Why do most special functions not have a multiple zero ? Let $z$ be a complex number. Let $f(z)$ be a special function meromorphic in the entire complex plane. Then the zero's $x_i$ such that $f(x_i) = 0$ rarely have multiplicity above $1$.
It seems we need to take powers or explicitly multiply to define the function to have multiplicity above $1$.
If we express $f(z) = exp(m(z))(x-x_1)(x-x_2)...$ then by a probabilistic argument , the probability that - assuming there are an infinite amount of zero's - , $x_i = x_j$ is equal to the probability of guessing a real value between $0$ and $1$ correctly.
Although this gives an intuive argument , I feel this is not very good.
The Riemann zeta function serves as a good example. Even if we assume RH to hold it is unknown if any of the zero's have multiplicity , but it appears that all have multiplicity 1.
We can test multiplicity for a certain zero or a finite region with theorems of cauchy such as the argument principle and basicly with integrals. ( contour integrals and restatements and generalizations of it )
Testing multiplicity locally seems quite easy , but why is creating multiplicity in a nontrivial way ( trivial = multiply by $(x-1)^2$ , take $f(x)^2$ etc ) so hard ?
Is it really a matter of statistics or did I overlooked something. Im a bit confused about this.
Can we construct example that have multiple zero's ?
Maybe I should think of this more in terms of polynomials and use another way to detect and create multiplicity ... such as considering the derivates at zero's that make polynomial asymptotics and thereby show multiplicity ?
Maybe I just need more tools ?
For a meromorphic g(z) : Is there something about integrals or integral transforms involving g(z) that decreases the probability at multiple zero's compared to that of g(z) ?