Re: A Formula for Pi

In article
<4a6c3352-a021-4f21-9009-bcdfff53043d@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Mensanator <mensanator@xxxxxxx> wrote:

On Jun 21, 11:37?pm, Michael Press <rub...@xxxxxxxxxxx> wrote:
In article
<41c3a2f3-d399-42f6-b21d-ce897b6ba...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





?Mensanator <mensana...@xxxxxxx> wrote:
On Jun 20, 2:35?pm, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <4f348778-0c41-45e4-bb51-9fcc3dca9...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Mensanator <mensana...@xxxxxxx> wrote:
On Jun 20, 10:57 am, Maury Barbato <mauriziobarb...@xxxxxxxx> wrote:
Jose Carlos Santos wrote:
On 20-06-2008 7:16, Maury Barbato wrote:

I found in the book "The Penguin Dictionary of
Curious and Interesting Numbers" by Wells the
following formula involving pi

(pi - 3)/4

= sum_{k=1 to infty}[(-1)^(k+1)]/[2k*(2k+1)*(2k+2)]

Is there anybody who knows a proof of this
wonderful
series?

You already got a reply. I only want to remark that
the formula is
equivalent to

pi - 3 = sum_{k = 1 to oo}(-1)^{k + 1}/(k(2k +
2k + 1)(k + 1)

= ?sum_{k = 1 to oo}(-1)^{k + 1}/(1^2 +
+ 1}/(1^2 + 2^2 + ... + k^2).

Best regards,

Jose Carlos Santos

A little slip. We have

pi - 3 =
sum_{k = 1 to oo}(-1)^{k + 1}/[k(2k+ 1)(k + 1)]
= (1/6)* sum_{k = 1 to oo}(-1)^{k + 1}/(1^2 + 2^2 + ... + k^2)

It's a very very beautiful formula!

Be that as it may, how fast does it converge?
How many terms do I have to sum to get 100 decimal
place accuracy?

He never claimed that it was an efficient way to compute pi,

I didn't say he claimed it was efficient. I'm _asking_
whether or not it is efficient.

simply
that it was a beautiful formula. ?

Fine. It's beautiful. No disagreemrnt there.

It is a personal opinion,

Which everyone is entitled to.

but I agree.

? ? oo
? ? --- ? ? k+1 ? ? ? 1
? ? > ? (-1) ? ?--------------
? ? --- ? ? ? ? 2k(2k+1)(2k+2)
? ? k=1

? ? ? ? oo
? ? ? 1 --- ? ? k-1 ? 1 ? ? 2 ? ? ?1
? ? = - > ? (-1) ? ?( -- - ---- + ---- ) ? ? [partial fractions]
? ? ? 2 --- ? ? ? ? ? 2k ? 2k+1 ? 2k+2
? ? ? ? k=1

? ? ? ? ? oo
? ? ? 1 ? --- ? ? k ?1
? ? = - + > ? (-1) ?---- ? ? ? ? ? ?[collapse telescoping terms]
? ? ? 4 ? --- ? ? ? 2k+1
? ? ? ? ? k=1

? ? ? 1 ? ? pi
? ? = - + ( -- - 1 ) ? ? ? ? ? ? ? ? ? ? ? ? ? ?[Gregory series]
? ? ? 4 ? ? 4

? ? ? pi - 3
? ? = ------
? ? ? ? 4

Being an alternating series with monotonically decreasing terms, the
error is less than 1/(8n^3) after n terms.

Some of us don't know how to do this. But given the series, I can
write a program to apply it.

But I would prefer a series that converges in ~300 terms to the
one that converges in ~10**34 terms.

How about Ramanujan's series that gets eight decimal places for each term?

1 ? ?2.sqrt{2} ? ? ? ? (4n)! ?[1103 + 26390n]
-- = --------- ?sum_n ------- ---------------
pi ? ? ?9801 ? ? ? ? ?(n!)^4 ? (4*99)^4n

Thanks for the suggestion, but I'm trying
to do this solely with unlimited precision
integers and rationals. No floats take part
in the calculations, only the creation of the
digit string used to test for convergence.

There is a simple recurrence for successive rational
approximations to sqrt{2} - 1:

{a_n} = 0, 1, 2, 5, 12, 29, 70, 169, ...

a_{n-1}
------- --> sqrt{2} - 1
a_n

a_{n+1} = 2.a_n + a_{n-1}

(a_n + a_{n-1})^2 - 2.(a_n)^2 = +/- 1

so

|a_n + a_{n-1} | 1
|------------- - sqrt{2} | < ---
| a_n | a_n

Or you could run a Newton iteration for successive rational approximations.

I tried Dave Seaman's suggestion of the Machin
formula and reduced the number of terms from 335
to 95 for 100 digit precision. That's probably
good enough for my purposes.

I actually don't care about pi as there's a
GMP function that's faster than anything I
could come up with. It was simply an exercise
in getting an arbitrary precicion float from
a series of unlimited precision rationals.
I thought the algorithm might come in handy
sometime in case I need this for something
other than pi.

At least one of the industrial strength programs
to calculate a high approximation of pi used Ramujan's series.

--
Michael Press
.