8
|
8570 Views
|
0 Replies
|
8 Total Likes
View groups...
Share
GROUPS:

# A million digits (feh) of ?

Posted 8 years ago

NOTE: All utility functions are defined at the end and also in the attached notebook.

You can compute the first million digits of $\pi$ without printing them in < 1 second (the 1st number in the the time of computation without printing, and the printed later image is actually only a minuscule part of the whole 10^6 digits):

tim[N[Pi, 10^6]]


0.314149, 0

and then print the last 99:

tim[Mod[Round[10^10^6*%], 10^99]]


0.021651,0

315614033321272849194418437150696552087542450598956787961303311646283996346460422090106105779458151

Far better than digits is a continued fraction:

longer[Pi, 9]


This expression = $\pi$

Simplify[ReleaseHold[%]]


?

and can be freely lengthened (or shortened):

longer[%%]


Notice that $e$, unlike $\pi$, has a pattern:

longer[E, 9]


But proving that ? will never develop a pattern is one of the great unsolved problems. For faster, but non-resumable continued fractions, Mathematica has, e.g.,

ContinuedFraction[Pi, 9]


{3, 7, 15, 1, 292, 1, 1, 1, 2}

Note the largish term at position 431:

Take[ContinuedFraction[Pi, 433], -4]


{1, 4, 20776, 1}

Around 1986, in a calculation taking several weeks, I found

tim[Take[ContinuedFraction[Pi, 11504932], -4]]


8.36905,4

{1, 1, 878783625, 6}

I thought Eric Weisstein found a bigger one, but Pi Continued Fraction page doesn't seem to say. Simple functions of e also have patterns:

longer[(E^2 + 1)/(E^2 - 1), 9]


$e^x$ has a nice infinite series:

SeriesCoefficient[E^x, {x, 0, n}]


This means

(#1 == Activate[#1] & )[Inactive[Sum][x^n/n!, {n, 0, Infinity}]]


Taking a few terms

ser = Activate[%[[1]] /. Infinity -> 9]


we can numerically test Euler's celebrated

(#1 == Activate[#1] & )[E^Inactivate[I*Pi]]


N[ser /. x -> I*Pi]


-0.976022 + 0.00692527 I

The square root of Euler's identity is

(#1 == Activate[#1] & )[E^(Inactivate[I*Pi]/2)]


N[ser /. x -> I*(Pi/2)]


0.0000247373 + 1. I

We can even use Euler's identity to calculate ? by solving

eq = E^(I*x) + 1 == 0


This has infinitely many solutions!

Simplify[Solve[eq]]


Unprotect[C]; Table[%[[1, 1]], {C[1], -2, 2}]


{x -> -3 ?, x -> -?, x -> ?, x -> 3 ?, x -> 5 ?}

Newton's iteration says: to solve $f(x)=0$, choose an initial guess for x and iterate $g(g(...g(x)))$ where $g = x - f/ df/dx$

g[x] = Simplify[x - eq[[1]]/D[eq[[1]], x]]


Starting with a very precise value of 3

NestList[I/E^(I*#1) + #1 + I & , 3.69., 7]


Each iteration of Newton's method typically doubles the number of correct digits. But isn't it slightly cockeyed to seek a real answer by applying a complex function to a real approximation? Suppose we just took the real part of $g$:

ComplexExpand[Re[g[x]]]


x + Sin[x]

(Remembering that Euler's identity generalizes to Euler's formula:)

(#1 == ComplexExpand[#1] & )[E^(I*x)]


Using this new $g$

NestList[#1 + Sin[#1] & , 3.69., 5]


NestList[#1 + Sin[#1] & , 9.69., 5]


Dividing the last approximation by 3

Last[%]/3


3.1415926535897932384626433832795028841971693993751058209749445923078

## Utility functions

These are definitions of all utility functions you will need for the above evaluations.

Clear[tim];
tim[xp_]:=(Print[#[[1]],",",Length[#[[2]]]];
If[#[[1]]>69,Speak[#[[1]]]];#[[2]])&@AbsoluteTiming[xp]

SetAttributes[tim,HoldAll]

Clear[shorter];
shorter[cf_,0]:=cf

shorter[cf_,n_: 1]:=shorter[cf/.(a_: 0)+1/HoldForm[r_]:>HoldForm@@{(a+1/r)},n-1]

Clear[longer];
longer[cf_,0]:=cf

longer[x_?NumericQ,n_: 1]:=longer[HoldForm@@{x},n]

longer[cf_,n_: 1]:=longer[cf/.HoldForm[r:Except[_Integer]]:>
Floor[r]+1/HoldForm@@{Simplify@Together[1/Mod[r,1]]},n-1]

tail[cf_]:=Cases[cf,HoldForm[x_]->x,\[Infinity]][[1]]
`
Attachments: