Searching for an explicit functional form Let $f:\mathbb N \to \mathbb R$ be a strictly decreasing function. Suppose $$\frac{f(x)}{f(x+1)}=2^x,\qquad \forall x\in\mathbb N.$$ Is it possible to find an explicit functional form of $f$? Any hint at finding a solution is appreciated.
 A: Hint: We compute a little to find out what's going on. 
Let us rewrite the equation as $f(n+1)=\frac{f(n)}{2^{n}}$. Let $f(1)=a$. 
We have $f(2)=\frac{a}{2^1}$, and therefore $f(3)=\frac{a}{2^{1}\cdot 2^2}$, and therefore $f(4)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3}$, and therefore $f(5)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3\cdot 2^4}$, and so on.
To get simpler-looking expressions, note that for example $2^1\cdot 2^2\cdot 2^3\cdot 2^4=2^{1+2+3+4}$, and use the formula for the sum of the first $k$ positive integers. 
A: From the condition we get $\ln f(x)-\ln f(x+1)=x\ln 2$. Note that $g(x)=x^2$ leads to $g(x+1)-g(x)=2x+1$; a correction and generalization to $g(x)=c(x-\frac12)^2+b$ leads to $g(x+1)-g(x)=2cx$, suggesting $$f(x)=\exp\left(-\frac12\ln 2\cdot (x-\frac12)^2+b\right)=\frac1{4^{x^2-x+1/4}}\cdot e^b$$
as possible solution. Note that $e^b>0$ but we can (summarize a $\frac1{4^{1/4}}$ into the constant and) instead take 
$$ f(x)=\frac a{4^{x(x-1)}}. $$
If $a$ is negative, this $f$ is strictly increasing for $x\in\mathbb N$.
