2

Let $f(a)=\frac{a}{(1-a)} \frac{2^\frac{k}{a}-1}{2^\frac{K}{1-a}-1}$, where $a \in I:=(0,1)$. Individuallly, $\frac{a}{(1-a)}$ is an increasing function $\forall a \in I$ because its first derivative $\frac{1}{(1-a)^2}$ is positive. But $\frac{2^\frac{k}{a}-1}{2^\frac{K}{1-a}-1}$ is a decreasing function $\forall a \in I$ and $k>0$. The first derivative of $\frac{2^\frac{k}{a}-1}{2^\frac{k}{1-a}-1}$ is negative for $k>0$. For $k>0$, simulation results of the multiplication function $f(a)$ show that it is a decreasing function. How to prove mathematically $f(a)$ is a decreasing function for $k>0$?

Sam
  • 364
abc
  • 39
  • 1
    Hint: your function is of the form $\frac{f(a)}{f(1-a)}$, where $f(a)$ is decreasing. –  May 04 '20 at 06:41
  • @YvesDaoust If we consider $k(a)=\frac{f(a)}{f(1-a)}$ then for decreasing function $k'(a)$ should be less than zero. How to prove $k'(a)<0$? – abc May 04 '20 at 19:44
  • write the derivative of the quotient. –  May 04 '20 at 19:47
  • @YvesDaoust $k'(a)=\frac{f'(a)f(1-a)+f'(1-a)f(a)}{f(1-a)^2}$[Differentiating $f(1-a)$ w.r.t. $a$ gives $-f'(1-a)$]. As the term $f'(a)<0$, the term $f'(a)f(1-a)<0$. But $f'(1-a)>0$. Then how to show $k'(a)<0$? – abc May 05 '20 at 13:49
  • Double check your formula. –  May 05 '20 at 13:50
  • @YvesDaoust Here the variable is $a$. So derivative of $f(1-a)$ w.r.t. $a$ should be $-f'(1-a)$. Please correct me if I am wrong. – abc May 05 '20 at 14:05
  • How is $f'(1-a)>0$ when $f'(a)<0$ ? –  May 05 '20 at 14:24
  • @YvesDaoust you have mentioned in the solution that $f(1-a)$ is an increasing function. So, $f'(1-a)>0$. – abc May 05 '20 at 14:57
  • This is wrong. If $f'(a)<0$, then $f'(\text{anything})<0$. $\dfrac{d}{da}f(1-a)$ is not $f'(1-a)$. –  May 05 '20 at 15:10
  • @YvesDaoust Sorry I had done one mistake $\frac{d}{da}f(1-a)$ should be $-f'(1-a)$. If the function is $\frac{a}{1-a}\frac{2^{\frac{k}{a}}-1}{2^{\frac{k_1}{1-a}}-1}$ then also this solution will work . Am I right? – abc May 06 '20 at 15:10

1 Answers1

2

For $t>0$, we have

$$\frac{e^t-1}t=1+\frac t2+\frac{t^2}{3!}+\cdots$$ and this is a growing function.

Hence

$$f(a)=a(2^{K/a}-1)$$ is decreasing for $a\in(0,\infty)$, for any $K>0$. (Take $t=\dfrac{K\log2}a$.)

Then $f(1-a)$ is increasing and as both functions are positive, the ratio

$$\frac{f(a)}{f(1-a)}$$ is decreasing.