Difference between revisions of "Ramanujan, S."
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I tried calculating Ramanujan's series (as suggested by [[Crow 1999]]) to estimate 2/π in a spreadsheet. | I tried calculating Ramanujan's series (as suggested by [[Crow 1999]]) to estimate 2/π in a spreadsheet. | ||
| − | <math>1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1\times3}{2\times4}\right)^3-13\left(\frac{1\times3\times5}{2\times4\times6}\right)^3+\ | + | <math>1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1\times3}{2\times4}\right)^3-13\left(\frac{1\times3\times5}{2\times4\times6}\right)^3+\cdots = \frac{2}{\pi}</math> |
The products quickly became so large the spreadsheet returned an error after the first 151 terms. | The products quickly became so large the spreadsheet returned an error after the first 151 terms. | ||
Revision as of 07:02, 4 September 2018
https://en.wikipedia.org/wiki/Srinivasa_Ramanujan
I tried calculating Ramanujan's series (as suggested by Crow 1999) to estimate 2/π in a spreadsheet.
[math]1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1\times3}{2\times4}\right)^3-13\left(\frac{1\times3\times5}{2\times4\times6}\right)^3+\cdots = \frac{2}{\pi}[/math]
The products quickly became so large the spreadsheet returned an error after the first 151 terms.
