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Suppose that $f$ is continuous on $[0, 1]$. Then calculate the following limit:
$$\lim_{n\to\infty} \frac{\displaystyle\int_{0}^{1} f(x) \sin^{2n} (2 \pi x) \space dx}{\displaystyle\int_{0}^{1} e^{x^2}\sin^{2n} (2 \pi x) \space dx}$$

What should i start with? (high school problem)

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    Are there any restrictions on $f$?2012-07-07
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    Well, what about starting to tell us what is $\,f\,$, what have you tried so far, where're you stuck...?2012-07-07
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    @nullUser: i added all we know about $f(x)$2012-07-07
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    Hint: [Laplace's method](http://en.wikipedia.org/wiki/Laplace%27s_method). Answer (under suitable assumptions soon to appear on this page): $\frac{f(1/4)+f(3/4)}{\mathrm e^{1/16}+\mathrm e^{9/16}}$.2012-07-07
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    @DonAntonio: no idea what i should do here.2012-07-07
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    did's idea is probably that the only remaining parts will be as the $\sin$ terms become $\pm 1$ (getting delta functions there and his result)...2012-07-07
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    @RaymondManzoni To see that what you describe is indeed what is going on, just follow the link...2012-07-07
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    Is this a *high-school* problem?2012-07-07
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    @Siminore: of course. Major part of the problems posted by me and my bro are high school problems. Maybe there is one or two exceptions involving gamma function.2012-07-07
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    What sort of answer do you expect? Take $f(x)=ae^{x^2}$ ...2012-07-07
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    @Mark Bennet: a correct answer.2012-07-07
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    What I mean by "what sort of answer" is whether you want a value, a function or a method ...2012-07-07
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    @Mark Bennet: i just wanted to know the way a problem like this one may be approached. This problem seems to be a bit complicated.2012-07-07
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    I'd love to know what high school, where and under what syllabus and/or under what teacher is this a high school problem...2012-07-08
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    Like @DonAntonio said.2012-07-08
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    @did: a student that learns things has no difficulty with these problems because in the essence they are all easy. I just need to learn more.2012-07-08
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    A student, *any* student, has to construct his learning building in such a way that (s)he can study stuff based on what (s)he already knows. This question is no high school stuff, not even close, in any educational system I know, and that's what I asked what's your high school, where and what's your mathematics syllabus: it really interests me.2012-07-08
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    Chris'sister: I simply cannot understand the first sentence of your last comment. It reads like a Zen koan to me, except I fail to see any meaning in it.2012-07-14
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    @did: don't worry about that. People usually fail when trying to understand all things. The aim is to do your best, and it's not from Zen ... it's from my experience.2012-07-14
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    *To do one's best* is not *not from Zen*. (But you did not explain your previous statement.)2012-07-14
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    did: when one is passionate with something and spend much time doing that thing, then that person becomes better and better and all problems in that area becomes easier and easier. The difficulty level of a problem is relative in the sense that is related to the knowledge and the experience of a person. Strictly referring at those problems i've faced so far they all seem very easy to me even if i didn't manage to solve them all. Probably i didn't learn all necessary things to solve them.I could write so much on this subject!2012-07-14
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    @did: i like to talk! :-)2012-07-14
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    @did: do you think i'm too optimistic? I saw so many nice, simple, elementary proofs for my problems. I cannot think that they are difficult problems.2012-07-14
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    What I think is that you use the phrase *difficult problem* in a somewhat unconventional way. But there is no problem with that, as far as I am concerned...2012-07-14

2 Answers 2

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For every continuous function $g$ defined on $(-1,1)$ and every $\delta$ in $[0,1]$, define $$ I_n(g)=\int_{-1}^1g(x)\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx,\qquad J_n(\delta)=\int_{-\delta}^{+\delta}\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx. $$ Fix a continuous function $g$ defined on $(-1,1)$. Since $g$ is continuous, there exists some finite $c$ such that $|g|\leqslant c$ uniformly on $(-1,1)$. Since $g$ is continuous at $0$, for every $(a,b)$ such that $a\lt g(0)\lt b$, there exists some $\delta$ in $[0,1]$ such that $a\lt g\lt b$ uniformly on $(-\delta,\delta)$. Hence, $$ aJ_n(\delta)-c(J_n(1)-J_n(\delta))\leqslant I_n(g)\leqslant bJ_n(\delta)+c(J_n(1)-J_n(\delta)). $$ Here is a basic but crucial fact:

For every $\delta$ in $[0,1]$, $J_n(1)-J_n(\delta)\ll J_n(\delta)$ when $n\to\infty$.

Now, fix some continuous functions $g$ and $h$ defined on $(-1,1)$ such that $h(0)\ne0$, say, $h(0)\gt0$. Assume without loss of generality that $g\gt0$ everywhere (if necessary, add to $g$ a large multiple of $h(0)$). Then, for every positive $(a,b,a',b')$ such that $a\lt g(0)\lt b$ and $a'\lt h(0)\lt b'$, $$ \frac{a}{b'}\leqslant\liminf\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\limsup\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\frac{b}{a'}. $$ Hence, $$ \lim\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}=\frac{g(0)}{h(0)}. $$ To solve the question asked, apply the result above to the functions $g$ and $h$ defined on $(-1,1)$ by $g(x)=f(\frac14(x+1))+f(\frac14(x+3))$ and $h(x)=e(\frac14(x+1))+e(\frac14(x+3))$ with $e(t)=\mathrm e^{t^2}$.

Note: To prove the basic but crucial fact mentioned above, one can show that $J_n(\delta)\sim2/\sqrt{\pi n}$ when $n\to\infty$, for every $\delta$ in $(0,1]$.

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    Thank you for your solution.2012-07-07
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    how did you think of this approach? Did you meet similar problems before?2012-07-07
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    Yes, as already mentioned in my comment (did you go and read the WP page?). The idea is that $\sin^{2n}(2\pi x)\to0$ except at $x=\frac14$ and $x=\frac34$ where $\sin^{2n}(2\pi x)=1$ identically, hence all that matters are the values of the function integrated, at these points.2012-07-07
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This isn't at all high-school level, but pursuing the delta function idea, if you let $I_n = \int_0^{\pi}\sin^{2n}(2\pi x)\,dx = \int_{\pi}^{2\pi}\sin^{2n}(2\pi x)\,dx$, then your limit is the same as $${\int_0^{\pi} f(x){\sin^{2n}(2\pi x) \over I_n} \, dx + \int_{\pi}^{2\pi} f(x){\sin^{2n}(2\pi x) \over I_n}\, dx \over \int_0^{\pi} e^{x^2} {\sin^{2n}(2\pi x) \over I_n} \, dx + \int_{\pi}^{2\pi} e^{x^2} {\sin^{2n}(2\pi x) \over I_n}\, dx }$$ The functions $\sin^{2n}(2\pi x) \over I_n$ converge in a distribution sense to $\delta(x - {1 \over 4})$ on $(0,\pi)$ and to $\delta(x - {3 \over 4})$ on $(\pi,2\pi)$ (Alternatively, you can talk about approximations to the identity). So as $n$ goes to infinity the above converges to ${f(1/4) + f(3/4) \over e^{1/16} + e^{9/16}}$.

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    Thank you for your solution.2012-07-07
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    the approach needed for this problem is totally different from the major part of the calculus problems i've seen so far.2012-07-07
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    @Chris'sister - then you've learned something ...2012-07-07