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Math Help - help with 2 questions?

  1. #1
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    help with 2 questions?

    --
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  2. #2
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    Quote Originally Posted by sbsite View Post
    --
    <br />
\lim_{n \to \infty} \frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n}}=<br />
\lim_{n \to \infty} \sqrt{1+2/n}-\sqrt{1+1/n}=0<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sbsite View Post
    --
    Find:

    <br />
\lim_{n \to \infty} \frac{1 \times 3 \times .. \times (2n-1)}{2\times 4 \times .. \times 2n}=<br />
\lim_{n \to \infty} \frac{(2n)!}{(2\times 4 \times .. \times 2n)^2}=<br />
\lim_{n \to \infty}\frac{(2n)!}{2^{2n} (n!)^2}<br />

    Now Stirlings formula for the factorial is:

    n! \sim \sqrt{2 \pi n}n^n e^{-n}

    so:

    <br />
\frac{(2n)!}{2^{2n} (n!)^2}\sim<br />
\frac{\sqrt{4 \pi n}(2n)^{2n}e^{-2n}}{2^{2n}(\sqrt{2 \pi n}n^n e^{-n})^2}=<br />
\frac{\sqrt{2}}{\sqrt{2 \pi n}}<br />

    so:

    <br />
\lim_{n \to \infty}\frac{(2n)!}{2^{2n} (n!)^2}=0<br />

    (I suppose to make this rigourus I should put in the error term for Stirlings formula
    but I guess I will leave this tpo the reader)

    RonL

    Supporting evidence is in the attachment)
    Attached Thumbnails Attached Thumbnails help with 2 questions?-gash.jpg  
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  4. #4
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    Let me do the second one, it looks more interesting.

    I hope you are familar with the Euler-Wallis product formula.
    Thus,
    \frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left( 1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2} \right)\cdot ...
    If,
    x=\pi/2
    We have,
    \frac{2}{\pi}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{6^2}\right)...
    Thus,
    \frac{2}{\pi}=\prod_{n=1}^{\infty}\frac{4n^2-1}{4n^2}
    Reciprocal time,
    \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1}
    Factor,
    \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)}=\frac{2\cdot 2\cdot 4\cdot 4\cdot...}{1\cdot 3\cdot 3\cdot 5\cdot 5...}
    ---
    Now we do thy problem
    ---
    The limit of,
    \frac{1\cdot 3\cdot 5\cdot ... (2n-1)}{2\cdot 4\cdot 6\cdot ... (2n)}

    If you multiply the numerator and denominator by the missing terms in order to obtain the Euler-Wallis product you will have the expression above. But since 2n+1 is missing there will be an extra division so,
    \mbox{ Wallis Product }\cdot \frac{1}{2n+1}
    Now the Wallis product will converge to \frac{2}{\pi} and \frac{1}{2n+1}\to 0 thus, the product of these two convergents tells us the limit is zero.
    Note, I was extremely lazy in checking whether its square will converge to Euler-Wallis product because the last term is missing. So I might be wrong and the limit is zero.
    The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Let me do the second one, it looks more interesting.

    I hope you are familar with the Euler-Wallis product formula.
    Thus,
    \frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left( 1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2} \right)\cdot ...
    If,
    x=\pi/2
    We have,
    \frac{2}{\pi}=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{4^2}\right)\left(1-\frac{1}{6^2}\right)...
    Thus,
    \frac{2}{\pi}=\prod_{n=1}^{\infty}\frac{4n^2-1}{4n^2}
    Reciprocal time,
    \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1}
    Factor,
    \frac{\pi}{2}=\prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)}=\frac{2\cdot 2\cdot 4\cdot 4\cdot...}{1\cdot 3\cdot 3\cdot 5\cdot 5...}
    ---
    Now we do thy problem
    ---
    The limit of,
    \frac{1\cdot 3\cdot 5\cdot ... (2n-1)}{2\cdot 4\cdot 6\cdot ... (2n)}

    If you multiply the numerator and denominator by the missing terms in order to obtain the Euler-Wallis product you will have the expression above. But since 2n+1 is missing there will be an extra division so,
    \mbox{ Wallis Product }\cdot \frac{1}{2n+1}
    Now the Wallis product will converge to \frac{2}{\pi} and \frac{1}{2n+1}\to 0 thus, the product of these two convergents tells us the limit is zero.
    Note, I was extremely lazy in checking whether its square will converge to Euler-Wallis product because the last term is missing. So I might be wrong and the limit is zero.
    The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.
    A bit gappy, but then so is my "proof", but I suppose because its mine
    I know how to fill those gaps (and that they do fill).

    (I agree that this is more interesting, took some thought to deduce
    what the limit must be - always an advantage before starting a proof)

    RonL
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    A bit gappy, but then so is my "proof", but I suppose because its mine
    I know how to fill those gaps (and that they do fill).

    (I agree that this is more interesting, took some thought to deduce
    what the limit must be - always an advantage before starting a proof)

    RonL
    Do not think that I came up with this full solution be realizing that the solutions for the sine functions form the infinite product. I have seen similar ones before, in fact, when I saw the evens and odd seperated by division a thought immediately occured to me, Wallis infinite product! I just manipulated it so that it worked. I think it would have been cooler is the user asked to find,
    \lim \frac{1\cdot 3\cdot ... \cdot (2n+1)}{2\cdot 4\cdot ... \cdot (2n)}
    In such a case the solution who have involved \pi! I was hoping for that badly, and assumed that was the case until I arrived that the conlusion that there is one more term in the demoninator which made a possibly amazing problem very boring.

    ---
    In fact one way Sirling's asympomatic approximation is derived is from what I said!
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