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Comments:
<0> yay Italy beat Germany!
<1> yay!
<2> italia!!1
<0> Yay yay! all italian girls, i e you!
<3> why is the cantor set compact?
<3> yo
<3> why is the cnator set compact?
<4> are subspaces of compact spaces compact?
<5> is ]0,1] compact?
<3> [0, 1] is compact
<6> integral: Closed subsets of compact spaces are compact.
<0> Cale: thanks for the link http://en.wikipedia.org/wiki/Cayley-****son_algebra , nice.
<7> Is it possible to represent pi as a limit and if so what is it
<8> magicwindow: its the second result on a google search -- http://functions.wolfram.com/Constants/Pi/09/
<0> magicwindow: http://en.wikipedia.org/wiki/Pi
<8> magicwindow: the first formula seems the simplest
<8> % Limit[(2^(4 n + 1) n!^4)/((2 n + 1) (2 n)!^2), n -> Infinity]
<9> [Shiba]: "Time limit exceeded for computation."
<8> lol
<0> the wikipedia is really becoming a amazing database of everything
<8> of course it can't compute that limit...
<10> hi could someone please give me a hint on this problem: Consider the tent sequence (t_n) given by t_{n+1} = 1 - 2|t_n - 1/2|, t_0 = a and the sequence of fractional parts (s_n) = (frac(2^n a)). I would like to prove that if (t_n) is uniformly distributed mod 1, then (s_n) is also uniformly distributed mod 1.
<7> I didn't understand the binomial part
<10> I also know that the general term of (t_n) is t_n = frac((-1)^floor(2^n a) 2^n a).
<7> But then again I think that's the fourth formula
<7> Oh on wikipedia I see
<10> perhaps it would help considering the binary expansion of the initial term a = 0.a1a2a3... where a_i \in {0,1} for all i>=1
<10> by considering two strictly increasing functions f, g:N -> N, where a_f(n) = 0, a_g(n) = 1, for all n>=1, one can write (t_n) = (frac(2^f(n) a)) U (1 - frac(2^g(n) a)), where U means that the two subsequences form a partition of the original sequence (t_n)
<10> would it help to consider by reductio ad absurdum that the sequence (s_n) is not uniformly distributed mod 1?
<10> a similar thing to prove would be that the initial term a is normal to base 2, since this is equivalent to the uniform distribution of (s_n)
<11> hello
<11> does anybody know a good method to calculate a division ?
<11> i want to implement it in a programming language
<11> for big numbers
<12> Subtract.
<10> koro: are you available by any chance?
<6> beck: What did you do in elementary school?
<6> :)
<10> Cale: are you familiar with uniform distribution theory by any chance?
<6> Fogg: um, maybe? Could you be more specific?
<10> I am trying to prove the problem I have stated above about the tent sequence
<10> more precisely, I would like to prove that if the tent sequence (t_n) is uniformly distributed mod 1 then the sequence (s_n) is u.d. mod 1
<0> Normality of will always depend on the infinite string of digits on the end, not on any finite computation.
<0> what does that mean?
<10> Cale: I have managed to prove vice-versa, if (s_n) is u.d. mod 1 then (t_n) is u.d. mod 1
<10> Cale: but I really believe this should be a "if and only if" statement
<6> xah: That if any tail of the digits of is normal, then is normal, but normality is not determined by any initial finite substring of the digits.
<0> well, how can there be a tail? since the digits are infinite?
<10> Cale: in the end I would like to be able to say: The tent sequence is u.d. mod 1 if and only if the initial term is normal to base 2.
<6> xah: by a tail, I mean the sequence {d_(n+k)} for n = 0 to infinity, for some k.
<10> Cale: I have seen a proof for: If (s_n) is u.d. mod 1 then the initial term is normal to base 2. This was easy since the substring b_1b_2...b_t appeared in the digits of the initial term starting from position k if and only if s_k \in [0.b_1b_2..b_t, 0.b_1b_2..b_t + 2^(-t)], and since (s_n) was u.d. mod 1, the asymptotic frequency of the substring b_1..b_t in the initial term would be 2^(-t), thus the normality
<10> Cale: but one could not say the same thing about the tent sequence since apart from the left bit-shifting it also does some bit-negation
<6> Fogg: yeah, it does
<6> Fogg: but only sometimes
<10> yes, iff the digit after the decimal point is 1
<6> hmm
<6> s/1/0/?
