December 2004 - mathfun@mailman.xmission.com

[math-fun] new sequence
by Schroeppel, Richard 18 Dec '04

18 Dec '04

I think the Nth term is very roughly N^c, for some c>1. Rich -----Original Message----- From: Schroeppel, Richard Sent: Fri 12/17/2004 5:04 PM To: njas(a)research.att.com Cc: Schroeppel, Richard; rcs(a)cs.arizona.edu Subject: new sequence I'm not sure if you want to include this sort of sequence, but it seemed like an interesting idea. 171 181 272 282 1531 1631 2532 2632 3151 3161 3252 3262 11711 11811 12712 12812 14171 14181 14271 14272 15171 15172 16171 17141 17161 17162 17261 17331 17910 18141 18161 18331 18910 21721 21821 22722 22822 24171 24172 24272 24282 25271 25272 26272 27162 27261 27262 27242 27332 27920 28262 28332 28920 31731 31831 32732 32832 33171 33181 33272 33282 91071 91081 91710 91810 92072 92082 92720 92820 Theorems from propositional calculus, translated into decimal digits. Blocks of 1s & 2s are variables: A = 1, B = 2, C = 11, D = 12, E = 21, ... Not = 3; And = 4; Xor = 5; Or = 6; Implies = 7; Equiv = 8; Left Parenthesis = 9; Right Parenthesis = 0. Operator binding strength is in numerical order, Not > And > ... > Equiv. The non-associative "Implies" is evaluated from Left to Right; A->B->C is interpreted (A->B)->C. Redundant parentheses are permitted. Example: 17162 is the theorem A->AvB. Rich rcs(a)cs.arizona.edu rschroe(a)sandia.gov

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RE: [math-fun] density
by Daniel Asimov 16 Dec '04

16 Dec '04

Gary McGuire writes: << ... Number theorists tend to work with another type of density, called Dirichlet density. This is often easier to compute. When both densities exist, they are equal. Sometimes the Dirichlet density exists when the above natural density does not. An example of such a case is S = {set of primes whose first digit is 1}. It follows easily from the prime number theorem (according to Serre's "A course in arithmetic") that the natural density does not exist. Can anyone explain why this is so? >> This suprises me about S. The primes become arbitrarily sparse, on average, so their natural density should exist and = 0. No? The # of primes <= x is asymptotic to x/ln(x), so the density of primes <= x is asymptotic to 1/ln(x), which -> 0 as x -> oo, If the primes' density is 0, then it seems that must also be true of any subset of the primes, such as the one menitoned above. --Dan Asimov

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[math-fun] Re: math-fun Digest, Vol 22, Issue 7
by Gary McGuire 15 Dec '04

15 Dec '04

>This suprises me about S. The primes become arbitrarily sparse, >on average, so their natural density should exist and = 0. No? > >The # of primes <= x is asymptotic to x/ln(x), so the density >of primes <= x is asymptotic to 1/ln(x), which -> 0 as x -> oo, > >If the primes' density is 0, then it seems that must also be true of >any subset of the primes, such as the one menitoned above. Apologies - I meant to say the natural density in the set of primes! (and not the density in the positive integers) To repeat the question, let S be the set of primes with first digit 1. Define the natural density of S in the primes to be the limit of number of elements in S that are <= n -------------------------------------- as n --> infinity. number of primes that are <= n This does not exist (stated in Serre "A course in arithmetic"). Why so? Gary McGuire

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[math-fun] density
by Gary McGuire 15 Dec '04

15 Dec '04

The recent discussion of the density of a subset S of the positive integers defines the density of S to be d if number of elements in S that are <= n ----------------------------------------- approaches d as n --> infinity. n This is usually called the "natural" density. Two problems with this are 1) the limit may not exist 2) the limit can be hard to compute. Number theorists tend to work with another type of density, called Dirichlet density. This is often easier to compute. When both densities exist, they are equal. Sometimes the Dirichlet density exists when the above natural density does not. An example of such a case is S = {set of primes whose first digit is 1}. It follows easily from the prime number theorem (according to Serre's "A course in arithmetic") that the natural density does not exist. Can anyone explain why this is so? Gary McGuire

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RE: [math-fun] Article on math education
by Schroeppel, Richard 15 Dec '04

