Closed form for the $n$-th derivative of $\csc^2$?

Question

To find a closed representation for the $n$-th derivative of $f(t)=\csc^2(t)$, I started with the Ansatz: $$ f^{(n)}(t) = \frac{p_n(\sin t, \cos t)}{\sin^{2+n}(t)} \tag1 $$ where $p_n$ is a polynomial (with integer coefficients) in two variables. This leads to the recursion $$ p_{n+1}=xy\frac{\partial p_n}{\partial x} -x^2\frac{\partial p_n}{\partial y} -(2+n)\,y\,p_n~;\qquad p_0=1 \tag2 $$ and I am stuck. Likely, someone has solved this before and has a pointer to the $n$-th derivative of $\csc^2$?

With "closed representation", I mean something like a finite sum and without recursion.

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Note: This question is on $\cot$ and only comes up with a recursion. (It's a polynomial in just 1 variable so maybe simpler to turn into an explcit form.)

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Edit

That question leads to Lemma 2.1 in: V.S. Adamchik, On the Hurwitz function for rational arguments, Applied Mathematics and Computation, Volume 187, Issue 1, 1 April 2007, Pages 3–12.

It states for the $n$-th derivative of $\cot z$, $n>1$:

$$ \frac{d^n}{dz^n} \cot z = (2i)^n (\cot(z)-i)\sum_{j=1}^n \frac{k!}{2^k} \left\{ {n \atop k} \right\} (i\cot(z)-1)^k \\ $$

$$ \text{where } \left\{ {j \atop k} \right\} \text{ is the Stirling subset}\dots $$

but after the $\dots$ the text is truncated

I cannot make 100% sense of it and how $k$ is computed etc. Maybe someone kows the missing bits.

Answer 1

Since $(-\cot x)'=(\csc^2 x)$, so $(\csc^2 x)^{(n)}=-(\cot x)^{(n+1)}$ and we only have to find the $n$-th derivative of cotangent.

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Since $\cot x=\frac1\pi \psi \left(1-\frac{x}{\pi }\right)-\frac1\pi\psi \left(\frac{x}{\pi }\right)$ (where $\psi$ is polygamma function), its n-th derivative is

$\cot^{(n)}(x)=(-1)^n \pi ^{-n-1} \psi ^{(n)}\left(1-\frac{x}{\pi }\right)-\pi ^{-n-1} \psi ^{(n)}\left(\frac{x}{\pi }\right)$

Wolfram Mathematica says you can express it in cotangent functions:

$\cot^{(n)}(x)=(-2 i)^n (\cot (x)+i) \underset{k=0}{\overset{n}{\sum }}\left(-\frac{1}{2}\right)^k (i \cot (x)+1)^k k! \mathcal{S}_n^{(k)}$, where $S$ are the Stirling numbers of the second kind.

There are also other formulas.

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Thus,

$(\csc^2 x)^{(n)}=(-1)^n \pi ^{-n-2} \psi ^{(n+1)}\left(1-\frac{x}{\pi }\right)+\pi ^{-n-2} \psi ^{(n+1)}\left(\frac{x}{\pi }\right)$

(in terms of polygamma function)

$(\csc^2 x)^{(n)}=-(-2 i)^{n+1} (\cot (x)+i) \underset{k=0}{\overset{n+1}{\sum }}\left(-\frac{1}{2}\right)^k (i \cot (x)+1)^k k! \mathcal{S}_{n+1}^{(k)}$

(in terms of elementary functions)

Mathematica code verifying the two formulas above:

n = 3;
Plot[{Evaluate[
   D[Csc[x]^2, {x, n}]], -(-2 I)^(
    n + 1) (I + Cot[x]) Sum[(-(1/2))^k (1 + I Cot[x])^
     k k! StirlingS2[n + 1, k], {k, 0, n + 1}], Pi^(-n - 2)
     PolyGamma[n + 1, x/Pi] + (-1)^n Pi^(-n - 2)
     PolyGamma[n + 1, 1 - x/Pi]}, {x, -10, 10}, 
 AspectRatio -> Automatic, PlotRange -> 10]

Answer 2

Here is a closed form using a different approach. Applying the binomial theorem, $$(a+b)^{-2}=\sum\limits_{k=1}^\infty\binom{-2}ka^{-k-2}b^k=\sum_{k=0}^\infty(k+1)a^{-k-2}(-b)^k$$ to $\csc^2(x)=-\frac4{(e^{-i x}-e^{ix})^2}$ and again on $(k+1)^n$ gives:

$$\frac{d^n}{dx^n}\csc^2(x)=-4\sum_{k=0}^\infty (k+1)\frac{d^n}{dx^n}e^{2i(k+1)x}= -4(2i)^n\sum_{k=0}^\infty (k+1)^{n+1}e^{2i (k+1)x}=-4(2i)^n\sum_{k=0}^\infty\sum_{m=0}^{n+1}\binom{n+1}mk^me^{2i (k+1)x}$$

Switching sums gives the following Hurwitz Lerch transcendent solution:

$$\boxed{\frac{d^n}{dx^n}\csc^2(x)=-i^n2^{n+2}e^{2 i x}\sum_{m=0}^{n+1}\binom{n+1}m\Phi(e^{2 i x},-m,0)}$$

shown here. However, converting the Lerch transcendent into a polylogarithm, in software, gives the wrong result.