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Re: Functional equation
Posted:
May 24, 2005 7:23 AM


Newsgrps: alt.algebra.help Attchmnt: Subject : Re: Functional equation
On Mon, 23 May 2005, Will wrote:
> A function f(x) has the following properties: > 1) f(1) = 1 > 2) f(2x) = 4f(x) + 6 > 3) f(x+2) = f(x) + 12x + 12 > f(2x + 2) = 4f(x + 1) + 6 f(2x + 2) = f(2x) + 12*2x + 12 = 4f(x) + 24x + 18 f(x + 1) = f(x) + 6x + 3
f(x + 1)  f(x) = 6x + 3 f(x) = 3x(x  1) + 3x + c = 3x^2 + c 1 = 3 + c; c = 2
f(x) = 3x^2  2; f(1) = 1 f(2x) = 12x^2  2 = 4(3x^2  2) + 6 f(x + 2) = 3x^2 + 12x + 12  2 = f(x) + 12x + 12
> Given the above 3 properties, determine f(x). > I found a function f(x)=3(x^2)2 that
Yuck,equationwithoutspaces.
> satisfies the above 3 properties, but how can I show whether or not > this solution is unique. > It is, if it's continuous.
f(x/2) = (f(x)  6)/4 f(x/2^n) = (f(x)  2(4^n  1))/4^n f(x/2^(n+1)) = (f(x/2^n)  6)/4 = ((f(x)  2(4^n  1))/4^n  6)/4 = ((f(x)  2(4^n  1)  6*4^n)/4^(n+1) = ((f(x)  2(4^(n+1)  1)/4^(n+1) f(x/2^n) = (3x^2  2  2(4^n  1))/4^n = 3x^2 / 4^n  2




