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I'm trying to classify (at least as fully as possible) the values of $t$ in $(0,\pi/2)$ for which the following equation has a solution for some natural number, $n$.

$$\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$$

Does anyone know of any results that will help me?

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@J.M.: I would suggest not using \displaystyle in question titles, as it is still typeset as large inline text, here and on the front page. –  Rahul Nov 24 '10 at 2:34
    
@Rahul: Done now. –  J. M. Nov 24 '10 at 2:36
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3 Answers

I don't know if there is any closed form for the solutions, but by replacing $\sin\big((n+2)t\big)$ with $\sin nt \cos 2t + \cos nt \sin 2t$ and rearranging, your problem is equivalent to finding the solutions of $$\tan nt = \frac{n \sin 2t}{n + 2 - n \cos 2t}.$$ For any $n > 0$, the RHS is bounded, while the LHS crosses from $-\infty$ to $\infty$ exactly $2n$ times over a $2 \pi$ interval. Some experimentation suggests that there are indeed exactly $2n$ solutions; possibly you could bound the slopes of the two expressions to make this rigorous. This would also let you find bounds for the locations of the solutions, and easily find them numerically through a bisection procedure.

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Maybe the Chebyshev polynomials are helpful here? –  Yuval Filmus Nov 24 '10 at 3:59
    
@Yuval: That's an interesting idea! I'm not experienced with working with Chebyshev polynomials, but glancing at Wikipedia, it seems to me that perhaps they might lead to a purely algebraic equation in the end. If so, that would be the really complete solution to the problem. But I haven't the energy to work out the details, so if you or someone else wants to go ahead they should please do so. –  Rahul Nov 24 '10 at 4:09
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If one is going the Chebyshev route, the OP's equation becomes $\frac{U_n(v)}{n}=\frac{U_{n+2}(v)}{n+2}$ where $v=\cos\;t$ . There are $n+2$ intersections within the interval $[-1,1]$ ; I've no time for further analysis so somebody might want to run away with this. –  J. M. Nov 24 '10 at 4:22
    
@J.M.: Very nice! I think there are $n+2$ intersections for $U_n(v) = U_{n+2}(v)$, but if you divide by $n$ and $n+2$ it seems to send the outermost roots outside $[-1,1]$. At least, $U_n(1)/n > U_{n+2}(1)/(n+2)$, so there is at least one root beyond 1, and similarly for the negative case. –  Rahul Nov 24 '10 at 4:39
    
@J.M. $\sin nt$ can be written as a polynomial of $\cos t$? That is not true for, say $n=2$. Or maybe you meant $U_n$ need not be polynomial? –  TCL Nov 24 '10 at 23:27
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EDIT: See J.M.'s comment to Rahul's answer: it's better to use the "other" Chebyshev polynomials.

Square your identity: $$(n+2)^2 \sin^2 nt = n^2 \sin^2 (n+2)t.$$ Convert from sines to cosines: $$(n+2)^2 + n^2 \cos^2 (n+2)t = n^2 + (n+2)^2 \cos^2 nt.$$ Now let $x = \cos t$ to get $$(n+2)^2 + n^2 T_{n+2}(x)^2 = n^2 + (n+2)^2 T_n(x)^2,$$ where the $T_m$ are the Chebyshev polynomials. We get a polynomial equation of degree $2(n+2)$ in $x$ which you can solve to find all solutions (you will need to check that the signs of the sines match).

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You're dealing with sines so $U$ is more "natural" to use than $T$ ; see my comment in Rahul's answer. –  J. M. Nov 24 '10 at 4:24
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EDIT: This solution is wrong! The error is highlighted below (thanks for Zaricuse and Rahul).

There is no solution.

Given $t$, define a function $$f(n) = \frac{\sin(nt)}{n}.$$ Its derivative is $$f'(n) = \frac{\cos(nt)nt - \sin(nt)}{n^2}.$$ If $f(n) = f(n+2)$ then the derivative must have a zero somewhere in between, say $f'(m) = 0$. That implies $$\tan(mt) = mt.$$ However, for $x \in (0,\pi)$ we have $\tan(x) > x$ (consider the Taylor expansion of $\tan$, for example). The unfortunate error is that although $t < \pi$, $x = mt$ is not necessarily less than $\pi$.

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$tan(x) = x$ does have solutions in its other branches, just not the one containing $(0,0)$. –  Zarrax Nov 24 '10 at 3:19
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I think $mt$ can be bigger than $\pi$ here. –  Zarrax Nov 24 '10 at 3:28
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Zaricuse is right. Consider that WolframAlpha can find solutions easily. –  Rahul Nov 24 '10 at 3:32
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