Background
My understanding is the input neurons seem to seem to compute a weighted sum moving from one layer to another.
But to compute this weighted sum the sum must be discrete. Is there any known method to compute the sum when the activation is a continuous function? Is the below formula of any consequence problems in artificial intelligence? Can anyone give a specific problem where it might be useful?
My Method
Let $b_r = \sum_{d \mid r} a_d\mu(\frac{m}{d})$. We prove that if the $b_r$'s are small enough, the result is true (where $\mu$ is the mobius function).
Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s}\right)\int_0^\infty f(x)dx.$$
I will not go into the proof of this over but for those who are interested: https://math.stackexchange.com/questions/2888976/a-rough-proof-for-infinitesimals I will merely state what the formula means:
Consider we have a curve $f(x)$ now if one wishes to perform a weighted sum in the limiting case of this function.
Consider the curve $f(x)$. Then splitting it to $k/n= h$ intervals then adding the first strip ($d_1$ times): $ f(h) \cdot d_1$. Then the second strip ($d_2$ times) $ f(2h) \cdot d_2$ times ... And so on . Hence. $d_r$ can be thought of as the weight at $f(rh)$.