Is it possible to write $\tan^{-1}(x)$ as a power series of $\tanh(x)$?

Question

$\tan^{-1}(x)$ looks very similar to $\tanh(x)$ if $x$ is small enough.

Look.

But they diverge from each other as $x$ grows.

And for very big $x$'s, They almost represent the constant functions $1$ and $\frac \pi 2$ (for $\tanh(x)$ and $\tan^{-1}(x)$, respectively).

Is it possible to write $\tan^{-1}(x)$ as a power expansion of $\tanh(x)$?

I mean can we say this?

$$\tan^{-1}(x)=\sum^{\infty}_{i=0} \alpha_i \tanh^i(x)$$

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The power series is the thing I want. Not the resemblance between them.

Answer 1

Let $u=\tanh x \iff \tanh^{-1}u=x$. Then it is enough to expand $\tan^{-1}\tanh^{-1}u$ around $u=0$.

You will find that $\tan^{-1}\tanh^{-1}u=u+\frac{u^5}{15}+\frac{u^7}{45}+\frac{64u^9}{2835}+O(u^{11})$, and thus

$$\tan^{-1}x=\tanh x+\frac{(\tanh x)^5}{15}+\frac{(\tanh x)^7}{45}+...$$

Answer 2

A lazy way would be linear regression $$\tan^{-1}(x)=\sum^{n}_{i=0} \alpha_i \tanh^{2i+1}(x)$$ and, for the range $-1 \leq x \leq 1$, you could get, using $y=\tanh(x)$,$$\tan^{-1}(x)=0.990525 y+0.062262 y^3$$ $$\tan^{-1}(x)=1.00139 y-0.0178282 y^3+0.117721 y^5$$ $$\tan^{-1}(x)=0.999706 y+0.00644068 y^3+0.0302925 y^5+0.0897757 y^7$$

Edit

In Will Sherwood's good solution, the cubic term is missing (and this is correct based on the used approach). However performing the last given linear regression, the corresponding term is highly significant $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.999706 & 0.000026 & \{0.999654,0.999758\} \\ b & 0.006441 & 0.000306 & \{0.005836,0.007045\} \\ c & 0.030293 & 0.001044 & \{0.028233,0.032352\} \\ d & 0.089776 & 0.001061 & \{0.087684,0.091867\} \\ \end{array}$$ and, over the considered range, the curve fit is better than the theoretical septic expansion as soon as $x >0.5$.