November 2014 - mathfun@mailman.xmission.com

Re: [math-fun] More 4th Grade Science
by Henry Baker 12 Nov '14

12 Nov '14

It isn't quite the same as jumping, but "During the mission, [Alan Shepard] hit two golf balls on the lunar surface" https://en.wikipedia.org/wiki/Alan_Shepard At 05:02 AM 11/12/2014, Bill Gosper wrote: ><Photo of Apollo astronaut about about 1ft above the lunar surface, > resembling those bad photoshops of Chinese officials floating above recently opened highways> > >"On the moon, an astronaut can jump higher than on Earth. Is this fact or opinion?" >Assuming that cumbersome spacesuits and safety concerns prevented anyone >from actually trying this, how can we call it fact? "Fact" means "done". >So isn't it still simply a very believable opinion? >--rwg

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[math-fun] Set theoretic development of numerals
by David Wilson 12 Nov '14

12 Nov '14

In set theoretical (ZF or ZFC) developments of the natural numbers, I have often seen the successor function defined Sn = n ∪ {n} and a few of the smaller numerals defined [1] 0 = ∅, 1 = S0, 2 = S1, 3 = S2, &c. The &c has no meaning within the theory, the student is expected to expand the &c to an infinitude of like definitions for the remaining numerals. Later, if the student encounters the string 23 within some statement of the theory, he is expected to interpret 23 as shorthand for SSSSSSSSSSSSSSSSSSSSSSS0, even though he was never given neither an explicit definition of the numeral 23 nor a general definition of numerals whereby he could deduce the definition of 23. My question is, can the set theoretic development of natural numbers be augmented with a set theoretic definition of numerals so that if 0 or 23 or 1093 or 6661661161 or any other string of digits has a definition as a natural number within the theory, without recourse to the student’s understanding of the values of the numerals?

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[math-fun] catch the cat
by David Wilson 11 Nov '14

11 Nov '14

http://www.infinitecat.com/games/cat-box.html Is there a strategy?

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[math-fun] Reboot attempt: The chance that N independent statistical tests fail simultaneously
by Warren D Smith 11 Nov '14

11 Nov '14

--I reiterate to all: READ THE WEB PAGE http://rangevoting.org/CombinedTestFail.html Has anybody? This web page begins with the problem description then solves the case N=2, which everybody here seems to be having immense trouble comprehending for some mysterious to me reason, in several ways, showing the solution corresponds exactly, word for freaking word, with the problem description. Once you follow the N=2 case, and agree that (2b-a)a is the only solution, hopefully your mental block will be overcome. I evidently have triggered some kind of mass mental block, which I unfortunately am also mentally blocked from comprehending what the block is. I guess it must be my fault, but not really seeing why. Perhaps my new answer to Greathouse below overcomes/identifies the communication difficulty. The good news is, this leads to astonishingly beautiful mathematics, and I still have not completely understood it, but I'm getting a good way there, as is explained on the web page. >Ch.Gr: >Let's say that we're testing a coin to see if it's unfairly biased towards >heads. The null hypothesis is that it's a fair coin. Test 1 consists of >flipping the coin twice and seeing if both times it comes up heads; p = >0.25. Test 2 consists of flipping it once and seeing if it comes up heads; >p = 0.5. The chance that a fair coin would fail both tests is 1/8 for a >p-value of 0.125. Your formula gives 1/4 * (2*1/2 - 1/4) = 3/16 = 0.1875. >Am I missing something? >> I don't know what you mean when you say that my tests are dependent -- I'm > not reusing coin flips. Would it be better if I tested two different coins > with the null hypothesis being "both coins are fair"? The numbers remain > the same, of course. > Charles Greathouse > Analyst/Programmer > Case Western Reserve University --oh. Sorry, I misunderstood you (Ch.Gr.); stupidly thought you were re-using flips.. Sorry, the real explanation is, your N tests (N=2 for you) are not interchangeable. My problem implicitly assumed they were.

