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Consider a sequence $x_n, n\ge1$ formed by positive solutions to $x \sin{x}=1$.

How can we find

$$\lim _{n\rightarrow \infty}(n(x_{2n+1}-2\pi n))= L$$


$$\lim _{n\rightarrow \infty}(n^3(x_{2n+1}-2\pi n- \frac{L}{n}))= L_2$$


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You mean $2x_{n+1}$ instead of $2_{n+1}$? –  Pedro Tamaroff Oct 11 '12 at 22:45
I think he means $x_{2n+1}$ where he writes "2_{n+1}", otherwise the limit won't exists, I think. –  fgp Oct 11 '12 at 22:50
@Peter Fixed. It was a typo. –  Ali Oct 11 '12 at 22:59
I wonder if there is an elementary approach for it ... –  Chris's sis Oct 25 '12 at 18:19
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5 Answers

Let $y$ be a variable tending to zero. Put

$$ \begin{array}{l} a=y-\frac{5}{6}y^3, \\ b=y-\frac{5}{6}y^3+\frac{169}{120}y^5 \end{array} $$

Using Taylor expansions, one finds that

$$ \begin{array}{l} \sin(a)\bigg(\frac{1}{y}+a\bigg)=1-\frac{169}{120}y^4+O(y^5), \\ \sin(b)\bigg(\frac{1}{y}+b\bigg)=1+\frac{5021}{1680}y^6+O(y^7) \end{array} $$

so for small enough $y$ there will be a $c\in ]a,b[$ such that $\sin(c)\bigg(\frac{1}{y}+c\bigg)=1$. This $c$ will satisfy $c=y-\frac{5}{6}y^3+O(y^5)$.

Applying this to $y=\frac{1}{2\pi n}$ yields

$$ L_1=\frac{1}{2\pi}, \ L_2=\frac{-5}{6(2\pi)^3}=\frac{-5}{48\pi^3} $$

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What does the function O represent? –  Ali Oct 25 '12 at 13:30
@Ali: this is big-O notation. –  robjohn Oct 25 '12 at 18:07
This does work out cleaner as a product than a difference (+1) –  robjohn Oct 25 '12 at 18:30
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Here is a solution for the first limit without expansions:

First we need to find some facts about the sequence $x_n$:

  • note that in each interval $(2n\pi,(2n+1)\pi)$ we have two solutions $x_{2n+1}<x_{2n+2}$.

  • therefore $x_{2n+1}=2\pi n +\varepsilon_n$ where $\varepsilon_n <\pi/2$. This proves at once that $\displaystyle \frac{x_{2n+1}}{n} \to 2\pi$

  • furthermore we have $\sin \varepsilon_n =\sin(x_{2n+1})=\frac{1}{x_{2n+1}} \to 0$, and therefore $x_{2n+1}-2n \pi=\varepsilon_n \to 0$.

Now we are able to attack the first limit noting that $\sin y/y \to 1$ as $y \to 0$.

$$\lim_{n \to \infty} n(x_{2n+1}-2\pi n)= \lim_{n \to \infty} \frac{n}{x_{2n+1}}\cdot \frac{x_{2n+1}-2\pi n}{\sin(x_{2n+1}-2\pi n)}\cdot \frac{\sin x_{2n+1}}{\frac{1}{x_n}}=\frac{1}{2\pi}. $$

For the second limint, however, any method you choose will get you to write the expansion of $\sin$, because you need the higher order terms.

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The points at which $(2\pi n+x)\sin(2\pi n+x)=1$ are the the points where

$$ \begin{align} 0 &=\frac1{2\pi n+x}-\sin(x)\\ &=\left(\frac1{2\pi n}-\frac x{4\pi^2 n^2}+\frac{x^2}{8\pi^3n^3}-\frac{x^3}{16\pi^4n^4}+\dots\right)-\left(x-\frac{x^3}6+\dots\right)\\ &=\frac1{2\pi n}-\frac{4\pi^2n^2+1}{4\pi^2n^2}x+\frac1{8\pi^3n^3}x^2+\frac{8\pi^4n^4-3}{48\pi^4n^4}x^3+O\left(x^4\right)\tag{1} \end{align} $$ Thus, $$ \frac{2\pi n}{4\pi^2n^2+1} =x-\frac1{8\pi^3n^3+2\pi n}x^2-\frac{8\pi^4n^4-3}{48\pi^4n^4+12\pi^2n^2}x^3+O\left(x^4\right)\tag{2} $$ The inverse series for the series on the right-hand-side of $(2)$ is $$ x=y+\frac1{8\pi^3n^3+2\pi n}y^2+\frac{32\pi^6n^6+8\pi^4n^4-12\pi^2n^2+3}{12(4\pi^3n^3+\pi n)^2}y^3+O\left(y^4\right)\tag{3} $$ Plugging $y=\frac1{2\pi n}-\frac1{8\pi^3n^3}+O\left(\frac1{n^5}\right)$ from the left-hand-side of $(2)$ into $(3)$ yields $$ x=\frac1{2\pi n}-\frac5{48\pi^3n^3}+O\left(\frac1{n^4}\right)\tag{4} $$ The root in question is located at $$ 2\pi n+x=2\pi n+\frac1{2\pi n}-\frac5{48\pi^3n^3}+O\left(\frac1{n^4}\right)\tag{5} $$ and $(5)$ says that $L=\frac1{2\pi}$ and $L_2=-\frac5{48\pi^3}$ .

