My initial thought was that this functional equation looks like a familiar quadratic equation. So it seems reasonable to find the values of x for which a quadratic equation is formed. i.e. Let \(\ f(x+1) = x f(x) \ , \ f(x+2) = x^2 f(x): \) \[\ f(x) (x^2 -5x+6) = f(x)(x-2)(x-3) = 0 \] From this equation it is clear that x=2 and x=3 satisfies the conditions. \[\therefore\ f(4) = 4 f(2)......(1) \ and \ f(3) = 2f(2)....(2) \] Now lets sum together the following 3 equation: \[\ f(2) -5f(1) +6 f(0) = 0 \] \[\ f(3) -5f(2) +6 f(1) = 0 \] \[\ f(4) -5f(3) +6 f(2) = 0 \] This gives \[\ 6f(0) +f(1) -4f(3) +f(4) = 0 \] Using equation (1) and (2): \[\ 6f(0) +f(1) = 4f(2) \] subtracting this equation from \(\ 6f(0) -5f(1) = -f(2) \) gives: \[\ 6f(1) =5 f(2) \] From looking at an earlier equation we have: \[\ f(3) = 5f(2) - 6f(1) = 0 \implies\ f(2) = 0 \implies\ f(4) = 0 \] \[\implies f(0) = 0 \] In fact we can use induction to show that \(\ f(x) = 0 \) for all non-negative integers.

I had a feeling that I may have found a unique solution, so thanks for pointing that out. I think I found the error, so here's my (attempt at a) correction... \[\ f(x)(x-2)(x-3) = 0 \Rightarrow\ f(x+1) = 2 . f(x) = 2^2 .f(x-1) \] \[\ =...= 2^x f(1) = C 2^x \] Similarily: \[\ f(x) = K. 3^x \] is also a solution. Furthermore, we could put \(\ f(x) = K a^x \) to show that only a = 2,3 produce solutions.

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TopNewestMy initial thought was that this functional equation looks like a familiar quadratic equation. So it seems reasonable to find the values of x for which a quadratic equation is formed. i.e. Let \(\ f(x+1) = x f(x) \ , \ f(x+2) = x^2 f(x): \) \[\ f(x) (x^2 -5x+6) = f(x)(x-2)(x-3) = 0 \] From this equation it is clear that x=2 and x=3 satisfies the conditions. \[\therefore\ f(4) = 4 f(2)......(1) \ and \ f(3) = 2f(2)....(2) \] Now lets sum together the following 3 equation: \[\ f(2) -5f(1) +6 f(0) = 0 \] \[\ f(3) -5f(2) +6 f(1) = 0 \] \[\ f(4) -5f(3) +6 f(2) = 0 \] This gives \[\ 6f(0) +f(1) -4f(3) +f(4) = 0 \] Using equation (1) and (2): \[\ 6f(0) +f(1) = 4f(2) \] subtracting this equation from \(\ 6f(0) -5f(1) = -f(2) \) gives: \[\ 6f(1) =5 f(2) \] From looking at an earlier equation we have: \[\ f(3) = 5f(2) - 6f(1) = 0 \implies\ f(2) = 0 \implies\ f(4) = 0 \] \[\implies f(0) = 0 \] In fact we can use induction to show that \(\ f(x) = 0 \) for all non-negative integers.

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Observe that \(f(x) = C2^x \) is a solution to the functional equation.

What is the error in your solution, in which you only found \(C=0\)?

Note: How can we classify the rest of the solutions?

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I had a feeling that I may have found a unique solution, so thanks for pointing that out. I think I found the error, so here's my (attempt at a) correction... \[\ f(x)(x-2)(x-3) = 0 \Rightarrow\ f(x+1) = 2 . f(x) = 2^2 .f(x-1) \] \[\ =...= 2^x f(1) = C 2^x \] Similarily: \[\ f(x) = K. 3^x \] is also a solution. Furthermore, we could put \(\ f(x) = K a^x \) to show that only a = 2,3 produce solutions.

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