Unanswered Questions

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

How can I compute F from G according to the following text? (I implemented this but I need a more efficient implementation.)

 

Given a set G of polynomials which are a subset of k[U, X] and a monomial order with U << X, we want to comput set F from G s.t.


1. F is subset of G and for any two distinct f1, f2 in F , neither lpp (f1) is a multiple of lpp (f2) nor lpp (f2) is a multiple of lpp (f1).


2. for every polynomial g in G, there is some polynomial f in F such that lpp (g) is a multiple of
lpp (f ), i.e. ⟨lpp (F )⟩ = ⟨lpp (G)⟩,

--------------------------------------------------------------------------------------

It is worth nothing that F is not unique.

Example:  Let us consider G = {ax^2 − y, ay^2 − 1, ax − 1, (a + 1)x − y, (a + 1)y − a} ⊂ Q[a, x, y], with the lexicographic order on terms with a < y < x.

Then F = {ax − 1, (a + 1)y − a} and F ′ = {(a + 1)x − y, (a + 1)y − a} are both considered set.

please not that K[U,X] is a parametric polynomial ring (U is e sequence of parameters and X is a sequence of variables).

lpp(f) is leading monomial of f w.r.t. variables X. For example lpp(a*x^2+b*y)= x^2.

In Maple 2015.1 we have

restart;

solve([sin(2*x)/cos(x+3*Pi/2)=1,  x>-4*Pi, x<-5*Pi/2], x, allsolutions, explicit);

solve([sin(2*x)/cos(x+3*Pi/2)=1, x>0, x<2*Pi], x, allsolutions, explicit);

 

 

In the first example, the error message is not clear (actually there exists a unique root  x=-11*Pi/3), in the second example, one root  (x=5*Pi/3) is lost.

 

I've got this huge chunk of code which leads to an optimiazation at the very last line (Bestangles:=minimize(maximize()-minimize))). This minization is taking a very long time (havent solved it yet) and I would very much like to reduce that time. As I've understood maple does optimization by differentiating and then finding all extremes and comparing. Would this mean that since I minimize and optimize within a minimization command, it differentiates a ton of times? And if this is the case, can I somehow do the differentiation beforehand, since it is the same function being differentiate all the time? Or is there some other way I can improve the code? 
Thanks alot!

Heres the full code:

 

Consider a taper steel plate of uniform thickness t := 25mm as shown in the Fig. In addition to its self weight, the plate is subjected to a point load P := 100N at its mid point. Find the global force vector [F] , global stiffness matrix [K] , displacement in each element (1 and 2) , stresses in each element  (1 and 2) and reaction force at the support.Take E := 2*10^5N/mm2; rho := 8.2*10^(-5)kg/m3;

restart

t__1 := 150:

t__3 := 75:

w := 25:

l := 600:

t__2 := (t__1-t__3)/l*((1/2)*l)+t__3 = 225/2

A__1 := t__1*w = 3750``

A__2 := t__2*w = 5625/2``

A__3 := t__3*w = 1875``

Revised areas:

A__1e := (A__1+A__2)*(1/2) = 13125/4``

A__2e := (A__2+A__3)*(1/2) = 9375/4``

  E := 2*10^11:m2; F__1 := R__1:is support reaction N; F__2 := 100:N;``

rho__1 := 82*10^(-6) = 41/500000  N/mm2

rho__2 := 82*10^(-6) = 41/500000 N/mm2

l := 600:``

Number of elements,

n__e := 2:

l__e := 300 = 300````

q__0 := 100:N/m ; l := 1: m; n__e := 4:  elementsl  l__e := l/n__e: m;

We shall consider a two element system as shown in the Fig.
For element 1 Stiffness matrix K is

                                           Vector[row](2, {(1) = 1, (2) = 2})
K__1 := A__1e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[2625000000000000, -2625000000000000], [-2625000000000000, 2625000000000000]])  Vector(2, {(1) = 1, (2) = 2})

