Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^4}-\dfrac{1}{6x^6}-\cdots$

Question

Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^4}-\dfrac{1}{6x^6}-\cdots$

$\log_e\sqrt{x^2-1}=\dfrac{1}{2}\log_e(x^2-1)=\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]$

I know:

$\log_e(x+1)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots$

$\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}- \cdots$

How would I get $\log_e(x-1)$?

Answer 1

Here is the solution.

$\log_e \sqrt{x^2 - 1} - \log_e x = \frac{1}{2}\log_e \left({1 - \frac{1}{x^2}}\right)$.

Now use the Taylor series expansion of $\log_e {(1 + x)}$ and replace $x$ by $\frac{-1}{x^2}$.

Answer 2

$\log(x-1) = \log(x(1-1/x)) = \log x + \log(1-1/x)$