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[math-fun] Stupid steam train question
by Henry Baker 08 Nov '17

08 Nov '17

I was watching a movie from the 1930's yesterday, and it showed a steam engine pulling a train out of the station. I noticed that the smoke coming out of the smokestack was in synchrony with the steam coming out of the cylinders that drive the wheels. There was definitely a "puff" "puff" coming out of the smokestack. Since this movie was from the 1930's, I assume that it was an authentic movie of a real steam train, and not some special effect. Forgive me, but I always thought that train smoke stacks were more analogous to ocean liner smoke stacks, where the smoke coming out is *not* under pressure. I can only speculate that some steam engines might have a *fan* (or equivalent) that forced cool air through the burning coal. Since early steam engines didn't have electricity, these "fans" would have to be something more akin to the old-style *belt-driven* engine fans on automobiles. Or perhaps there is some sort of reciprocating *air pump* attached to the steam engine that pumps air through the coals in synchrony with the main steam cylinders?

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[math-fun] More on (Moron?) Infineon primes bug
by Henry Baker 08 Nov '17

08 Nov '17

Dan Bernstein on the Infineon "high security" RSA bug: https://blog.cr.yp.to/20171105-infineon.html "This could even be a deliberate weakness with plausible deniability" "The authors notified Infineon at the beginning of February 2017. It seems that Infineon negotiated nine months of delay before the paper was posted. Did this much larger delay help security, or hurt security?"

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Re: [math-fun] More random polygons: fun puzzle (HALF-SPOILER)
by James Propp 07 Nov '17

07 Nov '17

Another hint: If you write the number of vertices as a sum of 0,1-valued random variables (aka indicator random variables), the expected value of the sum is expressed as the sum of a bunch of probabilities, each of which is a ratio of areas. Jim On Monday, November 6, 2017, James Propp <jamespropp(a)gmail.com> wrote: > I wrote: > > >> I actually know two different proofs of the result. One follows the >> approach to such problems that I learned from George Hart (see his email to >> math-fun from October 18), and the other follows the approach I learned >> from Warren Smith (see my email to math-fun from October 19). If you look >> at their solutions to the random-slices-of-a-cube problem you might be >> inspired, as I was, to see how to solve the random-intersecting-parallelog >> rams-puzzle. >> >> George's method can also be used to prove that if you take a random >> nonempty intersection of a regular hexagon with a translate of itself, the >> expected number of sides is exactly 5. >> > > Hint: instead of counting sides, count vertices. > > I'll sketch the proof tomorrow or Wednesday, if nobody beats me to it. > > Anyway, two days have passed since I posted the puzzle, so I think it's >> appropriate to have an unbridled conversation, with or without spoilers. >> > > Let me say, by way of motivation, that the "George-style" proof is really > sweet. You'll feel good when you find it. > > Jim >

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[math-fun] Back to square one (Archimedes type formula)
by François Mendzina Essomba 07 Nov '17

07 Nov '17

An omission of limits in two previous formulas 1/sqrt(Pi)=Limit((8/(m*(Gamma(1/4))^2))*sum((1-(k/m)^4)^(1/4),k=1..m-1),m=infinity); (Pi)^(3/2)=Limit((4*((Gamma(3/4))^2/(m)))*sum((1-(k/m)^4)^(1/4),k=1..m-1),m=infinity); Back to square one Archimedes type formula a_(n+1):= 2^(n+3/2)*(3-(3+(3*(3-a_(n)^2/(2^(2*n+1)))^2+9)^(1/2))^(1/2))^(1/2); a_(1):= 2*(6-2*3^(1/2))^(1/2); n---infinity; a_(n)--- Pi We obtain by developing this formula: Pi=2^(n+3/2)*sqrt(3-sqrt(3+3*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+_)))))))); we have n (*) times the number (2); For example: Pi~2^(6+3/2)*sqrt(3-sqrt(3+3*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2)))))))); I suspect that there are an infinity of his formulas FME...

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[math-fun] Vivacious fruit vendor proffering samples
by Bill Gosper 06 Nov '17

06 Nov '17

"Slice of apple?" "Um, no thanks." "It will raise your IQ." "Well, OK. <Munch>" "What do you think?" (Thoughtfully:) "I think that 37 times 37 is 1369." She just looked at me funny. --rwg

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[math-fun] Are you a Gammaturge?
by Bill Gosper 06 Nov '17

06 Nov '17

Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == GoldenRatio In[345]:= N[%, 11111] Out[345]= True --rwg

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[math-fun] some easy little formulas ...
by François Mendzina Essomba 06 Nov '17

