Weierstrass theorems and primes.

[tpm]

Does 'Weirstrass theorem' allow the existence of an entire function so:

$$f(z)= g(z) \prod _p(1- \frac{x}{p^{k}})$$

so for every prime p then f(p)=0 , and k>1 and integer??

the main question is to see if a function can have all the primes as its real roots

 

[matt grime]

Of course a function can have the primes, and only the primes, as its only roots, although you clearly want the function to be analytic, don't you, eljose?

 

[tacman]

The Weierstrass theorems are a set of theorems in complex analysis that deal with the properties of entire functions. An entire function is a function that is analytic (i.e. has a power series expansion) in the whole complex plane. The theorems state that any entire function can be represented as a product of its zeros, and that the zeros can be ordered in a way that the product converges uniformly on compact sets.

Based on this, it is possible for an entire function to have all primes as its real roots. This is because the Weierstrass theorem allows for the existence of an entire function that can be represented as a product of its zeros. In the case of the given function f(z), it is possible for the primes to be the zeros of the function, and for the function to have the desired form.

However, it is important to note that the Weierstrass theorem does not specifically state that all primes must be real roots of the function. It only guarantees the existence of an entire function that can be represented in such a form. Therefore, it is not a guarantee that every entire function will have all primes as its real roots.

In conclusion, the Weierstrass theorems do allow for the existence of an entire function that can have all primes as its real roots. However, this is not a guarantee for all entire functions and it is important to carefully consider the specific properties and conditions of the function in question.