# Blog Archives

## Functional equations challenge 1

This is a challenging problem !!

Find all continuous  functions      such that for all :

Solution : ( first try it , if not then you can read )

1) replacing the equation by a simpler one :
=============================================
Let     and so   and the equation becomes  and so :

is a continuous function such that
Q.E.D.

2)
=======================

Let
From 1) above, we get that  and so, since  is continuous :
Either , either  for some .

and so

Let then  :  the restriction of  to  :
and so  is a bijection, so is monotonous
implies then  is increasing and so  is increasing.

So,    and so  which is impossible and so
Same,    and so  which is impossible and so

So
Q.E.D.

3) general solution of  with  continuous
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Then the original equation  becomes  (since ) and so :

The general continuous solutions of  are any odd continuous function  such that
A simple way to build them is :
3.1) If  :

3.2) If  for some  :
Let any continuous fonction  from  such that  and define  as :
:
:
:

Notice that  gives the solution

4) general solution of  with  continuous
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The general solution of original equation is  where  is any function described in 3) above

5) examples of solutions
======================
gives the solution

gives the solution

and  gives the solution :
If  :
If  :

and  gives the solution :

Notice that the only differentiable solutions are  and

## Problem of the day 3

Nice problem !

Functional Equations !

Q)
Determine all functions ƒ satisfying  the functional relations

ƒ(x) + ƒ(1/(1-x)) =( 2(1-2x)/(x(1-x))

where x is a real number x ≠ 0 , x ≠ 1