[SciPy-user] Can SciPy compute ln(640320**3 + 744)/163**.5 to 30 places?

Dick Moores rdmoores at gmail.com
Mon Jan 15 01:06:03 CST 2007


Please show me how to do that in just plain python. I don't have or
know ipython.

Thanks

On 1/14/07, Bryan Van de Ven <bryanv at enthought.com> wrote:
> I'm going to go out on a limb here and say that when you do that print,
> the mpf object is first converted to a regular python float, which does
> not have enough precision, of course.
>
> Converting the mpf object to a clnum rational mpq object in ipython:
>
> In [17]: a = cl.log(640320**3 + 744)/cl.sqrt(163)
>
> In [18]: cl.mpq(a)
> Out[18]: mpq(181338285223186289202345004581,57721768930156197489438103340)
>
> And then just as check, evaluating this fraction to 50 digits in
> Mathematica:
>
> In[6]:= N[ 181338285223186289202345004581/57721768930156197489438103340,50]
>
> Out[6]= 3.1415926535897932384626433832797266193475498808835
>
> Seems to match up...
>
>
> Dick Moores wrote:
> > On 1/13/07, Fernando Perez <fperez.net at gmail.com> wrote:
> >
> >> In [16]: import clnum as n
> >>
> >> In [17]: n.set_default_precision(50)
> >>
> >> In [18]: n.log(640320**3 + 744)/n.sqrt(163)
> >> Out[18]: mpf('3.141592653589793238462643383279726619347549880883522422293',55)
> >>
> >
> > I assumed you did that in Python. So I tried it with this script,
> > adding a print statement:
> > ===================================
> > # clnumTest1-a.py
> >
> > import clnum as n
> >
> > n.set_default_precision(50)
> >
> > print "%.50f" % (n.log(640320**3 + 744)/n.sqrt(163))
> > ======================================
> > result: 3.141592653589793100000000000000
> >
> > So clarify, if you would, please.
> >
> > Thanks,
> >
> > Dick Moores
> > _______________________________________________
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> > SciPy-user at scipy.org
> > http://projects.scipy.org/mailman/listinfo/scipy-user
> >
>
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