Assignment 15

course Mth 174

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Physics II

06-23-2009

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17:28:33

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 -->

I En(.05^1/3) I = f(.05^1/3) - P3(.05^1/3) I < or equal to M/ (n + 1)! * I x - aI^n + 1

f(x) = .05^1/3 = .7937

f '(x)= 0

f ''(x)= 0

f '''(x) = 0

so, P3 = .05^1/3 = .7937

I En(.05^1/3) I = .7937 - .7937 < or equal to 1 / 4! * x^4

= 1 / 24 * (1/3)^4 = -.000541

How do you know what to put in for x for the right side of the inequality?

You use the value of x that gives you the biggest possible result.

** 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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17:28:39

What error did you estimate?

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17:28:48

What function did you compute the Taylor polynomial of?

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17:29:13

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

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17:44:54

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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17:52:05

explain how you proved the result.

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RESPONSE -->

sin(x) = x - x^3/3! + x^5/5! - x^7/7! +.......((-1) ^n/2 x^(n+2)) / (n+2)!

En(x) = sin(x) - Pn(x) = sin(x) - sin(x) = x - x^3/3! + x^5/5! - x^7/7! +.......((-1) ^n/2 x^(n+2)) / (n+2)!

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

En(x) = sin(x) - Pn(x) < or equal to x^(n + 1)/ (n + 1)!

With any constant x as n increases the result gets closer and closer to 0 because each increase in n multiplies the result by a smaller fraction.

This proves that the limit as n > infinity of sin(x) = 0, and therefore converges.

It's not obvious that x^(n + 1)/ (n + 1)! approaches zero for any x. If x gets large, x^(n+1) gets very, very large.

However a ratio test will show that x^(n+1) / (n+1)! does approach zero for any value of x, giving us the stated result.

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17:53:31

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

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RESPONSE -->

The error is = f(x) - Pn(x)

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17:53:47

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

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RESPONSE -->

With any constant x as n increases the result gets closer and closer to 0 because each increase in n multiplies the result by a smaller fraction.

This proves that the limit as n > infinity of sin(x) = 0, and therefore converges.

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17:56:02

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

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RESPONSE -->

M represents a positive constant that the function is bounded by for all positive values of x near 0. M/ (n + 1)! * (x - a) ^ (n+1) must be greater than the error.

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17:59:32

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

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RESPONSE -->

When you integrate the following expression twice

-M < or equal to En^ (n+1)(x) < or equal to M

you get En(x) < or equal to M/ (x + 1)! x^(n + 1)

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18:00:54

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

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RESPONSE -->

The limit is 0 because as n increases for a number x the result decreases and approaches 0.

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18:07:32

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 -->

I started with the sin(theta) series:

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

Then I added 1

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

Then I substituted theta^1/2

1 + theta^1/2 - (theta^1/2)^3/3! + (theta^1/2)^5/5! - (theta^1/2)^7/7!

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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18:07:50

what are the first four nonzero terms of the series?

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RESPONSE -->

1 + theta^1/2 - (theta^1/2)^3/3! + (theta^1/2)^5/5! - (theta^1/2)^7/7!

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18:08:02

Explain how you obtained these terms.

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18:10:50

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

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RESPONSE -->

f (x) = sqrt (z)

f(x) = z^1/2

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

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18:11:04

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

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RESPONSE -->

I started with the sin(theta) series:

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

Then I added 1

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

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18:12:03

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

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I wasn't sure how to substitute the sqrt(z) into the 1 + sin(theta) function

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18:13:04

Query Add comments on any surprises or insights you experienced

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