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.
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?
My guess is that no matter how much you scale the function horizontally, you will always achieve a countable infinite number of intervals such that sin(Ax) (with 'A' being the scaling factor) is less than -1/2 (resp. more than +1/2) on that interval, and that interval contains an integer. The actual construction of a_n and b_n would be harder, if not impossible, but the proof doesn't require a construction. I wouldn't be surprised if this was due to the fact that pi is transcendental over Q. The details are beyond me for the moment.
And once you have that countable infinite collection of such intervals, you just pick an integer from each one, and the same argument follows as with this video.
Some scales A for sin(Ax) do have a convergent limit for example sin(3pi x) has a period less than 1 but lim sin(3pi n) = 0. Where the scale A is not an integer multiple of pi the limit sin(An) doesn't exist. I.e., the period T = 2pi/A isn't of the form 1/N or 2/N for some integer N. Case 1: Assume the period T is rational and of the form p/q with p and q being integers that are coprime and p >= 3. Since p and q are coprime we can find an integer M such that q * M = 1 (mod p). Define the sequence a_k = M + pk and b_k = pk. Then sin(A*a_k) = sin(2pi/T * a_k) = sin( 2pi q/p * (M+pk) ) = sin( 2pi q*M/p + 2pi * qk) = sin(2pi/p). Similarly sin(A*b_k) = 0. Thus lim sin(A*a_k) = sin(2pi/p) 0 = lim sin(A*b_k) so in particular the lim sin(A*n) doesn't exist. Case 2: Assume the period T is irrational. Since the fractional parts of the multiples of 1/T, i.e. the set { {m/T} : m an integer }, is dense in the unit interval, we can find a sequence a_k and b_k of integers such that {a_k/T} and {b_k/T} converges to 1/4 and 3/4 respectively. By taking subsequences if necessary we can also assume 1/12 < {a_k/T} < 5/12. Then sin(A*a_k) >= 1/2 iff A*a_k mod 2pi in [pi/6 , 5pi/6] iff a_k/T = A*a_k / 2pi mod 1 in [1/12 , 5/12] iff {a_k/T} in [1/12 , 5/12] Similarly we can take subsequences for the b_k such that sin(A*b_k) = 1/2 and liminf sin(An)
I just wandering. The subsequence should contain the values of the main sequence (Un). And n is integers, we can't choose k2(pi) as n right. so we don't have points where sin(n) be zero or 1 neither -1. so how do we say it has converging subsequence. anyone have an idea?
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!
I agree. The ceiling (or floor) function, I think if clearer than the using min or max of an intersection. At the end you say sin(a_k) terms are all strictly > 1/2, as pi is irrational. I don’t see why pi being irrational is relevant. What am I missing?
@@bobh6728 Remember that a_k is an integer, since these represent a subsequence of all integers. But sin(a_k) is 1/2 only for some specific multiples of pi/6. For this to be possible would mean there is an integer n such that n = m x pi/6, where m is also an integer. But that implies pi = 6n/m, which is impossible since it expresses pi as the quotient of two integers.
@@RGP_Maths Thanks. So you were proving you can go from >= to just >, or the strictly part. That makes sense now. I thought you were saying since pi is irrational then sin(n_a) is > 1/2 which is not true by itself.
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
I'd imagine that, for any number in [-1, 1], it should be possible to find a subsequence which converges to it. Probably something you could prove using irrationality of pi and possibly other properties of pi?
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
I was thinking about this yesterday. Thanks
Hello The Beautiful Mind, dear *Dr. Barker* .
Thank you so much, I like it.
Waiting For Your Next Video.
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ua-cam.com/play/PLgwyp12I7OOCFyYpl-Qyr13r-xGdX3BfR.html
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.
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?
My guess is that no matter how much you scale the function horizontally, you will always achieve a countable infinite number of intervals such that sin(Ax) (with 'A' being the scaling factor) is less than -1/2 (resp. more than +1/2) on that interval, and that interval contains an integer. The actual construction of a_n and b_n would be harder, if not impossible, but the proof doesn't require a construction.
I wouldn't be surprised if this was due to the fact that pi is transcendental over Q. The details are beyond me for the moment.
And once you have that countable infinite collection of such intervals, you just pick an integer from each one, and the same argument follows as with this video.
Some scales A for sin(Ax) do have a convergent limit for example sin(3pi x) has a period less than 1 but lim sin(3pi n) = 0. Where the scale A is not an integer multiple of pi the limit sin(An) doesn't exist. I.e., the period T = 2pi/A isn't of the form 1/N or 2/N for some integer N.
Case 1: Assume the period T is rational and of the form p/q with p and q being integers that are coprime and p >= 3. Since p and q are coprime we can find an integer M such that q * M = 1 (mod p). Define the sequence a_k = M + pk and b_k = pk.
Then sin(A*a_k) = sin(2pi/T * a_k) = sin( 2pi q/p * (M+pk) ) = sin( 2pi q*M/p + 2pi * qk) = sin(2pi/p). Similarly sin(A*b_k) = 0.
Thus lim sin(A*a_k) = sin(2pi/p) 0 = lim sin(A*b_k) so in particular the lim sin(A*n) doesn't exist.
Case 2: Assume the period T is irrational. Since the fractional parts of the multiples of 1/T, i.e. the set { {m/T} : m an integer }, is dense in the unit interval, we can find a sequence a_k and b_k of integers such that {a_k/T} and {b_k/T} converges to 1/4 and 3/4 respectively.
By taking subsequences if necessary we can also assume 1/12 < {a_k/T} < 5/12.
Then sin(A*a_k) >= 1/2
iff A*a_k mod 2pi in [pi/6 , 5pi/6]
iff a_k/T = A*a_k / 2pi mod 1 in [1/12 , 5/12]
iff {a_k/T} in [1/12 , 5/12]
Similarly we can take subsequences for the b_k such that sin(A*b_k) = 1/2 and liminf sin(An)
When dealin with SIN:
It’s just SPOOKY how relatively little we KNOW:
E=MC2
I just wandering. The subsequence should contain the values of the main sequence (Un). And n is integers, we can't choose k2(pi) as n right. so we don't have points where sin(n) be zero or 1 neither -1. so how do we say it has converging subsequence. anyone have an idea?
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!
Good idea to use ceiling functions - that's much more concise than what I wrote!
I agree. The ceiling (or floor) function, I think if clearer than the using min or max of an intersection.
At the end you say sin(a_k) terms are all strictly > 1/2, as pi is irrational. I don’t see why pi being irrational is relevant. What am I missing?
@@bobh6728 Remember that a_k is an integer, since these represent a subsequence of all integers. But sin(a_k) is 1/2 only for some specific multiples of pi/6. For this to be possible would mean there is an integer n such that n = m x pi/6, where m is also an integer. But that implies pi = 6n/m, which is impossible since it expresses pi as the quotient of two integers.
@@RGP_Maths Thanks. So you were proving you can go from >= to just >, or the strictly part. That makes sense now. I thought you were saying since pi is irrational then sin(n_a) is > 1/2 which is not true by itself.
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
That interval should be written [-1,1]. This comment is confused. You seem to misunderstand this video.
I'd imagine that, for any number in [-1, 1], it should be possible to find a subsequence which converges to it. Probably something you could prove using irrationality of pi and possibly other properties of pi?
Oh, that was a little anti-climatic.
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
If the limit isn't unique, it means the limit doesn't exist
The concept of "possible values for the limit" is incoherent.