Ramanujan's beastly expressions for pi

Molinaro · Apr 19, 2012

[latex]$$ \frac{1}{\pi} = \sqrt{8} \sum_ {n=0}^{\infty} \frac{(1103+26390n)(2n-1)!!(4n-1)!! } { (99^{4n+2})(32^{n})(n!)^{3} } $$[/latex]

and

[latex]$$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_ {k=0}^{\infty} \frac{(4k)!(1103+26390k)} {(k!)^{4}(396^{4k})} $$[/latex]

These equations never cease to amaze me. At first sight they seemed akin to magic. I could not even begin to guess how they were derived, or I should say, discovered.

His wiki page mentions the basis for the derivation of the 2nd equation.

I wonder how fast they converge on the digits of pi. Is one faster than the other? Or how many terms of the sum can be calculated before the factorials require the use of a big numbers math library.

jasonpatterson · Apr 19, 2012

I don't really have a good way to calculate the value, but just throwing the numbers into excel, using k = 0 alone gives 3.14159273 and k=1 takes the value out to 3.1415926535897939 rather than 3.1415926535897932. Beyond that I can't make excel handle the number of digits required. It may be that the k=1 value is off a bit because of rounding in excel as well, I don't know. In any case, the second one converges fairly quickly, it seems.

I am amazed at the interesting ways people have found to calculate pi. One of my favorites is Buffon's Needles. The mathematics aren't complicated in any way, but it's a very simple physical method that uses probability. In short, if you drop N needles of length L onto a flat surface with lines evenly spaced at 2L such that they land randomly then count the number of needles, n, that cross a line, N/n = pi as N goes to infinity.

This was recently done by Numberphile (an excellent, fun math channel) on youtube, if anyone is interested; they explain the math as well.

Molinaro · Apr 19, 2012

I remember being taught that method in school. We did it by dropping short pencils on a hardwood floor.

I don't think the 1st equation can be calculated for n=2 without writing a program that handles very large numbers as strings or arrays of digits, with a math library written to work with them. The term (4n-1)!! for n=2 is 7!! Which is 5040!

I found an online factorial calculator that works for numbers up to 5000! which is equal to:

To get 5040! involves multiplying the above number by a roughly 150 digit number.

So.... ya, maybe I'll make this a programming project and see if I can work out the term for n=2 and see just how many digits of pi it gives correctly.

I Ratant · Apr 19, 2012

There's only one digit in pi!
It is a three! (3).
Ain't nothing on the other side of that!
I read that in a book, about a king and a sea.

Vorticity · Apr 19, 2012

Molinaro said:
I don't think the 1st equation can be calculated for n=2 without writing a program that handles very large numbers as strings or arrays of digits, with a math library written to work with them. The term (4n-1)!! for n=2 is 7!! Which is 5040!

That's not what !! means.

7!! = 1*3*5*7 = 105

Vorticity · Apr 19, 2012

Molinaro said:
[latex]$$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_ {k=0}^{\infty} \frac{(4k)!(1103+26390k)} {(k!)^{4}(396^{4k})} $$[/latex]

I've taken the reciprocal of this expression using arbitrary-precision arithmetic in Mathematica to get estimates of pi to 30 or 40 decimal places, for various upper limits kmax on the sum:

kmax = 0: pi ~ 3.14159273001330566031399618903 (6 good digits)
kmax = 1: pi ~ 3.14159265358979387799890582631 (15 good digits)
kmax = 2: pi ~ 3.14159265358979323846264906570 (23 good digits)
kmax = 3: pi ~ 3.141592653589793238462643383279555273160 (31 good digits)

For reference, pi = 3.141592653589793238462643383279502884197...

W.D.Clinger · Apr 19, 2012

Vorticity said:
That's not what !! means.

7!! = 1*3*5*7 = 105

Ah, thanks.

Molinaro said:
I wonder how fast they converge on the digits of pi. Is one faster than the other? Or how many terms of the sum can be calculated before the factorials require the use of a big numbers math library.

They appear to converge at almost exactly the same rate. With either formula, here are the first 75 digits of the result when the result is truncated at 1 term, 2 terms, et cetera:

Code:

3.141592730013305660313996189025215518599581607110033559656536290128551455441
3.141592653589793877998905826306013094216645029322848879173963791505784400649
3.141592653589793238462649065702758898156677480462334781168399595644739794559
3.141592653589793238462643383279555273159974210420379911216703896006945787943
3.141592653589793238462643383279502884197663818133030623976165590998553105508
3.141592653589793238462643383279502884197169399379846832743512507281953256921
3.141592653589793238462643383279502884197169399375105821020933242469729074078
3.141592653589793238462643383279502884197169399375105820974944592757814327091
3.141592653589793238462643383279502884197169399375105820974944592307816410719
3.141592653589793238462643383279502884197169399375105820974944592307816406286
3.141592653589793238462643383279502884197169399375105820974944592307816406286

Ladewig · Apr 19, 2012

Vorticity said:
That's not what !! means.

7!! = 1*3*5*7 = 105

OK, I believe you; but Molinaro should still get acknowledgement for best use of spoiler tags in a math thread.

NutCracker · Apr 19, 2012

I Ratant said:
There's only one digit in pi!
It is a three! (3).
Ain't nothing on the other side of that!
I read that in a book, about a king and a sea.

Rumour has it that the author of said book found 1/pi = 3 * sigma(k=1,infinity, 1/10^k) rather silly, so he changed pi to something a little more exciting.

