 mersenneforum.org A "difference of two squares" conundrum
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 Register FAQ Search Today's Posts Mark Forums Read 2019-01-07, 16:13 #1 Dr Sardonicus   Feb 2017 Nowhere 22·3·499 Posts A "difference of two squares" conundrum It is a well-known identity that a^2 - b^2 = (a - b)*(a + b). Once upon a time, long long ago, I happened to notice a non-trivial 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?   2019-01-09, 06:39 #2 LaurV Romulan Interpreter   "name field" Jun 2011 Thailand 2·5,059 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.    2019-01-09, 12:44   #3
Dr Sardonicus

Feb 2017
Nowhere

10111011001002 Posts Quote:
 Originally Posted by LaurV 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. We?

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 End-of-Tracks Town.   2019-01-09, 15:38 #4 uau   Jan 2017 2·3·52 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=a-e, 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 2019-01-09 at 16:03 Reason: forum ate paragraph divisions  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MooMoo2 Other Chess Games 5 2016-10-22 01:55 wildrabbitt Miscellaneous Math 11 2015-03-06 08:17 jinydu Programming 6 2005-10-01 00:09 nitai1999 Software 7 2004-08-26 18:12

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