July 2010 - mathfun@mailman.xmission.com

Re: [math-fun] reciprocal mod 100
by Dan Asimov 08 Jul '10

08 Jul '10

I wasn't really clear. I meant Rich's "whole algorithm" whereby the inverse of an element x of (Z/100Z)* is always of the form x^(-1) = x + K(x mod 10). So I wondered whether, by analogy, for any primes p and q, the inverse of an element x of (Z/((pq)^2 Z))* is of the form x^(-1) = x + L(x mod pq). --Dan << 33*3 = 99, which is -1 mod 100. Since 33*3 = -1, 3 = -1/33, and so 1/33 = -3 = 97. >> On Thu, Jul 8, 2010 at 2:44 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > Rich wrote: > > << > > Q: What's the reciprocal of 33 mod 100? > > Hint: Add 64. > > When N ends in 3, the (mod 100) reciprocal of N is N+64. > So the reciprocal of 33 is 97. 33*97 = 3201 = 1 (mod 100). > > The whole algorithm ... > Suppose the tens digit of N is T. > If N = 10T+1, then 1/N = 102-N > 10T+3, N+64 > 10T+7, N+36 > 10T+9, 98-N > >> > > So I wondered if this kind of thing always happens for the > square of a product of two primes, and programmed it up > for 441 = (3*7)^2. But my eye can't pick out any pattern > as simple as this one for 100. > > Hmm, does something like Rich's observation always work if one of the > primes is 2, > i.e., for numbers of the form 4 p^2, p prime ? I sleep as fast as possible so I can get more rest in the same amount of time.

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Re: [math-fun] Probability puzzle
by Mike Speciner 08 Jul '10

08 Jul '10

OK. Let me try to explain a bit more... [And, yes, I'm simplifying, but I think this is a sufficient exposition for the purpose of the problem.] There are a fixed number of types of sexually hereditable objects ("chromosomes"). A chromosome of a given type can come in many versions. A given individual has exactly two chromosomes of each type, one chosen randomly from its mother's corresponding pair and one chosen randomly from its father's pair. There is one gender-determining chromosome type with two subtypes, X and Y. A male individual has one X and one Y; a female has two Xs. So, for this pair, the gender of the offspring is determined by the chromosome subtype inherited from the father. (The father can contribute either his X or his Y, the mother always contributes one of her Xs.) [Since males inherit their Y chromosome from their fathers, this genetic information can be used to trace paternal ancestry.] There is a small amount of additional hereditable information that is passed from mother to child, her mitochondrial genome, which also comes in many versions. [So this can be used to trace maternal ancestry.] I'll call an individual's specific combination of chromosomes and mitochondrial genome its Hereditable Information (HI). The actual traits of an individual are determined by its HI. Simple exercise: Prove that the proportion of a given chromosome version in the population remains the same in future generations. Assume that no particular combination of chromosome versions has any survival advantage. Now, for a given couple, what would determine the probability of having a boy? From the above, it would seem that boys and girls are equally likely. It must be that one of the genders is more likely to be conceived (conception being the process that combines the father's and mother's hereditable contributions) or to survive to birth, and this would be determined by the specific pair of HIs possessed by the couple. (It might also depend on the specific HI being contributed by each parent for the potential offspring in question, what follows can be easily extended to include this as well.) So, can we invent a universe of HIs and a mapping from father-mother pairs of HIs to gender probabilities (at birth) that produces Andy's hypothesized bimodal gender ratio AND continues to produce it in succeeding generations? Or perhaps more easily, can we invent a universe and mapping that produces a stable overall gender ratio in the absense of termination rules, AND where application of the gender termination rules alters that steady state. Note that we may also have to specify coupling rules (i.e. how coparents are paired). --ms On Thursday 08 July 2010 12:42:31 Dan Asimov wrote: > Mike Speciner wrote: > > << > OK, but what are the plausible heredity rules that preserve the relevant > ratios in the steady state (i.e., without the termination rule)? > (By "plausible", I mean conforming to standard human genome kinds of > things--gender determination from XvY sperm, mitochondrial genome from > mother, mother-father gene pairs for everything else.) Presumably, the > children of the two kinds of couples are genetically different in some > way. What are the coupling rules that preserve the ratios, remembering > that you need a boy to couple with a girl, and we probably need to assume > monogamy to apply the "stop at first boy" rule (although it could be > applied to just one parent, taking just that parent out of the > reproducing pool). Assume a finite population, so at least some > "interracial" coupling is required in the steady state. > > > I just don't know enough biology to translate this question into a > mathematical one. > > But if you restated it in math language, that would help (at least) me. > > --Dan > > I sleep as fast as possible so I can get more rest in the same amount of > time. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] reciprocal mod 100
by Dan Asimov 08 Jul '10

