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Thread: A limit

  1. #1
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    A limit

    Compute $\displaystyle \lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$
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    Re: A limit

    Just toying with it a bit, I note:

    n = 4 results in something between 1 and 2 half-periods
    n = 7 results in something between 2 and 3 half-periods
    n = 13 results in something between 4 and 5 half-periods
    n = 26 results in something between 8 and 9 half-periods

    Seems like this could lead somewhere useful, at least getting bounds on it.
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    Re: A limit

    Okay, I haven't quite talked myself into it, yet, but let's see how far it gets me.

    For an arbitrary 'n', the half period is $\displaystyle \frac{\pi}{n}$. If we restrict the evaluation of the integral to the first half period, the Absolute Values become unnecessary.

    Given the half-period of $\displaystyle \frac{\pi}{n}$, quite obviously there are $\displaystyle \frac{n}{\pi}$ such half periods in [0,1].

    Somewhat to my surprise, this makes the result independent of 'n'.

    $\displaystyle \frac{n}{\pi}\cdot\int_{0}^{\frac{\pi}{n}}[\sin(n\cdot x)]^{3}\;dx\;=\;\frac{4}{3\cdot \pi}$

    It's not obvious what I'm missing. Clearly, the limit it greater than that, but I figured I'd get it out there for ridicule.
    Last edited by TKHunny; Jun 24th 2011 at 02:16 PM.
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    Re: A limit

    $\displaystyle \dfrac4{3\pi}$ is the answer.
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    Re: A limit

    Hmmm...Maybe I was looking at rounding error, but I definitely exceeded that in numerical examples.

    Some other way to get there?
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    Re: A limit

    Quote Originally Posted by Krizalid View Post
    Compute $\displaystyle \lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$
    Here's my idea:

    Spoiler:



    Let


    $\displaystyle \displaystyle I=\lim \int_0^1 |\sin(nx)|^3\text{ }dx$


    Make the obvious substitution to get that


    $\displaystyle \displaystyle I=\lim \frac{1}{n}\int_0^n |\sin(x)|^3\text{ }dx$


    I can show it converges if that's what you really want, but one can show it converges, so we know that


    $\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\int_0^{\pi n}|\sin(x)|^3\text{ }dx$


    But, of course we can rewrite this as


    $\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$


    But, by the $\displaystyle \pi$-peridocity of $\displaystyle |\sin(x)|$ we know that


    $\displaystyle \displaystyle \int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$


    Is a constant function of $\displaystyle j$, and one can easily check by standard methods that


    $\displaystyle \displaystyle \int_0^{\pi}|\sin(x)|^3\text{ }dx=\frac{4}{3}$


    Thus,


    $\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx=\lim \frac{4n}{3\pi n}=\frac{4}{3\pi}$



    Remark: Of course, this method applied to a more general scenario shows that for sufficiently well-behaved functions ($\displaystyle C^0(\mathbb{R})$ should suffice) of period $\displaystyle T$ one has that


    $\displaystyle \displaystyle \lim \int_a^b f(nx)\text{ }dx=\frac{b-a}{T}\int_0^T f(x)\text{ }dx$

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    Re: A limit

    Quote Originally Posted by Krizalid View Post
    Compute $\displaystyle \lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$
    $\displaystyle sin^3x=\frac{1}{4}\left(3sinx-sin3x\right)$

    As n increases, the function

    $\displaystyle |sin^3(nx)|$

    oscillates from 0 to 1 more rapidly.
    In the limit

    $\displaystyle \lim_{n\rightarrow\infty}\int_{0}^1|sin^3(nx)|dx$

    is the same as the fraction of the area of the rectangle $\displaystyle (1)\pi$
    given by

    $\displaystyle \int_{0}^{\pi}sin^3xdx$

    Hence

    $\displaystyle \lim_{n\rightarrow\infty}\int_{0}^1|sin^3(nx)|dx= \frac{1}{4\pi}\int_{0}^{\pi}\left(3sinx-sin3x\right)dx$

    $\displaystyle =\frac{1}{4\pi}\left[\left(-3cos\pi+\frac{cos3\pi}{3}\right)-\left(-3cos(0)+\frac{cos3(0)}{3}\right)\right]$

    $\displaystyle =\frac{1}{4\pi}\left[\left(-3(-1)-\frac{1}{3}\right)-\left(-3+\frac{1}{3}\right)\right]$

    $\displaystyle =\frac{1}{4\pi}\left(6-\frac{2}{3}\right)=\frac{1}{12\pi}(18-2)=\frac{4}{3\pi}$
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