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RE: [math-fun] Calculating pi
by Schroeppel, Richard 12 Jul '06

12 Jul '06

A straightforward approach is to calculate the area of the quarter circle in the first quadrant. Split the circle into 10 vertical segments, and use the trapezoid rule to get inside and outside estimates for the area. This requires nothing worse than square roots and slopes of tangent lines -- a minimum of analytic geometry. It's easy to see that bumping up the number of slices to 100 will reduce the error by a factor of around 10. This part can be programmed numerically. The next step is deeper: Writing out the binomial for sqrt(1-x^2), believing it converges (at least out to x=1/2), rearranging the terms to make an integral, and accepting integral(x^K) = x^(K+1)/(K+1), may take some selling. But after that, you actually have a reasonably convergent series for pi, related to arcsin(1/2) and the area of the portion of the unit circle between x=0 and x=1/2 and y>=0. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Jason Holt Sent: Wednesday, July 12, 2006 2:05 PM To: math-fun Subject: [math-fun] Calculating pi A friend of mine recently asked me what pi is all about. This was rather remarkable, as he's failing out of high school, and people treat him like a failure. Turns out he's fascinated by physics and math. Anyway. So I described pi in terms of the experimental method of drawing circles and noticing that their circumference always seems a little more than 3 times their diameter. That led to a great discussion about experimental vs. theoretical science, and eventually he asked me how we went from experimental methods of calculating pi to computational approaches. I checked the wikipedia, but its approaches all seem to assume advanced math to prove that the approximations describe pi, and he hasn't gotten past algebra and geometry. At this point, the simplest approach I can think of is to write a program to choose random points in a circumscribed circle, then calculate the proportion of the points which fall within it. But that still has a heuristic element to it. Are there any plain arguments for the deterministic approximations? (For that matter, how about doing the same thing for the trig functions? I took the same approach for sin/cos/tan, talking about building a table of angle/ratio measurements, but I had no idea how to describe computational approaches.) -J _______________________________________________ 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] Calculating pi
by Cordwell, William R 12 Jul '06

12 Jul '06

Buffon's needle experiment combines, in a sense, experiment and trig. Although the derivation might be too advanced, it could be presented heuristically, I think. There are lots of simulations available on the web. --Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com] On Behalf Of Jason Holt Sent: Wednesday, July 12, 2006 3:06 PM To: math-fun Subject: Re: [math-fun] Calculating pi On Wed, 12 Jul 2006, Richard Guy wrote: > Surely best to follow Archimedes and > calculate the perimeters of inscribed > and circumscribed regular 6-, 12-, > 24-, ...-gons? R. Working that out computationally seems to require sines, so that the problem of calculating pi reduces to the problem of calculating sines, yes? -J _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] System of 3 equations
by Christian Boyer 12 Jul '06

12 Jul '06

Do you know if it is possible to solve this system of equations, with n=3 or 4: a^n + b^n = c^n + d^n e^n + f^n = g^n + h^n (ae)^n + (bf)^n = (cg)^n + (dh)^n a,b,c,d,e,f,g,h being distinct integers. Solutions with n=1 or 2 are easy to find. For example: 1 + 4 = 2 + 3 6 + 7 = 5 + 8 (1*6) + (4*7) = (2*5) + (3*8) 2² + 11² = 5² + 10² 79² + 46² = 86² + 31² (2*79)² + (11*46)² = (5*86)² + (10*31)² But what about n=3? I am unable to find at least one solution. And n=4 seems hopeless. Christian.

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RE: [math-fun] Prime "number system" ? (Goldbach, etc.)
by Schroeppel, Richard 10 Jul '06

10 Jul '06

The Prime Number Theorem, pi(X) ~ X/logX, implies that the ratio of consecutive primes approaches 1 as a limit. In fact, a bit more is proven - roughly, for large enough numbers, there's always a prime between any two cubes. So by taking the representation of N to be (largest prime <= N) + (recurse to represent the balance), each prime will reduce the size of the leftover from N down to O(N^(2/3)), and the total number of terms is around O(loglogN). Of course, if you allow Goldbach, then 3 primes is likely always enough. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Henry Baker Sent: Monday, July 10, 2006 2:19 PM To: Michael Reid Cc: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] Prime "number system" ? (Goldbach, etc.) Interesting. Is it then true that there is a prime of every length when expressed in the binary system? Then I suppose we could always choose the smallest such for each length, and we would never need more than ~log_2(n) such numbers in the representation. So this representation is as compact as the usual binary system? At 01:08 PM 7/10/2006, Michael Reid wrote: >> In an analogy with Zeckendorf (Fibonacci) number systems, can every >> positive integer be expressed as a _sum_ of _distinct_ (positive) primes ? >> >> If not, what is the smallest integer not so expressible? >> >> For this purpose, I'll allow "1" as a "prime", although I suspect >> it's only needed a constant number of times. > >yes, with your convention of considering 1 to be prime. >you've already shown how to do the first few cases. >for a larger value of N , let p be the largest prime <= N . >then p > N/2 , by the theorem usually called "Bertrand's postulate". >express N - p as a sum of distinct "primes", and add p to the >result. since p > N - p , the primes in this sum are distinct. > >it is probably easy to modify this argument to show that 1 is only >needed a few times (twice?) > >mike _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] sqrt( |Rubik| )
by Steve Witham 10 Jul '06

