[SciPy-user] Can SciPy compute ln(640320**3 + 744)/163**.5 to 30 places?
fperez.net at gmail.com
Sun Jan 21 23:22:14 CST 2007
On 1/21/07, Dick Moores <rdm at rcblue.com> wrote:
> OK, I'll reclarify. Of course, I'm trying to learn clnum. I thought
> that was understood. Please show me how to use clnum to compute
> (5/23)**(2/7) (that's 5/23 to the 2/7 power), with a precision of
> 50. Surely the answer is not
> Just by inspection this is clear, is it not? A number between 0 and 1
> raised to a power between 0 and 1 will get closer to 1, right? (5/23
> is 0.217391304348) I believe you've given me (5/23.)*(2/7.).
Sorry, my bad: one misplaced parenthesis and me being terribly sloppy
in not actually checking the numerical value (I just hastily saw
0..6.. and didn't look further).
In : import clnum as n
In : n.set_default_precision(50)
This is what I repeated to you like a mindless idiot:
In : n.exp(n.log(n.mpq('5/23')*n.mpq('2/7')))
This is the correct form (note position of parenthesis):
In : n.exp(n.log(n.mpq('5/23'))*n.mpq('2/7'))
Again, my apology for the confusion, it was entirely my fault.
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