# Homework Help: How would I solve this? (real analysis I think)

1. Jul 16, 2013

### e^(i Pi)+1=0

1. The problem statement, all variables and given/known data

f is a function defined on natural numbers such that

2f(n) * f(2n+1) = f(2n) *[2f(n) + 1] and

8f(n) > f(2n) > 4f(n)

Find value of f(12)

3. The attempt at a solution

$f(n)=\frac{f(2n)}{2f(2n+1)-2f(2n)}$

$f(12)=\frac{f(24)}{2f(25)-2f(24)}\frac{}{}$

2f(n) > f(n) so f(n) > 0 and f(2n) > 0

Because $\frac{f(2n)}{2f(2n+1)-2f(2n)} > 0$ and f(2n) > 0 we know that

f(2n+1) > f(2n) so $f'(n) > 0$

Other than that, I have no idea. Math level: ODEs

2. Jul 16, 2013

### Mandelbroth

Do you mean that the function f is given by $f:\mathbb{N}\rightarrow\mathbb{R}$? If so, why are we using derivatives?

3. Jul 16, 2013

### haruspex

The problem is underspecified. I have a family of solutions with two partly arbitrary constants. Requiring f:N→N fixes one of them; knowing f(1) fixes the other. Whether there are other solutions, I don't know.