08 Jul '10

Rich wrote: << Q: What's the reciprocal of 33 mod 100? Hint: Add 64. When N ends in 3, the (mod 100) reciprocal of N is N+64. So the reciprocal of 33 is 97. 33*97 = 3201 = 1 (mod 100). The whole algorithm ... Suppose the tens digit of N is T. If N = 10T+1, then 1/N = 102-N 10T+3, N+64 10T+7, N+36 10T+9, 98-N >> So I wondered if this kind of thing always happens for the square of a product of two primes, and programmed it up for 441 = (3*7)^2. But my eye can't pick out any pattern as simple as this one for 100. Hmm, does something like Rich's observation always work if one of the primes is 2, i.e., for numbers of the form 4 p^2, p prime ? --Dan I sleep as fast as possible so I can get more rest in the same amount of time.

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Re: [math-fun] Probability puzzle
by Dan Asimov 08 Jul '10

08 Jul '10

Mike Speciner wrote: << OK, but what are the plausible heredity rules that preserve the relevant ratios in the steady state (i.e., without the termination rule)? (By "plausible", I mean conforming to standard human genome kinds of things--gender determination from XvY sperm, mitochondrial genome from mother, mother-father gene pairs for everything else.) Presumably, the children of the two kinds of couples are genetically different in some way. What are the coupling rules that preserve the ratios, remembering that you need a boy to couple with a girl, and we probably need to assume monogamy to apply the "stop at first boy" rule (although it could be applied to just one parent, taking just that parent out of the reproducing pool). Assume a finite population, so at least some "interracial" coupling is required in the steady state. >> I just don't know enough biology to translate this question into a mathematical one. But if you restated it in math language, that would help (at least) me. --Dan I sleep as fast as possible so I can get more rest in the same amount of time.

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Re: [math-fun] Probability puzzle
by Dan Asimov 08 Jul '10

08 Jul '10

Okay, now suppose the same question but suppose for given p+q = 1, a large number of people flip their own biased coin, with Prob(H) = p and Prob(T) = q = 1-p, until reaching the first T. What will be the asymptotic proportion of H's and T's flipped, in simplest form? --Dan P.S. I had to solve this problem before I appreciated the reasoning behind Gene's beautiful solution, which didn't register with me at first. << PUZZLE: Suppose that in a very large population, each family has children until the first boy is born, and then has no more. What is the percentage of girls in the population? (Of course, make all the usual simplifying assumptions in this kind of problem.) >> I sleep as fast as possible so I can get more rest in the same amount of time.

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Re: [math-fun] math-fun Digest, Vol 89, Issue 6
by Mark Niemiec 08 Jul '10

08 Jul '10

Re: [math-fun] Probability puzzle Veit Elser <ve10(a)cornell.edu> wrote: > I'll try channeling von Neumann on this one: > > B = 1/2 + 1/4 + 1/8 + 1/16 + ... = 1 > > G = 0/2 + 1/4 + 2/8 + 3/16 + ... = 1 > > => B = G There is an even simpler solution: Totally ignoring the identities of the mothers of any of the children, each child has a 50% chance of being a boy. How many children any particular family has is totally irrelevant to the gender of any particular child, and hence the probability of any specific gender. -- Mark D. Niemiec <mniemiec(a)gmail.com>

