Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

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This update also include a fix to the problem with the simplify extension mechanism, as first reported on MaplePrimes. Thanks, as always, for helping us make Maple better.

This update is available through Tools>Check for Updates in Maple, and is also available from the Maple 2024.2 download page, where you can find more details.

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When I export images as SVG the resulting file always has the image much larger than the viewbox resulting in only part of it showing. My normal workflow is to then load in Inkscape and correct the error, however, I would love to not have an extra step. Is there any other fix for this?

How do you convert this system of equations into matrix form? The decoupling process is performed.Convert the equations into photographs.fuxian1030.mw

What I do at the moment

u-> changecoords(u,[x,y,z],spherical_physics,[r,theta,phi]):
map(%,[x,y,z])
  [r cos(phi) sin(theta), r sin(phi) sin(theta), r cos(theta)]

Any other (preferably shorter) ways to look up transformations defined in ?coords

I am studying linear operators that have vanishing Nijenhuis Torsion, that is a (1-1) tensor L whose corresponding (1-2) tensor N given by a tensor equation of L (see attached) is identically zero.

I am new to maple, i have used it to plot vector fields and solve systems of equations in the past but i am unfamiliar with the DifferentialGeometry and Physics packages.

Attached is my best effort at solving this problem directly for a simple 2d case without the use of any packages and i am wondering if it is possible to do it all in one line without having to define tensor components one by one.

nijenhuis_torsion.mw

How can I use Maple to solve a system of nonlinear equations symbolically and display the steps in the solution?

Can i get this ode in a "standardform"  ?

verg:= (-delta*eta^2 + alpha*eta)*diff(diff(U(xi), xi), xi) - U(xi)*(2*eta*gamma*theta*(delta*eta - alpha)*U(xi)^2 + eta^2*delta*k^2 + (-alpha*k^2 - 2*delta*k)*eta + 2*k*alpha + delta) = 0;

Since the puzzle task "A circle is to be disturbed ..." makes no fun, here is a Maple task:
The term to be simplified step by step:
(2+10/(3*sqrt(3)))^(1/3)+(2-10/(3*sqrt(3)))^(1/3)

I have learned that the SPECTRA.mla library can solve SDP problems. I have tried to download and use it, but I am still missing the FGb module

> with(SPECTRA);
> M := Matrix(6, 6, [[10, 1, 0, m[1], -m[3], m[2]], [1, -2*m[1]+27, m[3], -27/2, -m[4], -m[5]], [0, m[3], -2*m[2], m[4], m[5], 0], [m[1], -27/2, m[4], 10, 0, m[6]], [-m[3], -m[4], m[5], 0, -2*m[6], 0], [m[2], -m[5], 0, m[6], 0, 1]]);
> SolveLMI(M);
=> Error, (in SPECTRA:-SolveLMI) `FGb` does not evaluate to a module

 

How can I resolve this issue on a Windows environment, beause I don't see install file for Windows, only MacOS and Linux:

We are working to obtain a fully symbolic dynamic model for a robot. 

Using the CPU (I9-12900K) and 128 GB of DDR5 RAM (5600 MHZ) did not compute a 7x7 Inverse of a symbolic matrix

Is is possible to exploit the CUDA functions to compute it on the GPU? I have a NVidia RTX A6000 (48 GB of DDR6 GPU memory)

I tried this: 

CUDA:-Enable(true)

CUDA: -MatrixInverse(D_Q):

But it does not use the GPU to compute this. 

Maybe I'm doing smth wrong. 

Thank you, 

Calin

Hi

Dear friends, I am a relatively new user and I have a problem in entering and calculating the 2F1 hypergeometric function. My question is how to enter this function in Maple in equations so that Maple recognizes it? Because hypergeom ([1], [2], [3]) Maple itself is a 3-element function, while 2F1 hypergeometric ([1], [2], [3], [4]) is a four-element!

Almost i did 10 method for this ode equation all of them are succes but this one is giving me some confusing and i am looking for  get my answer, the mothod say if we have the auxilary equation if substitute the solution of this auxilary equation in our series solution then substitute in ode equation must be satisfy but it is not satisfy so when he did assumption for the auxilary equation he say it satisfy if we sabstitute this assumption in our series solution!

