Re: [math-fun] dumb question about general relativity
It seems to me that if we on the outside of a black hole never see things fall past the event horizon, then a white hole is just a black hole with stuff already rushing out from the event horizon. In other words, nothing can escape from inside a black hole, but then, nothing ever gets that far in anyway. An observer falling into a black hole continues past the event horizon onto a piece of paper that branches off from our history; I don't see why white hole ejecta can't have started on a similar side track-- except that both those tracks seem "non-falsifiable". I guess a singularity is a place where a scientific theory continues majestically past science? What I really want are gray holes, randomly absorbing and ejecting large chunks of matter contrary to all probability but consistent with "classical GR". It's hard to imagine a black hole forming after the big bang and *then* ejecting something. You have to imagine a very red-shifted thing accelerating from +.999...c to -.999...c -- how did it get up the gumption in such a short time? Was it actually in a tight chaotic orbit all along?
From: Rowan Hamilton <rowanham@gmail.com>
Classical mechanics is invariant under translations in time - this results in conservation of energy via Noether's Theorem.
Looking at http://en.wikipedia.org/wiki/Noether%27s_theorem it looks like the only kind of symmetries it applies to are transformations that you can have continuous versions of, like translation and rotation. Is this true? It makes me sad! I would have liked a conservation law to pop out of *every* kind of symmetry!
Classical mechanics is *not* invariant under time reversal, as others have pointed out, since this violates the Second Law of Thermodynamics.
I second the motion to marginalize the second law to statistics and revert to time-reversible physics. Surely a time-reversed black hole is allowed under a time-reversed Second Law. I'm curious how to tell whether the Second Law is working in forward or reverse. I can take two measurements of entropy, but looking at my log, how do I know whether the 1:01pm measurement was taken before or after the 1:03pm measurement? --Steve
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n, (where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial) has the limit, with n=7, (x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 , (giving that the product of these three tans is the negative reciprocal of their sum). Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives 4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169) and the correct root can be selected numerically from the eight. Can someone suggest a method more likely to be Mathematica's? --rwg
tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of x^3 - sqrt(7) x^2+ x + 1/sqrt(7) and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. I found the polynomial after the fact, so no proof here. I have no idea how Mathematica simplifies the sum. On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Like others I have no idea how mathematica does it, but here's a possibility: The first observation is that the expression in question is positive. Second, each of them are elements of the cyclotomic field of 28th roots of unity (e.g. if z = exp(2 pi i/28), you can write them all as rational functions in z). After standard simplification (taking the 28-th cyclotomic polynomial Phi_28 for the irreducible relation that z satisfies), you can apply the generators of the galois group (there are two of them, 3 and -1, since (Z/28 Z)* has rank 2), you can see that the expression in question is a quadratic real irrational. One then easily finds its irreducible polynomial being x^2 -7, and then using the first observation picks out sqrt(7). Victor On Thu, Jul 30, 2009 at 12:30 AM, James Buddenhagen <jbuddenh@gmail.com>wrote:
tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of
x^3 - sqrt(7) x^2+ x + 1/sqrt(7)
and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. I found the polynomial after the fact, so no proof here.
I have no idea how Mathematica simplifies the sum.
On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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James Buddenhagen associated the polynomial: x^3 - sqrt(7)*x^2 + x + 1/sqrt(7) to the identity: tan(3*pi/14) + cot(pi/7) - tan(pi/14) = sqrt(7). Is there a polynomial that we can associate to the fact that tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)? Warut On Thu, Jul 30, 2009 at 11:30 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of
x^3 - sqrt(7) x^2+ x + 1/sqrt(7)
and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. I found the polynomial after the fact, so no proof here.
I have no idea how Mathematica simplifies the sum.
On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On Tue, Aug 18, 2009 at 8:05 AM, Warut Roonguthai<warut822@gmail.com> wrote:
James Buddenhagen associated the polynomial:
x^3 - sqrt(7)*x^2 + x + 1/sqrt(7)
to the identity:
tan(3*pi/14) + cot(pi/7) - tan(pi/14) = sqrt(7).
Is there a polynomial that we can associate to the fact that
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)?
Warut
I'm not sure this helps, but tan(3*pi/11) is a root of: x^5-3*x^4*sqrt(11)+22*x^3-2*x^2*sqrt(11)-11*x+sqrt(11) and 2*sin(2*pi/11) is a root of: y^5-y^4*sqrt(11)+3*y^2*sqrt(11)-11*y+sqrt(11) and the relationship x + 2y = sqrt(11) transforms either polynomial into the other. James
Thanks so much, James. Never expected the solution to look like this, but this is exactly what I wanted. :) Warut On Tue, Aug 18, 2009 at 10:12 PM, James Buddenhagen<jbuddenh@gmail.com> wrote:
On Tue, Aug 18, 2009 at 8:05 AM, Warut Roonguthai<warut822@gmail.com> wrote:
James Buddenhagen associated the polynomial:
x^3 - sqrt(7)*x^2 + x + 1/sqrt(7)
to the identity:
tan(3*pi/14) + cot(pi/7) - tan(pi/14) = sqrt(7).
