More Taylor series stuff, HELP!

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.

 

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.

 

Is it 4?

 

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

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

Is that what you're talking about?

 

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

 

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?

 

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?

 

So basically, I want to find where

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

is less than 0.001?

 

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

 

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