More Taylor series stuff, HELP

[the7joker7]

Homework Statement

Let T_(4)(x): be the Taylor polynomial of degree 4 of the function ` f(x) = ln(1+x) ` at `a = 0 `.


Suppose you approximate ` f(x) ` by ` T_(4)(x) `, find all positive values of x for which this approximation is within 0.001 of the right answer. (Hint: use the alternating series approximation.)

The Attempt at a Solution

I got to be honest, I'm not entirely sure where to even start with this...I've gotten my sealegs when it comes to these series and how to find them and stuff, and I know how to find the error in an alternating series, but I don't know how to do it with an ln function.

The answer is 0.346572421578, fyi.

 

[Dick]

If you know it, tell me, how DO you find the error in an alternating series?

 

[the7joker7]

Well, if you want to find the maximum possible error in an alternating series, you'd just take the next term. For example if you want to find the error in a series after 10 terms, the max possible errror is the 11th term.

 

[Dick]

Right. Good! So what's the appropriate term in that taylor series?

 

[the7joker7]

Is it 4?

 

[Dick]

Is it 4?

I think the 4th degree taylor polynomial has a last term with x^4 in it. What's the term after that? And not just x^5, I want the coefficient too.

 

[the7joker7]

If I'm on the same page as you...

(f'''''(x) * (x - a))/x!

Is that what you're talking about?

 

[Dick]

If you mean (ln(1+x))'''''*x^5/5!, yes. You are expanding around a=0. And f(x)=ln(1+x).

 

[the7joker7]

So the 5th derivative of ln(1 + x) would be 1/((1 + x)^5), correct?

 

[Dick]

No, the coefficient is wrong. When you differentiate 1/(1+x)^2 don't you get -2/(1+x)^3? What happened to factors like the 2?

 

[the7joker7]

So, let me see...using that as a template, it would be...

f(x) = ln(1 + x)

f'(x) = 1/(1 + x)^2

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

f'''(x) = 6/(1 + x)^4

f''''(x) = -24/(1 + x)^5

f'''''(x) = 120/(1 + x)^6

Does that work?

 

[Dick]

Doh. f'(x)=1/(1+x). Shuffle your results around a bit. You are on the right track though.

 

[the7joker7]

f(x) = ln(1 + x)

f'(x) = 1/(1 + x)

f''(x) = 1/(1 + x)^2

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

f''''(x) = 6/(1 + x)^4

f'''''(x) = -24/(1 + x)^5

How bout that?

 

[Dick]

The sign is wrong. That went wrong in the f'' step. But that's ok, we don't need that. So now, ta da, the term you want if f'''''(a)*x^5/5!, where a=0. Do you agree?

 

[the7joker7]

So basically, I want to find where

(24/(1 + x)^2) * x^5/5!

is less than 0.001?

 

[Dick]

No, f'''''(a), not f'''''(x). a=0. Look at the definition of taylor series. The variable in the derivative is not x. It's the point you are expanding around, a. It's a CONSTANT. It's easier than you are making it out to be.

 

[the7joker7]

If a is 0 it would just simplify to 0...shouldn't a = 4 since it's the 4th term?

 

[Dick]

What book are you getting your taylor series information out of? NO! The x part is x^n/n!. With n=5. The derivative part is f''''''(a) where a=0. That's what your problem statement says!? For example, f'(x)=1/(1+x). So f'(a) where a=0 is 1. It doesn't just simplify to 0!?

 

[the7joker7]

I'm sorry, you're starting to lose me...sorry if I'm being a bother, but don't you insert a for x?

 

[Dick]

The taylor series for f expanded around x=a is
f(a)+f'(a)*(x-a)+f''(a)*(x-a)^2/2!+f'''(a)*(x-a)^3/3!+...
Does that ring a bell? x and a play two completely different roles. x is the expansion parameter and a is the point you are expanding around. a is a CONSTANT, not a variable.

 

[the7joker7]

So, if I'm understanding correctly...the term I'm looking for will be...

((24/(1 + 4)^4) * (4 - 0)^5)/5!

That gives me something really close to the right answer...ugh, I know I'm so close.

 

[Dick]

You aren't understanding me. f'''''(a)=24/(1+a)^5. a=0. f'''''(a)=24. Why are you putting a=4? I TOLD YOU NOT TO. I've said a=0, a=0, a=0 over and over again.

 

[the7joker7]

Alright...

24 * x^5/5!

And then solve for what x makes the error less than 0.001?

 

[Dick]

Yes. Yes. Yes. Yes. Yes. Thank you!