How Does Taylor's Theorem Apply to Logarithmic Series?

[kidsmoker]

Homework Statement

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

$$log(1+x) = \sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}x^{k}$$

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:

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

where

$$R(n,x) = \frac{x^{n+1}}{(n+1)!}f^{(n+1)}(t)$$

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

$$log(1-x) = \sum^{n}_{k=1}\frac{-1}{k}x^{k} + \frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}$$

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!

 

[Born2bwire]

Take the magnitude of

$$\frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}$$

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

 

[Architeuthis Dux]

Dear student,

First of all, great job on part (a) of the problem! You have correctly applied Taylor's theorem with the Lagrange remainder to show that the given series converges to log(1+x) for 0<x<1. Now, for part (b), there seems to be some confusion regarding the remainder term. In your attempt, you have written the remainder term as:

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

However, the correct form of the remainder term for Taylor's theorem with the Lagrange remainder is:

R(n,x) = \frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(t)

where t is a number between a and x. In this case, a = 0 and x is between -1 and 0. So, we can write:

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

for some t in (-1,x). This means that the denominator of the remainder term will always be negative since x is negative. Therefore, as n tends to infinity, the remainder term will tend to zero since the numerator will also be negative. This shows that the given series also converges to log(1-x) for -1<x<0.

I hope this helps clarify any confusion you had. Keep up the good work!