Maclaurin series for ln

Question

I need to find the Maclaurin series for $\ln\sqrt{\frac{1+x}{1-x}} $. I found it is $\frac{x^{2n+1}}{{2n+1}}$ ,is it correct? I got the 1/2 outside and solved the maclaurin for the normal log. If it is okay how can I find a relationship between this and the maclaurin series artg.I know it is $(-1)^n$ of my series but how can I write it?

Answer 1

The Taylor series for $\log\sqrt{\frac{1+x}{1-x}}$ at $0$ is $\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$, because$$\begin{align}\log\sqrt{\frac{1+x}{1-x}}&=\frac12\log\left(\frac{1+x}{1-x}\right)\\&=\frac12\left(\log(1+x)-\log(1-x)\right)\\&=\frac12\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots-\left(-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\cdots\right)\right)\\&=x+\frac{x^3}3+\frac{x^5}5+\cdots\end{align}$$And the Taylor series for $\arctan(x)$ at $0$ is quite similar: $\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$. That's because$$\arctan'(x)=\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$

Answer 2

It’s very helpful to have memorized that $$\ln(u+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}u^n}{n}$$

So taking $$u=\sqrt{\frac{1+x}{1-x}}-1$$

you can have

$$\ln\sqrt{\frac{1+x}{1-x}} = \sum_{n=1}^\infty\frac{(-1)^{n+1} \left(\sqrt{\frac{1+x}{1-x}}-1\right) ^n}{n}$$

Answer 3

$$\log\left(\sqrt{\frac{1+x}{1-x}}\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)=\frac{1}{2}\left(\log(1+x)-\log(1-x)\right)=\frac{1}{2}\left(\sum\limits_{n\geq 1}(-1)^{n+1}\frac{x^n}{n}-\sum\limits_{n\geq 1}(-1)\frac{x^n}{n}\right)=\frac{1}{2}\left(\sum\limits_{n\geq 1}(-1)^{n+1}\frac{x^n}{n}+\sum\limits_{n\geq 1}\frac{x^n}{n}\right)=\frac{1}{2}\left(\sum\limits_{n\geq 1}((-1)^{n+1}+1)\frac{x^n}{n}\right)=\frac{1}{2}\left(\sum\limits_{n\geq 1}2\frac{x^{2n-1}}{2n-1}\right)=\sum\limits_{n\geq 1}\frac{x^{2n-1}}{2n-1}=:f(x)$$ And the arctan is: $$\tan^{-1}(x)=\sum\limits_{n\geq 1}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}=:g(x)$$ One relation can be: $$\tan^{-1}(x)=-\log\left(\sqrt{\frac{1+ix}{1-ix}}\right)$$ Or alternatively: $$g(x)=-f(ix)$$

Answer 4

As the derivative of this function is $$\frac12\bigl(\ln(1+x)'-\ln(1-x)'\bigr)=\frac1{1-x^2}=\sum_{n\ge0}x^{2n},$$ integrating term by term, we indeed obtain $$\ln\biggl(\sqrt{\frac{1+x}{1-x}}\biggr)=\sum_{n\ge0}\frac{x^{2n+1}}{2n+1}. $$ On the other hand, this function is also the inverse hyperbolic function $$\arg\tanh x$$ which explains why this expansion is very close to the expansion of $\arctan x$.

Answer 5

As already commented, if $f(x)=\log \sqrt{\frac{1+x}{1-x}}$, then $f(x)=\frac{1}{2}(\log(1+x)-\log(1-x))$. Note $\log(1+x)=\sum_{n \ge 1}(-1)^{n+1}\frac{x^n}{n}$ and $-\log(1-x)=\sum_{n \ge 1}\frac{x^n}{n}$. Thus $\log(1+x)-\log(1-x)=2(\sum_{n \ge 0}\frac{x^{2n+1}}{2n+1})$.