Definitions

# Functional equation

In mathematics or its applications, a functional equation is an equation expressing a relation between the value of a function (or functions) at a point with its values at other points. Properties of functions can for instance be determined by considering the types of functional equations they satisfy. The term functional equation is usually reserved for equations that are not in a simple sense reducible to algebraic equations, often because two or more known functions are substituted as arguments into another function.

## Examples

• The functional equation


f(s) = 2^spi^{s-1}sinleft(frac{pi s}{2}right)Gamma(1-s)f(1-s)

is satisfied by the Riemann zeta function ζ. The capital Γ denotes the gamma function.

• These functional equations are satisfied by the gamma function. Gamma function is the unique solution of the system of all the three equations:

$f\left(x\right)=\left\{f\left(x+1\right) over x\right\},!$
$f\left(y\right)fleft\left(y+frac\left\{1\right\}\left\{2\right\}right\right)=frac\left\{sqrt\left\{pi\right\}\right\}\left\{2^\left\{2y-1\right\}\right\}f\left(2y\right)$
$f\left(z\right)f\left(1-z\right)=\left\{pi over sin\left(pi z\right)\right\},!,,,$ (Euler's reflection formula)

• The functional equation

$fleft\left(\left\{az+bover cz+d\right\}right\right) = \left(cz+d\right)^k f\left(z\right),!$

where a, b, c, d are integers satisfying adbc = 1, i.e. 
begin{vmatrix} a & bc & dend{vmatrix},=1, which means that $begin\left\{bmatrix\right\} a & bc & dend\left\{bmatrix\right\},$ is a unitary matrix (i.e. having determinant 1), defines f to be a modular form of order k.

• Miscellaneous examples not necessarily involving "famous" functions:

$f\left(x + y\right) = f\left(x\right)f\left(y\right), ,!$ satisfied by all exponential functions

$f\left(xy\right) = f\left(x\right) + f\left(y\right),!$, satisfied by all logarithmic functions

$f\left(x + y\right) = f\left(x\right) + f\left(y\right),!$ (Cauchy functional equation)

$f\left(x + y\right) + f\left(x - y\right) = 2\left[f\left(x\right) + f\left(y\right)\right],!$ (quadratic equation or parallelogram law)

$f\left(\left(x + y\right)/2\right) = \left(f\left(x\right) + f\left(y\right)\right)/2,!$ (Jensen)

$g\left(x + y\right) + g\left(x - y\right) = 2\left[g\left(x\right) g\left(y\right)\right],!$ (d'Alembert)

$f\left(h\left(x\right)\right) = cf\left(x\right),!$ (Schröder equation)

$f\left(h\left(x\right)\right) = f\left(x\right) + 1,!$ (Abel equation).

One such example of a recurrence relation is

$a\left(n\right) = 3a\left(n-1\right) + 4a\left(n-2\right),!$

• The commutative and associative laws are functional equations. When the associative law is expressed in its familiar form, one lets some symbol between two variables represent a binary operation, thus:

$\left(a*b\right)*c = a*\left(b*c\right).,$

But if we write $f\left(a, b\right)$ instead of $a * b,$ then the associative law looks more like what one conventionally thinks of as a functional equation:

$f\left(f\left(a, b\right),c\right) = f\left(a, f\left(b, c\right)\right).,!$

One thing that all of the examples listed above share in common is that in each case two or more known functions (sometimes multiplication by a constant, sometimes addition of two variables, sometimes the identity function) are substituted into the unknown function to be solved for.

• The b-integer and b-decimal parts of real numbers were introduced and studied by M.H.Hooshmand . The b-parts real functions have many interesting number theoretic explanations, analytic and algebraic properties, and satisfy the functional equation:

$f\left(f\left(x\right) + y - f\left(y\right)\right) = f\left(x\right).,!$

The following functional equations are as a generalization of the b-parts functional equation for semigroups and groups, even in a binary system (magma), that are introduced by him:

Associative equations ;

$f\left(f\left(xy\right)z\right)=f\left(xf\left(yz\right)\right); ,; f\left(f\left(xy\right)z\right)=f\left(xf\left(yz\right)\right)=f\left(xyz\right)$

Decomposer equations ;

$f\left(f^*\left(x\right)f\left(y\right)\right)=f\left(y\right); ,; f\left(f\left(x\right)f_*\left(y\right)\right)=f\left(x\right)$

Strong decomposer equations ;

$f\left(f^*\left(x\right)y\right)=f\left(y\right); ,; f\left(xf_*\left(y\right)\right)=f\left(x\right)$

Canceler equations ;

$f\left(f\left(x\right)y\right)=f\left(xy\right); ,; f\left(xf\left(y\right)\right)=f\left(xy\right); ,; f\left(xf\left(y\right)z\right)=f\left(xyz\right)$

where $f^*\left(x\right)f\left(x\right)=f\left(x\right)f_*\left(x\right)=x$. In , the general solution of the decomposer and strong decomposer equations are introduced in the sets with a binary operation and semigroups respectively and also associative equations in arbitrary groups. In that paper it is proven that the associative equations and the system of strong decomposer and canceler equations do not have any nontrivial solutions in the simple groups.

When it comes to asking for all solutions, it may be the case that conditions from mathematical analysis should be applied; for example, in the case of the Cauchy equation mentioned above, the solutions that are continuous functions are the 'reasonable' ones, while other solutions that are not likely to have practical application can be constructed (by using a Hamel basis for the real numbers as vector space over the rational numbers). The Bohr-Mollerup theorem is another well-known example.

## Solving functional equations

Solving functional equations can be very difficult but there are some common methods of solving them.

A discussion of involutary functions is useful. For example, consider the function

$f\left(x\right) = frac\left\{1\right\}\left\{x\right\}$.

Then consider

f(f(x)) = x,

if we continue the pattern we end up with x for an even number of compositions and f(x) for an odd number. This same idea applies to many other functions, i.e.

$f\left(x\right) = frac\left\{1\right\}\left\{1-x\right\}, f\left(x\right) = 1-x$ and many others.

Example 1: Solve $f\left(x+y\right)^2 = f\left(x\right)^2 + f\left(y\right)^2$ for all $x,y in mathbb\left\{R\right\},$ assuming f is a real-valued function.

Let $x=y=0$: $f\left(0\right)^2=f\left(0\right)^2+f\left(0\right)^2$. So $f\left(0\right)^2=0$ and $f\left(0\right)=0$.

Now, let $y=-x$:

$f\left(x-x\right)^2=f\left(x\right)^2+f\left(-x\right)^2$
$f\left(0\right)^2=f\left(x\right)^2+f\left(-x\right)^2$
$0=f\left(x\right)^2+f\left(-x\right)^2$

A square of a real number is nonnegative, and a sum of nonnegative numbers is zero iff both numbers are 0. So $f\left(x\right)^2=0$ for all x and $f\left(x\right)=0$ is the only solution.