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Re: Functional equation
Posted:
May 25, 2005 12:35 AM


From: Vincent Johns <vjohns@alumni.caltech.edu> > 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 > > > > But now I want to solve a more general problem. > > Given the above 3 properties, determine f(x).
> It's unique if it's continuous, but I'm not sure of the converse;
The converse is, it's continuous, if it's unique, which is obvious as 3x^2  2 is a solution.
> you may be able to use some of the following ideas in your proof.
To elaborate upon your ideas.
f(x + 1) = f(x) + 6x + 3. Use (4) 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. Use f(x  1) = f(x)  6(x  1)  3
> By induction, we can show that f(x) is unique for all values of x > that are integers, and for all that are integer multiples of powers > of 2. Also, if f(x) is defined for all real x such that (0 <= x < 2), > then it's unique for all real x (because of the f(x + 2) condition, > and using induction).
If it's defined over [0,1), it's determined over R.
> As ticbol showed, f(x) = 3*x^2  2 is a solution, > and this is defined for all real numbers x.
By calculus of finite differences from (4) only f(x) = (6/2)x(x  1) + 3x + c = 3x^2 + c and with f(1) = 1 c = 2 by f(1) = 1 showing f(x) over the integers is 3x^2  2
> If the function be continuous, then those (negative) powers > of 2 would be enough to make this a unique solution.
Using only f(x + 1) = f(x) + 6x + 3, f(0) = 1 (5) one then has when x is an integer f(x) = 3x^2  2 f(nx) = 3n^2 x^2  2 = n^2 (3x  2) + 2n^2  2 = n^2 f(x) + 2n^2  2 As f is assumed defined over R, for x in Z f(x) = f(n * x/n) = n^2 f(x/n) + 2n^2  2 f(x/n) = f(x)/n^2  2 + 2/n^2 = (3x^2  2)/n^2  2 + 2/n^2 = 3(x/n)^2  2 which determines f over the rationals.
> I'm not sure if there are any discontinuous functions that might > satisfy all the given conditions. I thought I'd found some, but > what I'd found didn't work. Even though the domain includes a lot > of numbers, it omits some (such as x = pi/10) for which we can't > derive a value from the conditions. So there may be a way to > construct such a function that does not equal (3*x^2  2) for some > values of x, but I wasn't able to do that.
As above for x in Q, n in N f(x^n) = x^(2n2) f(x) + 2^(2n1)  2 f(x^(1/n)) = f(x)/x^(2n2)  2 + 2/x^(2n2)
f(x) = 3x^2  2; f(1/x) = 3/x^2  2 x^2 = (f(x) + 2)/3; f(1/x) = 3*3/(f(x) + 2)  2
Thus (5) or (1)(3) determine f for numbers constructed of roots.
One specualates if they determine f for all algebraic numbers. Yet without continuity, how could they determine f for transcendal numbers? I think not.
Thus the equations seem to leave unsettled f(pi) and others. So set f(pi) = r /= 3pi^2  2 and you've a discontinuous f sporting (1)  (3), have you not?
To define f over all of R in a similar manner, let B be a Hamos base for the reals over the rationals as a vector space with rational scalars, and for each x in B, set f(x) = any real.




