Jee functions help

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

domain= $R$

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

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Comments

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.

Curtis Clement - 5 years, 10 months ago

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?

Calvin Lin Staff - 5 years, 10 months ago

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.

Curtis Clement - 5 years, 10 months ago

@Curtis Clement – Are those the only solutions?

Calvin Lin Staff - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$