Solve for F: Find f(x+y)=f(x)(a-y) + f(y)f(a-x), f(0)=1/2

[Penultimate]

Find all the fumction F: R$$\rightarrow$$R So that :
1) f(x+y)=f(x)(a-y) + f(y)f(a-x)
2)f(0)=1/2

I don`t know if i posted in the right place , but if not please moderators send me an email.
Thanks...

 

[ChaoticRoman]

Find all the fumction F: R$$\rightarrow$$R So that :
1) f(x+y)=f(x)(a-y) + f(y)f(a-x)
2)f(0)=1/2

I don`t know if i posted in the right place , but if not please moderators send me an email.
Thanks...

I suppose a is a given constant and first rule is for all x, y from R.

First assume x=y=0: 1/2 = 1/2 a + 1/2 f(a) $$\Rightarrow$$ f(a) = 1-a (1)
Then x=0 and y=a: f(a) = 1/2 . 0 + f(a)² $$\Rightarrow$$ f(a) = 0 or 1 (2)
f(a) can't be 1, because then according (1) a=0 and f(0) is 1/2.
Let's assume x=a and y=0: f(a) = f(a)² + 1/4 and f(a)=0 $$\Rightarrow$$ 0=1/4

This function can't exist.