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For those unfamiliar, a question explaining the definition is in the question here. The definition itself is

$$\lim_{m\to\infty}\lim_{n\to\infty}\cos^{2n}\left(m!\pi x\right)$$ and evaluates to 1 for rational $x$ and 0 for irrational $x$. The part I don't totally understand is why the exponent is $2n$ as opposed to just $n$.

This is a very-not rigorous question, but my concern stems from the fact that as $n\to\infty$, $2n$ isn't necessarily even, let alone an integer, which negates the whole point of making rational values become $1$ instead of $\pm1$. It seems like you could, on the other hand, write it as

$$\lim_{m\to\infty}\lim_{n\to\infty}\left(\cos^{2}\left(m!\pi x\right)\right)^n$$ although I'm not entirely sure if this is the same expression.

My other question is that if the first equation is 'right', it seems to like it logically follows that

$$\lim_{n\to\infty}(-1)^{2n}=1$$ as well. Is this true?

Calvin Godfrey
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  • yes, $(-1)^{2n}=((-1)^2)^n=1^n=1$ for all natural numbers $n$ – J. W. Tanner Dec 31 '19 at 22:00
  • I agree with that, but when you're taking a limit, isn't it not necessarily over natural numbers/integers? – Calvin Godfrey Dec 31 '19 at 22:10
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    Wikipedia says, in the definition of Dirichlet function, the indices are integers – J. W. Tanner Dec 31 '19 at 22:14
  • Whoops! Missed that in the reading :P if you post that as an answer with a little more explanation I can accept it. – Calvin Godfrey Dec 31 '19 at 22:18
  • I could, but at the moment I'm confused why the exponent is $2n$ rather than $n$; for large enough $m$, when $x$ is rational, isn't $m!\pi x$ an even multiple of $\pi$? – J. W. Tanner Dec 31 '19 at 23:01
  • @J.W.Tanner : But the inner limit will not exist for some $x$, even if that failure is in the boring sense of oscillating between $\pm 1$. With the even powers, such oscillations do not occur for any $x$. – Lutz Lehmann Dec 31 '19 at 23:17
  • Thanks, @LutzLehmann, I see now; for example, when $x=1$ (or any odd integer) and $m=1$, $(\cos(m!\pi x))^n=\cos(\pi)^n=(-1)^n$, so the inner limit does not exist, but it does exist for all $m>1$, so does that really matter since we'll be taking $m\to\infty$? like $\lim\limits_{m\to\infty} \lim\limits_{n\to\infty} \dfrac1{m-1}$ doesn't make sense when $m=1$ – J. W. Tanner Dec 31 '19 at 23:26
  • It might matter if the outer limit is in some uniform, function space sense, not just point-wise. – Lutz Lehmann Dec 31 '19 at 23:32

2 Answers2

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There's going to be a lot of weasel words in this answer but bear with me.

Generally speaking variables that are named $i,j,k$ or $m,n$ are always integers. Also often limits that go to infinity are limits over sequences (indexed by integers) rather than limits over real numbers. Therefore, any time I see "$\lim_{n \to \infty}$" I assume that it is a limit over a sequence indexed by $n$ and hence $n$ is an integer. In many cases it doesn't matter if you consider a limit over the integers or over the real numbers but sometimes it does, as in this example.

You will also notice that "$m!$" isn't defined when $m$ is non-integral. So we must interpret $m$ to be an integer as well. (You can define $!$ at non-integer input but that isn't normally part of the definition.)

Trevor Gunn
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You were concerned that as $n\to \infty$, $2n$ is not necessarily even, let alone an integer,

but in the definition of the Dirichlet function in Wikipedia, $n$ is an integer.

Therefore, at rational numbers, the Dirichlet function becomes $1$, not $\pm1$,

and you could write $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\left(\cos^{2}\left(m!\pi x\right)\right)^n$ instead of $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\cos^{2n}\left(m!\pi x\right),$

and $\lim\limits_{n\to\infty}(-1)^{2n}=\lim\limits_{n\to\infty}((-1)^2)^n=\lim\limits_{n\to\infty}1^n=\lim\limits_{n\to\infty}1=1.$

J. W. Tanner
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