$n$-th derivative of $\sin^k(x)$

Question

I would like to know if there is a general formula to calculate the $n$-th derivative of $\sin^k(x)$ evaluated at $x=0$, that is,

$$\left.\frac{d^n}{d x^n} (\sin^k(x))\right|_{x=0}$$

with $0\leq k \leq n$.

Answer 1

You have a composition of two functions. To take the $n$th derivative, apply the Faa di Bruno formula outlined here, with $f(x)=x^k$ and $g(x)=\sin(x)$, and the divided $n$th derivative $f^{[n]}=\frac{1}{n!}f^{(n)}$.

$$\begin{align} \left(f\circ g\right)^{[n]}(x)&=\sum_{P\in\mathcal{P}_n}f^{[\left\lvert P\right\rvert]}(g(x))g^{[P]}(x)\\ &=\sum_{P\in\mathcal{P}_n}f^{[\left\lvert P\right\rvert]}(\sin(x))\sin^{[P]}(x)\\ &=\sum_{P\in\mathcal{P}_n}\binom{k}{\left\lvert P\right\rvert}\sin^{k-\left\lvert P\right\rvert}(x)\sin^{[P]}(x)\\ \end{align}$$

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Since you are only interested at evaluation at $x=0$, then since $\sin(0)=0$, you will only need to consider the terms where $|P|=k$.

$$\begin{align} \left.\left(f\circ g\right)^{[n]}(x)\right|_{x=0}&=\left.\sum_{\left\lvert P\right\rvert=k}\sin^{[P]}(x)\right|_{x=0}\\ \left.\left(f\circ g\right)^{(n)}(x)\right|_{x=0}&=n!\left.\sum_{\left\lvert P\right\rvert=k}\sin^{[P]}(x)\right|_{x=0}\\ \end{align}$$

And actually the only terms that give a nonzero contribution are those that come from a partition consisting of only odd numbers (since otherwise $\left.\sin^{[P]}(x)\right|_{x=0}$ will have a factor of $\left.\sin(x)\right|_{x=0}$). For example, with $k=8$, and $n=12$, the only ordered partitions of $12$ that use $8$ odd numbers are $$(1+1+1+1+1+1+1+5)\times 8\quad(1+1+1+1+1+1+3+3)\times \binom{8}{2}$$ so $$\left.\frac{d^{12}}{dx^{12}}\sin^8(x)\right|_{x=0}=12!\cdot\left(8\cdot\frac{1}{5!}+28\cdot\left(\frac{-1}{3!}\right)^2\right)=404490240$$

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Here is a summary formula: $$\begin{align} \left.\frac{d^n}{dx^n}\sin^k(x)\right|_{x=0} &=n!\sum_{\begin{array}{c}j_1+j_2+\cdots j_k=n\\j_i\text{ odd},\; j_i\geq1\end{array}}\prod_{i=1}^k\frac{(-1)^{(j_i-1)/2}}{j_i!}\\ &=(-1)^{(n-k)/2}\sum_{\begin{array}{c}j_1+j_2+\cdots j_k=n\\j_i\text{ odd},\; j_i\geq1\end{array}}\binom{n}{j_1\;j_2\;\cdots\;j_k} \end{align}$$

Some corollaries:

$$\left.\frac{d^n}{dx^n}\sin^k(x)\right|_{x=0}=\begin{cases}0&n<k\\ n!&n=k\\ 0&n\not\equiv k\mod{2}\\ (-1)^{(n-1)/2}&k=1\\ -n!k/6&n=k+2 \end{cases}$$

Answer 2

Probably the easiest way to do this is to use the exponential form of $\sin$ and then the binomial theorem: $$ \sin^k(x) = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^k=\frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j} e^{jix}(-1)^{k-j}e^{-(k-j)ix}=\frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j} (-1)^{k-j}e^{(2j-k)ix} $$

The $n$th derivative of $e^{ax}$ evaluated at zero is $a^n$, so the formula you're looking for is $$ \frac{1}{(2i)^k}\sum_{j=0}^k \binom{k}{j}(-1)^{k-j}[i(2j-k)]^n $$

You could turn this into a purely real formula, but it'd require a certain amount of case analysis; this is probably the most compact form unless there's some simplifying binomial identity I'm missing.

(The other possible approach would be to examine the Taylor series of $\sin^k x$, but that requires the multinomial theorem instead of the binomial theorem, so it's likely to be a lot messier...)

Answer 3

In Wikipedia you find the formulas for the transformation of a power of a sine to a sum of sines of multiples. Taking the derivatives at $x=0$ is then straightforward.

For instance, for odd $k$ and odd $n$

$$2^{1-k}(-1)^{(k-1)/2}\sum_{j=0}^{(k-1)/2}(-1)^{j+(n-1)/2}\binom kj(k-2j)^n.$$

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Due to the renaming of the indexes "on the fly", typos are likely.

Answer 4

First using the identity,

$$ \sin x=\frac{1}{2 i}\left(e^{x i}-e^{-x i}\right) $$

to expand $$\begin{aligned} \sin ^{k} x &=\frac{1}{(2i)^ k } \sum_{j=0}^{k}(-1)^j\left(\begin{array}{l} k \\ j \end{array}\right) e^{x(k-j) i} e^{-x i j} \\ &=\frac{1}{2^{k} i^{k}} \sum_{j=0}^{k}(-1)^j\left(\begin{array}{l} k \\ j \end{array}\right) e^{x(k-2 j) i} \end{aligned} $$

Differentiating it $n$ times w.r.t. $x$ yields $$ \frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right)=\frac{1}{2^{k} i ^k} \sum_{j=0}^{k}(-1)^{j} \left(\begin{array}{l} k \\ j \end{array}\right)[(k-2 j) i]^{n} e^{x(k-2 j) i}= \boxed{\frac{i^{n-k}}{2 ^ k} \sum_{j=0}^{k}(-1)^{j} \left(\begin{array}{l} k \\ j \end{array}\right)(k-2 j)^{n} e^{x(k-2 j)i} }$$

Putting $x=0$ yields $$ \left.\frac{d^{n}}{d x ^{n}}\left(\sin ^{k} x\right)\right|_{x=0}=\frac{i^{n-k}}{2^{k}} \sum_{j=0}^{k}(-1)^{j}\left(\begin{array}{c} k \\ j \end{array}\right)(k-2 j)^{n} $$

Consequently, assuming $x$ is real, we have

A. When $n$ and $k$ are of different parity,

$$ \left.\frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right) \right|_{x=0} =0 $$

B. When $n$ and $k$ are of same parity,

$$ \left.\frac{d^{n}}{d x^{n}}\left(\sin ^{k} x\right) \right|_{x=0} =\frac{(-1)^{\frac{n-k}{2}}}{2^{k}} \sum_{j=0}^{k}(-1)^{j}\left(\begin{array}{c} k \\ j \end{array}\right)(k-2 j)^{n} $$