What is the $n$-th difference of $\sin(ax)$

Question

It is proven in the textbook that I'm working out of, that in theory of difference calculus,

$$\Delta \sin{ax}=2\sin{\frac{a}{2}}\cos{a}\left(x+\frac{1}{2}\right)$$

$$\Delta \cos{ax}=-2\sin{\frac{a}{2}}\sin{a}\left(x+\frac{1}{2}\right)$$

$$\Delta \sin{(ax+b)}=2\sin{\frac{a}{2}}\cos\left(ax+b+\frac{a}{2}\right)$$

$$\Delta \cos{(ax+b)}=-2\sin{\frac{a}{2}}\sin\left(ax+b+\frac{a}{2}\right)$$

Note here that the Delta notation $\Delta f(x)=f(x+1)-f(x)$. From there it can be shown that the above identity is true. Now, I'm looking for higher order differences. So I was tasked with determining,

$$\Delta^n \sin{ax}$$

under the definition of $\Delta^n=\Delta^{n-1}(\Delta)$ Now, using these previous identities, it would appear that the $n$-th difference formula depends on $n$. From what I have done algebraically,

$$\Delta \sin{ax}=2\sin\frac{a}{2}\cos\left(ax+\frac{a}{2}\right)$$

$$\Delta^2 \sin{ax}=-2^2\sin^2\frac{a}{2}\sin\left(ax+\frac{2a}{2}\right)$$

$$\Delta^3 \sin{ax}=-2^3\sin^3\frac{a}{2}\cos\left(ax+\frac{3a}{2}\right)$$

$$\Delta^4 \sin{ax}=2^4\sin^4\frac{a}{2}\sin\left(ax+\frac{4a}{2}\right)$$

So it would appear that if we consider this for any $n$, then we get

$$\Delta^n \sin{ax}=(-1)^{(n-1)/2}2^n\sin^n\frac{a}{2}\cos\left(ax+\frac{na}{2}\right)$$ if $n$ is odd, and

$$\Delta^n \sin{ax}=(-1)^{n/2}2^n\sin^n\frac{a}{2}\sin\left(ax+\frac{na}{2}\right)$$

First question. Is this correct? Second question. Is there a way to write this as a single expression instead of by cases? The alternating sin and cos seemingly makes it impossible to write as a single expression.

Answer 1

Note that the complex function $\mathrm{e}^{\mathrm{i}ax}$ is an eigenvector of $\Delta$: $$\Delta\mathrm{e}^{\mathrm{i}ax}=(\mathrm{e}^{\mathrm{i}a}-1)\mathrm{e}^{\mathrm{i}ax}\text{.}$$ Consequently, $$\begin{split}\Delta^n\mathrm{e}^{\mathrm{i}ax}&=(\mathrm{e}^{\mathrm{i}a}-1)^n\mathrm{e}^{\mathrm{i}ax}\\ &=(2\sin\tfrac{\alpha}{2})^n\mathrm{e}^{\mathrm{i}(ax +n(\tfrac{\mathrm{\pi}}{2}+\tfrac{a}{2})) }\text{.}\end{split}$$ Taking real and imaginary parts gives $$\begin{align} \Delta^n\cos ax&=(2\sin\tfrac{\alpha}{2})^n\cos\left(ax+\tfrac{na}{2}+\tfrac{n\pi}{2}\right)\\ \Delta^n\sin ax&=(2\sin\tfrac{\alpha}{2})^n\sin\left(ax+\tfrac{na}{2}+\tfrac{n\pi}{2}\right)\text{.} \end{align}$$ The last line is the same as your result.