06 Nov '17

Hello some easy little formulas .. Pi=4*product((16384*n^2*(2*n-1)^3*(4*n+1)^4)/((2*n+1)*(8*n-3)^4*(8*n+1)^4),n=1..infinity); 1/sqrt(Pi)=(8/(Gamma(1/4))^2)*product(((4*n+1)^2-1)/(4*n+1)^2,n=1..infinity); 1/sqrt(Pi)=(8/(m*(Gamma(1/4))^2))*sum((1-(k/m)^4)^(1/4),k=1..m-1); (Pi)^(3/2)=(4*((Gamma(3/4))^2/(m)))*sum((1-(k/m)^4)^(1/4),k=1..m-1); (Pi)^(3/2)=(4*(Gamma(3/4))^2)*product(((4*n+1)^2-1)/(4*n+1)^2,n=1..infinity); Pi=((Gamma(1/4))^4/64)*(1+sum(((-1)^(n)/(4*n+1))*((2*n-1)!)/(2^(2*n-1)*(n!)*((n-1)!)),n=1..infinity))^(-2); (Pi)^3=(16*(Gamma(3/4))^4)*(1+sum(((-1)^(n)/(4*n+1))*((2*n-1)!)/(2^(2*n-1)*(n!)*((n-1)!)),n=1..infinity))^(2); Best regards

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Re: [math-fun] Are you a Gammaturge?
by Bill Gosper 05 Nov '17

05 Nov '17

On 2017-11-05 09:42, Dan Asimov wrote: > That sure is a lot of digits of accuracy for a Boolean variable. > > —Dan I found it numerically and didn't believe it at first. --rwg > > > -----Original Message----- >>From: Bill Gosper <billgosper(a)gmail.com> >>Sent: Nov 5, 2017 4:12 AM >>To: math-fun(a)mailman.xmission.com >>Subject: [math-fun] Are you a Gammaturge? >> >>Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == >>GoldenRatio >> >>In[345]:= N[%, 11111] >> >>Out[345]= True >>--rwg

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Re: [math-fun] sockrilege
by Bill Gosper 05 Nov '17

05 Nov '17

On 2017-11-03 18:09, Neil Sloane wrote: > Quicker to look it up! https://oeis.org/A051288. > Duh, a month in the laboratory can often save an afternoon in the library. > I added the socksual interpretation. > Wow, I had never connected socksuality with foot-fetishism. But (sniff), all my socks are achiral, in more ways than one! "Think socks (those weird "chiral" kind.)" --rwg > Best regards > Neil > > Neil J. A. Sloane, President, OEIS Foundation. > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. > Phone: 732 828 6098; home page: http://NeilSloane.com > Email: njasloane(a)gmail.com > > > On Fri, Nov 3, 2017 at 8:38 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > >> Our old friend STAN.K bought n pairs of white "chiral" (podal!) socks >> clearly stamped L and R, >> which his housekeeper nevertheless paired willy-nilly. Stan wondered, for >> n pairs, how likely >> were 0, 2, 4, ... mismatches. >> I couldn't remember how to map a washer-dryer onto an urn, so I just >> tabulated >> >> 2 >> >> 4 2 >> >> 8 12 >> >> 16 48 6 >> >> 32 160 60 >> >> 64 480 360 20 >> >> 128 1344 1680 280 >> >> 256 3584 6720 2240 70 >> >> 512 9216 24192 13440 1260 >> >> and fit a formula with Mathematica. The nth row represents n pairs, >> row sum = binomial(2n,2). Thus the 2nd row, 4 2, represents 2 pairs, >> LRLR,LRRL,RLLR,RLRL = 4 perfectos, and >> LLRR,RRLL = 2 mismatches. >> (The left column is how many perfectos for n pairs.) >> The empirical formula I got was >> StringReverse@"]j2-n,j,j[laimonitluM)j2-n(^2" >> which then Julian promptly derived in a single line: >> >> StringReverse@".LR dna RL neewteb esoohc ot syaw )j2-n(^2 eht yb >> deilpitlum >> neht si hcihw ,)!)j2-n(*2^!j(/!n si LR ro RL rehtie eb ot j2-n dna ,RR eb >> ot j ,LL eb ot sriap j esoohc ot syaw fo rebmun eht" >> >> I guess that makes it fairly clear why the left (perfecto) column is 2^n. >> And, why the right diagonal is 2n choose n. --rwg

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Re: [math-fun] Are you a Gammaturge?
by Dan Asimov 05 Nov '17

05 Nov '17

That sure is a lot of digits of accuracy for a Boolean variable. —Dan -----Original Message----- >From: Bill Gosper <billgosper(a)gmail.com> >Sent: Nov 5, 2017 4:12 AM >To: math-fun(a)mailman.xmission.com >Subject: [math-fun] Are you a Gammaturge? > >Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == >GoldenRatio > >In[345]:= N[%, 11111] > >Out[345]= True >--rwg >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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