In mathematics or its applications, a functional equation is an equation expressed in terms of both independent variables and unknown functions, which are to be solved for. 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 an unknown function, which is to be solved for.

Examples

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• The functional equation

$$

\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)

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

• The functional equation

$\backslash Gamma(x)=\{\backslash Gamma(x+1)\; \backslash over\; x\}$

is satisfied by the gamma function.

• Euler's reflection formula:

$\backslash Gamma(z)\backslash Gamma(1-z)=\{\backslash pi\; \backslash over\; \backslash sin(\backslash pi\; z)\}$

is satisfied by the gamma function.

• The functional equation

$f\backslash left(\{az+b\backslash over\; cz+d\}\backslash right)\; =\; (cz+d)^k\; f(z)$

where a, b, c, d are integers satisfying ad − bc = 1, defines f to be a modular form of order k.

• Miscellaneous examples not necessarily involving "famous" functions:

f(x + y) = f(x)f(y), satisfied by all exponential functions

f(xy) = f(x) + f(y), satisfied by all logarithmic functions

f(x + y) = f(x) + f(y) (Cauchy functional equation)

f(x + y) + f(x − y) = 2f(x) + 2f(y) (quadratic equation or parallelogram law)

F(az) = aF(z)(1 − F(z)) (Poincaré equation)

G(z) = λ⁻¹ G(G(λz)) (chaos theory, scaling)

f((x + y)/2) = (f(x) + f(y))/2 (Jensen)

g(x + y) + g(x − y) = 2g(x)g(y) (d'Alembert)

f(h(x)) = cf(x) (Schröder equation)

f(h(x)) = f(x) + 1 (Abel equation).

• A simple form of functional equation is a recurrence relation. This, formally speaking, involves an unknown function on integers, and also translation operators.

One such example of a recurrence relation is

$a(n)\; =\; 3a(n-1)\; +\; 4a(n-2)$

• 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:

(a * b) * c = a * (b * c),

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

f(f(a, b), c) = f(a, f(b, c)).

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.

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

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Solving functional equations can be very difficult but in this section we will discuss some common methods of solving them. A discussion of involutary functions is crucial to our ability to solve functional equations. For example, consider the function $f(x)\; =\; \backslash frac\{1\}\{x\}$. 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. $f(x)\; =\; \backslash frac\{1\}\{1-x\},\; f(x)\; =\; 1-x$ and many others. Keep involutary functions in mind as you solve functional equations as they are often key elements.

Example 1: $f(x\; +\; y)\; =\; f(x)\; +\; f(y)\; +\; 2\backslash sqrt\{f(x)f(y)\},\; \backslash forall\{x,y\; \backslash in\; \backslash mathbb\{R\}\}$ (Taken from M&IQ 1991)

A convenient values for y, y = 0, tells that $f(x)\; =\; f(x)\; +\; f(0)\; +\; 2\backslash sqrt\{f(x)f(0)\}\; \backslash Leftrightarrow\; f(x)\; =\; f(0)/4$. So f(x) is a constant divided by 4 (because f(0) is a constant), and so we test this solution to see if it is valid. Plugging in k/4 into the expression for f(x) we have that k/4 = k/4 + k/4 + k/4. This is true only for k = 0, so that is the only solution to this functional equation.

Solve $f(xy)\; =\; f(x)\; +\; f(y).$

Here we could approach the problem as we did in the previous fashion, by making substitutions and trying to solve for f(x) but there is a much simpler way, we simply recognize that this function is satisfied by ln(x), so that ln(xy) = ln(x) + ln(y). Note this can be generalized further by throwing in constants but we have already done almost all of the work.

See also

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• Functional equation (L-function)

External links

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• Functional Equations: Exact Solutions at EqWorld: The World of Mathematical Equations.
• Functional Equations: Index at EqWorld: The World of Mathematical Equations.

Mathematical analysis

Funktionalgleichung | Équation fonctionnelle | Equazione funzionale | Funktionaaliyhtälö | Funktionalekvation