A limit

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June 24th 2011, 08:47 AM
Krizalid

A limit

Compute $\lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$
June 24th 2011, 09:40 AM
TKHunny

Re: A limit

Just toying with it a bit, I note:

n = 4 results in something between 1 and 2 half-periods
n = 7 results in something between 2 and 3 half-periods
n = 13 results in something between 4 and 5 half-periods
n = 26 results in something between 8 and 9 half-periods

Seems like this could lead somewhere useful, at least getting bounds on it.
June 24th 2011, 02:05 PM
TKHunny

Re: A limit

Okay, I haven't quite talked myself into it, yet, but let's see how far it gets me.

For an arbitrary 'n', the half period is $\frac{\pi}{n}$ . If we restrict the evaluation of the integral to the first half period, the Absolute Values become unnecessary.

Given the half-period of $\frac{\pi}{n}$ , quite obviously there are $\frac{n}{\pi}$ such half periods in [0,1].

Somewhat to my surprise, this makes the result independent of 'n'.

$\frac{n}{\pi}\cdot\int_{0}^{\frac{\pi}{n}}[\sin(n\cdot x)]^{3}\;dx\;=\;\frac{4}{3\cdot \pi}$

It's not obvious what I'm missing. Clearly, the limit it greater than that, but I figured I'd get it out there for ridicule.
June 24th 2011, 05:50 PM
Krizalid

Re: A limit

$\dfrac4{3\pi}$ is the answer.
June 24th 2011, 07:06 PM
TKHunny

Re: A limit

Hmmm...Maybe I was looking at rounding error, but I definitely exceeded that in numerical examples.

Some other way to get there?
June 25th 2011, 01:51 AM
Drexel28

Re: A limit

Quote:

Originally Posted by Krizalid

Compute $\lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$

Here's my idea:

Spoiler:

Let

$\displaystyle I=\lim \int_0^1 |\sin(nx)|^3\text{ }dx$

Make the obvious substitution to get that

$\displaystyle I=\lim \frac{1}{n}\int_0^n |\sin(x)|^3\text{ }dx$

I can show it converges if that's what you really want, but one can show it converges, so we know that

$\displaystyle I=\lim \frac{1}{\pi n}\int_0^{\pi n}|\sin(x)|^3\text{ }dx$

But, of course we can rewrite this as

$\displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$

But, by the $\pi$ -peridocity of $|\sin(x)|$ we know that

$\displaystyle \int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$

Is a constant function of $j$ , and one can easily check by standard methods that

$\displaystyle \int_0^{\pi}|\sin(x)|^3\text{ }dx=\frac{4}{3}$

Thus,

$\displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx=\lim \frac{4n}{3\pi n}=\frac{4}{3\pi}$

Remark: Of course, this method applied to a more general scenario shows that for sufficiently well-behaved functions ( $C^0(\mathbb{R})$ should suffice) of period $T$ one has that

$\displaystyle \lim \int_a^b f(nx)\text{ }dx=\frac{b-a}{T}\int_0^T f(x)\text{ }dx$
June 25th 2011, 02:23 PM
Archie Meade

Re: A limit

Quote:

Originally Posted by Krizalid

Compute $\lim_{n\to\infty}\int_0^1|\sin nx|^3\,dx,\,n\in\mathbb N.$

$sin^3x=\frac{1}{4}\left(3sinx-sin3x\right)$

As n increases, the function

$|sin^3(nx)|$

oscillates from 0 to 1 more rapidly.
In the limit

$\lim_{n\rightarrow\infty}\int_{0}^1|sin^3(nx)|dx$

is the same as the fraction of the area of the rectangle $(1)\pi$
given by

$\int_{0}^{\pi}sin^3xdx$

Hence

$\lim_{n\rightarrow\infty}\int_{0}^1|sin^3(nx)|dx= \frac{1}{4\pi}\int_{0}^{\pi}\left(3sinx-sin3x\right)dx$

$=\frac{1}{4\pi}\left[\left(-3cos\pi+\frac{cos3\pi}{3}\right)-\left(-3cos(0)+\frac{cos3(0)}{3}\right)\right]$

$=\frac{1}{4\pi}\left[\left(-3(-1)-\frac{1}{3}\right)-\left(-3+\frac{1}{3}\right)\right]$

$=\frac{1}{4\pi}\left(6-\frac{2}{3}\right)=\frac{1}{12\pi}(18-2)=\frac{4}{3\pi}$