Functional equation

A functional equation is an equation expressing the relationship between the value of a function at one point and its values at other points. Many properties of functions can be determined by examining the functional equations that these functions satisfy. The term “functional equation” is commonly used for equations that are not reducible by simple methods to algebraic equations . This irreducibility is most often due to the fact that the arguments of the unknown function in the equation are not the independent variables themselves, but some of these functions from them.

Examples

Functional equation:

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

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

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

,

Where $\Gamma (z)$ ${\ displaystyle \ Gamma (z)}$ $\Gamma(z)$ - Euler gamma function , satisfies Riemann zeta function $\zeta$ ${\ displaystyle \ zeta}$ $\zeta$ .

The gamma function is the only solution to this system of three equations:

f(x)={f(x+1) \over x}

{\ displaystyle f (x) = {f (x + 1) \ over x}}

f(x)={f(x+1) \over x}

f(y)f\left(y+{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2^{2y-1}}}f(2y)

{\ displaystyle f (y) f \ left (y + {\ frac {1} {2}} \ right) = {\ frac {\ sqrt {\ pi}} {2 ^ {2y-1}}} f (2y )}

f(y)f\left(y+{\frac {1}{2}}\right)={\frac {{\sqrt {\pi }}}{2^{{2y-1}}}}f(2y)

f(z)f(1-z)={\pi  \over \sin(\pi z)}

{\ displaystyle f (z) f (1-z) = {\ pi \ over \ sin (\ pi z)}}

f(z)f(1-z)={\pi  \over \sin(\pi z)}

( Euler addition formula )

Functional equation:

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

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

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

,

Where $a, b, c, d$ ${\ displaystyle a, b, c, d}$ $a,b,c,d$ are integers satisfying equality $ad-bc=1$ ${\ displaystyle ad-bc = 1}$ $ad-bc=1$ , i.e:

{\begin{vmatrix}a&b\\c&d\end{vmatrix}}\,=1

{\ displaystyle {\ begin {vmatrix} a & b \\ c & d \ end {vmatrix}} \, = 1}

{\begin{vmatrix}a&b\\c&d\end{vmatrix}}\,=1

,

determines $f$ ${\ displaystyle f}$ $f$ as a modular form of order $k$ ${\ displaystyle k}$ $k$ .

Functional Cauchy Equations:

$f(x+y)=f(x)+f(y)$ ${\ displaystyle f (x + y) = f (x) + f (y)}$ $f(x+y)=f(x)+f(y)$ - all linear homogeneous functions satisfy $f(x)=ax$ ${\ displaystyle f (x) = ax}$ $f(x)=ax$ ,
$f(x+y)=f(x)f(y)$ ${\ displaystyle f (x + y) = f (x) f (y)}$ $f(x+y)=f(x)f(y)$ - satisfy all exponential functions $f(x)=\exp \left(\alpha x\right)=a^{x}$ ${\ displaystyle f (x) = \ exp \ left (\ alpha x \ right) = a ^ {x}}$ $f(x)=\exp \left(\alpha x\right)=a^{x}$ ,
$f(xy)=f(x)+f(y)$ ${\ displaystyle f (xy) = f (x) + f (y)}$ $f(xy)=f(x)+f(y)$ - satisfy all logarithmic functions $f(x)=\alpha \log \left(x\right)=\log _{a}\left(x\right)$ ${\ displaystyle f (x) = \ alpha \ log \ left (x \ right) = \ log _ {a} \ left (x \ right)}$ $f(x)=\alpha \log \left(x\right)=\log _{a}\left(x\right)$ ,
$f(xy)=f(x)f(y)$ ${\ displaystyle f (xy) = f (x) f (y)}$ - satisfy all power functions $f(x)=\exp \left(\alpha \log \left(x\right)\right)=x^{a}$ ${\ displaystyle f (x) = \ exp \ left (\ alpha \ log \ left (x \ right) \ right) = x ^ {a}}$ .

