assignment 15

course Mth 174

Please explain number 10.4.8

jډ׋pᖙٙassignment #015

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Physics II

08-03-2009

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20:45:57

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.

What error did you estimate?

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

The present 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).

All these derivatives are undefined at x = 0.

Since all the derivatives are easily evaluated at x = 1, we expand about x = 1.

The maximum possible magnitude 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. We will still have a valid limit on the error if we use the slight overestimate M = 13.

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

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20:54:46

What function did you compute the Taylor polynomial of?

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20:55:01

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

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17:07:18

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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17:30:29

explain how you proved the result.

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Since we know that the lim. as n ->inf. of x^n/x! = 0, we can find that the series x-x^3/3!+x^5/5!+...(-1)^(n/2)(x^n/n!) for odd values of n is equal to sin(x) as En(x) approaches 0.

The answers you give in subsequent questions are pretty much equivalent to the solution given here:

** For even values of 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.

None of the Taylor coefficients exceeds the maximum magnitude M = 1.

For any x, lim(n -> infinity} (x^n / n!) = 0:

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.

Putting this together formally in terms of the definition of the error term.

En(x) = M / (n+1)! * x^(n+1)

Since M = 1, En(x) becomes = 1 / (n+1)! * x^(n+1)

As n -> infinity this approaches zero.

If the error term approaches zero as n -> infinity, the series converges for all x.

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. The series does converge for all values of x.**

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17:51:16

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

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For degree n, the error term is Abs.(sin(x)-Pn(x) inf., the error approaches 0.

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17:58:05

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

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The error term is less than or equal to Abs.(x)^(n+1)/((n+1)!) which approaches 0 as n approaches infinity. Therefore, as n->inf, En(x)->0.

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19:02:41

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

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Since f(x)=sin(x), any derivative is +,- sin(x) or +,- cos(x), no matter which derivative it is. So, for every n, f^(n+1)(x) is less than or equal to 1 between the interval 0 to x, thus giving an M value of 1.

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19:09:05

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

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The error is = Abs.(sin(x) - Pn(x)) which is always less than l x l^(n+1)/((n+1)!).

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20:52:05

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

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Because as n gets bigger and bigger, i.e. approaches inf, the denominator grows much faster than the numerator does, causing the fraction to be equal to 0.

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21:27:42

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

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21:29:46

what are the first four nonzero terms of the series?

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

This being a composite function, we will have to form the composite of one Taylor series with another.

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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21:34:28

Explain how you obtained these terms.

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The series for 1+sin(x) = 1+x-(x^3)/3!+(x^5)/5!+...and the series for 1/sqrt(1+x) = 1-(1/2)x+(((-1/2)(-3/2))/2!)x^2+...So I merely added the desired elements of both series.

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21:40:18

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

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1+(1/2)z+((1/2)/2!)z^2+(((1/2)(3/2))/3!)z^3+...

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21:41:11

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

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1+theta-theta^3/3!+theta^5/5!+...

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21:42:29

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

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Take the odd numbers and alternating signs from the 1+sin(theta) series and multiply them by the increasing fractions from the sqrt(z) series to obtain the desired series.

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21:42:34

Query Add comments on any surprises or insights you experienced

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&#Good responses. See my notes and let me know if you have questions. &#