From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Trig Problem
Date: 27 Apr 1995 19:22:49 GMT

In article <3nj3rp$7ld$1@mhadg.production.compuserve.com>,
Mark Whittington  <100566.3172@CompuServe.COM> wrote:
>I wonder if any can help with a seemingly intractable pron^?^?^?^?^?^?^[[Dblem^R^R^R.

You really need to fix that back-space key. This is how your post looked
(I edited it to appear just as odd on your screen).

>We need to know the maximum possible amplitude of two superimposed sine waves
>of variable phase(p), amplitude(A) and angular frequency(f).
>
>In other words, the maximum possible amplitude (between 0 and A1+A2) of
>
>A1.SIN(f1.t + P1) + A2.SIN(f2.t + P2)

It depends on whether the ratio  f2/f1  is rational or not, but essentially
the answer is   A1+A2.

If  f2/f1 = n/m  in lowest terms, then  f1/m = f2/n  is some number,  f, say.
In that case, your function looks like

	A1.SIN(m.u + P1) + A2.SIN(n.u + P2)

where  u = f.t. In this case, the function is periodic with period  2 pi
(for  u); you can compute the maximum for all  t  just by finding the maximum
for  u  in  [0, 2 pi]. There are only a finite number of local maxima here
(something like  n*m,  I guess), so you can look for them graphically and
compute them numerically. Alternatively, you can go algebraic: use
the angle-addition formulas to write the above expression as a (very large)
polynomial in  sin(u)  and  cos(u)  (involving  A1, A2, and  sin  and  cos
of  P1  and P2). Maple will expand these for you, for example. Then from
cos^2(x) = 1 - sin^2(x), you can write the whole thing as a function
	Q1(s) + sqrt(1-s^2) Q2(s)
where  Q1  and  Q2  are polynomials in  s  (=sin(u)). This should be
even easier to maximize numerically, since there are no trig
function-calls.

If  f2/f1  is irrational, then there are values of  t
you can choose which will make _both_ of  f1.t + P1  and  f2.t + P2  be
as close as you like to  pi/2 + 2.k.pi  for some integer k  (i.e. k1 or k2);
thus, you can make each of the  SIN  terms as close to  1  as you like,
giving a value for the sum close to  A1+A2.

Indeed, in this case the set of integer multiples  (f2/f1).k  is dense in
the unit circle, so you can choose an integer  k  making this product
be as close as desired to the following magic number:
	-(P2/2pi - 1/4) + (f2/f1) (P1/2pi - 1/4)
(mod  1), say
	(f1/f2).k = n - (P2/2pi - 1/4) + (f2/f1) (P1/2pi - 1/4) + epsilon
for some integer  n  and very small epsilon. Then we let
	t = (1/f1).( -P1 + pi/2 + 2.k.pi ),   
so that  SIN(f1.t + P1) = SIN(pi/2) = 1. On the other hand,
	f2.t+P2 = (f2/f1).k.2pi + (f2/f1)(-P1+pi/2) + P2
	        = n.2pi + pi/2 + 2pi.epsilon
so SIN(f2.t+P2) = SIN(pi/2 + 2pi.epsilon)  is as close to  1  as you want.

The case in which  f2/f1 = n/m  is rational but with  m  and  n  large 
resembles the case with  f2/f1  irrational: the multiples  (f2/f1).k will
occupy  m  evenly-spaced points on the unit circle  R/Z, so that you
can get rather close to the magic number, though you can't make epsilon
as small as you want even by taking large  k.

You can discover a relation between  epsilon  and  k  by working with
rational approximations to  f2/f1, for example using the continued-fraction
expansion of  f2/f1.

dave