The functional equations of Cauchy are reduced to each other. So, the equation $f(x_{1}x_{2})=f(x_{1})f(x_{2})$ ${\ displaystyle f (x_ {1} x_ {2}) = f (x_ {1}) f (x_ {2})}$ reduced to equation $g(y_{1}+y_{2})=g(y_{1})+g(y_{2})$ ${\ displaystyle g (y_ {1} + y_ {2}) = g (y_ {1}) + g (y_ {2})}$ after replacement $g(y)=\log \left|f(\exp y)\right|$ ${\ displaystyle g (y) = \ log \ left | f (\ exp y) \ right |}$ (for this, of course, you need to $f(x)$ ${\ displaystyle f (x)}$ was not identical zero). In the class of continuous functions and in the class of monotone functions, the given solutions are the only ones, except for the degenerate solution. $f(x)\equiv 0$ ${\ displaystyle f (x) \ equiv 0}$ . However, in a wider class of functions, quite exotic solutions are possible, see the article “Hamel Basis” .

Others:

$f(x+y)+f(x-y)=2[f(x)+f(y)]$ ${\ displaystyle f (x + y) + f (xy) = 2 [f (x) + f (y)]}$ - quadratic equation or parallelogram identity , satisfies $f(x)=kx^{2}$ ${\ displaystyle f (x) = kx ^ {2}}$ ,
$f\left({\frac {x+y}{2}}\right)={\frac {f(x)+f(y)}{2}}$ ${\ displaystyle f \ left ({\ frac {x + y} {2}} \ right) = {\ frac {f (x) + f (y)} {2}}}$ - Jensen equation, satisfy all linear functions $f(x)=ax+b$ ${\ displaystyle f (x) = ax + b}$ ,
$f(x+y)f(x-y)=f(x)^{2}$ ${\ displaystyle f (x + y) f (xy) = f (x) ^ {2}}$ - the Lobachevsky equation (version of the Jensen equation), satisfies $f(x)=ac^{x}$ ${\ displaystyle f (x) = ac ^ {x}}$ ,
$f(x+y)+f(x-y)=2[f(x)f(y)]$ ${\ displaystyle f (x + y) + f (xy) = 2 [f (x) f (y)]}$ - d'Alembert equation,
$f(h(x))=f(x)+1$ ${\ displaystyle f (h (x)) = f (x) +1}$ - ,
$f(h(x))=cf(x)$ ${\ displaystyle f (h (x)) = cf (x)}$ - , the solution is the Koenigs function related to the function $\textstyle h(x)$ ${\ displaystyle \ textstyle h (x)}$ .

Recurrence Relationships

A particular type of functional equations is a recurrent relation containing an unknown function of integers and a shift operator .

Linear recurrence relations:

a(n)=\sum _{i=1,k}c_{i}\cdot a(n-i)

{\ displaystyle a (n) = \ sum _ {i = 1, k} c_ {i} \ cdot a (ni)}

(Where $c_{1},c_{2},\dots ,c_{k}$ {\ displaystyle c_ {1}, c_ {2}, \ dots, c_ {k}} - constants not dependent on $n$ ${\ displaystyle n}$ ) have a theory, the analogue of which is the theory of linear differential equations. For example, for a linear recurrence relation:

a(n)=3a(n-1)+4a(n-2)

{\ displaystyle a (n) = 3a (n-1) + 4a (n-2)}

,

it is enough to find two linearly independent solutions; all other solutions will be their linear combinations.

To find these solutions, one has to substitute the trial function in the recurrent relation $a(n)=\lambda ^{n}$ ${\ displaystyle a (n) = \ lambda ^ {n}}$ with undefined parameter $\lambda$ ${\ displaystyle \ lambda}$ and try to find those $\lambda$ ${\ displaystyle \ lambda}$ for which this recurrence relation will be satisfied. For the given example, we obtain a quadratic equation $\lambda ^{2}=3\lambda +4$ ${\ displaystyle \ lambda ^ {2} = 3 \ lambda +4}$ with two different roots $\lambda =-1$ ${\ displaystyle \ lambda = -1}$ and $\lambda =4;$ ${\ displaystyle \ lambda = 4;}$ therefore, the general solution for a given recurrence relation is the formula $a(n)=d_{1}4^{n}+d_{2}(-1)^{n}$ ${\ displaystyle a (n) = d_ {1} 4 ^ {n} + d_ {2} (- 1) ^ {n}}$ (constants $d_{1}$ ${\ displaystyle d_ {1}}$ and $d_{2}$ ${\ displaystyle d_ {2}}$ are selected so that when $n=1$ ${\ displaystyle n = 1}$ and $n=2$ ${\ displaystyle n = 2}$ the formula gave the desired values for the values $a(1)$ ${\ displaystyle a (1)}$ and $a(2)$ ${\ displaystyle a (2)}$ ). In the case of multiple roots of a polynomial, additional test solutions are functions $n\lambda ^{n},$ ${\ displaystyle n \ lambda ^ {n},}$ $n^{2}\lambda ^{n}$ ${\ displaystyle n ^ {2} \ lambda ^ {n}}$ and so on.

