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Hi community.

I'm interested in manipulating a formal expression

$$\int \mathrm{d}t \; e^{- 2 \int \mathrm{d}t \; h(t)}.$$

My notebook contains the following code

var('t')
h = function('h')(t)
integrate(exp(integrate(-2*h,t)),t)

but the result is +Infinity.

QUESTION(S):

• Why I am getting a result if the function h is not explicit?
• Where does the +Infinity come from?

Thanks to the comment by @fredericc I found the answer! The solution is to use the algorithm='maxima' flag, e.g. var('t

Integrating an integral Hi community. I'm interested in manipulating a formal expression $$\int \mathrm{d}t \; e^{-

Dear all,

A long time ago I was trying to implement a SAGE code for working with Differential Forms with values in a certain Lie algebra, but due to my lack of programming knowledge, I couldn't.

This kind of objects are important for working with non-Abelin gauge theories.

Question

Is it possible to define and work with those objects?

So far there is no reference of it in the manual.

Thank you!

dear all:

I'd like to know if there is any way to change the size of the font of labels (in a plot) without changing the size of the numbers on the ticks.

Thank you

Thank you @dsejas Yes, your comment was incredible useful! It drove me to a nice solution. Since I had a lot of assumptions, I was trying to find a way to drop the last assumption (still don't know it possible), and looking for that I found the with assuming(condition): code statement, which execute the code assuming the condition without registering in the assumptions() list! Thank you very much!

Hi. I'm solving a differential equation which depends on two parameters $c1$ and $c2$. They are both positive

assume(c1 > 0)
assume(c2 > 0)

but in order to solve the equation, maxima ask me whether c2-3 is positive, negative or zero. With the condition

assume(c2 - 3 > 0)

the equation is solved; But imposing c2 - 3 < 0 triggers and inconsistency... I believe that it is due to the initial assumption.

Question

Is it possible to set a condition like assume(0 < c2 < 3)?

I'm interested in solving the differential equation $$3 h' + 3 h^2 = c_1,$$ where $c_1$ is a positive real number.

var('t')
var('c1', latex_name=r'c_1')
h = function('h')(t)
eq = -3*h^2 + c1 - 3*diff(h, t)
eq_sol = desolve(eq, h, ivar=t, contrib_ode=True)

The above code works, but it's not solved explicitly for $h$, so

h_sol = solve(eq_sol, h)
h_sol = h_sol[0]
h_sol

This gives something like $$h\left(t\right) = \frac{\sqrt{3} \sqrt{c_{1}} {\left(e^{\left(\frac{2}{3} \, \sqrt{3} C \sqrt{c_{1}} + \frac{2}{3} \, \sqrt{3} \sqrt{c_{1}} t\right)} + 1\right)}}{3 \, {\left(e^{\left(\frac{2}{3} \, \sqrt{3} C \sqrt{c_{1}} + \frac{2}{3} \, \sqrt{3} \sqrt{c_{1}} t\right)} - 1\right)}},$$

in sage notation (non-LaTeX) it starts like

h(t) == 1/3*sqrt(3)*sqrt(c1)* ...

Question 1: Is there a way to allocate to the solution (i.e. h_sol) the RHS of the above? without the h(t) == part.

I had to set by hand (it is ease, but it would be nice to automatize the allocation)

var('C')    # the integration constant introduced above
h_sol = 1/3*sqrt(3)*sqrt(c1)* ...

Then, by simply looking at the solution it is clear that it can be simplified. I tried things like

h_sol = h_sol.canonicalize_radical()
h_sol = h_sol.collect_common_factors()
h_sol = h_sol.simplify_rectform(complexity_measure = None)

but none of them returns the expected result, which could be obtained from Mathematica's kernel

mathematica("DSolve[3*h'[t] + 3*h[t]^2 == C[1], h[t], t]//FullSimplify")

$$ \sqrt{\frac{c_1}{3}} \tanh\left( \sqrt{\frac{c_1}{3}} (t - 3 c_2) \right) $$

Question 2: How could the expression h_sol be manipulated to obtain the hyperbolic tangent?

Dear community.

This might sound dump, but I'm trying to determine whether a tensor satisfy the properties of a metric (under certain conditions). Of course it is a (0,2)-symmetric tensor, call it $S$, but I cannot (to my understanding) calculate the (Levi-Civita-like) connection components that would be associated to $S$... unless I declare it as a metric.

The way it is implemented makes sense... and it's solid!

What I did...?

I defined like a metric and calculate the associated connection (and curvatures)

Why should I do something else?

In the file src/sage/manifolds/differentiable/metric.py the metric is defined (as it should) to be symmetric, but it does not allow to consider extensions of General Relativity like say Einstein--Strauss model. Thus, I need an instance to calculate the derived quantities of a tensor that is a generalization of a metric.

Question:

Is this possible?

Thanks for the theory!

hehehehe... Thank you again! I almost got it... but I tried .expression().

Dear community,

I have a differential equation that depends on a function $\xi(t)$, but is a component of a tensor (calculated with sagemanifold)

M = Manifold(4, 'M', latex_name=r"\mathcal{M}")
U.<t,r,th,ph> = M.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')
xi = function('xi')(t)
f = function('f')(t)
h = function('h')(t)
g = function('g')(t)
Ric = M.tensor_field(0,2, 'Ric')
Ric[0,0] = 3/2*f*h - 3/4*h^2 + 3/2*f*xi + 3/4*xi^2 - 3/2*diff(h, t) - 3/2*diff(xi, t)
Ric.display()

I'd like to define the restriction to $\xi = 0$, and assign it to a new tensor

Ric0 = M.tensor_field(0,2, 'Ric0')
Ric0[0,0] = Ric[0,0].substitute_expression({xi:0, diff(xi, t):0})
Ric0.display()

but I get an AttributeError because the

AttributeError: 'ChartFunctionRing_with_category.element_class' object has no attribute 'substitute_expression'

I know that it works for functions

var('t')
xi = function('xi')(t)
f = function('f')(t)
h = function('h')(t)
ode = 3/2*f*h - 3/4*h^2 + 3/2*f*xi + 3/4*xi^2 - 3/2*diff(h, t) - 3/2*diff(xi, t)
ode0 = ode.substitute_expression({xi:0, diff(xi, t):0})

Question

Is there a way to substitute functions that are not 𝚜𝚊𝚐𝚎.𝚜𝚢𝚖𝚋𝚘𝚕𝚒𝚌.𝚎𝚡𝚙𝚛𝚎𝚜𝚜𝚒𝚘𝚗.𝙴𝚡𝚙𝚛𝚎𝚜𝚜𝚒𝚘𝚗 but ChartFunctionRing_with_category.element_class?

Excellent! Thank you for the answer, and a wonderful use of the lambda function!

Hi, I find the option only_nonredundant=True really helpful, but it is annoying to type over and over the option on the notebook. Is it possible to define the option globally (within a notebook)?