Math Is Fun Forum

Shelled

Guest

Functions

For any function f,  f(x+y) = f(x) +f(y).
I've been asked to state whether the above is a true or false statement. I haven't seen it before, but I googled it and found out that it's called Cauchy's functional equation; so I guess the statement is true?

But I don't understand how it works, or what it's saying. Could you explain it with an example?

eigenguy

Member

Registered: 2014-03-18

Posts: 78

Re: Functions

Just because someone has studied functions satisfying an equation does not mean every function satisfies it. Very, very few functions satisfy this one. The ONLY functions you know that satisfy this equation are those of the form f(x) = Ax for some constant A. Even constant non-zero functions fail it.

---

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

eigenguy

Member

Registered: 2014-03-18

Posts: 78

Re: Functions

Some examples, to help you realize what is going on:
Let f(x) = 2x
Then, for example: f(3 + 1) = f(4) = 8. f(3) = 6. f(1) = 2, and 2 + 6 = 8, so indeed f(3+1) = f(3) + f(1).  More generally f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)

But now, let f(x) = x^2
Then f(3 + 1) = f(4) = 16, but f(3) = 9 and f(1) = 1, so f(3+1) ≠ f(3) + f(1)

---

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

Shelled

Guest

Re: Functions

Thanks for clearing that up

So, does this function have more conditions that need to be satisfied to work?

eigenguy

Member

Registered: 2014-03-18

Posts: 78

Re: Functions

It doesn't need conditions to make it work. It is just a condition that some functions can satisfy. If we add a very common additional condition to it, namely that the function f be continuous, then the only functions that satisfy it (assuming f is a function from real numbers to real numbers) are f(x) = Ax, for constants A.

Without adding the continuity condition, then it is possible to break the real numbers into a collection of subsets, each of which is closed under addition, and which intersect only in that they all contain 0, so that on each of these sets f(x) = Ax for some A, but the value A differs from set to set. These sets are intermixed with each other in much the same way that the rationals and irrational intermix, so the graph of such a function doesn't look like anything you are used to, but rather it looks like a cloud of points.

---

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich