On 3 Set, 23:40, Marshall Hampton hampto...@gmail.com wrote:
Here's a slightly different workaround that doesn't require any lambda
expressions:
var('z,z0,x,y')
f=abs(1/(z-z0)).subs({z:x+I*y,z0:1/2+i})
ff = fast_callable(f,vars=[x,y],domain=CDF)
plot3d(ff,(x,-2,2),(y,-2,2))
-Marshall
Hi,
On Sat, Sep 4, 2010 at 7:01 PM, sps debernasave...@libero.it wrote:
But what is fast_callable function ? Where can I find adaguate
documentation?
See this page in the reference manual
What about to use polynomial division to get polynomial q=Q/p and then
return P/q ?
I do not remember the command for polynomial division, but should be
in the manual or help.
Robert
On 4 zář, 05:54, Cary Cherng cche...@gmail.com wrote:
I have a rational function P(x)/Q(x) with numerators and
How do u stick a matrix on the bottom of antoher martix, in particular
the identity matrix
so if had M (l,k) how do I stick Id(k) on the bottom to produce a new
matrix
N=M and N is dimension (l+k,k)
Id
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On Sat, Sep 4, 2010 at 8:15 AM, andrew ewart aewartma...@googlemail.com wrote:
How do u stick a matrix on the bottom of antoher martix, in particular
the identity matrix
so if had M (l,k) how do I stick Id(k) on the bottom to produce a new
matrix
N=M and N is dimension (l+k,k)
Id
Look
On Sat, Sep 4, 2010 at 3:25 AM, ma...@mendelu.cz ma...@mendelu.cz wrote:
What about to use polynomial division to get polynomial q=Q/p and then
return P/q ?
I do not remember the command for polynomial division, but should be
in the manual or help.
If Q and p are polynomials, polynomial
if a have a matrix M of dimension (m,n)
how do i first check for j columns deep (from left), j=n
for when the ith row will be all zeros
eg
M=([1,0,1],[0,0,1],[1,1,1])
check j=1 gives output of 1
j=2 gives output of 1 and j=3 gives output of 3
also how do i take out a matrix N of dimension (k,l)
I should have given the original full context. These polynomials P,Q,
and p are all in Z[x1,...,xn]. They are all multivariate.
On Sep 3, 8:54 pm, Cary Cherng cche...@gmail.com wrote:
I have a rational function P(x)/Q(x) with numerators and denominators
of very large degree. From the context I
HI Jason, I tried what you posted and get some errors , I am using
the latest Sage version 4.5.2
--
| Sage Version 4.5.2, Release Date: 2010-08-05 |
| Type notebook() for the GUI, and license() for
And thats another problem. How do I tell sage to give me the
denominator of this rational function?
On Sep 4, 3:25 am, ma...@mendelu.cz ma...@mendelu.cz wrote:
What about to use polynomial division to get polynomial q=Q/p and then
return P/q ?
I do not remember the command for polynomial
On Sep 4, 2010, at 13:54 , Cary Cherng wrote:
And thats another problem. How do I tell sage to give me the
denominator of this rational function?
In general, the .denominator and .numerator methods will (should) give
you the components of a rational element.
HTH
Justin
--
Justin C.
On 9/4/10 12:56 PM, andrew ewart wrote:
if a have a matrix M of dimension (m,n)
how do i first check for j columns deep (from left), j=n
for when the ith row will be all zeros
eg
M=([1,0,1],[0,0,1],[1,1,1])
check j=1 gives output of 1
j=2 gives output of 1 and j=3 gives output of 3
I realize
.denominator() gave me 1. Sage prints out the rational function as R1
+ .. + R2 where Ri is a rational function and each having a different
denominator. Sage doesn't seem to be automatically writing this as one
fraction with a common denominator, could that be the reason it is
returning 1 for the
Hi!
On 4 Sep., 22:52, tvn nguyenthanh...@gmail.com wrote:
HI Jason, I tried what you posted and get some errors , I am using
the latest Sage version 4.5.2
--
| Sage Version 4.5.2, Release Date: 2010-08-05
Ok i think I've resolved my problems by avoiding var for declaring
variables and instead using
R.g17,g19,g27,g29,g37,g38,g47,g48,g58,g59,g68,g69 =
PolynomialRing(QQ)
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On Sat, Sep 4, 2010 at 7:00 PM, Cary Cherng cche...@gmail.com wrote:
Ok i think I've resolved my problems by avoiding var for declaring
variables and instead using
R.g17,g19,g27,g29,g37,g38,g47,g48,g58,g59,g68,g69 =
PolynomialRing(QQ)
Yes, that should be orders of magnitude faster than doing
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