# Double limit.

• May 25th 2011, 08:56 AM
Also sprach Zarathustra
Double limit.
Prove the following:

$\lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! x)=\begin{cases} 1 \text{ if }x \in\mathbb{Q} \\ 0 \text{ if } x\in \mathbb{R}/\mathbb{Q}\end{cases}$ and show that:

$\lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! e)\neq \lim_{k\to\infty}\lim_{n\to\infty}\cos^{2k}(\pi n! e)$
• May 25th 2011, 09:06 AM
TheEmptySet
Hint if

$x \in \mathbb{Q}$

Then there exists an N such that for all n > N

$xn! \in \mathbb{Z}$

the above fails if x is irrational. See where this takes you.
• May 25th 2011, 11:26 AM
Also sprach Zarathustra
Quote:

Originally Posted by TheEmptySet
Hint if

$x \in \mathbb{Q}$

Then there exists an N such that for all n > N

$xn! \in \mathbb{Z}$

the above fails if x is irrational. See where this takes you.

Yes, I know this part. My trouble is writing the proof or the "explanation" in a right way.

Is the following will consider a mathematical proof:

When x is rational the double limit can be written as:

$\{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot \{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot ...=\{(-1)(-1)\}\{(-1)(-1)\}...=1\cdot 1\cdot 1 ...=1$

or:

$\{cos(2t\pi)cos(2t\pi)\}\cdot \{cos(2t\pi)cos(2t\pi)\}\cdot ...=1\cdot 1\cdot 1\cdot ... =1$

and if x irrational:

The double limit can be understood like that(?):

$\prod_{n = 0}^\infty x_n^2$ when $|x_i|<1$ for all $i$, which is equals (the product) to 0.

$\lim_{k\to\infty}\lim_{n\to\infty}cos^{2k}(\pi n!e)=?$