Closed-form expressions for coefficients $a_k$ in $2^{2n-1}\sinh^{2n}(x)=\sum_{k=0}^n a_k \cosh(2kx)$

Question

It is known that $$2^{2-1}\sinh^{2}(x)=\cosh(2x)-1$$ $$2^{4-1}\sinh^{4}(x)=\cosh(4x)-4\cosh(2x)+3$$

What are the closed-form expression for coefficients $a_k$ ($k=0,1,\cdots,n$) in the expression below?

$$2^{2n-1}\sinh^{2n}(x)=\sum_{k=0}^n a_k \cosh(2kx)$$

Thanks- mike

Answer 1

$a_0 = (-1)^n\frac{1}{2} \left( \begin{array}{c} 2n \\ n \end{array}\right)$, $a_n = 1$ and for $0<k<n$ $$ \begin{array}{l} a_k = (-1)^{n-k} \left( \begin{array}{c} 2n \\ n-k \end{array}\right) \end{array} $$ To see this, start from $2 \sinh x = (e^x - e^{-x})$ and expand using the binomial theorem $$ 2^{2n} \sinh^{2n} x = (e^x - e^{-x})^{2n} = \sum_{r=0}^{n} (-1)^r \left( \begin{array}{c} 2n \\ r \end{array}\right) e^{rx} e^{-(2n-r)x} = \sum_{r=0}^{n} (-1)^r \left( \begin{array}{c} 2n \\ r \end{array}\right) e^{(2r-2n)x} $$ Then break this into three parts: $\sum_{r=0}^{n-1}$, $\sum_{s=n+1}^{2n}$, and the $r=n$ term which is $(-1)^n \left( \begin{array}{c} 2n \\ n \end{array}\right)$.

In the second sum, change the summation index to $r = 2n - s$ and then those two sums group for like values of $r$, giving (after a bit of inde shifting) twice the expression appearing above.

Finally divide out that factor of 2, obtaining the $a_k$ for $2^{2n-1} \sinh^{2n} x$