Is it possible express $\sinh(nx)$ in terms of $\sinh^k(x)$?

Question

I wonder if it’s possible express $\sinh(nx)$ in terms of $\sinh^k(x)$, that is

$$\sinh(nx)=\sum_{k=0}^{A(n)} A_k\sinh^k(x)$$

Thanks in advance!

Answer 1

It is possible if $n$ is odd, and it is impossible if $n$ is even.

To see the first recall that $\sinh x=\dfrac{e^x-e^{-x}}2$ and observe applying binomial theorem that for odd $n$: $$ \sinh^nx=\frac{1}{2^{n-1}}\sum_{m=0}^{\frac{n-1}2}(-1)^m\binom nm \sinh{(n-2m)x}. $$ As the first result $\sinh3x=4\sinh^3x+3\sinh x$ will be obtained. Then apply induction.

To see the second observe that since $\sinh nx$ is an odd function the sum in rhs of your equation runs in fact only over odd $k$ and thus cannot produce the even exponents $e^{\pm nx}$.

Answer 2

Its possible for odd integer $n$:

$$\sinh((2k+1)x) = \sum_{m=0}^k \binom{2k+1}{2m+1}\cosh^{2k-2m} (x) \sinh^{2m+1}(x) $$ and we can express that purely in powers of $\sinh x$ by using $$\cosh^2x = \sinh^2 x +1 $$ so that $$\sinh((2k+1)x) = \sum_{m=0}^k \binom{2k+1}{2m+1} \sum_{t=0}^{k-m}\binom{k-m}{t} \sinh^{2m+2t+1}(x) $$ and here we simplify by letting $r = m+t$: $$\sinh((2k+1)x) = \sum_{m=0}^k \binom{2k+1}{2m+1} \sum_{r=m}^{k}\binom{k-m}{r-m} \sinh^{2r+1}(x) $$ The binomial coefficient sums can even be simplified into a closed form without sums, to get a power series in $\sin x$ with closed form coefficients! $$ \sinh((2k+1)x) = \sum_{r=0}^k \frac{4^r(1+2k) \binom{k+r}{2r}}{1+2r} \sinh^r x $$ For example, $$ \sinh (9x) = 256 \sinh^9x + 576 \sinh^7 x + 432 \sinh^5 x + 120 \sinh^3 x + 9 \sinh x $$ For even integer $n$ it is also almost possible, but in the form $$ \sinh(2kx) = \cosh x \sum a_r\sinh^r x $$ and you can't express that $\cosh x$ as a polynomial in $\sinh x$.

Answer 3

Not a complete solution.

We have

$$\sinh(nx)=\dfrac{\exp(nx)-\exp(-nx)}{2}=\dfrac{\exp(x)^{2n}-1}{2\exp(x)^n}.$$

As

$$\sinh(x) = \dfrac{\exp(x)-\exp(-x)}{2}=\dfrac{\exp(x)^2-1}{2\exp(x)}$$ $$\implies 2\exp(x)\sinh(x)=\exp(x)^2-1$$ $$\implies 0=\exp(x)^2-2\exp(x)\sinh(x)+\sinh^2(x)-\sinh^2(x)-1$$ $$1+\sinh^2(x)=(\exp(x)-\sinh(x))^2$$ $$\exp(x)=\sinh(x)+\sqrt{1+\sinh^2(x)}.$$

I selected the positive sign of the square root because of $\exp(x)>\sinh(x)$.

Hence,

$$\sinh(nx)=\dfrac{\left[\sinh(x)+\sqrt{1+\sinh^2(x)}\right]^{2n}-1}{2\left[\sinh(x)+\sqrt{1+\sinh^2(x)}\right]^n}.$$

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Edit: You can use the multiple-angle formula for the sine function

$$\sin(nx) = \sum_{k=0}^{n}\dfrac{n!}{k!(n-k)!}\cos^k x\sin^{n-k}x\sin\left[ \dfrac{1}{2}(n-k)\pi \right]$$

evaluated at $x=-iu$ to obtain the expression that the OP wanted (with an additional complex scaling factor). Note as already mentioned by other users this is only possible for odd $n$. Only then you will be able to rewrite the $\cos x$ terms by using $1-\sin^2 x$.