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Math Help - Limit (5)

  1. #1
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    Limit (5)

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

    Quote Originally Posted by NonCommAlg View Post
    Define the sequence \{x_n \} by x_1=\frac{\pi}{2} and x_{n+1}=\sin x_n, \ n \geq 1. Evaluate \lim_{n\to\infty}\sqrt{n} x_n.
    Assume that the limit exists and is equal to A. Substitute A in the recurrence relation \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

    {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.

    \mathop {\lim }\limits_{n \to \infty } \sqrt n {x_n} = \sqrt 3 .
    Last edited by DeMath; July 31st 2009 at 12:53 PM. Reason: typo
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  3. #3
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    Define y_n=\sqrt{n} x_n, then  y_{n+1}=\sqrt{n+1} \sin \frac {y_n}{\sqrt{n}}

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

    z_{n+1}=\sqrt{n+1} \sin \frac{y_n}{\sqrt{n}}-y_n
    = \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 O(\frac{1}{n^2}). The second term is \frac{y_n(3-y_n^2)}{6}\frac{1}{n}+O(\frac{1}{n^2}).

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

    This method also applies to limit (1).


    Last edited by mr fantastic; September 19th 2009 at 01:33 AM. Reason: Restored original reply and fixed some latex
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    It is clear that x_n>0 and that x_{n+1}=\sin x_n<x_n for all n\geq1. Therefore x_n is decreasing, bounded below and converges to L, say.

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

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

    Now consider \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, \sin^2 x=x^2-\frac{x^4}3+o(x^5) and so

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

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

    \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} =\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; August 8th 2009 at 08:01 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 x_n>0 and that x_{n+1}=\sin x_n<x_n for all n\geq1. Therefore x_n is decreasing, bounded below and converges to L, say.

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

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

    Now consider \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, \sin^2 x=x^2-\frac{x^4}3+o(x^5) and so

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

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

    \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} =\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  x is in complex domain?
    Last edited by mr fantastic; September 19th 2009 at 01:30 AM. Reason: Restored original reply
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    Define the sequence \{x_n \} by x_1=\frac{\pi}{2} and x_{n+1}=\sin x_n, \ n \geq 1. Evaluate \lim_{n\to\infty}\sqrt{n} x_n.
    It would be interesting to investigate when x is defined in complex domain, i.e.

    Define the sequence \{z_n \} by z_1=1+i and z_{n+1}=\sin z_n, \ n \geq 1. Assuming the limit exists, evaluate \lim_{n\to\infty}\sqrt{n} z_n, and find how the initial value of  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 z_1. If the limit does exist, then it is either \sqrt{3} or -\sqrt{3} for z_1 \neq 0, depending on the initial value of z_1, even when z_n is a complex number.
    Last edited by mr fantastic; September 19th 2009 at 01:31 AM. Reason: Restored original reply
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