This question is from Ponnusamy and Silvermann Complex Variables pg 436, subsection wietestrauss product theorem
Question: Construct an entire function whose only zeroes are z=ln n.
Weierstrauss Product theorem guarentees that such a function exists, I thought of sin ($\pi z$) but it's zeroes will also be -1,-2,... . Similar problem is also for cos z and tan z.
Also, there are similar questions in same exercise wich ask for zeroes only at z=n of multiplicity n , zeroes only at z=$n^{3/4}$ .
Is there any general approach to these type of questions?
If yes kindly share. If no just answer 1 series asked.
Thanks!
Edit:there are similar questions in same exercise wich ask for zeroes only at z=n of multiplicity n.
There is an example before the exercise in which It is proved that an entire function woth zeroes only at z=+ve integers is $\prod_{n=1}^{\infty} (1-z/n) e^{z/n}$ and I adapted that proof to see if $\prod_{n=1}^{\infty} (1-z/n)^n e^{z/n}$ will be the polynomial asked in this but the mentioned product is not uniformly convergent product. So, I need an convergence producing factor which was $e^{z/n}$ in case of roots only at z=+ve integers. Can you please tell me how can I find such a convergence producing factor.
Also, I have solved the entire function with zeroes only at z=$n^{3/4}$ but not the above mentioned part in which I am unable to find the convergence producing factor.
Also ,this method used for entire function whose zeroes are only +ve integers can't be applied on ln (n) as the series $1\(ln n)^p $ diverges for all p . So, No progress can be made on this part.
Kindly help.
Since you changed your question.. Take $k_n$ such that $$\sup_{|z|\le \frac12 \log n} |\log(1-z/\log n)+\sum_{m=1}^{k_n} \frac{(z/\log n)^m}{m}|\le 2^{-n}$$ (can you remind us why it exists ?)
Then $$\prod_{n\ge 2} (1-z/\log n) \exp(\sum_{m=1}^{k_n} \frac{(z/\log n)^m}{m})$$ is entire.
(can you remind us why ?)