MaplePrimes Questions

when evalhf is used it shows error. please help

 

evalhf.mw

Hello everyone, Greetings!

I am facing a really strange problem. I need to write an expression, however, maple out of nowhere assigns values to the variable used. only to those which are written inside sin (). In previous versions the out put is fine. Is there a new way to write expressions in maple 2019? I am not sure.


 

restart

96*sin(2*beta*y)*cos(2*beta*y)*beta^4 + 96*sin(2*beta*y)*beta^4

(0.525982730176588e-113+0.525982730176588e-113*I)*beta^4

(1)

``


 

Download strngmpl.mw

 

I get the following when MAPLE starts...

Warning, .hdb help databases are deprecated, 'C:\Program Files\Maple 2018\lib\OrthogonalExpansions.hdb' will not be used, see ?HelpTools,Migrate help page for more information

I have recently loaded an orthogonal expansions package created for earlier versions of MAPLE.

How can I remove the message?

MRB
 

Hello,

How I can take Laplace inverse?

Thank you

LAPLACE

Maple Worksheet - Error

Failed to load the worksheet //convert/LAPLACE
 

Download LAPLACE

 

Hi, im working on integration methods and i would like to make a routine that put to left side from a equation only the terms that have 'j+1'. So for example, we have the heat-conduction equation and we want to develope a approximated solution by an implicit method, so:

at this point i have tryed different ways in order to select only the terms that T [any,j+1], because we want T[i,j+1],T[i+1,j+1] and T[i-1,j+1] to the lhs of the equation and the rest to the rhs

I would like to make Maple read only the second index as a criteria to select or not. 

Someone have any idea on how to make maple deal with it?

I have encountered the situation frequently where I want to simplify an equation by cancelling out terms on both sides.  I have tried simplify() with a variety of assumptions(J,L>0,etc) and I haven't been able to get it to work.  On a simple equation, one can use 'solve' however there are situations where solve doesn't work and I just want to simplify the equation not solve it.

The script below shows the situation.  I cancel out JL and the complex exponential by manually identifying that they are common factors.  Is there an automatic way of doing this type of simplification?

If I use expand() it clearly shows the common factors on both sides but I haven't found the command that removes any common terms.


 

E2 := (sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j+1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j-1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)-4*(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l+1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l-1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L) = h^2*(sum(sum(`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)

(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j+1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j-1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)-4*(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l+1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l-1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L) = h^2*(sum(sum(`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)

(1)

E3 := E2*J*L; E4 := simplify(lhs(E3)) = simplify(rhs(E3))

J*L*((sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j+1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(j-1)*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)-4*(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l+1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)+(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*(l-1)*n/L), n = 0 .. L-1), m = 0 .. J-1))/(J*L)) = h^2*(sum(sum(`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*j*m/J)*exp(-(2*I)*Pi*l*n/L), n = 0 .. L-1), m = 0 .. J-1))

 

sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*(j+1)*L+l*n*J)*Pi/(J*L)), n = 0 .. L-1), m = 0 .. J-1)+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*L*(j-1)+l*n*J)*Pi/(J*L)), n = 0 .. L-1), m = 0 .. J-1)-4*(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1))+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l+1)*J+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1)+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l-1)*J+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1) = h^2*(sum(sum(`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1))

(2)

 

subsindets(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*(j+1)*L+l*n*J)*Pi/(J*L)), n = 0 .. L-1), m = 0 .. J-1)+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*L*(j-1)+l*n*J)*Pi/(J*L)), n = 0 .. L-1), m = 0 .. J-1)-4*(sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1))+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l+1)*J+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1)+sum(sum(`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l-1)*J+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1) = h^2*(sum(sum(`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L)), n = 0 .. L-1), m = 0 .. J-1)), specfunc({Sum, sum}), proc (S) options operator, arrow; op(1, S) end proc)

`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*(j+1)*L+l*n*J)*Pi/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*L*(j-1)+l*n*J)*Pi/(J*L))-4*`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l+1)*J+L*j*m)/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l-1)*J+L*j*m)/(J*L)) = h^2*`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L))

(3)

 

simplify((`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*(j+1)*L+l*n*J)*Pi/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*(m*L*(j-1)+l*n*J)*Pi/(J*L))-4*`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l+1)*J+L*j*m)/(J*L))+`#mover(mi("u"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(n*(l-1)*J+L*j*m)/(J*L)) = h^2*`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]*exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L)))*(1/exp(-(2*I)*Pi*(J*l*n+L*j*m)/(J*L))))

2*`#mover(mi("u"),mo("ˆ"))`[m, n]*(-2+cos(2*Pi*m/J)+cos(2*Pi*n/L)) = h^2*`#mover(mi("ρ",fontstyle = "normal"),mo("ˆ"))`[m, n]

(4)

``


 

Download common_factors.mw

How can I expand the series in multi-variable, where all variable tends to infinity.

Suppose I have a function F(x,y,z) and I want the series of F in all three variable at {x=Infinity, y=infinity, z=infinity} 

 

Does MAPLE have capabability to do multidimensional FTs i.e. (x,y)->(u,v)? If not, are there any links to MAPLE packages which meet this requirement that can be recommended?

