On Mon, Dec 22, 2008 at 3:37 PM, John Cremona wrote:
>
> 2008/12/22 William Stein :
>>
>> On Mon, Dec 22, 2008 at 1:51 PM, M. Yurko wrote:
>>>
>>> First, about my issues with PARI's precision. I just tested the
>>> following:
>>> -pari(-10).eint1().n(150)
>>> and got 2492.22897624187775411641609
2008/12/22 William Stein :
>
> On Mon, Dec 22, 2008 at 1:51 PM, M. Yurko wrote:
>>
>> First, about my issues with PARI's precision. I just tested the
>> following:
>> -pari(-10).eint1().n(150)
>> and got 2492.2289762418777541164160993503173813223839 which is
>> inaccurate after the 14th decimal p
On Mon, Dec 22, 2008 at 1:51 PM, M. Yurko wrote:
>
> First, about my issues with PARI's precision. I just tested the
> following:
> -pari(-10).eint1().n(150)
> and got 2492.2289762418777541164160993503173813223839 which is
> inaccurate after the 14th decimal place. As Stein said, Its quite
> like
First, about my issues with PARI's precision. I just tested the
following:
-pari(-10).eint1().n(150)
and got 2492.2289762418777541164160993503173813223839 which is
inaccurate after the 14th decimal place. As Stein said, Its quite
likely that I didn't call PARI correctly, as this is the first time
On Dec 22, 2008, at 8:28 AM, M. Yurko wrote:
> When I timed these functions over 1 through 700 at 53 bits of
> precision, the python one took 5.46 sec., the double precision cython
> one took only .02 sec., and the arbitrary precision one took 6.77 sec.
> After looking at the .c file that cython
2008/12/22 William Stein :
>
> On Mon, Dec 22, 2008 at 9:57 AM, John Cremona wrote:
>>
>> 2008/12/22 M. Yurko :
>>>
>>> It is the exponential integral, but I want greater than double
>>> precision. I tried PARI, and it reports the number in arbitrary
>>> precision, but it is only seems to be accu
On Mon, Dec 22, 2008 at 9:57 AM, John Cremona wrote:
>
> 2008/12/22 M. Yurko :
>>
>> It is the exponential integral, but I want greater than double
>> precision. I tried PARI, and it reports the number in arbitrary
>> precision, but it is only seems to be accurate to double precision as
>> all di
2008/12/22 M. Yurko :
>
> It is the exponential integral, but I want greater than double
> precision. I tried PARI, and it reports the number in arbitrary
> precision, but it is only seems to be accurate to double precision as
> all digits after are completely wrong. Also, I'm trying to compare a
It is the exponential integral, but I want greater than double
precision. I tried PARI, and it reports the number in arbitrary
precision, but it is only seems to be accurate to double precision as
all digits after are completely wrong. Also, I'm trying to compare a
few a acceleration ideas for the
Thanks a lot, both for the explanation and the solution! globals() did
the trick for me.
Georg.
On Dec 22, 4:19 pm, "Mike Hansen" wrote:
> On Mon, Dec 22, 2008 at 6:12 AM, Timothy Clemans
>
> wrote:
>
> > sage: sage0("var('a b c')")
> > (a, b, c)
> > sage: sage0("a = b/c")
>
> > b
>
> That's n
On Mon, Dec 22, 2008 at 8:40 AM, John Cremona wrote:
>
> That looks very like the exponential integral you are computing. If
> so, you can use Sage's function Ei() which calls scipy's
> special.exp1().
Watch out, since scipy is double precision only.
Pari has a real-only exponential integral t
That looks very like the exponential integral you are computing. If
so, you can use Sage's function Ei() which calls scipy's
special.exp1().
John Cremona
2008/12/22 M. Yurko :
>
> Alright, below is the original python implementation of the function:
>
> def python(x,bits):
>i = 1
>sum_c
On Sun, Dec 21, 2008 at 6:26 PM, evans wrote:
>
> All these are identical (unless there are typos)...
>
> x^2 + ((2-y)/3)x + 3 = 0
>
> x^2 + (2/3-y/3)x + 3 = 0
>
> x^2 + (\frac{2}{3}-\frac{y}{3}) x + 3 = 0
>
> x^2 + \frac{(2-y)}{3}x + 3 = 0
>
> x^2 + \frac{(2-y)}{3} * x + 3 = 0
>
> 3 + x^2 + \fra
Alright, below is the original python implementation of the function:
def python(x,bits):
i = 1
sum_current = RealNumber(x,min_prec=bits)
sum_last = 0
term = sum_current
add_term = (ln(x)+euler_gamma).n(bits)
while sum_current != sum_last:
i+=1
term = term*
Hello,
On Mon, Dec 22, 2008 at 6:10 AM, M. Yurko wrote:
>
> Thanks for your help. I tried your first and last suggestions, but
> they yielded code that was slower than the original python
> implementation. However, I'll take a look at sage.rings.real_mpfr and
> try to use mpfr directly.
Well, I
On Mon, Dec 22, 2008 at 6:12 AM, Timothy Clemans
wrote:
>
> sage: sage0("var('a b c')")
> (a, b, c)
> sage: sage0("a = b/c")
>
> b
That's not quite right as it creates 'a' in a different session.
There are a few variations depending on exactly what you want to do.
If you want to assign something
sage: sage0("var('a b c')")
(a, b, c)
sage: sage0("a = b/c")
b
-
c
On Mon, Dec 22, 2008 at 9:03 AM, Georg Muntingh
wrote:
>
> A quick question: If one is given a variable name as a string s, how
> can one assign a sy
Thanks for your help. I tried your first and last suggestions, but
they yielded code that was slower than the original python
implementation. However, I'll take a look at sage.rings.real_mpfr and
try to use mpfr directly.
On Dec 22, 1:44 am, "William Stein" wrote:
> On Sun, Dec 21, 2008 at 1:44
A quick question: If one is given a variable name as a string s, how
can one assign a symbolic expression to it? Here is what I try that
doesn't work:
var('a b c')
s = "a"
eval(s + " = b/c")
Traceback (most recent call last):
...
a = b/c
^
SyntaxError: invalid syntax
--~--~--
On Dec 22, 3:26 am, evans wrote:
> What I need is something that could take any two of these equations
> and tell me if they are equal or not. ...
One thing i can think of is to store them in expressions, the left
hand sides, and take differences. Then a test if it is zero. (or store
the entire
All these are identical (unless there are typos)...
x^2 + ((2-y)/3)x + 3 = 0
x^2 + (2/3-y/3)x + 3 = 0
x^2 + (\frac{2}{3}-\frac{y}{3}) x + 3 = 0
x^2 + \frac{(2-y)}{3}x + 3 = 0
x^2 + \frac{(2-y)}{3} * x + 3 = 0
3 + x^2 + \frac{(2-y)}{3}x + 5 = 2
What I need is something that could take any tw
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