1. Limits problem

Hi guys,

I'm having real difficulty with this

Show that

$\displaystyle \lim_{k\to\infty} \prod^k_{r=1}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \lim_{k\to\infty}\prod^{k}_{r=1} \left( 1 - \frac{x^2}{r^2} \right)$, where $\displaystyle n = 2k+1$.

Now, I know by L'Hopital's rule, that

$\displaystyle \lim_{n\to\infty}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = 1 - \frac{x^2}{r^2}$

But the relation of n with k has really thrown me off.

Any help would be much appreciated.

HTale.

2. Originally Posted by HTale
Hi guys,

I'm having real difficulty with this

Show that

$\displaystyle \lim_{k\to\infty} \prod^k_{r=1}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \lim_{k\to\infty}\prod^{k}_{r=1} \left( 1 - \frac{x^2}{r^2} \right)$, where $\displaystyle n = 2k+1$.

Now, I know by L'Hopital's rule, that

$\displaystyle \lim_{n\to\infty}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = 1 - \frac{x^2}{r^2}$

But the relation of n with k has really thrown me off.

Any help would be much appreciated.

HTale.
Your problem amounts to interchanging a limit and an infinite product (or a series), can you see why?

You need to show $\displaystyle \lim_k \prod_{r=1}^k \left(1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}\right) = \prod_{r=1}^\infty \left(1 - \frac{x^2}{r^2}\right)$. Due to your remark, this can be rewritten as
$\displaystyle \lim_k \prod_{r=1}^\infty a(r,k) = \prod_{r=1}^\infty \lim_k a(r,k)$
where $\displaystyle a(r,k)=1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}$ if $\displaystyle r=1,\ldots,k$ and $\displaystyle a(r,k)=1$ if $\displaystyle r>k$.

You need a theorem to justify that. A good one would be the "dominated convergence theorem" (you may know another name):
Dominated convergence theorem (for series):
If $\displaystyle (u(r,k))_{r,k\geq 0}$ is such that

• for every $\displaystyle r$, $\displaystyle u(r,k)\to_{k\to\infty} u(r)$;
• there is $\displaystyle (v(r))_{r\geq 0}$ such that $\displaystyle |u(r,k)|\leq v(r)$ for every $\displaystyle r,k$, and $\displaystyle \sum_{r=0}^\infty v(r)<\infty$,

Then, for every $\displaystyle k$, the series $\displaystyle \sum_r u(r,k)$ and $\displaystyle \sum_r u(r)$ converge, and we have $\displaystyle \lim_{k\to\infty} \sum_{r=0}^\infty u(r,k)=\sum_{r=0}^\infty u(r)$.
I wrote the theorem for series, so you must let $\displaystyle u(r,k)=\log a(r,k)$ in order to have $\displaystyle \sum_r u(r,k)=\log\prod_r a(r,k)$ and apply the theorem.

Thus you must find $\displaystyle v(r)\geq 0$ such that $\displaystyle \sum_r v(r)$ converges and, for every $\displaystyle 1\leq r\leq n$,
$\displaystyle \left|\log\left(1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}\right)\right|\leq v(r)$.
(And the inequality $\displaystyle |u(r,k)|\leq v(r)$ is obvious if $\displaystyle r>n$ since we would have $\displaystyle u(r,k)=0$)

To this aim, you can use the following bounds: if $\displaystyle 0\leq u\leq \frac{\pi}{2}$, then $\displaystyle \frac{2}{\pi}u\leq \sin u\leq u$. I let you carry on from there.

Note: I assumed $\displaystyle 0<x<1$ so that the factors of the product are positive and the logarithm makes sense.

3. Originally Posted by Laurent
Your problem amounts to interchanging a limit and an infinite product (or a series), can you see why?
Thank you for help, it's like a mist has been cleared! But I can't see why the interchange is justified. What are the required conditions on the products for this to be the case?

HTale

4. Originally Posted by HTale
Thank you for help, it's like a mist has been cleared! But I can't see why the interchange is justified. What are the required conditions on the products for this to be the case?
This is the point of the "dominated convergence theorem" I mentioned; I'd bet you have a theorem looking like this in your course. It is given for series, but products can be turned into series via logarithm, cf. my post.

5. Originally Posted by Laurent
Due to your remark, this can be rewritten as
$\displaystyle \lim_k \prod_{r=1}^\infty a(r,k) = \prod_{r=1}^\infty \lim_k a(r,k)$
where $\displaystyle a(r,k)=1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}$ if $\displaystyle r=1,\ldots,k$ and $\displaystyle a(r,k)=1$ if $\displaystyle r>k$.
Sorry to bring this up again, but do you mean

$\displaystyle \lim_k \prod_{r=1}^k a(r,k) = \prod_{r=1}^\infty \lim_k a(r,k)$?