Madalch · Apr 19, 2012

That gives me an idea for a question for a math exam...."Determine the last digit of 88!"

TubbaBlubba · Apr 19, 2012

Madalch said:
That gives me an idea for a question for a math exam...."Determine the last digit of 88!"

On second thought, I'd better not- half the students would answer, "IT'S AN 8, OF COURSE!!"

1. There is no last digit.
2. Were there a last digit, it ought to be either 0 or 9, depending on what decimal expansions you choose.

Meridian · Apr 19, 2012

Molinaro said:
[latex]$$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_ {k=0}^{\infty} \frac{(4k)!(1103+26390k)} {(k!)^{4}(396^{4k})} $$[/latex]

It's actually pretty easy to work out roughly how quickly an expression like this converges (but not what it converges to!) using Stirling's formula: [latex]$n!\sim \sqrt{2\pi n}(n/e)^n$ [/latex] as [latex]$n\to\infty$[/latex].
The point is that the terms are getting smaller quickly, so the difference between the partial sum and the limit is essentially the next term.

Going for a rough calculation, ignore polynomial factors in (constant powers of) [latex]$k$[/latex] and approximate [latex]$k!$[/latex] by [latex]$(k/e)^k$[/latex]. The factorial part [latex]$(4k)!/k!^4$[/latex] is very roughly [latex]$(4k/e)^{4k}/(k/e)^{4k} = (4k)^{4k}/k^{4k} = 4^{4k}$[/latex], so the kth term is (up to polynomial factors) [latex]$(4/396)^{4k}=1/99^{4k}$[/latex] and you can pretty much expect 8 extra digits per term.

Incidentally, this phenomenon that the [latex]$e^n$[/latex] parts cancel out is very common, as is that the [latex]$k^k$[/latex] bits do (leaving the parts like [latex]$4^k$[/latex] as here).

You can do the same thing for the other one, after writing [latex]$(2n-1)!!=(2n-1)\times (2n-3)\times\cdots\times 3\times 1 = (2n)!/(n!2^n)$[/latex], or using the direct approximation [latex]$(2n/e)^n$[/latex].

Madalch · Apr 20, 2012

TubbaBlubba said:
1. There is no last digit.
2. Were there a last digit, it ought to be either 0 or 9, depending on what decimal expansions you choose.

Why would you say there is no last digit? The product of the first 88 natural numbers will also be a natural number, and it will be zero, because one of those natural numbers will be 10. In fact, there will be eight zeros at the end of the number, because 10*20*30*40*50*60*70*80 = 8!*10^8.

Molinaro · Apr 20, 2012

Vorticity said:
That's not what !! means.

7!! = 1*3*5*7 = 105

Thanks for that. I had not seen it before now.

ctamblyn · Apr 20, 2012

Madalch said:
Why would you say there is no last digit? The product of the first 88 natural numbers will also be a natural number, and it will be zero, because one of those natural numbers will be 10. In fact, there will be eight zeros at the end of the number, because 10*20*30*40*50*60*70*80 = 8!*10^8.

In the product 88!, consider the factors which are multiples of 5:

5 * 10 * 15 * 20 * 25 * 30 * 35 * 40 * 45 * 50 * 55 * 60 * 65 * 70 * 75 * 80 * 85

Keeping just the powers of 5 in each factor, you get:

5 * 5 * 5 * 5 * (5^2) * 5 * 5 * 5 * 5 * (5^2) * 5 * 5 * 5 * 5 * (5^2) * 5 * 5 = 5^20

Each factor of 5 can be paired with a factor of 2 from somewhere else in the original product, and you get a factor of 10^20. So, you should have 20 zeroes at the end (unless I made an error, which is not unlikely!)

Molinaro · Apr 20, 2012

Madalch said:
Why would you say there is no last digit? The product of the first 88 natural numbers will also be a natural number, and it will be zero, because one of those natural numbers will be 10. In fact, there will be eight zeros at the end of the number, because 10*20*30*40*50*60*70*80 = 8!*10^8.

Actually, there's quite a few more than 8 zeroes at the end:

88! = 185482642257398439114796845645546284380220968949399346684421580986889562184028199319100141244804501828416633516851200000000000000000000

ctamblyn said:
Each factor of 5 can be paired with a factor of 2 from somewhere else in the original product, and you get a factor of 10^20. So, you should have 20 zeroes at the end (unless I made an error, which is not unlikely!)

Correct!

ctamblyn · Apr 20, 2012

Molinaro said:
Correct!

*pats self on back*

Madalch · Apr 20, 2012

ctamblyn said:
*pats self on back*

Adds a pat from me, too.

ctamblyn · Apr 20, 2012

Madalch said:
Adds a pat from me, too.

Received with gratitude

TubbaBlubba · Apr 20, 2012

Madalch said:
Why would you say there is no last digit? The product of the first 88 natural numbers will also be a natural number, and it will be zero, because one of those natural numbers will be 10. In fact, there will be eight zeros at the end of the number, because 10*20*30*40*50*60*70*80 = 8!*10^8.

Oh jesus... That's not cool, man.

Ramanujan's beastly expressions for pi

Illuminator

Philanthropic Misanthrope

Illuminator

Penultimate Amazing

Fluid Mechanic

Fluid Mechanic

Philosopher

I lost an avatar bet.

Muse

The Jester

Knave of the Dudes

Critical Thinker

The Jester

Illuminator

Data Ghost

Illuminator

Data Ghost

The Jester

Data Ghost

Knave of the Dudes