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Old 2019-01-07, 16:13   #1
Dr Sardonicus
 
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Default 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?
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Old 2019-01-09, 06:39   #2
LaurV
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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.
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Old 2019-01-09, 12:44   #3
Dr Sardonicus
 
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Quote:
Originally Posted by LaurV View Post
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.
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Old 2019-01-09, 15:38   #4
uau
 
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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
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