April 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Reciprocals in Q(2^(1/3))
by Dan Asimov 05 Apr '18

05 Apr '18

I'm curious how the expressions for the coefficients of the reciprocal change if the question had been about the field Q(tau) if tau = 2^(1/3) * exp(2πi/3), rather than about the field Q(2^(1/3)). —Dan Rich Schroeppel wrote: ----- The product of (all) the conjugates is the norm. The reciprocal of one field element is the product of the other conjugates, divided by the norm. For a cube root w, the other two conjugates of a0 + a1w + a2ww are, (using u^3=1 and recalling that u+uu = -1): a0 + a1wu + a2wwuu and a0 + a1wuu + a2wwu. The norm is a0^3 + 2a1^3 + 4a2^3 - 6a0a1a2. [side puzzle: prove this is never zero for rational a0,a1,a2 unless a0=a1=a2=0.] The "rational" part (the w^0 term) of the product of the other two conjugates is (selecting three terms from the nine term product) a0^2 + a1wu * a2wwu + a1wuu * a2wwuu, which, after using w^3=2, uuuu=u, and u+uu = -1, becomes a0^2 + a1 a2 w^3 (uu + uuuu), and then a0^2 - 2a1a2. b0 is just this expression divided by the norm. b1w and b2ww come from a circular substitution a0 -> a2ww -> a1w -> a0, which, of course, is equivalent to Henry's answer. Higher roots follow the same principles, but are more complicated. Full polynomials, with intermediate degree terms, are another step up in complexity, but the same method will work. I was working with the 2^(1/5) field, and needed to compute reciprocals. I worked out the complete formula, which came out fairly large. I found a better computational method was to work with only the first half of the radical terms, using F + G 5rt2 + H 5rt4, which has a much simpler reciprocal formula and norm formula. For a general number with all five terms, A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16, it was easy to come up with a multiplier X + Y 5r2 + Z 5r4 which zeroed out the 5r8 and 5r16 terms in the product. This gives the equation (A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16) * (X + Y 5r2 + Z 5r4) = P + Q 5r2 + R 5r4. Then I used the shorter reciprocal formula to compute 1 / (P + Q 5r2 + R 5r4), and multiply that by X + Y 5r2 + Z 5r4, getting the reciprocal of the full width field element A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16. This idea also works for the cbrt2 field, but doesn't give any computing advantage. It is helpful for tribonacci fields. Rich ----------- Quoting Henry Baker <hbaker1(a)pipeline.com>: > BTW, "maxima" is free... > > 2 > 2 a1 a2 - a0 > (%o2) [[b0 = - --------------------------------, > 3 3 3 > 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 > 2 > 2 a2 - a0 a1 > b1 = --------------------------------, > 3 3 3 > 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 > 2 > a0 a2 - a1 > b2 = - --------------------------------]] > 3 3 3 > 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 > > > At 08:32 AM 4/5/2018, Allan Wechsler wrote: >> Suppose w is the real cube root of 2, w^3 = 2. >> >> Then for a0, a1, a2 rational, numbers of the form a0 + a1*w + a1*w^2 form a >> field. What I want is the formula for the multiplicative inverse in this >> field. It's elementary algebra, but it is voluminous and I keep making >> errors when I try to do it by hand. Any computer algebra system could >> probably instantly solve for b0, b1, b2 subject to (a0 + a1w + a2ww)(b0 + >> b1w + b2ww) = 1. Can somebody do me a favor and give me formulas for the >> b's in terms of the a's? -----

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[math-fun] Two identities
by françois mendzina essomba2 05 Apr '18

