Re: [math-fun] Turing -- world's fastest mathematician?
Be extremely careful about such analyses; the mere existence of Usain Bolt obliterated 100 years of scientific speculation about what makes a good human sprinter. I think that I will see (at least on TV!) someone (probably a Kenyan) do a sub-2 hour marathon in my lifetime (I'm currently 66). (Turing was 1 hour faster in the marathon, but he also had 30 years on me!) At 10:01 AM 12/26/2013, Warren D Smith wrote:
This page by Brian Skinner gives an interesting statistical analysis which concludes the fastest achievable mile time is 3min 39.6sec. http://gravityandlevity.wordpress.com/2009/04/22/the-fastest-possible-mile/
He then redid it for marathon finding 2hr 02min 43sec: http://gravityandlevity.wordpress.com/2011/04/11/the-fastest-possible-marath... In the comments he says he thinks this estimate has an error bar of +-15 seconds, i.e. making the bold prediction that 2hr 2min will never happen.
if these true, the best athletes so far were already quite near to best possible performances.
In[1]:= Limit[Abs[Sin[x]]^(1/x), x -> Infinity] Out[1]= 1 But on the RHS of In[1] inside the limit, it takes the value 0 for arbitrarily large x. --Dan
P.S. On the subject, a friend challenged me to determine with proof the limit as n -> oo of |sin(n)|^(1/n), where n takes integer values (or show the limit does not exist). I haven't solved this yet, but maybe you will. --Dan On 2013-12-26, at 10:45 PM, Dan Asimov wrote:
In[1]:= Limit[Abs[Sin[x]]^(1/x), x -> Infinity]
Out[1]= 1
But on the RHS of In[1] inside the limit, it takes the value 0 for arbitrarily large x.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If the expression has a limit it must be 1. This is the same as looking for the limit of (1/n)*log |sin(n)|. Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1]. In particular, there's a positive probability (i.e. the set of n with this property has positive asymptotic density) that its uniformly bounded away from 0, so that the limit over that subsequence must be 0 (and thus the original is 1). The only fly in the ointment are for those n which are close to m*pi for some integer m. That is n/m is a good approximation to pi. We do know about the irrationality measures for pi. There is a positive constant C, and exponent e, so that | pi - n/m | > C/m^e The best known e is due to Hata (see Wadim Zudlin's nice survey http://arxiv.org/abs/math/0404523) and is a little more than 8. Nevertheless, it's thought (conjecture) that the "right" value is e=2 (though this is somewhat controversial). So |m*pi - n| > C/m^(e-1) or we have |sin(n)| is approximately C'/n^(e-1) (for some other C', since n <= 4m). Taking logs says that log |sin(n)| is approximately log C' - (e-1)*log(n). So dividing by n and taking the limit gives 0. So I'd say that the limit in the original problem exists and is 1. Victor On Fri, Dec 27, 2013 at 11:19 AM, Dan Asimov <dasimov@earthlink.net> wrote:
P.S. On the subject, a friend challenged me to determine with proof the limit as n -> oo of |sin(n)|^(1/n), where n takes integer values (or show the limit does not exist).
I haven't solved this yet, but maybe you will.
--Dan
On 2013-12-26, at 10:45 PM, Dan Asimov wrote:
In[1]:= Limit[Abs[Sin[x]]^(1/x), x -> Infinity]
Out[1]= 1
But on the RHS of In[1] inside the limit, it takes the value 0 for arbitrarily large x.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.) Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2). No? --Dan On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
You're right. Weyl's criterion says that m*pi mod 1 is uniformly distributed. So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform, You then get the formula that you gave. However, that doesn't invalidate my rough argument since this is just a mild distortion of uniform. Victor On Fri, Dec 27, 2013 at 1:01 PM, Dan Asimov <dasimov@earthlink.net> wrote:
The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.)
Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be
frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2).
No?
--Dan
On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
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I said
So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform,
I meant So you should have Prob(|sin(n)| in [a,b]) = Prob(n mod pi in [arcsin(a),arcsin(b)]) which isn't unform, On Fri, Dec 27, 2013 at 1:09 PM, Victor Miller <victorsmiller@gmail.com>wrote:
You're right. Weyl's criterion says that m*pi mod 1 is uniformly distributed. So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform, You then get the formula that you gave. However, that doesn't invalidate my rough argument since this is just a mild distortion of uniform.
