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

$\tan^{-1}(x)$ 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?

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

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

## marked as duplicate by J. M. is a poor mathematician, Claude Leibovici, Alex M., астон вілла олоф мэллбэрг, kingW3Jan 6 '17 at 12:07

• Is there any "Implicit function theorem" for analytic functions? – ajotatxe Jan 6 '17 at 6:38
• @ajotatxe I totally didn't understand your comment. – AHB Jan 6 '17 at 6:39
• $\arctan$ and $\tanh$ are analytic. Moreover the function $f(x,y)=\tan x-\text{arctanh} y$ is analytic. And $f(\arctan t, \tanh t)$ is $0$. If there is some "implicit function theorem" for analytic functions, then there is an analytic function $g$ (locally, at least) such that $g(\tanh t)=\arctan t$. – ajotatxe Jan 6 '17 at 6:47
• @ajotatxe yes. I am exactly looking for that $g$. – AHB Jan 6 '17 at 6:49
• @JohnMa Well, it is $f(\arctan t,\tanh t)=\tan\arctan t-\text{arctanh}\tanh t=t-t=0$. – ajotatxe Jan 6 '17 at 6:52

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
$$\tan^{-1}x=\tanh x+\frac{(\tanh x)^5}{15}+\frac{(\tanh x)^7}{45}+...$$
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 $y=\tanh(x)$,$$\tan^{-1}(x)=0.990525 y+0.062262 y^3$$ $$\tan^{-1}(x)=1.00139 y-0.0178282 y^3+0.117721 y^5$$ $$\tan^{-1}(x)=0.999706 y+0.00644068 y^3+0.0302925 y^5+0.0897757 y^7$$
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 $x >0.5$.