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Let $x_n$ be the sequence of increasing solutions to $x\sin{x} = 1$. Define $$a = \lim_{n \to \infty} n(x_{2n+1} - 2\pi n) $$

and $$b = \lim_{n \to \infty} n^3 \left( x_{2n+1} - 2\pi n - \frac{a}{n} \right) $$

These limits can be evaluated, but they appear to be the first limits in some sequence. How can we find a general formula for all the limits in this sequence?

(For example, $$c = \lim_{n \to \infty} n^5\left( x_{2n + 1} - 2\pi n - \frac{a}{n} - \frac{b}{n^3} \right) $$ might be the next in the sequence).

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  • $\begingroup$ Are you meaning that, for any $n$, $x_n$ is the solution of $x_n \sin (x_n)=1$ ? $\endgroup$ Mar 20, 2014 at 3:55
  • $\begingroup$ Ayesha, you say that the limits can be evaluated. Do u have a reference for these evaluations? $\endgroup$
    – Paramanand Singh
    Mar 20, 2014 at 5:31
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    $\begingroup$ The equation has an infinite number of positive solutions - $x_n$ denotes the $n$th positive solution. $\endgroup$
    – Ayesha
    Mar 20, 2014 at 12:16
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    $\begingroup$ You may also be interested in this question. $\endgroup$ Mar 22, 2014 at 14:34

1 Answer 1

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Because you are always evaluating the limit, this is an asymptotic expansion of the explicit expression for the solutions. Write

$$x=2\pi n +\epsilon$$ You get $$\sin \epsilon=\frac{1}{2\pi n +\epsilon}$$

Your first limit in this notation is $$a=\lim_{n\to\infty}n\epsilon$$ We are seeking the series for $\epsilon$ expanded in inverse powers of $n$.

$$z=\frac{1}{2\pi n} = \frac{1}{\frac{1}{\sin \epsilon}-\epsilon}=f(\epsilon)$$

Formally, this is done. Use the Lagrange Inversion Theorem to express the inverse of $f(\epsilon)$ as a power series of $z$:

$$\epsilon(z)=\sum \frac{z^k}{k!} \lim_{x\to 0}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{x}{f(x)}\right)^k$$ $$=\sum \frac{z^k}{k!} \lim_{x\to 0}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{x}{\sin x}-x^2\right)^k$$

The first term (k=1): $$\lim_{x\to 0}\left(\frac{x}{\sin x}-x^2\right)=1$$ The second term (k=2): $$\lim_{x\to 0}\frac{d}{dx}\left(\frac{x}{\sin x}-x^2\right)^2=$$ $$\lim_{x\to 0}2\left(\frac{x}{\sin x}-x^2\right)\left(\frac{\sin x-x \cos x}{\sin^2 x}-2x\right)=0$$

Mathematica says that the third term is $-5$ and fourth is again zero. You can write

$$\epsilon=\frac{1}{2\pi n}-\frac{5}{3!}\frac{1}{(2\pi n)^3}+\frac{169}{5!}\frac{1}{(2\pi n)^5}-\frac{15063}{7!}\frac{1}{(2\pi n)^7}\cdots$$

I realize that this is not a very elegant solution and the terms in expansion don't seem to follow any simple pattern - the function $f$ is too ugly. I did half of this on paper so if I made any stupid mistakes, please point them out.

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    $\begingroup$ Hmmm, this is great! I understand that the $n$th term in the sequence is simply the coefficient of the $n$th term in the asymptotic expansion, yes? So we have, for example, $$a = \frac{1}{2\pi}, b = -\frac{5}{48\pi^3}, c = \frac{169}{3840\pi^5}. . .$$ (Is this right?) So there's no way to find a recursion or some other pattern? $\endgroup$
    – Ayesha
    Mar 22, 2014 at 23:36
  • $\begingroup$ Exactly. You can confirm it by putting the expansion in your limits. You actually have a closed-form expression for each coefficient (just evaluate the limit of the derivative of the power for each $k$). If you manage to evaluate that limit for a general $k$, you'll get your pattern. However, Wolfram Alpha says "no result found in terms of standard functions" so it seems the function $f$ is too complicated to return anything reasonable. Just look at those factors. $\endgroup$
    – orion
    Mar 23, 2014 at 0:18

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