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Saved to my thinkbank.

I understand the formula but not the implications, other than determining whether a given number is rational 😅

The case when x is rational is easier to see as m!x will be an integer from some value of m. Cool formula 🤓

I choose many server to save my images

The other side, I guess the cos^2 is less than 1 so the limit is 0.

I can live with it.

I mean … yeah …

Can x at least be reasonable?

ITS ALL GREEK TO ME :)

If X is rational, there will be at least one (actually, infinitely many) Ms for which MX is an integer, and therefore the cosine is either -1 or +1 M! includes many even numbers, therefore the cosine will be +1, not -1 For irrational X, the cosine is always strictly between -1 and +1, and therefore tends to zero as n increases (I still suspect in missing something, no pen and paper at hand) And now I'm curious about complex X too

So -1<=cos(stuff)<=1 So if we have an even power of cos(stuff) then 0<=cos^(2n)(stuff)<=1 Now let then power tend to infinity and you'll see the curves' peaks get steeper so that as n gets larger, most values are close to zero whereas the peaks shoot up to 1, WHEREVER THERE IS A MULTIPLE OF PI. Use geogebra to convince yourself of this (see below) . Now consider the 'stuff' inside the function. Now if x is a rational number, then it can be expressed as p/q where p and q are integers. After multiplying by m!, as m gets arbitrarily large, you're bound to cancel the q. Hence we are dealing with a multiple of pi and its therefore a peak and equals 1. Otherwise it equals zero. Not a proof, just an attempt to explain intuitively. Hope that helps.

The reason it cannot take a value of -1 is simply because its an even power of cos(x).

I've subsequently realised that you almost certainly knew the power was even and hence it takes a value of 1 in any case. Apologies!