This is probably a question to Edgardo: In another thread, the following quantity is considered [Eqs. (5) and (6) combined]:

expand(gamma_[definition]);

This is all very well, but it seems to depend on the metric loaded: if the Schwarzschild metric g_[sc] is loaded, then the above output results, but if the Minkowski metric g_[minkowski] is loaded, then the output of the above expansion is identically zero. Does that make sense? Is that intentional? The explicitly evaluated Ricci rotation coefficients vanish identically for the Minkowski metric, of course, but if evaluation is performed for that case, then why not also for the Schwarschild metric [not meaning to say that I want evaluation]?

In the following codes I am very sure exp(-3*q) is a factor both at numerator and denominator. However, I dont know how to annihilate this factor. Can someone kindly help to ensure it cancels out? Thanks and kind regards.

restart:
s:=(sum(a[j]*x^j,j=0..2)+sum(a[j]*exp(-(j-2)*x),j=3..4)):
F:=diff(s,x):
p1:=simplify(eval(s,x=q))=y[n]:
p2:=simplify(eval(F,x=q))=f[n]:
p3:=simplify(eval(F,x=q+h))=f[n+1]:
p4:=simplify(eval(F,x=q+3*h/2))=f[n+3/2]:
p5:=simplify(eval(F,x=q+2*h))=f[n+2]:

vars:= seq(a[i],i=0..4):
Cc:=eval(<vars>, solve({p||(1..5)}, {vars})):
for i from 1 to 5 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
T:y[n+2]=collect(simplify(eval(Cf,x=q+2*h)), [y[n],f[n],f[n+1],f[n+3/2],f[n+2]], recursive);

Does anyone know why Maple doesn't simplify the following expression?

Thanks!

Hi dears,

I hope that my request (question) is appropriate for Mapleprimes.

I know Gröbner bases and Buchberger's algorithm and I want to understand  the F4-algorithm. However, I know that  the corresponding paper can be found:

https://www.sciencedirect.com/science/article/pii/S0022404999000055 

Could you please state the sketch and main parts of the algorithm s.t. I can understand it?
Is there any primary Maple implementation of F4-algorithm?

Thanks in advance.

Hello All,

Please any sugestions on mathmateical projects. They assigned me to do a project for my math class for something I should be interested about. But since the deadline is one week and I don't have anything in mind right now. So I just need y'all help t suggest for me any ideas 

I believe I've found a bug, where Eigenvectors returns two eigenvectors which are the same (even though the eigenvalues are different). The expressions involves RootOf's, and it seems that the eigenvalues use "index" to distinguish themselves, yet the eigenvectors do not.

with(LinearAlgebra);

x := RootOf(z^2-t, z);
m := Matrix(2, (i,j) -> evala(add(a[j, k]*((-1)^(i-1))^k*x^k, k = 0..1)));
ev := Eigenvectors(m);

Download calculation.mws

can anyone explain to me this result ?

in the book "maple by example" they said that sin(Pi/12);
sin (1/12 π)
returns sin(π/12) because it does not know a formula for the explicit value of sin(π/12). but when i'm calculated this in maple i had got , this result, what does this result mean ?

hello,

this is my first post here so sorry if it's done wrong...

I'm having difficulties with the following command because maple only returns a solution if the first two boundary conditions of the "if" arguments are true. If the the first or the second condition is false then maple doesn't give a solution solution.

for i to n do if H[i] < 2.7 then if A[i, f] < 12 then if A[i, o] < 1.2 then Q[i, foo] := evalf(610*(A[i, o]*sqrt(H[i])*h[k]*A[i, T])^(1/2)) else Q[i, foo] := evalf(7.8*A[i, t]+378*A[i, o]*sqrt(H[i])) end if end if end if; print(Q[i, foo]); end do;

I've also tried with the elif command but it gives a similar problem

Anyone knows how to solve this?

