

36

or

limxaf(x)=± and limxag(x)=±

( The±'s don't have to be the same sign). Then we cannot just 'plug in' to evaluate the limit, and these are traditionally called indeterminate forms. The unexpected trick that works often is that (amazingly) we are entitled to take the derivative of both numerator and denominator:

limxaf(x)g(x)=limxaf(x)g(x)

No, this is not the quotient rule. No, it is not so clear why this would help, either, but we'll see in exam- ples.

Example: Find limx0(sinx)/x: both numerator and denominator have limit 0, so we are entitled to ap- ply L'Hospital's rule:

limx0sinxx=limx0cosx1

In the new expression, neither numerator nor denominator is 0 at x=0, and we can just plug in to see that the limit is 1.

Example: Find limx0x/(e2x-1) : both numerator and denominator go to 0, so we are entitled to use

L'Hospital's rule:

limx0xe2x-1=limx012e2x

In the new expression, the numerator and denominator are both non-zero when x=0, so we just plug in 0 to get

limx0xe2x-1=limx012e2x=12e0=12

Example: Find limx0+xlnx: The 0+ means that we approach 0 from the positive side, since otherwise we won't have a real-valued logarithm. This problem illustrates the possibility as well as necessity of rear- ranging a limit to make it be a ratio of things, in order to legitimately apply L'Hospital's rule. Here, we rearrange to

limx0+xlnx=limx0lnxl/x

In the new expressions the top goes to - and the bottom goes to + as x goes to 0 (from the right). Thus, we are entitled to apply L'Hospital's rule, obtaining

limx0+xlnx=limx0lnxl/x=limx01/x-1/x2

Now it is very necessary to rearrange the expression inside the last limit: we have

lim1/x=limx0-x

x0-1/x2

The new expression is very easy to evaluate: the limit is 0.

It is often necessary to apply L'Hospital's rule repeatedly: Let's find limx+x2/ex: both numerator and denominator go to as x+, so we are entitled to apply L'Hospital's rule, to turn this into

limx+2xex

But still both numerator and denominator go to , so apply L'Hospital's rule again: the limit is

limx+2ex=0

37

since now the numerator is fixed while the denominator goes to +.

Example: Now let's illustrate more ways that things can be rewritten as ratios, thereby possibly making L'Hospital's rule applicable. Let's evaluate

limx0xx

It is less obvious now, but we can't just plug in 0 for x: on one hand, we are taught to think that x0=1, but also that 0x=0; but then surely 00 can't be both at once. And this exponential expression is not a ratio.

The trick here is to take the logarithm:

ln(limx0+xx)=limx0+ln(xx)

The reason that we are entitled to interchange the logarithm and the limit is that logarithm is a continu- ous function (on its domain). Now we use the fact that ln(ab)=blna, so the log of the limit is

limx0+xlnx

Aha! The question has been turned into one we already did! But ignoring that, and repeating ourselves, we'd first rewrite this as a ratio

limx0+xlnx=limx0+lnxl/x

and then apply L'Hospital's rule to obtain

limx0+1/x-1/x2=limx0+-x=0

But we have to remember that we've computed the log of the limit, not the limit. Therefore, the actual limit is

limx0+xx=elog of the limit =e0=1

This trick of taking a logarithm is important to remember.

Example: Here is another issue of rearranging to fit into accessible form: Find

limx+x2+x+1-x2+1

This is not a ratio, but certainly is indetermiate', since it is the difference of two expressions both of

which go to +. To make it into a ratio, we take out the largest reasonable power of x:

limx+x2+x+1-x2+1=limx+x(1+1x+1x2-1+1x2)

=limx+1+1x+1x2-1+1x21/x

The last expression here fits the requirements of the L'Hospital rule, since both numerator and denomina- tor go to 0. Thus, by invoking L'Hospital's rule, it becomes

=limx+12-1x2-2x31+1x+1x2--2x31+1x2-1/x2

38

This is a large but actually tractable expression: multiply top and bottom by x2, so that it becomes

=limx+12+1x1+1x+1x2+1+1x2-1x

At this point, we can replace every 1x by 0, finding that the limit is equal to

12+01+0+0+01+0=12

It is important to recognize that in additional to the actual application of L'Hospital's rule, it may be nec- essary to experiment a little to get things to settle out the way you want. Trial-and-error is not only ok, it is necessary.

#26.85 Find limx0(sinx)/x

#26.86 Find limx0(sin5x)/x

#26.87 Find limx0(sin(x2))/x2

#26.88 Find limx0x/(e2x-1)

#26.89 Find limx0xlnx

#26.90 Find

limx0+(ex-1)lnx

#26.91 Find

limx1lnxx-1

#26.92 Find

limx+lnxx

#26.93 Find

limx+lnxx2

#26.94 Find limx0(sinx)x

27. Exponentialgrowth and decay. a differen Tia/equa tion

This little section is a tiny introduction to a very important subject and bunch of ideas: solving differen- tial equations. We'll just look at the simplest possible example of this.

The general idea is that, instead of solving equations to find unknown numbers, we might solve equations to find unknown functions. There are many possibilities for what this might mean, but one is that we

have an unknown function y of x and are given that y and its derivative y (with respect to x) satisfy a relation

y=ky

39

where k is some constant. Such a relation between an unknown function and its derivative (or derivatives) is what is called a differential equation. Many basic 'physical principles' can be written in such terms, using 'time' t as the independent variable.

Having been taking derivatives of exponential functions, a person might remember that the function f(t)=ekt has exactly this property:

ddtekt=kekt

For that matter, any constant multiple of this function has the same property:

ddt(cekt)=kcekt

And it turns out that these really are all the possible solutions to this differential equation.

There is a certain buzz-phrase which is supposed to alert a person to the occurrence of this little story: if

a function f has exponential growth or exponential decay then that is taken to mean that f can be written in the form

f(t)=cekt

If the constant k is positive it has exponential growth and if k is negative then it has exponential decay.

Since we've described all the solutions to this equation, what questions remain to ask about this kind of thing? Well, the usual scenario is that some story problem will give you information in a way that requires you to take some trouble in order to determine the constants c, k. And, in case you were wondering where you get to take a derivative here, the answer is that you don't really: all the 'calculus work' was done at the point where we granted ourselves that all solutions to that differential equation are given in the form f(t)=cekt.

First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that

c=f(0)

that is, that the constant c is the value of the function at time t=0. This is true simply because

f(0)=cek0=ce0=c1=c

from properties of the exponential function.

More generally, suppose we know the values of the function at two different times:

y1=cekt1

y2=cekt2

Even though we certainly do have 'two equations and two unknowns', these equations involve the un-

known constants in a manner we may not be used to. But it's still not so hard to solve for c, k: dividing the first equation by the second and using properties of the exponential function, the c on the right side cancels, and we get

y1y2=ek(t1-t2)

Taking a logarithm (base e, of course) we get

lny1-lny2=k(t1-t2)

Dividing by t1-t2, this is

k=lny1-lny2t1-t2

40

Substituting back in order to find c, we first have

y1=celny1-lny2t1-t2t1

Taking the logarithm, we have

lny1=lnc+lny1-lny2t1-t2t1

Rearranging, this is

lnc=lny1-lny1-lny2t1-t2t1=t1lny2-t2lny1t1-t2

Therefore, in summary, the two equations

y1=cekt1

y2=cekt2

allow us to solve for c, k, giving

k=lny1-lny2t1-t2

c=et1lny2-t2lny1t1-t2

A person might manage to remember such formulas, or it might be wiser to remember the way of deriving them.

Example: A herd of llamas has 1000llamas in it, and the population is growing exponentially. At time t=4 it has 2000 llamas. Write a formula for the number of llamas at arbitrary time t.

Here there is no direct mention of differential equations, but use of the buzz-phrase ζgrowing exponentially' must be taken as indicator that we are talking about the situation

f(t)=cekt

where here f(t) is the number of llamas at time t and c, k are constants to be determined from the infor- mation given in the problem. And the use of language should probably be taken to mean that at time

t=0 there are 1000llamas, and at time t=4 there are 2000. Then, either repeating the method above or plugging into the formula derived by the method, we find

c= value of f at t=0=1000

k=lnf(t1)-lnf(t2)t1-t2=ln1000-ln20000-4

=ln10002000-4=ln12-4=(ln2)/4

Therefore,

f(t)=1000eln24t=10002t/4

This is the desired formula for the number of llamas at arbitrary time t.

Example: A colony of bacteria is growing exponentially. At time t=0 it has 10 bacteria in it, and at time t=4 it has 2000. At what time will it have 100, 000 bacteria?

Even though it is not explicitly demanded, we need to find the general formula for the number f(t) of bac- teria at time t, set this expression equal to 100, 000, and solve for t. Again, we can take a little shortcut

here since we know that c=f(O) and we are given that f(O)=10. (This is easier than using the bulkier more general formula for finding c). And use the formula for k:

k=lnf(t1)-lnf(t2)t1-t2=ln10-ln2,0000-4=ln102,000-4=ln2004

41

Therefore, we have

f(t)=10eln2004t=10200t/4

as the general formula. Now we try to solve

100, 000=10eln2004t

for t: divide both sides by the 10 and take logarithms, to get

ln10,000=ln2004t

Thus,

t=4ln10,000ln2006.953407835

#27.95 A herd of llamas is growing exponentially. At time t=0 it has 1000llamas in it, and at time t=4 it has 2000 llamas. Write a formula for the number of llamas at arbitrary time t.

#27.96 A herd of elephants is growing exponentially. At time t=2 it has 1000 elephants in it, and at time t=4 it has 2000 elephants. Write a formula for the number of elephants at arbitrary time t.

