12

I have to find out the following limit $$\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!πx)^{2m}]$$ for $x$ rational and irrational. for $x$ rational $x$ can be written as $\frac{p}{q}$ and as $n!$ will have $q$ as its factor the limit should be equal to 1. the second part of irrational is giving me problems. I first thought that limit should be zero as absolute value of cosine term is less than 1 and power it to infinity you should get $0$. But then I realised that it was wrong. I brought the limit down to this form. $$e^{-\sin^2(n!πx)m}$$ after this I find the question quite ambiguous as they have just said $x$ is irrational. If I take $x$ as $\frac{1}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e}$ but if I take $x$ as $\frac{2}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e^4}$. please help me and tell me where have I gone wrong?

Martin Sleziak
  • 53,687
avz2611
  • 3,658
  • It doesn't make sense to "take $x$ as $\frac{1}{n! x \sqrt{m}}$---by definition, you're evaluating a limit in $m$ and $n$, i.e., studying the behavior for fixed $x$ as $m, n$ vary. Anyway, I don't think the limit exists for irrational $x$, at least not for this ordering of the limits. For fixed $m$, $\cos(n! \pi x)$ with vary in $(-1 , 1)$ without converging, and thus so will $\cos^{2m}(n! \pi x)$. – Travis Willse Oct 16 '14 at 07:02
  • 3
    related discussion – Petite Etincelle Oct 16 '14 at 07:30
  • See this and this and other questions shown there among linked questions. – Martin Sleziak Oct 16 '14 at 08:03
  • It says it is 0, haven't I just shown that is wrong @martin – avz2611 Oct 16 '14 at 08:24
  • D'oh! I noticed only now that this is a different questions. Order of the limit is different. Your question is about $\lim_{m \to \infty} \lim_{n \to \infty} \cos^{2m}(n! \pi x)$, the questions I linked are about $\lim_{n \to \infty} \lim_{m \to \infty} \cos^{2m}(n! \pi x)$ – Martin Sleziak Oct 16 '14 at 08:29
  • My guess would be that for irrational $x$ the inner limit $\lim_{n \to \infty} \cos^{2m}(n! \pi x)$ does not exist. (Or, equivalently, the limit $\lim_{n \to \infty} \cos(n! \pi x)$ does not exist.) – Martin Sleziak Oct 16 '14 at 08:35
  • Maybe this one doesn't, but you can have a limit nevertheless when m->∞, using considerations like cos(n!$\pi$x) <1 when x is irrational. I don't know, but it is not impossible – mvggz Oct 16 '14 at 08:38
  • In fact the one annoying thing would be that $U_n = sin(\pin!x) $ is dense in ]-1,1[. In such case you would have a hard time proving that the limit is 0, since you could not majorate your sequence by something constant and smaller than 1 – mvggz Oct 16 '14 at 08:43
  • possible duplicate of Dirichlet's function – CiaPan Oct 16 '14 at 08:44
  • @CiaPan It's not a duplicate of that question - the limits are taken in a different order. (See my comment above.) – Martin Sleziak Oct 16 '14 at 12:02
  • 2
    By the "related discussion" link, the inner limit for $x=e$ is $1$, and so is the outer limit. Presumably, you can extend that to all integer powers of $e$ and many other reals, using series with a faster decrease than $1/i!$. –  Feb 20 '15 at 09:23

1 Answers1

1

For $m>0$ the limit of $cos^{2m}(n!\pi.x)$ as $n$ goes to infinity does not exist for some $x$. For example, for natural number k, let $f(k)=1/2$ if $k$ is an integer power of $2$, otherwise let $f(k)=0$. Let $x=f(0)/0!+f(1)/1!+f(2)/2!+...$

DanielWainfleet
  • 57,985