example of solving a functional equation
Let’s determine all twice differentiable real functions which satisfy the functional equation
(1) |
for all real values of and .
Substituting first in (1) we see that or . The substitution gives , whence . So is an odd function.
We differentiate both sides of (1) with respect to and the result with respect to :
The result is simplified to , i.e.
Denoting , we obtain the equation
for all real values of and . This is not possible unless the proportion has a on . Thus the homogeneous linear differential equation or
with some , is valid.
There are three cases:
-
1.
. Now and consequently . If one especially equal to 1, the solution is the identity function (http://planetmath.org/IdentityMap) . This yields from (1) the well-known “memory formula”
-
2.
with . According to the oddness one obtains for the general solution the sine function . The special case means in (1) the
which is easy to verify by using the addition and subtraction formulae (http://planetmath.org/AdditionFormula) of sine.
-
3.
with . According to the oddness we obtain for the general solution the hyperbolic sine (http://planetmath.org/HyperbolicFunctions) function . The special case gives from (1) the
The solution method of (1) is due to andik and perucho.
Title | example of solving a functional equation |
Canonical name | ExampleOfSolvingAFunctionalEquation |
Date of creation | 2013-03-22 15:30:00 |
Last modified on | 2013-03-22 15:30:00 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 14 |
Author | pahio (2872) |
Entry type | Example |
Classification | msc 34A30 |
Classification | msc 39B05 |
Related topic | ChainRule |
Related topic | AdditionFormula |
Related topic | SubtractionFormula |
Related topic | DefinitionsInTrigonometry |
Related topic | GoniometricFormulae |
Related topic | DifferenceOfSquares |
Related topic | AdditionFormulas |