#27.97 A colony of bacteria is growing exponentially. At time t=0 it has 10 bacteria in it, and at time t=4 it has 2000. At what time will it have 100, 000 bacteria?

#27.98 A colony of bacteria is growing exponentially. At time t=2 it has 10 bacteria in it, and at time t=4 it has 2000. At what time will it have 100, 000 bacteria?

28. Th e second and higher deriva tives

The second derivative of a function is simply the derivative of the derivative. The third derivative of a function is the derivative of the second derivative. And so on.

The second derivative of a function y=f(x) is written as

y=f(x)=d2dx2f=d2fdx2=d2ydx2

The third derivative is

y =f(x)=d3dx3f=d3fdx3=d3ydx3

And, generally, we can put on a 'prime' for each derivative taken. Or write

dndxnf=dnfdxn=dnydxn

for the nth derivative. There is yet another notation for high order derivatives where the number of 'primes' would become unwieldy:

dnfdxn=f(n)(x)

as well.

The geometric interpretation of the higher derivatives is subtler than that of the first derivative, and we won't do much in this direction, except for the next little section.

42

#28.99 Find f(x) for f(x)=x3-5x+1.

#28.100 Find f(x) for f(x)=x5-5x2+x-1. #28.101 Find f(x) for f(x)=x2-x+1.

#28.102 Find f(x) for f(x)=x.

29. In fle ction points, conca vity upward and downward

A point of inflection of the graph of a function f is a point where the second derivative f is 0. We have to wait a minute to clarify the geometric meaning of this.

A piece of the graph of f is concave upward if the curve 'bends' upward. For example, the popular parabola y=x2 is concave upward in its entirety.

A piece of the graph of f is concave downward if the curve 'bends' downward. For example, a 'flipped' version y=-x2 of the popular parabola is concave downward in its entirety.

The relation of points of inflection to intervals where the curve is concave up or down is exactly the same as the relation of critical points to intervals where the function is increasing or decreasing. That is, the points of inflection mark the boundaries of the two different sort of behavior Further, only one sample value of f need be taken between each pair of consecutive inflection points in order to see whether the curve bends up or down along that interval.

Expressing this as a systematic procedure: to find the intervals along which f is concave upward and con- cave downward:

Compute the second derivative f of f, and solve the equation f(x)=0 for x to find all the inflection points, which we list in order as x1<x2<...<xn. (Any points of discontinuity, etc., should be added to the list!)

We need some auxiliary points: To the left of the leftmost inflection point x1 pick any convenient point to, between each pair of consecutive inflection points xi, xi+1 choose any convenient point ti, and to the right of the rightmost inflection point xn choose a convenient point tn.

Evaluate the second derivative f at all the auxiliary points ti.

Conclusion: if f(ti+1)>0, then f is concave upward on (xi, xi+1) , while if f(ti+1)<0, then f is concave downward on that interval.

Conclusion: on the 'outside' interval (-, xo) , the function f is concave upward if f(to)>0 and is concave downward if f(to)<0. Similarly, on (xn, ) , the function f is concave upward if f(tn)>0 and is concave downward if f(tn)<0.

Find the inflection points and intervals of concavity up and down of

f(x)=3x2-9x+6

First, the second derivative is just f(x)=6. Since this is never zero, there are not points of inflection. And the value of f is always 6, so is always >0, so the curve is entirely concave upward.

43

Find the inflection points and intervals of concavity up and down of

f(x)=2x3-12x2+4x-27

First, the second derivative is f(x)=12x-24. Thus, solving 12x-24=0, there is just the one inflection point, 2. Choose auxiliary points to=0 to the left of the inflection point and t1=3 to the right of the inflection point. Then f(0)=-24<0, so on (-, 2) the curve is concave downward. And f(2)=12> 0, so on (2, ) the curve is concave upward.

Find the inflection points and intervals of concavity up and down of

f(x)=x4-24x2+11

the second derivative is f(x)=12x2-48. Solving the equation 12x2-48=0, we find inflection points ±2. Choosing auxiliary points -3,0,3 placed between and to the left and right of the inflection points, we evaluate the second derivative: First, f(-3)=129-48>0, so the curve is concave upward on (-, -2) . Second, f(0)=-48<0, so the curve is concave downward on (-2,2) . Third, f(3)=129-48>0, so the curve is concave upward on (2, ) .

#29.103 Find the inflection points and intervals of concavity up and down of f(x)=3x2-9x+6.

#29.104 Find the inflection points and intervals of concavity up and down of f(x)=2x3-12x2+4x-27.

#29.105 Find the inflection points and intervals of concavity up and down of f(x)=x4-2x2+11.

30. Another differen Tia/equa Tion. projectile motion

Here we encounter the fundamental idea that if s=s(t) is position, then s˙ is velocity, and s is accelera- tion. This idea occurs in all basic physical science and engineering.

In particular, for a projectile near the earth's surface travelling straight up and down, ignoring air resis- tance, acted upon by no other forces but gravity, we have

acceleration due to gravity =-32 feet/sec 2

Thus, letting s(t) be position at time t, we have

s¨(t)=-32

We take this (approximate) physical fact as our starting point.

From s¨=-32 we integrate (or anti-differentiate) once to undo one of the derivatives, getting back to ve- locity:

v(t)=s˙=s˙(t)=-32t+vo

where we are calling the constant of integration vo'. (No matter which constant vo we might take, the

derivative of -32t+vo with respect to t is -32.)

Specifically, when t=0, we have

v(0)=vo

Thus, the constant of integration vo is initial velocity. And we have this formula for the velocity at any time in terms of initial velocity.

44

We integrate once more to undo the last derivative, getting back to the position function itself:

s=s(t)=-16t2+vot+so

where we are calling the constant of integration so'. Specifically, when t=0, we have

s(0)=so

so so is initial position. Thus, we have a formula for position at any time in terms of initial position and initial velocity.

Of course, in many problems the data we are given is not just the initial position and initial velocity, but something else, so we have to determine these constants indirectly.

#30.106 You drop a rock down a deep well, and it takes 10 seconds to hit the bottom. How deep is it? #30.107 You drop a rock down a well, and the rock is going 32 feet per second when it hits bottom. How deep is the well?

#30.108 If I throw a ball straight up and it takes 12 seconds for it to go up and come down, how high

did it go?

31. Graphing ra tional functions, asymptotes

This section shows another kind of function whose graphs we can understand effectively by our methods. There is one new item here, the idea of asymptote of the graph of a function.

A vertical asymptote of the graph of a function f most commonly occurs when f is defined as a ratio f(x)=g(x)/h(x) of functions g, h continuous at a point xo, but with the denominator going to zero at

that point while the numerator doesn't. That is, h(xo)=0 but g(xo)0. Then we say that f blows up at

xo, and that the line x=xo is a vertical asymptote of the graph of f.

And as we take x closer and closer to xo, the graph of f zooms off (either up or down or both) closely to the line x=xo.

A very simple example of this is f(x)=1/(x-1) , whose denominator is 0 at x=1, so causing a blow-up at that point, so that x=1 is a vertical asymptote. And as x approaches 1 from the right, the values of the function zoom up to +. When x approaches 1 from the left, the values zoom down to -.

A horizontal asymptote of the graph of a function f occurs if either limit

limx+f(x)

or

limx- f(x)

exists. If R=limx+f(x) , then y=R is a horizontal asymptote of the function, and if L=

limx-f(x) exists then y=L is a horizontal asymptote.

As x goes off to + the graph of the function gets closer and closer to the horizontal line y=R if that limit exists. As x goes of to - the graph of the function gets closer and closer to the horizontal line y= L if that limit exists.

So in rough terms asymptotes of a function are straight lines which the graph of the function approaches at infinity. In the case of vertical asymptotes, it is the y-coordinate that goes off to ifinity, and in the case of horizontal asymptotes it is the x-coordinate which goes off to ifinity.

45

Find asymptotes, critical points, intervals of increase and decrease, inflection points, and intervals of con- cavity up and down of f(x)=x+32x-6: First, let's find the asymptotes. The denominator is 0 for x=3 (and this is not cancelled by the numerator) so the line x=3 is a vertical asymptote. And as x goes to ±, the function values go to 1/2, so the line y=1/2 is a horizontal asymptote.

The derivative is

f(x)=1(2x-6)-(x+3)2(2x-6)2=-12(2x-6)2

Since a ratio of polynomials can be zero only if the numerator is zero, this f(x) can never be zero, so

there are no critical points. There is, however, the discontinuity at x=3 which we must take into ac-

count. Choose auxiliary points 0 and 4 to the left and right of the discontinuity. Plugging in to the deriva- tive, we have f(O)=-12/(-6)2<0, so the function is decreasing on the interval (-, 3) . To the right, f(4)=-12/(8-6)2<0, so the function is also decreasing on (3, +) .

The second derivative is f(x)=48/(2x-6)3. This is never zero, so there are no inflection points. There is the discontinuity at x=3, however. Again choosing auxiliary points 0,4 to the left and right of the

discontinuity, we see f(0)=48/(-6)3<0 so the curve is concave downward on the interval (-, 3) . And f(4)=48/(8-6)3>0, so the curve is concave upward on (3, +) .

Plugging in just two or so values into the function then is enough to enable a person to make a fairly good qualitative sketch of the graph of the function.

#31.109 Find all asymptotes of f(x)=x-1x+2. #31.110 Find all asymptotes of f(x)=x+2x-1. #31.111 Find all asymptotes of f(x)=x2-1x2-4. #31.112 Find all asymptotes of f(x)=x2-1x2+1.

32. Basic integration formulas

The fundamental use of integration is as a continuous version of summing. But, paradoxically, often inte- grals are computed by viewing integration as essentially an inverse operation to differentiation. (That fact is the so-called Fundamental Theorem of Calculus.)

