June 2017 - mathfun@mailman.xmission.com

[math-fun] Isaac Newton Invents Logarithms
by Henry Baker 18 Jun '17

18 Jun '17

Isaac Newton Invents Logarithms [A speculative dream; by Henry Baker, 6/18/2017. Another in the series of: "Things I wish my high school math teacher had told me".] Suppose you are Isaac Newton, and it is your Annus Mirabilis 1666. You've just invented calculus and power series. You've discovered that the antiderivative of x^n is x^(n+1)/(n+1). You know nothing of logarithms -- having never heard of Alphonse Anonio de Sarasa (Google him). Something is bothering you terribly; what is the form of the integral of x^n when n=-1? The problem is, the denominator of x^(n+1)/(n+1) vanishes when n=-1, so what exactly does this mean? You know that the definite integral of x^n from x=a to x=b is = b^(n+1)/(n+1) - a^(n+1)/(n+1) = (b^(n+1)-a^(n+1))/(n+1) In particular, you know that the definite integral of x^n from x=1 to x=b is = (b^(n+1)-1^(n+1))/(n+1) = (b^(n+1)-1)/(n+1) You quickly grow tired of dealing with "n+1", so you replace "n+1" by "m", so the definite integral of x^(m-1) from x=1 to x=b is = (b^m-1)/m You, being Isaac Newton, believe in the fundamental continuity of the universe; in fact, you believe in continuity so much that you cannot even comprehend the possibility of non-continuous functions. So you investigate what happens as m "approaches" 0 (i.e., as n approaches -1). But since we will be investigating small numbers, it will be more convenient to change variables once more so that m=1/p, and p will be an integer we will make very large. Yes, we have changed from integer m (and hence n) to non-integer m,n, but Newton did this all the time, as it is a necessary consequence of his belief that everything is continuous. So your formula for the integral of x^n (= x^(1/p-1)) from 1 to b is now: = (b^(1/p)-1)/(1/p) = p*(b^(1/p)-1) You (or actually, your assistant, Max Gnuplot) now plots these functions for p=1,2,4,8,16,32,64,128: plot2d([ev((b^(1/p)-1)*p,p=1), ev((b^(1/p)-1)*p,p=2), ev((b^(1/p)-1)*p,p=4), ev((b^(1/p)-1)*p,p=8), ev((b^(1/p)-1)*p,p=16), ev((b^(1/p)-1)*p,p=32), ev((b^(1/p)-1)*p,p=64), ev((b^(1/p)-1)*p,p=128)], [b,1,5]); You can hardly believe your eyes: the curves are converging! As p grows and grows, the plots become more and more similar, which convinces Newton that he may be onto something. But you're still suspicious, so you also check negative values of p: plot2d([ev((b^(1/p)-1)*p,p=-1), ev((b^(1/p)-1)*p,p=-2), ev((b^(1/p)-1)*p,p=-4), ev((b^(1/p)-1)*p,p=-8), ev((b^(1/p)-1)*p,p=-16), ev((b^(1/p)-1)*p,p=-32), ev((b^(1/p)-1)*p,p=-64), ev((b^(1/p)-1)*p,p=-128)], [b,1,5]); Aha! you say. The positive p's converge from above, the negative p's converge from below, so there must be a specific curve in between the upper and lower bounds that is the answer to our question: what happens to x^(n+1)/(n+1) as n approaches -1? Now that Newton knows that a function exists which answers his question, he wants to give it a name. Having just participated in a recent misadventure with an apple tree, now reduced to firewood, Newton decides to call the function "log(b)": log(b) = lim (b^(1/p)-1)*p p->oo [p->-oo also works.] Newton, being Newton, wants to better understand this function, so he asks about its inverse function, which he temporarily calls "ilog". If c=log(b), then ilog(c)=b, so after simple algebra, ilog(c) = b = lim (1+c/p)^p p->oo Now Newton immediately notices the connection between ilog(c) and continuously compounding interest; Newton didn't become warden of the Royal Mint for nothing! (Yes, Newton was "Mint Newton", not "Fig Newton".) Given the relationship with compound interest, Newton decides to call this inverse function "exp(c)", as he fully EXPects to reap a reward of exp(c) from an interest rate of c. exp(1)=2.71828... is thus the "Mark of the Beast" that one might expect from a year numbered 1666. exp(c) = b = lim (1+c/p)^p p->oo Newton knows about power series (including power series inversion), so he starts laboriously computing the power series for exp(c): exp(c) = 1 + c + c^2*approx(1/2) + c^3*approx(1/6) + c^4*approx(1/24) + ... where "approx(x)" = very close approximation in *base p* (!) of x. Furthermore, Newton notices that as p grows larger, this approximation becomes better and better. Newton leaps (parabolically, as was his wont!) to the conclusion that exp(c) = 1/0! + c/1! + c^2/2! + c^3/3! + ... Now, Newton was quite familiar with simplifying formulae like x^2*(x-1)*(x-2)2/(x-1) = x^2*(x-2), where a potential problem at x=1 disappears when taking the limit as x approaches 1, so he finally concludes that his new function log(b) = (b^0)/0 so that his antiderivative formula becomes unexceptional. QED

