Functional equations

[Darya]

I've encountered functional equations for the first time and having trouble with trying to solve this one

[MATH]f(xy)+f(xz)+f(yz)=xyz[/MATH]
I've so far tried to substitute different values, such is x=y=z=0, then f(0)=0, but im not even sure it's a valid thing to do because we have so many variables.
Or if x=1 we get f(y)+f(z)+f(yz)=yz , similarly for y=1 or z=1 but it gave nothing.

Please, does anybody have any tips or methods of solving such an equation?


(P.S if it's too tricky maybe there's an easier one, which is similar I could attempt to solve?)

 

[Dr.Peterson]

I've encountered functional equations for the first time and having trouble with trying to solve this one

[MATH]f(xy)+f(xz)+f(yz)=xyz[/MATH]
I've so far tried to substitute different values, such is x=y=z=0, then f(0)=0, but im not even sure it's a valid thing to do because we have so many variables.
Or if x=1 we get f(y)+f(z)+f(yz)=yz , similarly for y=1 or z=1 but it gave nothing.

Is this the entire statement of the problem? Were you told anything about the domain, or whether it is continuous?

You're right that f(0) = 0. Also, if you set only z=0, you get f(xy) = 0 for any x and y, which would seem to imply that f is identically zero. Yet if you set x=y=z=1, you find that f(1) = 1/3.

On the other hand, if you let x=y=z=u, you can get a formula for f(u). And you get a different formula if you let y=z=1.

Did you just make this up randomly? Not all equations have solutions!

 

[HallsofIvy]

Yes, given that f(xy)+ f(xz)+ f(yz)= xyz for all x, y, and z, it is correct to put in specific values for x, y, and z. If x= y= z= 0 then f(0)+ f(0)+f(0)= 3f(0)= 0 so f(0)= 0. Also, setting x= y= z= 1 we have f(1)+ f(1)+ f(1)= 3f(1)= 1 so f(1)= 1/3.

However, setting x= y= 1 and z= 0, f(1)+ f(0)+ (0)= 1+ 0+ 0= 1= 1(0)(0)= 0 which is not true! There is NO function satisfying this equation.