def test():
for b in range(5,6):
for c in range(b+1,b+2):
print(b,c,n(c/b))
print(5,6,n(6/5))
And the output
sage: test()
(5, 6, 1.00)
(5, 6, 1.20)
sage: version()
'SageMath version 8.0, Release Date: 2017-07-21'
And the question: why are the two lines not identical
def demo():
var('N,x')
test = ((N*(3*I*sqrt(3) + 9) + N*(3*I*sqrt(3) - 3)))*x
print("test = ")
print(test)
print("test.full_simplify() = ")
print(test.full_simplify())
Here is the output
sage: demo()
test =
(N*(3*I*sqrt(3) + 9) + N*(3*I*sqrt(3) - 3))*x
test.full_simplify() =
(36*I*sqrt(3)*N
def nov13():
var('M,N,z')
f = (M^2-3*N)*(-i *sqrt(3)-1) *z^3
f = f + (M^2 *(-i *sqrt(3) +3) + 3*N*(-i *sqrt (3) - 1))*z^2
f = f + (M^2 *(i *sqrt(3)+3) + 3*N* (i* sqrt(3)-1))*z + (M^2-3*N)* (i*
sqrt(3)-1)
g = f.substitute(M=6,N=11)
complex_plot(g, (-3, 3), (-3, 3))
if this code is put in a file
the drawing command inside a loop and draw a lot of graphs, I'll need to
do it with "show"
rather than "return". Thank you.
Michael
On Tuesday, November 13, 2018 at 8:38:52 PM UTC-8, Michael Beeson wrote:
>
> def nov13():
> var('M,N,z')
> f = (M^2-3*
Oops, "243" in my post should have been "k". I don't know how to edit a
post after I've posted it.
On Wednesday, November 14, 2018 at 10:31:34 PM UTC-8, Michael Beeson wrote:
>
> After quite some searching I did not succeed to find documentation for
&
After quite some searching I did not succeed to find documentation for sage
functions to work with complex numbers as much as I would like.
For example if I have a complicated rational expression, how can I tell
Sage "bring this to the form a + bi". It seems real() and imag() only
work
Thank you, that was very instructive to see the "right way" to start by
using an appropriate ring.
I guess you can go on to divide out by the other linear factor and get the
quadratic equation,
and solve it, but I got that far myself, and then could not work with the
solutions of the quadrat
def test(p,q):
t = p/q;
print(p,q,p/q,t)
def test2():
for p in range(1,4):
for q in range(1,4):
test(p,q)
sage: test2()
(1, 1, 1, 1)
(1, 2, 0, 0)
(1, 3, 0, 0)
(2, 1, 2, 2)
(2, 2, 1, 1)
(2, 3, 0, 0)
(3, 1, 3, 3)
(3, 2, 1, 1)
(3, 3, 1, 1)
sage: test(*2*,*3*)
(2, 3, 2/3, 2/3)
sage: versio
I see by inserting print(type(p)) that range produces objects of type int,
while when it's called from top-level the types of p and q are
sage.rings.integer.Integer.
So that answers "why" but not "how to prevent".
On Friday, February 15, 2019 at 2:56:14 PM UT
M UTC-8, Michael Beeson wrote:
>
> def test(p,q):
> t = p/q;
> print(p,q,p/q,t)
> def test2():
> for p in range(1,4):
> for q in range(1,4):
> test(p,q)
>
> sage: test2()
>
> (1, 1, 1, 1)
>
> (1, 2, 0, 0)
>
> (1, 3, 0, 0)
>
> (2, 1, 2, 2)
>
&g
Well, the real cure is to use IntegerRange instead of range.
Sorry to bother those of you who actually can read a manual.
On Friday, February 15, 2019 at 2:56:14 PM UTC-8, Michael Beeson wrote:
>
> def test(p,q):
> t = p/q;
> print(p,q,p/q,t)
> def test2():
> for p in rang
sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x)
[x == -sqrt(-x^2 + 1) + 7/2]
sage: version()
'SageMath version 8.0, Release Date: 2017-07-21'
That doesn't look like a solution to me because x still appears on the
right.
Is this the intended behavior?
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ution.
So this example shows that (a) is sometimes false. And when is (b) true?
