$\frac{d^m}{dx^m}\left(\sin x\right)^n$

Question

Is there a simple formula for

\begin{align} \frac{d^m}{dx^m}\left(\sin x\right)^n\Bigg|_{x=0}, \end{align}

with $m\geq 1$ and $n\geq 1$ integers?

If we expand the $n$-the power of the sine we have to consider the even and odd cases, and then when evaluating $x=0$ we have to further consider the cases of $m$ even or odd (with ugly coefficients), so I wonder if a compact formula is possible.

notice that $\sin(0)=0$, so you only care about the parts that contain only $\cos(x)$ — Exodd, Nov 20 '22 at 13:10
Take the formal series $$h(x) = x -x^3+x^5-x^7+\dots$$ What you're looking for is the coefficient of $x^m$ in $h(x)^n$. — Exodd, Nov 20 '22 at 13:23
Notice in particular that it is zero if $m-n$ is negative or odd — Exodd, Nov 20 '22 at 13:29
There is a simpler formula using the complex expansion of $\sin(x)$ — Тyma Gaidash, Nov 20 '22 at 15:15

Тyma Gaidash · Accepted Answer · 2022-11-20T17:01:08.680

Here is a similar one from:

Does sinc function have any special inverse function defined?

$$\frac{d^m}{dx^m}(\sin(x)^n)=\frac{d^m}{dx^m}\left(\left(\frac i2\right)^n\sum_{k=0}^n\binom nk(e^{-ix})^{n-k}(-e^{ix})^k\right)=\left(\frac i2\right)^n\sum_{k=0}^n\binom nk(-1)^k \frac{d^m}{dx^m}e^{i(2k-n)x}= \left(\frac i2\right)^n\sum_{k=0}^n\binom nk(-1)^k (i(2k-n))^me^{i(2k-n)x}$$

testable here

At $x=0$, you get:

$$\boxed{\left.\frac{d^m}{dx^m}(\sin(x)^n)\right|_{x=0}=\left(\frac i2\right)^n\sum_{k=0}^n\binom nk(-1)^k (i(2k-n))^m}$$

There is a hypergeometric result and these simplifications for it, but how do you apply them here for a non sum form?

$\frac{d^m}{dx^m}\left(\sin x\right)^n$

1 Answers1