Optimizing Simpson's Rule for Error Bound: Finding the Minimum Value of n

[SherlockOhms]

Homework Statement

Calculate the value of n so that the approximation is within 0.0001. b = 2, a = 1. f(x) = 1/x.

Homework Equations

f⁴(x) = 24/x^5 (Think this is correct)
Error <= (b-a)^5/180n^4(MAX_{x [a,b]}(f⁴(x))

The Attempt at a Solution

Well, 24/x^5 obtains it's max at x =1. Thus (MAX_{x [a,b]}(f⁴(x)) = 24.
I subbed in all the given values and keep getting 6 as my answer. The correct answer is 8 though. Could somebody point out where I'm going wrong?

 

[Mark44]

Homework Statement

Calculate the value of n so that the approximation is within 0.0001. b = 2, a = 1. f(x) = 1/x.

Homework Equations

f⁴(x) = 24/x^5 (Think this is correct)
Error <= (b-a)^5/180n^4(MAX_{x [a,b]}(f⁴(x))

What you wrote is ambiguous.
Is it ((b - a)⁵/180) * n⁴ or
(b - a)⁵/(180 * n⁴)?

The Attempt at a Solution

Well, 24/x^5 obtains it's max at x =1. Thus (MAX_{x [a,b]}(f⁴(x)) = 24.
I subbed in all the given values and keep getting 6 as my answer. The correct answer is 8 though. Could somebody point out where I'm going wrong?

 

[SherlockOhms]

Apologies! It's (b-a)^5/(180*n^4).

 

[Mark44]

I get 6 as well. Is 8 the answer in the back of the book? It's possible they have the wrong answer.

One way to check is to do Simpson's with n = 6, and compare the answer you get with the integral itself,
$$\int_1^2 \frac{dx}{x} = ln(2) \approx. .69315$$

You should have agreement in either 2 or 3 decimal places.

 

[SherlockOhms]

Thanks for this too. There's most likely a mistake alright. I'll be sure to double check it in the morning though.