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[math-fun] The chance that N statistical tests fail simultaneously
by Warren D Smith 11 Nov '14

11 Nov '14

Perhaps this nicely typeset page will help you understand the problem: http://rangevoting.org/CombinedTestFail.html The answer is NOT a*b*c*...* despite the tests being independent. This is a trap too many fall into. Indeed, if there were N uniform(0,1) independent randoms, and somebody told you "there exists an ordering such that x_j<j/N for each j=1,2,3,...,N" you do NOT want to be an idiot and conclude "my god, that is an amazingly exponentially improbable miracle." Actually, it is quite likely even for N=100. Eh? So hopefully I've piqued your interest now. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] The chance that N independent statistical tests fail simultaneously
by Warren D Smith 11 Nov '14

11 Nov '14

1. The OEIS strikes again! A conjectured closed form is now found from A124774 which is explained on the web page http://rangevoting.org/CombinedTestFail.html along with everything else I found out, which page I strongly suggest math-funners read, since they seem to keep not comprehending the problem, and maybe this page un-mysterifies. (I'm actually boggled at how much everybody cannot understand the problem, which proves it must be a great problem, and/or I am crazy.) 2. To answer James Propp > I do not understand Warren's example. Specifically, I do not know what he > has in mind as the independent events whose conjunction is being > considered. > > Note that the event > "there exists an ordering such that x_j<j/N for each j=1,2,3,...,N" > is not the conjunction of the N events > "there exists an ordering such that x_1<1/N", > "there exists an ordering such that x_2<2/N", > ..., > "there exists an ordering such that x_N<N/N". --No....: All N! orderings such as x1,x2,xN, and x2,x1,...,xN, and ... xN...x2,x1 are considered. What is the freaking chance, that at least one of these orderings obeys (after reordering) xj<j/N for each j=1,2,3,...,N? [Where each x_j is an iid uniform(0,1)]

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[math-fun] The chance that N statistical tests fail simultaneously
by Warren D Smith 11 Nov '14

11 Nov '14

The chance that N statistical tests fail simultaneously =========Warren D. Smith====Nov 2014======== Suppose you perform N statistical tests. One fails with p-level a. Another fails with p-level b. ... Another fails with p-level z. What is the p-level for this combined event? Call the answer F_N(a,b,c,...,z). We may assume wlog that 0<a<b<c<...<z<1. F_1(a)=a, of course. It is NOT the case that F_2(a,b)=ab. Actually, F_2(a,b) = 2ab-aa = a(2b-a). Because that is the chance that if x and y are independent uniform01 randoms, then either x<a and y<b, or x<b and y<a. What about F_3(a,b,c), F_4(a,b,c,d), and so on? It is possible to determine these by working out certain N-dimensional volumes via something like an inclusion-exclusion argument. But it gets more and more painful. As an easy upper bound, F_N(a,b,c,...,z) < N! abc...z. But this bound can be pretty weak. For example it is easy to exactly work out the special case F_N(x,x,x,...,x) = x^N. I would like the exact answer in general. The following recurrence F_{N+1}(a,b,c,...z) = (N+1) * integral(from x=0..a) F_N( (b-x)/(1-x), (c-x)/(1-x), ..., (z-x)/(1-x) ) * (1-x)^N dx seems an easier way to get the answer. From this I compute: F_3(a,b,c) = [3b(2c-b)+a(a-3c)]a F_4(a,b,c,d) = [4(b-3d)b^2+6c(4bd+(a-2b)c-2da)+(4d-a)a^2]a F_5(a,b,c,d,e) = [20bc^3-5b^4+30b^2d^2-60bcd^2+20b^3e-60bc^2e-60b^2de+120bcde-10c^3a+30cd^2a+30c^2ea-60cdea-10d^2a^2+20dea^2-5ea^3+a^4]a. These keep getting messier. But there might be some simple general form for the answer (e.g. if we go back to the original inclusion-exclusion idea and do not get confused), and/or there might be some algorithm for computing answer whose runtime grows only polynomially with N. Can you find them? At present, the best algorithm I thought of runs in time exponential in N, roughly like 4^N in fact.

