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I want to prove $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$ First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.

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What have you tried? –  Matthew Conroy Nov 6 '12 at 18:37
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You want to prove. We want to help students who show effort. Supply any ideas you tried before now. –  rschwieb Nov 6 '12 at 18:57
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4 Answers 4

up vote 4 down vote accepted

HINT

In fact, it is not hard to compute the limit exactly. Let $I_n = \displaystyle \int_0^{\pi} \cos^n(t/2) dt$. Compute $I_n$ and use Stirling's formula to obtain asymptotics of $I_n$.

Move your cursor over the gray area for the complete solution.

Setting $t = 2x$, we get $$I_n = 2 \int_{0}^{\pi/2} \cos^n(x) dx$$ For $n = 2k+1$, we get that \begin{align} I_{2k+1} & = 2 \left( \dfrac{4^k (k!)^2}{(2k+1)!} \right)\\ & \sim 2 \cdot 4^k \times (2 \pi k) \times \left(\dfrac{k}e \right)^{2k} \times \dfrac1{\sqrt{2 \pi (2k+1)}} \left(\dfrac{e}{2k+1} \right)^{2k+1}\\ & = \dfrac{2^{2k+2} \pi k^{2k+1} e^{2k+1}}{e^{2k} \sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{2 \pi (2k)^{2k+1} e}{\sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(\dfrac{2k}{2k+1} \right)^{2k+1}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(1-\dfrac1{2k+1} \right)^{2k+1}\\ & \sim \dfrac{\sqrt{2\pi}e}{\sqrt{2k+1}} \times e^{-1}\\ & = \sqrt{\dfrac{2 \pi}{2k+1}} \end{align} Hence, $$\sqrt{2k+1} I_{2k+1} \sim \sqrt{2 \pi}$$ You will get the same result for even $n$ as well courtesy Wallis formula.

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@downvoter: I am amazed that you have downvoted a valid and a correct answer without providing a reason. –  user17762 Nov 6 '12 at 19:46
    
Marvis, I think you also missed something around. The $e^{2k}$ and the $e^{2k+1}$ simplify to $e$, unless I'm missing something. –  Pedro Tamaroff Nov 6 '12 at 23:15
    
@PeterTamaroff Yes. But then the remaining $e$ gets cancelled by the term $$(2k/(2k+1))^{2k+1} = (1-1/(2k+1))^{2k+1}$$ which is asymptotically $e^{-1}$. –  user17762 Nov 6 '12 at 23:19
    
Oh, well. I think it is worth including some notes and steps. People sometimes downvote when the answer is not helpfull, however correct it might be. –  Pedro Tamaroff Nov 6 '12 at 23:27
    
@PeterTamaroff Thanks. I have now added the intermediate steps. –  user17762 Nov 6 '12 at 23:41
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By a change of variables $t\mapsto 2x$, you want to show $$\mathop {\lim }\limits_{n \to + \infty } 2\sqrt n \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx > 0.$$

Now, you need to evaluate $$I\left( n \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx$$

For $n\geq 2$ we can reduce this integral with integration by parts

$$\eqalign{ & {\cos ^{n - 1}}x = u \cr & \cos xdx = dv \cr} $$

then $$\eqalign{ & - \left( {n - 1} \right){\cos ^{n - 2}}x\sin xdx = du \cr & \sin x = v \cr} $$

thus $$\int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx = \int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx} $$ for the other term vanishes. But $$\displaylines{ \int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx} = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right)dx} \cr = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}xdx} - \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x{{\cos }^2}xdx} \cr = \left( {n - 1} \right)I\left( {n - 2} \right) - \left( {n - 1} \right)I\left( n \right) \cr} $$

which means $$I\left( n \right) = \left( {\frac{{n - 1}}{n}} \right)I\left( {n - 2} \right)$$ If $n=2m$ is even, or if $n=2m+1$ is odd we have

$$\displaylines{ I\left( n \right) = \frac{{2m - 1}}{{2m}}\frac{{2m - 3}}{{2m - 2}} \cdots \frac{3}{4}\frac{1}{2}I\left( 0 \right) \cr I\left( n \right) = \frac{{2m}}{{2m + 1}}\frac{{2m - 2}}{{2m - 1}} \cdots \frac{4}{5}\frac{2}{3}I\left( 1 \right) \cr} $$

for we reduce the number by $2$ each time, so if we start with an odd number, we'll be left with $1$.