05 Apr '18

Hello For some values of a and b, especially values taken between 0 and 1 with a greater than b First identity ; theta=sum(((-1)^(n+1)*cos(a*theta)^(1-2*n)*sin(b*theta)^(2*n-1)*cos((b+a)*(2*n-1)*theta))/(2*n-1),n,1,inf)/b+sum(((-1)^(n+1)*sin(b*theta)^(2*n)*sin(2*(b+a)*n*theta))/(n*cos(a*theta)^(2*n)),n,1,inf)/(2*b); second identity; Pi/2-a*theta=sum((cos(a*theta)^n*sin((b+a)*n*theta))/(n*cos(b*theta)^n),n=1..infinity); With the second indentity, we can deduce many formulas: Pi/6=sum(sin((7*Pi*n)/12)/(n*2^(n/2)),n=1..infinity); Pi/6=sum(sin((2*Pi*n)/3)/n,n=1..infinity); Pi/6=sum(sin((8*Pi*n)/15)/(cos(Pi/5)^n*n*2^n),n=1..infinity); Pi/4=sum((2^(n/2)*sin((5*Pi*n)/12))/(n*3^(n/2)),n=1..infinity) ; We find the Madhava-Gregory-Leibniz formula, with a=1/4 and b=1/4 ; Pi/4=sum(sin((Pi*n)/2)/n,n=1..infinity); FME...

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Re: [math-fun] Reciprocals in Q(2^(1/3))
by Henry Baker 05 Apr '18

05 Apr '18

BTW, "maxima" is free... 2 2 a1 a2 - a0 (%o2) [[b0 = - --------------------------------, 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 2 2 a2 - a0 a1 b1 = --------------------------------, 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 2 a0 a2 - a1 b2 = - --------------------------------]] 3 3 3 4 a2 - 6 a0 a1 a2 + 2 a1 + a0 At 08:32 AM 4/5/2018, Allan Wechsler wrote: >Suppose w is the real cube root of 2, w^3 = 2. > >Then for a0, a1, a2 rational, numbers of the form a0 + a1*w + a1*w^2 form a >field. What I want is the formula for the multiplicative inverse in this >field. It's elementary algebra, but it is voluminous and I keep making >errors when I try to do it by hand. Any computer algebra system could >probably instantly solve for b0, b1, b2 subject to (a0 + a1w + a2ww)(b0 + >b1w + b2ww) = 1. Can somebody do me a favor and give me formulas for the >b's in terms of the a's? > >Thank you!

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[math-fun] Reciprocals in Q(2^(1/3))
by Allan Wechsler 05 Apr '18

05 Apr '18

Suppose w is the real cube root of 2, w^3 = 2. Then for a0, a1, a2 rational, numbers of the form a0 + a1*w + a1*w^2 form a field. What I want is the formula for the multiplicative inverse in this field. It's elementary algebra, but it is voluminous and I keep making errors when I try to do it by hand. Any computer algebra system could probably instantly solve for b0, b1, b2 subject to (a0 + a1w + a2ww)(b0 + b1w + b2ww) = 1. Can somebody do me a favor and give me formulas for the b's in terms of the a's? Thank you!

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[math-fun] special value of my identity
by françois mendzina essomba2 03 Apr '18

03 Apr '18

a special value of my identity Pi/4=sum(sin((Pi*n)/4)/(n*2^(n/2)),n=1..infinity);

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[math-fun] The fummin problem
by Dan Asimov 03 Apr '18

03 Apr '18

(Please delete my e-mail address from any such documents before distributing them — thanks.) The multi-author muffin paper begins as follows: ----- You have m muffins and s students. You want to divide the muffins into pieces and give the shares to students such that every student has m muffins. Find a divide-and-distribute protocol that maximizes the minimum piece. ----- Apologies for not remembering whether I posted the following already, or merely thought it: Suppose we change the problem by changing the objective criterion, from maximizes the minimum piece. (the muffin problem) to minimizes the *number* of pieces (the fummin problem). (I think now I proposed this criterion for finding nice geometric dissections like regular polygon-to-regular polygon ones, but not for a muffin problem.) Minor simplification: Given M disjoint circles of unit circumference, What is the smallest number of pieces to chop them into that can be reassembled into S disjoint congruent circles? (Of course, each would have circumference M/S.) First question: What is the smallest case, if any, where the solution to this problem uses a different dissection from that used in the muffin problem for the same M and S ? —Dan

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Re: [math-fun] Is mathematical truth "real" ?
by Henry Baker 03 Apr '18

03 Apr '18

Actually, that statement is itself subject to experimental verification. :-) At 03:12 PM 4/2/2018, Dan Asimov wrote: >Experiments don't *change* physical truth, either.

