sin(n) does not converge - the subsequence approach

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We look at the limit, as n tends to infinity, of sin(n), where n are integers.

00:00 Intro
00:07 Continuous limit
01:41 Discrete limit - idea of the proof
03:18 Proof that sin(n) does not converge

Dr Barker

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I was thinking about this yesterday. Thanks

aweebthatlovesmath

Hello The Beautiful Mind, dear *Dr. Barker* .
Thank you so much, I like it.
Waiting For Your Next Video.

wuyqrbt

Nothing like a little real analysis in the morning!

comicrelief

What about if you horizontally scale the function such that the sine cycle is less than 1? How would you modify the proof for the integers in that case?

bscutajar

My thought was to use the continued fraction approximations of pi. If you take a(j) to be twice the numerator of each successive approximation, it gets converges on 2*k*pi, so sin(a(j)) converges on 0. So set b(j)=a(j)+1. Then sin(b(j)) will converge on sin(1). These don't converge on the same value, so sin(n) doesn't converge, just like your argument.

jursamaj

I could see where you were going once you highlighted the intervals based on multiples of pi/6, but expected you would use the ceiling function to express integers in those intervals more conveniently. E.g. a_1 = Ceiling(pi/6), b_1 = Ceiling(7pi/6), etc., having shown that the width of each interval accommodates an integer as you did. Incidentally, the sin(a_k) terms are all strictly >1/2, as pi is irrational. Similarly for the sin(b_k)s. Not that this makes a scrap of difference to a nice proof!

RGP_Maths

When dealin with SIN:
It’s just SPOOKY how relatively little we KNOW:
E=MC2

glennfrick

The natural question to ask is what are thr possible values of thr limit. For sin(x) we know that the limits are any no. Between (-1, +1). What happens for the discrete case. The fact that the limit isnt unique isnt shocking

shohamsen

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