Is there a polynomial that we can associate to the fact that
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)?
Warut
I'm not sure this helps, but tan(3*pi/11) is a root of:
x^5-3*x^4*sqrt(11)+22*x^3-2*x^2*sqrt(11)-11*x+sqrt(11)
and 2*sin(2*pi/11) is a root of:
y^5-y^4*sqrt(11)+3*y^2*sqrt(11)-11*y+sqrt(11)
and the relationship x + 2y = sqrt(11) transforms either polynomial into the other.
James
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Thanks to all the info I learned from this thread, I now know how to find/construct interesting trig puzzles. The one I find beautiful and not close to anything I found on the Internet (in particular, not on http://mathworld.wolfram.com/TrigonometryAnglesPi13.html ) is tan(4*pi/13) + 4*sin(pi/13) = sqrt(13 - 2*sqrt(13)). Warut :)
On Tue, 18 Aug 2009, Warut Roonguthai wrote:
Is there a polynomial that we can associate to the fact that
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)?
How about x^2 - 11? Using Maple 13: r:=tan(3*Pi/11) + 4*sin(2*Pi/11): simplify(r^2); 11 By the way, applying PolynomialTools[MinimalPolynomial] to a numerical approximation of r yields the polynomial x^2 - 11. I.e., PolynomialTools[MinimalPolynomial](evalf(r),4); -11+_X^2
James Buddenhagen associated the polynomial:
x^3 - sqrt(7)*x^2 + x + 1/sqrt(7)
to the identity:
tan(3*pi/14) + cot(pi/7) - tan(pi/14) = sqrt(7).
Is there a polynomial that we can associate to the fact that
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)?
Warut
rwg> BCCee Dick Askey replied:
Here are two similar identities. tan(3pi/11) + 4sin(2pi/11) = 11^(1/2) sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = (1/2)*7^(1/2). These are in Chapter 7 of Hobson's "A Treatise on Plane and Advanced Trigonometry". In case you do not know this book, it has a number of gems. Dover has it back in print. Dick
Mma 7.0 FullSimplify does these, too. Maybe it just guesses them numerically? No, it has a ferocious algebraic number simplifier:
In[39]:= ToRadicals[Tan[3*Pi/11] + 4*Sin[2*Pi/11]] Out[39]= 5/11 7/22 4/11 I (1 + (-1) ) -2 (-1) (-1 + (-1) ) + ---------------- 5/11 -1 + (-1)
Perhaps I should have moved the problem to Macsyma and interpolated some missing steps: (c1) (domain:complex, mma2mac( -2*(-1)^(7/22)*(-1+(-1)^(4/11))+(I*(1+(-1)^(5/11)))/(-1+(-1)^(5/11)))) 5/11 ((- 1) + 1) %i 7/22 4/11 (d1) ------------------ - 2 (- 1) ((- 1) - 1) 5/11 (- 1) - 1 Find polynomials for these two terms separately: (c2) multiply_conjugates(x-args(%)) 11 9 7 5 3 (d2) [x - 55 x + 330 x - 462 x + 165 x - 11 x, 22 20 18 16 14 12 x - 88 x + 3344 x - 71808 x + 957440 x - 8200192 x 10 8 6 4 + 45101056 x - 154632192 x + 309264384 x - 317194240 x 2 + 126877696 x ] The polynomial satisfied satisfied by the sums of pairs of roots of these two polynomials is (c3) factor(resultant(subst(y,x,%[1]),subst(x-y,x,%[2]),y)) 2 2 10 (d3) x (x - 11) 10 8 6 4 2 2 (x - 143 x + 2970 x - 15246 x + 29909 x - 20339) 10 8 6 4 2 2 (x - 143 x + 3674 x - 25102 x + 41877 x - 20339) 10 8 6 4 2 2 (x - 143 x + 4906 x - 60126 x + 224389 x - 49379) 10 8 6 4 2 2 (x - 143 x + 5962 x - 87934 x + 514789 x - 1036739) 10 8 6 4 2 2 (x - 143 x + 7194 x - 161678 x + 1594197 x - 5131379) 10 8 6 4 2 2 (x - 55 x + 330 x - 462 x + 165 x - 11) 10 8 6 4 2 4 (x - 55 x + 858 x - 5214 x + 11605 x - 5819) 10 8 6 4 2 4 (x - 55 x + 1034 x - 7502 x + 14949 x - 11) 10 8 6 4 2 2 (x - 44 x + 704 x - 4928 x + 14080 x - 11264)
In[40]:= RootReduce[%]
Out[40]= Sqrt[11]
I confess ignorance of how, other than numerically, to choose the correct factor. Perhaps that 10th power is trying to tell me something? --rwg
On Thu, Jul 30, 2009 at 11:30 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
tan(3pi/14), cot(pi/7), and -tan(pi/14) are the roots of
x^3 - sqrt(7) x^2+ x + 1/sqrt(7)
and so their sum is sqrt(7) and their product is -1/sqrt(7) and the sum of their products in pairs is 1. �I found the polynomial after the fact, so no proof here.
