• Support PF! Buy your school textbooks, materials and every day products Here!

More Taylor series stuff, HELP!

  • Thread starter the7joker7
  • Start date
  • #1
113
0

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
If you know it, tell me, how DO you find the error in an alternating series?
 
  • #3
113
0
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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
Right. Good! So what's the appropriate term in that taylor series?
 
  • #5
113
0
Is it 4?
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #7
113
0
If I'm on the same page as you...

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

Is that what you're talking about?
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
618
If you mean (ln(1+x))'''''*x^5/5!, yes. You are expanding around a=0. And f(x)=ln(1+x).
 
  • #9
113
0
So the 5th derivative of ln(1 + x) would be 1/((1 + x)^5), correct?
 
  • #10
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #11
113
0
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?
 
  • #12
Dick
Science Advisor
Homework Helper
26,258
618
Doh. f'(x)=1/(1+x). Shuffle your results around a bit. You are on the right track though.
 
  • #13
113
0
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?
 
  • #14
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #15
113
0
So basically, I want to find where

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

is less than 0.001?
 
  • #16
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
Last edited:
  • #17
113
0
If a is 0 it would just simplify to 0...shouldn't a = 4 since it's the 4th term?
 
  • #18
Dick
Science Advisor
Homework Helper
26,258
618
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!!!!?????
 
  • #19
113
0
I'm sorry, you're starting to lose me...sorry if I'm being a bother, but don't you insert a for x?
 
  • #20
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #21
113
0
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.
 
  • #22
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
Last edited:
  • #23
113
0
Alright...

24 * x^5/5!

And then solve for what x makes the error less than 0.001?
 
  • #24
Dick
Science Advisor
Homework Helper
26,258
618
Yes. Yes. Yes. Yes. Yes. Thank you!
 

Related Threads for: More Taylor series stuff, HELP!

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
16
Views
967
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
23
Views
3K
  • Last Post
Replies
3
Views
25K
  • Last Post
Replies
6
Views
964
  • Last Post
Replies
13
Views
849
Top