The $n$th derivative of the hyperbolic cotangent

Asked Jan 04 '17 at 12:27

Active Jan 07 '17 at 11:38

Viewed 92 times

While working on a problem, I came to this:

What is the $n$th derivative of the hyperbolic cotangent?

For simplicity, let $c=\coth(x)$.

$c^{(0)}=c$

$c^{(1)}=-c^2+1$

$c^{(2)}=2c^3-2c$

$c^{(3)}=-6c^4+5c^2-2$

$c^{(4)}=24c^5-34c^3+10c$

Etc. It appears to be representable as a polynomial of $c$. Any ideas on what the coefficients are?

Update:

It appears the leading coefficient is trivially given by $(-1)^nn!$.

I am not entirely sure about the next non-zero coefficient, but I believe it is given as follows:

$$\sum_{p=0}^n(-1)^p\frac{(n-2)!}{(n-2-p)!}(n-p)!$$

Or something along these lines, where $\frac1{k!}=0$ if $k<0$.

Update:

Perhaps we should first look at the derivatives of hyperbolic tan? As there is no alternating sign...

edited Jan 07 '17 at 11:38

asked Jan 04 '17 at 12:27

You can easily see by the chain rule and induction that it does indeed always become a polynomial. What polynomial, I can't immediately say. – Arthur Jan 04 '17 at 12:41
Maybe this can help you find a pattern. – Zubzub Jan 04 '17 at 12:44
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If the coefficient of $c^k$ in the $n^{\text{th}}$ derivative of $c$ is $a_{n,k}$ then $a_{n,k}=(k-1)a_{n-1,k-1} -(k+1)a_{n-1,k+1}$. In particular $a_{n,n}= (-1)^{n-1}(n-1)!$ while $(c-1)(c+1)$ is a factor of the derivatives. The coefficients are signed versions of OEIS A155100 – Henry Jan 04 '17 at 12:55
hmm... I think I've seen something like before. Do you have any more of the derivatives calculated in terms of $c$? – Brevan Ellefsen Jan 07 '17 at 03:02
@BrevanEllefsen I can make some... – Simply Beautiful Art Jan 07 '17 at 11:32
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Have you seen this? – J. M. ain't a mathematician Jan 07 '17 at 11:55
@J.M.isn'tamathematician Thank you so much! :D – Simply Beautiful Art Jan 07 '17 at 11:58

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