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Topic: Functional equation
Replies: 1   Last Post: May 24, 2005 7:23 AM

 William Elliot Posts: 4,698 Registered: 3/18/05
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

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