The notation, which we're stuck with for historical reasons, is as peculiar as the notation for derivatives:

the integral of a function f(x) with respect to x is written as

f(x)dx

The remark that integration is (almost) an inverse to the operation of differentiation means that if

ddxf(x)=g(x)

then

g(x)dx=f(x)+C

The extra C, called the constant of integration, is really necessary, since after all differentiation kills off constants, which is why integration and differentiation are not exactly inverse operations of each other.

46

Since integration is almost the inverse operation of differentiation, recollection of formulas and processes for differentiation already tells the most important formulas for integration:

xndx = 1n+1xn+1+C unless n=-1

exdx = ex+C

1xdx = lnx+C

sinxdx = -cosx+C

cosxdx = sinx+C

sec2xdx = tanx+C

11+x2dx = arctanx+C

And since the derivative of a sum is the sum of the derivatives, the integral of a sum is the sum of the inte- grals:

f(x)+g(x)dx=f(x)dx+g(x)dx

And, likewise, constants 'go through' the integral sign:

cf(x)dx=cf(x)dx

For example, it is easy to integrate polynomials, even including terms like x and more general power functions. The only thing to watch out for is terms x-1=1x, since these integrate to lnx instead of a power of x. So

4x5-3x+11-17x+3xdx=4x66-3x22+11x-17x3/23/2+3lnx+C

Notice that we need to include just one 'constant of integration'.

Other basic formulas obtained by reversing differentiation formulas:

axdx = axlna+C

logaxdx = 1lna1x+C

11-x2dx = arcsinx+C

1xx2-1dx = arcsec x+C

Sums of constant multiples of all these functions are easy to integrate: for example,

52x-23xx2-1+5x2dx=52xln2-23 arcsec x+5x33+C

#32.113 4x3-3cosx+7x+2dx=? #32.114 3x2+e2x-11+cosxdx=? #32.115 sec2xdx=?

#32.116 71+x2dx=?

#32.117 16x7-x+3xdx=?

#32.118 23sinx-21-x2dx=?

47

33. Th e simplest substitutions

The simplest kind of chain rule application

ddxf(ax+b)=af(x)

(for constants a, b) can easily be run backwards to obtain the corresponding integral formulas: some and illustrative important examples are

cos(ax+b)dx = 1asin(ax+b)+C

eax+bdx = 1aeax+b+C

ax+bdx = 1a(ax+b)3/23/2+C

1ax+bdx = 1aln(ax+b)+C

Putting numbers in instead of letters, we have examples like

cos(3x+2)dx = 13sin(3x+2)+C

e4x+3dx = 14e4x+3+C

-5x+1dx = 1-5(-5x+1)3/23/2+C

17x-2dx = 17ln(7x-2)+C

Since this kind of substitution is pretty undramatic, and a person should be able to do such things by re- flex rather than having to think about it very much.

#33.119 e3x+2dx=?

#33.120 cos(2-5x)dx=?

#33.121 3x-7dx=?

#33.122 sec2(2x+1)dx=?

#33.123 (5x7+e6-2x+23+2x)dx=? #33.124 cos (7-llx) dx=?

34. Substitutions

The chain rule can also be 'run backward', and is called change of variables or substitution or some- times u-substitution. Some examples of what happens are straightforward, but others are less obvious. It is at this point that the capacity to recognize derivatives from past experience becomes very helpful.

Example Since (by the chain rule)

ddxesinx=cosxesinx

then we can anticipate that

cosxesinxdx=esinx+C

48

Example Since (by the chain rule)

ddxx5+3x=12(x5+3x)-1/2(5x4+3)

then we can anticipate that

12(5x4+3)(x5+3x)-1/2dx=x5+3x+C

Very often it happens that things are off by a constant. This should not deter a person from recognizing the possibilities. For example: since, by the chain rule,

ddx5+ex=12(5+ex)-1/2ex

then

ex(5+ex)-1/2dx=212ex(5+ex)-1/2dx=25+ex+C

Notice how for 'bookkeeping purposes' we put the 12 into the integral (to make the constants right there) and put a compensating 2 outside.

Example: Since (by the chain rule)

ddxsin7(3x+1)=7sin6(3x+1)cos(3x+1)3

then we have

cos(3x+1)sin6(3x+1)dx

=12173cos(3x+1)sin6(3x+1)dx=121sin7(3x+1)+C

#34.125 cosxsinxdx=?

#34.126 2xex2dx=?

#34.127 6x5ex6dx=?

#34.128 cosxsinxdx=?

#34.129 cosxesinxdx=?

#34.130 12xexdx=?

#34.131 cosxsin5xdx=?

#34.132 sec2xtan7xdx=?

#34.133 (3cosx+x)e6sinx+x2dx=?

#34.134 exex+1dx=?

49

35. Area and definite integrals

The actual definition of 'integral' is as a limit of sums, which might easily be viewed as having to do with area. One of the original issues integrals were intended to address was computation of area.

First we need more notation. Suppose that we have a function f whose integral is another function F:

f(x)dx=F(x)+C

Let a, b be two numbers. Then the definite integral of f with limits a, b is

abf(x)dx=F(b)-F(a)

The left-hand side of this equality is just notation for the definite integral. The use of the word 'limit' here has little to do with our earlier use of the word, and means something more like 'boundary', just like it

does in more ordinary English.

A similar notation is to write

[g(x)]ab=g(b)-g(a)

for any function g. So we could also write

abf(x)dx=[F(x)]ab

For example,

05x2dx=[x33]05=53-033=1253

As another example,

233x+1dx= [3x22+x]23= (3322+3)-(3222+2)=212

All the other integrals we had done previously would be called indefinite integrals since they didn't have 'limits' a, b. So a definite integral is just the difference of two values of the function given by an indefinite integral. That is, there is almost nothing new here except the idea of evaluating the function that we get by integrating.

But now we can do something new: compute areas:

For example, if a function f is positive on an interval [a, b], then

abf(x)dx= area between graph and x-axis, between x=a and x=b

It is important that the function be positive, or the result is false.

For example, since y=x2 is certainly always positive (or at least non-negative, which is really enough), the area 'under the curve' (and, implicitly, above the xhyphenaxis) between x=0 and x=1 is just

01x2dx=[x33]01=13-033=13

50

More generally, the area below y=f(x) , above y=g(x) , and between x=a and x=b is

area =abf(x)-g(x)dx

=leftliml˙lrightlimit (upper curve- lower curve) dx

It is important that f(x)g(x) throughout the interval [a, b].

For example, the area below y=ex and above y=x, and between x=0 and x=2 is

02ex-xdx=[ex-x22]02=(e2-2)-(e0-0)=e2+1

since it really is true that exx on the interval [0,2].

As a person might be wondering, in general it may be not so easy to tell whether the graph of one curve is above or below another. The procedure to examine the situation is as follows: given two functions f, g, to find the intervals where f(x)g(x) and vice-versa:

Find where the graphs cross by solving f(x)=g(x) for x to find the x-coordinates of the points of inter- section.

Between any two solutions x1, x2 of f(x)=g(x) (and also to the left and right of the left-most and right- most solutions!), plug in one auxiliary point of your choosing to see which function is larger.

Of course, this procedure works for a similar reason that the first derivative test for local minima and

maxima worked: we implicitly assume that the f and g are continuous, so if the graph of one is above the graph of the other, then the situation can't reverse itself without the graphs actually crossing.

As an example, and as an example of a certain delicacy of wording, consider the problem to find the area between y=x and y=x2 with 0x2. To find where y=x and y=x2 cross, solve x=x2: we find solutions x=0,1. In the present problem we don't care what is happening to the left of 0. Plugging in the value 1/2 as auxiliary point between 0 and 1, we get 12 (12)2, so we see that in [0,1] the curve y=x is the higher. To the right of 1 we plug in the auxiliary point 2, obtaining 222, so the curve y=x2 is higher there.

Therefore, the area between the two curves has to be broken into two parts:

area =01(x-x2)dx+12(x2-x)dx

since we must always be integrating in the form

leftright higher-lower dx

In some cases the 'side' boundaries are redundant or only implied. For example, the question might be to find the area between the curves y=2-x and y=x2. What is implied here is that these two curves themselves enclose one or more finite pieces of area, without the need of any 'side' boundaries of the form x=a. First, we need to see where the two curves intersect, by solving 2-x=x2: the solutions are x=-2,1. So we infer that we are supposed to find the area from x=-2 to x=1, and that the two curves close up around this chunk of area without any need of assistance from vertical lines x=a. We need to find which curve is higher: plugging in the point 0 between -2 and 1, we see that y=2-x is higher. Thus, the desired integral is

area =-21(2-x)-x2dx

51

#35.135 Find the area between the curves y=x2 and y=2x+3.

#35.136 Find the area of the region bounded vertically by y=x2 and y=x+2 and bounded horizontally by x=-1 and x=3.

#35.137 Find the area between the curves y=x2 and y=8+6x-x2.

#35.138 Find the area between the curves y=x2+5 and y=x+7.

36. Lengths of Curves

The basic point here is a formula obtained by using the ideas of calculus: the length of the graph of y= f(x) from x=a to x=b is

arc length =ab1+(dydx)2dx

Or, if the curve is parametrized in the form

x=f(t) y=g(t)

with the parameter t going from a to b, then

arc length =ab(dxdt)2+(dydt)2dt

This formula comes from approximating the curve by straight lines connecting successive points on the

curve, using the Pythagorean Theorem to compute the lengths of these segments in terms of the change in x and the change in y. In one way of writing, which also provides a good heuristic for remembering the formula, if a small change in x is dx and a small change in y is dy, then the length of the hypotenuse of the right triangle with base dx and altitude dy is (by the Pythagorean theorem)

hypotenuse =dx2+dy2=1+(dydx)2dx

Unfortunately, by the nature of this formula, most of the integrals which come up are difficult or impossi- ble to 'do'. But if one of these really mattered, we could still estimate it by numerical integration.

