Assignment 15

course Mth 174

Section 10.4 of the text (Lagrange error bounds) seems to me to be very poorly written and inadequate. It provides insufficient examples for various types of situations. I would recommend that additional explanations and examples be added to the class notes.

W鲩ð~䈅ǵassignment #015

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Physics II

03-31-2007

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09:48:40

Query 10.4.8 (was 10.4.6 3d edition) (was p. 487 problem 6) magnitude of error in estimating .5^(1/3) with a third-degree Taylor polynomial about 0.

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09:59:53

What error did you estimate?

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Using f(x) = x^(1/3) doesn't work because the fourth derivative is undefined at x=0

I used f(x) = (1-x)^(1/3) on the range of 0<=x<=0.5 (1-x at x=0.5 = 0.5 as required by the problem)

f4(x) = -80/[81*(1-x)^(11/3)]

Max value of |f4(x)| occurs at x=0.5 = 12.54243

Therefore, we use M = 12.54243

E3(x) = |f(x) - P3(x)| <= [M/(n+1)!]*x^(n+1)

E3(x) <= (12.54243/4!)*(0.5^4)

E3(x) <= 0.0327

This agrees with my estimate:

** The maximum possible error of the degree-3 polynomial is based on the fourth derivative and is equal to the maximum possible value of the n = 4 term.

The function is x^(1/3).

Its derivatives are

f'(x) = 1/3 x^(-2/3),

f''(x) = -2/9 x^(-5/3),

f'''(x) = 10/27 x^(-8/3),

f''''(x) = -80/81 x^(-11/3).

We think in terms of expanding about x = 1, since all these derivatives are undefined at x = 0, and since it is easy to substitute 1.

The maximum possible absolute value of the fourth derivative f''''(x) = -80/81 x^(-11/3) between x = .5 and x = 1 occurs at x = .5, where we obtain | f '''' (x) | = 80/81 * (.5^-(11/3) ) = 12.54 approx; to be sure we have a valid limit on the error we will use the slight overestimate M = 13.

So the maximum possible error is | 13 / 4! * (.5 - 1)^4 |, or about .034. **

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10:01:38

What function did you compute the Taylor polynomial of?

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See previous response.

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10:01:53

What expression did you use in finding the error limit, and how did you use it?

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See previous response.

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10:02:27

Query 10.4.20 (was 10.4.16 3d edition) (was p. 487 problem 12) Taylor series of sin(x) converges to sin(x) for all x (note change from cos(x) in previous edition)

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10:31:23

explain how you proved the result.

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En(x) = |sin x - Pn(x)| = sinx - (x - (x^3/3!) + (x^5/5! -...+ (-1)^n*[x^(2n+1)/(2n+1)!])

sin x = x - (x^3/3!) + (x^5/5! -...+ (-1)^n*[x^(2n+1)/(2n+1)!] + En(x)

|fn(x+1)| must be one of |sin x|, |-sin x|, |cos x|, or |-cos x| for all n. Therefore, |fn(x+1)| <= 1 for all x. Thus, M = 1

|En(x)| = |sin x - Pn(x)| <= [1*|x|^(n+1)]/(n+1)!

lim (n>>infinity) [1*|x|^(n+1)]/(n+1)! =

lim (n>>infinity) x^n/n! = 0 (as shown on page 500)

To show that the Taylor polynomial for sin x also converges:

lim(n>>infinity) |a n+1| / |a n| = lim (n>>infinity) |C n+1| / |C n|

= |x| * lim(n>>infinity) |[(-1)^(n+1)/(2n+2)!] / [(-1)^n/(2n+1)!]|

= |x| * lim(n>>infinity) |-1/(2n+1)|

lim(n>>infinity) |-1/(2n+1)| = 0, so |x| * lim(n>>infinity) |-1/(2n+1)| = 0, therefore the radius of convergence is infinity and the Taylor polynomial converges for all x.

Therefore the Taylor series for sin x converges to sin x

** For even n the nth derivative is sin(x); when expanding about 0 this will result in terms of the form 0 * x^n / n!, or just 0.

