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 gotta 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, lemme 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!