On Monday, February 18, 2019 at 12:56:56 PM UTC-8, Michael Beeson wrote:
>
> sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x)
>
> [x == -sqrt(-x^2 + 1) + 7/2]
>
>
> sage: version()
>
> 'SageMa
When I try to reproduce Eric's post, I get an error message about an
unexpected keyword argument
(maybe my version of Sage is too old.) But look at this:
sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x,explicit_solutions=True)
[1/4*I*sqrt(41) + 7/4 == -1/2*sqrt(7/2*I*sqrt(41) + 2), 1/4*I*sqrt(41)
after solving an equation (or not) for x, I can check if the answer still
contains x by ans.has(x).
That should weed out any non-explicit solutions.
But still: am I guaranteed for any class of equations, e.g. polynomial
equations of degree <= 4,
that if solve produces an empty list there re
sage: t
tan(1/2*arctan(12/5))
sage: t.trig_simplify()
sin(1/2*arctan(12/5))/cos(1/2*arctan(12/5))
sage: n(t)
0.667
But trig_simplify couldn't get 2/3. Maybe there
is a fancier command that will do it?
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The solution of a cubic or quartic may require the use of complex numbers.
(Indeed that's how the complex numbers were first discovered.)
Below I exhibit a long expression for such a number that solve() found for
me.
It evaluates using n(t) to a real (decimal) number, and it passes " t in
RR"
Oh, and range(0,n(t)) also crashes.
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In the following example I would like to make Sage realize that (p,q,r)
are constants and (a,b) are variables
so in the end everything should be expressed as a polynomial in a,b. In
particular b^2 should be rewritten as 1-a^2
(b and a are actually sin and cosine of something) but b should no
I appreciate Eric's post, and I do use subs sometimes, but it makes me
nervous since
it will happily substitute any old thing you tell it to, even an incorrect
thing. So, if your idea
is to check a computation, it is a dangerous thing. True, if you put only
correct equations in,
you'll
SageMath has built-in functions to compute the regulator and the
fundamental unit of a quadratic field. The regulator of a quadratic field
is the log of the absolute value of the fundamental unit.
So, the following code should print out the same number on each line.
But, as you can check, it
sage: n(log(*2*,*408*/*370*))
7.08999206263157
sage: log(*2*,*2*)
1
sage: *2*^*7*
128
sage: n(log(*2*,*408.0*/*370*))
7.08999206263157
sage: version()
'SageMath version 8.7, Release Date: 2019-03-23'
sage: n(log(*2*,*1.001*))
693.493696416899
sage:log(2,1) gives an error instead of 0
And also
sage: log(*2*,x)
log(2)/log(x)
which goes a long ways to explain the previous results. So log(2,x) is not
actually log to the base 2 of x.
On Saturday, July 18, 2020 at 11:45:06 AM UTC-7, Michael Beeson wrote:
>
> sage: n(log(*2*,*408*/*370*))
>
> 7.08999206263157
&
I am just learning Sage. I tried to define a polynomial and then
find the polynomial remainder upon division by the
cyclotomic_polynomial(18), which is 1-x^3+x^6.This is easily
accomplished in Mathematica using the PolynomialRemainder function.
But I could not find the analog of that function
Michael
On Jan 3, 12:42 pm, bump wrote:
> On Jan 3, 10:59 am, Michael Beeson wrote:
>
> > I am just learning Sage. I tried to define a polynomial and then
> > find the polynomial remainder upon division by the
> > cyclotomic_polynomial(18), which is 1-x^3+x^6. This is
I want to take, for example, x^2 + 1 mod a*x and get quotient (1/a)*x
and remainder 1.It doesn't work if I work in PolynomialRing because
then you can't have 1/a.
It doesn't work in the quotient field because then you always get remainder
0.To have f.quo_rem(g) work, I must a
sage: K. =
FractionField(PolynomialRing(QQ,10,'depgmfuvjN'))
sage: R. = K[]
sage: w=u
sage: u=0
sage: w
u
Why doesn't Sage answer 0 for the value of w here? More generally, if I
have some complicated expression and I assign a value to one of its
variables,
Sage knows the value I've just ass
sage: K. = FractionField(PolynomialRing(QQ,4,'pdeN'))
sage: R. = K[]
sage: a = x^3-x^-3
sage: b = x^5-x^-5
sage: c = x^8-x^-8
sage: X = p*a + d*b + e*c
sage: f = x^16 *(X^2- N*b*c)
and Sage does not answer. It just hangs and I have to kill the session.
