November 2017 - mathfun@mailman.xmission.com

[math-fun] Tropical summation formulae
by Tom Duff 09 Nov '17

09 Nov '17

I got this from Dan Piponi (@sigfpe on twitter): sum (a+b)^-2 (c+d)^-2 (a+b+c+d)^-2 = 1/24 where the sum is over all integers (a,b,c,d) with 1<=a<=b, 1<=c<=d Paper: Tropical formulae for summation over a part of SL(2, Z) Nikita Kalinin, Mikhail Shkolnikov https://arxiv.org/abs/1711.02089 -- Tom Duff. Branches broken off trees.

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Re: [math-fun] Stupid steam train question
by Mike Beeler 09 Nov '17

09 Nov '17

I asked a train fan friend, who had this answer that seems to explain it. — Mike > Begin forwarded message: > > From: Claudio Topolcic <claudio(a)sunsetharrier.com> > Subject: Re: [math-fun] Stupid steam train question > Date: November 8, 2017 at 10:29:30 PM EST > To: Mike Beeler <mikebeeler2(a)gmail.com> > Cc: Claudio Topolcic <claudio(a)sunsetharrier.com> > > Yes, you guys are figuring it out. This was worked out in the early days of steam locomotives. The first design was more like a tea pot, where the hot gasses went around the boiler. This is very inefficient since most of the hot gasses don't come anywhere near the boiler. So they ran the hot gasses through tubes in the boiler, like your house furnace. But you want to force the cold air through the burning coal to burn faster. I think this is what tall smoke stacks are for in power stations. So you could use a fan, and I think this design was used. But there is a simpler design. The spent steam from the cylinders goes up in the "blast pipe" in the smoke box at the front of the boiler, and that pipe ends somewhat below the bottom of the stack. So the exhaust steam passing out of that pipe and to the stack causes a Venturi effect that sucks the hot gasses through the pipes in the boiler through the smoke box and out the stack. So yes indeed the puffing of the smoke and steam exhausted out the stack are in synch with the cylinders. > > I assume that below, Henry was seeing steam coming out of the cylinders, and this is what he was taking about. I assume he figured that was the spent steam being exhausted from the cylinders. That is not the case. There are various reasons why some steam would be exhausted directly from the cylinders. I don't know all, but an example that I do know is that there is a valve at the bottom of each cylinder that the engineer can open remotely. They open the valves when the cylinders are still cold because some of the steam condenses in the cold cylinders and if they left liquid water in the cylinders then that could break the pistons when they move to their extreme. But that is not the bulk of the exhausted spent steam, that goes up the stack. > > I don't know about steam ships, but I would expect that they do the same thing with the spent steam. But there is a lot of plumbing between the boiler and the funnel, and I don't know what happens to puffs. Also big steam ships have been using turbines for a while, with no puffs. But I don't know. > > Check out the Wikipedia page https://en.wikipedia.org/wiki/Steam_locomotive <https://en.wikipedia.org/wiki/Steam_locomotive>. It has this drawing, check out #12: [ I think math-fun is only text, so I removed the illustration. It shows the outside and a cut-away of a steam loco. Part # 12 is the “blast pipe”. ]

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[math-fun] Stupid steam train question
by Henry Baker 09 Nov '17

09 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 07 Nov '17

07 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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