The large-distance profile is proportional to $[\sqrt{2\beta-1},x]^{-1}$. At $x=2$, the exact profile is proportional to $(4\beta-1)^{-1/2}$. Thus
$$
\boxed{\frac{\Sigma_{\rm actual}(2a_p)}
{\Sigma_{r^{-1}\ {
m extrapolation}}(2a_p)}
=\frac{2\sqrt{2\beta-1}}{\sqrt{4\beta-1}}
\longrightarrow\sqrt2}
$$
as $\beta\to\infty$. The enhancement is the finite-radius remnant of the source-ring pile-up.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 2 / vii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/2/vii/solution