If it would answer I would like to continue
If I break the computation into smaller pieces it works OK:
sage: K. = FractionField(PolynomialRing(QQ,4,'pdeN'))
> sage: R. = K[]
> sage: a = x^3-x^-3
> sage: b = x^5-x^-5
> sage: c = x^8-x^-8
> sage: X = p*a +d*b + e*c
> sage: H = R(x^8 * X)
> sage: f = H - N*b*c*x^16
> sage: f
> -N*x^29 + N*x^
oh, never mind, this isn't the same computation as I didn't square X.
On Monday, January 14, 2013 2:54:08 PM UTC-8, Michael Beeson wrote:
>
> If I break the computation into smaller pieces it works OK:
>
>
> sage: K. = FractionField(PolynomialRing(QQ,4,'pdeN'))
So one problem with the original post was that the thing I was trying to
cast to a polynomial isn't a polynomial.
I should have multiplied by x^32, not x^16. The correct input works
correctly (see below). Still, attempting
to cast a rational function with too big a denominator to a polynomi
Sage hangs on the following input:
sage: K. = FractionField(PolynomialRing(QQ,4,'pdeN'))
sage: R. = K[]
sage: a = x-x^-1
sage: b = x^6-x^-6
sage: c = x^7-x^-7
sage: X = p*a + d*b + e*c
sage: F = N*a*b*c - X^2*(x-x^-1)
and also on this closely related input
sage: K. = FractionField(PolynomialRing
K. = FractionField(PolynomialRing(QQ,10,'pdegmfhlrN'))
R. = K[]
x = s + h/(4*l)
G = e*l*x^4 - e*h*x^3 + (-3*e*l-e*r+N)*x^2 + 2*e*h*s + 2*e*l + 2*e*r-N
print G
The response looks fishy:
e*l*s^4 + ((-30423614405477505635920876929024*e*h^2 -
243388915243820045087367015432192*e*l^2 -
8112963841460
Here I attempt to solve Pell's equation with d = 1621 following the method
on page 93 of Stein's book.
But the solution produced is instead a solution of the negative Pell
equation x^2-y^2 = -1 (instead of 1).
Actually, the example on page 93 (after correcting the typo "v" to "u") has
the same
ts will give me the solutions of x^2 - 5 y^2 =
plus or minus 1, not just 1.
I'm sorry for taking your time to point that out.
On Friday, October 31, 2014 9:14:02 AM UTC-7, Michael Beeson wrote:
>
> Here I attempt to solve Pell's equation with d = 1621 following the method
> on
sage: def testPlot():
: u = [[1,2],[3,4],[5,7]]
: list_plot(u)
:
sage: testPlot()
sage:
doesn't produce any plot, but executing the body of the function directly
at the sage prompt does produce a plot.
What am I doing wrong?
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I tried to follow Guninski's instructions. I found eulerprod.py and put
it in the right directory
and then I got an error due to not having psage.ellcurve.lseries.helper.
I googled for that file and put it in helper.py but I still get an error.
Maybe the file
needs to have a complicated path
after declaring variables make these definitions
sage: a = Z - Z^-1
sage: b = L - L^-1
sage: c = Z^2L-Z^-2L^-1
sage: f = (p*a + q*b + r *c) *a + (n*a + m *b + l*c) * a*b
Now I can tell it assume(Z^3 * L^2 == -1)but I can't get it to use
that assumption
in something like simplify(expand(Z^3*
The following sage session shows a call to compute the resultant of
two polynomials that fails, and
another call that seems quite similar in syntax and semantics that
succeeds just fine. What's going on here? It seems to not
realize that f is a polynomial, since it says it's a
FractionFieldEle
another failed attempt to compute a resultant, that doesn't involve
any quotient fields:
sage: R.=QQ[]
sage: F = i*m*z^14 + (d+g)*z^13 + i*(f-p)*z^12 + e*z^11 + i*f*z^10 +
(e-g)*z^9 + i*(m+p)*z^8 + 2*d*z^7 - i*(m+p)*z^6 + (e-g)*z^5 -i*f*z^4 +
e*z^3 - i*(f-p)*z^2 + (d+g)*z - i*m
sage: G = d*z^16 -
"i" is sqrt(-1), which sage seems usually to realize without being
told.