HTale.

6. Limits problem

Hi guys,

I'm having real trouble proving that

$\displaystyle \displaystyle \prod^k_{r=1}\left(\frac{1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}}{1-\frac{x^2}{r^2}}\right) \rightarrow 0$ as $\displaystyle n=2k+1 \rightarrow \infty$

The following hint was provided,

HINT: choose an m to be much smaller than n, which approaches infinity (for example, $\displaystyle m=\sqrt{n}$ or $\displaystyle m=\epsilon n$) and separately consider the part of the product from which r goes from 1 to m and the part from m+1 to n. Taking the logarithm and bounding the sum from r=m+1 to r=n one should be able to prove that the second product is close to 1. As for the first product ranging from 1 to m, one must show that the numerator and denominator of each term has ratio very close to 1.

So, does that mean starting off with something like,

$\displaystyle \prod^m_{r=1}\left(\frac{1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}}{1-\frac{x^2}{r^2}}\right)\prod^n_{r=m+1}\left(\frac{ 1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}}{1-\frac{x^2}{r^2}}\right)$

I have no idea how to continue with this.

Thank you all so very very much in advance,

Regards,
HTale.

PS: Just incase it's too small to read, the bracketed part with the sin's read

$\displaystyle \sin^2\left(\frac{\pi x}{n}\right)$ for the numerator, and

$\displaystyle \sin^2\left(\frac{\pi r}{n}\right)$ for the denominator.

7. Ok, I have not made any sense of the hint (it was provided by the author of the book from the exercise I'm trying to solve), but if we take the logarithm of the above, and manipulating it a bit, we get

$\displaystyle \lim_k \sum^k_{r=1}\left(1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \lim_k \sum^k_{r=1}\left(1-\frac{x^2}{r^2}\right)$,

which is the same as saying,

$\displaystyle \lim_k \sum^k_{r=1}\left(1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \sum^{\infty}_{r=1}\left(1-\frac{x^2}{r^2}\right)$.

Now, here's the interesting bit. By L'Hopitals rule

$\displaystyle \lim_k\left(1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right)=1-\frac{x^2}{r^2}$.

So now I'm appealing to any analysts here; is there some theorem out there which allows me to do

$\displaystyle \lim_k \sum^k_{r=1}\left(1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \sum^{\infty}_{r=1}\lim_k\left(1-\frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right)=\sum^{\infty}_{r=1}\left(1-\frac{x^2}{r^2}\right).$

Any help would be much appreciated.

HTale.

8. First of all, your manipulation result is incorrect: $\displaystyle log(\prod(x)) = \sum (log(x))$. You dropped the "log"-s in the process.

Then, l'Hopital's rule does not make sense here, since the RHS is also a limit on $\displaystyle k \rightarrow \infty$. Remember: r is an integer between 1 and k ! So the r-s close to k tend to $\displaystyle \infty$ too.

However, $\displaystyle \frac {\sin^2\left(\frac{\pi x}{n}\right)} {\sin^2\left(\frac{\pi r}{n}\right)} \sim \frac{x^2}{r^2}$

when $\displaystyle \frac{\pi x}{n} \ll 1$ and $\displaystyle \frac{\pi r}{n} \ll 1$. (Because $\displaystyle sin(y) \sim y$ when $\displaystyle y \rightarrow 0$)

The first condition is satisfied when $\displaystyle k \rightarrow \infty$, cause $\displaystyle n = 2k+1 \rightarrow \infty$.
But the second condition is not, cause r goes from 1 to k and when r = k for instance , $\displaystyle \frac{\pi k}{2 k+1}\rightarrow 0.5 \pi$ which is not $\displaystyle \ll 1$.

But that's when separating the sum / product comes into play.
if you limit the range of r to 1 to m = integer_part($\displaystyle \sqrt{n}$), then $\displaystyle \forall r \leq m$, you get to have $\displaystyle \frac{\pi r}{n} \ll 1$ when $\displaystyle k \rightarrow \infty$.

So the 1st part of the sum / product vanishes out smoothly. -> wrong

No ! in fact the 2nd part of the product vanishes (tends to 1).
The first part does not. Even if every r-term tends individually to 1, their product doesn't. (Give me some time to find a counter-example).

The first part tends to 0, because intuitively,
as $\displaystyle r \rightarrow \infty$, there are more and more r-terms with $\displaystyle \sin^2\left(\frac{\pi r}{n}\right)$ "close" to $\displaystyle \sin^2\left(\frac{\pi x}{n}\right)$, which makes the denominator of the r-term close to 0. More and more of these r-terms accumulate, dragging the product close to 0.

I guess I'm being a bit unclear with the explanation ... Can someone help me out ?