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[math-fun] two problems from a Facebook thread
by Thane Plambeck 18 Jun '17

18 Jun '17

(1) (via Peter Winkler) Here's a (seemingly) new hat puzzle for you: 100 prisoners will be fitted with red or blue hats according to fair coinflips. Then the lights are turned on; each prisoner sees every other hat color and must write down a guess for his own. If a majority (at least 51) get their own hat color right, all the prisoners will be freed. As usual the prisoners can't communicate once the hats are placed, but have time to strategize beforehand. Can they achieve a 90% probability of being freed? How about 95%? (2) (Johan Wästlund) I'll just throw in here a question of (what turns out to be) similar flavor: Suppose you play a type of one-person tic-tac-toe where in every "move" you select a set of unoccupied squares, and then flip a coin to decide whether you get all those squares, or your "opponent" gets them. What winning probability can you achieve? -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] report on proof of the Kepler Conjecture
by rcs＠xmission.com 18 Jun '17

18 Jun '17

The Kepler Conjecture is that the usual sphere packing is optimal. A report on the machine-verified proof of the Kepler Conjecture has been published. The proof is available, and can be checked with tolerable computing resources. https://www.cambridge.org/core/journals/forum-of-mathematics-pi/article/ formal-proof-of-the-kepler-conjecture/78FBD5E1A3D1BCCB8E0D5B0C463C9FBC The effort to make a formal machine-checkable proof turned into a major project. The author list for the report looks like it's for a big-science paper. The Kepler Conjecture is a very expensive theorem. Rich

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[math-fun] Grad school question
by Michael Collins 14 Jun '17

14 Jun '17

I have a student who just completed a Master's degree and hopes to start a Ph.D. in mathematics in fall 2018. He is interested in cryptography and we were wondering about mathematics departments that have cryptographic research groups. As far as I can tell the great majority of crypto research, even the very theoretical work, is done in CS departments. Any suggestions? -Mike Collins

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[math-fun] (no subject)
by Fred Lunnon 14 Jun '17