A second approach (similar to Ewan Delanoy)

There are two sequences of roots; one near $2n\pi$ $$ \begin{align} 1&=(2n\pi+x)\sin(2n\pi+x)\\ \frac1{2n\pi}&=\left(1+\frac{x}{2n\pi}\right)\sin(x)\tag{6} \end{align} $$ and one near $(2n+1)\pi$ $$ \begin{align} 1&=((2n+1)\pi+x)\sin((2n+1)\pi+x)\\ -\frac1{(2n+1)\pi}&=\left(1+\frac{x}{(2n+1)\pi}\right)\sin(x)\tag{7} \end{align} $$ To solve either $(6)$ or $(7)$, we will use the series $$ (1+px)\sin(x)=x+px^2-\frac16x^3-\frac p6x^4+O\left(x^5\right)\tag{8} $$ and its inverse $$ x=y-py^2+\frac{1\color{#C00000}{+12p^2}}{6}y^3\color{#C00000}{-\frac{2p+15p^3}{3}y^4}+O\left(y^5\right)\tag{9} $$ For $(6)$, we will use $y=p=\frac1{2n\pi}$, and for $(7)$, we will use $y=-p=-\frac1{(2n+1)\pi}$. For each of these, the terms in red contribute no more than the error term, so can be ignored.

Thus, for the $x$ in $(6)$, we get $$ x=p-\frac56p^3+O\left(p^5\right)\tag{10} $$ and for the $x$ in $(7)$, we get $$ x=-p-\frac76p^3+O\left(p^5\right)\tag{11} $$ Thus, the roots for $(6)$ are $$ 2n\pi+x=2n\pi+\frac1{2n\pi}-\frac5{48n^3\pi^3}+O\left(\frac1{n^5}\right)\tag{12} $$ and the roots for $(7)$ are $$ (2n+1)\pi+x=(2n+1)\pi-\frac1{(2n+1)\pi}-\frac7{6(2n+1)^3\pi^3}+O\left(\frac1{n^5}\right)\tag{13} $$ The roots in $(12)$ are the same as those in $(5)$ only with better error term. We can continue from there.

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I'll given you a informal derivation of $L$, and leave it to you to formalize that. Since you need $\sin x = \frac{1}{n}$ for $x\sin x = 1$ to hold, the $x_n$ will generally lie close to some zero of $\sin x$, and get closer the larger $n$ gets. To figure out to which zero of $\sin x$ some $x_n$ lies close, look at the first few $x_n$. You get (a plot of $x\sin x$ helps!) $$ \begin{eqnarray} x_1 &\leq& \frac{\pi}{2} \\ x_2 &\leq& \pi \\ x_3 &\approx& 2\pi \\ x_4 &\approx& 3\pi \\ \ldots \end{eqnarray} $$ Thus, for large n, you have $x_n \approx (n-1)\pi$. If you were just interested in $\lim_{x\to\infty} (x_{2n+1} - 2\pi n)$, that approximation would suffice to see that the limit is zero!

But since you need $\lim_{x\to\infty} (n(x_{2n+1} - 2\pi n))$, we need to add a first-order term to our approximation of $x_n$. Since the limit contains only odd $x_n$, we're interested only in zeros of the form $2\pi n$. The slope of $\sin x$ at those zeros is +1 which yields the following first-order approximation $$ \sin (2\pi n + \epsilon) \approx \epsilon $$ From that, you get the following updated approximation of $x_n$ for odd n $$ x_n \approx (n-1)\pi + \frac{1}{(n-1)\pi} $$ Observe that if you compute $x_n\sin x_n$ (again, for odd n) with the approximation of $\sin x$ above, you get $((n-1)\pi + \frac{1}{(n-1)\pi})\frac{1}{(n-1)\pi)} \approx 1$. If you put that approximation into your limit, you get $$ \lim_{n \to \infty} (n\frac{1}{2n\pi}) = \frac{1}{2\pi} $$

For the second problem, you'll need to add third-order terms to those approximations.

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it would be fantastic if you could complete the solution and formalize the definition. –  Ali Oct 12 '12 at 0:07
@Ali You should be able to do that easily. Instead of my hand-wavering "approximation" use the Taylor expansion of $\sin x$. –  fgp Oct 12 '12 at 0:09
Rather, I meant to ask whether you'd be able to write the second problem's solution out explicitly. It isn't coming together for whatever reason... –  Ali Oct 12 '12 at 0:14
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Interesting. . . What was the source for this?

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Doesn't answer the question, this is a comment. –  Simon Hayward Nov 23 '12 at 14:50
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  martini Nov 23 '12 at 15:13
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