Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

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);

i have got alot of mixed and high degree derivatives. For example:

u[x]*u[x,t]*eta[x,t]+u[]^2*u[x]*eta[x]+kis(x,y)u[x,t]^2*u[]+eta(x, y)*u[]*u[x]^2+ksi[x,t]*u[x]^2*u[x,t]+......

like this alot of terms

my question is how can i solve divided by the derivative of the u(x,t) partial differential equations system and so  how can i find eta(x,t,u) and ksi(x,t,u) 

what is the name of this decomposition?

Y = S^-1*X*S, where S is unitary fourier transform

do maple have function to solve this?

Hi MaplePrimers!

I have a simulation in MapleSIM, exported as a compiled procedure in maple using -LinkModel(), and -GetCompiledProc.

I'm trying to do parameter estimation on my MapleSIM model.  Within a optimization scheme, I call the MapleSIM model, and it will output a curve.  Using a least squares method, I compare this measurements to synthetic experimental data (I know the actual values), and generate an objective function.  The optimization algorithm will try different parameter values, and try to minimze the objective function.  When the curves are exactly the same, the objective function will be zero.

The problem I am having is certain parameter sets will cause the model to require very small steps.  I wish to put a timeout on these experiments, because speed is important.  However, I would also like to see the results up to the point of requiring very small steps.  For timeout, I was using code along the lines of:

out:= timelimit(30,cProc(params = PData)); #simulate with 30s limit

where PData are the parameter guessses, and cProc is the compiled MapleSim model.

I would like 'out' to be assigned whatever the results were after 30 seconds, even if the model had not finished integrating.

 

Thanks in advance for any help!

I want to define a band matrix using n vectors. Detial given maple sheet. please help me

Help.mw

I want to define a matrix of M by M which have some condition. detail given in maple file. please help me.

Help_Maple-1.mw

I wan to define a square matrix for M and r detial given in Maple sheet. Please Help....Help_Maple.mwHelp_Maple.mw

FirstEigenVector := Matrix(3, 1, {(1, 1) = -.736895432967255+0.*I, (2, 1) = -.588906969844997+0.*I, (3, 1) = -.331924240964690+0.*I});
SecondEigenVector := Matrix(3, 1, {(1, 1) = -.589856901397123+0.*I, (2, 1) = .320280857681335+0.*I, (3, 1) = .741275257969058+0.*I});
ThirdEigenVector := Matrix(3, 1, {(1, 1) = .330233185410229+0.*I, (2, 1) = -.742030156443046+0.*I, (3, 1) = .583384341736151+0.*I});
LHS := ProjOfEigenVector;
LHS := Matrix(3, 3, {(1, 1) = -.736895432967255+0.*I, (1, 2) = -.589856901397123+0.*I, (1, 3) = .330233185410229+0.*I, (2, 1) = -.588906969844997+0.*I, (2, 2) = .320280857681335+0.*I, (2, 3) = -.742030156443046+0.*I, (3, 1) = -.331924240964690+0.*I, (3, 2) = .741275257969058+0.*I, (3, 3) = .583384341736151+0.*I});
RHS := c1*FirstEigenVector+c2*SecondEigenVector+c3*ThirdEigenVector;
RHS := Matrix(3, 1, {(1, 1) = (-.736895432967255+0.*I)*c1+(-.589856901397123+0.*I)*c2+(.330233185410229+0.*I)*c3, (2, 1) = (-.588906969844997+0.*I)*c1+(.320280857681335+0.*I)*c2+(-.742030156443046+0.*I)*c3, (3, 1) = (-.331924240964690+0.*I)*c1+(.741275257969058+0.*I)*c2+(.583384341736151+0.*I)*c3});
solve([LHS[1][1] = RHS[1][1], LHS[2][2] = RHS[2][1], c1^2+c2^2+c3^2 = 1], [c1, c2, c3]);

 

after calculated the projection matrix, 

it is a 3*3 matrix on left hand side

however, combination of eigenvectors on right hand side is 3*1 matrix

when calculated c1,c2,c3 under the condition c1^2+c2^2+c3^2 = 1

how to know whether LHS[1][1] = RHS[1][1], or LHS[1][2] = RHS[1][1] or

LHS[1][3] = RHS[1][1]

First 245 246 247 248 249 250 251 Last Page 247 of 334