Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

restart;

kernelopts(version);

`Maple 2020.0, X86 64 LINUX, Mar 4 2020, Build ID 1455132`

convert([1,2,3], Vector[row]);

Vector[row](3, {(1) = 1, (2) = 2, (3) = 3})

convert([1,2,3], Vector[col]);

This worksheet shows an unexpected behavior of pdsolve().  It solves the heat equation on the interval (−1,1) but fails on the interval (0,1).  I see no technical reason for this happening — it's probably due to a little bug lurking somewhere.

restart;

kernelopts(version);

`Maple 2020.0, X86 64 LINUX, Mar 4 2020, Build ID 1455132`

Physics:-Version();

`The "Physics Updates" package is not installed`

Solve the heat equation on the interval -1 < x and x < 1 and

boundary conditions u(-1, t) = 0, u(1, t) = 0.

pde := diff(u(x,t),t) = diff(u(x,t),x,x);

diff(u(x, t), t) = diff(diff(u(x, t), x), x)

bc := u(-1,t)=0, u(1,t)=0;

u(-1, t) = 0, u(1, t) = 0

ic := u(x,0) = cos(Pi/2*x);

u(x, 0) = cos((1/2)*Pi*x)

pdsol := expand(pdsolve({pde,bc,ic}));

u(x, t) = cos((1/2)*Pi*x)*exp(-(1/4)*t*Pi^2)

That's good.  Apply pdetest to verify it:

pdetest(pdsol, [pde,bc,ic]);

[0, 0, 0, 0]

Now, solve the same problem on the interval 0 < x and x < 1 and

boundary condition u__x(0, t) = 0 and u(1, t) = 0.  It should be

evident that the solution remains the same as the one calculated

above, due to symmetry, and here is the verification:

bc_new := D[1](u)(0,t)=0, u(1,t)=0;

(D[1](u))(0, t) = 0, u(1, t) = 0

pdetest(pdsol, [pde,bc_new,ic]);

[0, 0, 0, 0]

But for some reason Maple's pdsolve fails to solve the PDE.  Actually

its response is somewhat erratic -- sometimes it returns nothing at all,

and sometimes it exits with an error message.

pdsolve({pde,bc_new,ic});

 

Download pdsolve-fails.mw

 

I need to solve general linear programming problem with multi-precision support. What are my options with Maple 2020?

 

What have I done!

Whatever I have created - Maple doesn't think it is a list.

The question is - what is it then?

proclist.mw

I'm trying to obtain the dynamical response of a simply-supported beam with a cantilever extension, coupled to a spring-mass system. In mathematical terms, this system is ruled by three PDEs (relative to each bare part of the main structure) and one ODE (relative to the spring-mass system). I think my mathemical model is finely formulated, but Maple keeps telling me this:

Error, (in pdsolve/numeric/process_IBCs) improper op or subscript selector

I believe it is because my PDEs depend on "x" and "t", while the ODE depends solely on "t". I have tried to transform my ODE into a "PDE", making it also dependent of "x", but without imposing any boundary conditions relative to "x". However, after this Maple points a new error message:

Error, (in pdsolve/numeric) initial/boundary conditions must be defined at one or two points for each independent variable

Could someone help me finding a solution? My algorythm in shown in the attached file below.

Worksheet.mw

I'm trying to plot this integral but without success. Could someone help me?  I tried it the following way.

Thanks in advance for any help!

integral_test.mw

 

Hello everyone, we are doing some excercises on graphs. The commands PetersenGraph and GeneralisedPetersenGraph only draw an indirected one, and i was wondering, is there a way to draw a directed Petersen graph?

Here's a very simple example, working as intended:

 

 

Now, if I try to use the exception indexing function:

 

 

We can notice that b[1] isn't using exception anymore (as if the fourth input overwrote the second one) and that b[1][2] isn't linked to the value "two". However, if I define the tables first, I get the expected result:

 

 

Why does the exception indexing function prevent me from getting an existing table entry in the second case?

Hi,

how to generate a graph of ellipsoid from ellipse equation ( revolution x axis or y axis) ?

