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Thread: limit sinus-sqrt

  1. #1
    Super Member dhiab's Avatar
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    limit sinus-sqrt

    limit sinus-sqrt-44079011.jpg
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    Re: limit sinus-sqrt

    Quote Originally Posted by dhiab View Post
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    I highly doubt a limit exists. Why would you think it does? I would imagine you can come up with subsequences that could have any limit you desire between 0 and 1.
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    Forum Admin topsquark's Avatar
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    Re: limit sinus-sqrt

    I agree with SlipEternal. On the other hand we can provide an argument that it's 0. (Clearly there will be other methods that say otherwise.)

    Anyway, for very large n, and that $\displaystyle n^2 >> n$, the limit becomes $\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n)$. Since n is an integer and $\displaystyle sin(\pi n) = 0$ for all integer values of n then:

    $\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n) = 0$.

    -Dan
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    Re: limit sinus-sqrt

    We can write
    \begin{align} \lim_{n \to \infty} \sqrt{n^2+n} &= \lim_{n \to \infty} \sqrt{\tfrac14(4n^2 + 4n + 1-1)} \\ &= \lim_{n \to \infty}\tfrac12\sqrt{(2n+1)^2 - 1} \end{align}
    So clearly, \begin{equation} \sqrt{n^2+n} \approx \tfrac12 (2n+1) \end{equation} and this is a very slight over-estimate for large $n$. (A better over-estimate is $\sqrt{n^2+n} \approx \tfrac12\left((2n+1)-\frac{1}{2(2n+1)}\right)$).

    Thus $$\sin{\big(\pi\sqrt{n^2+n}\big)} \approx \sin {\big(\pi(n+\tfrac12)\big)}$$ and this means that $\sin{\big(\pi\sqrt{n^2+n}\big)}$ alternates between approximately $1$ and approximately $-1$. So there is no limit.

    You'll find that my approximation of $\sqrt{n^2+n}$ is closer than topsquark's.
    Last edited by Archie; Oct 15th 2018 at 08:00 PM.
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    Re: limit sinus-sqrt

    Quote Originally Posted by dhiab View Post
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    I find it unbelievable that this is even a multiple thread.
    LOOK HERE.
    It should be clear that there is no limit. Rather the function oscillates between 0 & 1.
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    Re: limit sinus-sqrt

    I've just realised that I forgot that the $\sin$ function is squared, so the limit is actually $1$.

    Plato's comment would make sense for a continuous (real) variable $x$, but not for the discrete (natural) $n$.
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    Re: limit sinus-sqrt

    More formally, we can demonstrate that $$\frac12\left( (2n+1) - \frac1{2n+1} \right) < \sqrt{n^2 + n} < \frac12(2n+1)$$ and then, perhaps via the identity $\sin{(A-B)} = \sin{(A)}\cos{(B)} - \cos{(A)}\sin{(B)}$, that $$\lim_{n \to \infty} \left|\sin{\left(\pi\sqrt{n^2+n}\right)}\right| = \left| \sin {\left(\tfrac12(2n+1)\pi\right)} \right| = 1$$
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    Re: limit sinus-sqrt

    Quote Originally Posted by Archie View Post
    We can write
    \begin{align} \lim_{n \to \infty} \sqrt{n^2+n} &= \lim_{n \to \infty} \sqrt{\tfrac14(4n^2 + 4n + 1-1)} \\ &= \lim_{n \to \infty}\tfrac12\sqrt{(2n+1)^2 - 1} \end{align}
    So clearly, \begin{equation} \sqrt{n^2+n} \approx \tfrac12 (2n+1) \end{equation} and this is a very slight over-estimate for large $n$. (A better over-estimate is $\sqrt{n^2+n} \approx \tfrac12\left((2n+1)-\frac{1}{2(2n+1)}\right)$).

    Thus $$\sin{\big(\pi\sqrt{n^2+n}\big)} \approx \sin {\big(\pi(n+\tfrac12)\big)}$$ and this means that $\sin{\big(\pi\sqrt{n^2+n}\big)}$ alternates between approximately $1$ and approximately $-1$. So there is no limit.

    You'll find that my approximation of $\sqrt{n^2+n}$ is closer than topsquark's.
    Meh. What's a few millions of terms between friends?

    -Dan
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    Re: limit sinus-sqrt

    $1$ in the limit apparently. But I can be picky like that. But you are within an engineer's approximation.
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  10. #10
    Super Member dhiab's Avatar
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    Re: limit sinus-sqrt

    Hello here is a solutionlimit sinus-sqrt-44081580_2758423584183007_991960239320334336_o.jpgClick image for larger version. 

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