Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for a solution for the equation $$\frac{\sin(\alpha \cdot x)}{\alpha}=\frac{\sin(\beta \cdot x)}{\beta}$$ where $\alpha$ and $\beta$ are constants.

How do I approach this?

share|improve this question
Even Wolfram is getting nowhere!! [1]:wolframalpha.com/input/… –  Eisen Nov 8 '11 at 17:51
Might be a silly suggestion, but if you're just looking for a solution, $x = 0$ fits the bill. –  Joel Cohen Nov 8 '11 at 18:26
add comment

2 Answers

Flip it around and plot contour lines of $\alpha$ vs. $\beta$ for various values of $x$. Ignoring the obvious $\alpha=\beta$ solutions, the resulting plot is a real beauty.

enter image description here

share|improve this answer
This might be more illuminating: goo.gl/tVf4R –  leonbloy Nov 8 '11 at 21:26
@leonbloy - you mean this: goo.gl/dxcxB The equation you have is not of the same form as the one from the original posting. Check the sin(x*y) vs. x*sin(y) –  ja72 Nov 9 '11 at 16:06
add comment

This answer assumes you are interested in the solutions of this equaton in the real line.

Using $u=\alpha x$ and $\gamma=\beta/\alpha$, the question is equivalent to finding the zeroes of the function $g$ defined by $$ g(u)=\gamma\sin(u)-\sin(\gamma u). $$ Assume without loss of generality that $\gamma>1$. Then, $\sin(u)=1$ implies $g(u)\geqslant\gamma-1>0$ and $\sin(u)=-1$ implies $g(u)\leqslant-\gamma+1<0$, hence $g$ has at least one zero in each interval $I_k$ of length $\pi$ around $k\pi$, for $k$ in $\mathbb Z$. A zero in $I_0$ is $u=0$.

There might be exactly one zero in each interval $I_k$. This is so in the simulations I performed but these are based on a very limited number of parameters $\gamma$ and of intervals $I_k$.

If $\gamma$ is a rational number $\gamma=m/n$ with $m$ and $n$ integers, then $2n\pi$ is a period of $g$ hence the zeroes in $I_{k+2n}$ are exactly those in $I_k$ translated by $2n\pi$. If $\gamma$ is irrational, I see no reason to suspect a similar regularity.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.