[math-fun] Three flake frames
(Easy for YOU to say.) gosper.org/flakes.png First frame is swing angle π/3, confirming DanA's guess about trisected areas. 2nd frame is implausible swing angle 7π/33, or VERY nearly, showing subflake at a peculiar angle. 3rd frame is even weirder angle that seems to make a triskelion out of ten tiny flakes. Doubt this one. --rwg Note that I answered DMakin about when Mike Stay did, but mine was delayed three hours. Death to spammers. Also, an embarrassing Fibonacho postscript: In[997]:= Sum[Binomial[n - k, k], {k, 0, n}] + Sum[Binomial[n - k, k], {k, n + 1, n + 1}] Out[997]= Binomial[-1, 1 + n] + Fibonacci[1 + n] In[998]:= Table[%, {n, 0, 9}] Out[998]= {0, 2, 1, 4, 4, 9, 12, 22, 33, 56} should be the same as In[999]:= Sum[Binomial[n - k, k], {k, 0, n + 1}] Out[999]= ((3 + n) Hypergeometric2F1[1/2 - n/2, -(n/2), -n, -4] + Binomial[-2, 2 + n] (-1 + Hypergeometric2F1[(2 + n)/2, (3 + n)/2, 2 + n, -4]))/(3 + n) In[1001]:= Simplify[Table[%%, {n, 0, 9}]] Out[1001]= {1/10 (5 - Sqrt[5]), 5/2 - 3/(2 Sqrt[5]), 2 - 2/Sqrt[5], 1/10 (55 - 7 Sqrt[5]), 1/10 (65 - 11 Sqrt[5]), 13 - 9/Sqrt[5], 37/2 - 29/(2 Sqrt[5]), 65/2 - 47/(2 Sqrt[5]), 50 - 38/Sqrt[5], 1/10 (835 - 123 Sqrt[5])} I.e., how to compute√5 as a finite sum of integers.
On Fri, Oct 4, 2013 at 9:52 PM, Bill Gosper <billgosper@gmail.com> wrote:
(Easy for YOU to say.) gosper.org/flakes.png First frame is swing angle π/3, confirming DanA's guess about trisected areas. 2nd frame is implausible swing angle 7π/33, or VERY nearly, showing subflake at a peculiar angle. 3rd frame is even weirder angle that seems to make a triskelion out of ten tiny flakes. Doubt this one.
Especially since they number thirteen.
--rwg
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Adam wrote: Bill wrote: (Easy for YOU to say.) gosper.org/flakes.png First frame is swing angle π/3, confirming DanA's guess about trisected areas. 2nd frame is implausible swing angle 7π/33, or VERY nearly, Actual value = ArcTan[(5 Sqrt[3])/11] ~ 0.666946 whereas 7 Pi/33 ~ 0.666398 That's a pretty massive difference. Indeed, it's closer to the even more implausible value of 2/3. In general, all of your swing angles will be arctangents of things in the quadratic field Q[Sqrt[3]], since they can be expressed as arguments of solutions to linear equations, the coefficients of which are complex numbers in the imaginary quadratic field Q[Sqrt[-3]]. The process of obtaining the equation and solution is pretty easy and boring. Sincerely, Adam P. Goucher http://cp4space.wordpress.com OMG, there really ARE 13 of them little goobers? gosper.org/flake13.png That's not so boring. --rwg On Fri, Oct 4, 2013 at 10:06 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Oct 4, 2013 at 9:52 PM, Bill Gosper <billgosper@gmail.com> wrote:
(Easy for YOU to say.) gosper.org/flakes.png First frame is swing angle π/3, confirming DanA's guess about trisected areas. 2nd frame is implausible swing angle 7π/33, or VERY nearly, showing subflake at a peculiar angle. 3rd frame is even weirder angle that seems to make a triskelion out of ten tiny flakes. Doubt this one.
Especially since they number thirteen.
--rwg
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Bill Gosper