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Consider the following transcendental equation:

$$\phi = \beta \sin \phi . \qquad (*)$$

How does one generate a description of how $\phi$ depends on $\beta$?

My attempt

From inspection (i.e. drawing pictures) one finds that for $\beta<1$ there is a single solution: $\phi=0$. At $\beta=1$ the solution at $\phi=0$, $\beta=0$ splits into three solutions, and then as $\beta$ increases additional solutions continue to appear in pairs.

The only technique I know is to assume that $\phi$ is implicitly given as a function of $\beta$. One then writes this function as a Taylor series and solves for the coefficients up to desired order.

Supposing $\beta$ is a function of $\phi$ we write a Taylor series $$\beta(\phi) = a\phi + b\phi^2 + \cdots.$$ Using this and expanding the $\sin$ function in $(*)$ we find

$$ \begin{align} \phi &= (a\phi + b\phi^2 + \cdots)(\phi - \phi^3/6 \cdots) \\ \phi &= a\phi^2 + b\phi^3 - \frac{a}{6}\phi^4 - \frac{b}{6}\phi^5 + \cdots \end{align} $$

There are no solutions other than $\phi=0$, $\beta=0$, so we have not captured the new pair of solutions which appears as $\beta$ increases past 1. This is strange. The Taylor series could have failed because $d\beta/d\phi$ is diverging at our expansion point. To check this, differentiate $(*)$ with respect to $\phi$ to find $$\frac{d\beta}{d\phi} = \frac{1 - \beta(\phi) \cos\phi}{\sin\phi}.$$ As $\phi$ approaches 0 from the right, we have $d\beta/d\phi = (1-\beta(\phi))/\phi$. This may or may not diverge; it depends on the functional form of $\beta(\phi)$.

The same ambiguity shows up if one instead regards $\phi$ as an implicit function of $\beta$.

From a numerical solution I can see that in fact $d\phi/d\beta = \infty$ at $\beta=1$. That suggests that the failure of the Taylor series for $\beta$ in terms of $\phi$ was not due to diverging derivative. This suggests that $\beta$ cannot be written as a Taylor series in $\phi$ for a different reason, e.g. fractional powers are needed. How then, does one proceed?

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You are essentially looking for an inverse to the $\operatorname{sinc}$ function ($\operatorname{sinc}(z)=\frac{\sin z}{z}$), since then you could write $\frac{1}{\beta}=\frac{\sin\phi}{\phi}\implies\phi=\operatorname{sinc}^{-1}\left(\frac{1}{\beta}\right)$. But see this SE post for people's thoughts on that.

You can generally find solutions to equations like this to arbitrary precision using numerical methods though.

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  • $\begingroup$ Any idea why the Taylor series didn't work? $\endgroup$ – DanielSank Oct 11 '14 at 16:44
  • $\begingroup$ For several reasons. Starting small, you aren't using the correct series for $\sin$ (alternating terms). But that doesn't matter because there are bigger issues. The graph of $\operatorname{sinc}$ should make it clear that $\operatorname{sinc}$ is not invertible on $(-a,a)$. It is, however, invertible on $[0,a)$ if $a$ doesn't get too large. But it's derivative at $0$ is $0$, so it's inverse at $y=1$ would be vertical. So if you set out to find a Taylor series for $\operatorname{sinc}^{-1}$ centered at $0$, you will not be able to find one that converges beyond radius $1$. $\endgroup$ – alex.jordan Oct 11 '14 at 18:19
  • $\begingroup$ Actually, perhaps the biggest issues is that you will only have one branch of a function that has infinitely many branches. $\endgroup$ – alex.jordan Oct 11 '14 at 18:38
  • $\begingroup$ Sure, but suppose I only care about following one branch's evolution as $\beta$ increases. For $\beta<1$ there is one solution, $\phi=0$. Then at $\beta=1$ this bifurcates into three branches. One remains at $\phi=0$ and the other two are mirrored about the origin. Suppose I just care about following one of those branches... $\endgroup$ – DanielSank Oct 11 '14 at 18:57
  • $\begingroup$ How do you get from $\phi=a\phi^2 + b\phi^3 - \frac{a}{6}\phi^4 - \frac{b}{6}\phi^5 + \cdots$ to "There are no solutions other than $\phi=0$"? Depending on $a,b,$ etc. there could be more solutions. And for $a,b,$ etc. that do give a local Taylor series for other branches of $\operatorname{sinc}^{-1}$, there would be other solutions. Also, those other branches' Taylor series would have a nonzero constant term. $\endgroup$ – alex.jordan Oct 11 '14 at 19:41

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