# Jee functions help

If $f(x+2)-5f(x+1)+6f(x)=0$

domain=$$R$$

Find the value of $$f(0),f(1),f(2),f(3)$$

Note by Aakash Mandal
2 years, 11 months ago

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## Comments

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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.

- 2 years, 11 months ago

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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?

Staff - 2 years, 11 months ago

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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.

- 2 years, 11 months ago

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Are those the only solutions?

Staff - 2 years, 11 months ago

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