1. ## taylor series

$\log(1-x) = -x-\frac{x^2}{2}-\frac{x^2}{3}-\frac{x^4}{4}- \ldots -\frac{x^n}{n} - R_n$ for $-1 \leq x < 1$.

Why isn't this the case for $-1 < x < 1$? Instead, why do we replace $x$ with $-x$ to get $\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} +\ldots + (-1)^{n-1} \frac{x^n}{n} - R_{n}'$?

2. Originally Posted by particlejohn
$\log(1-x) = -x-\frac{x^2}{2}-\frac{x^2}{3}-\frac{x^4}{4}- \ldots -\frac{x^n}{n} - R_n$ for $-1 \leq x < 1$.

Why isn't this the case for $-1 < x < 1$? Instead, why do we replace $x$ with $-x$ to get $\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} +\ldots + (-1)^{n-1} \frac{x^n}{n} - R_{n}'$?

Ok

I will go in depth with this,

We know (hopefully) that

$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

We also know

$\int\frac{-dx}{1-x}=\ln(1-x)$

So now making a substitution we can see that

$\ln(1-x)=-\int\sum_{n=0}^{\infty}x^n=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$

So now we know that integrating/differentiating a power series only changes the endpoint behavior

So we know that since

$\sum_{n=0}^{\infty}x^n$ converges $\forall{x}\in(-1,1)$

So now we must check the endpoints.

At $x=-1$ we have

$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}$

Which diverges by the Alternating series test.

Since $\frac{1}{n+2}\leq\frac{1}{n+1}\quad\forall{n}\in\m athbb{N}$

Now we need to check at

$x=1$

We have

$-\sum_{n=0}^{\infty}\frac{1}{n+1}$

Which diverges by the integral test

$\int_0^{\infty}\frac{dx}{x+1}=\ln(x+1)\bigg|_{0}^{ \infty}=\infty$

So we have that

$\ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad\forall {x}\in[-1,1)$
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Now for your second question we can see the by direct substitution into either the log or the $\frac{1}{1-x}$ equation, or seeing that

$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^{n}$

This implies that

$\ln(1+x)=\int\frac{dx}{1+x}$

is found by

$\int\sum_{n=0}^{\infty}(-1)^nx^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}$

Now once again we must check the endpoints,

so at $x=-1$

We have

$-\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(-1)^n}{n+1}=-\sum_{n=0}^{\infty}\frac{1}{n+1}$

Which as I showed earlier is divergent by the integral test.

But at

$x=1$

We have

$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}$

Which as was shown earlier is convergent by the alternating series test.

Hope this helps.

3. Originally Posted by Mathstud28

So now we must check the endpoints.

At $x=-1$ we have

$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}$

Which diverges by the Alternating series test.

Since $\frac{1}{n+2}\leq\frac{1}{n+1}\quad\forall{n}\in\m athbb{N}$
It converges, and that is what the alternating series test tells us.

RonL

4. Originally Posted by CaptainBlack
It converges, and that is what the alternating series test tells us.

RonL
Yeah, I that is what I meant to say (as you can see by it's inclusion in the Radius of Convergence)

5. $-\ln (1-x)=\int_{0}^{x}{\frac{dy}{1-y}}=\sum\limits_{n\,=\,0}^{\infty }{\int_{0}^{x}{y^{n}\,dy}}=\sum\limits_{n\,=\,0}^{ \infty }{\frac{x^{n+1}}{n+1}}=\sum\limits_{n\,=\,1}^{\inf ty }{\frac{x^{n}}{n}}.$