Anyway there is no "i" in my first post on the resultant, and also
I get the same error with "CC" in place of "QQ".
On Jan 31, 3:38 pm, William Stein wrote:
> 2010/1/31 Micha
So taking your suggestion to use a quadratic number field, I get rid
of syntax errors at last. But I guess the problem is too difficult
as no answer comes back in a few minutes.
I let Mathematica run a similar problem for 36 hours with no reply;
but I don't understand why it's too difficult. Se
me of them.
On Jan 31, 6:37 pm, Michael Beeson wrote:
> So taking your suggestion to use a quadratic number field, I get rid
> of syntax errors at last. But I guess the problem is too difficult
> as no answer comes back in a few minutes.
> I let Mathematica run a similar problem for
Oh, gcd is the multivariate gcd, not the gcd as a polynomial in z.
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On the page
http://www.sagemath.org/doc/reference/sage/symbolic/expression.html
the first example has eqn.subs(x==5) and I think it should be
eqn.subs(x=5)
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sage: var('z'); var('a');var('b');
sage: F = a*z + i*b*z^2
sage:
Now F.norm() or something should give me (a*z)^2 + (b*z^2)^2, but I
can't find a command to do that.
I want to do this when F is a polynomial of degree 6 with complex
rational coefficients to eliminate i and produce
a polynomial o
I want to create a vector space of dimension three, over the field
whose elements are symbolic expressions. (The reason is that then I
can do vector calculus on such objects, which represent surfaces in 3-
space if the expressions depend on 2 parameters.) How can I do
this? The following, for
--
| Sage Version 4.2.1, Release Date: 2009-11-14 |
| Type notebook() for the GUI, and license() for information.|
--
WARNING: There
bolic computations, and since I did many symbolic
computations successfully in spite of that message,
I have grown used to ignoring that message.
On Jan 30, 3:03 pm, William Stein wrote:
> On Sun, Jan 30, 2011 at 1:54 PM, Michael Beeso
I downloaded the latest 64-bit dmg of Sage for OS X and installed it.
It works, but I get a strange error message about sage-64.txt.
Specifically, if I put the following lines
in a sage file and run sage with that file for input, (here are the
lines)
L = [[cos(pi*i/100),sin(pi*i/100)] for i in
Here are the contents of a sage file.
var('z')
C = ComplexField()
i = C(0,1)
X = real(integral(z,z))
print X
X = simplify(X)
print X
Here is the corresponding output
1/2*real_part(z)^2 - 1/2*imag_part(z)^2
1/2*z^2
The first line is already surprising as I expected 1/2 *
real_part(z^2). But i
It seems that var('z') makes sage think z is real.
simplify(real(z)) returns z and simplify(imag(z)) returns 0.
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Sage version 4.6.1 (I know it's old, new one is downloading now, but I
don't think this is a version problem.)
Given: polynomial f in x with some letters for the coefficients, and
polynomial psi of lower degree in x with constant coefficients.
Wanted: remainder of f on division by psi as
E = EllipticCurve([sage: E = EllipticCurve([0,40,0,300,0])
sage: E.mwrank()
Then mwrank reports that the rank is 0, which is good, but it also has a
cryptic remark that "points of rank 0 were found". I
independently wrote a short C program that uses the Nagel-Lutz theorem to
verify there a
u quote doesn't
contain that phrase. I'm sorry to have taken so much of your time
with an essentially trivial question--you certainly don't need to reply
with any further explanation. I have understood very clearly what
mwrank is telling me, thanks to your first letter.
Michael Be
I want to compute the regulator of a real quadratic field Q(sqrt d) to
high precision,
accurately enough to compute the fundamental unit. The default
breaks at d = 331 where fundamental unit needs more than 53 bits (the
precision of doubles). The documentation says that Pari computes to a
sage: d = *6*
sage: p = d/*2*
sage: p
3
sage: is_prime(p)
False # Huh?!!
sage: is_prime(*3*)
True
sage: p==*3*
True
This happens in version 8.7 and also in the current version (installed
yesterday)
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The following seems fishy:
``
sage: K. = QuadraticField(-*1*)
sage: K.factor(*13*)
(Fractional ideal (-3*a - 2)) * (Fractional ideal (2*a + 3))
``
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