10 Jul '06

>From: "Schroeppel, Richard" <rschroe(a)sandia.gov> > >We now have enough computing power to find & prove the distance- >to-start for any particular position, using the obvious square-root >meet-in-the-middle algorithm and a big disk. http://en.wikipedia.org/wiki/Rubik%27s_Cube_group says that the number of positions is ( 8! 3^8 12! 2^12 ) / 12 = 43,252,003,274,489,856,200 so the square root is about 6,576,625,523 Each position needs about 5 bytes, and I guess having two copies of the set would make it go a lot faster, so that's 66 terabytes...by my amateur guess. This NYT article about Google's new server farm complex in The Dalles, Oregon http://tinyurl.com/pfvz8 says, "The best guess is that Google now [pre-The Dalles] has more than 450,000 servers spread in at least 25 locations around the world. " If those servers each have 2x10^8 ips, that's about 10^14 ips, which matches Hans Moravec's estimate of the effective power of a human brain (though not in a tight enough cluster). --Steve

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Re: [math-fun] Prime "number system" ? (Goldbach, etc.)
by Michael Reid 10 Jul '06

10 Jul '06

> In an analogy with Zeckendorf (Fibonacci) number systems, > can every positive integer be expressed as a _sum_ of _distinct_ (positive) primes ? > > If not, what is the smallest integer not so expressible? > > For this purpose, I'll allow "1" as a "prime", although I suspect it's only needed a > constant number of times. yes, with your convention of considering 1 to be prime. you've already shown how to do the first few cases. for a larger value of N , let p be the largest prime <= N . then p > N/2 , by the theorem usually called "Bertrand's postulate". express N - p as a sum of distinct "primes", and add p to the result. since p > N - p , the primes in this sum are distinct. it is probably easy to modify this argument to show that 1 is only needed a few times (twice?) mike

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[math-fun] Prime "number system" ? (Goldbach, etc.)
by Henry Baker 10 Jul '06

10 Jul '06

(I hope I am not embarrassing myself with this one...) In an analogy with Zeckendorf (Fibonacci) number systems, can every positive integer be expressed as a _sum_ of _distinct_ (positive) primes ? If not, what is the smallest integer not so expressible? For this purpose, I'll allow "1" as a "prime", although I suspect it's only needed a constant number of times. 2=2 3=3 4=3+1 5=5 6=5+1 7=7 8=5+3 9=7+2 10=7+3 11=11 12=7+5 13=13 14=11+3 15=13+2 16=13+3 17=17 18=13+5 19=19 20=17+3 21=19+2 22=19+3 23=23 24=19+5 25=23+2 26=23+3 27=19+5+3 28=23+5 29=29 etc. I'm not (yet) worried about the smallest number of such primes in the "representation".

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Re: [math-fun] sqrt( |Rubik| ), Google, and humans
by The Purple Streak, Hilarie Orman 10 Jul '06

10 Jul '06

There are about 6x10^9 human brains spread in at least 25 locations around the world, though not in a tight enough cluster. Hilarie

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[math-fun] diameter of Rubik's cube puzzle
by Schroeppel, Richard 10 Jul '06

10 Jul '06

Has the diameter of the Rubik's cube puzzle been established yet? We now have enough computing power to find & prove the distance- to-start for any particular position, using the obvious square-root meet-in-the-middle algorithm and a big disk. We can search for edge positions by generating random positions and using the exact distance algorithm to guide a hill-climb away from the start position. But this is only heuristic. Rich

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Re: [math-fun] 3-dimensional rhombus
by dasimov＠earthlink.net 10 Jul '06

10 Jul '06

Michael Kleber wrote: << Russ Cox wrote: > complete the analogy: > > square : rhombus :: cube : ____ ? Do you mean to require that the twelve edges all have the same length, or that the six faces all have the same area? Or both? Or perhaps you want all faces to be congruent? Now I'm not even sure which Gareth's "equilateral parallelepiped" would best fit -- either the first or the last, I think. >> Technically there are several ways to generalize the rhombus to 3D. But the most natural and symmetric way seems to me for it to be a parallepiped all of whose edges are the same length <=> all of whose faces are rhombi <=> all of whose faces are congruent rhombi. The word for this 3D object is "rhombohedron". (It generalizes neatly to n dimensions. Not sure if there is a word for the n-dimensional object, but perhaps it ought to be "rhombotope".) All such rhombotopes can be obtained from a cube by applying an arbitrary dilatation to one main diagonal while keeping its orthogonal directions fixed, I ween. --Dan

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