Are there known closed formula expressions for their power series expansion at the origin? I couldn't find anything online.

Edit: (to clarify) Of course we could simply take the series expansion of

$$\tanh(x) = \sum_{n=1}^{\infty} \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1} $$

And square it with a Cauchy product:

$$\tanh(x)^2 = \sum_{n=2}^{\infty} \Big[\sum_{i+j=n}\frac{2^{2i}(2^{2i}-1)B_{2i}}{(2i)!}\frac{2^{2j}(2^{2j}-1)B_{2j}}{(2j)!}\Big] x^{2n-2} $$

The question is whether or not one can simplify the coefficient term

$$ \sum_{i+j=n}\frac{2^{2i}(2^{2i}-1)B_{2i}}{(2i)!}\frac{2^{2j}(2^{2j}-1)B_{2j}}{(2j)!} $$

in any meaningful way.

I tried to do the following:

$$ \tanh^2 =\frac{\sinh^2}{\cosh^2} \iff \tanh^2(1+\sinh^2) = \sinh^2$$

Now letting $\tanh(x)^2 = \sum_{n=0}^\infty a_{2n} x^{2n}$, and using the known formula $\sinh(x)^2 = \sum_{n=1}^{\infty} \frac{2^{2n-1}}{(2n)!}x^{2n}$ yields a Volterra type difference equation:

$$a_{2n} = \frac{2^{2n}}{(2n)!} - \sum_{k=0}^{n} a_{2k}\frac{2^{2n-2k}}{(2n-2k)!} \qquad\text{for }n\ge1 $$

which I had no luck solving so far (it's a hairy business!)

  • $\begingroup$ Do you know the Maclaurin expansions for $\tan$ and $\tanh$? $\endgroup$ – Simply Beautiful Art Aug 2 '18 at 16:28
  • $\begingroup$ @SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible.... $\endgroup$ – Hyperplane Aug 2 '18 at 16:30
  • 2
    $\begingroup$ I don't see how you could. $\endgroup$ – Simply Beautiful Art Aug 2 '18 at 16:31

The series expansions of $\tan^2 x$ and $\tanh^2 x$ can be obtained by differentiating the series expansions of $\tan x$ and $\tanh x$ and utilising $\sec^2 x = 1+\tan^2 x$ and $\DeclareMathOperator{\sech}{sech} \sech^2 x = 1-\tanh^2 x$ respectively

As $\frac{d \; \tan x}{dx}=\sec^2x$ and $ \frac{d \; \tanh x}{dx}=\sech^2x$

$$\tan^2 x=\frac{d \; \tan x}{dx}-1 \tag{1}$$


$$\tanh^2 x=1-\frac{d \; \tanh x}{dx} \tag{2}$$

I am not sure this is what you meant though as you mentioned closed form expressions in your question.

[Corrected hyperbolic identity]


This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $\ \tan(x). \ $ Because of the identity $$ \frac{d}{dx} \tan (x) = \sec^2(x) = 1 + \tan^2(x) = 1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +\dots $$ the function $\ 1+\tan^2(x) \ $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $\ x. \ $ A very similar relationship holds for the hyperbolic tangent function since $$ \frac{d}{dx} \tanh (x) = \sech^2(x) = 1 - \tanh^2(x) = 1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +\dots . $$ As a consequence of this, we get the recursion $\ a_n = \sum_{k=1}^n a_{k-1}a_{n-k} {2n \choose 2k-1} $ for $\ n>0. \ $


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