#36.139 Find the length of the curve y=1-x2 from x=0 to x=1.

#36.140 Find the length of the curve y=14(e2x+e-2x) from x=0 to x=1.

#36.141 Set up (but do not evaluate) the integral to find the length of the piece of the parabola y=x2 from x=3 to x=4.

52

37. Numerical integration

As we start to see that integration 'by formulas' is a much more difficult thing than differentiation, and

sometimes is impossible to do in elementary terms, it becomes reasonable to ask for numerical approxima- tions to definite integrals. Since a definite integral is just a number, this is possible. By constrast, indefi- nite integrals, being functions rather than just numbers, are not easily described by 'numerical approxima- tions'.

There are several related approaches, all of which use the idea that a definite integral is related to area. Thus, each of these approaches is really essentially a way of approximating area under a curve. Of course, this isn't exactly right, because integrals are not exactly areas, but thinking of area is a reasonable heuris- tic.

Of course, an approximation is not very valuable unless there is an estimate for the error, in other words, an idea of the tolerance.

Each of the approaches starts the same way: To approximate abf(x)dx, break the interval [a, b] into

smaller subintervals

[x0, x1], [x1, x2], . . . , [xn-2, xn-1], [xn-1, xn]

each of the same length

x=b-na

and where x0=a and xn=b.

Trapezoidal rule: This rule says that

abf(x)dxx2[f(x0)+2f(x1)+2f(x2)+...+2f(xn-2)+2f(xn-1)+f(xn)]

Yes, all the values have a factor of '2' except the first and the last. (This method approximates the area under the curve by trapezoids inscribed under the curve in each subinterval).

Midpoint rule: Let x¯i=12(xi-xi-1) be the midpoint of the subinterval [xi-1, xi]. Then the midpoint rule says that

abf(x)dxx[f(x¯1)+...+f(x¯n)]

(This method approximates the area under the curve by rectangles whose height is the midpoint of each subinterval).

Simpson's rule: This rule says that

abf(x)dx

x3[f(x0)+4f(x1)+2f(x2)+4f(x3)+...+2f(xn-2)+4f(xn-1)+f(xn)]

Yes, the first and last coefficients are '1', while the 'inner' coefficients alternate '4' and '2'. And n has to be an even integer for this to make sense. (This method approximates the curve by pieces of parabolas). In general, the smaller the x is, the better these approximations are. We can be more precise: the er- ror estimates for the trapezoidal and midpoint rules depend upon the second derivative: suppose that

|f(x)|M for some constant M, for all axb. Then

error in trapezoidal rule M(b-a)312n2

53

error in midpoint rule M(b-a)324n2

The error estimate for Simpson's rule depends on the fourth derivative: suppose that |f(4)(x)|N for some constant N, for all axb. Then

error in Simpson's rule N(b-a)5180n4

From these formulas estimating the error, it looks like the midpoint rule is always better than the trape- zoidal rule. And for high accuracy, using a large number n of subintervals, it looks like Simpson's rule is the best.

38. A verages and Weigh Ted A verages

The usual notion of average of a list of n numbers x1, ..., xn is

average of x1, x2, ..., xn=x1+x2+...+xnn

A continuous analogue of this can be obtained as an integral, using a notation which matches better: let f be a function on an interval [a, b]. Then

average value of f on the interval [a, b]=abf(x)dxb-a

For example the average value of the function y=x2 over the interval [2, 3] is

average value of f on the interval [a, b]=23x2dx3-2=[x3/3]233-2=33-233(3-2)=19/3

A weighted average is an average in which some of the items to be averaged are ζmore important' or

ζless important' than some of the others. The weights are (non-negative) numbers which measure the rela- tive importance.

For example, the weighted average of a list of numbers x1, ..., xn with corresponding weights w1, ..., wn is

w1 . x1+w2x2+...+wnxn

w1+w2+...+wn

Note that if the weights are all just 1, then the weighted average is just a plain average.

The continuous analogue of a weighted average can be obtained as an integral, using a notation which matches better: let f be a function on an interval [a, b], with weight w(x) , a non-negative function on [a, b]. Then

weighted average value of f on the interval [a, b] with weight w=abw(x)f(x)dxabw(x)dx

Notice that in the special case that the weight is just 1 all the time, then the weighted average is just a plain average.

For example the average value of the function y=x2 over the interval [2, 3] with weight w(x)=x is

average value of f on the interval [a, b] with weight x

54

=23xx2dx23xdx=[x4/4]23[x2/2]23=14(34-24)12(32-22)

39. Centers of Mass (Centroids)

For many (but certainly not all!) purposes in physics and mechanics, it is necessary or useful to be able to consider a physical object as being a mass concentrated at a single point, its geometric center, also called its centroid. The centroid is essentially the ζaverage ' of all the points in the object. For simplicity, we will just consider the two-dimensional version of this, looking only at regions in the plane.

The simplest case is that of a rectangle: it is pretty clear that the centroid is the 'center' of the rectangle. That is, if the corners are (0,0), (u, 0), (0, v) and (u, v) , then the centroid is

(uv2¯'2¯)

The formulas below are obtained by 'integrating up' this simple idea:

For the center of mass (centroid) of the plane region described by f(x)yg(x) and axb, we have

x-coordinate of the centroid = average x-coordinate

=abx[g(x)-f(x)]dxab[g(x)-f(x)]dx

leftrightx[upper-lower] dx leftrightx[upper-lower]dx

=leftright[upper-lower]dx¯=area¯of the region

And also

y-coordinate of the centroid = average y-coordinate

=ab12[g(x)2-f(x)2]dxab[g(x)-f(x)]dx

leftright12[uppe2-lower2]dx leftright12[uppe2-lower2]dx

=leftright[upper-lower]dx¯=area¯of the region

Heuristic: For the x-coordinate: there is an amount (g(x)-f(x))dx of the region at distance x from the y- axis. This is integrated, and then averaged dividing by the total, that is, dividing by the area of the entire region.

For the y-coordinate: in each vertical band of width dx there is amount dxdy of the region at distance y from the x-axis. This is integrated up and then averaged by dividing by the total area.

For example, let's find the centroid of the region bounded by x=0, x=1, y=x2, and y=0.

x-coordinate of the centroid =01x[x2-0]dx01[x2-0]dx

=[x4/4]01[x3/3]01=1/4-01/3-0=34

55

And

y-coordinate of the centroid =0112[(x2)2-0]dx01[x2-0]dx

=12[x5/5]01[x3/3]01=12(1/5-0)1/3-0=310

#39.142 Find the center of mass (centroid) of the region 0x1 and 0yx2.

#39.143 Find the center of mass (centroid) of the region defined by 0x1,0y1 and x+y1. #39.144 Find the center of mass (centroid) of a homogeneous plate in the shape of an equilateral trian- gle.

40. Volum es by Cross Sections

Next to computing areas of regions in the plane, the easiest concept of application of the ideas of calculus is to computing volumes of solids where somehow we know a formula for the areas of slices, that is, areas of cross sections. Of course, in any particular example, the actual issue of getting the formula for the cross section, and figuring out the appropriate limits of integration, can be difficult.

The idea is to just 'add them up':

volume =1eftlimitrightlimit (area of cross section at x) dx

where in whatever manner we describe the solid it extends from x=left limit to x=right limit. We must suppose that we have some reasonable formula for the area of the cross section.

For example, let's find the volume of a solid ball of radius 1. (In effect, we'll be deriving the formula for this). We can suppose that the ball is centered at the origin. Since the radius is 1, the range of x coordi- nates is from -1 to +1, so x will be integrated from -1 to +1. At a particular value of x, what does the cross section look like? A disk, whose radius we'll have to determine. To determine this radius, look at how the solid ball intersects the x, y-plane: it intesects in the disk x2+y21. For a particular value of x, the values of y are between ±1-x2. This line segment, having x fixed and y in this range, is the in- tersection of the cross section disk with the x, y-plane, and in fact is a diameter of that cross section disk. Therefore, the radius of the cross section disk at x is 1-x2. Use the formula that the area of a disk of radius r is πr2: the area of the cross section is

cross section at x=π(1-x2)2=π(1-x2)

Then integrate this from -1 to +1 to get the volume:

volume =1eftright area of cross-section dx

=-1+1π(1-x2)dx=π[x-x33]-1+1=π[(1-13)-(-1-(-1)33)]=23+23=43

56

#40.145 Find the volume of a circular cone of radius 10 and height 12 (not by a formula, but by cross sections).

#40.146 Find the volume of a cone whose base is a square of side 5 and whose height is 6, by cross- sections.

#40.147 A hole 3 units in radius is drilled out along a diameter of a solid sphere of radius 5 units. What is the volume of the remaining solid?

#40.148 A solid whose base is a disc of radius 3 has vertical cross sections which are squares. What is the volume?

41. Solids ofRe volution

Another way of computing volumes of some special types of solid figures applies to solids obtained by ro- tating plane regions about some axis.

If we rotate the plane region described by f(x)yg(x) and axb around the x-axis, the volume of the resulting solid is

volume =abπ(g(x)2-f(x)2)dx

=1eftlimitrightlimit π (upper curve2-lower curve2)dx It is necessary to suppose that f(x)0 for this to be right.

This formula comes from viewing the whole thing as sliced up into slices of thickness dx, so that each slice is a disk of radius g(x) with a smaller disk of radius f(x) removed from it. Then we use the formula

area of disk =π radius2

and 'add them all up'. The hypothesis that f(x)0 is necessary to avoid different pieces of the solid 'overlap' each other by accident, thus counting the same chunk of volume twice.

