Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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<r<0,\hspace{3mm} 0.5<a<1,\hspace{3mm} \mbox{and}\hspace{3mm} K>k^{*}? \end{equation}

share|improve this question

1 Answer 1

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 $$ f(a)=\frac{\left(-\frac{1}{ar}\right)^K-\left(-1-\frac{1}{ar}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})}{\left(\frac{1}{a}\right)^K-\left(-1+\frac{1}{a}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})}. $$

Now, one has $r<0$ and $|r|a<a$ always. This implies that numerator and denominator have always the same signs. But there is an important difference that is $\frac{1}{a|r|}>\frac{1}{a}$. Hypergeometric functions are also helpful in this direction. This can be easily seen with some numerical check (I did it with Mathematica) and one see that, in the given intervals, it is always $\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})<\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})$. I think, but I have not done it, that checking the properties of this function one can avoid a numerical check.

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

enter image description here

share|improve this answer
    
It 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 $a$? –  Seyhmus Güngören Apr 27 '12 at 10:48
2  
+1 for most terrifying formula I've seen today. –  MathematicalOrchid Apr 27 '12 at 11:22
    
@SeyhmusGüngören: The idea is quite simple here. If you have a number $0<a<1$ then you will surely get $\frac{1}{a}>1$. 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. –  Jon Apr 27 '12 at 11:55
    
yes 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))) –  Seyhmus Güngören Apr 27 '12 at 12:08
    
@SeyhmusGüngören: I have improved the answer. I hope this will help you. –  Jon Apr 27 '12 at 12:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.