If n is odd, the nth derivative is +- cos(x). Expanding about 0 this derivative has magnitude 1 for all n. So the nth term, for n odd, is just 1 / n! * x^n.

In terms of the theorem for the error term, we see that none of the coefficients exceeds the maximum magnitude M = 1.

For any x, lim(n -> infinity} (x^n / n!) = 0, because lim { n -> infinity} ( [ x^(n+1) / (n+1)! ] / [ x^n / n! ] ) = lim (n -> infinity) ( x / n ) = 0; the limit is zero since x is fixed and n increases without bound.

This needs to be put together formally in terms of the definition of the error term.

We get En(x) = M / (n+1)! * x^(n+1) = 1 / (n+1)! * x^(n+1) with M + 1, which as n -> infinity approaches zero. Since the error term approaches zero as n -> infinity, the series converges for all x. **

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10:31:48

What is the error term for the degree n Taylor polynomial?

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See previous response.

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10:32:09

Can you prove that the error term approaches 0 as n -> infinity?

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See previous response.

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10:32:20

What do you know about M in the expression for the error term?

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See previous response.

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10:56:27

How do you know that the error term must be < | x | ^ n / ( n+1)! ?

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Each subsequent term in the sequence is produced by multiplying the previous term by |x|/(n+1): [|x|^n+1)/(n+1)!] / [|x|^n/n!] = |x|/(n+1)

For arbitrary |x|, as long as n > 2|x|, |x|/(n+1) < (1/2), so each subsequent term is obtained by multipling the previous term by a number < (1/2). Therefore, as n>>infinity, for any x, the error term converges to 0 and it must be less than |x|^n / (n+1)!

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10:57:34

How you know that the limit of | x | ^ n / ( n+1)! is 0?

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See previous response.

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10:58:02

Query problem 10.3.10 (3d edition 10.3.12) (was 9.3.12) series for `sqrt(1+sin(`theta))

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11:08:36

what are the first four nonzero terms of the series?

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Rewrite as sqrt (1+sin theta) = (1+sin theta)^(1/2)

Produce Taylor series for (1+x)^(1/2) using binomial expansion with p = 1/2:

(1+x)^(1/2) = 1 + x/2 + [(.5)(.5-1)x^2]/2! + [(.5)(.5-1)(.5-2)x^3]/3! ...

= 1 + x/2 + x^2/8 + x^3/16 ...

Produce Taylor series for sin theta

sin theta = theta - theta^3/3! + theta^5/5! - theta^7/7! ...

Substitute sin theta for x in series for (1+x)^(1/2):

sqrt (1+ sin theta) = 1 + (1/2)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...) - (1/8)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^2 + (1/16)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^3 ...

sqrt (1+ sin theta) = 1 + theta/2 - (theta^2)/8 - (theta^3)/48 ...

Your approach should work; you got -1/48 for the theta^3 term, whereas I believe the correct coefficient is -1/16. Otherwise your expansion agrees with that obtained below. However a different approach, which doesn't involve the binomial expansion, is used below:

Expanding y = sqrt(x) about x = 1 we get derivatives

y ' = 1/2 x^(-1/2)

y '' = -1/4 x^(-3/2)

y ''' = 3/8 x^(-5/2)

y '''' = -5/16 x^(-7/2).

Substituting 1 we get values 1, 1/2, -1/4, 3/8, -5/16.

The degree-4 Taylor polynomial is therefore

sqrt(x) = 1 + 1/2 (x-1) - 1/4 (x-1)^2 / 2 + 3/8 (x-1)^3/3! - 5/16 (x-1)^4 / 4!.

It follows that the polynomial for sqrt(1 + x) is

sqrt( 1 + x ) = 1 + 1/2 ( 1 + x - 1) - 1/4 (1 + x-1)^2 / 2 + 3/8 (1 + x-1)^3/3! - 5/16 (1 + x-1)^4 / 4!

= 1 + 1/2 (x) - 1/4 (x)^2 / 2 + 3/8 (x)^3/3! - 5/16 (x)^4 / 4!.

