Here's a hint

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- May 5th 2011, 01:31 PM #1

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- May 5th 2011, 02:22 PM #2

- May 5th 2011, 02:22 PM #3
Fun problem! I'm not sure I've solved it, but I have made some progress, or at least found a few things out. First, suppose you set y = 0. Then you get f(x) = f(x) f(0) for all x. Then either f(0) = 1 or f(x) = 0 for all x. Suppose f(0) = 1. Then we could set x = 0, and get f(-y) = f(0) f(y) = f(y), thus making f even. But the f(x) = 0 case was also even. Hence, we conclude that f is even.

If we set x = y = 0, then we get that f(0) = f(0) f(0), which implies that f(0) is in the set {0,1}. Letting x = 4 and y = 4 shows us that f(0) = f(4) f(4), which implies that f(4) is in the set {-1, 0, 1}. Setting x = 4 and y = 0 gives us f(4) = f(4) f(0).

I think your problem is underdetermined. Why? f(x) = 0 satisfies the function equation, and f(x) = 1 satisfies the function equation, for all x. Therefore, with the information you have, you cannot uniquely determine f.

Reply to pickslides: unfortunately, the exponential function is neither even nor odd, so I don't think the exponential function fits here.

- May 5th 2011, 05:35 PM #4
I agree with Ackbeet. I leave it to you to prove the details. Consider the following:

f(0) = f(0)*f(0) implies f(0) = 0 or f(0) = 1. We can show that f(n) = 0 for all n if f(0) = 0.

So let's assume that f(0) = 1. Then

f(1) = f(2)*f(1) implies f(1) = 0 or f(2) = 1. Again we can show that f(n) = 0 for all n > 1 if f(1) = 0.

So let's assume that f(2) = 1. Then

f(2) = f(4)*f(2), etc.

I get either f(4) = 0 or f(4) = 1.

-Dan