Calculating the number of intersections between sine and cosine curves

ShiningArcanine · 2006-07-29, 03:52

In my math classes, I tend to think of better methods of calculating things than the methods my teachers have shown me. When I have described my methods to math teachers, they said that someone else probably thought of them before I did, but whenever I read about math history, I find that people who came up with surprisingly simple things (like the child who came up with n2 = 2n1 - n0 + 2 for finding perfect squares) were the first to think about them, so today I figured that it might be worth a try to post on a math message board one of the formulas I have created and ask if anyone has done it before and whether or not it is useful.

My math teacher said that to find the number of intersections for sine and cosine curves in a given range, one should plot them, circle the intersections and then count them to find the answer. I did not like that method, so I came up with the formula f1 * f2 * (r / 2 Pi)^2 to calculate it. f1 is frequency one, f2 is frequency 2 and r is the range in radians. It only works when the result is an integer and the range is above a certain size; it has been a while so I cannot recall what the exact range was although I think that 2 Pi was the minimum. The formula is also expandable to utilize mutiple frequencies over mutiple ranges as (r / 2 Pi) is raised to the power of the number of frequencies, so it becomes possible to calculate the number of intersections between 4 sine and/or cosine curves with the derivative f1 * f2 * f3 * f4 * (r / 2 Pi)^4. In reflection, I must state that all curves must be unique such that you cannot have two sine or cosine curves with the same frequency.

Has this been done before and is it useful?

alpertron · 2006-07-31, 20:43

You are trying to find:

$\large \cos \,\frac{t f_1}{2\pi}\, - \,\sin \frac{t f_2}{2\pi} = 0$

in the range [0, 2*pi)

Since $\cos \,u \,- \,\cos v = -2 \sin\frac{u+v}{2} \,\sin\frac{u-v}{2}$

(see for example http://aleph0.clarku.edu/~djoyce/jav...dentities.html )

we get:

$\large -2 sin(t\frac{f_1+f_2}{4\pi}) \,sin(t\frac{f_1-f_2}{4\pi} - \frac{\pi}{4}) = 0$

The number of zeros (counting multiplicity) is the sum of zeros of both sines.

The number of zeros of the first sine is $2(f_1+f_2)$ .
The number of zeros of the second sine is $2|f_1-f_2|$ .

Adding them we have: $2 max(f_1, f_2)$ .

For example, when the frequences are 7 and 5, the number of intersections in [0, 2*pi) is 14.

ShiningArcanine · 2006-08-06, 21:55

I guess that it has been done before. Thanks for demonstrating that.

2006-07-29, 03:52	#1
ShiningArcanine Dec 2005 2²·23 Posts	Calculating the number of intersections between sine and cosine curves In my math classes, I tend to think of better methods of calculating things than the methods my teachers have shown me. When I have described my methods to math teachers, they said that someone else probably thought of them before I did, but whenever I read about math history, I find that people who came up with surprisingly simple things (like the child who came up with n2 = 2n1 - n0 + 2 for finding perfect squares) were the first to think about them, so today I figured that it might be worth a try to post on a math message board one of the formulas I have created and ask if anyone has done it before and whether or not it is useful. My math teacher said that to find the number of intersections for sine and cosine curves in a given range, one should plot them, circle the intersections and then count them to find the answer. I did not like that method, so I came up with the formula f1 * f2 * (r / 2 Pi)^2 to calculate it. f1 is frequency one, f2 is frequency 2 and r is the range in radians. It only works when the result is an integer and the range is above a certain size; it has been a while so I cannot recall what the exact range was although I think that 2 Pi was the minimum. The formula is also expandable to utilize mutiple frequencies over mutiple ranges as (r / 2 Pi) is raised to the power of the number of frequencies, so it becomes possible to calculate the number of intersections between 4 sine and/or cosine curves with the derivative f1 * f2 * f3 * f4 * (r / 2 Pi)^4. In reflection, I must state that all curves must be unique such that you cannot have two sine or cosine curves with the same frequency. Has this been done before and is it useful? Last fiddled with by ShiningArcanine on 2006-07-29 at 04:00 Reason: Removed details concerning my current level of education; Clarifications

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2006-07-31, 20:43	#2
alpertron Aug 2002 Buenos Aires, Argentina 1446₁₀ Posts	You are trying to find: $\large \cos \,\frac{t f_1}{2\pi}\, - \,\sin \frac{t f_2}{2\pi} = 0$ in the range [0, 2pi) Since $\cos \,u \,- \,\cos v = -2 \sin\frac{u+v}{2} \,\sin\frac{u-v}{2}$ (see for example http://aleph0.clarku.edu/~djoyce/jav...dentities.html ) we get: $\large -2 sin(t\frac{f_1+f_2}{4\pi}) \,sin(t\frac{f_1-f_2}{4\pi} - \frac{\pi}{4}) = 0$ The number of zeros (counting multiplicity) is the sum of zeros of both sines. The number of zeros of the first sine is $2(f_1+f_2)$ . The number of zeros of the second sine is $2\|f_1-f_2\|$ . Adding them we have: $2 max(f_1, f_2)$ . For example, when the frequences are 7 and 5, the number of intersections in [0, 2pi) is 14.

2006-08-06, 21:55	#3
ShiningArcanine Dec 2005 1011100₂ Posts	I guess that it has been done before. Thanks for demonstrating that.