Now the first four nonzeo terms of the expansion of sin(theta) give us the polynomial

sin(theta) = theta-theta^3/3!+theta^5/5!-theta^7/7!.

We note that since sin(theta) contains theta as a term, the first four powers of sin(theta) will contain theta, theta^2, theta^3 and theta^4, so our expansion will ultimately contain these powers of theta.

We will expand the expressions for sin^2(theta), sin^3(theta) and sin^4(theta), but since we are looking for only the first four nonzero terms of the expansion we will not include powers of theta which exceed 4:

sin^2(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^2 = theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4

sin^3(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^3 = theta^3 + powers of theta exceeding 4

sin^4(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^4 = theta^4 + powers of theta exceeding 4

Substituting sin(theta) for x in the expansion of sqrt(1 + x) we obtain

sqrt(1 + sin(theta)) = 1 + 1/2 (sin(theta)) - 1/4 (sin(theta))^2 / 2 + 3/8 (sin(theta))^3/3! - 5/16 (sin(theta))^4 / 4!

= 1 + 1/2 (theta-theta^3/3! + powers of theta exceeding 4) - 1/4 (theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4)/2 + 3/8 ( theta^3 + powers of theta exceeding 4) / 3! - 5/16 (theta^4 + powers of theta exceeding 4)/4!

= 1 + 1/2 theta - 1/4 theta^2/2 - 1/2 theta^3 / 3! + 3/8 theta^3 / 3! + 2 / ( 4 * 3! * 2) theta ^4 - 5/16 theta^4 / 4!

= 1 + 1/2 theta - 1/8 theta^2 + 1/16 theta^3 - 5/128 theta^4.

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11:08:45

Explain how you obtained these terms.

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See previous response.

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13:59:09

What is the Taylor series for `sqrt(z)?

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Using f(x) = sqrt (z) will not work because f(0) and all derivatives of f(x) evaluated at 0 = 0.

Let f(x) = sqrt (1+x)

Using binomial expansion:

sqrt (1+x) = 1 + x/2 - x^2/8 + x^3/16 + ...

Substituting z-1 for x (since 1 + (z-1) = z)

sqrt (z) = 1 + (z-1)/2 - (z-1)^2/8 + (z-1)^3/16 + ...

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14:11:52

What is the Taylor series for 1+sin(`theta)?

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Since 1 is a constant, its Taylor series is itself (if f(x) =1, f(0) =1 and all derivatives of f(x) = 0

Taylor series for sin (theta) = theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

Adding the two series gives

1 + sin (theta) = 1 + (theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

= 1 + theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,,

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14:20:49

How are the two series combined to obtain the desired series?

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See previous response. Since a Taylor series is a power series it follows the rules of a power series, including the rule that power series can be added term by term.

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14:22:14

Query Add comments on any surprises or insights you experienced

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The Taylor series for sqrt(z) provides an interesting way to manually calculate approximations of square roots.

Along with a table containing some reference values, your calculator uses Taylor expansions to evaluate functions.

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"

This looks good. See my notes and let me know if you have questions.

I've noted your comments on Section 10.4. It sound like a bit of a cliche, but it's not: I value comments from all students, and particularly from those who are doing well in the course. This topic is tough enough for most students that they have trouble telling me what's wrong, though it's clear that a student who doesn't understand the functions very well will have trouble with the just essential idea of a bound over an interval.

I'm always glad to answer questions on specific problems and point out how the error bound is used. If you have questions related to any of the problems, or to anything else, send me the specifics and I'll be glad to respond.

Another possible problem, for which I apologize, is that Class Notes #28 are not mentioned until Asst 16. In fact the class notes correspondences for Chapters 9 and 10 were not good; the class notes were for the first edition of the text, and there has been some 'drift' of sections, as well as wholesale reordining of some chapters, as editions changed. My problem assignments and queries have pretty much kept pace, but the ordering of the Class Notes has suffered. I've just made some related corrections to the Assignments Page.