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Please how to find order of $$ f_k(z) = \prod\limits_{n=1}^{\infty} \left(1-\frac{z}{n^k}\right) .$$

Let $M(r) = \max \{|f_k(z)|:|z| = r\}.$ Then order of $f_k(z)$ is defined as : $$\lambda = \limsup_{r\to \infty} \frac{\log \log M(r)}{\log r}.$$

Can someone help to solve this. Thank you.

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4  
What have you done? I claim it's obvious which $z$ maximizes $f_k(r)$ to get an explicit formula for $M(r)$. Taylor/Laurent series are often useful; did you get anywhere with those? –  Hurkyl Mar 17 '12 at 17:19
    
@Hurkyl: I fairly new to this...can u help me start. –  Yuri Mar 17 '12 at 17:47

1 Answer 1

The order of this function is $\frac{1}{k}$, and we must have $k>1$.

Proof: First, note that $\prod_{n=1}^{\infty}\left(1-\frac{z}{n^{k}}\right)$ will attain its maximum modulus on the circle of radius $r$ at $z=-r$. Taking the logarithm of this, we need to examine $$h(r)=\log\prod_{n=1}^{\infty}\left(1+\frac{r}{n^{k}}\right)=\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right).$$ If we can prove that this is bounded below and above by something close to $r^\frac{1}{k}$, then we will have shown that $\lambda=\frac{1}{k}$ since $$\lambda = \lim_{r\rightarrow \infty} \frac{\log h(r)}{\log r}.$$

Lower Bound: We have that $$\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\geq\sum_{n\leq r^{\frac{\epsilon}{k}}}\log\left(1+\frac{r}{n^{k}}\right)\geq r^{\frac{\epsilon}{k}}\log\left(1+r^{1-\epsilon}\right)$$ for any $0<\epsilon<1.$ Taking $\epsilon=1-\frac{1}{\log r},$ we see that $r^{1-\epsilon}=e,$ and hence the above is $$\geq e^{-\frac{1}{k}}r^{\frac{1}{k}}\log\left(1+e\right)\geq e^{-\frac{1}{k}}r^{\frac{1}{k}}.$$ This then implies that $\lambda\geq\frac{1}{k}.$

Upper Bound: We split the sum based on the size of $n$. If $n>r^{\frac{1}{k}},$ by using the bound $\log\left(1+\frac{r}{n^{k}}\right)\leq\frac{r}{n^{k}},$ we have $$\sum_{n>r^{\frac{1}{k}}}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\leq r\sum_{n>r^{\frac{1}{k}}}^{\infty}\frac{1}{n^{k}}\ll\frac{r}{r^{\frac{1}{k}k-1}}=r^{\frac{1}{k}}.$$ Next, for $n^{k}\leq r,$ we have that $\log\left(1+\frac{r}{n^{k}}\right)\leq\log\left(r+1\right),$ which implies that $$\sum_{n\leq r^{\frac{1}{k}}}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\leq r^{\frac{1}{k}}\log\left(r+1\right).$$ Thus $$\sum_{n=1}^{\infty}\log\left(1+\frac{r}{n^{k}}\right)\ll r^{\frac{1}{k}}\log r,$$ which implies that $\lambda\leq\frac{1}{k}.$

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Thank for the answer. What is $\ll$? –  Yuri Mar 17 '12 at 23:30
    
@Yuri: Writing $f(x)\ll g(x)$ is just another notation for $f(x)=O(g(x))$. See Big-O notation. This just means there exists a constant $M$ such that $|f(x)|\leq M|g(x)|$ for all large $x$. –  Eric Naslund Mar 18 '12 at 8:17

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