One of the widely known recurrence relations is $a(n)=a(n-1)+a(n-2)$ ${\ displaystyle a (n) = a (n-1) + a (n-2)}$ defining the Fibonacci sequence .

Solving Functional Equations

There are some general methods for solving functional equations.

In particular, it may be useful to use the concept of involution , that is, the use of the properties of functions for which $f(f(x))=x$ ${\ displaystyle f (f (x)) = x}$ ; the simplest involutions:

f(x)=-x

{\ displaystyle f (x) = - x}

,

f(x)={\frac {1}{x}}

{\ displaystyle f (x) = {\ frac {1} {x}}}

,

f(x)={\frac {1}{1-x}}+1

{\ displaystyle f (x) = {\ frac {1} {1-x}} + 1}

,

f(x)=1-x

{\ displaystyle f (x) = 1-x}

.

For example, to solve the equation:

f^{2}(x+y)=f^{2}(x)+f^{2}(y)

{\ displaystyle f ^ {2} (x + y) = f ^ {2} (x) + f ^ {2} (y)}

for all $x,y\in \mathbb {R}$ ${\ displaystyle x, y \ in \ mathbb {R}}$ and $f:\mathbb {R} \to R$ ${\ displaystyle f: \ mathbb {R} \ to R}$ put $x=y=0$ ${\ displaystyle x = y = 0}$ : $f^{2}(0)=f^{2}(0)+f^{2}(0)$ ${\ displaystyle f ^ {2} (0) = f ^ {2} (0) + f ^ {2} (0)}$ . Then $f^{2}(0)=0$ ${\ displaystyle f ^ {2} (0) = 0}$ and $f(0)=0$ ${\ displaystyle f (0) = 0}$ . Next, putting $y=-x$ ${\ displaystyle y = -x}$ :

f^{2}(x-x)=f^{2}(x)+f^{2}(-x)

{\ displaystyle f ^ {2} (xx) = f ^ {2} (x) + f ^ {2} (- x)}

f^{2}(0)=f^{2}(x)+f^{2}(-x)

{\ displaystyle f ^ {2} (0) = f ^ {2} (x) + f ^ {2} (- x)}

0=f^{2}(x)+f^{2}(-x)

{\ displaystyle 0 = f ^ {2} (x) + f ^ {2} (- x)}

The square of a real number is non-negative, and the sum of non-negative numbers is zero if and only if both numbers are equal to 0. So $f^{2}(x)=0$ ${\ displaystyle f ^ {2} (x) = 0}$ for all $x$ ${\ displaystyle x}$ and $f(x)\equiv 0$ ${\ displaystyle f (x) \ equiv 0}$ is the only solution to this equation.

Literature

Golovinsky, I. A. Early history of analytical iterations and functional equations. // Historical and mathematical research. M .: Science, vol. XXV, 1980, p. 25-51.
Kuczma M. On the functional equation φn (x) = g (x). Ann. Polon. Math 11 (1961) 161-175 .
The functional theory of the relationship between the laws and inequalities. Warszawa - Kraków - Katowice: Polish Scientific Publishers & Silesian University, 1985.
Likhtarnikov L.M. Elementary introduction to functional equations. St. Petersburg: Lan, 1997.

Links

Functional Equations: Exact Solutions at EqWorld: The World of Mathematical Equations.
Functional Equations: Index at EqWorld: The World of Mathematical Equations.
IMO Compendium for problem solving.