Victor
On Fri, Dec 27, 2013 at 1:01 PM, Dan Asimov <dasimov@earthlink.net> wrote:
The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.)
Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be
frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2).
No?
--Dan
On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
What follows is a less hand-wavy proof that lim |sin(n)|^(1/n) = 1. The Weyl criterion was a bit of a red herring, though it lead me to realize that if the limit existed, then the answer had to be 1. The only way that the limit couldn't be 1 would be if pi had an unusually close sequence of rational approximations. More specifically 1) For all real x we have |sin(x)| <= |x| (pretty easy exercise) If || x || denotes the distance to the nearest integer from x, then we have 2) |sin(x)| <= pi* ||x/pi|| (applying (1) to x mod pi) Thus, if n is an integer we have 0 <= - log |sin(n)| <= -log pi - log || n/pi || However, by what I said before there is a positive real C, and a positive e (known to be around 8, is sufficient) such that ||n/pi || >= C/n^e Taking logs, and plugging into the above we get 0 <= -log|sin(n)| <= \log pi - log C + e*log(n) We then divide by n and take limits, and see that -(1/n) log |sin(n)| is squeezed to become 0. On Fri, Dec 27, 2013 at 1:36 PM, Victor Miller <victorsmiller@gmail.com>wrote:
I said
So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform,
I meant
So you should have Prob(|sin(n)| in [a,b]) = Prob(n mod pi in [arcsin(a),arcsin(b)]) which isn't unform,
On Fri, Dec 27, 2013 at 1:09 PM, Victor Miller <victorsmiller@gmail.com>wrote:
You're right. Weyl's criterion says that m*pi mod 1 is uniformly distributed. So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform, You then get the formula that you gave. However, that doesn't invalidate my rough argument since this is just a mild distortion of uniform.
Victor
On Fri, Dec 27, 2013 at 1:01 PM, Dan Asimov <dasimov@earthlink.net>wrote:
The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.)
Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be
frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2).
No?
--Dan
On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
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Whoops, for (1) I meant that for -pi/2 <= x <= pi/2, that |sin(x)| >= |x| or that |sin(x)| >= pi || x/pi || for *all* real x. The rest of what I said follows without change. On Fri, Dec 27, 2013 at 2:59 PM, Victor Miller <victorsmiller@gmail.com>wrote:
What follows is a less hand-wavy proof that lim |sin(n)|^(1/n) = 1.
The Weyl criterion was a bit of a red herring, though it lead me to realize that if the limit existed, then the answer had to be 1. The only way that the limit couldn't be 1 would be if pi had an unusually close sequence of rational approximations. More specifically
1) For all real x we have |sin(x)| <= |x| (pretty easy exercise) If || x || denotes the distance to the nearest integer from x, then we have 2) |sin(x)| <= pi* ||x/pi|| (applying (1) to x mod pi)
Thus, if n is an integer we have
0 <= - log |sin(n)| <= -log pi - log || n/pi ||
However, by what I said before there is a positive real C, and a positive e (known to be around 8, is sufficient) such that
||n/pi || >= C/n^e
Taking logs, and plugging into the above we get
0 <= -log|sin(n)| <= \log pi - log C + e*log(n)
We then divide by n and take limits, and see that -(1/n) log |sin(n)| is squeezed to become 0.
On Fri, Dec 27, 2013 at 1:36 PM, Victor Miller <victorsmiller@gmail.com>wrote:
I said
So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform,
I meant
So you should have Prob(|sin(n)| in [a,b]) = Prob(n mod pi in [arcsin(a),arcsin(b)]) which isn't unform,
On Fri, Dec 27, 2013 at 1:09 PM, Victor Miller <victorsmiller@gmail.com>wrote:
You're right. Weyl's criterion says that m*pi mod 1 is uniformly distributed. So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform, You then get the formula that you gave. However, that doesn't invalidate my rough argument since this is just a mild distortion of uniform.
Victor
On Fri, Dec 27, 2013 at 1:01 PM, Dan Asimov <dasimov@earthlink.net>wrote:
The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.)
Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be
frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2).
No?
--Dan
On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
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participants (3)
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Dan Asimov -
Henry Baker -
Victor Miller