What Is the Value of f(12) Given These Function Conditions?

[e^(i Pi)+1=0]

Homework Statement

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)

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

 

[Mandelbroth]

Homework Statement

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)

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

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

 

[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.