Thanks

QEllipsoide.mw

Hi Guys,

I am trying to intgrate a function involving hyperbolic functions in a range of 0 to 1 and it is giving me very large value of 10^94. However, on doing integration terms terms i can see that some large terms involving 10^191 cancel out with each other and I can have a fnite value of this integration. It would be really helpful if someone can help me out why it is happening with int function and how can I solve this case involving hyperbolic function. For reference maple file is attached. Thanks in advance and much appreciated.

With Regards

Sunit

question.mw

I've drawn shape with the Graph function and all the lines are drawn as the same length even though the shape is weighted and when I do the WeightMatrix function I get the right weights.  What do I need to do to have the lines drawn to the correct length?

THE ISSUE: It's returning a list of [1] when it should be returning [1,1,1,7]

##Find remainder##
rm := proc(a, b) local n; n := 0; while 0 <= b - n*a do n := n + 1; end do; b - (n - 1)*a; end proc;
rm := proc (a, b) local n; n := 0; while 0 <= b-n*a do n := n+1 

   end do; b-(n-1)*a end proc


rm(8, 3657);
                               1

rm(16, 12345);
                               9

##FINDING THE WHOLE NUMBER PORTION ##
whole := proc(a, b) local r, i; r := 0; i := 0; if a < b then r := rm(a, b); i := (b - r)/a; else 0; end if; end proc;
whole := proc (a, b) local r, i; r := 0; i := 0; if a < b then 

   r := rm(a, b); i := (b-r)/a else 0 end if end proc


whole(8, 3657);
                              457

whole(16, 12345);
                              771

 


j = whole(8, 3657);
                            j = 457

k = rm(8, 3657);
                             k = 1

L := [];
                            L := []

L := [op(L), rm(8, 8657)];
                            L := [1]

j = whole(8, 457);
                             j = 57


##GETTING THE LIST OF DIGITS (BEFORE REVERSING)##
HELPER := proc(a, b, L) local j; j := whole(a, b); [op(L), rm(a, b)]; if 0 < j then HELPER(a, j, L); else ; end if; L; end proc;
HELPER := proc (a, b, L) local j; j := whole(a, b); [op(L), 

   rm(a, b)]; if 0 < j then HELPER(a, j, L) else  end if; L end 

   proc


CNS := proc(a, b) HELPER(a, b, L); end proc;
          CNS := proc (a, b) HELPER(a, b, L) end proc

 

CNS(8, 3657);
                              [1]

 

HELPER(8, 3657, L);
                              [1]

 

There is an interesting preprint here, on a method for integration of some pseudo-elliptic integrals.

It was mentioned in a (sci.math.symbolic) usenet thread, which can be accessed via Google Groups here.

Inside the paper, the author mentions that the following is possible for some cases -- to first convert to RootOf form.

restart;

kernelopts(version);

`Maple 2020.0, X86 64 LINUX, Mar 4 2020, Build ID 1455132`

ig:=((-3+x^2)*(1-6*x^2+x^4)^(-1/4))/(-1+x^2);

(x^2-3)/((x^4-6*x^2+1)^(1/4)*(x^2-1))

int(ig,x);

int((x^2-3)/((x^4-6*x^2+1)^(1/4)*(x^2-1)), x)

infolevel[int] := 2:

S:=simplify([allvalues(int(convert(ig,RootOf),x))],size):

