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I have problem with showing that the limit of the following function

$$\frac{ \sqrt{\frac{3 \pi}{2n}} - \int_0^{\sqrt 6}( 1-\frac{x^2}{6} +\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$

equal to $1$, with $n \to \infty$.

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    For what it counts, Wolphram alpha gives http://www.wolframalpha.com/input/?_=1325710547024&i=integral%5b(+1-%7bx%5e2%7d%2f%7b6%7d+%2b%7bx%5e4%7d%2f%7b120%7d)%5en%5d%2c+0%3Cx%3Csqrt(6)&fp=1&incTime=true $\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx= \sqrt6 F_1(\frac{1}{2};-n,-n; -\frac{3}{2};\frac{3i}{5i+\sqrt5},-\frac{3i}{-5i+\sqrt5})$ Where $F_1(a;b_1,b_2;c;x,y)$ is Appell Hypergeometric function of two variables.2012-01-04
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    I thought that the definite integral could be treated as a constant but Nikhil Bellarykar clarified that it would be a function of n, so using L'Hospital rule is not possible.2012-01-04
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    Wolphram alpha further gives the series expansion of the Appell function at $n=\infty$ as ${(\frac{3}{10})^n}(\frac{64n^2+56n}{27}+1)$. http://www.wolframalpha.com/input/?i=F1%281%2F2%3B%E2%88%92n%2C%E2%88%92n%2C%E2%88%923%2F2%2C+3i%2F%285i%2B%E2%88%9A5%29%2C%E2%88%923i%2F%28%E2%88%925i%2B%E2%88%9A5%29%29+series+ But if we plug in this value in the original function, the answer does not come as 1. Am I wrong to simply plug in this value? maybe. Even if I just try to plug in a simple series expansion, the result does not seem to be 1. Someone please clarify what is missing from my approach.2012-01-04

1 Answers 1

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Hint: First note that for $x\in[0,\sqrt{6}]$, $1-\frac{x^2}{6}+\frac{x^4}{120}$ monotonically decreases from $1$ to $\frac{3}{10}$ and that $$ 1-\frac{x^2}{6}+\frac{x^4}{120}\le1-\frac{x^2}{9}\le e^{-x^2/9}\tag{1} $$ You might try the change of variables $x\mapsto x/\sqrt{n}$ so that $$ \int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x =\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\tag{2} $$ and $(1)$ then says that for $x\in[0,\sqrt{6n}]$ $$ \left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\le e^{-x^2/9}\tag{3} $$ Thus, we have $$ \begin{align} \int_a^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x &\le\int_a^\infty e^{-x^2/9}\;\mathrm{d}x\\ &\le\frac{9}{2a}e^{-a^2/9}\tag{4} \end{align} $$ Notice that the integrand on the right in $(2)$ tends to $e^{-x^2/6}$, so consider $$ \begin{align} \left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n &=e^{-x^2/6}\exp\left(n\log\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)+\frac{x^2}{6}\right)\\ &=e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}+O\left(\frac{x^8}{n^3}\right)\right)\tag{5} \end{align} $$ Applying $(5)$ to $(2)$ yields $$ \begin{align} &\left|\;\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x -\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x\;\right|\\ &\le\frac{1}{\sqrt{n}}\int_0^{\log(n)} e^{-x^2/6}O\left(\frac{x^8}{n^3}\right)\;\mathrm{d}x\\ &+\frac{1}{\sqrt{n}}\int_{\log(n)}^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x +\frac{1}{\sqrt{n}}\int_{\log(n)}^\infty e^{-x^2/6}\left|1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right|\;\mathrm{d}x\\ &=\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\tag{6} \end{align} $$ Therefore, $$ \begin{align} \int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x &=\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\\ &=\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x+\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\\ &=\sqrt{\frac{3\pi}{2n}}\left(1-\frac{3}{20n}+\frac{11}{160n^2}+O\left(\frac{1}{n^3}\right)\right)\tag{7} \end{align} $$

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    Truly amazing effort!2012-01-07
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    @David: I was really hoping that there was a simpler solution. If the instructor shows you a simpler solution, please post it here.2012-01-08
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    Sure. Thank you very much, robjohn.2012-01-08
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    @robjohn: you have interesting proof. But how to show that there are no other terms with $1/n$, $1/n^2$ so on?2012-06-18
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    Sorry, I still don't see it. Could you elaborate please. Also, how to be in the situation when one would like to get result with $O(1/n^2)$ or $O(1/n)$. Which argument one should use to show that there are finite number of terms? Thank you.2012-06-18
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    @robjohn: I've got how binomial theorem works for the integral $\int_0^{\sqrt {6n}}(1-t^2/(6n)+t^4/(120n^2))$. But when we represented as an integral $\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{6480n^2}\right)\;\mathrm{d}x$, we multiplied by the series of exponential function. And the series is infinite, so there are infinite number of terms for each $(1/n)^k, k=0,1,2,3, \ldots$. How to show that there are infinite number of terms, say with $(1/n^3)$? Or, it follows just from the equality of the two integrals? Thank you.2012-06-23
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    @Aleks.K: I have deleted my previous comment which was in error. Look back at equation $(5)$: $$ \left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n =e^{-x^2/6}\exp\left(n\log\left(1-\frac{x^2}{6n} +\frac{x^4}{120n^2}\right)+\frac{x^2}{6}\right) $$ This is simply rewriting things in terms of $\log$ and $\exp$. Using the first three terms of the power series for $\log(1+u)$, where $u=-\frac{z^2}{6n}+\frac{x^4}{120n^2}$ gives2012-06-23
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    @Aleks.K: $$ \begin{align} &e^{-x^2/6}\exp\left(n\left(u-\frac12u^2+\frac13u^2 +O\left(u^4\right)\right)+\frac{x^2}{6}\right)\\ &=e^{-x^2/6}\exp\left(n\left(-\frac{x^2}{6n} -\frac{x^4}{180n^2}-\frac{x^6}{6480n^3} +O\left(\frac{x^8}{n^4}\right)\right)+\frac{x^2}{6}\right)\\ &=e^{-x^2/6}\exp\left(-\frac{x^2}{6}-\frac{x^4}{180n} -\frac{x^6}{6480n^2}+O\left(\frac{x^8}{n^3}\right) +\frac{x^2}{6}\right)\\ &=e^{-x^2/6}\exp\left(-\frac{x^4}{180n}-\frac{x^6}{6480n^2} +O\left(\frac{x^8}{n^3}\right)\right) \end{align} $$2012-06-23
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    @Aleks.K: Then using the fist three terms for $\exp(u)$ where $u=-\frac{x^4}{180n}-\frac{x^6}{6480n^2} +O\left(\frac{x^8}{n^3}\right)$ gives $$ e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2} +O\left(\frac{x^8}{n^3}\right)\right) $$ Does that answer your question?2012-06-23
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    robjohn: Thank you. But it does not explained why we have the finite number of terms for each summand $1/n, 1/n^2, ...$2012-06-23
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    @Aleks.K: If you are computing a power series $a_0+a_1x+a_2x^2+\dots$ where $x=\frac{b_1}{n}+\frac{b_2}{n^2}+\frac{b_3}{n^3}+\dots$, the $a_kx^k$ term is only going to have terms with $\frac1n$ raised at least to the power $k$. So to get *all* the terms larger than $\frac{1}{n^4}$, we just need to include $a_0+a_1x+a_2x^2+a_3x^3$ since all the terms $a_4x^4+a_5x^5+a_6x^6+\dots$ will have $\frac1n$ raised at least to the power 4.2012-06-23