- #1

RChristenk

- 49

- 4

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

- Relevant Equations
- Logarithm Rules, Logarithmic Series

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

##\Rightarrow \log_e(x-1)=\log_e[x(1-\dfrac{1}{x})]=\log_ex+\log_e(1-\dfrac{1}{x})##

We know:

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

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

Hence:

##\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]=\dfrac{1}{2}[\log_e(1+x)+\log_ex+\log_e(1-\dfrac{1}{x})]##

##\Rightarrow \dfrac{1}{2}(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots+\log_ex-\dfrac{1}{x}-\dfrac{(\dfrac{1}{x})^2}{2}-\dfrac{(\dfrac{1}{x})^3}{3}-\cdots)##

##\Rightarrow \dfrac{1}{2}(\log_ex+x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots-\dfrac{1}{x}-\dfrac{1}{2x^2}-\dfrac{1}{3x^3}-\cdots)##

This is as far as I got. I'm not sure how to perform the necessary algebra to get to

##\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^2}-\dfrac{1}{6x^6}-\cdots##.

##\Rightarrow \log_e(x-1)=\log_e[x(1-\dfrac{1}{x})]=\log_ex+\log_e(1-\dfrac{1}{x})##

We know:

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

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

Hence:

##\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]=\dfrac{1}{2}[\log_e(1+x)+\log_ex+\log_e(1-\dfrac{1}{x})]##

##\Rightarrow \dfrac{1}{2}(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots+\log_ex-\dfrac{1}{x}-\dfrac{(\dfrac{1}{x})^2}{2}-\dfrac{(\dfrac{1}{x})^3}{3}-\cdots)##

##\Rightarrow \dfrac{1}{2}(\log_ex+x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots-\dfrac{1}{x}-\dfrac{1}{2x^2}-\dfrac{1}{3x^3}-\cdots)##

This is as far as I got. I'm not sure how to perform the necessary algebra to get to

##\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^2}-\dfrac{1}{6x^6}-\cdots##.