<6> oh, in the end
<6> yeah
<13> % Solve[x+2==3]
<0> Cale: ... sorry, but still don't unstand.
<9> bucky1: {{x -> 1}}
<10> Cale: yes.. considering t_k = .10111.., t_{k+1} =.1000...
<6> xah: It means that you can't determine that is normal, for instance, by computing it to n places and looking at those n digits.
<6> xah: no matter how large n may be
<10> Cale: do you think a proof by reduction to the absurd would be good? I mean suppose that the initial term is not normal to base 2, thus the sequence (s_n) is not u.d. mod 1. Would this imply that (x_n) is not u.d. mod 1, thus a contradiction?
<13> % Simplify[w^3 - (D +2*C+2*H)*w^2 + (2*D*H+H*H)*w + D*H*H]
<9> bucky1: D*H^2 + H*(2*D + H)*w - (2*C + D + 2*H)*w^2 + w^3
<14> xah: in fact you couldn't even prove that 0 is *not* normal that way.
<13> heh
<10> Cale: sorry, that should be (t_n) instead of (x_n)
<13> % Factor[w^3 - (D +2*C+2*H)*w^2 + (2*D*H+H*H)*w + D*H*H]
<9> bucky1: D*H^2 + 2*D*H*w + H^2*w - 2*C*w^2 - D*w^2 - 2*H*w^2 + w^3
<14> Fogg: do you have a reference that defines that tent sequence?
<10> int-e: the tent sequence is simply given by: t_{n+1} = 1 - 2|t_n - 1/2|, t_n = a \in [0,1)
<10> int-e: or in closed form: t_n = frac((-1)^floor(2^n a) 2^n a)
<0> Cale: ok, but what's with the tail thing?
<14> ok. I understand the name now.
<10> :)
<0> int-e: 0??
<6> xah: that you can tell whether (or really any number) is normal even if you disregard the first n digits completely, for any n.
<14> xah: 0 = 0.00000000000...
<14> xah: by looking at the first k digits you'll conjecture that that number isn't normal, but you could never proof it that way.
<13> % Solve[w^3 - (D +2*C+2*H)*w^2 + (2*D*H+H*H)*w + D*H*H,w]
<9> bucky1: Solve[D*H^2 + (2*D*H + H^2)*w - (2*C + D + 2*H)*w^2 + w^3, w]
<0> Cale: yeah ok. But still, one can't really know the tail of a infinite sequence can one?
<6> bucky1: Solve is for solving equations
<14> xah: in the example of 0 you actually know all digits ...
<10> could I even prove that if the tent sequence is uniformly distributed mod 1 then the initial term is simply normal to base 2? this would be a starting point.
<6> xah: Why not?
<13> Cale: how can i have mbot output that equation in terms of w
<13> or at least factor that equation
<6> bucky1: I don't see an equals sign
<0> int-e: your 0.0000... sound beserk. Zero is 0 all the way by definition.
<6> bucky1: So it's not an equation :)
<13> oh haha true
<15> ........
<14> xah: yes, but just by looking at the digits you'll never know that.
<13> % Solve[w^3 - (D +2*C+2*H)*w^2 + (2*D*H+H*H)*w + D*H*H==0,w]
<9> bucky1: {{w -> (2*C + D + 2*H)/3 - (2^(1/3)*(-(-2*C - D - 2*H)^2 + 3*(2*D*H + H^2)))/(3*(16*C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*C^2*H + 12*C*D*H - 6*D^2*H + 30*C*H^2 - 48*D*H^2 - 2*H^3 + Sqrt[(16*
<9> C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*C^2*H + 12*C*D*H - 6*D^2*H + 30*C*H^2 - 48*D*H^2 - 2*H^3)^2 + 4*(-(-2*C - D - 2*H)^2 + 3*(2*D*H + H^2))^3])^(1/3)) + (16*C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*
<9> C^2*H + 12*C*D*H - 6*D^2*H + 30*C*H^2 - 48*D*H^2 - 2*H^3 + Sqrt[(16*C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*C^2*H + 12*C*D*H - 6*D^2*H + 30*C*H^2 - 48*D*H^2 - 2*H^3)^2 + 4*(-(-2*C - D - 2*H)^2 + 3*(2*
<9> D*H + H^2))^3])^(1/3)/(3*2^(1/3))}, {w -> (2*C + D + 2*H)/3 + ((1 + I*Sqrt[3])*(-(-2*C - D - 2*H)^2 + 3*(2*D*H + H^2)))/(3*2^(2/3)*(16*C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*C^2*H + 12*C*D*H - 6*D^2*
<9> H + 30*C*H^2 - 48*D*H^2 - 2*H^3 + Sqrt[(16*C^3 + 24*C^2*D + 12*C*D^2 + 2*D^3 + 48*C^2*H + 12*C*D*H - 6*D^2*H + 30*C*H^2 - 48*D*H^2 - 2*H^3)^2 + 4*(-(-2*C - D - 2*H)^2 + 3*(2*D*H + H^2))^3])^(1/3)) -
<9> [5 @more lines]
<6> heh
<15> please, god, no @more lines
<13> wow sorry everyone..
