For $f=\mathbf1_{[-\pi,\pi)}$, its $2\pi$-periodization is one almost everywhere. The refinement mask is the periodic function equal to one on $[-\pi/2,\pi/2)$ and zero on the rest of $[-\pi,\pi)$. Fourier inversion gives the <Shannon scaling function>
$$
\boxed{\phi(x)=\frac{\sin\pi x}{\pi x}}.
$$
Since $m(t)=\tfrac12\sum_na_ne^{-int}$, its Fourier coefficients give
$$
\boxed{a_0=1,\qquad a_n=\frac{2\sin(n\pi/2)}{\pi n}\quad(n\ne0)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 5 / c / Solution

{c}

The Shannon scaling function is $\phi(x)=\sin(\pi x)/(\pi x)$ and has a rectangular Fourier transform.


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