How can I show that \begin{equation} f(a)=\frac{\sum_{i=1}^{k^*-1} \left(\begin{array}{c} K \\ i \\ \end{array} \right) \left(-1-\frac{1}{ar}\right)^i+1}{\sum_{i=1}^{k^*-1} \left(\begin{array}{c} K \\ i \\ \end{array} \right) \left(-1+\frac{1}{a}\right)^i+1} \end{equation} is an increasing funtion of $a$ for \begin{equation} -1
Does division of polynomials give an increasing function?
1 Answers
This function can be cast in a closed form as $ f(a)=\frac{\left(-\frac{1}{ar}\right)^K-\left(-1-\frac{1}{ar}\right)^{k^*}\frac{\Gamma(K+1)\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})}{\Gamma(k^*+1)\Gamma(K-k^*+1)}}{\left(\frac{1}{a}\right)^K-\left(-1+\frac{1}{a}\right)^{k^*}\frac{\Gamma(K+1)\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})}{\Gamma(k^*+1)\Gamma(K-k^*+1)}}. $ The function $_2F_1$ is the hypergeometric function. Now, we note that $a<1$ and $|r|<1$ and so, for $K>k^*$ and the dependence on the inverse of $a$ make this an increasing function in the given intervals.
We recognize a simpler rewriting of this formula as $ f(a)=\frac{\left(-\frac{1}{ar}\right)^K-\left(-1-\frac{1}{ar}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})}{\left(\frac{1}{a}\right)^K-\left(-1+\frac{1}{a}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})}. $
Looking at this formula at increasing $a$, we note at the numerator $a$ appears always multiplied by a reducing factor $r$ that maintains it always smaller than the quantity appearing at the denominator and so, increasing it with the given interval and properly multiplying it for increasing but negative $r$, this function can only increase. For the sake of completeness, I give here a graph of this for $k^*=100$ and $K=150$. For $r=-0.3$ you will get
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0thank you very much for the explanation! – 2012-04-30