A Collection of Functional Equations
by SomeonecoolLovesMaths, Dec 3, 2024, 1:09 PM
Here are a few problems of Functional Equations:
satisfy the equation 
.
is bijective.
such that 
.
But a function can't give two distinct values, unless 
this is not true so 
,
To obtain any 
just let 
.
Thus 
which satisfy for all 
:
for 
![\[f(x)=x^2\cdot f\left(\frac{1}{x}\right)=x^2\cdot f\left(1+\frac{1-x}{x}\right)=x^2+x^2\cdot f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot \frac{(1-x)^2}{x^2}\cdot f\left(\frac{x}{1-x}\right)=\]](http://latex.artofproblemsolving.com/4/d/a/4da809b44b08f413bde69c4df27874c14bc7e4b8.png)
![\[x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}-1\right)=x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}\right)-(1-x)^2=x^2-(1-x)^2+f(1-x)=2x+f(-x)=2x-f(x)\]](http://latex.artofproblemsolving.com/3/d/d/3ddf8024f0a68b7279bee29f51e58e913fd1d3aa.png)
and hence 
holds for all 
which easily extends to 
such that
and 
be the assertion 
and so 
and 
since 
is non constant;
and so :
and so 
implies 
which gives immediately 
,
and 
which indeed is a solution.
implies 
which gives immediately 
and 
and 
which indeed is a solution.
implies 
which gives immediately 
which indeed is a solution.This post has been edited 3 times. Last edited by SomeonecoolLovesMaths, Dec 3, 2024, 1:22 PM
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