# Root of a special rational function with positive coefficients

During my research I came across the following problem:

I need to find a root of the following function: $$\Gamma_{N}(x) = \sum\limits_{i=0}^{M}\left(\frac{\sum\limits_{n=0}^{n_F}n\ \alpha_{i,n} x^n}{\sum\limits_{n=0}^{n_F}\alpha_{i,n} x^n}\right)- N,$$ where $\alpha_{i,n}$ are positive coefficients such that in general, $\alpha_{i,n} \in [0,1]$. Parameter $M$ and $n_F$ are integer numbers and $N$ is real. I want to consider only positive $x$, so I am searching for root in the domain $x \in [0, +\infty]$. I will do all calculations numerically, but I was wondering if this function possesses some properties?

Question: Is this function monotonic, such that there exists only a single root?

• Please check your formula. Is the summation index $i$ or $n$? – Alexandre Eremenko Apr 1 '16 at 13:42
• Sorry, I missed extra index in $\alpha$. Coefficients $\alpha_{i,n}$ depend on both $n$ and $i$. – WoofDoggy Apr 1 '16 at 14:27
• It would be good if you try first with a bunch of sets of random numbers to see if there is some numerical evidence. If it is monotonic, M=0 is enough. – Wolfgang Apr 1 '16 at 16:06

it suffices to prove that if $a_k\geqslant 0$ for all $k$ and $x>y>0$, then $$\frac{\sum na_nx^n}{\sum a_n x^n}\geqslant \frac{\sum na_ny^n}{\sum a_n y^n}.$$ Multiplying by common denominator this reduces to $$\sum_{n,k}a_na_k(n-k)(x^ny^k-x^ky^n)\geqslant 0,$$ each summand is non-negative.
Yes, it is monotonic for positive $x$. $N$ is irrelevant for monotonicity. Set $$P_i(x)=\sum_n\alpha_{i,n}x^n, \quad P=\prod_iP_i.$$ By assumption, these are polynomials with positive coefficients. For such polynomials $xP'(x)/P(x)$ is increasing. This is a simple consequence of the Hadamard Three circles Theorem. One only has to notice that $$M(r):=\max_{|z|\leq r}|P(z)|=P(r)$$ for polynomials with positive coefficients. Hadamard's theorem says that for every entire function $rM'(r)/M(r)$ is increasing; strictly increasing unless the function is a monomial. Now it remains to notice that $$\Gamma(x)=xP'(x)/P(x)=x\sum_i(\log P(x))'.$$