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Re: [math-fun] Probability puzzle
by Dan Asimov 08 Jul '10

08 Jul '10

Mike Speciner wrote: << Yes. I suppose a simple rephrasing of the problem... Suppose you continually flip a coin, with [probabilities of] p and q as you describe. Notice every time you get H. What is the fraction of Ts? Another question is, are there plausible rules of heredity and reproduction that would alter the Boy/Girl ratio if "families" ceased reproducing at the first boy, but not if they stopped randomly? >> Yes, exactly. That point of view fully justifies Gene's Book solution, even for a biased coin. And shows that any "stopping" condition like H or THT is equally a red herring. I sleep as fast as possible so I can get more rest in the same amount of time.

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[math-fun] eta(-q)
by Bill Gosper 08 Jul '10

08 Jul '10

I often cite Jacobi's "aequatio identica satis abstrusa" polynomial relating the 8th powers of eta(q), eta(q^2), and eta(q^4). Solving for eta(q)^8 (c112) solve(gimme(1,2,4),eta(q)^8) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) + 8 eta (q ) (d112) [eta (q) = - --------------------------------------------, 4 4 eta (q ) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) - 8 eta (q ) eta (q) = --------------------------------------------] 4 4 eta (q ) leads to the apparent absurdity that it is an even function, which is quickly refuted by numerical experiment: (c113) expand(subst(-0.5d0,q,%)) (d113) [3.17357586985114d0 %i + 1.83226488275225d0 = - 3.66452976550451d0, 3.17357586985114d0 %i + 1.83226488275225d0 = 3.83964397049972d-5] eta(<negative>) is not even real. So what is eta(-q)? Massaging the infinite product definition gives eta(q^2)^3/eta(q)/eta(q^4) times some 24th root of 1. But empirically, there is no such root that works for the whole unit disk. Rather, the power is the crazy four-valued step-function(arg q) with a point discontinuity (mainly derived by tutor Julian) in http://gosper.org/etaminus.pdf . If this is an example of Truth = Beauty, the corresponding formula for eta(i q) must be just unbearably gorgeous. And what I'm really after is eta(cis(2*pi/5) q) for nice R-R CF valuations. However, once we know that CF = algebraic/q^(1/5), we can just LatticeReduce for hugepolynomial(CF*q^(1/5)). --rwg

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Re: [math-fun] Probability puzzle
by Tom Karzes 08 Jul '10

08 Jul '10

The proportion of H's to total flips is simply p: Expected H's per run = p + p^2 + p^3 + ... = p/(1-p) = p/q Expected T's per run = 1 Expected flips per run = p/q + 1 Proportion of H's to total flips = (p/q)/(p/q + 1) = p/(p + q) = p Tom > Okay, now suppose the same question but suppose for given p+q = 1, a large > number of people flip their own biased coin, with > > Prob(H) = p > and > Prob(T) = q = 1-p, > > until reaching the first T. > > What will be the asymptotic proportion of H's and T's flipped, in simplest > form? > > --Dan > > P.S. I had to solve this problem before I appreciated the reasoning behind > Gene's beautiful solution, which didn't register with me at first. > > << > PUZZLE: Suppose that in a very large population, each family has children > until the first boy is born, and then has no more. What is the percentage > of girls in the population? > > (Of course, make all the usual simplifying assumptions in this kind of > problem.) >>> > > I sleep as fast as possible so I can get more rest in the same amount of > time. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >

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[math-fun] Probability puzzle
by Dan Asimov 08 Jul '10

08 Jul '10

(This problem is said to be one that Google asked prospective employees in job interviews. My phrasing is almost identical to how it was presented to me.) PUZZLE: Suppose that in a very large population, each family has children until the first boy is born, and then has no more. What is the percentage of girls in the population? (Of course, make all the usual simplifying assumptions in this kind of problem.) --Dan I sleep as fast as possible so I can get more rest in the same amount of time.

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