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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) ?

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  • $\begingroup$ Sorry, I am not familiar with the notion of special function. Maybe an aspect of interest: A multiple zero of $f$ is also a common zero of $f$ and $f'$. $\endgroup$ – Hagen von Eitzen Nov 17 '12 at 21:52
  • $\begingroup$ Special function is a mathematical term and is well known. For starters it has intresting properties and is not an elementary function. Good examples are Euler's Gamma function and Riemann's Zeta function. Other examples are Bessel functions and elliptic integrals. Note however that I required meromorphic on the complex plane. $\endgroup$ – mick Nov 17 '12 at 21:58
  • $\begingroup$ I think your heuristic argument about probability is actually pretty good. A "random" meromorphic function does not likely have multiple zeros, so in order to have multiple zeros, we have to design the function especially for that purpose. This means that a function designed independently for another purpose (which all the "special" functions are) is not likely to have multiple zeroes. $\endgroup$ – Ted Nov 17 '12 at 22:01
  • $\begingroup$ About your comment of f and f' that was known to me. But It seems that leads to another statistical argument. However it intresting for solving polynomials ... maybe I will post a question related to that later ( Im going off subject here so I quit ) $\endgroup$ – mick Nov 17 '12 at 22:03
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    $\begingroup$ @mick: Ah, I found something on Wikipedia, but there "is no general formal definition" and the informal one contains the phrase "more or less", so I personally won't call it a mathematical term. Thus I think it just means "function that deserves special interest because it has nice properties that are useful in various mathematical contexts". But precisely that makes the random argument less appealing: The zeroes are not randomly distributed. For $\sin$ they are periodic and for $\zeta$, well there's RH. $\endgroup$ – Hagen von Eitzen Nov 17 '12 at 22:09

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