How would I solve this? (real analysis I think)

  • #1
247
1

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



[itex]f(n)=\frac{f(2n)}{2f(2n+1)-2f(2n)}[/itex]

[itex]f(12)=\frac{f(24)}{2f(25)-2f(24)}\frac{}{}[/itex]

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

Because [itex]\frac{f(2n)}{2f(2n+1)-2f(2n)} > 0[/itex] and f(2n) > 0 we know that

f(2n+1) > f(2n) so [itex]f'(n) > 0[/itex]

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

Answers and Replies

  • #2
611
24

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



[itex]f(n)=\frac{f(2n)}{2f(2n+1)-2f(2n)}[/itex]

[itex]f(12)=\frac{f(24)}{2f(25)-2f(24)}\frac{}{}[/itex]

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

Because [itex]\frac{f(2n)}{2f(2n+1)-2f(2n)} > 0[/itex] and f(2n) > 0 we know that

f(2n+1) > f(2n) so [itex]f'(n) > 0[/itex]

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?
 
  • #3
haruspex
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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.
 

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