Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

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

How can we find

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

and

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

?

share|improve this question
1  
You mean $2x_{n+1}$ instead of $2_{n+1}$? –  Pedro Tamaroff Oct 11 '12 at 22:45
1  
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

5 Answers 5

Interesting. . . What was the source for this?

share|improve this answer
    
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

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.

share|improve this answer

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.

share|improve this answer

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

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

share|improve this answer
    
What does the function O represent? –  Ali Oct 25 '12 at 13:30
1  
@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

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.