[math-fun] Rockets and arccos identities
The USA did indeed beat the Russians in creating the first artificial satellite, they exploded a rocket nose containing a lot of ball bearings and explosive. Result, some of the balls went into orbit. This was well known and was announced. It was, however, an extremely stupid idea for the future of space-trash. It is impressive how awesomely irresponsible a lot of power-mad US scientists and/or those employing them, were, another example being the mega bombing of, and leaving of radioactive waste on, various Pacific atolls for absolutely no good reason. The story about the manhole cover being the first thing humans made escape the earth is amusing, but could it really survive passage thru the atmosphere and exposure to the radiation from the nuke? ----- All the arccos identities (as I explained in my first post) can be proven using complex number arithmetic involving integers and sqrt(integers) only, no trig or arc-trig needed. Schroeppel then posted one explicit example of such a proof. It is trivial to get mathematica to prove them in this way, although it almost always is not smart enough to prove these identities if they are given in the form I gave them involving acos (or maybe it is capable of that, but if so you'd have to know more about mathematica than I do; naively using their usual simplifiers cannot do it). I was aware of the Machin-like arctan identities, although I do not understand the latest greatest technology by Arndt, Nimbran, et al for finding those of very high order. Unlike the arccos identities, those can be proven using complex arithmetic involving integers only, no sqrt(integer) needed. Also of course the arccos identities can be re-expressed using arccos(a/b)=arcsin(sqrt(1-(a/b)^2)) if desired, or arccos(a/b)=arctan(sqrt(b^2-a^2)/a). For this reason I suspect my arccos identities are a wider class than the Machin-like arctan identities. The reason I was interested in them and the rational-rational triangles ( https://dl.dropboxusercontent.com/u/3507527/RatRatTriangles.txt ) was because they arose inside gear-configurations by William Somsky. In some cases the arccos identities can be proven without need of any sqrt(integer), such as arccos(a/b) when a^2+c^2=b^2 arises from pythagorean triple. However such all-integer-proofs usually are not applicable. I conjecturally know all rational-side-ratio, rational-angled triangles in all 3 geometries, but currently only am able to prove the conjecture in the Euclidean geometry. For some reason I do not currently understand, the arccos identities like to employ powers of 2, 3 or 5 as denominators: 1*acos(-31/81)+2*acos(-5/9) = 360.00000000000 1*acos(-17/81)+4*acos(-1/3) = 540.00000000000 1*acos(17/81)+2*acos(-7/9) = 360.00000000000 1*acos(47/81)+2*acos(-8/9) = 360.00000000000 1*acos(49/81)+2*acos(4/9) = 180.00000000000 1*acos(73/81)+2*acos(2/9) = 180.00000000000 1*acos(79/81)+4*acos(-2/3) = 540.00000000000 1*acos(-47/128)+2*acos(-9/16) = 360.00000000000 1*acos(-7/128)+6*acos(-1/4) = 720.00000000000 1*acos(7/128)+3*acos(-7/8) = 540.00000000000 1*acos(41/128)+2*acos(-13/16) = 360.00000000000 1*acos(47/128)+6*acos(-3/4) = 900.00000000000 1*acos(79/128)+2*acos(7/16) = 180.00000000000 1*acos(97/128)+2*acos(-15/16) = 360.00000000000 1*acos(103/128)+2*acos(5/16) = 180.00000000000 1*acos(115/128)+3*acos(5/8) = 180.00000000000 1*acos(117/128)+3*acos(-3/8) = 360.00000000000 1*acos(119/128)+2*acos(3/16) = 180.00000000000 1*acos(127/128)+2*acos(1/16) = 180.00000000000 4*acos(-117/125)+6*acos(-24/25) =1620.00000000000 2*acos(-71/125)+3*acos(-23/25) = 720.00000000000 1*acos(-44/125)+3*acos(-4/5) = 540.00000000000 2*acos(44/125)+6*acos(-3/5) = 900.00000000000 1*acos(71/125)+3*acos(-1/5) = 360.00000000000 2*acos(117/125)+6*acos(-4/5) = 900.00000000000 1*acos(118/125)+3*acos(-2/5) = 360.00000000000 I currently have no idea why that is, but it would be nice to understand it, e.g. recognize pattern, produce closed forms. I'll remark that in some ways, the "acos miracle" identities are analogous to Fermat's last theorem (Wiles) and the Catalan conjecture (proven by Preda Mihailescu). Except that those were about nonexistence, while the acos miracles do exist, plentifully. A table of AcosMiracle identities (as well as some "near miss" inexact examples I computed mainly to get an idea of how much precision was needed) along with some theory, is now posted here: https://dl.dropboxusercontent.com/u/3507527/AcosMiracles128.out One of the most impressively close near misses is 3*acos(89/119)+99*acos(38/53)-25*pi=4.333...*10^(-13).
As far as I can tell, that was October 16, 1957 — 12 days after Sputnik. —Dan
The USA did indeed beat the Russians in creating the first artificial satellite, they exploded a rocket nose containing a lot of ball bearings and explosive. Result, some of the balls went into orbit. This was well known and was announced. It was, however, an extremely stupid idea for the future of space-trash. It is impressive how awesomely irresponsible a lot of power-mad US scientists and/or those employing them, were, another example being the mega bombing of, and leaving of radioactive waste on, various Pacific atolls for absolutely no good reason.
