Large powers of sine appear Gaussian -- why?

As part of approximating an integral, I have noticed that $\sin^k(x), x\in[0, \pi]$ look almost identical to $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ once $k$ is large enough (in practice, the two equations are visually identical for $k\geq 15$). As an example, see this picture of the two functions for $k=10$:

Can anybody explain why these functions practically are identical?

Edit: I should mention that the expression $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ is derived by a 2nd order Taylor expansion of $\log\sin^k(x)$ around the mode $x_0 = \frac{\pi}{2}$, i.e. a Laplace approximation.

Solution 1:

It's easier to shift $x$ by $\pi/2$ and establish the property for the cosine.

From Taylor and the limit definition of the exponential, for small $x$

$$\cos^kx\approx\left(1-\frac{x^2}2\right)^k=\left(1-\frac{kx^2}{2k}\right)^k\approx e^{-kx^2/2}.$$

The plot shows good convergence of $(1-t/k)^k$ to $e^{-t}$.

Solution 2:

This is too long for a comment.

Semiclassical made a very good comment.

Let us use Taylor series around $x=\frac \pi 2$ $$\sin(x)=1-\frac{1}{2} \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ $$\sin^k(x)=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} k (3 k-2) \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ On the other hand $$e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{8} k^2 \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ Computing the difference $$\sin^k(x)-e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=-\frac{1}{12} k \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ AT least, they are very close in the area around the maximum.