Functional equation

In mathematics, a functional equation[1][2][3][4] is any equation that specifies a function in implicit form.[5] Often, the equation relates the value of a function (or functions) at some point with its values at other points. For instance, properties of functions can be determined by considering the types of functional equations they satisfy. The term functional equation usually refers to equations that cannot be simply reduced to algebraic equations.

Examples


f(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)f(1-s)
is satisfied by the Riemann zeta function. The capital Γ denotes the gamma function.
f(x)={f(x+1) \over x}\,\!
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)}\,\!\,\,\,       (Euler's reflection formula)
f\left({az+b\over cz+d}\right) = (cz+d)^k f(z)\,\!
where a, b, c, d are integers satisfying ad bc = 1, i.e. 
\begin{vmatrix} a & b\\c & d\end{vmatrix} = 1, defines f to be a modular form of order k.
f(x + y) = f(x) + f(y)\,\! (Cauchy functional equation)

Exponentiating,

f(x + y) = f(x)f(y), \,\! satisfied by all exponential functions
f(xy) = f(x) + f(y)\,\!, satisfied by all logarithmic functions
f(xy) = f(x) f(y)\,\!, satisfied by all powers
f(x + y) + f(x - y) = 2[f(x) + f(y)]\,\! (quadratic equation or parallelogram law)
f((x + y)/2) = (f(x) + f(y))/2\,\! (Jensen)
g(x + y) + g(x - y) = 2[g(x) g(y)]\,\! (d'Alembert)
f(h(x)) = h(x + 1)\,\! (Abel equation)
f(h(x)) = cf(x)\,\! (Schröder's equation).
f(h(x)) = (f(x))^c\,\! (Böttcher's equation).
f(h(x)) = h' (x) f (x) (Julia's equation).
\omega(  \omega(x,u),v)=\omega(x,u+v) (Translation equation)
f(x+y) = f(x)g(y)+f(y)g(x)\,\! (sine addition formula).
g(x+y) = g(x)g(y)-f(y)f(x)\,\! (cosine addition formula).
f(xy) = \sum g_l(x) h_l(y)\,\! (Levi-Civita).
a(n) = 3a(n-1) + 4a(n-2)\,\!
(a\circ b)\circ c = a\circ (b\circ c)~.

But if we wrote ƒ(a, b) instead of a  b then the associative law would look more like what one conventionally thinks of as a functional equation,

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

One feature 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 inside the argument of the unknown functions 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

Solving functional equations can be very difficult, but there are some common methods of solving them. For example, in dynamic programming a variety of successive approximation methods[6][7] are used to solve Bellman's functional equation, including methods based on fixed point iterations.

A main method of solving elementary functional equations is substitution. It is often useful to prove surjectivity or injectivity and prove oddness or evenness, if possible. It is also useful to guess possible solutions. Induction is a useful technique to use when the function is only defined for rational or integer values.

A discussion of involutory functions is topical. For example, consider the function

f(x) = 1-x \, .

Composing f with itself gives Babbage's functional equation (1820),[8]

f(f(x)) = 1-(1-x) = x \, .

Several other functions also satisfy this functional equation,

f(f(x)) = x ~,

including, beyond f(x) = −x,

 f(x) = \frac{a}{x}\, ,

and

 f(x) = \frac{b-x}{1+cx} ~ ,

which includes the previous three as special cases or limits.

Example 1. Find all functions f that satisfy

f(x+y)^2 = f(x)^2 + f(y)^2\,

for all x,y ∈ ℝ, assuming ƒ is a real-valued function.

Let x = y = 0,

f(0)^2=f(0)^2+f(0)^2.\,

So ƒ(0)2 = 0 and ƒ(0) = 0.

Now, let y = x,

f(x-x)^2=f(x)^2+f(-x)^2\,
f(0)^2=f(x)^2+f(-x)^2\,
0=f(x)^2+f(-x)^2~.

A square of a real number is nonnegative, and a sum of nonnegative numbers is zero iff both numbers are 0.

So ƒ(x)2 = 0 for all x and ƒ(x) = 0 is the only solution.

See also

Notes

  1. Rassias, Themistocles M. (2000). Functional Equations and Inequalities. 3300 AA Dordrecht, The Netherlands: Kluwer Academic Publishers. p. 335. ISBN 0- 7923-6484-8.
  2. Hyers, D. H.; Isac, G.; Rassias, Th. M. (1998). Stability of Functional Equations in Several Variables. Boston: Birkhäuser Verlag. p. 313. ISBN 0-8176-4024-X.
  3. Jung, Soon-Mo (2001). Hyers-Ulam-Rassias Stability of Functional Equations in Mathematical Analysis. 35246 US 19 North # 115, Palm Harbor, FL 34684 USA: Hadronic Press, Inc. p. 256. ISBN 1-57485-051-2.
  4. Czerwik, Stephan (2002). Functional Equations and Inequalities in Several Variables. P O Box 128, Farrer Road, Singapore 912805: World Scientific Publishing Co. p. 410. ISBN 981-02-4837-7.
  5. Cheng, Sui Sun; Wendrong Li (2008). Analytic solutions of Functional equations. 5 Toh Tuck Link, Singapore 596224: World Scientific Publishing Co. ISBN 978-981-279-334-8.
  6. Bellman, R. (1957). Dynamic Programming, Princeton University Press.
  7. Sniedovich, M. (2010). Dynamic Programming: Foundations and Principles, Taylor & Francis.
  8. Ritt, J. F. (1916). "On Certain Real Solutions of Babbage's Functional Equation". The Annals of Mathematics 17 (3): 113. doi:10.2307/2007270. JSTOR 2007270.

References

External links

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