Functional equations satisfied by Typical functions

The differential equation satisfied by the function y=sqrt(sin x+sqrt(sin x+sqrt(sin x+............oo)))

Functional equations resulting from properties of functions

When does a production function satisfy increasing returns to scale?

A functional equation is an equation, which relates the values assumed by a function at two or more points, which are themselves related in a particular manner. For example, we define an odd function by the relation f(-x) = -f(x) for all x.This defination can be paraphrased to say that it is a function f(x), which satisfies the functional relation f(x) + f(y) = 0, whenever x+y = 0. Of course this does not identify the function uniquely, sometimes with some additionl information, a function satisfying a given functional equation can be identified uniquely. Suppose a functional equation has a relation between f(x) and f(1/x) , then due to the reason that reciprocal of a reciprocal gives back the original number, we can substitute 1/x for x. This will result into another equation and solving these two, we can find f(x) uniquely. Similarly, we can solve an equation which contains f(x) and f(-x). Such equations are of repetitive nature . If for every x in R , the function f(x) satisfies the relation af(x) + bf(-x) = g(x), then

A functional equation is an equation, which relates the values assumed by a function at two or more points, which are themselves related in a particular manner. For example, we define an odd function by the relation f(-x) = -f(x) for all x.This defination can be paraphrased to say that it is a function f(x), which satisfies the functional relation f(x) + f(y) = 0, whenever x+y = 0. Of course this does not identify the function uniquely, sometimes with some additionl information, a function satisfying a given functional equation can be identified uniquely. Suppose a functional equation has a relation between f(x) and f(1/x) , then due to the reason that reciprocal of a reciprocal gives back the original number, we can substitute 1/x for x. This will result into another equation and solving these two, we can find f(x) uniquely. Similarly, we can solve an equation which contains f(x) and f(-x). Such equations are of repetitive nature . In the functional equation af(x) + bf(-x) = g(x) , if a + b = 0, then f(x) is equal to

A function phi(x) satisfies the functional equation x^(2)phi(x)+phi(1-x)=2x-x^(4) for all real x. Then phi(x) is given by