# Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.

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• Feb 18th 2013, 10:22 PM
Rick66
Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Hello everyone,

I'm new to this forum so thanks for having me. I'm a musician and wave theorist currently working on a problem relating to tonality in music.

I need to find the extrema of the sum of 3 sine waves of whole-numbered frequencies A, B and C in lowest form i.e. f(t) = sin(2piAt) +sin(2piBt) +sin(2piCt) is periodic with a GCD of 1Hz.
The 1st derivative test for extrema gives

f'(t) = 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0

where I need to solve for t. Since the wave is periodic we can restrict the domain to 0 <= t <= 1. I've tried finding trig substitutions, reframing the problem in complex form etc but this one truly has me stumped. I would have pulled all my hair out by now if I had any. (Headbang)

If somebody could help me with this it would be greatly appreciated.

Thanks

Rick66
• Feb 19th 2013, 03:36 PM
topsquark
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Quote:

Originally Posted by Rick66
Hello everyone,

I'm new to this forum so thanks for having me. I'm a musician and wave theorist currently working on a problem relating to tonality in music.

I need to find the extrema of the sum of 3 sine waves of whole-numbered frequencies A, B and C in lowest form i.e. f(t) = sin(2piAt) +sin(2piBt) +sin(2piCt) is periodic with a GCD of 1Hz.
The 1st derivative test for extrema gives

f'(t) = 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0

where I need to solve for t. Since the wave is periodic we can restrict the domain to 0 <= t <= 1. I've tried finding trig substitutions, reframing the problem in complex form etc but this one truly has me stumped. I would have pulled all my hair out by now if I had any. (Headbang)

If somebody could help me with this it would be greatly appreciated.

Thanks

Rick66

Unless there is a geometry I'm not seeing here I think you are going to have to resort to numeric approximation.

-Dan
• Feb 20th 2013, 05:53 AM
Rick66
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Thanks Dan,

Yes I think you're right. Do you have any suggested methods of approximation I should look at? I've been thinking Taylor series of cos but I'm not sure if it would give other than the few points around zero.

-Rick
• Feb 20th 2013, 05:56 AM
Rick66
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
PS: Taking the 2nd degree Taylor for each cos term gave me only the two points at both sides of the origin.
• Feb 23rd 2013, 09:27 AM
Fumbles
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
This may be really stupid, but what about this:

$2 \pi A \cos (2 \pi A t) + 2 \pi B \cos (2\pi Bt)+ 2 \pi C \cos ( 2 \pi Ct)=0$

So then we can divide through by $2 \pi$:

$A \cos (2 \pi A t) + B \cos (2\pi Bt)+ C \cos ( 2 \pi Ct)=0$

This is 0 when each of the cos terms is $\frac{\pi}{ 2}$. ie when:

$2 \pi At = \frac{\pi}{2}$

$2 \pi Bt = \frac{\pi}{2}$

$2 \pi Ct = \frac{\pi}{2}$

Which gives:

$At= \frac{1}{4}$

$Bt= \frac{1}{4}$

$Ct= \frac{1}{4}$

So these are all equal to each other:

$At=Bt=Ct$

So $A=B=C$ which makes things a bit easier.
• Feb 23rd 2013, 09:51 AM
ILikeSerena
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Hi Rick66! :)

Since you only have sines with no phase shift, your absolute extreme is at a predictable location.
This is where all the sines simultaneously take their maximum value.
The absolute maximum is where the fractional parts of At, Bt, and Ct are simultaneously 1/4.
I have to think about how to solve that exactly with a neat formula.

A typical approximation method is Newton-Raphson - see wiki.

@Fumbles: I believe that A, B, and C are numbers that are given - they are not supposed to be solved. We want to solve only for t.
Furthermore, you are forgetting the (mod 2pi) term.
• Feb 23rd 2013, 10:12 AM
Rick66
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Thanks Fumbles,

Unfortunately, the three frequencies of the problem in question are not equal. It is true that setting for eg 2piAt = (2N+1)pi/2, N = 0,1,2..., gives t = (2N+1)/4A which is close to your method. But the problem then is that finding a particular N for each of the three frequencies is not guaranteed i.e. A/B might not equal odd/odd. It also doesn't give the other zero's.

But thanks anyway,

Rick66
• Feb 23rd 2013, 10:33 AM
Rick66
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Thanks ILikeSerena,

Yes you're correct about the absolute extreme. Since the frequencies are whole then this occurs at every GCD period of 1 sec. Its the other values that are eluding me.
Now when you say "combined frequency is the LCM" do you have something specific in mind? It's just that I tried to work with the LCM and didn't get very far. At any rate, thanks for the heads up concerning Newton-Raphson method which I'll look into right now.

Btw, by "eyeballing" the plots I realised that the zero's of the simpler function cos (2piAt) + cos (2piBt) + cos (2piCt) = 0 would probably work just as well. So if you or anyone knows the solution to this it'd be much appreciated.

Thanks

Rick66
• Feb 23rd 2013, 10:46 AM
ILikeSerena
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
Quote:

Originally Posted by Rick66
Thanks ILikeSerena,

Yes you're correct about the absolute extreme. Since the frequencies are whole then this occurs at every GCD period of 1 sec. Its the other values that are eluding me.
Now when you say "combined frequency is the LCM" do you have something specific in mind? It's just that I tried to work with the LCM and didn't get very far. At any rate, thanks for the heads up concerning Newton-Raphson method which I'll look into right now.

Btw, by "eyeballing" the plots I realised that the zero's of the simpler function cos (2piAt) + cos (2piBt) + cos (2piCt) = 0 would probably work just as well. So if you or anyone knows the solution to this it'd be much appreciated.

Thanks

Rick66

I just realized that the sines do not necessarily take on a maximum value simultaneous anywhere.
For instance sin(2pi.3t) and sin(2pi.5t) are never simultaneously both 1.
Now with cosines, that is a different matter.
The sum of 3 cosines take on their maximum value at t=0,1,2,...
The sum of 3 sines are simultaneously zero at t=0,1,2,...

I deleted the part of comment mentioning LCM, since it's slightly more complex.
What we have is that if the GCD(A,B,C)=1, that the period T is 1.
Generally, the period T = 1 / GCD(A,B,C) = LCM(A,B,C) / (ABC).
• Feb 23rd 2013, 10:57 AM
ILikeSerena
Re: Please help to solve; 2piAcos(2piAt) +2piBcos(2piBt) +2piCcos(2piCt) = 0.
To properly find all extrema in the interval [0,1], I think you should divide the interval in 2.A.B.C equal-length sub intervals.
In each sub interval [a,b] you should check if f'(a) and f'(b) have a different sign.
If they do, then f' has a root on [a,b].

With Newton-Raphson you can find the root of the derivative on each sub interval [a,b] as follows:

$t_0 = (a+b)/2$

$t_{k+1} = t_k - \frac {f'(t_k)} {f''(t_k)}$