Also rendered fine for me, via gmail and Mac Mail v 8.2.
On Oct 17, 2018, at 7:41 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
Hmm, Bill, it rendered fine for me, from gmail.
On Wed, Oct 17, 2018 at 3:44 PM Bill Gosper <billgosper@gmail.com> wrote:
On 2018-10-17 11:20, Henry Baker wrote:
Please stick with ASCII, as it's impossible to read these UTF8 characters (see below).
GAA! What I sent was more legible than ASCII, and pastable into Mathematica to boot! (From which you could extract ASCII if you really wanted it.) Duck-sucking, dog-bleeping xmission.com totally vandalized it. If they allowed, I could have attached Gosper.org/screwxmission.pdf I think I'll stick to mathfuneavesdroppers. Shall I add your name?
What about time around an elliptical orbit? Any closed form solutions for -- e.g., approximations to #days from solstice?
Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill
At 11:09 AM 10/17/2018, Bill Gosper wrote:
(Mail to some kids.)
Beginners may find it discouraging that the circumference of an ellipse
requires
this unfamiliar EllipticE function, but it is actually well worth familiarizing! For example, it provides the world's most rapidly convergent π formulas, and has fascinating properties. Actually, there are infinitely many eccentricities where the circumference EllipticE is expressible in terms radicals and factorials, but they're factorials of fractions!
You may already know that
In[210]:= (1/2)!
Out[210]= √π/2
but halves are the only known fractions whose factorials are familiar.
Somewhat amazingly, the 1 × 1/√2 ellipse (bounding box 2×√2) has circumference 9 π^(3/2)/(16 (3/4)!^2) + 32 (3/4)!^2/(9√π) ~ 5.4025755241907 (compared with the perimeter of the bounding box = 6.82842712474619.)
But this is perhaps the nicest case.
The circumference of a 1 by (1/ð“ + 1/√ð“)/√2 ellipse is (where ð“ := (1+√5)/2, the Golden Ratio)
ArcLength[Circle[{,}, {(1/ð“ + 1/√ð“)/√2, 1}]] == 9 Ï€^(3/2)/(10 √2 5^(7/8) ð“^(1/4) (1/20)! (9/20)!) + 2 √2 5^(3/8) (4 √5 + 10 √ð“) (1/20)! (9/20)!/(9 ð“^(1/4) √π) ~ 6.26092807313208, 2Ï€-ish because
In[255]:= N[(1/GoldenRatio + 1/√GoldenRatio)/√2]
Out[255]= 0.992908994700242
That's rounder than an Indiana circle.
—rwg
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