14 Jun '17

A family of functions f(), g(),... is defined on integer pairs in |Z^2 . They are nonzero in general `on land' but have zeros which congregate in n x n lattice square regions `in ponds'. I want to establish that they satisfy a certain system of linear relations `bridges' between quartets of values f(w), f(x), f(y), f(z) at arguments w, x, y, z , lying one on each bank of a pond. An inductive argument exists which constructs a bridge over an n-pond of f() from one over an (n-2)-pond of a related family member g() ; normally all that should then be necessary is progressively shrinking the pond until it dries up when n passes zero. But there is a snag. The construction of g() from f() proceeds stepwise: starting from (say) x on the bank, it fans out point-by-point around the pond and thence over the entire land (essentially common to both functions); however it breaks down inside ponds, which must be filled separately once their banks are built. As a result, after a full circuit skirting the pond, g(x) might conceivably become inconsistent with its previous value; at this stage g() may be assumed single-valued only within simply-connected regions of land. Now if only I could somehow prove g() globally single-valued, I could go ahead and build the bridge; or if only I already had the bridge, I could deduce g() must be single-valued. One way to break into this cycle might be to assume that g() branches around ponds, then formulate the induction on n to avoid reliance on consistency by cutting the plane from pond centre away from currently selected w, x, y, z , in order finally to deduce that the entire flapdoodle was unnecessary! I would be happier about resorting to this bizarre stratagem if I knew of a similar situation elsewhere. For example, is there an instance of an identity involving some function of a complex variable, constructed via iterated analytic extension around a singularity, such the proof requires one initially to assume a branch point; despite which the identity finally implies that the function is single-valued? Fred Lunnon

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[math-fun] Bill & Rohan's G12 talk
by Bill Gosper 13 Jun '17

13 Jun '17

On 2017-06-05 08:13, Jon Ziegler wrote: > Yes, I enjoyed watching the talk. Well done, Bill & Rohan! > > The Minskys & Trinskys book is available at blurb.com here > <http://www.blurb.com/b/2172660-minskys-trinskys-3rd-edition>. The > preview provides the entire book for your perusal. You can buy a > hardcopy too if you want. Our talk was very remiss in failing to recognize that the book was almost entirely the work of Jack Holloway, including the image in my penultimate "slide". (But don't miss Julian's discussion of accumulation points in Minskyspace.) Also, we had a miscue: an orbit consisting of four skewed rectangles vs four axis-aligned squares. I IMovie-patched in the correct animation (silent) at gosper.org/G4G12Minsky.m4v . Unfortunately, most players lose the rest of the audio as well. JVC doesn't. --rwg > > I very much doubt there will be a new edition, at least not from Corey > or Julian. Although since Julian is well on his way to becoming a > mathematician (graduating from Cambridge later this month, and grad > school at UCLA thereafter) I suppose he might get interested in the > topic again. Much more likely would be if Rohan wants to take it on. > > Regards, > Jon Ziegler > >> On Jun 5, 2017, at 6:31 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> Agreed! >> >> Where can I get Minsky's & Trinsky's today ? (Digital version preferred) >> >> Will there be a Minsky's & Trinsky's II ? >> >> At 09:34 PM 6/1/2017, Robert Smith wrote: >>> Great talk! >>> >>> On Thu, Jun 1, 2017 at 19:56 Bill Gosper <billgosper(a)gmail.com> wrote: >>> >>>> https://youtu.be/lXsVWwPa7bc --rwg >>>> http://www.gathering4gardner.org/g4g12-videos/ >

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Re: [math-fun] Geometry question
by Fred Lunnon 13 Jun '17

13 Jun '17

<< Huh? Distance between the two points is sqrt(2) in both cases. >> On third thoughts --- must've been a bit too late at night for me --- apologies, everybody! WFL > On 12-Jun-17 21:25, Fred Lunnon wrote: >> Similar, not congruent. WFL >> >> On 6/13/17, Mike Speciner <ms(a)alum.mit.edu> wrote: >>> But that's isomorphic to {<0,0>,<1,1>} >>> >>> On 12-Jun-17 19:55, Fred Lunnon wrote: >>>> I doubt that this is blond; I'm pretty certain it lacks hairy ears. >>>> >>>> Regarding the next question, it ain't so: take n = 1 , X = { 0, >>>> sqrt(2) } . >>>> >>>> WFL >>>> >>>> >>>> On 6/12/17, Dan Asimov <dasimov(a)earthlink.net> wrote: >>>>> This may be trivial, hard, or auropileous; I'm not sure: >>>>> >>>>> Suppose we are given a finite subset X of R^n such that >>>>> >>>>> * The dot product of any two vectors between points of X is an >>>>> integer. >>>>> >>>>> (I.e., for all x,y,z,w in X, the real number >>>>> >>>>> <x-y, z-w> >>>>> >>>>> lies in Z.) >>>>> >>>>> Question: >>>>> --------- >>>>> Does it follow that for any sufficiently high dimension d, >>>>> there is a subset X' of the integer lattice Z^d such that >>>>> X' is congruent to X ??? >>>>> >>>>> —Dan >>>>> >>>>> _______________________________________________ >>>>> math-fun mailing list >>>>> math-fun(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>> >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> > >

