Derivation of a function over $\frac{1}{\sinh(t)}\frac{d }{dt}$

Question

I can not calculate the next derivative, someone has an idea $$\left( \frac{1}{\sinh(t)}\frac{d }{dt} \right)^n \left( e^{z t} \right)$$ Where $n\in \mathbb N$, $t>0$ and $z\in \mathbb C$.

Thanks in advance

Answer 1

$$\left(\frac{1}{\sinh(t)}\frac{\text{d}}{\text{d}t}\right)^n e^{zt} = \left(\frac{1}{\sinh^n(t)}\right)z^n e^{zt}$$

You simply have to derive $n$ times the exponential.

Then if you want you can write in the end, using $e^{zt} = \sinh(zt) + \cosh(zt)$ the result as:

$$z^n \frac{\sinh(zt) + \cosh(zt)}{\sinh^n(t)}$$

EDIT (because what I wrote above is a malarkey)

Containing the parentheses an operator, the way to treat it is to watch if some useful expression arises from a $n$ times products:

$$\left(\frac{1}{\sinh(t)}\frac{\text{d}}{\text{d}t}\right)^2 = \left(\frac{1}{\sinh(t)}\frac{\text{d}}{\text{d}t}\right)\cdot\left(\frac{1}{\sinh(t)}\frac{\text{d}}{\text{d}t}\right)$$

Calling $\sinh(t) = a(t)$ we would have:

$$a(t)\frac{d}{dt} \left(a(t) \frac{d}{dt} \right) = a(t)a'(t) + a^2(t)\dfrac{d^2}{dt^2}$$

where $a'(t)$ means differentiation with respect upon $t$.

Going on and we find:

$$a(t)\frac{d}{dt} \left(a(t) \frac{d}{dt} \right)\left(a(t) \frac{d}{dt} \right) = \left(a(t)a'(t) + a^2(t)\dfrac{d^2}{dt^2}\right)\left(a(t) \frac{d}{dt} \right) = a^2(t)a'(t)\frac{d}{dt} + a^2(t)a''(t) \frac{d}{dt} + a^3(t)\frac{d^3}{dt^3}$$

One should verify if a sort of recurrence does exist..

Answer 2

HINT

Take $ z=1 , x= \cosh t $

$$ \rightarrow \frac{d^n}{d x^n} ({\sqrt{x^2+1}-x} ) $$

$$

( you can modify for $ z \ne 1 $)

Answer 3

$x=\cosh(t) \Rightarrow$ $\frac{1}{\sinh(t)}\frac{d }{dt} = \frac{d}{dx}$ and

$x=\cosh(t) \Rightarrow$ $arcosh(x)=t \Rightarrow$ $z\,t =z \,arcosh(x)\Rightarrow$ $z\,t =z \left(\ln(x+\sqrt{x^2-1})\right) $.

So \begin{align} \left( \frac{1}{\sinh(t)}\frac{d }{dt} \right)^n \left( e^{z t} \right) &= \frac{d^n}{d x^n} \left( e^{z \left(\ln(x+\sqrt{x^2-1})\right) } \right) = \frac{d^n}{d x^n} \left( e^{\ln\left(x+\sqrt{x^2-1}\right)^z } \right) \\ &= \frac{d^n}{d x^n} \left( \left(x+\sqrt{x^2-1}\right)^z \right) \\ &= ??? \end{align} Then I can not continue !!!

Answer 4

Not a closed form, but we can derive a simple recursion formula to ease the calculation of the expression. We notice that the result can be viewed as a polynomial in $z$ with some $t$-dependent coefficients. If we take $$\left(\frac{1}{\sinh(t)}\frac{d}{dt}\right)^n e^{zt} \equiv \sum_{k=0}^n a_k^{(n)}(t) z^k e^{zt}$$

and apply the differential operator we find the following recursion

$$a_{k}^{(n+1)}(t) = \frac{1}{\sinh(t)}\left(\frac{da_k^{(n)}(t)}{dt} + a_{k-1}^{(n)}(t)\right)$$

In the formula above we take $a_{k}^{(n)} \equiv 0$ if $k < 0$ or $k > n$. The initial conditions for the recursion is $a_{0}^{0}(t) = 1$.