Homework Help: More Taylor series stuff, HELP!

the7joker7 · May 28, 2008

1. The problem statement, all variables and given/known data

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.)

3. 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 · May 28, 2008

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

the7joker7 · May 28, 2008

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 · May 28, 2008

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

the7joker7 · May 28, 2008

Is it 4?

Dick · May 28, 2008

the7joker7 said: ↑

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 · May 28, 2008

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

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

Is that what you're talking about?

Dick · May 28, 2008

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

the7joker7 · May 28, 2008

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

Dick · May 28, 2008

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 · May 28, 2008

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 · May 28, 2008

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

the7joker7 · May 28, 2008

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 · May 28, 2008

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 · May 29, 2008

So basically, I want to find where

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

is less than 0.001?

Dick · May 29, 2008

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 · May 29, 2008

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

Dick · May 29, 2008

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 · May 29, 2008

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 · May 29, 2008

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.