<6> You're solving a cubic equation with polynomial coefficients, it's going to be a mess
<13> yea
<6> xah: so you know all the tails
<6> xah: In general, if you have the definition for a number, you know all its digits in some sense of the word.
<6> (supposing that it's computable, I guess)
<10> Cale: I've got as far as finding: lim_{n -> oo} 1/n \sum_{k=1}^n K_{a_ka_{k+1}} = 1/2 = lim_{n -> oo} 1/n \sum_{k=1}^n K_{(1-a_k)a_{k+1}}, therefore the asymptotic frequency of finding two consecutive equal or distinct digits is 1/2. Does this imply that the initial term a = 0.a_1a_2... is simply normal to base 2?
<14> I think you've just defined 'computable', sort of.
<6> Yeah, kind of
<10> Cale: where K_xy is the Kronecker delta
<0> so, the wikp statement: Normality of will always depend on the infinite string of digits on the end, not on any finite computation. basically says computation of digits is futile.
<0> very bad way to express it. Misleading.
<6> Fogg: Is 0.001100110011... normal?
<10> no it isn't
<6> xah: I thought it was pretty obvious what it said...
<15> it is fairly obvious, if you know a definition for normality
<15> i guess it's harder if you're learning from wikipedia as the source
<10> the group of digits 010 does not ever appear
<6> Fogg: right
<0> HiLander: exactly. That's why it's very bad.
<6> Fogg: but it has the property that the probabilities that two consecutive digits are equal or not equal are both 1/2
<0> and Cale made it worse by chalking it up to some tail of a sequence.
<10> Cale: well spotted.. thanks for the example
<10> that's even a rational number, it can never be normal :)
<15> Cale didn't "make it worse"
<10> Cale: but isn't it simply normal to base 2?
<6> I just reexpressed what it was saying in the hopes that another viewpoint on what it said would make it clearer
<10> Cale: since both 0 and 1 occur with equal asymptotic frequency
<12> http://eom.springer.de/
<6> Fogg: oh, is that all that's meant by 'simply'?
<10> Cale: yes
<6> Okay then, sure
<14> Fogg: uh ... why don't you just map all numbers t_i to t'_i min(t_i, 1-t_i) (if t_i is uud in [0,1], t'_i is uud in [0,1/2]? I think that gets rid of all negations.
<14> t'_i = min(t_i, 1-t_i)
<10> hmm.. let me think about that
<0> the gist of the reason that wikip is bad, is because it states the obvious. Normally this isn't a problem, but it becomes a serious problem, when the subject is complicated.
<16> Olathe, nice
<16> Olathe, will be nicer if they add a search feature too:)
<15> when the subject is complicated, i don't know why you'd expect random people on the internet to be a good source to begin with
<14> Fogg: I probably messed something up though.
<10> int-e: I am trying to see if min(t_i,1-t_i) actually gives the sequence of fractional parts (frac(2^n a))
<6> If for every n, the probability that d_(n+1) = d_n and that d_(n+1) = 1 - d_n are both 1/2 where d_n are the binary digits of the number, then it doesn't really matter what d_n actually is. d_(n+1) is 0 or 1 with probability 1/2 either way.
<14> Fogg: no, it doesn't.
<10> Cale: that is correct
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