Proving Recurrence Relation for f(x)

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SUMMARY

The discussion centers on proving the recurrence relation for the function f(x) defined by the equation log(f(x+2)) = log(f(x+1)) + log(f(x)). This leads to the simplified form f(x+2) = f(x+1)f(x). The goal is to demonstrate that f(x) can be expressed as f(x) = e^{c_1 L_x + c_2 F_x}, where L_x and F_x are specific sequences. Participants emphasize the need to establish the uniqueness of the solution derived from the recurrence relation.

PREREQUISITES
  • Understanding of logarithmic properties and their applications in recurrence relations
  • Familiarity with exponential functions and their representations
  • Basic knowledge of sequences and series in mathematics
  • Experience with mathematical proofs and uniqueness arguments
NEXT STEPS
  • Study the properties of logarithms in depth to facilitate manipulation of recurrence relations
  • Explore the concept of unique solutions in the context of functional equations
  • Investigate the application of exponential functions in solving recurrence relations
  • Learn about generating functions as a method for solving recurrence relations
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Mathematics students, researchers in combinatorics, and anyone interested in solving and proving properties of recurrence relations.

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Homework Statement


Let's say I had this recurrence relation:

[tex]log\left(f\left(x+2\right)\right) = log\left(f\left(x+1\right)\right) + log\left(f\left(x\right)\right)[/tex]

How do I prove, then, that...

[tex]f\left(x\right) = e^{c_1 L_x + c_2 F_x}[/tex]

?

Homework Equations



There probably are some, but I don't know any.

The Attempt at a Solution



I've gotten the equation to remove the logs, but I just get...

[tex]f\left(x+2\right) = f\left(x+1\right)f\left(x\right)[/tex]

I don't know where to go from there.
 
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First, use the properties of the logarithm to get rid of the logarithm:
[tex]log(f(x+ 2))= log(f(x+1))+ log(f(x))= log(f(x+1)f(x))[/tex]
and, since log is one-to-one, f(x+2)= f(x+1)f(x).

It's certainly true that the formula you gives satisfies that. Can you prove the solution is unique?
 

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