If we rotate the plane region described by f(x)yg(x) and axb around the y-axis (instead of the x-axis), the volume of the resulting solid is

volume =ab2πx(g(x)-f(x))dx

=1eftright2πx ( upper - lower) dx

This second formula comes from viewing the whole thing as sliced up into thin cylindrical shells of thick- ness dx encircling the y-axis, of radius x and of height g(x)-f(x) . The volume of each one is

(area of cylinder of height g(x)-f(x) and radius x) . dx=2πx(g(x)-f(x))dx

and 'add them all up' in the integral.

As an example, let's consider the region 0x1 and x2yx. Note that for 0x1 it really is the case that x2yx, so y=x is the upper curve of the two, and y=x2 is the lower curve of the two. Invoking the formula above, the volume of the solid obtained by rotating this plane region around the x-axis is

volume =1eftrightπ(upper2-1ower2)dx

57

=01π((x)2-(x2)2)dx=π[x3/3-x5/5]01=π(1/3-1/5)

On the other hand, if we rotate this around the y-axis instead, then

volume =1eftright2πx (upper-lower) dx

=012πx(x-x2)dx=π012x33-2x44dx= [2x33-2x44]01=23-12=16

#41.149 Find the volume of the solid obtained by rotating the region 0x1,0yx around the y-axis.

#41.150 Find the volume of the solid obtained by rotating the region 0x1,0yx around the x-axis.

#41.151 Set up the integral which expresses the volume of the doughnut obtained by rotating the region (x-2)2+y21 around the y-axis.

42. Surfaces ofRe volution

Here is another formula obtained by using the ideas of calculus: the area of the surface obtained by rotat- ing the curve y=f(x) with axb around the x-axis is

area =ab2πf(x)1+(dydx)2dX

This formula comes from extending the ideas of the previous section the length of a little piece of the curve is

dx2+dy2

This gets rotated around the perimeter of a circle of radius y=f(x) , so approximately give a band of width dx2+dy2 and length 2πf(x) , which has area

2πf(x)dx2+dy2=2πf(x)1+(dydx)2dx

Integrating this (as if it were a sum!) gives the formula.

As with the formula for arc length, it is very easy to obtain integrals which are difficult or impossible to evaluate except numerically.

Similarly, we might rotate the curve y=f(x) around the y-axis instead. The same general ideas apply to compute the area of the resulting surface. The width of each little band is still dx2+dy2, but now the length is 2πx instead. So the band has area

width × length =2πxdx2+dy2

58

Therefore, in this case the surface area is obtained by integrating this, yielding the formula

area =ab2πX1+(dydx)2dx

#42.152 Find the area of the surface obtained by rotating the curve y=14(e2x+e-2x) with 0x1 around the x-axis.

#42.153 Just set up the integral for the surface obtained by rotating the curve y=14(e2x+e-2x) with 0x1 around the y-axis.

#42.154 Set up the integral for the area of the surface obtained by rotating the curve y=x2 with 0 x1 around the x-axis.

#42.155 Set up the integral for the area of the surface obtained by rotating the curve y=x2 with 0 x1 around the y-axis.

43. Integration by parTs

Strangely, the subtlest standard method is just the product rule run backwards. This is called integration by parts. (This might seem strange because often people find the chain rule for differentiation harder to get a grip on than the product rule). One way of writing the integration by parts rule is

f(x)g(x)dx=f(x)g(x)-f(x)g(x)dx

Sometimes this is written another way: if we use the notation that for a function u of x,

du=dudxdx

then for two functions u, v of x the rule is

udv=uv-vdu

Yes, it is hard to see how this might be helpful, but it is. The first theme we'll see in examples is where we could do the integral except that there is a power of x 'in the way':

The simplest example is

xexdx=xd(ex)=xex-exdx=xex-ex+C

Here we have taken u=x and v=ex. It is important to be able to see the ex as being the derivative of itself.

A similar example is

xcosxdx=xd(sinx)=xsinx-sinxdx=xsinx+cosx+C

59

Here we have taken u=x and v=sinx. It is important to be able to see the cosx as being the derivative of sinx.

Yet another example, illustrating also the idea of repeating the integration by parts:

x2exdx=x2d(ex)=x2ex-exd(x2)

=x2ex-2xexdx=x2ex-2xex+2exdx

=x2ex-2xex+2ex+C

Here we integrate by parts twice. After the first integration by parts, the integral we come up with is xexdx, which we had dealt with in the first example.

Or sometimes the theme is that it is easier to integrate the derivative of something than to integrate the thing:

lnxdx=lnxd(x)=xlnx-xd(lnx)

=xlnx-x1xdx=xlnx-1dx=xlnx-x+C

We took u=lnx and v=x.

Again in this example it is easier to integrate the derivative than the thing itself:

arctanxdx=arctanxd(x)=xarctanx-x d(arctan x)

=xarctanx-x1+x2dx=xarctanx-122x1+x2dx

=xarctanx-12ln(1+x2)+C

since we should recognize the

2x1+x2

as being the derivative (via the chain rule) of ln(1+x2) .

#43.156 lnxdx=?

#43.157 xexdx=?

#43.158 (lnx)2dx=?

#43.159 xe2xdx=?

#43.160 arctan3xdx=? #43.161 x3lnxdx=?

#43.162 ln3xdx=?

#43.163 xlnxdx=?

60

44. Partial Fractions

Now we return to a more special but still important technique of doing indefinite integrals. This depends on a good trick from algebra to transform complicated rational functions into simpler ones. Rather than try to formally describe the general fact, we'll do the two simplest families of examples.

Consider the integral

1x(x-1)dx

As it stands, we do not recognize this as the derivative of anything. However, we have

1x-1-1x=x-(x-1)x(x-1)=1x(x-1)

Therefore,

1x(x-1)dx=1x-1-1xdx=ln(x-1)-lnx+C

That is, by separating the fraction 1/x(x-1) into the 'partial' fractions 1/x and 1/(x-1) we were able to do the integrals immediately by using the logarithm. How to see such identities?

Well, let's look at a situation

cx+d(x-a)(x-b)=Ax-a+Bx-b

where a, b are given numbers (not equal) and we are to find A, B which make this true. If we can find the A, B then we can integrate (cx+d)/(x-a)(x-b) simply by using logarithms:

cx+d(x-a)(x-b)dx=Ax-a+Bx-bdx=Aln(x-a)+Bln(x-b)+C

To find the A, B, multiply through by (x-a)(x-b) to get

cx+d=A(x-b)+B(x-a)

When x=a the x-a factor is 0, so this equation becomes

ca+d=A(a-b)

Likewise, when x=b the x-b factor is 0, so we also have

cb+d=B(b-a)

That is,

A=ca+da-b B=cb+db-a

So, yes, we can find the constants to break the fraction (cx+d)/(x-a)(x-b) down into simpler 'partial' fractions.

Further, if the numerator is of bigger degree than 1, then before executing the previous algebra trick we must first divide the numerator by the denominator to get a remainder of smaller degree. A simple example is

x3+4x2-x+1x(x-1)=?

61

We must recall how to divide polynomials by polynomials and get a remainder of lower degree than the divi- sor. Here we would divide the x3+4x2-x+1 by x(x-1)=x2-x to get a remainder of degree less than 2 (the degree of x2-x). We would obtain

x3+4x2-x+1x(x-1)=x+5+4x+1x(x-1)

since the quotient is x+5 and the remainder is 4x+1. Thus, in this situation

x3+4x2-x+1x(x-1)dx=x+5+4x+1x(x-1)dx

Now we are ready to continue with the first algebra trick.

In this case, the first trick is applied to

4x+1x(x-1)

We want constants A, B so that

4x+1x(x-1)=Ax+Bx-1

As above, multiply through by x(x-1) to get

4x+1=A(x-1)+Bx

and plug in the two values 0,1 to get

4 0+1=-A 4 1+1=B

That is, A=-1 and B=5.

Putting this together, we have

x3+4x2-x+1x(x-1)=x+5+-1x+5x-1

Thus,

x3+4x2-x+1x(x-1)dx=x+5+-1x+5x-1dx

=x22+5x-lnx+5ln(x-1)+C

In a slightly different direction: we can do any integral of the form

ax+b1+x2dx

because we know two different sorts of integrals with that same denominator:

11+x2dx=arctanx+C 2x1+x2dx=ln(1+x2)+C

where in the second one we use a substitution. Thus, we have to break the given integral into two parts to do it:

ax+b1+x2dx=a22x1+x2dx+b11+x2dx

62

=a2ln(1+x2)+barctanx+C

And, as in the first example, if we are given a numerator of degree 2 or larger, then we divide first, to get a remainder of lower degree. For example, in the case of

x4+2x3+x2+3x+11+x2dx

we divide the numerator by the denominator, to allow us to write

x4+2x3+x2+3x+11+x2=x2+2x+x+11+x2

since the quotient is x2+2x and the remainder is x+1. Then

x4+2x3+x2+3x+11+x2dx=x2+2x+x+11+x2

=x33+x2+12ln(1+x2)+arctanx+C

These two examples are just the simplest, but illustrate the idea of using algebra to simplify rational func- tions.

#44.164 1x(x-1)dx=?

#44.165 1+x1+x2dx=?

#44.166 2x3+4x(x+1)dx=?

#44.167 2+2x+x21+x2dx=? #44.168 2x3+4x2-1dx=?

#44.169 2+3x1+x2dx=?

#44.170 x3+1(x-1)(x-2)dx=? #44.171 x3+1x2+1dx=?

