2
$\begingroup$

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!)

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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$

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

and

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

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

[Corrected hyperbolic identity]

$\endgroup$
0
$\begingroup$

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. \ $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.