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[math-fun] 0,0,1,0,0,1,...
by Bill Gosper 11 Nov '14

11 Nov '14

I was showing the kids an infinite series with a "three phase" summand, i.e. a trinomial with different parts switched on or off depending on n mod 3. In[61]:= FindSequenceFunction[{0, 0, 1, 0, 0, 1, 0, 0, 1}, n] Out[61]= Mod[1 + 2 n^2, 3] That's pretty clever, but it befuddles Sum when it appears in multiple phases. "Luckily", Sum isn't much good with DirichletCharacter either, so I didn't have to figure out how to explain that to them. I expected In[65]:= FullSimplify[Sum[(-1)^(2*k*n/3), {n, 3}]/3, k \[Element] Integers] Out[65]= 1/3 (1 + (-1)^(2 k/3) + (-1)^(4 k/3)) In[67]:= Table[Simplify[%%], {k, 9}] Out[67]= {0, 0, 1, 0, 0, 1, 0, 0, 1} to work, but no. Not even In[69]:= FullSimplify[Re[%65], k \[Element] Integers] Out[69]= 1/3 (1 + Cos[(2 k \[Pi])/3] + Cos[(4 k \[Pi])/3]) I wound up having to trisect series into a nine term summand. Interestingly, FindSequenceFunction of every sixth term produces a trig expression equivalent to %69 if n -> 2n. But FullSimplify can't reduce it to %69. Trying to write the "general decimator" function exposed a serious issue with Sum: For a period of d-1 zeroes then 1, In[20]:= Assuming[d \[Element] Integers && k \[Element] Integers && d > 0, Sum[(-1)^(2*k*n/d), {n, d}]/d] Out[20]= ((-1)^((2 k)/d) (-1 + (-1)^(2 k)))/((-1 + (-1)^((2 k)/d)) d) This is either 0, or 0/0 ! Throwing in a Simplify In[21]:= Assuming[d \[Element] Integers && k \[Element] Integers && d > 0, Simplify[Sum[(-1)^(2*k*n/d), {n, d}]/d]] Out[21]= 0 ! Yet specializing d to 4, In[22]:= Sum[(-1)^(2*k*n/4), {n, 4}]/4 Out[22]= 1/4 ((-1)^k + (-1)^(2 k) + I^k + I^(3 k)) In[23]:= Table[%, {k, 9}] Out[23]= {0, 0, 0, 1, 0, 0, 0, 1, 0} I'm not even sure how to complain about this. What should Sum do? On top of all this, either my laptop is flaking, or there's a nonreproducible horror someplace: When I copied and pasted the section appearing as In|Out[61] above, $Line was actually 57. But what got pasted instead was some previous clipboard junk. When I went back to refetch In|Out[57], they had vanished! --rwg

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[math-fun] catch the cat
by Warren D Smith 10 Nov '14

10 Nov '14

see http://home.broadpark.no/~oddvark/angel/index.html -

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Re: [math-fun] Undecidable propositions
by Henry Baker 09 Nov '14

09 Nov '14

This type of discussion is why the terms "recursive" v. "recursively enumerable" (RE) were defined. A set S which is _recursive_ can have its elements and non-elements determined in a finite number of steps; i.e., a finite number of steps will conclude whether x in S or not (x in S). A recursively enumerable set can be enumerated, but given an x, there is no finite procedure to determine conclusively if x in S or not. The best you can do is to list all of the elements using your enumeration procedure, and hope that x shows up in the list. But since there is no way to list all of the non-elements of S, you can never conclude that not (x in S) in a finite amount of time. Most standard theories start with axioms & then make deductions willy-nilly to produce a stream of true statements. But if the theory is suitably powerful, then there will be statements that never show up in the stream. Unfortunately, listing all of the false statements (which appear as "not S"" in the stream) won't help, because there will be statements that won't show up in either stream -- either the "true" stream or the "false" stream. These statements that don't appear in the stream _ever_ are "undecidable". At 08:07 PM 11/9/2014, Michael Kleber wrote: >Given a single Turing machine T, is there a decider D which takes the >description of T as its input, and returns "true" if T halts and "false" if >T does not? Well certainly there is: either the constant machine D_t that >always returns "true", or else the constant machine D_f which always >returns "false", does the job. There's no algorithm to decide which of >these is actually the right D for the given T, but one of them definitely >is. > >On Sun, Nov 9, 2014 at 10:31 PM, Warren D Smith <warren.wds(a)gmail.com> wrote: > >> I do not see the basis for APG's claim for for any particular turing >> machine T, it is decidable whether it halts. Au contraire, it is >> decidable IF it halts, but if it does not halt, then why is >> it decidable? >> >> Similarly, particular propositions like the RH are decidable if false, >> but might be undecidable if true. >> >> So I do not see APG's claim that only CLASSES of propositions undecidable. >> >> -- >> Warren D. Smith >> http://RangeVoting.org <-- add your endorsement (by clicking >> "endorse" as 1st step)

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