13 thoughts on “Ramanujan pi approximation

  1. The first factor (2 Sqrt[2] + Sqrt[10]) inside the logarithm is equal to Sqrt[2] Phi^3 where Phi = (1+Sqrt[5])^/2 is the Golden Ratio.

  2. The formula itself is probably inspired from the similar formula for the Ramanjuan Constant:

    Exp[Pi Sqrt[163]] ~ 640320^3 + 744

    because there is a similar “identity”

    Exp[Pi Sqrt[190]] ~ (Sqrt[2] Phi^3 (3 + Sqrt[10]))^12 + 24

    Dropping the 24, taking logarithms and “solving” for Pi, gives your posted approximation.

    The algebraic foundations (modular functions and all that) for this kind of magic is explained in http://www.oocities.org/titus_piezas/Ramanujan_a.pdf

  3. Unfortunately, I cannot locate a digital version of Ramanujan’s original 1914 paper: Ramanujan, S. “Modular Equations and Approximations to Pi.” Quart. J. Pure Appl. Math. 45, 350-372, 1913-1914.

    The Göttinger Digitalisierungszentrum has only caught up to 1900: http://gdz.sub.uni-goettingen.de/dms/load/toc/?IDDOC=646353, so you might need a goold old-fashioned library if you want to go back to the original source.

  4. I’ve read the paper you mentioned, but it was hardly clear from the paper where the approximation came from other than “it has something to do with modular functions.” The paper you linked to first seems more promising.

  5. I think he was having fun. The operand of the logarithm can be rewritten as (sqrt(8)+sqrt(10))*(sqrt(9)+sqrt(10)) . People who like to play with numbers like to be able to do something creative with numbers in succession, with patterns. As for myself, it’ll probably be a long time before I forget this formula…

    pi = (3*4 / sqrt (190)) * log ((sqrt(8)+sqrt(10))*(sqrt(9)+sqrt(10)))

  6. I’m sorry but as far as I can see, this equation is inaccurate. If I’m missing something or if there is a trick to this, please, please let me know.

  7. Chloe: Maybe you’re trying to take logarithms base 10. Logs on this site are always natural logs.

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