example of solving a functional equation
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:
. 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”
The solution method of (1) is due to andik and perucho.
|Title||example of solving a functional equation|
|Date of creation||2013-03-22 15:30:00|
|Last modified on||2013-03-22 15:30:00|
|Last modified by||pahio (2872)|