So finally since

$$\displaylines{ I\left( 0 \right) = \frac{\pi }{2} \cr I\left( 1 \right) = 1 \cr} $$

we get for $n$ even $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}\frac{\pi }{2}$$ and for $n$ odd $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}$$

Thus we have two sequences to consider

$$\displaylines{ {a_{2m}} = \sqrt {2m} \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \cr {a_{2m + 1}} = \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr} $$

But Wallis' approximation says $$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{2n} = \frac{\pi }{2}$$

so that

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} = \sqrt \pi $$

All things considered $$\displaylines{ \mathop {\lim }\limits_{m \to \infty } {a_{2m + 1}} = \mathop {\lim }\limits_{m \to \infty } \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr = \mathop {\lim }\limits_{m \to \infty } \frac{{\sqrt {2m + 1} }}{{2m + 1}}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr = \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt {2m + 1} }}\frac{{\sqrt m }}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr = \mathop {\lim }\limits_{m \to \infty } \sqrt {\frac{m}{{2m + 1}}} \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} = \frac{1}{{\sqrt 2 }}\sqrt \pi = \sqrt\frac\pi2 \cr} $$


$$\mathop {\lim }\limits_{m \to \infty } {a_{2m}} = \frac{\pi }{{\sqrt 2 }}\mathop {\lim }\limits_{m \to \infty } \sqrt m \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}} = \frac{\pi }{{\sqrt 2 }}\frac1{\sqrt\pi} = \sqrt\frac\pi2 $$

Hence, the limit exists and is positive, as desired.

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I'll correct this later. –  Pedro Tamaroff Nov 6 '12 at 19:55
    
This is almost the same as my answer above and you are missing a factor of $\sqrt{2}$ somewhere. Another nitpick, it is Wallis' and not Walli's. –  user17762 Nov 6 '12 at 19:58
    
@Marvis Hehe, yes, Wallis'(s) formula and Stirling's approximation are almost equivalent. If one proves the first, one proves the other, and vice versa. –  Pedro Tamaroff Nov 6 '12 at 23:05
    
@Peter Tamaroff: I think this is a tad unclear. –  Chris's sis Nov 17 '12 at 14:07
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I think I found the missing $\sqrt 2$. I believe it was supposed to be $\frac 1{2n}$ in the quote of Wallis' approximation rather than $\frac 1n$. –  Karl Kronenfeld May 11 '13 at 13:01
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Your first attempt is a good idea, in that it reduces the problem to something you can compute explicitly. Too bad it didn't work out. Perhaps you should instead try to analyze the problem from a different angle. After all, $\cos(t/2)<1$ most places, so the $n$th power goes to zero really fast (geometrically, that is) as $n\to\infty$. That is why you might expect the limit to be zero. You only have the $\sqrt{n}$ factor to stand against this. But it can't win against a geometric descent to zero! So you need to concentrate on the places where $\cos(t/2)$ is close to $1$, that is, the very small $t$. Now you may see why your attempt failed, for $\sin(t/2$ is also very small where $t$ is small, and that blows it for you.

So maybe you should look for another function that is smaller than $\cos(t/2)^n$, yet is close to it when $t$ is small, and which you can integrate explicitly.

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I don't guarantee that this answer will lead you to a solution, but at least I hope it will point your efforts in a useful direction. –  Harald Hanche-Olsen Nov 6 '12 at 19:40
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This integral can be analyzed using Laplace's Method. Write

$$\cos{\left(\frac{t}{2}\right)}^n = e^{n \log{\cos{(t/2)}}}$$

Note that $\log{\cos{(t/2)}}$ has a maximum at $t=0$; as $n \to \infty$ there is an increasingly narrow and high peak about $t=0$. Thus we may Taylor expand about $t=0$ and ignore contributions away from a small neighborhood of $t=0$:

$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\epsilon} dt \, e^{n \log{(1-t^2/8)}} $$

where $\epsilon = O\left(n^{-1/2}\right)$. Because contributions away from $[0,\epsilon]$ are exponentially small, we may extend the integral out to infinity within the error of the leading-order approximation. Taylor expanding the log term, we find

$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\infty} dt \, e^{-n t^2/8} = \frac12 \sqrt{\frac{8 \pi}{n}}$$

Therefore

$$\lim_{n \to \infty} \sqrt{n} \int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n =\sqrt{2 \pi}$$

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