15 Dec '04

I'm not sure you *are* allowed to say that. My approach is to take square roots of both sides, so we're now comparing sqrt7 ^ sqrt8 <=> sqrt8 ^ sqrt7. Then transpose the powers (take both roots of both sides) to get sqrt7th root of sqrt7 <=> sqrt8th root of sqrt8. Now we have a question about values of the Xth root of X. We know from prior calculus work that this has a smooth maximum at X=e=2.718. sqrt7 = 2.645, while sqrt8 = 2.828; this makes sqrt7 substantially closer to e, deltas of .073 versus .110. If we assume the Xth-root-of-X function has a reasonably symmetrical hump around e, then the sqrt7 expression is larger. Rich PS: Puzzle: Estimate how many times a day someone wishes that someone else knew more math? I was at the Borders bookstore over the weekend, in the science aisle. A couple of teen girls walked through the next aisle (math) and one spontaneously wondered "Why would someone buy a MATH book?" I restrained my urge to comment, but it left me wondering "Why would someone make such a useless remark?" -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com on behalf of Richard Guy Sent: Wed 12/15/2004 8:40 AM To: math-fun Subject: Re: [math-fun] Article on math education Am I allowed to say that sqrt is increasing more rapidly than log, so that sqrt 8 sqrt 7 ------ > ------ ln 8 ln 7 sqrt 8 sqrt 7 and hence 7 > 8 ? [ I've multiplied by ln 7 ln 8 and taken antilogs] R. On Wed, 15 Dec 2004, Bill Dubuque wrote: > I wonder how those prospective teachers would have fared > > comparing 7^sqrt(8) vs. 8^sqrt(7) instead of 1/13 vs. 0.13 > > It seems this problem is deviously difficult. When it was posted > to the newsgroup sci.math only two people offered correct proofs. > I encourage fellow funsters to give it a try first before looking > below to see my quick arithmetic proof [1]; it requires < 2 minutes > of purely mental arithmetic (no pencil, paper or calculator needed!). > In fact, only arithmetic of two-by-one digit integers is employed. > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > SPOILER BELOW > > Note 8*29^2 > 7*31^2 via x=30 in 8(x-1)^2-7(x+1)^2 = x^2-30x+1 > > so sqrt(8) > 31/29 sqrt(7) by taking sqrt of above. > > Thus to prove > > sqrt(8) sqrt(7) > 7 > 8 > > it suffices to prove > > 31/29 sqrt(7) sqrt(7) > 7 > 8 > > or 7^30/8^28 > 8/7 > but > (7^5/2^14)^6 > (42/41)^6 since 7^5 = 7(50-1)^2 > 7(2500-100) = 16800 > and 2^14 = 2^4 2^10 < 2^4 1025 = 16400 > > 1 + 6/41 since (1+x)^n > 1+nx [binomial theorem, x>0] > > > 1 + 1/7 since 7*6 > 41 > > Another proof [2] proceeds via calculus, namely the inequality > > (2n+1) > All n: x f (a) > 0 => f(a+x) > f(a-x) > > Does anyone else know any other "interesting" proofs? > > Below are said sci.math threads. > > [1] Bill Dubuque, sci.math, 1996/06/08 > http://google.com/groups?threadm=WGD.96Jun8060426@berne.ai.mit.edu > [2] Don Davis, sci.math, 2000/11/24 > http://google.com/groups?threadm=dtd-2411002306240001@ppp0c005.std.com > > --Bill Dubuque > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ 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] Article on math education
by MCKAY john 15 Dec '04

15 Dec '04

I was recently asked by a senior manager of purchasing for a large school district - What is 3/8 + 3/8 ? I expect this manager deals in many millions of dollars annually. It makes me wonder how important math is for the non- mathematician. John McKay

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[math-fun] three integers relatively prime
by Gary McGuire 14 Dec '04

14 Dec '04

>NOW, how about three random integers K, L, M? > >By same reasoning, for each p there must be at most one of the three random >integers that p divides. The probability of this is > >Prob(p divides none of K,L,M) + Prob(p divides just one of K,L,M) = > >(1-1/p)3 + 3(1/p)*(1-1/p)2 = > >1 - 3/(p2) + 2/(p3). > >Thus Prob(K,L,M) are relatively prime = > >Product over all prime p of (1 - 3/(p2) + 2/(p3)). > >I don't know how to sum this in closed form, but >numerically it turns out that > >Prob( (K,L,M) = 1 ) ~ 0.1284 > >(taking the product over the first 1000 primes). Am I making some mistake here: Prob(p divides all three of K, L, M)=1/p^3 Prob(p does not divide all three) = 1 - 1/p^3 giving the final probability as the product of these terms over all p, which is 1/zeta(3) = 0.8319... Gary McGuire

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Re: [math-fun] Relatively Prime; math education
by Daniel Asimov 13 Dec '04

13 Dec '04

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[math-fun] Relatively Prime; math education
by Schroeppel, Richard 13 Dec '04

13 Dec '04

There are two interpretations of "relatively prime" for three (or more) numbers A,B,C. a) Coprime (in pairs): All pairs have GCD=1. GCD(A,B)=GCD(A,C)=GCD(B,C)=1. b) No divisor >1 for all the numbers. GCD(A,B,C)=1. The triple 2,3,4 satisfies (b) but not (a). (a) -> (b). Naturally, the likelihoods of (a) and (b) are different. ----- Math Education, Miscellaneous Thoughts If we want to teach logical thinking, let's design a Logical Thinking course. We can probably come up with something that makes sense to 8th graders. Does it make sense to spend class time teaching, say, long division? There seems to be some conflation of arithmetic with mathematics. There are several different ways to solve two linear equations in two unknowns, and the underlying ideas can be explained in ten minutes. The reporter's inability to assist his daughter with her math homework is curious. This might be evidence that some intelligent people really can't do math. There's a wide range of skill level and interest in math: Maybe we should abandon teaching the same math to everyone in fourth grade. There's no shortage of smart people. We don't need to find all of them. Much of higher math is only justifiable as art, with a very small audience. As an airplane passenger, would you rather your designer had studied number theory or fluid mechanics? What if he plays the flute? Or windsurfs? I was pleased to see that US math students have reached par with the Hungarians, although perhaps the Hungarians are less pleased. Is there an accurate way to do a meanigful test valid across countries? Technical question: Design a method for balancing a checkbook, something more than "recheck all your arithmetic". Is the influx of smart people into the US making the rest of the world dumber? It's more fun to teach a class where the students want to be there. Why have we organized the world to teach algebra to the bored? One problem with this discussion is we've heard most of it many times before. I'll offer a prize of $1 for the first new thought on the subject. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] three integers relatively prime
by Daniel Asimov 13 Dec '04

13 Dec '04

<< Am I making some mistake here: Prob(p divides all three of K, L, M)=1/p^3 Prob(p does not divide all three) = 1 - 1/p^3 giving the final probability as the product of these terms over all p, which is 1/zeta(3) = 0.8319... >> I don't understand the reason you choose to multiply those particular factors (for each prime p). The criterion is that either 0 or 1 of K,L,M can be divisible by p, but no more than 1 of them. --Dan Asimov

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