20190107, 16:13  #1 
Feb 2017
Nowhere
2·2,687 Posts 
A "difference of two squares" conundrum
It is a wellknown identity that
a^2  b^2 = (a  b)*(a + b). Once upon a time, long long ago, I happened to notice a nontrivial function f for which R) f^2(a)  f^2(b) = f(a  b)*f(a + b), for all real (or complex) a and b [The function was f(x) = sin(x); also f(x) = sin(k*x)/k, k not 0 also works. Multiplying f(x) by a (nonzero) constant) doesn't affect the relationship.] Given the relationship, it is easy to prove (can you?) that 1) f(0) = 0, and 2) f(x) is an odd function of x If you assume that f(x) is differentiable (or even analytic), it is not hard to show (can you?) that 3) 2 * f'(x) * f(x) = f'(0) * f(2*x) Note that from this, it follows that if f'(0) = 0, f is identically 0. So, since we can multiply by a nonzero constant, we may assume that f'(0) = 1. Also, 4) If f'(0) = 1, f(x) is a polynomial, and f(x) satisfies (R), then f(x) = x. Obviously by replacing k with k*i above, k real, we obtain f(x) = sinh(k*x)/k also satisfies (R). So much for puzzles. I also have a vague recollection of having shown that the sine function was pretty much it. But I can't remember how. Can anyone shed some light on this? 
20190109, 06:39  #2 
Romulan Interpreter
"name field"
Jun 2011
Thailand
5^{3}·79 Posts 
We bet you reached this stuff following our discussion on PM related to the current IMB puzzle (Jan 2019) hehe. You are on the right track.

20190109, 12:44  #3  
Feb 2017
Nowhere
2×2,687 Posts 
Quote:
I'd encountered the functional identity ages ago as I said. But it was the IBM puzzle that brought the old chestnut back to mind, I am sure. I just don't remember the argument I concocted all those decades ago. Perhaps I'll wake up at O'Dark Thirty some night recollecting the answer. Unfortunately, WRT the IBM puzzle, I'm stuck in EndofTracks Town. 

20190109, 15:38  #4 
Jan 2017
11^{2} Posts 
The following argument should work for at least nicely behaved classes of f:
Consider f at the set of points 0, e, 2e, 3e... and so on. If f is nicely behaved, we can assume that for a small enough e these points determine all of f. Consider three consecutive points. Set x=ae, y=a, z=a+e. Then you get f^2(y)  f^2(e) = f(x)*f(z) which you can solve for f(z) as f(z) = (f^2(y)  f^2(e))/f(x). So you can calculate the next point if you know the last two. This means there are only two degrees of freedom, and f(e) and f(2e) determine everything. One degree of freedom is multiplication by a constant. The other is a bit trickier: if three consecutive points curve toward the x axis (assume they're in a section where sign of f doesn't change), you get functions like sin(x*c)/c (larger values of c for increasing curvature). If they're in a line, you get f(x)=x. For curvature away from x axis, sinh(x*c)/c. Since you can fit one of these types to any possible points (0,0), (e, f(e)), (2e, f(2e)), there are no more classes. Last fiddled with by uau on 20190109 at 16:03 Reason: forum ate paragraph divisions 
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