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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0It is a very good idea to put everything in a closed form but I am not able to see easily that $f(a)$ is increasing after all. What do you mean by dependence on the inverse of $a$? – 2012-04-27
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0@SeyhmusGüngören: The idea is quite simple here. If you have a number $01$. Besides, if $K>k^*$, $\frac{1}{a^K}>\frac{1}{a^{k^*}}$. In this case, the hypergeometric function does help and your function, with the given intervals, is seen to increase. Also note that, in your case, $a|r|< a$ and so the numerator gives a larger number than the denominator. – 2012-04-27
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0yes but there is a difference term inbetween and the other gamma and hypergeometric terms are also confusing as what happens finally. I am also surprised that it is simply correct for you while I am still trying to understand what is going on))) – 2012-04-27
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0@SeyhmusGüngören: I have improved the answer. I hope this will help you. – 2012-04-27
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0Could you fill in some of the details of your answer? I don't think "hypergeometric functions are also helpful in this direction" could be more vague. – 2012-04-27
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0I accept that nominator is always greater than the denominator and nominator and the denominator have the same signs. These two information do not suffice to deduce that $f(a)$ is increasing with $a$. I can also have the same information from the original description. If the derivative could be shown to be greater than zero than I would say okay. In its current version I cannot understand any improvement. – 2012-04-27
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0@AntonioVargas: I will provide more thorough information, thanks for your comment. For Seyhmus: I think you are right and I am in need for a better clarification. Of course, having a formula in a closed form is far better than the ratio of sums! – 2012-04-28
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0My friend is interested in how you found the very first formula with $\Gamma$s thanks alot. – 2012-04-30
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0@SeyhmusGüngören: Indeed, your sums can be given in a closed form. This can be obtained using Wolfram Alpha or Mathematica. The fact that appears Gamma products is just due to the fact that one does not assume $k^*$ and $K$ integers otherwise these are just $\binom{K}{k^*}$. It is interesting to note that the reason why a hypergeometric function appears is that your sums do not run to infinity otherwise one can us the Newton binomial formula and end the story. – 2012-04-30
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0thank you very much for the explanation! – 2012-04-30