Stage1: first-stage indefinite integration

Stage2: second-stage indefinite integration

Norman: enter Risch-Norman integrator

Norman: exit Risch-Norman integrator

int/algrisch/int: Risch/Trager's algorithm for algebraic function

int/algrisch/int: entering integrator at time 9.029

int/algrisch/int: function field has degree 4

int/algrisch/int: computation of an integral basis: start time 9.031

int/algrisch/int: computation of an integral basis: end time 9.038

int/algrisch/int: normalization at infinity: start time 9.039

int/algrisch/int: normalization at infinity: end time 9.048

int/algrisch/int: genus of the function field 3

int/algrisch/int: computation of the algebraic part: start time 9.059

int/algrisch/int: computation of the algebraic part: end time 9.060

int/algrisch/int: computation of the transcendental part: start time 9.063

int/algrisch/transcpar: computing a basis for the residues at time 9.068

int/algrisch/residues: computing a splitting field at time 9.068

int/algrisch/transcpar: basis for the residues computed at time 9.103

int/algrisch/transcpar: dimension is 2

int/algrisch/transcpar: building divisors at time 9.300

int/algrisch/transcpar: testing divisors for principality at time 9.605

int/algrisch/goodprime: searching for a good prime at time 9.606

int/algrisch/goodprime: good prime found at time 9.704

int/algrisch/goodprime: searching for a good prime at time 9.704

int/algrisch/goodprime: good prime found at time 9.762

int/algrisch/areprinc: the divisor is principal: time 10.084

int/algrisch/areprinc: the divisor is principal: time 11.833

int/algrisch/transcpar: divisors proven pincipal at time at time 11.833

int/algrisch/transcpar: generators computed at time 11.834

int/algrisch/transcpar: orders are [1 1]

int/algrisch/transcpar: check that the candidate is an actual antiderivative

int/algrisch/transcpar: the antiderivative is elementary

int/algrisch/transcpar: antiderivative is (1/2)*ln((RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^3*_z^3-RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^2*_z^4+RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^2*_z^2-3*_z^3*RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)+RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)*_z-5*_z^2+1)/((_z+1)*(_z-1)*_z^2))+(1/2)*RootOf(_Z^2+1)*ln((RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^2*RootOf(_Z^2+1)*_z^4+RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^3*_z^3-RootOf(_Z^2+1)*RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)^2*_z^2+3*_z^3*RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)-5*RootOf(_Z^2+1)*_z^2-RootOf(_Z^4*_z^4-_z^4+6*_z^2-1 index = 1)*_z+RootOf(_Z^2+1))/((_z+1)*(_z-1)*_z^2))

int/algrisch/int: computation of the transcendental part: end time 12.040

int/algrisch/int: exiting integrator for algebraic functions at time 12.041

S[1];

(1/2)*ln(-((x^4-6*x^2+1)^(3/4)*x+(x^4-6*x^2+1)^(1/2)*x^2+(x^4-6*x^2+1)^(1/4)*x^3+x^4-(x^4-6*x^2+1)^(1/2)-3*(x^4-6*x^2+1)^(1/4)*x-5*x^2)/((x+1)*(x-1)))+((1/2)*I)*ln((-(x^4-6*x^2+1)^(3/4)*x+(I*x^2-I)*(x^4-6*x^2+1)^(1/2)+(x^3-3*x)*(x^4-6*x^2+1)^(1/4)-I*x^2*(x^2-5))/((x+1)*(x-1)))

S[2];

(1/2)*ln(-((x^4-6*x^2+1)^(3/4)*x+(x^4-6*x^2+1)^(1/2)*x^2+(x^4-6*x^2+1)^(1/4)*x^3+x^4-(x^4-6*x^2+1)^(1/2)-3*(x^4-6*x^2+1)^(1/4)*x-5*x^2)/((x+1)*(x-1)))-((1/2)*I)*ln((-(x^4-6*x^2+1)^(3/4)*x+(-I*x^2+I)*(x^4-6*x^2+1)^(1/2)+(x^3-3*x)*(x^4-6*x^2+1)^(1/4)+I*x^2*(x^2-5))/((x+1)*(x-1)))

simplify( diff(S[1],x) - ig ), simplify( diff(S[2],x) - ig );

0, 0

 

Download int_pe.mw

Let E all triplets as X=(p,q,r) such as p^2+q^2=r^2. We define the application f of E dans C complex as X in E f(X)=(p+Iq)/r=Z. Calculate abs(Z). Show that in E the law noted * defined by
X1*X2=(p1*p2-q1*q2,p2*q1+p1*q2,r1*r2) is an internal law. Calculate f(X1*X2). Then if X0=(3,4,5), find
X0*X0, X0*(X0*X0).Thank you for the help.

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