tolga's profile - activity

I think there was something like alarm and AlarmInterrupt. Maybe you can use them with try and except.

The default Maxima solver needs assumptions to set floor((%i+1)/(2*%pi)) as shown in the message. You can either give su

Actually, I am not very happy but I think this is a fair solution to the problem.

I would like to define new derivative operators dd1 and dd2. They should follow the following rules (dd being either dd1 or dd2 - They follow the same rules.):

dd(a+b)=dd(a)+dd(b);
dd(a*b)=b*dd(a)+a*dd(b);
dd(-a)=-dd(a);
dd(1/a)=-1/a^2*dd(a);
dd(c)=0;
dd(c*a)=c*dd(a);
dd(a**c)=c*a**(c-1)*dd(a);

where c is a symbolic variable or a number of any kind and, a and b are scalar field functions on a manifold.

If the code is something like the following:

# First operator
def dd1(x):
    ...

# Second operator
def dd2(x):
    ...

Man = Manifold(4, 'Man', r'\mathcal{Man}')
CO.<t,r,th,ph> = Man.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')
a=Man.scalar_field(function('a')(*CO))
b=Man.scalar_field(function('b')(*CO))
c=var('c')

Then, it should produce these sample outputs for the following inputs:

Input> dd1(a)
Outpt> dd1(a)

Input> dd1(dd2(a))
Outpt> dd1(dd2(a))

Input> dd2(a*b)
Outpt> b*dd2(a)+a*dd2(b)

Input> dd1(c*a)
Outpt> c*dd1(a)

Input> dd2(dd1(c*a))
Outpt> c*dd2(dd1(a))

Input> dd1(-a)
Outpt> -dd1(a)

Input> dd2(c)
Outpt> 0

Input> dd1(7)
Outpt> 0

The following code seems to do the job. However, it looks very clumsy and I would be very happy to receive comments. ##

The version on the repository is rather old. I would suggest that you install the latest version using conda/mamba easil

The following code seems to do the job. However, it looks very clumsy and I would be very happy to receive comments. fr

The following seems to do the job. However, it looks very clumsy and I would be very happy to receive comments. from sa

Defining new operators with specific rules I would like to define new derivative operators dd1 and dd2. They should foll

I do not think that you will need loops. Please check this page to see that the Kretschmann scalar is defined and calcul

Please check the types in the following code: print("Before var:",type(r)) var('r') print("After var:",type(r)) print("

Set S=0 In the for loop: k=1, S=0+1=1 k=2, S=1+2=3 k=3, S=3+3=6 Set S=2 * S = 2 * 6 = 12

Set S=0 In the for loop: k=1, S=0+1=1 k=2, S=1+2=3 k=3, S=3+3=6 Set S=2S=26=12

Set S=0 In the for loop: k=1, S=0+1=1 k=2, S=1+2=3 k=3, S=3+3=6 Set S=2S=26=12

Set S=0 In the for loop: k=1, S=0+1=1 k=2, S=1+2=3 k=3, S=3+3=6 Set S=2S=26=12

We can discuss the substitution and simplification if @eric_g 's solution does not fit your needs.

Please see my answer below.

You actually put the functions Delta, rho2, and Sigma in the metric at the beginning. You should use them in a closed fo

Please check the answer of @eric_g below. If it does not work out, you can gather your code in one cell and paste it her

As general advice, try to give your metric in a closed form as much as you can. For example, instead of (r^2 + a^2*(cosθ

Also try this: def f(x): var('t') return numerical_integral(exp(t^2)/sqrt(t),1,x^2) f(4) As you can see from

This can be a workaround: def f(x): var('t') return real(integral(exp(t^2)/sqrt(t),t,1,x^2)) f(4).numerical_ap

Please try C=matrix(QQ,n,n).

What is your SageMath version? https://sagecell.sagemath.org/ gives the right answer.

Try canonicalize_radical(): var('alpha,beta,x,y') f=exp(alpha*log(x)+beta*log(y)) print(f.canonicalize_radical())

This answer might be helpful: https://ask.sagemath.org/question/37814/how-to-draw-three-pyramids-inside-a-right-prism/

Please check this: v=[1/n^2 for n in range(1,51)] print(v)

Please check this (): v=[1/n^2 for n in range(1,51)] print(v)

Please check this: v=[1/n^2 for n in range(1,50)] print(v)

Uninstalling and reinstalling might be the easiest solution.

Thank you @achrzesz. This is a solution to my problem.

That solves my problem. Thank you.

I have some scalar fields and a matrix whose elements are these scalar fields. A sample code is the following:

reset()
M=Manifold(4, 'M', r'\mathcal{M}')
BL.<t,r,th,ph> = M.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')
f=M.scalar_field(function('f')(*BL))
g=M.scalar_field(function('g')(*BL))
h=M.scalar_field(function('h')(*BL))
h=0
mymatrix=matrix([[f+g,g,h],[f,h,g],[h,g,f]])

f, g, or h can be zero or nonzero (according to some calculations).

When I try to display my matrix with show(mymatrix) or print(mymatrix), I see Scalar field on the 4-dimensional differentiable manifold M instead of the non-zero elements.

What should I do to display my matrix as

[f+g g 0]
[f 0 g]
[0 g f]

preferably without using loops, etc?

Displaying a matrix with scalar fields I have some scalar fields and a matrix whose elements are these scalar fields. A

You can try canonicalize_radical()

Keeping everything in h is easier: (Qi*R).subs(relation).simplify_full()

Maybe it is trivial but can (Qi*R).subs(h==log(q)).simplify_full() solve your problem?

Can you post the codes here? They are written in SageMath, right?

First, without tensor fields:

reset()
H=function('H',imag_part_func=0)(x)
term1=(1/2)*sqrt(2)*sqrt(I*H)
term2=(1/2)*sqrt(2)*sqrt(-I*H)
result=(term1*term2).canonicalize_radical()
print(result)

This gives the result: -1/2*H(x)

However, when term1 and term2 are the elements of tensors as in

reset()
Man=Manifold(4, 'Man', r'\mathcal{M}')
BL.<t,r,th,ph> = Man.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')
H=function('H',imag_part_func=0)(x)
term1=(1/2)*sqrt(2)*sqrt(I*H)
term2=(1/2)*sqrt(2)*sqrt(-I*H)
T1=Man.tensor_field(0,1,[0,0,term1,0])
T2=Man.tensor_field(0,1,[0,0,term2,0])
result=T1[2]*T2[2]
print(result)

I get, 1/2*I*H(x)

I need to get -1/2*H(x) as in the first case. I can solve my problem by using .expr() for the elements of the tensor:

reset()
Man=Manifold(4, 'Man', r'\mathcal{M}')
BL.<t,r,th,ph> = Man.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')
H=function('H',imag_part_func=0)(x)
term1=(1/2)*sqrt(2)*sqrt(I*H)
term2=(1/2)*sqrt(2)*sqrt(-I*H)
T1=Man.tensor_field(0,1,[0,0,term1,0])
T2=Man.tensor_field(0,1,[0,0,term2,0])
result=(T1[2].expr()*T2[2].expr()).canonicalize_radical()
print(result)

But is there a way to get the result I needed without .expr()? I think I am missing something about the tensor fields.

Last question: Is there a way to define the function H as a differentiable function (i.e. diff(H,r) is not zero) of only