Assignment 15

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course MTH 174

11:49 a.m. 7/19/10

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015. `query 15

Physics II

07-19-2010

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10:56:31

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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RESPONSE -->

We find a bound for the fourth derivative using the function x^(1/3). The fourth derivative of this function is : -80/81 x^(-11/3). We have to find the error about x=1 instead of x=0 because all of the derivatives are also 0 at x=0. So between x=1 and x=.5 the fourth derivative has a maximum absolute value at x=.5. The value of the fifth derivative at x=.5 is approximately 12.54. With this can make the bound an easy number like 13. From theorem 10.1 we have 13/(3+1)! (1-.5)^4= -0.03385. So our maximum error is about 0.034.

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10:56:58

What error did you estimate?

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0.03385 or approximately 0.034.

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10:57:17

What function did you compute the Taylor polynomial of?

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The function was x^(1/3).

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10:58:21

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

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RESPONSE -->

I used M/(n+1)! (x-a)^n+1. With M=13, n=3, x=1 and a=.5.

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11:16:44

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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RESPONSE -->

This problem is very similar to the example in the book. Here we have: |f^(n+1)(x)|<= 1 for all n and all x. The Lagrange error bound is thus M=1 and this gives:

|x|^(n+1)/(n+1)!. Using the Lagrange error function and rearranging gives the expression: sin x= Pn(x)+En(x). So we want to show that as n -> inf, En(x) -> 0, which means that we want to show: lim n->inf |x|^n+1/(n+1)!= 0.

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11:19:14

explain how you proved the result.

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RESPONSE -->

To show that L=0 we use the ratio test. The setup is then: lim n->inf [(x^n+2)/(n+2)!]/[(x^n+1)/(n+1)!]. This can be reduced to: x/(n+1). We can see very clearly that this expression goes to 0 as n goes to infinity.

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11:19:53

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

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RESPONSE -->

We have M=1 and so: |x|^n+1/(n+1)!.

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11:20:21

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

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Yes and I did so above using the ration test.

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11:21:17

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

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I knew M=1 because all of the derivative are either +-sin x or +-cos x all of which will be less than 1.

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11:21:58

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

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This is obtained from the error function.

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11:22:31

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

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RESPONSE -->

I used the ratio test but with slightly different expression.

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11:39:01

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

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RESPONSE -->

Here we use x= sin theta and have: sqrt(1+x) or (1+x)^1/2 and we can use the binomial series equation here to get: 1 + x/2 -x^2/8 + x^3/16 - 5x^4/128. Then we can undo the substitution and end up with: 1 + sin theta / 2 - (sin theta)^2 / 8 + (sin theta)^3 / 16 - 5(sin theta)^4 /128.

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11:39:14

what are the first four nonzero terms of the series?

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Those are given above.

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11:39:39

Explain how you obtained these terms.

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For my solution I used a substitution along with the binomial theorem.

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11:42:58

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

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RESPONSE -->

I didn't have to find this for my solution but it is a simple series. We can't expand this around x=0 because the derivatives are all undefined. So we use x=1 and the formula for the Taylor series about x=a and this makes: 1+ 1/2(x-1)- 1/8 (x-1)^2 + 1/16 (x-1)^3 - 5/128 (x-1)^4. We could just as easily use this formula and obtain the same results as I did above.

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11:46:05

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

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RESPONSE -->

Here again, I did not work my solution this way but it is also a simple series and actually works out to the same as sin theta except that f(0)= 1 instead of 0. After the first term the one no longer figures into the series as its derivative is 0. So all of this leads to: 1 + theta - theta^3/3! + theta^5/5! - theta^7/7!.

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11:47:16

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

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Just as I did a simple substitution can be used however substituting the latter into the former can be a little trickier than my method.

Very nice solutions.

For comparison:

We substitute the Taylor expansion of sin(x) into the Taylor expansion of 1 / sqrt(x) and ignore terms with powers exceeding 4:

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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11:48:01

Query Add comments on any surprises or insights you experienced

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RESPONSE -->

Some of the error problems really gave me trouble on this one.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#

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