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Re: [math-fun] Is mathematical truth "real" ?
by Dan Asimov 03 Apr '18

03 Apr '18

Except you can think of math as the mapping from axiom systems to the set of true propositions under that system. Then the only axiom system one is using across the board is the language of logic, and then you don't have to keep redefining what is true. —Dan ----- Except in some cases you can choose different axioms in mathematics and change what was 'true' to 'false' or 'provable' to 'unprovable'. Brent On 4/2/2018 3:08 PM, Dan Asimov wrote: > This was basically my point in the attempted conversation with the > molecular biologist: Why do I believe in *physical* truth? Because > as instruments improved we have been able to independently confirm > physical measurements to a higher and higher degree of accuracy. > > Similarly with math: We can independently confirm mathematical truth > (at least of the kind corresponding to a theorem). > > —Dan > > > Mike Stay wrote: > ----- > I think mathematical truth is "real" in the sense that it is > objective; it can be independently verified. Mathematicians talk > about discovering mathematical truth rather than inventing it. We can > build mathematical gadgets in a similar way to building devices, but > there are laws that govern their behavior in much the same way that > physics governs the devices' behavior. > > On Mon, Apr 2, 2018 at 10:42 AM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> If this common discussion subject among mathematicians has come up in >> math-fun previously, it must have been a while ago: >> >> Is mathematical truth real? I was reminded of this question when I >> happened to sit next to an esteemed molecular biologist who has strong >> opinions on the matter: He thinks math is "all in the brain" of humans >> who think about it. I could not convince him that math has any kind of >> independent existence — though I certainly believe it myself. >> >> What do other think about this? I would say that mathematical truth >> is *at least* as real of a thing as physical truth. > ----- -----

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Re: [math-fun] Is mathematical truth "real" ?
by Dan Asimov 03 Apr '18

03 Apr '18

This was basically my point in the attempted conversation with the molecular biologist: Why do I believe in *physical* truth? Because as instruments improved we have been able to independently confirm physical measurements to a higher and higher degree of accuracy. Similarly with math: We can independently confirm mathematical truth (at least of the kind corresponding to a theorem). —Dan Mike Stay wrote: ----- I think mathematical truth is "real" in the sense that it is objective; it can be independently verified. Mathematicians talk about discovering mathematical truth rather than inventing it. We can build mathematical gadgets in a similar way to building devices, but there are laws that govern their behavior in much the same way that physics governs the devices' behavior. On Mon, Apr 2, 2018 at 10:42 AM, Dan Asimov <dasimov(a)earthlink.net> wrote: > If this common discussion subject among mathematicians has come up in > math-fun previously, it must have been a while ago: > > Is mathematical truth real? I was reminded of this question when I > happened to sit next to an esteemed molecular biologist who has strong > opinions on the matter: He thinks math is "all in the brain" of humans > who think about it. I could not convince him that math has any kind of > independent existence — though I certainly believe it myself. > > What do other think about this? I would say that mathematical truth > is *at least* as real of a thing as physical truth. -----

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Re: [math-fun] Is mathematical truth "real" ?
by Dan Asimov 02 Apr '18

02 Apr '18

Experiments don't *change* physical truth, either. On the other hand, I do mathematical experiments all the time, like calculating something to see what its value is, something that can lead to observing a pattern not previously seen. —Dan ----- You can do experiments from now til kingdom come, and none of them will change any mathematical truths. -----

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