45. Trigonometric Integrals

Here we'll just have a sample of how to use trig identities to do some more complicated integrals involv- ing trigonometric functions. This is 'just the tip of the iceberg'. We don't do more for at least two rea- sons: first, hardly anyone remembers all these tricks anyway, and, second, in real life you can look these things up in tables of integrals. Perhaps even more important, in 'real life' there are more sophisticated viewpoints which even make the whole issue a little silly, somewhat like evaluating 26 'by differentials' without your calculator seems silly.

The only identities we'll need in our examples are

cos2x+sin2x=1 Pythagorean identity

63

sinx=1-cos2x2 half-angle formula

cosx=1+cos2x2 half-angle formula

The first example is

sin3xdx

If we ignore all trig identities, there is no easy way to do this integral. But if we use the Pythagorean identity to rewrite it, then things improve:

sin3xdx=(1-cos2x)sinxdx=-(1-cos2x sinx)dx

In the latter expression, we can view the -sinx as the derivative of cosx, so with the substitution u= cosx this integral is

-(1-u2)du=-u+u33+C=-cosx+cos3x3+C

This idea can be applied, more generally, to integrals

sinmxcosnxdx

where at least one of m, n is odd. For example, if n is odd, then use

cosnx=cosn-1xcosx=(1-sin2x)n-12cosx

to write the whole thing as

sinmxcosnxdx=sinmx(1-sin2x)n-12cosxdx

The point is that we have obtained something of the form

(polynomial in sinx) cosxdx

Letting u=sinx, we have cosxdx=du, and the integral becomes

(polynomial in u) du

which we can do.

But this Pythagorean identity trick does not help us on the relatively simple-looking integral

sin2xdx

since there is no odd exponent anywhere. In effect, we 'divide the exponent by two', thereby getting an odd exponent, by using the half-angle formula:

sin2xdx=1-cos2x2dx=x2-sin.2x22+C

64

A bigger version of this application of the half-angle formula is

sin6xdx=(1-cos2x2)3dx=18-38cos2x+38cos22x-18cos32xdx

Of the four terms in the integrand in the last expression, we can do the first two directly:

18dx=x8+C -38cos2xdx=-316sin2x+C

But the last two terms require further work: using a half-angle formula again, we have

38cos22xdx=316(1+cos4x)dx=3x16+364sin4x+C

And the cos32x needs the Pythagorean identity trick:

18cos32xdx=18(1-sin22x)cos2xdx=18[sin2x-sin32x3]+C

Putting it all together, we have

sin6xdx=x8+-316sin2x+3x16+364sin4x+18[sin2x-sin32x3]+C

This last example is typical of the kind of repeated application of all the tricks necessary in order to treat all the possibilities.

In a slightly different vein, there is the horrible

secxdx

There is no decent way to do this at all from a first-year calculus viewpoint. A sort of rationalized-in-

hindsight way of explaining the answer is:

secxdx=secx(secx+tanx)secx+tanxdx

All we did was multiply and divide by secx+tanx. Of course, we don't pretend to answer the question of how a person would get the idea to do this. But then (another miracle?) we 'notice' that the numerator is the derivative of the denominator, so

secxdx=ln(secx+tanx)+C

There is something distasteful about this rationalization, but at this level of technique we're stuck with it. Maybe this is enough of a sample. There are several other tricks that one would have to know in order to claim to be an 'expert' at this, but it's not really sensible to want to be 'expert' at these games, because

there are smarter alternatives.

#45.172 cos2xdx=?

#45.173 cosxsin2xdx=? #45.174 cos3xdx=?

65

#45.175 sin25xdx=?

#45.176 sec(3x+7)dx

#45.177 sin2(2x+1)dx=? #45.178 sin3(1-x)dx=?

46. Trigonome tric Substitutions

This section continues development of relatively special tricks to do special kinds of integrals. Even

though the application of such things is limited, it's nice to be aware of the possibilities, at least a little bit.

The key idea here is to use trig functions to be able to 'take the square root' in certain integrals. There are just three prototypes for the kind of thing we can deal with:

1-x2 1+x2 x2-1

Examples will illustrate the point.

In rough terms, the idea is that in an integral where the 'worst' part is 1-x2, replacing x by sinu (and, correspondingly, dx by cosudu), we will be able to take the square root, and then obtain an integral in

the variable u which is one of the trigonometric integrals which in principle we now know how to do. The point is that then

1-x2=1-sin2x=cos2x=cosx

We have 'taken the square root'.

For example, in

1-x2dx

we replace x by sinu and dx by cosudu to obtain

1-x2dx=1-sin2ucosudu=cos2ucosudu=

=cosucosudu=cos2udu

Now we have an integral we know how to integrate: using the half-angle formula, this is

cos2udu=1+cos2u2du=u2+sin2u4+C

And there still remains the issue of substituting back to obtain an expression in terms of x rather than u. Since x=sinu, it's just the definition of inverse function that

u=arcsinx

To express sin2u in terms of x is more aggravating. We use another half-angle formula

sin2u=2sinucosu

Then

14sin2u=142sinucosu=14x1-x2

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where 'of course' we used the Pythagorean identity to give us

cosu=1-sin2u=1-x2

Whew.

The next type of integral we can 'improve' is one containing an expression

1+x2

In this case, we use another Pythagorean identity

1+tan2u=sec2u

(which we can get from the usual one cos2u+sin2u=1 by dividing by cos2u). So we'd let

x=tanu dx=sec2udu

(mustn't forget the dx and du business!).

For example, in

1+x2xdx

we use

x=tanu dx=sec2udu

and turn the integral into

1+x2xdx=1+tan2utanusec2udu=

=sec2utanusec2udu=secutanusec2udu=1sinucos2udu

by rewriting everything in terms of cosu and sinu.

For integrals containing x2-1, use x=secu in order to invoke the Pythagorean identity

sec2u-1=tan2u

so as to be able to 'take the square root'. Let's not execute any examples of this, since nothing new really happens.

Rather,, let's examine some purely algebraic variants of these trigonometric substitutions, where we can

get some mileage out of completing the square. For example, consider

-2x-x2dx

The quadratic polynomial inside the square-root is not one of the three simple types we've looked at. But, by completing the square, we'll be able to rewrite it in essentially such forms:

-2x-x2=-(2x+x2)=-(-1+1+2x+x2)=-(-1+(1+x)2)=1-(1+x)2

Note that always when completing the square we 'take out' the coefficient in front of x2 in order to see what's going on, and then put it back at the end.

So, in this case, we'd let

sinu=1+x cosudu=dx

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In another example, we might have

8x-4x2dx

Completing the square again, we have

8x-4x2=-4(-2+x2)=-4(-1+1-2x+x2)=-4(-1+(x-1)2)

Rather than put the whole -4' back, we only keep track of the ±, and take a +4' outside the square root entirely:

8x-4x2dx=-4(-1+(x-1)2)dx

=2-(-1+(x-1)2)dx=21-(x-1)2)dx

Then we're back to a familiar situation.

#46.179 Tell what trig substitution to use for x8x2-1dx #46.180 Tell what trig substitution to use for 25+16x2dx #46.181 Tell what trig substitution to use for 1-x2dx

#46.182 Tell what trig substitution to use for 9+4x2dx #46.183 Tell what trig substitution to use for x9x2+1dx #46.184 Tell what trig substitution to use for x8x2-1dx

47. EisTorica /and th eoretical comments. Mea n Valu e Theorem

For several reasons, the traditional way that Taylor polynomials are taught gives the impression that the ideas are inextricably linked with issues about infinite series. This is not so, but every calculus book I

know takes that approach. The reasons for this systematic mistake are complicated. Anyway, we will not make that mistake here, although we may talk about ifinite series later.

Instead of following the tradition, we will immediately talk about Taylor polynomials, without first tiring ourselves over ifinite series, and without fooling anyone into thinking that Taylor polynomials have the ifinite series stuff as prerequisite!

The theoretical underpinning for these facts about Taylor polynomials is The Mean Value Theorem, which itself depends upon some fairly subtle properties of the real numbers. It asserts that, for a function f dif- ferentiable on an interval [a, b], there is a point c in the interior (a, b) of this interval so that

f(c)=f(b)-f(a)b-a

Note that the latter expression is the formula for the slope of the 'chord' or 'secant' line connecting the two points (a, f(a)) and (b, f(b)) on the graph of f. And the f(c) can be interpreted as the slope of the tangent line to the curve at the point (c, f(c)) .

In many traditional scenarios a person is expected to commit the statement of the Mean Value Theorem to memory. And be able to respond to issues like 'Find a point c in the interval [0,1] satisfying the conclu- sion of the Mean Value Theorem for the function f(x)=x2.' This is pointless and we won't do it.

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48. Taylor polynomials. formulas

Before attempting to illustrate what these funny formulas can be used for, we just write them out. First, some reminders:

The notation f(k) means the kth derivative of f. The notation k! means k-factorial which by definition is

k!=1234...(k-1)k

Taylor's Formula with Remainder Term first somewhat verbal version: Let f be a reasonable func- tion, and fix a positive integer n. Then we have

f (input) =f(basepoint)+f(basepoint)1! (input- basepoint)

+f(basepoint)2!(input-basepoint)+f(basepoint)3!(input-basepoint)

. . . +f(n)(basepoint)n!(input-basepoint)+f(n+1)(c)(n+1)!(input-basepoint)

for some c between basepoint and input.

That is, the value of the function f for some input presumably 'near' the basepoint is expressible in terms of the values of f and its derivatives evaluated at the basepoint, with the only mystery being the precise nature of that c between input and basepoint.

