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

A taylor series question

  • Thread starter kidsmoker
  • Start date
  • #1
88
0

Homework Statement



(a) Use Taylor's theorem with the Lagrange remainder to show that

[tex]log(1+x) = \sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}x^{k}[/tex]

for 0<x<1.

(b) Now apply Taylor's theorem to log(1-x) to show that the above result holds for -1<x<0.


Homework Equations



Taylor's theorem w/ Lagrange remainder:

[tex]f(x) = \sum^{n}_{k=0}\frac{(x-a)^{k}}{k!}f^{(k)}(a) + R(n,x)[/tex]

where

[tex]R(n,x) = \frac{x^{n+1}}{(n+1)!}f^{(n+1)}(t)[/tex]

for some t in (0,x).

The Attempt at a Solution



I seem to have done part (a) okay by just writing it all out and then showing that the remainder tends to zero as n tends to infinity.

Then when I do the same thing with log(1-x) I get

[tex]log(1-x) = \sum^{n}_{k=1}\frac{-1}{k}x^{k} + \frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}[/tex]

for some t in (0,x). But we're considering the case here were 0<x<1, so the denominator in the remainder will be between -1 and 0. Consequently, as n tends to infinity, this part will tend to zero and the fraction as a whole will tend to infinity. I've checked my working a few times and tried fiddling about with different methods, including trying the Cauchy remainder, but I can't seem to show that R->0 as n->inf.

Any help appreciated!
 

Answers and Replies

  • #2
Born2bwire
Science Advisor
Gold Member
1,779
18
Take the magnitude of

[tex]\frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}[/tex]

and work to replace the (t-1) with x. Just keep in mind that x and t are both (0,1).
 
Last edited:

Related Threads for: A taylor series question

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
588
  • Last Post
Replies
1
Views
5K
Replies
8
Views
5K
  • Last Post
Replies
4
Views
3K
Top