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[math-fun] Geometry question
by Dan Asimov 12 Jun '17

12 Jun '17

This may be trivial, hard, or auropileous; I'm not sure: Suppose we are given a finite subset X of R^n such that * The dot product of any two vectors between points of X is an integer. (I.e., for all x,y,z,w in X, the real number <x-y, z-w> lies in Z.) Question: --------- Does it follow that for any sufficiently high dimension d, there is a subset X' of the integer lattice Z^d such that X' is congruent to X ??? —Dan

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[math-fun] Distributing and discussing puzzles online
by James Propp 12 Jun '17

12 Jun '17

Any suggestions (based on your experience as a puzzle-poser or puzzle-solver) on best practices for sharing puzzles in a way that allows solvers to see what other solvers have done? Three people I can think of who do this are Matt Parker, James Tanton, and Alex Bellos (though there may be others who are doing it too that I just don't happen to know about). Keeping in mind that I'm often a sub-optimal user of newfangled media, here's my impression of what they're doing: Parker posts videos on YouTube (though not as many of them feature puzzles as formerly), and discussion of problems takes place through the amorphous YouTube comments feature. For his really popular videos, hundreds of people post comments (many of them of the uninformative "Cool puzzle!" variety), which makes the comments feature of limited value to someone as impatient as I am. In two-parters he'll sometimes comment on people's comments. Tanton uses Twitter, but I haven't figured out how to follow threads efficiently (sometimes it appears that there are no replies, but then it turns out I just didn't know how to access them). I'm not sure what Bellos does. I know that he posts puzzles on Monday mornings and solutions on Monday evenings (UK time), so I'm guessing he gets emails during the day and summarizes them, which seems insanely hectic to me. I know that he always reminds people not to post spoilers in the comments, so I infer that this is an ongoing problem. Which brings me to one of my major concerns: how to create a forum that's helpful to those who want a hint but doesn't inadvertently give away solutions? Since I'm only just getting into this game now --- my first (not-quite-ready-for-distribution) video is at http://faculty.uml.edu/jpropp/barefootmath/Swine-in-a-line-1a.mov --- I'd like to know what I should be doing! My current plan is to invite people to post comments to http://barefootmath.wordpress.com, which I will lightly moderate, urging people to label spoilers as spoilers. This seems workable as long as I don't attract too large a following (which seems unlikely to happen anytime soon if ever) and I don't post too many puzzles at once (which seems unlikely to happen since I'm still trying to keep active as a researcher, which limits the amount of time I can spend on blogging and tweeting and sharing puzzles and whatnot). Any thoughts? Thanks, Jim Propp

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Re: [math-fun] Cantor set overlaps
by Dan Asimov 11 Jun '17

11 Jun '17

It's kind of a generalized pigeonhole principle, if we view the pigeonhole principle as an exercise in elementary measure theory, a.k.a. counting. Two disjoint measurable sets such as the ones below will always have a combined measure equal to the sum of their measures. If that's greater than the total measure of the space you're in, then the disjoint sets weren't. —Dan ----- >> My intuition says that since those two sets are each in the unit >> interval, which of course has measure 1, and since they each have a >> measure of more than 1/2, that by the pigeonhole principle overlaps >> are inevitable. -----

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