I have no idea how Mathematica simplifies the sum.
On Wed, Jul 29, 2009 at 10:33 PM, <rwg@sdf.lonestar.org> wrote:
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. �Later I found a tricky proof: �The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) �= 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) �= x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal �of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
� � � �4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg
This is settled.
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n, (where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial) has the limit, with n=7, (x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 , (giving that the product of these three tans is the negative reciprocal of their sum). Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives 4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169) and the correct root can be selected numerically from the eight. Can someone suggest a method more likely to be Mathematica's? --rwg
BCCee Dick Askey replied: Here are two similar identities. tan(3pi/11) + 4sin(2pi/11) = 11^(1/2) sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = (1/2)*7^(1/2). These are in Chapter 7 of Hobson's "A Treatise on Plane and Advanced Trigonometry". In case you do not know this book, it has a number of gems. Dover has it back in print. Dick Mma 7.0 FullSimplify does these, too. Maybe it just guesses them numerically? No, it has a ferocious algebraic number simplifier: In[39]:= ToRadicals[Tan[3*Pi/11] + 4*Sin[2*Pi/11]] Out[39]= 5/11 7/22 4/11 I (1 + (-1) ) -2 (-1) (-1 + (-1) ) + ---------------- 5/11 -1 + (-1) In[40]:= RootReduce[%] Out[40]= Sqrt[11] --rwg
Last night, Mathematica embarrassed Macsyma and me by simplifying tan(3*pi/14)+cot(pi/7)-tan(pi/14) to sqrt(7) in front of dozens of impressionable children. Later I found a tricky proof: The quotient of two specializations of the very handy formula
prod(a*sin(2*j*pi/n+g)+b*cos(2*j*pi/n+g),j,1,n) = 2*(b^2+a^2)^(n/2)*(T[n](0)-T[n](-(a*sin(g)+b*cos(g))/sqrt(b^2+a^2)))/2^n,
(where T[n](x):=cos(n*acos(x)) = then nth Chebychev polynomial)
has the limit, with n=7,
(x^2-tan(pi/14)^2)*(x^2-cot(pi/7)^2)*(x^2-tan(3*pi/14)^2) = x^6-5*x^4+3*x^2-1/7 ,
(giving that the product of these three tans is the negative reciprocal of their sum).
Equating coefficients of x, adjoining tan(3*pi/14)+cot(pi/7)-tan(pi/14) = y, and eliminating the trigs gives
4*(y^2-7)*(7*y^6-91*y^4+245*y^2-169)
and the correct root can be selected numerically from the eight.
Can someone suggest a method more likely to be Mathematica's? --rwg
rwg@sdf.lonestar.org wrote:
BCCee Dick Askey replied: Here are two similar identities. tan(3pi/11) + 4sin(2pi/11) = 11^(1/2) sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = (1/2)*7^(1/2). These are in Chapter 7 of Hobson's "A Treatise on Plane and Advanced Trigonometry". In case you do not know this book, it has a number of gems. Dover has it back in print. Dick
Mma 7.0 FullSimplify does these, too. Maybe it just guesses them numerically? No, it has a ferocious algebraic number simplifier:
In[39]:= ToRadicals[Tan[3*Pi/11] + 4*Sin[2*Pi/11]] Out[39]= 5/11 7/22 4/11 I (1 + (-1) ) -2 (-1) (-1 + (-1) ) + ---------------- 5/11 -1 + (-1)
In[40]:= RootReduce[%]
Out[40]= Sqrt[11] --rwg
Gosper is mentioned prominently and favorably in the latest issue of the American Mathematical Monthly.
participants (7)
-
Edwin Clark -
James Buddenhagen -
rwg＠sdf.lonestar.org -
Stephen B. Gray -
Steve Witham -
victor miller -
Warut Roonguthai