What are all even  functions in real set which are true in relation f(x+y)=(f(x)+f(y))/(1+f(x)*f(y))?

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Find all even functions `f:RR->RR` such that `f(x+y)=(f(x)+f(y))/(1+f(x)f(y))`

There are three such functions:

Let x=y=0. Then `f(0+0)=f(0)=(f(0)+f(0))/(1+f(0)f(0))` so


`=>f(0)^3-f(0)=0 ==> f(0)=0,+-1`

(1) Suppose f(0)=0. Let y=-x. Note that f(x)=f(-x) since the function is even.

Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`

then `0=(2f(x))/(1+f(x)^2)` . The denominator is never zero so this is true when 2f(x)=0 ==>f(x)=0.

The first function is `f(x)=0 AA x in RR`

(2) Suppose f(0)=1. Let y=-x.

Then `f(x+y)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`

and `1=(2f(x))/(1+f(x)^2)` ==> `f(x)^2-2f(x)+1=0`

`==> (f(x)-1)^2=0 ==> f(x)=1`

The second function is `f(x)=1 AA x in RR`

(3) Suppose f(0)=-1. Let y=-x.

Then `f(x+y)=f(x-x)=f(0)=(f(x)+f(-x))/(1+f(x)f(-x))`

so `-1=(2f(x))/(1+f(x)^2) ==> f(x)^2+2f(x)+1=0`

then `(f(x)+1)^2=0 ==> f(x)=-1`

The third function is `f(x)=-1 AA x in RR`