1
$\begingroup$

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...

  • 0
    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. – 2017-01-04
  • 0
    Maybe [this](http://www.wolframalpha.com/input/?i=Table%5BD%5BCoth%5Bx%5D,%7Bx,n%7D%5D,%7Bn,1,10%7D%5D) can help you find a pattern. – 2017-01-04
  • 1
    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](http://oeis.org/A155100) – 2017-01-04
  • 0
    hmm... I think I've seen something like before. Do you have any more of the derivatives calculated in terms of $c$? – 2017-01-07
  • 0
    @BrevanEllefsen I can make some... – 2017-01-07
  • 1
    Have you seen [this](http://functions.wolfram.com/ElementaryFunctions/Coth/20/02/)? – 2017-01-07
  • 0
    @J.M.isn'tamathematician Thank you so much! :D – 2017-01-07

0 Answers 0