Taylor's Formula with Remainder Term second somewhat verbal version: Let f be a reasonable func- tion, and fix a positive integer n.

f(basepoint +increment)=f(basepoint)+f(basepoint)1! (increment)

+f(basepoint)2!(increment)+f(base3!point) (increment )3

. . . +f(n)(basepoint)n!(increment)+f(n+1)(c)(n+1)!(increment)

for some c between basepoint and basepoint +increment.

This version is really the same as the previous, but with a different emphasis: here we still have a base- point, but are thinking in terms of moving a little bit away from it, by the amount increment.

And to get a more compact formula, we can be more symbolic: let's repeat these two versions:

Taylor's Formula with Remainder Term: Let f be a reasonable function, fix an input value xo, and fix a positive integer n. Then for input x we have

f(x)=f(xo)+f(xo)1!(x-xo)+f(xo)2!(x-xo)2+f(xo)3!(x-xo)3+...

. . . +f(n)(xo)n!(x-xo)n+f(n+1)(c)(n+1)!(x-xo)n+1

for some c between xo and x.

Note that in every version, in the very last term where all the indices are n+1, the input into f(n+1) is not the basepoint xo but is, instead, that mysterious c about which we truly know nothing but that it lies

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between xo and x. The part of this formula without the error term is the degree-n Taylor polynomial for f at xo, and that last term is the error term or remainder term. The Taylor series is said to be expanded at or expanded about or centered at or simply at the basepoint xo.

There are many other possible forms for the error/remainder term. The one here was chosen partly be- cause it resembles the other terms in the main part of the expansion.

Linear Taylor's Polynomial with Remainder Term: Let f be a reasonable function, fix an input

value xo. For any (reasonable) input value x we have

f(x)=f(xo)+f(xo)1!(x-xo)+f(c)2!(x-xo)2

for some c between xo and x.

The previous formula is of course a very special case of the first, more general, formula. The reason to in- clude the 'linear' case is that without the error term it is the old approximation by differentials formula, which had the fundamental flaw of having no way to estimate the error. Now we have the error estimate.

The general idea here is to approximate 'fancy' functions by polynomials, especially if we restrict ourselves to a fairly small interval around some given point. (That 'approximation by differentials' circus was a very crude version of this idea).

It is at this point that it becomes relatively easy to 'beat' a calculator, in the sense that the methods here can be used to give whatever precision is desired. So at the very least this methodology is not as silly and obsolete as some earlier traditional examples.

But even so, there is more to this than getting numbers out: it ought to be of some intrinsic interest that pretty arbitrary functions can be approximated as well as desired by polynomials, which are so readily

computable (by hand or by machine)!

One element under our control is choice of how high degree polynomial to use. Typically, the higher the degree (meaning more terms), the better the approximation will be. (There is nothing comparable to this in the 'approximation by differentials

Of course, for all this to really be worth anything either in theory or in practice, we do need a tangible error estimate, so that we can be sure that we are within whatever tolerance/error is required. (There is nothing comparable to this in the 'approximation by differentials', either).

And at this point it is not at all clear what exactly can be done with such formulas. For one thing, there are choices.

#48.185 Write the first three terms of the Taylor series at 0 of f(x)=1/(1+x) . #48.186 Write the first three terms of the Taylor series at 2 of f(x)=1/(1-x) . #48.187 Write the first three terms of the Taylor series at 0 of f(x)=ecosx.

49. Classic examples of Taylor polynomials

Some of the most famous (and important) examples are the expansions of 11-x, ex, cosx, sinx, and log(1+ x) at 0: right from the formula, although simplifying a little, we get

11-x=1+x+x2+x3+x4+x5+x6+...

ex=1+x1!+x22!+x33!+x44!+...

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cosx=1-x22!+x44!-x66!+x88!...

siu x=x1!-x33!+x55!-x77!+...

log(1+x)=x-x22+x33-x44+x55-x66+...

where here the dots mean to continue to whatever term you want, then stop, and stick on the appropriate remainder term.

It is entirely reasonable if you can't really see that these are what you'd get, but in any case you should do the computations to verify that these are right. It's not so hard.

Note that the expansion for cosine has no odd powers of x (meaning that the coefficients are zero), while the expansion for sine has no even powers of x (meaning that the coefficients are zero).

At this point it is worth repeating that we are not talking about infinite sums (series) at all here, although we do allow arbitrarily large finite sums. Rather than worry over an ifinite sum that we can never truly evaluate, we use the error or remainder term instead. Thus, while in other contexts the dots would mean ifinite sum', that's not our concern here.

The first of these formulas you might recognize as being a geometric series, or at least a part of one. The other three patterns might be new to you. A person would want to be learn to recognize these on sight, as if by reflex!

50. Computa tional tricks regarding Taylor polynomials

The obvious question to ask about Taylor polynomials is 'What are the first so-many terms in the Taylor polynomial of some function expanded at some point?'.

The most straightforward way to deal with this is just to do what is indicated by the formula: take how- ever high order derivatives you need and plug in. However, very often this is not at all the most efficient.

Especially in a situation where we are interested in a composite function of the form f(xn) or

f (polynomial in x) with a 'familiar' function f, there are alternatives.

For example, looking at f(x)=ex3, if we start taking derivatives to expand this at 0, there will be a big mess pretty fast. On the other hand, we might start with the 'familiar' expansion for ex

ex=1+x+x22!+x33!+ec4!x4

with some c between 0 and x, where our choice to cut it off after that many terms was simply a whim. But then replacing x by x3 gives

ex3=1+x3+x62!+x93!+ec4!x12

with some c between 0 and x3. Yes, we need to keep track of c in relation to the new x.

So we get a polynomial plus that funny term with the 'c' in it, for the remainder. Yes, this gives us a different-looking error term, but that's fine.

So we obtain, with relative ease, the expansion of degree eleven of this function, which would have

been horrible to obtain by repeated differentiation and direct application of the general formula. Why

'eleven'?: well, the error term has the x12 in it, which means that the polynomial itself stopped with a

x11 term. Why didn't we see that term? Well, evidently the coefficients of x11, and of x10 (not to mention x, x2, x4, x5, x7, x8!) are zero.

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As another example, let's get the degree-eight expansion of cosx2 at 0. Of course, it makes sense to use

cosx=1-x22!+x44!+-sinc5!x5

with c between 0 and x, where we note that -sinx is the fifth derivative of cosx. Replacing x by x2, this becomes

cosx2=1-x42!+x84!+-sinc5!x10

where now we say that c is between 0 and x2.

#50.188 Use a shortcut to compute the Taylor expansion at 0 of cos(x5) . #50.189 Use a shortcut to compute the Taylor expansion at 0 of e(x2+x). #50.190 Use a shortcut to compute the Taylor expansion at 0 of log(11-x) .

51. Pysqrt. More serious qu estions about Taylor polynomials

Beyond just writing out Taylor expansions, we could actually use them to approximate things in a more serious way. There are roughly three different sorts of serious questions that one can ask in this context. They all use similar words, so a careful reading of such questions is necessary to be sure of answering the question asked.

(The word 'tolerance' is a synonym for 'error estimate', meaning that we know that the error is no worse than such-and-such)

Given a Taylor polynomial approximation to a function, expanded at some given point, and given a re- quired tolerance, on how large an interval around the given point does the Taylor polynomial achieve that tolerance?

Given a Taylor polynomial approximation to a function, expanded at some given point, and given an in- terval around that given point, within what tolerance does the Taylor polynomial approximate the function on that interval?

Given a function, given a fixed point, given an interval around that fixed point, and given a required tol- erance, find how many terms must be used in the Taylor expansion to approximate the function to within the required tolerance on the given interval.

As a special case of the last question, we can consider the question of approximating f(x) to within a given tolerance/error in terms of f(xo), f(xo), f(xo) and higher derivatives of f evaluated at a given point xo. In 'real life' this last question is not really so important as the third of the questions listed above, since

evaluation at just one point can often be achieved more simply by some other means. Having a polyno-

mial approximation that works all along an interval is a much more substantive thing than evaluation at a single point.

It must be noted that there are also other ways to approach the issue of best approximation by a polyno- mial on an interval. And beyond worry over approximating the values of the function, we might also want the values of one or more of the derivatives to be close, as well. The theory of splines is one approach to approximation which is very important in practical applications.

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52. Determining Tolera nce/Error

This section treats a simple example of the second kind of question mentioned above: 'Given a Taylor

polynomial approximation to a function, expanded at some given point, and given an interval around that given point, within what tolerance does the Taylor polynomial approximate the function on that interval?' Let's look at the approximation 1-x22+x44! to f(x)=cosx on the interval [hyphen 12, 12]. We might ask ζWithin what tolerance does this polynomial approximate cosx on that intervali

To answer this, we first recall that the error term we have after those first (oh-so-familiar) terms of the expansion of cosine is

-sinc5!x5

For x in the indicated interval, we want to know the worst-case scenario for the size of this thing. A sloppy but good and simple estimate on sinc is that |sinc|1, regardless of what c is. This is a very happy kind of estimate because it's not so bad and because it doesn't depend at all upon x. And the biggest that x5 can be is (12)50.03. Then the error is estimated as

|-sinc5!x5|1255!0.0003

This is not so bad at all!

We could have been a little clever here, taking advantage of the fact that a lot of the terms in the Taylor expansion of cosine at 0 are already zero. In particular, we could choose to view the original polynomial 1-x22+x44! as including the fifth-degree term of the Taylor expansion as well, which simply happens to be zero, so is invisible. Thus, instead of using the remainder term with the '5' in it, we are actually entitled to use the remainder term with a '6'. This typically will give a better outcome.

That is, instead of the remainder we had must above, we would have an error term

-cosc6!x6

Again, in the worst-case scenario |-cosc|1. And still |x|12, so we have the error estimate

|-cosc6!x6|1266!0.000022

This is less than a tenth as much as in the first version.

