Dear friends,
I'm trying to solve the following problem: to know which is the greater value - and minimum - of the function f(x) = sin(a x) + sin(b x).
a and b will be the result of dividing 2 PI by different integers, both positive and different of zero.
I can't find the roots of the derivative of the function sadly...
I will eventually also try to find the same parameters of a function of the type: g(x) = sin(a x) + sin(b x) + sin(c x)
Is there a way around this problem?
Kind regards,
Kepler
The only way to do this, I believe, is just to go through the cases. Since the same basic techniques work for finding zeros of derivatives as for the function [you are just working with cosines instead of sines], I'll just go through a couple of steps for finding some zeros of the function. First, lets take advantage of the fact that a and b are 2\(\displaystyle \pi\)/n for some integer n and define a=1/N and b=1/M and write our function as
f(x) = sin(2\(\displaystyle \pi\)x/N) + sin(2\(\displaystyle \pi\)x/M)
First, the only way for the function to be zero is for (a) both of the sines to be zero or (b)for the sines to be the negative of one another.
(a) Both zero. For sin(ax) to be zero, ax [and bx] must be (ai)zero plus a multiple of 2\(\displaystyle \pi\) or (aii) \(\displaystyle \pi\) plus a multiple of 2\(\displaystyle \pi\). This gives the following four cases
CASE(ai for a - ai for b):
ax=2\(\displaystyle \pi\)x/N=2n\(\displaystyle \pi\) and bx=2\(\displaystyle \pi\)x/M=2m\(\displaystyle \pi\)
or
x = Nn = Mm or m and n are multiples of the least common multiples of N and M. Example N = 7, M = 5, then case (ai for a - ai for b) is for x = 35 n, n\(\displaystyle \epsilon\, \mathbb{Z}\).
...
Note: When it comes to the derivative, the zero's will involve ratios of a and b so that you will have initial angles you need to find which may not be so simple but the periodicy is still there. You might look at
http://www.wolframalpha.com/input/?i=7+cos(2+pi+x+/+5)+++5+cos(2+pi+x+/+7)
to get some ides.
EDIT:Fix some typos
