It is worth noticing that this sequence is the chaotic case of the logistic map, which is related to the bit-shift map exactly by the transformation t_n = sin^2(2*pi*a_n). The bit shift map applied to a number x in [0,1] yields 2x if x
It is interesting that the solutions for t, as the number of iterations goes to infinity, converges in distribution to a beta(1/2,1/2) distribution which is the "stationary distribution" of the logistic map for r=4
Bonjour ! I don't undestand all words in englisrh but your explination still "clair" . j'ai appris beaucoup d'astuces avec vous. . Merci beaucoup. Continuez !!!
I like the closed form solution found for the sequence by clever use of the sine function and proof by induction More generally, given any value of t the formula can be used to calculate the result of nth iteration by first finding theta = arcsine(sqrt(t)) and plugging this angle into the sin^2(2^n*theta). Now this obviously works for all values of t on the closed interval [0,1] but it also works for all real t outside this interval using complex numbers. That is, the arcsine(x) for values of x outside the closed interval [-1,1] is given by -i*ln(i*x + sqrt(1 - x^2)) which can be derived from Euler’s formula. Even though the resulting angle theta is complex, substitution into sin^2(2^n*theta) returns the expected real value for nth iteration of the sequence. The formula I derived for the nth iteration when t > 1 is 1 - ((sqrt(t) + sqrt(t - 1))^(2^n ) + (sqrt(t) + sqrt(t -1))^(-2^n))^2/4. For example, if t = 4 and n = 3 the result is -354,079,488 the same as the direct calculation. Notice that this formula does not use either the natural log or the trig functions. They dropped out after simplification.
It seems pretty easy to get iterated radicals instead of sines. Solving y=4x(1-x) for x, gives x=(1±sqrt(1-y))/2. Starting from y=0 we get 0 and 1, y=1 gives 1/2 and y=1/2 gives (2±sqrt(2))/4. And here is where the fun begins - if a=2±sqrt(2±sqrt(2+...) iterated finitely many times, then y=a/4 gives x=(2±sqrt(a))/4. So, after checking that "a/4s" don't get larger than 1, we see that t = 0,1, and all the numbers of the form a/4 with up to 2022 nested surds.
What's wrong with this reasoning: if a(n+1) = 4 * a(n) * (1-a(n)), and a(n+1) = 0, then either a(n) = 0 or (1-a(n)) = 0. So a(n) = 0 or a(n) = 1. By induction, the same is true of a(n-1), a(n-2), ... a(0). ???
If a(n) = 1, then setting a(n-1)=a for simplicity, 4a(1-a)=1 giving 4a^2-4a+1=0 or (2a-1)^2=0. Hence a(n-1)=a=1/2. So you could say that a(n-1) can be 0, 1/2, or 1. Then a(n-2) could be any of these same values plus two more for the case when a(n--1) = 1/2. And so on.
@@KAS_ACADEMY So you will be calling France by it's official name "France" in Turkish? I don't think so! You want to determine Turkeys official name in a language like English, even with a letter that doesn''t exist in English, why not even use Chinese characters or Kanji? Tell me, that's not silly! I think, I'll call it rather 土耳其! That's my decision, live with it!
The trick of t= sin^2 theta is really amazing
It is worth noticing that this sequence is the chaotic case of the logistic map, which is related to the bit-shift map exactly by the transformation t_n = sin^2(2*pi*a_n). The bit shift map applied to a number x in [0,1] yields 2x if x
The value 4 for logistic map is a well-known case in chaos theory that has a analytic solution.
It is interesting that the solutions for t, as the number of iterations goes to infinity, converges in distribution to a beta(1/2,1/2) distribution which is the "stationary distribution" of the logistic map for r=4
I love solutions that make you say "how did they come up with this"
The values of t giving a[n]=0 can be expressed as:
0, 1,
1/2,
1/2 ± √(2)/4
1/2 ± √(2 ± √2)/4,
1/2 ± √(2 ± √(2 ± √2))/4,
…
1/2 ± √(2 ± √( … ± (2 ± √2) … )) / 4
where the deepest nesting of square roots is n-3.
