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I would like to solve the following quadratic equation to get a "nice" analytic solution for $\rho$.

$\rho^2(r\sin\theta-2nr^2)+\rho(2nr^3-2r^2\sin\theta-2\sin\theta+2nr)-2nr^2+3r\sin\theta=0$

where $r=1-\cos\theta$ (I cannot see how this could be used to simplify the quadratic equation.)

I would hope that the solution to be found which be of a simple form as that would correspond nicely to the larger problem I am working on.

Also $\theta=\pi/n$

EDIT: I would hope that the solution $\displaystyle \rho=1-\frac{sin(\frac{\pi}{n})}{n}$ would be one of the solutions to the quadratic.

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    $\begingroup$ One sort-of nice thing about $r=1-\cos\theta$ is that it's also equal to $2\sin^2\frac{\theta}{2}$. That may be easier to juggle. $\endgroup$ Aug 5, 2014 at 18:34
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    $\begingroup$ @Semiclassical even better, put everything into half-angle form, it helps a bit. Having both $r$ and $\sin \theta$ makes it impossible to group them correctly. $\endgroup$
    – orion
    Aug 5, 2014 at 18:48
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    $\begingroup$ Let $\displaystyle\;z = \frac{\sin\theta}{2nr - 2\sin\theta}\;$, we can rewrite the equation as $$\rho^2 (1 + z) - \rho (r+1/r) + (1-z) = 0$$ The two roots have the form: $$\rho_{\pm} = \frac{r + 1/r \pm \sqrt{ (r+1/r)^2 - 4(1-z^2)}}{2(1+z)} = \frac{r + 1/r \pm \sqrt{ (r-1/r)^2 + 4z^2}}{2(1+z)} $$ Let $t = \tan\frac{\theta}{2}$, we have: $$r = 1 - \frac{1-t^2}{1+t^2} = \frac{2t^2}{1+t^2} \quad\text{ and }\quad z = \frac{\frac{2t}{1+t^2}}{2n\frac{2t^2}{1+t^2} - \frac{4t}{1+t^2}} = \frac{1}{2(nt-1)}$$ Whether you consider the final expression in $t$ is simple is up to you. $\endgroup$ Aug 9, 2014 at 19:51
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    $\begingroup$ If $\rho = 1-\sin(\pi/n)/n$ were a solution, then you would be able to substitute the value into the equation and get zero. However, what you get is a complicated relation involving $n$ and trigonometric functions of $\pi/n$. This relation may be true for some $n$, but it's not true for all $n$ (for instance, $n=1$ and $n=2$ already fail). $\endgroup$
    – Blue
    Aug 15, 2014 at 9:41
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    $\begingroup$ The value $\rho = 1 - \sin(\pi/n)/n$ is not a root for $n=3$, $4$, $6$, $17$, $55$, or $123$, either. I'd be rather surprised if it were a root for any particular integer $n$. $\endgroup$
    – Blue
    Aug 15, 2014 at 13:14

2 Answers 2

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It is not a particularly simple form.

Solution:

$\rho = (R \pm S)/Q$

Where

$Q = \textrm{ sin}( \frac{\pi}{2 n} )^3( 32 n - 16 \textrm{ cot}(\frac{\pi}{2 n}))$

$R = -12 \textrm{ cos}(\frac{\pi}{2 n}) + 6 \textrm{ cos}(\frac{3 \pi}{2 n} ) - 2 \textrm{ cos}( \frac{5 \pi}{2 n}) + 2 n \left(14 \textrm{ sin}( \frac{\pi}{2 n}) - 5 \textrm{ sin}( \frac{3\pi}{2 n}) + \textrm{ sin}( \frac{5\pi}{2 n}) \right)$

$S = \sqrt{2} \sqrt{22 - 58 n^2 + 6 (1 + 17 n^2) \textrm{ cos}( \frac{\pi}{n} ) - 56 n^2 \textrm{ cos}(\frac{2\pi}{n}) + (9 + 11 n^2) \textrm{ cos}(\frac{3\pi}{n}) + 2 (-3 + n^2) \textrm{ cos}(\frac{4\pi}{n}) - (-1 + n^2) \textrm{ cos}(\frac{5\pi}{n}) - 20 n \textrm{ sin}( \frac{\pi}{n}) + 8 n \textrm{ sin}( \frac{2\pi}{n}) - 22 n \textrm{ sin}(\frac{3\pi}{n}) + 12 n \textrm{ sin}( \frac{4\pi}{n} ) - 2 n \textrm{ sin}( \frac{5\pi}{n})} $

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I don't think that it is so easy as to solve it, it would be possible if we had the expression leading to this which is relatively simple and we could get simple closed expression,Anyways if you're asking for a solution:


Solving in wolfram alpha at here I got the following two roots:

Let $$\cos(\lambda\pi/n)=c_{\lambda}$$ And similiarly $$\sin(\lambda\pi/n)=s_{\lambda}$$ Now let $$\color{red}{S_Q}=62n^2-106n^2c_1+72n^2c_2-37n^2c_3\\+10n^2c_4-n^2c_5-36ns_1+32ns_2-38ns_3\\+16ns_4-2ns_5+6c_1+9c_3+c_5+22$$ And $$\color{red}{D_N}=-12ns_1+4ns_3+c_1-c_3$$ And $$\color{red}{R_D}=-14ns_1+5ns_3-ns_5+6c_1-3c_3+c_5$$ Then roots are: $$\rho_1=\frac{\pm1}{\sqrt{2}\color{red}{D_N}}[\sqrt{\color{red}{S_Q}}-\color{red}{R_D}]$$

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