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Hi everyone

I am currently trying to make my own simple package including a few procedures. So far I have been able to write some "code" that actually works when I open the document and hit "enter". I would, however, like to save the package so it can be accessed during any Maple session using the command "with". I have unsuccesfully tried to comprehend the Maple help pages regarding this question but I definitely don't want to mess things up.

This is what I have written:

mat := module ()
description "useful procedures for mathematics, physics and chemistry";
export AtomicWeight;
option package;

   AtomicWeight := proc (x) description "returns the average atomic mass of the naturally ocurring element";
   Units:-AddSystem(NewSystem, Units:-GetSystem(SI), u);
   return evalf(ScientificConstants:-Element(x, atomicweight, system = NewSystem, units))
   end proc

end module;

What should I do to save it correctly?

Thank in advance,
Mads


 

 

 

 

I received an unexpected error message when trying to minimize a function: evaluating

returns the error message

Error, (in @) too many levels of recursion

Why am I getting this message?  It's hard for me to see how minimizing a function involves recursion, unless Maple is trying to iteratively approximate a solution.

How to open batch maple output as mws in Maple Player?  I was used to doing this daily at work before retirement in my university office on a campus computer.  There was an "open" command in a file menu of some sort that I could click on, I seem to remember.   Am just fresh afte downloading Maple Player on my iMac and have it installed.  The file from my batch Maple output is form a run of batch Maple on the office machine that I can logon to remotely from home, as I did in this case.  I get rough looking graphic output (using the alphabet style graphics methods).  It would be good to have a more detailed and fine grained graphics to answer some basic questions that I am analyzing.  Thanks . . !!

Hello;
        i am wording on fluid dynamics, in which i can up a system of nonlinear partial differential equation with i am suppose to solve using implicit keller box method. i need an asistance on how to implement this in maple.

 

error module is maple.dll_unloaded

 

i have already called stopmaple(kv);

in

testfunction(string hello){....stopmaple(kv));

 

testfunction(1);

testfunction(2);

when call testfunction again , it got error

Hello everybody,

 

This one is an easy one but I cannot find the answers.  Suppose that I clik on the template for evaluating a function.  But mine have two (or more) variables (f=f(x,y)).  I have the x=a as one placeholder.  But I need to have an another inplut placeholder for y=b.  Can somenone tell me how to do it under Windows and under Linux.

 

Thank you in advance for your help.

 

--------------------------------------
Mario Lemelin
Maple 17 Ubuntu 13.10 - 64 bits
Maple 17 Win 7 - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987
 u_{tt} = c^2 u_{xx}, \,
 u(t,0)=0, \quad u(t,L)=0,

as well as the initial conditions

 u(0,x)=f(x), \quad u_t(0,x)=g(x).

 

Trace  on computing the distinct-degree decomposition of the squarefree polynomial
f=x^17+2x^15+4x^13+x^12+2x^11+2x^10+3x^9+4x^4+3x^3+2x^2+4x belong to F_5[x].tell from the output only how many irreducible factors of degree i the polynomial f has, for all i.

 

Mapleprimes_Integral.mw

I have a question regarding following problem:

assume(a > 0, a < 1, t > 0, Z0 > 0, z > 0)

f1 := proc (z) options operator, arrow; 1/z end proc

proc (z) options operator, arrow; 1/z end proc

(1)

I_1 := int(f1(z)*ln((a*z+1)/(1+z/a)), z = 0 .. Z0); 1; MultiSeries:-asympt(%, Z0, 3)

2*ln(Z0)*ln(a)+(a^2-1)/(a*Z0)-(1/4)*(a^4-1)/(a^2*Z0^2)+O(1/Z0^3)

(2)

Using the representation which should hold for all a>0 and z>0

int(z*exp(t)*(a^2-1)/((exp(t)+a*z)*(exp(t)*a+z)), t = 0 .. infinity); 1; combine(%)

ln((a*z+1)*a/(z+a))

(3)

I'm calculating the result the other way around

int(z*exp(t)*(a^2-1)*f1(z)/((exp(t)+a*z)*(exp(t)*a+z)), z = 0 .. Z0); 1; I_2 := int(%, t = 0 .. infinity); 1; MultiSeries:-asympt(%, Z0, 3)

2*ln(Z0)*ln(a)+(a^2-1)/(a*Z0)-(1/4)*(a^4-1)/(a^2*Z0^2)+O(1/Z0^3)

(4)

plot(eval([I_1, I_2], a = 1/2), Z0 = 0 .. 10)

 

So the results are the same.

