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Topic: Functional equation
Replies: 1   Last Post: May 25, 2005 12:35 AM

 William Elliot Posts: 4,698 Registered: 3/18/05
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^(2n-2) f(x) + 2^(2n-1) - 2
f(x^(1/n)) = f(x)/x^(2n-2) - 2 + 2/x^(2n-2)

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.

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