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### Solving Functional Equations

```Date: 08/11/2005 at 20:31:32
From: Dmitry
Subject: a*f(f(x))+b*f(x)+c=g(x)

I am currently a 9th grade student and have been teaching myself
calculus.  During my studies, I came across the chain rule.  I already
knew how to use functions to the second power as substitution.
However, if one would have an equation like...

f(f(x)) + 5f(x) + 7 = x^2 + 6x;

How would one go about solving something like this?

I have no idea how to approach this.  If a function, for instance, is
in the form of a*f^2(x) + b*f(x) + c = 0, it is simple to solve by
using f as a variable.  However, in this case, I really am clueless.

My idea was to differentiate both sides, obtaining...

f'(f(x))*f'(x) + f'(x) = 2x + 6...

I presume this would be a differential equation which I do not yet
have sufficient knowledge to solve.  My question is: is there an
easier way?  And, if not, can you show me how one would solve
something like this?

```

```
Date: 08/12/2005 at 20:11:26
From: Doctor Vogler
Subject: Re: a*f(f(x))+b*f(x)+c=g(x)

Hi Dmitry,

Thanks for writing to Dr. Math.  That's not an easy question to
answer.  That kind of an equation is called a "functional equation,"
and finding all functions that satisfy a particular functional
equation is often very difficult.  Some equations are easy, such as
the one you mentioned, which is simply a quadratic polynomial in f(x).
Some are harder.  Let's look at a couple which are in between, and see
what we can do.

f(f(x)) = r*f(x)

for some real number r.  Then if we let y = f(x), this becomes

f(y) = ry.

You might then jump to the conclusion that f(x) = rx is the only
solution.  But actually, we only need f(x) = rx for every x in the
range of f.  For example, you might have

rx  if x = r^n for some integer n > 0
f(x) = {
0  otherwise

Then this function also works.

f(f(x)) = r

for some real number r.  We can do the same as above, and we conclude
that f(y) = r for every y in the range of f.  For example, we can
choose any subset S of the real numbers, and then set f(y) = r for
every y in the set S, and then, for every x not in S, we can set f(x)
to be anything in S.

f(f(x)) = x.

It turns out that all of the solutions to this equation can be found
as follows:  Pair up all of the real numbers into pairs {a, b} and
singletons {c}.  Then define f(a) = b and f(b) = a for each pair, and
define f(c) = c for each singleton.  For example, you can pair up x
with x+1 when floor(x) is even, and with x-1 when floor(x) is odd.

But what happens if we also make a few extra conditions?  It takes
some challenging complex analysis to prove that every analytic
function of a complex variable that satisfies f(f(x)) = x is, in fact,
a rational function.  And it doesn't take as much work to show that
every rational function that satisfies that equation is either

f(x) = x,

or

f(x) = k/x,

for some constant a, or

f(x) = (x + a)/(bx - 1)

for some constants a and b.

Along similar lines, it is challenging to show, and requires the Axiom
of Choice, that there are real functions of a real variable which satisfy

f(x+y) = f(x) + f(y)

for which

f(x)/x

is not a constant.  On the other hand, if the function f is
differentiable, then it's a simple matter to show that f(x)=rx for
some constant r.  If the function is continuous, then you can prove
the same thing.  But there are other functions (discontinuous ones)
that satisfy the functional equation.

Next we should try to solve

f(f(x)) = x+1

or perhaps

f(f(x)) = 2x.

And then things start to get difficult.  Of course, one solution to
the first is

f(x) = x + 1/2,

and one solution to the second is

f(x) = x*sqrt(2),

but are there other solutions?  What are they?  That's not an easy

The best ideas that I can suggest are to guess a form for the
function, such as a first-degree rational function

ax + b
f(x) = ------
cx + d

and then try to see what coefficients (if any) will solve the
equation.  Another form is a power series

f(x) = a + bx + cx^2 + dx^3 + ...

and any analytic function can be expressed in this form.  These
methods don't work, however, for discontinuous solutions.

But here is one way to analyze your equation.  Let's find all rational
functions that solve it.  (A "rational function" means a polynomial in
x divided by another polynomial in x.)  It doesn't take too much work
to show that if the degree of the function f(x) is d (the degree of a
rational function means the larger of the degrees of the numerator and
the denominator), then the degree of f(f(x)) is d^2.  Now our equation is

f(f(x)) = x^2 + 6x - 7 - 5f(x).

So the degree of the left side is d^2.  What is the degree on the
right side?  Well, if the degree of the numerator of f(x) is less than
or equal to the degree of the denominator, then the degree of the
right side will be d+2.  If the degree of the numerator of f(x) is one
more than that of the denominator, then the degree of the right side
will be d+1.  If the degree of the numerator is at least two more than
that of the denominator, then the degree of the right side will be d,
with one exception:  If the degree of the numerator is exactly two
more, and the leading coefficients are opposites, then the x^2 will
cancel with the largest terms, and you will get a degree smaller than d.

Now consider that the degree of the right side has to equal the degree
of the left side.  So that means that

d^2 = d+2

or

d^2 = d+1

or

d^2 <= d.

But since d is a positive integer, the last equation only holds when d
is 1, and the first equation only holds when d is 2, and the middle
equation never holds.  But then if we have d = 1, then the degree of
the numerator certainly can't be two less than that.  So we have only
one possibility, which is that f(x) is a degree-two rational function

ax^2 + bx + c
f(x) = -------------.
x^2 + dx + e

Now substitute this expression into the equation

f(f(x)) + 5f(x) = x^2 + 6x - 7,

multiply by denominators to get a single big polynomial in x, and then
since they have to be the same functions on both sides of the
equation, that means that the coefficients have to be the same.  That
will give you several equations in the coefficients a, b, etc.  Then
see if there are any solutions to these equations.  If not, then there
are no rational functions that solve your functional equation.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus

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