# Using Taylor series expansion to solve the equation $\frac{\tanh^{-1}(x)}{\beta} -2x =0$

I want to use the Taylor series epansion of $$\tanh$$ to get an approximate solution $$\tilde{x}(\beta)$$ for the equation $$\frac{\tanh^{-1}(x)}{\beta} -2x =0$$ for $$\beta >\frac{1}{2}$$ such that I can calculate the limit $$\lim_{\beta \downarrow \frac{1}{2}}\frac{\tilde{x}(\beta)}{\left(\beta-\frac{1}{2}\right)^{\frac{1}{2}}}.$$ Unfortunately I have no clue on how to appropriately use the Taylor expansion to solve this. I guess it is possible to use the $$O$$-notation and break the problem down to a polynomial equation. Thank you very much in advance for any help.

The positive solution $$\tilde{x} (\beta)$$ to your equation tends to zero as $$\beta \downarrow \frac{1}{2}$$. Therefore, we can rewrite the equation and use the Taylor series $$\frac{\operatorname{artanh}(x)}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4)$$ to obtain $$2 \beta = \frac{\operatorname{artanh}(\tilde{x} (\beta))}{\tilde{x} (\beta)} = 1 + \frac{\tilde{x}^2 (\beta)}{3} + \mathcal{O}(\tilde{x}^4 (\beta)) \, .$$ We solve for the term with the lowest power of $$\tilde{x} (\beta)$$ and find $$\tilde{x} (\beta) = \sqrt{\frac{6 \left(\beta - \frac{1}{2}\right)}{1 + \mathcal{O}(\tilde{x}^2 (\beta))}} = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 + \mathcal{O}(\tilde{x}^2 (\beta))\right] = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 + \mathcal{O}\left(\beta - \frac{1}{2}\right)\right] \, .$$ This expansion is sufficient to compute the desired limit.

Using one more term of the above Taylor series leads to the better result $$\tilde{x} (\beta) = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 - \frac{9}{5} \left(\beta - \frac{1}{2}\right) + \mathcal{O}\left(\left(\beta - \frac{1}{2}\right)^2\right)\right] \, ,$$ which agrees with the first-order expansion of the more accurate Padé approximant provided by Awe Kumar Jha.

• Thank you very much. This in combination with the first answer helped me a lot. – Rupert R Mar 16 at 13:51

You should first see that, $$\tanh^{-1} (x) = \frac {1}{2} \ln \left( \frac {1+x}{1-x} \right)$$ Then, you can use the second order Pade approximant, $$\ln (1±x) ≈ \frac {±x(6±x)}{(6±4x)}$$ I hope it gives desired results.

# Edit

The Pade approximant of the original function is, $$\tanh^{-1} (x) ≈ \frac {x(15-4x^2)}{(15-9x^2)}$$ The solution to the resulting quadratic is, $$x=± \sqrt \frac {15(2\beta -1)}{2(9\beta -2)}, \beta ≠ \frac {2}{9}$$ What remains to do is the limit, $$\lim_{\beta → \frac {1}{2}} \sqrt {\frac {15}{9\beta -2}} = \sqrt {6}$$ I hope it helps.

• Thank you very much! This in combination with the second answer really helped me a lot. – Rupert R Mar 16 at 13:51
• It is a very good idea to have used the Padé approximant. $\to +1$ – Claude Leibovici Mar 17 at 3:37

Write the problem first as $$\beta=\frac{\tanh ^{-1}(x)}{2 x}$$ Now, expand the rhs using the usual $$\tanh ^{-1}(x)=\sum_{n=1}^\infty \frac {x^{2n+1}} {2n+1}$$ Using a few terms, we then have $$\beta=\frac{1}{2}+\frac{x^2}{6}+\frac{x^4}{10}+O\left(x^{6}\right)$$ Now, using series reversion $$\tilde{x} (\beta)=\sqrt{6} \sqrt{\beta -\frac{1}{2}}-\frac{9}{5} \sqrt{6} \left(\beta -\frac{1}{2}\right)^{3/2}+O\left(\left(\beta -\frac{1}{2}\right)^{5/2}\right)$$

as already given by ComplexYetTrivial.