I know how to convert $\sinh^{-1}{x}$ and $\cosh^{-1}{x}$ to $\ln{|x+\sqrt{x^2 \pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$\tanh^{-1}{\frac{x}{2}}$$
Can someone please show me how to convert it to a $\ln$ form? thanks!
$\endgroup$3 Answers
$\begingroup$Set $ y = \tanh^{-1} t $ and take $\tanh$ to take both sides so we have $$ \tanh y = t .$$
Now convert the $\tanh$ term into it's definition in terms of exponentials: $$ t = \frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = \frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= \frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
$\endgroup$ $\begingroup$Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=\tanh^{-1}x$
$\tanh y=x$
$\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=\frac{1+x}{1-x}$
$y=\frac{1}{2}\ln(\frac{1+x}{1-x})$
$\endgroup$ $\begingroup$$$\tanh x=\frac{e^{2x}-1}{e^{2x}+1} \Rightarrow \tanh \frac{x}{2}=\frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanh\frac{x}{2}$ we have to solve following equation :
$$\frac{x}{2}= \frac {e^{2artanh \frac{x}{2}}-1}{e^{2artanh \frac{x}{2}}+1} \Rightarrow x \cdot e^{2artanh \frac{x}{2}}+x=2e^{2artanh \frac{x}{2}}-2 \Rightarrow$$
$$x+2=(2-x) \cdot e^{2artanh \frac{x}{2}} \Rightarrow e^{2artanh \frac{x}{2}}=\frac{2+x}{2-x} \Rightarrow artanh \frac{x}{2} =\frac{1}{2} \ln \left(\frac{2+x}{2-x}\right) ; |x| < 2$$
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