[SciPy-user] Can SciPy compute ln(640320**3 + 744)/163**.5 to 30 places?
rdmoores at gmail.com
Mon Jan 15 02:35:31 CST 2007
On 1/15/07, Robert Kern <robert.kern at gmail.com> wrote:
> Dick Moores wrote:
> > Well if you mean for me to do:
> > =============================
> > # clnumTest1-b.py
> > import clnum as n
> > n.set_default_precision(50)
> > print n.log(640320**3 + 744)/n.sqrt(163)
> > ==================================
> > That gives me 3.1415926535897932385 .
> > So I STILL don't understand.
> It looks like the __str__ representation of clnum numbers (i.e. what you get by
> printing them, or by calling str() on them) is restricted to that many digits
> regardless of what the actual precision of the object is. Look at the example at
> the bottom of clnum's webpage:
> However, looking at the __repr__ representation (i.e. what was printed in the
> Out in the above example, and what can be obtained by calling repr() on the
> object), you can see all of the digits that are available according to the
> precision of the object.
OK! I tried
import clnum as n
print repr(n.log(640320**3 + 744)/n.sqrt(163))
Great! Thanks to you all!
BTW what's the 55 at the end? There are 58 digits including the initial 3.
> > Also, what are those In : things all you people use? Is that line
> > 16? If so, line 16 of what?
> It's just the IPython input prompt, like the >>> of the regular interpreter.
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