But if I calculate this with another function

f2 := proc (z) options operator, arrow; 1/(z*(z+a)) end proc

proc (z) options operator, arrow; 1/(z*(z+a)) end proc

(5)

I_3 := int(f2(z)*ln((a*z+1)/(1+z/a)), z = 0 .. Z0); 1; MultiSeries:-asympt(%, Z0, 3)

-dilog(a^2)/a-2*ln(a)/Z0+(1/2)*(2*ln(a)*a^2+a^2-1)/(a*Z0^2)+O(1/Z0^3)

(6)

int(z*exp(t)*(a^2-1)*f2(z)/((exp(t)+z*a)*(exp(t)*a+z)), z = 0 .. Z0); 1; I_4 := IntegrationTools:-Change(int(%, t = 0 .. infinity), t = ln(z)); 1; MultiSeries:-asympt(%, Z0, 3); 1; simplify(convert(convert(MultiSeries:-series(I_4, Z0, 1), polynom), polynom))

ln(a)*(ln(Z0)+ln(a))/a

(7)

I get another result :-/ The Integral doesn't even vanish in the limit Z0 -> 0

Though if I take the limit prior:

int(z*exp(t)*(a^2-1)*f2(z)/((exp(t)+z*a)*(exp(t)*a+z)), z = 0 .. infinity);

-dilog(a^2)/a

(8)

the result is correct. What is the problem here?



Download Mapleprimes_Integral.mw

 

I converted this comment into a question since it was unrelated to the question in which it was put.
That way also other MaplePrimes users can have a shot at this new question.
Preben Alsholm
 

Dear Preben Alsholm 

Hi;

Hope you to be healthy and have nice times,

I have another problem and if it doesn't consuming your valuable times, please guide me.

I have some Basis function,say W1(t),W2(t),W3(t),W4(t) which are orthonormal and i want to write a program that can approximate the integral of W(t)=[W1(t),W2(t),W3(t),W4(t)] again by Wi's, in other word
int(W(t')dt',t'=0..t)≈PW(t), where P is knows as integral operational matrix. the following is my attempt and unfortunately has no real solution!!!!

 

restart:
>              # Definition of 3th B-Spline
>
> piecewise(x>=0 and x<=1,1,0):
> h[1]:=unapply(%,x):
>
>             # Definition of 3th B-Spline
>
> h[2]:=simplify(int(h[1](x-t),t=0..1)):
> hh:=unapply(%,x):
>
>              #Definition of 3th B-Spline

int(hh(x-t),t=0..1): 
> simplify(%):
> N:=unapply(%,x):
>
> J:=1:     # Number of base function is 2^J+2
> phi:=linalg[matrix](2^J+2,1):
> for i from -2 to 2^J-1
> do
> N(2^J*x-i)*h[1](x): #for deleting out side of[0,1]
> simplify(%):
> phi[i+3,1]:=unapply(%,x):
> od:
>
> w[1]:=phi[1,1](x):
> w[1]/sqrt(int(w[1]^2,x=0..1)):
> W[1]:=unapply(%,x):     
>
> for i from 2 to 2^J+2
> do
> kk:=0:
> for j1 from 1 to i-1
> do
> aa[j1]:=int(phi[i,1](x)*w[j1],x=0..1):
> bb[j1]:=int(w[j1]^2,x=0..1):
> kk:=kk+aa[j1]/bb[j1]*w[j1]:
> od:
> w[i]:=simplify(phi[i,1](x)-kk):
> w[i]/sqrt(int(w[i]^2,x=0..1)):
> W[i]:=unapply(%,x):                 #Orthonormality process
>
>
> for j from 1 to 2^J+2 do
> for io from 1 to 2^J+2 do
> f[j]:=int(W[j](s),s=0..x);
> c[io][j]:=int(f[j](x)*W[io](x),x=0..1);
> od;
> od;

 

 

Best Wishes

Hi,

I want to compute a formula which is too complicated and it contains some variables. So I divide it into several parts. But it always turns out kernel connection has been lost. I looked maplesoft online help system and change the ConnectionType from 0 to 2. But it does not work. So how does this happen and how to solve?

