Can Series Expansion Prove the Relation Between Inverse Coth and ln(x+1)/(x-1)?

[ognik]

Hi - my sometimes surprising set-book asks to show by series expansion, that $ \frac{1}{2}ln\frac{x+1}{x-1} =coth^{-1} (x) $

I get LHS = $ x+\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}+... $, which I think $= tanh^{-1} $ but I have found different expansions for the hyperbolic inverses, so I'd appreciate confirmation (or not) please?

 

[Prove It]

Hi - my sometimes surprising set-book asks to show by series expansion, that $ \frac{1}{2}ln\frac{x+1}{x-1} =coth^{-1} (x) $

I get LHS = $ x+\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}+... $, which I think $= tanh^{-1} $ but I have found different expansions for the hyperbolic inverses, so I'd appreciate confirmation (or not) please?

You shouldn't need to use series to prove this:

$\displaystyle \begin{align*} y &= \textrm{arcoth}\,{(x)} \\ \textrm{coth}\,{(y)} &= x \\ \frac{\cosh{(y)}}{\sinh{(y)}} &= x \\ \frac{\frac{1}{2}\,\left( \mathrm{e}^{y} + \mathrm{e}^{-y} \right) }{\frac{1}{2} \left( \mathrm{e}^y - \mathrm{e}^{-y} \right) } &= x \\ \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{\mathrm{e}^y - \mathrm{e}^{-y}} &= x \\ \mathrm{e}^y + \mathrm{e}^{-y} &= x \, \left( \mathrm{e}^y - \mathrm{e}^{-y} \right) \\ \mathrm{e}^y + \mathrm{e}^{-y} &= x\,\mathrm{e}^y - x\,\mathrm{e}^{-y} \\ \mathrm{e}^y \,\left( \mathrm{e}^y + \mathrm{e}^{-y} \right) &= \mathrm{e}^y\,\left( x\,\mathrm{e}^y - x\,\mathrm{e}^{-y} \right) \\ \left( \mathrm{e}^y \right) ^2 + 1 &= x\,\left( \mathrm{e}^y \right) ^2 - x \\ 1 + x &= x\,\left( \mathrm{e}^y \right) ^2 - \left( \mathrm{e}^y \right) ^2 \\ 1 + x &= \left( x - 1 \right) \,\left( \mathrm{e}^y \right) ^2 \\ \frac{x + 1}{x - 1 } &= \left( \mathrm{e}^y \right) ^2 \\ \left( \frac{x + 1}{x - 1} \right) ^{\frac{1}{2}} &= \mathrm{e}^y \\ \ln{ \left[ \left( \frac{x + 1}{x - 1} \right) ^{\frac{1}{2}} \right] } &= y \\ y &= \frac{1}{2} \ln{ \left( \frac{x + 1}{x - 1} \right) } \end{align*}$

Thus $\displaystyle \begin{align*} \textrm{arcoth}\,{(x)} = \frac{1}{2} \ln{ \left( \frac{x+1}{x -1} \right) } \end{align*}$.

 

[ognik]

You shouldn't need to use series to prove this:

As you have ably demonstrated :-), but its in a section on series, so we do what we must.

Is it possible that the series for both \(\displaystyle coth^{-1}\) and \(\displaystyle Tanh^{-1}\) are the same?

I start from $\d{coth^{-1}}{x}-\frac{1}{1-x^2}$

But I don't get the same series for $coth^{-1}$, my 1st problem is that my first term in the series should be f(0) but my calculator says $\frac{1}{Tanh^{-1}(0)} = 0$, so $coth^{-1}$ is singular?

My 2nd term f'''(0), I get $x^3$ instead of $\frac{x^3}{3}$ - so I think I'm doing something wrong...

 

[ognik]

To summarize what I'm asking:

I think the 1st term of the taylor series for arc coth x is singular, so I can't find a series. Wolfram seems to agree.

If I ignore the 1st term (but cannot justify that), I get something similar to the LHS series I got - but really, I'm just fiddling.

How else can I use series expansion to prove the relation?