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Thread: Limit (5)

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    Limit (5)

    Define the sequence $\displaystyle \{x_n \}$ by $\displaystyle x_1=\frac{\pi}{2}$ and $\displaystyle x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty}\sqrt{n} x_n.$
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    Non-rigorous proff (solution)

    Quote Originally Posted by NonCommAlg View Post
    Define the sequence $\displaystyle \{x_n \}$ by $\displaystyle x_1=\frac{\pi}{2}$ and $\displaystyle x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty}\sqrt{n} x_n.$
    Assume that the limit exists and is equal to $\displaystyle A$. Substitute $\displaystyle A$ in the recurrence relation $\displaystyle \mathop{\lim }\limits_{n \to \infty } \frac{A}{{\sqrt {n + 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin A}}{{\sqrt n }}$. Then use the Taylor's formula for sinus.
    So we have

    $\displaystyle {A^2} = 6\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }} - \frac{1}{{\sqrt {n + 1} }}} \right)\sqrt {{n^3}} = \ldots = 3.$

    $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt n {x_n} = \sqrt 3 .$
    Last edited by DeMath; Jul 31st 2009 at 11:53 AM. Reason: typo
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  3. #3
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    Define $\displaystyle y_n=\sqrt{n} x_n$, then $\displaystyle y_{n+1}=\sqrt{n+1} \sin \frac {y_n}{\sqrt{n}}$

    If
    $\displaystyle \lim_{n\to\infty}y_n$ exists, it would indicate:
    (1)
    $\displaystyle \lim_{n\to\infty} \frac {y_n}{\sqrt{n}} = 0$
    (2) Series $\displaystyle z_1=y_1$, $\displaystyle z_{n+1}=y_{n+1}-y_n$ converges, since $\displaystyle \sum_{i=1}^{n+1} z_i=$$\displaystyle y_{n+1}$.

    $\displaystyle z_{n+1}=\sqrt{n+1} \sin \frac{y_n}{\sqrt{n}}-y_n$
    $\displaystyle = \sqrt{n+1} \{ \sin \frac{y_n}{\sqrt{n}} - \frac{y_n}{\sqrt{n}} + \frac{y_n^3}{6n\sqrt{n}} \}+\frac{y_n}{n} \{\sqrt{n^2+n}-n-\frac{y_n^2}{6}\sqrt{\frac{n+1}{n}} \}$

    The first term converges since it is $\displaystyle O(\frac{1}{n^2})$. The second term is $\displaystyle \frac{y_n(3-y_n^2)}{6}\frac{1}{n}+O(\frac{1}{n^2})$.

    For the second term to converge, it requires $\displaystyle \lim_{n\to\infty}y_n(3-y_n^2)=0$ (limit comparison test against the p-series), so $\displaystyle \lim_{n\to\infty}y_n=\sqrt{3}$, noting $\displaystyle \lim_{n\to\infty}y_n>0$. This also confirms $\displaystyle \lim_{n\to\infty}y_n$ does exist.

    This method also applies to limit (1).


    Last edited by mr fantastic; Sep 19th 2009 at 12:33 AM. Reason: Restored original reply and fixed some latex
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    It is clear that $\displaystyle x_n>0$ and that $\displaystyle x_{n+1}=\sin x_n<x_n$ for all $\displaystyle n\geq1$. Therefore $\displaystyle x_n$ is decreasing, bounded below and converges to $\displaystyle L$, say.

    Since $\displaystyle L=\sin L$ follows from the recurrence we must have $\displaystyle L=0$. Thus $\displaystyle x_n\to 0$.

    Now $\displaystyle x_n\to 0$ implies that $\displaystyle \frac{x_{n+1}}{x_n}=\frac{\sin x_n}{x_n}\to 1$.

    Now consider $\displaystyle \frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^2-\sin^2 x_n}{x_n^2x_{n+1}^2}$. By Taylor's theorem, $\displaystyle \sin^2 x=x^2-\frac{x^4}3+o(x^5)$ and so

    $\displaystyle
    \frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^4/3+o(x_n^5)}{x_n^2x_{n+1}^2}=\frac{x_n^2}{x_{n+1}^2 }\left(\frac13+o(x_n)\right)\to\frac13$.

    Thus $\displaystyle \frac{(n+1)-n}{\frac1{x_{n+1}^2}-\frac1{x_n^2}}\to 3$. By the Stolz-Cesaro theorem, $\displaystyle \frac n{\frac1{x_n^2}}\to 3$, i.e. $\displaystyle nx_n^2\to 3$. Thus $\displaystyle \lim_{n\to\infty}\surd n x_n=\surd 3$.