Bonjour ! I don't undestand all words in englisrh but your explination still "clair" . j'ai appris beaucoup d'astuces avec vous. . Merci beaucoup. Continuez !!!
Thank you! I've been looking for the solution to this problem, off and on, for around thirty years.
Really nice demonstration
I like the closed form solution found for the sequence by clever use of the sine function and proof by induction More generally, given any value of t the formula can be used to calculate the result of nth iteration by first finding theta = arcsine(sqrt(t)) and plugging this angle into the sin^2(2^n*theta). Now this obviously works for all values of t on the closed interval [0,1] but it also works for all real t outside this interval using complex numbers. That is, the arcsine(x) for values of x outside the closed interval [-1,1] is given by -i*ln(i*x + sqrt(1 - x^2)) which can be derived from Euler’s formula. Even though the resulting angle theta is complex, substitution into sin^2(2^n*theta) returns the expected real value for nth iteration of the sequence. The formula I derived for the nth iteration when t > 1 is 1 - ((sqrt(t) + sqrt(t - 1))^(2^n ) + (sqrt(t) + sqrt(t -1))^(-2^n))^2/4. For example, if t = 4 and n = 3 the result is -354,079,488 the same as the direct calculation. Notice that this formula does not use either the natural log or the trig functions. They dropped out after simplification.
It seems pretty easy to get iterated radicals instead of sines. Solving y=4x(1-x) for x, gives x=(1±sqrt(1-y))/2. Starting from y=0 we get 0 and 1, y=1 gives 1/2 and y=1/2 gives (2±sqrt(2))/4. And here is where the fun begins - if a=2±sqrt(2±sqrt(2+...) iterated finitely many times, then y=a/4 gives x=(2±sqrt(a))/4. So, after checking that "a/4s" don't get larger than 1, we see that t = 0,1, and all the numbers of the form a/4 with up to 2022 nested surds.
As bayrakları as as as İstiklal Marşı'nın kabulünde Michael Penn
If only there was a way to find a closed form for this recurrence.
You can use a generating function to show that t = 0
You were computing a2024, which is an+1. Shouldn’t you have stopped the sequence of theta at 2^2023 pi?
pls derive ramanujans equations of near misses to fermats last theorem and post a video
Michael, is there a place to send you problems?
Just saw the google form in the description! Never mind
audio is almost mute
what
Was the reason you uploaded this today it's our anthem's anniversary? (103rd year!)
Can't you just say that t is in the interval [0;1]?
What's wrong with this reasoning: if a(n+1) = 4 * a(n) * (1-a(n)), and a(n+1) = 0, then either a(n) = 0 or (1-a(n)) = 0. So a(n) = 0 or a(n) = 1. By induction, the same is true of a(n-1), a(n-2), ... a(0). ???
You are stuck with the case a(n)=1.
@@t567698 Doh. Thanks.
If a(n) = 1, then setting a(n-1)=a for simplicity, 4a(1-a)=1 giving 4a^2-4a+1=0 or (2a-1)^2=0. Hence a(n-1)=a=1/2. So you could say that a(n-1) can be 0, 1/2, or 1. Then a(n-2) could be any of these same values plus two more for the case when a(n--1) = 1/2. And so on.
Thanks to you both. Don't know what I was thinking.
neat
First view first comment
Please,Türkiye not Turkey.
Best Regards
France, not Fransa! Are you okay with that? This is silly!
@@rainerzufall42 not silly; The official name is Türkiye
@@KAS_ACADEMY Then you'll call France "France" instead of "Fransa" in Turkish in future?
@@rainerzufall42 İt is nit about how you call them; it is about how to write official name
@@KAS_ACADEMY So you will be calling France by it's official name "France" in Turkish? I don't think so! You want to determine Turkeys official name in a language like English, even with a letter that doesn''t exist in English, why not even use Chinese characters or Kanji? Tell me, that's not silly! I think, I'll call it rather 土耳其! That's my decision, live with it!