suppose for example i am working on the function

exp(-I*Pi*(n+2*n*(m-1))/m);

and i wanted to for what ever reason assign a unique symbol to each of the two times the indeterminant 'n' occurs:

exp(-I*Pi*(n[1]+2*n[2]*(m-1))/m)

How could i accomplish this for any function F?

i have tried the method of

map(op, [op(op(exp(-I*Pi*(n+2*n*(m-1))/m)))]);
                [-1, Pi, n, 2 n (m - 1), m, -1]
 

And though i could then use algsubs on the original function for each in the above list, but then ran into problems arising that for any function the number of times i need to map op in iteration is not known, i will actually work this out by the end of the night so i dont know why im bothering asking

Edit 2: yep its ok i discovered subsop

Hello All,

This is Alex, I'm doing a master degree in computer sience at UMICH. They assigned me to do a math project and deadlined by next week, and I need y'all help to suggest some ideas for me if you guys having any 

Thanks 

Ali

Why input fraction to eigenvector and then evalf output are all the same when input different ?

i would like to see more decimal numbers

digits := 36 

command can not show more decimal numbers

why can not see the difference?

if start from fraction

After set round screen display to 36 digits in options

If start from floating value

the result different from start from fraction

start from which is the most accurate and correct?

if fraction is correct, why all result are the same even if input are different?

I'm absolutely new in maple, but I need to solve PDE and I don't understand why maple do not solve it

restart; with(PDEtools);
U := diff_table(u(x, t));

pde[1] := U[t, t] = U[x, x]+5*sin(3*x);

bc[1] := eval(U[], x = 0) = 0; bc[2] := eval(U[], x = Pi) = 0;
ic[1] := eval(U[], t = 0) = 0; ic[2] := eval(D[2]*U[], t = 0) = 1;

sys[1] := [pde[1], bc[1], bc[2], ic[1], ic[2]];
pdsolve(sys[1]);

Error: Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {u(0, t), u(Pi, t), u(x, 0)} in the given boundary conditions {D[2]*u(x, 0) = 1, u(0, t) = 0, u(Pi, t) = 0, u(x, 0) = 0}

If I solve it  with only boundary conditions or without any conditions, maple gives me an answer. 

Please, help me to understand, how to solve this pde system with boundary and initials conditions. This is my firs use of maple so I hardly understand how to write code correctly.

Real part + complex part

but 

sometimes it display

complex part + real part 

how to consistent display real part + complex part?

Sometimes we have questions that apply to every execution we make, regardless of what the actual nature of the commands included in that execution are.

For this reason, I propose that in the same way that we have the feature of "start up" code, which automatically executes upon opening the specified worksheet, we also implement an option of automatically executed code for every execution that is carried out within the specified worksheet. 

for example, code that serves purposes of output such as 

" This execution took xxxx minutes and xx.x seconds to complete, and required an average of x % of your computer's total processor utilization capacity across all cores "

I know how i will be providing this feature in my worksheets, but i just feel that it would be a beneficial feature for those that use maple for purposes that do not involve a knowledge of how maple is programmed, and would not be able to do so.

Any discussion about this would be welcomed, which is why i am posting about it of course.

I am trying to see if Maple can solve Laplace PDE inside the disk in polar coordinates. Standard textbook problem. Radius of disk is `a`. The boundary conditions on the disk is `f(theta)`. One of the conditions needed also is that the solution is finite in the center of the disk.

I do not know how to tell Maple that the solution should be finite in the center of the disk. If I do not give this conditions, Maple gives me strange looking solution, which does not look like anything close to the standard series solution one gets from hand solution. There is not even a series solution.

This is what I tried

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta);
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

Now, how to tell it that `u(0,theta)` is bounded? So that the `ln(r)` solution do not show up? Adding `u(0,theta)<infinity` to the boundary conditions, gives error

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta),u(0,theta)<infinity;
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

The standard solution to this PDE is

Where `c0` and `cn` and `kn` above are found from boundary conditions at $u(a,\theta)$.

How can one get Maple to give the above solution? How to tell it that $u$ is bounded at $r=0$?