course Mth 174

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Question: 

Question: Query 10.4.8 (was 10.4.1 3d edition  formerly p. 487 problem 6) magnitude of error in estimating .5^(1/3) with a third-degree Taylor polynomial about 0.

Your solution:

Confidence Assessment:

Given Solution:

Given Solution: ** 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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Question: 

Question: 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)

explain how you proved the result.

Your solution:

Confidence Assessment:

Given Solution:

** 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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Question: 

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

what are the first four nonzero terms of the series?

Your solution:

Confidence Assessment:

Given Solution:

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.

STUDENT SOLUTION:

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 ...

INSTRUCTOR RESPONSE:

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, was used above.

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

RESPONSE -->

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 + ...

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

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RESPONSE -->

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! + ,,,

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

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RESPONSE -->

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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STUDENT COMMENT:

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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