Melvin Brown

 

 

I used thread for a loop and found that it takes more time than evalf. Please help to decrease the computational cost. Thank you.

 

thread_mp.mw

Consider the finite field G:=GF(p,k) for some prime p and a positive integer k. Let H be an nxm matrix over G.

My question: How to obtain the minimum number of linearly dependent columns of the H over G. 

Thanks in advance  

-Can we solve symbolically matrix equations where the matrices are ungiven by MAPLE?

For a simple example;

Let A.B+2B=C.

where A nxn is matrix and B, C and nx1 matrices. (all matrices are ungiven)

Question: Find symbolically matrix B in terms of C and A?

My try:

We can quickly calculate Matrix B by hand.  We have (A+2I)B=C and so B=(A+2I)^(-1).C   ( where (A+2I) is invertible and I is nxn identity matrix)

Briefly: Can we do for the more complex and difficult equations by MAPLE? 

Hello, i wish to get dot displayed for time derivative and prime displayed for space derivative. I put typesetting in extended mode but it seems to only work for variable that depend only of time. So how can i get (in 2D math) dot displayed for derivative in time and prime displayed for space? 

Example u(x,t):
 

diff(u(x,t),x) --->  u'(x,t) output displayed in prime

diff(u(x,t),t)  --->  u^. (x,t) output displayed in dot

Thanks

hi guys i have a question for you,

can anyone help me to find the correct code to mantain the structure of vel_qD list and simplify the term diff(s1(t),t)??

i'm driving crazy for that beacuse i'm not able to do it...

thanks, down here i post the code...

 

vel_qD := [diff(theta1(t), t) = -(diff(s1(t), t))*cos(theta3(t))/(LAD*sin(theta1(t)+theta3(t))), diff(theta2(t), t) = (L1*sin(theta1(t)+theta3(t))*(diff(s1(t), t))+L1*sin(theta1(t)-theta3(t))*(diff(s1(t), t))-2*LAD*sin(theta1(t)+theta3(t))*(diff(s1(t), t)))/(LAD*LBC*cos(theta1(t)+theta3(t)-theta2(t))-LAD*LBC*cos(theta1(t)+theta3(t)+theta2(t))), diff(theta3(t), t) = (diff(s1(t), t))*cos(theta1(t))/(L3*sin(theta1(t)+theta3(t))), diff(s2(t), t) = (L1*sin(theta1(t)+theta3(t)-theta2(t))*(diff(s1(t), t))+L1*sin(-theta3(t)-theta2(t)+theta1(t))*(diff(s1(t), t))-LAD*sin(theta1(t)+theta3(t)-theta2(t))*(diff(s1(t), t))-LAD*sin(theta1(t)+theta3(t)+theta2(t))*(diff(s1(t), t)))/(LAD*cos(theta1(t)+theta3(t)-theta2(t))-LAD*cos(theta1(t)+theta3(t)+theta2(t)))]


 

with(Physics)

Setup(mathematicalnotation = true)

[mathematicalnotation = true]

(1)

Setup(dimension = 3, coordinates = (X = [x1, x2, x3]), metric = 2*F6(X)*dx2*dx3+2*F5(X)*dx1*dx3+2*F4(X)*dx1*dx2+F1(X)*dx1^2+F2(X)*dx2^2+F3(X)*dx3^2)

`* Partial match of  'coordinates' against keyword 'coordinatesystems'`

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (x1, x2, x3)}

 

`Systems of spacetime Coordinates are: `*{X = (x1, x2, x3)}

 

[coordinatesystems = {X}, dimension = 3, metric = {(1, 1) = F1(X), (1, 2) = F4(X), (1, 3) = F5(X), (2, 2) = F2(X), (2, 3) = F6(X), (3, 3) = F3(X)}]

(2)

g_[]

Physics:-g_[mu, nu] = Matrix(%id = 18446746167581484926)

(3)

Weyl[alpha, beta, mu, nu, nonzero]

`[Length of output exceeds limit of 1000000]`

(4)

Weyl tensor is identically equal to zero in 3D and I have tried to show this by inputting an arbitrary metric and calculating components of Weyl tensor. Thus this answer should give me the empty set but I am not getting that, thanks in advance.

CODE:

with(Physics):

Setup(mathematicalnotation = true)

Setup(dimension = 3, coordinates = (X = [x1, x2, x3]), metric = 2*F6(X)*dx2*dx3+2*F5(X)*dx1*dx3+2*F4(X)*dx1*dx2+F1(X)*(dx1^2)+F2(X)*(dx2^2)+F3(X)*(dx3^2))


g_[]

Weyl[alpha, beta, mu, nu, nonzero]

 

Download Weyltensor0in3d.mw

Hi all, I have a problem but can't solve, anyone help me. Example we have json link http://myjson.com/16knzm

How to read procedure getName and getAge from json link above to be

Thank you very much. Sory because my english is bad.

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