A

functional

equation is like an ordinary equation, but instead of trying to solve it in terms of

reals (or

complexes, or

vectors etc) you try to find all functions from one

set to another that satisfy the given conditions.
This tends to be rather difficult.

The most typical functional equation is probably where you are trying to solve for some function f:

**R** →

**R** (i.e. a real function).

The most basic thing you can do with a functional equation is to substitute whatever you want for the free variables. If you are lucky this will give you the solution straight away.

*Example*: f(x) = x*f(-x) + 1.

Substituing x = -x gives f(-x) = -x*f(-x) + 1. Thus we have two ordinary equations in two unknowns (f(x) and f(-x)), and we can find the solution f(x) = (1+x)/(1+x^{2}).

Sometimes inductive arguments can be used to find f(x) for rational x. Then we can use that if f(x) = g(x) for all rational x, and f, g are both one of monotonic, continuous or bounded on every bounded interval, then f = g.

*Example*: f(x + y) = f(x) + f(y), f continuous.

Let k = f(1). By induction we find that for all integers n, f(nx) = n*f(x). Hence for integer q, f(q*1/q) = q*f(1/q) ⇒ f(1/q) = k/q. Hence for any integers p, q, f(p/q) = kp/q. Since f(x) = kx for all rational x and f is continuous we must have f(x) = kx for all x.

Sometimes we know that f behaves nicely with some operations (eg addition or inversion). It might then be possible to express other operations in terms of those operations and conclude that f behaves nicely with that operation too. Hopefully this can be used to solve the equation.

*Example*: f(x + y) = f(x) + f(y), f(x^{2}) = f(x)^{2}.

We can write 4xy = (x + y)^{2} - (x - y)^{2}, and hence f(4xy) = 4f(x)f(y) ⇒ f(xy) = f(x)f(y). So f is multiplicative, which is a useful property.

*Example*: f(1/x) = 1/f(x), f(x + 1) = f(x) + 1, f(2x) = 2f(x).

We can write x^{2} = 2/(1(1+x) + 1/(1-x) - 2) + 1, and hence f(x^{2}) = f(x)^{2}.