On 2015-07-17 10:21, Warren D Smith wrote:
The USA did indeed beat the Russians in creating the first artificial satellite, they exploded a rocket nose containing a lot of ball bearings and explosive. Result, some of the balls went into orbit. This was well known and was announced. It was, however, an extremely stupid idea for the future of space-trash. It is impressive how awesomely irresponsible a lot of power-mad US scientists and/or those employing them, were, another example being the mega bombing of, and leaving of radioactive waste on, various Pacific atolls for absolutely no good reason.
The story about the manhole cover being the first thing humans made escape the earth is amusing, but could it really survive passage thru the atmosphere and exposure to the radiation from the nuke?
-----
All the arccos identities (as I explained in my first post) can be proven using complex number arithmetic involving integers and sqrt(integers) only, no trig or arc-trig needed. Schroeppel then posted one explicit example of such a proof. It is trivial to get mathematica to prove them in this way, although it almost always is not smart enough to prove these identities if they are given in the form I gave them involving acos (or maybe it is capable of that, but if so you'd have to know more about mathematica than I do; naively using their usual simplifiers cannot do it).
They should. I've complained. By way of unusual simplifiers, In[338]:= Unprotect[Shallow]; Block[{Shallow = (FullSimplify[ ReleaseHold[#]] &)}, ClearSystemCache[]; Reduce[n*Pi == ArcCos[47/128] + 6*ArcCos[-3/4], n, Integers]] During evaluation of In[338]:= Reduce::ztest1: Unable to decide whether numeric quantity 0 is equal to zero. Assuming it is. >> Out[338]= n == 5 Meanwhile, we have a great source of crypto-integers. --rwg
I was aware of the Machin-like arctan identities, although I do not understand the latest greatest technology by Arndt, Nimbran, et al for finding those of very high order. Unlike the arccos identities, those can be proven using complex arithmetic involving integers only, no sqrt(integer) needed. Also of course the arccos identities can be re-expressed using arccos(a/b)=arcsin(sqrt(1-(a/b)^2)) if desired, or arccos(a/b)=arctan(sqrt(b^2-a^2)/a). For this reason I suspect my arccos identities are a wider class than the Machin-like arctan identities. The reason I was interested in them and the rational-rational triangles ( https://dl.dropboxusercontent.com/u/3507527/RatRatTriangles.txt ) was because they arose inside gear-configurations by William Somsky.
In some cases the arccos identities can be proven without need of any sqrt(integer), such as arccos(a/b) when a^2+c^2=b^2 arises from pythagorean triple. However such all-integer-proofs usually are not applicable.
I conjecturally know all rational-side-ratio, rational-angled triangles in all 3 geometries, but currently only am able to prove the conjecture in the Euclidean geometry.
For some reason I do not currently understand, the arccos identities like to employ powers of 2, 3 or 5 as denominators:
1*acos(-31/81)+2*acos(-5/9) = 360.00000000000 1*acos(-17/81)+4*acos(-1/3) = 540.00000000000 1*acos(17/81)+2*acos(-7/9) = 360.00000000000 1*acos(47/81)+2*acos(-8/9) = 360.00000000000 1*acos(49/81)+2*acos(4/9) = 180.00000000000 1*acos(73/81)+2*acos(2/9) = 180.00000000000 1*acos(79/81)+4*acos(-2/3) = 540.00000000000
1*acos(-47/128)+2*acos(-9/16) = 360.00000000000 1*acos(-7/128)+6*acos(-1/4) = 720.00000000000 1*acos(7/128)+3*acos(-7/8) = 540.00000000000 1*acos(41/128)+2*acos(-13/16) = 360.00000000000 1*acos(47/128)+6*acos(-3/4) = 900.00000000000 1*acos(79/128)+2*acos(7/16) = 180.00000000000 1*acos(97/128)+2*acos(-15/16) = 360.00000000000 1*acos(103/128)+2*acos(5/16) = 180.00000000000 1*acos(115/128)+3*acos(5/8) = 180.00000000000 1*acos(117/128)+3*acos(-3/8) = 360.00000000000 1*acos(119/128)+2*acos(3/16) = 180.00000000000 1*acos(127/128)+2*acos(1/16) = 180.00000000000
4*acos(-117/125)+6*acos(-24/25) =1620.00000000000 2*acos(-71/125)+3*acos(-23/25) = 720.00000000000 1*acos(-44/125)+3*acos(-4/5) = 540.00000000000 2*acos(44/125)+6*acos(-3/5) = 900.00000000000 1*acos(71/125)+3*acos(-1/5) = 360.00000000000 2*acos(117/125)+6*acos(-4/5) = 900.00000000000 1*acos(118/125)+3*acos(-2/5) = 360.00000000000
I currently have no idea why that is, but it would be nice to understand it, e.g. recognize pattern, produce closed forms. I'll remark that in some ways, the "acos miracle" identities are analogous to Fermat's last theorem (Wiles) and the Catalan conjecture (proven by Preda Mihailescu). Except that those were about nonexistence, while the acos miracles do exist, plentifully.
A table of AcosMiracle identities (as well as some "near miss" inexact examples I computed mainly to get an idea of how much precision was needed) along with some theory, is now posted here: https://dl.dropboxusercontent.com/u/3507527/AcosMiracles128.out
One of the most impressively close near misses is 3*acos(89/119)+99*acos(38/53)-25*pi=4.333...*10^(-13).
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participants (3)
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Dan Asimov -
rwg -
Warren D Smith