I attach my maple file which appears error.
Thanks a lot!

Regards,

Yan

I am trying to reduce a tensor expression: ωiωjUi,j 

For which I have tried the following

restart; with(Physics)

Setup(dimension = [3, `+`], coordinatesystems = X, spacetimeindices = lowercaselatin):

`The dimension and signature of the tensor space are set to: [3, +] `

 

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

 

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

 

`Defined objects with tensor properties`

(1)

omega[i] := d_[j](U[k](X))*ep_[i, j, k];

Physics:-LeviCivita[i, j, k]*Physics:-d_[`~j`](U[`~k`](X), [X])

 

Physics:-LeviCivita[l, m, n]*Physics:-d_[`~m`](U[`~n`](X), [X])

 

Physics:-LeviCivita[i, j, k]*Physics:-LeviCivita[l, m, n]*Physics:-d_[`~j`](U[`~k`](X), [X])*Physics:-d_[`~m`](U[`~n`](X), [X])*Physics:-d_[`~l`](U[`~i`](X), [X])

 

-(-Physics:-d_[n](U[i](X), [X])*Physics:-d_[k](U[`~n`](X), [X])+Physics:-d_[m](U[i](X), [X])*Physics:-d_[`~m`](U[k](X), [X])-Physics:-d_[i](U[k](X), [X])*Physics:-d_[l](U[`~l`](X), [X])+Physics:-d_[k](U[i](X), [X])*Physics:-d_[l](U[`~l`](X), [X]))*Physics:-d_[`~i`](U[`~k`](X), [X])

(2)

continuity := [D_[l](U[l](X)) = 0]

[Physics:-d_[l](U[`~l`](X), [X]) = 0]

(3)

red_eq := subs(continuity, expr)

-(-Physics:-d_[n](U[i](X), [X])*Physics:-d_[k](U[`~n`](X), [X])+Physics:-d_[m](U[i](X), [X])*Physics:-d_[`~m`](U[k](X), [X]))*Physics:-d_[`~i`](U[`~k`](X), [X])

(4)

Question 1. Am I using the continuity condition correctly? How do I use this condition correctly? If I change the index for the expression, substitution does not work correctly. 

Question 2. How do I expand the red_eq term in terms of the basis to give out the full expression? 

Question 3. I would like to eventually replace U by (A-Amean) in the current expression. How do I implement this? 

Download term8.mwterm8.mw

with(Groebner):
with(LinearAlgebra):
T := lexdeg([x1,x2,x3],[e1,e2,e3]);
hello1 := proc(xx,yy)
return MatrixMatrixMultiply(xx,yy);
end proc:
hello2 := proc(xx,yy)
return xx+yy- MatrixMatrixMultiply(xx,yy);
end proc:
m1 := Matrix(3, 3, {(1, 1) = -.737663975994461+0.*I, (1, 2) = -.588973463383001+0.*I, (1, 3) = .330094104689369+0.*I, (2, 1) = -.588012653178741+0.*I, (2, 2) = .320157823261769+0.*I, (2, 3) = -.742792089286083+0.*I, (3, 1) = -.331802619371428+0.*I, (3, 2) = .742030476217061+0.*I, (3, 3) = .582492741708719+0.*I});
m2 := Matrix(3, 3, {(1, 1) = -.742269137704830+0.*I, (1, 2) = -.590598631673326+0.*I, (1, 3) = .316590877121441+0.*I, (2, 1) = -.593533033362923+0.*I, (2, 2) = .360143915024171+0.*I, (2, 3) = -.719732518911068+0.*I, (3, 1) = -.311054762892221+0.*I, (3, 2) = .722142379823161+0.*I, (3, 3) = .617863510611693+0.*I});
m3 := Matrix(3, 3, {(1, 1) = -.751491355856820+0.*I, (1, 2) = -.574908634018322+0.*I, (1, 3) = .323636840615627+0.*I, (2, 1) = -.575794245520782+0.*I, (2, 2) = .332066412772496+0.*I, (2, 3) = -.747123071744916+0.*I, (3, 1) = -.322058579916187+0.*I, (3, 2) = .747804760642505+0.*I, (3, 3) = .580574121936877+0.*I});
AA := hello1(m1, m2);
BB := hello2(m1, m2);
GB := Basis([e1- AA,e2- BB],T):
NormalForm(m3, GB, T);

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