Timeline for Compute the first N digits of e
Current License: CC BY-SA 2.5
11 events
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Mar 13, 2011 at 20:33 | comment | added | Eelvex | @Jesse: Use 450x then. It's enough for every n<1000 and takes just a few seconds (~9s here). | |
Mar 13, 2011 at 19:46 | comment | added | Jesse Millikan |
@Eelvex: This brings up a point... This performs horribly for larger values of n because of the resulting denominator, which is (for example) */!i.999x if it's not reduced. I wasn't even patient enough to let +/%!i.999x finish. Hints here, possibly...
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Mar 13, 2011 at 18:20 | comment | added | Eelvex |
Ah, you're right, that's unfortunate. (That "mess" is shorter than anything else I could come up with - unless, off course, you use a fixed number for i. like: (0 j.<:n)":+/%!i.999x ; then tacitly:(+/%!i.999x)":~0 j.<: )
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Mar 13, 2011 at 18:04 | comment | added | Jesse Millikan | @Eelvex: But I need extra terms up until N=30 or so. (Any hints on reducing that mess of composes, or is that about right?) | |
Mar 13, 2011 at 17:32 | history | edited | Jesse Millikan | CC BY-SA 2.5 |
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Mar 13, 2011 at 14:49 | comment | added | aaaaaaaaaaaa | Would you care to explain your code? | |
Mar 13, 2011 at 10:09 | comment | added | Eelvex |
For 1000 digits this: (0 j.<:1000)":+/%!i.x:450 takes just a few seconds.
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Mar 13, 2011 at 10:05 | comment | added | Eelvex |
This converges very fast (for 100 digits you need "just" 70 terms) so you can drop that +9 .
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Mar 13, 2011 at 6:45 | history | edited | Jesse Millikan | CC BY-SA 2.5 |
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Mar 13, 2011 at 5:52 | history | edited | Jesse Millikan | CC BY-SA 2.5 |
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Mar 13, 2011 at 4:18 | history | answered | Jesse Millikan | CC BY-SA 2.5 |