Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to construct an entire function which vanishes at points $n+in$ for all $n$ integers. I'm looking for the most simple entire function which satisfies this condition, in the sense that the integers $p_n$ in weierstrass elementary factor should be as small as possible.

If i enumerate non-zero points of the form $n+in$ as $\{z_n\}$ i get $$f(z)=z\prod_{n=1}^{+\infty}E(\frac{z}{z_n},p_n)$$

which is entire with zeros where i want, for suitable integers $p_n$.

But i don't know hot to determine $p_n$'s. Moreover i don't know which is the right way to enumerate $n+in$ in this case. I need integers $p_n$ such that $$\sum_{n=1}^{+\infty}\frac{1}{|z_n|^{p_n+1}}<+\infty$$ How can i find them?

share|improve this question
6  
$$f(z) = \sin \left(\dfrac{\pi}{1+i}z \right)$$ –  user17762 Jan 9 '13 at 22:12

2 Answers 2

If you indeed want $\sum_{n}|z_n|^{-p_n-1}<\infty$, then enumeration makes no difference. Just set $z_n=n+in$. Now you see that $p_n\equiv 0$ does not work, while $p_n\equiv 1$ does. What is the "minimal" sequence of $p_n$? Well, there isn't one. Whatever sequence you come up with can be made "smaller" by replacing one of the nonzero numbers $p_n$ with a $0$.

However, it is not necessary to have $\sum_{n}|z_n|^{-p_n-1}<\infty$ in order for the product to converge. This is what the example of Marvis and Jonas demonstrates: in it $p_n\equiv 0$, yet the product converges conditionally with the enumeration $1,-1,2,-2,3,-3\dots $ (times $1+i$).

Moral: the Weierstrass theorem does not exhaust all possibilities for constructing entire functions with prescribed zeros.

share|improve this answer
    
ok,thank you very much, but suppose i want a sequence p_n such that we have convergence of weierstrass product, and p_n equal to a unique integer p. Is there a minimal p such that i can find such an entire function using weierstrass factorization? –  bateman Jan 9 '13 at 22:31
1  
@bateman Then $p_n\equiv 1$, of course. Because $1/n^2$ converges while $\sum 1/n$ does not. –  user53153 Jan 9 '13 at 22:53
    
tell me if you think this is right: using W factorization, our function is $f(z)=z\prod_{z_n} E(\frac{z}{z_n},1)=z(\prod_{n=1}^{\infty}(1-\frac{z}{n+in})e^{\frac{z}{n+in}}\pr‌​od_{n=1}^{\infty}(1+\frac{z}{n+in})e^{-\frac{z}{n+in}})=z\prod (1-\frac{z^2}{(1+i)^2 n^2})$ Now, from product factorization of sine (see for example Lang p. 379) i get that this is a constant times $sin(\frac{\pi z}{1+i})$ which is Jarvis example –  bateman Jan 10 '13 at 8:46

i tried to post a comment but there were problems, do you think the following is correct?

UUsing Weierstrass factorization, our function is $$f(z)=z\prod_{z_n} E(\frac{z}{z_n},1)=z\prod_{n=1}^{\infty}(1-\frac{z}{n+in})e^{\frac{z}{n+in}}\prod_{n=1}^{\infty}(1+\frac{z}{n+in})e^{-\frac{z}{n+in}}=z\prod (1-\frac{z^2}{(1+i)^2 n^2})$$Now, from product factorization of sine (see for example Lang p. 379) i get that this is a constant times $sin(\frac{\pi z}{1+i})$ which is Jarvis example

share|improve this answer
    
Yes, that works. –  user53153 Jan 10 '13 at 13:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.