My question is this how we get thus assumption ? and why finding exact  solution of auxilary equation not satisfy?

restart

with(PDEtools)

with(LinearAlgebra)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

Fode := (-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*eta*gamma*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(2)

NULL

F := sum(a[i]*G(xi)^i, i = 0 .. 1)

a[0]+a[1]*G(xi)

(3)

``

(4)

D1 := diff(F, xi)

a[1]*(diff(G(xi), xi))

(5)

NULL

S := (diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(6)

S1 := diff(G(xi), xi) = sqrt(G(xi)^4+A[2]*G(xi)^2+A[1])

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(7)

E1 := subs(S1, D1)

a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(8)

D2 := diff(E1, xi)

(1/2)*a[1]*(4*G(xi)^3*(diff(G(xi), xi))+2*A[2]*G(xi)*(diff(G(xi), xi)))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(9)

E2 := subs(S1, D2)

(1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(10)

K := U(xi) = F

U(xi) = a[0]+a[1]*G(xi)

(11)

K1 := diff(U(xi), xi) = E1

diff(U(xi), xi) = a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(12)

K2 := diff(U(xi), xi, xi) = E2

diff(diff(U(xi), xi), xi) = (1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(13)

L := eval(Fode, {K, K1, K2})

(1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(14)

L1 := normal((1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0)

 

collect(L1, {G(xi)})

(-2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1])*G(xi)^3+(-6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2)*G(xi)^2+(-6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1])*G(xi)-2*gamma*delta*eta^2*theta*a[0]^3+2*gamma*alpha*eta*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0

(15)

eq0 := -2*delta*eta^2*gamma*theta*a[0]^3+2*alpha*eta*gamma*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0

eq1 := -6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1] = 0

eq2 := -6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2 = 0

eq3 := -2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1] = 0

COEFFS := solve({eq0, eq1, eq2, eq3}, {alpha, eta, a[0], a[1]}, explicit)

case1 := COEFFS[4]

{alpha = delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k), eta = eta, a[0] = 0, a[1] = 1/(-gamma*theta)^(1/2)}

(16)

NULL

S

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(17)

S1

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(18)

S2 := dsolve(S, G(xi))

G(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = (1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = -(1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = (1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = JacobiSN((1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2)*xi+c__1, (-2*(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1])*A[1])^(1/2)/(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1]))*A[1]*2^(1/2)/(A[1]*(-A[2]+(A[2]^2-4*A[1])^(1/2)))^(1/2)

(19)

K

U(xi) = a[0]+a[1]*G(xi)

(20)

K4 := subs(case1, K)

U(xi) = G(xi)/(-gamma*theta)^(1/2)

(21)

NULL

K5 := subs(S2, K4)

U(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2)/(-gamma*theta)^(1/2)

(22)

NULL

F2 := subs(case1, Fode)

(-eta^2*delta+delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)*eta/(eta*k^2+eta*A[2]-2*k))*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k))*U(xi)^2+eta^2*delta*k^2+(-k^2*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)-2*k*delta)*eta+2*k*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)+delta) = 0

(23)

``

(24)

NULL

odetest(K5, F2)

-(1/2)*delta*eta*(A[2]^2-4*A[1])^(1/2)*(-2*(A[2]+(A[2]^2-4*A[1])^(1/2))/gamma)^(1/2)/((eta*k^2+eta*A[2]-2*k)*(-theta)^(1/2))

(25)


and i hope mapleprimes don't delete this question becuase of this pictures also it help for undrestanding

 

there is other picture for different auxilary equation just  add one multiply term for G(xi)^4 in case anyone needed i will upload

Download odetest.mw

how fixed that?

NULL

restart

NULL

A := Vector[row]([4, 0, 5])

Vector[row](%id = 36893490313998292860)

(1)

B := Vector[row]([3, 1, 0])

Vector[row](%id = 36893490313998279372)

(2)

C := Vector[row]([2, 1, 1])

Vector[row](%id = 36893490313998273820)

(3)

C.`&x`(A, B)

9

(4)

NULL

Download we.mw

In many papers, I've noticed that the solution to an ODE (ordinary differential equation) often emerges directly when there's only a single function involved. My question is: is there a way to generate solutions to an ODE by producing specific parameters?

 

restart

with(PDEtools)

with(LinearAlgebra)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

S := diff(F(xi), xi) = sqrt(P*F(xi)^4+Q*F(xi)^2+R)

diff(F(xi), xi) = (P*F(xi)^4+Q*F(xi)^2+R)^(1/2)

(2)

S1 := dsolve(S, F(xi))

xi-Intat(1/(P*_a^4+Q*_a^2+R)^(1/2), _a = F(xi))+c__1 = 0

(3)

S2 := (diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(4)

S3 := dsolve(S2, F(xi))

F(xi) = -(1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = -(1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = JacobiSN((1/2)*(2*(-4*P*R+Q^2)^(1/2)-2*Q)^(1/2)*xi+c__1, (-2*(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2)*R*P)^(1/2)/(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2))*R*2^(1/2)/(R*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)

(5)
 

NULL

Download get_all_solution_of_ode_by_generation.mw

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