But what happened here? Are there two different answers to the question of how well that polynomial ap- proximates the cosine function on that interval? Of course not. Rather, there were two approaches taken by us to estimate how well it approximates cosine. In fact, we still do not know the exact error!

The point is that the second estimate (being a little wiser) is closer to the truth than the first. The first estimate is true, but is a weaker assertion than we are able to make if we try a little harder.

This already illustrates the point that 'in real life' there is often no single 'right' or 'best' estimate of an error, in the sense that the estimates that we can obtain by practical procedures may not be perfect, but represent a trade-off between time, effort, cost, and other priorities.

#52.191 How well (meaning 'within what tolerance') does 1-x2/2+x4/24-x6/720 approximate cosx on the interval [-0.1,0.1] ?

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#52.192 How well (meaning 'within what tolerance') does 1-x2/2+x4/24-x6/720 approximate cosx on the interval [-1, 1]?

#52.193 How well (meaning 'within what tolerance') does 1-x2/2+x4/24-x6/720 approximate cosx on the interval [-π2, π2]?

53. Eo large an inte rval with given tolerance.7

This section treats a simple example of the first kind of question mentioned above: 'Given a Taylor poly- nomial approximation to a function, expanded at some given point, and given a required tolerance, on how large an interval around the given point does the Taylor polynomial achieve that tolerance?'

The specific example we'll get to here is ζFor what range of x25 does 5+110(x-25) approximate x to within.001 i

Again, with the degree-one Taylor polynomial and corresponding remainder term, for reasonable functions f we have

f(x)=f(xo)+f(xo)(x-xo)+f(c)2!(x-xo)2

for some c between xo and x. The remainder term is

remainder term =f(c)2!(x-xo)2

The notation 2! means 2-factorial', which is just 2, but which we write to be 'forward compatible' with other things later.

Again: no, we do not know what c is, except that it is between xo and x. But this is entirely reasonable, since if we really knew it exactly then we'd be able to evaluate f(x) exactly and we are evidently presum- ing that this isn't possible (or we wouldn't be doing all this!). That is, we have limited information about what c is, which we could view as the limitation on how precisely we can know the value f(x) .

To give an example of how to use this limited information, consider f(x)=x (yet again!). Taking xo= 25, we have

x=f(x)=f(xo)+f(xo)(x-xo)+f(c)2!(x-xo)2=

=25+12125(x-25)-12!141(c)3/2(x-25)2=

=5+110(x-25)-181c3/2(x-25)2

where all we know about c is that it is between 25 and x. What can we expect to get from this?

Well, we have to make a choice or two to get started: let's suppose that x25 (rather than smaller). Then we can write

25cx

From this, because the three-halves-power function is increasing, we have

253/2c3/2x3/2

Taking inverses (with positive numbers) reverses the inequalities: we have

25-3/2c-3/2x-3/2

So, in the worst-case scenario, the value of c-3/2 is at most 25-3/2= 1/125.

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And we can rearrange the equation:

x-[5+110(x-25)]=-181c3/2(x-25)2

Taking absolute values in order to talk about error, this is

|x-[5+110(x-25)]|=|181c3/2(x-25)2|

Now let's use our estimate |1c3/2| 1/125 to write

|x-[5+110(x-25)]||181125(x-25)2|

OK, having done this simplification, now we can answer questions like For what range of x25 does 5+110(x-25) approximate x to within.001 i) We cannot hope to tell exactly, but only to give a range of values of x for which we can be sure based upon our estimate. So the question becomes: solve the in- equality

|181125(x-25)2|.001

(with x25) . Multiplying out by the denominator of 8125 gives (by coincidece?)

|x-25|21

so the solution is 25x26.

So we can conclude that x is approximated to within.001 for all x in the range 25x26. This is a worthwhile kind of thing to be able to find out.

#53.194 For what range of values of x is x-x36 within 0.01 of sinx?

#53.195 Only consider -1x1. For what range of values of x inside this interval is the polynomial 1+x+x2/2 within.Ol of ex?

#53.196 On how large an interval around 0 is 1-x within 0.01 of 1/(1+x) ?

#53.197 On how large an interval around 100 is 10+x-10020 within 0.01 of x?

54. Achieving desired tolerance on desired inTerva /

This third question is usually the most difficult, since it requires both estimates and adjustment of num- ber of terms in the Taylor expansion: ζGiven a function, given a fixed point, given an interval around that fixed point, and given a required tolerance, find how many terms must be used in the Taylor expansion to approximate the function to within the required tolerance on the given interval.

For example, let's get a Taylor polynomial approximation to ex which is within 0.001 on the interval [hyphen 12, +12]. We use

ex=1+x+x22!+x33!+...+xnn!+ec(n+1)!xn+1

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for some c between 0 and x, and where we do not yet know what we want n to be. It is very convenient here that the nth derivative of ex is still just ex! We are wanting to choose n large-enough to guarantee that

|ec(n+1)!xn+1|0.001

for all x in that interval (without knowing anything too detailed about what the corresponding c's are!).

The error term is estimated as follows, by thinking about the worst-case scenario for the sizes of the parts of that term: we know that the exponential function is increasing along the whole real line, so in any

event c lies in [hyphen 12, +12] and

|ec|e1/22

(where we've not been too fussy about being accurate about how big the square root of e is!). And for x in that interval we know that

|xn+1|(12)n+1

So we are wanting to choose n large-enough to guarantee that

|ec(n+1)!(12)n+1|0.001

Since

|ec(n+1)!(12)n+1|2(n+1)!(12)n+1

we can be confident of the desired inequality if we can be sure that

2(n+1)!(12)n+10.001

That is, we want to 'solve' for n in the inequality

2(n+1)!(12)n+10.001

There is no genuine formulaic way to 'solve' for n to accomplish this. Rather, we just evaluate the left- hand side of the desired inequality for larger and larger values of n until (hopefully!) we get something smaller than 0.001. So, trying n=3, the expression is

2(3+1)!(12)3+1=11216

which is more like 0.01 than 0.001. So just try n=4:

2(4+1)!(12)4+1=160320.00052

which is better than we need.

The conclusion is that we needed to take the Taylor polynomial of degree n=4 to achieve the desired tolerance along the whole interval indicated. Thus, the polynomial

1+x+x22+x33+x44

approximates ex to within 0.00052 for x in the interval [hyphen 12, 12].

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Yes, such questions can easily become very difficult. And, as a reminder, there is no real or genuine claim that this kind of approach to polynomial approximation is 'the best'.

#54.198 Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate ex to within 0.001 on the interval [-1, +1].

#54.199 Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate cosx to within 0.001 on the interval [-1, +1].

#54.200 Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate cosx to within 0.001 on the interval [-π2, π2].

#54.201 Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate cosx to within 0.001 on the interval [-0.1, +0.1].

#54.202 Approximate e1/2=e to within.01 by using a Taylor polynomial with remainder term, ex- panded at 0. (Do NOT add up the finite sum you get

#54.203 Approximate 101=(101)1/2 to within 10-15 using a Taylor polynomial with remainder term. (Do NOT add up the finite sum you get.' One point here is that most hand calculators do not easily give 15 decimal places. Hah

55. Integrating Taylor polynomials. first example

Thinking simultaneously about the difficulty (or impossibility) of 'direct' symbolic integration of compli- cated expressions, by contrast to the ease of integration of polynomials, we might hope to get some mileage out of integrating Taylor polynomials.

As a promising example: on one hand, it's not too hard to compute that

0Tdx1-xdx=[-log(1-x)]0T=-log(1-T)

On the other hand, if we write out

11-x=1+x+x2+x3+x4+...

then we could obtain

0T(1+x+x2+x3+x4+...)dx=[x+x22+x33+...]0T=

=T+T22+T33+T44+...

Putting these two together (and changing the variable back to 'x') gives

-log(1-x)=x+x22+x33+x44+...

(For the moment let's not worry about what happens to the error term for the Taylor polynomial).

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This little computation has several useful interpretations. First, we obtained a Taylor polynomial for

-log(1-T) from that of a geometric series, without going to the trouble of recomputing derivatives. Sec- ond, from a different perspective, we have an expression for the integral

0Tdx1-xdx

without necessarily mentioning the logarithm: that is, with some suitable interpretation of the trailing dots,

0Tdx1-xdx=T+T22+T33+T44+...

56. Integrating th e error Term. example

Being a little more careful, let's keep track of the error term in the example we've been doing: we have

11-x=1+x+x2+...+xn+1(n+1)1(1-c)n+1xn+1

for some c between 0 and x, and also depending upon x and n. One way to avoid having the 1(1-c)n+1 'blow up' on us, is to keep x itself in the range [0,1) so that c is in the range [0, x) which is inside [0,1), keeping c away from 1. To do this we might demand that 0T<1.

For simplicity, and to illustrate the point, let's just take 0T12. Then in the worst-case scenario

|1(1-c)n+1|1(1-12)n+1=2n+1

Thus, integrating the error term, we have

|0T1n+11(1-c)n+1xn+1dx|1n+12n+1xn+1dx=2n+1n+10Txn+1dx

=2n+1n+1[xn+2n+2]0T=2n+1Tn+2(n+1)(n+2)

Since we have cleverly required 0T12, we actually have

|0T1n+11(1-c)n+1xn+1dx|2n+1Tn+2(n+1)(n+2)

2n+1(12)n+2(n+1)(n+2)=12(n+1)(n+2)

That is, we have

|-log(1-T)-[T+T22+...+Tnn]|12(n+1)(n+2)

for all T in the interval [0, 12]. Actually, we had obtained

|-log(1-T)-[T+T22+...+Tnn]|2n+1Tn+22(n+1)(n+2)

and the latter expression shrinks rapidly as T approaches 0.

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