[math-fun] Identity (5)
in http://gosper.org/fst.pdf, namely Sum[((-1)^n* Cos[Sqrt[(n + 1/2)^2 + a^2]*Pi])/((n + 1/2)*(n - a + 1/2)*(n + a + 1/2)), {n, 0, Infinity}] == (Pi*(Sin[a*Pi]^2 - 2*Sin[(a*Pi)/Sqrt[2]]^2))/(2*a^2*Cos[a*Pi]) gives oo=oo for a = 1/2+ (integer k), but gives (6*(-1)^k*Cos[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^3 + ((-1)^k*Sqrt[2]*Pi*Sin[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^2 if you skip the n=k term. --rwg With help from Corey and Julian, and buffoonery from Mathematica.
Consider the standard binary tree with infinitely many levels.
Suppose each edge is colored green with probability = p.
What is the probability f(p) that there exists an infinite green path starting at the root?
--Dan
----------------------------------------------------------------------- f(p) = (2p - 1)/p^2 for p > 1/2, f(p) = 0 for p <= 1/2. -- Gene Yow, slope 8 at p=1/2+ ! Maybe building a tree is a good way to test coins for fairness. --rwg
On Sun, Mar 4, 2012 at 6:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
in http://gosper.org/fst.pdf, namely
Sum[((-1)^n* Cos[Sqrt[(n + 1/2)^2 + a^2]*Pi])/((n + 1/2)*(n - a + 1/2)*(n + a + 1/2)), {n, 0, Infinity}] == (Pi*(Sin[a*Pi]^2 - 2*Sin[(a*Pi)/Sqrt[2]]^2))/(2*a^2*Cos[a*Pi])
gives oo=oo for a = 1/2+ (integer k), but gives
(6*(-1)^k*Cos[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^3 + ((-1)^k*Sqrt[2]*Pi*Sin[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^2
if you skip the n=k term.
Correction: n = |k+1/2| - 1/2
--rwg With help from Corey and Julian, and buffoonery from Mathematica.
Identity (5) generalizes to the bilateral sum (op. cit. last page) Sum[(((-1)^n*Cos[Sqrt[(phi + n + a/b)*(phi + n + a*b)]*Pi])/(phi + n))*(phi + n - a), {n, -Infinity, Infinity}] == (Pi*((Sin[a*Pi]*Cos[(phi - a)*Pi])/Sin[phi*Pi] - 2*Sin[(a*(b + 1)*Pi)/(2*Sqrt[b])]^2))/(a*Sin[(phi - a)*Pi]) which blows up when phi approaches any integer k, unless the sum skips n=-k, whereupon both sides equal (4*b*Cos[a*Pi] - 2*a*Pi*(2*b*Cos[(a*(1 + b)*Pi)/Sqrt[b]]*Csc[a*Pi] + (1 + b^2)*Sin[a*Pi]))/(4*a^2*b) (independent of k). Similarly, when phi approaches k+a, the sum must skip n=-k to give the finite value (2*Cos[(a*(1 + b)*Pi)/Sqrt[b]] + a*Pi*(-2*Cot[a*Pi] + ((1 + b)*Sin[(a*(1 + b)*Pi)/Sqrt[b]])/Sqrt[b]))/(2*a^2) Identity (5) came from specializing parameters and then "folding the sum in half". This somehow makes it sensitive to which term blows up. --rwg
Consider the standard binary tree with infinitely many levels.
Suppose each edge is colored green with probability = p.
What is the probability f(p) that there exists an infinite green path starting at the root?
--Dan
-----------------------------------------------------------------------
f(p) = (2p - 1)/p^2 for p > 1/2, f(p) = 0 for p <= 1/2.
-- Gene
Yow, slope 8 at p=1/2+ ! Maybe building a tree is a good way to test coins for fairness. --rwg
mrob>There is also an intentional surcharge for oddly-shaped packages, for the packing reason you mentioned and for other reasons. rwg>Shouldn't I get a discount for shipping twice as many regular tetrahedra as regular octahedra of the same edge length? --rwg
From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Sent: Sunday, March 4, 2012 6:45 PM Subject: [math-fun] Identity (5)
Consider the standard binary tree with infinitely many levels.
Suppose each edge is colored green with probability = p.
What is the probability f(p) that there exists an infinite green path starting at the root?
--Dan
-----------------------------------------------------------------------
f(p) = (2p - 1)/p^2 for p > 1/2, f(p) = 0 for p <= 1/2.
-- Gene
Yow, slope 8 at p=1/2+ ! Maybe building a tree is a good way to test coins for fairness. --rwg
A coin can be characterized by a quantity q, depending on its geometry, internal density distribution, and the prescribed tossing process, which is experimentally manifested as the probability of heads. When a coin has been repeatedly tossed, resulting in h heads and t tails, there is no better way to estimate q than by the standard application of Bayes' theorem. Before the coin is tossed, our state of knowledge is given by some prior probability p0(q), and after the tosses, our updated knowledge is given by the posterior probability p(q | h, t) = N p0(q) q^h (1 - q)^t. The normalization constant N is chosen to make int(p(q | h, t), q = 0..1) = 1. If we take the prior to be uniform, p(q | h, t) = ((h + t + 1)! / (h! t!)) q^h (1 - q)^t. -- Gene
participants (2)
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Bill Gosper -
Eugene Salamin