    By the way, the limit $\displaystyle \lim_{n\to\infty}\left(\frac1{\sin^2 x_n}-\frac1{x_n^2}\right)$ can be obtained from l'Hopitals rule. Since $\displaystyle \lim_{x\to 0}\frac{\sin x}x=1$ we only need to find $\displaystyle \lim_{x\to 0}\left(\frac{x^2-\sin^2 x}{x^4}\right)$. Thus

    $\displaystyle \lim_{x\to 0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\lim_{x\to 0}\frac{x^2-\sin^2 x}{x^4}=\lim_{x\to 0}\frac{2x-\sin 2x}{4x^3}$ $\displaystyle =\lim_{x\to 0}\frac{2-2\cos 2x}{12x^2}=\lim_{x\to 0}\frac{4\sin 2x}{24x}=\lim_{x\to 0}\frac{8\cos 2x}{24}=\frac13$.
    Last edited by halbard; Aug 8th 2009 at 07:01 AM. Reason: Additional thought
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  5. #5
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    Quote Originally Posted by halbard View Post
    It is clear that $\displaystyle x_n>0$ and that $\displaystyle x_{n+1}=\sin x_n<x_n$ for all $\displaystyle n\geq1$. Therefore $\displaystyle x_n$ is decreasing, bounded below and converges to $\displaystyle L$, say.

    Since $\displaystyle L=\sin L$ follows from the recurrence we must have $\displaystyle L=0$. Thus $\displaystyle x_n\to 0$.

    Now $\displaystyle x_n\to 0$ implies that $\displaystyle \frac{x_{n+1}}{x_n}=\frac{\sin x_n}{x_n}\to 1$.

    Now consider $\displaystyle \frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^2-\sin^2 x_n}{x_n^2x_{n+1}^2}$. By Taylor's theorem, $\displaystyle \sin^2 x=x^2-\frac{x^4}3+o(x^5)$ and so

    $\displaystyle
    \frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^4/3+o(x_n^5)}{x_n^2x_{n+1}^2}=\frac{x_n^2}{x_{n+1}^2 }\left(\frac13+o(x_n)\right)\to\frac13$.

    Thus $\displaystyle \frac{(n+1)-n}{\frac1{x_{n+1}^2}-\frac1{x_n^2}}\to 3$. By the Stolz-Cesaro theorem, $\displaystyle \frac n{\frac1{x_n^2}}\to 3$, i.e. $\displaystyle nx_n^2\to 3$. Thus $\displaystyle \lim_{n\to\infty}\surd n x_n=\surd 3$.

    By the way, the limit $\displaystyle \lim_{n\to\infty}\left(\frac1{\sin^2 x_n}-\frac1{x_n^2}\right)$ can be obtained from l'Hopitals rule. Since $\displaystyle \lim_{x\to 0}\frac{\sin x}x=1$ we only need to find $\displaystyle \lim_{x\to 0}\left(\frac{x^2-\sin^2 x}{x^4}\right)$. Thus

    $\displaystyle \lim_{x\to 0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\lim_{x\to 0}\frac{x^2-\sin^2 x}{x^4}=\lim_{x\to 0}\frac{2x-\sin 2x}{4x^3}$ $\displaystyle =\lim_{x\to 0}\frac{2-2\cos 2x}{12x^2}=\lim_{x\to 0}\frac{4\sin 2x}{24x}=\lim_{x\to 0}\frac{8\cos 2x}{24}=\frac13$.
    Great. Limit (5) and Limit (1) are problems of the same type. So can be solved in the same way.

    Can Stolz-Cesaro theorem be used for complex numbers? What happens if $\displaystyle x $ is in complex domain?
    Last edited by mr fantastic; Sep 19th 2009 at 12:30 AM. Reason: Restored original reply
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    Quote Originally Posted by NonCommAlg View Post
    Define the sequence $\displaystyle \{x_n \}$ by $\displaystyle x_1=\frac{\pi}{2}$ and $\displaystyle x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty}\sqrt{n} x_n.$
    It would be interesting to investigate when $\displaystyle x$ is defined in complex domain, i.e.

    Define the sequence $\displaystyle \{z_n \}$ by $\displaystyle z_1=1+i$ and $\displaystyle z_{n+1}=\sin z_n, \ n \geq 1.$ Assuming the limit exists, evaluate $\displaystyle \lim_{n\to\infty}\sqrt{n} z_n$, and find how the initial value of $\displaystyle z_1 $ influences of the value of the limit.

    Actually it can be found that the limit also exists except for some initial values of $\displaystyle z_1$. If the limit does exist, then it is either $\displaystyle \sqrt{3}$ or $\displaystyle -\sqrt{3}$ for $\displaystyle z_1 \neq 0$, depending on the initial value of $\displaystyle z_1$, even when $\displaystyle z_n$ is a complex number.
    Last edited by mr fantastic; Sep 19th 2009 at 12:31 AM. Reason: Restored original reply
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