For element 2 Stiffness matrix K is

                                         Vector[row](2, {(1) = 2, (2) = 3})
K__2 := A__2e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[1875000000000000, -1875000000000000], [-1875000000000000, 1875000000000000]])  Vector(2, {(1) = 2, (2) = 3})

Global stiffness matrix obtained by adding all the elemental stiffness matrices and given b

           Vector[row](3, {(1) = 0, (2) = 0, (3) = 0})

K__g := Matrix(3, 3, {(1, 1) = K__1[1, 1], (1, 2) = K__1[1, 2], (1, 3) = 0, (2, 1) = K__1[2, 1], (2, 2) = K__1[1, 2]+K__2[1, 1], (2, 3) = K__2[1, 2], (3, 1) = 0, (3, 2) = K__2[2, 1], (3, 3) = K__2[2, 2]}) = Matrix([[K__1[1, 1], K__1[1, 2], 0], [K__1[2, 1], K__1[1, 2]+K__2[1, 1], K__2[1, 2]], [0, K__2[2, 1], K__2[2, 2]]])  Vector(3, {(1) = 0, (2) = 0, (3) = 0})

For element 1 Load matrix F is

  F__1e := (1/2)*`&rho;__1`*A__1e*l__e*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[861/25600], [861/25600]]) Vector(2, {(1) = 1, (2) = 2})

``

For element 2 Load matrix F isNULL

F__2e := (1/2)*A__2e*l__e*`&rho;__2`*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[123/5120], [123/5120]]) 

``

 

Download wrong_answers.mwwrong_answers.mwwrong_answers.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

hello everyone

can any one tell me what is this anti reduction method. In the paper of serdal palmuk,the link is given bellow

http://www.hindawi.com/journals/mpe/2009/202307/

in this paper question #4 is first solved by anti reduction method for  exact solution.

but i dont understand this method,

if anybody know this then please also tell me how to solve this,

and in the next  (6 & 7 ) examples "in the pourus media equation" they first find its particular exact solution.i also dont understand this,so please tell me

actually i know how to solve ODE to find its exact solution but  i dont know how we find exact solutions of partial differtial equations,

so please help me to solve this problem

thanks

 

How to find the integral
,

assuming k and n  integer?
It is known (McCrea W. H., Whipple F. J. W.Random paths in two and three dimensions, Proc. Roy. Soc. Edinburgh. 1940. V. 60. P. 281–298) that

G(n,n)=2/Pi*sum(1/(2*k-1),k=1..n).

The general case is reduced to the case k=n.
This is not a creature of pure reason: the one appears in electric circuits
(see M. Skopenkov, A. Paharev, A. Ustinov, Through resistor net, Mat. pros. Issue 18 (2014), 33-65, in Russian, http://www.mccme.ru/free-books/matpros/pdf/mp-18.pdf).
I found G(8,8) = 182144/(45045*Pi) in 657.797 s and G(9,9) = 3186538/(765765*Pi) in 4157.687 s on my comp by

restart; s := time():(1/2)*VectorCalculus:-int((1-cos(9*Pi*x)*cos(9*Pi*y))/(sin((1/2)*Pi*x)^2+sin((1/2)*Pi*y)^2), [x, y] = Rectangle(0 .. 1, 0 .. 1)); time()-s;
Mathematica 10.3.0 does G(9,9) in 250.391 s on my comp.

 

Hello,

I have a question about poincare sections. I have this piece of code i need to analyse and I want to use a poincare section in order to so. How could I do it? I am interested in theta and omega. Any help is greatly appreciated! Thank you in advance!

Kind regards,

Gambia Man

with(plots):

a := 1.501*10^9:

Th := sqrt(4*Pi^2*a^3/(G*(Mh+Msat)));

1876321.326

 

0.3348672330e-5

(1)

HyperionOrbit := proc (`&theta;IC`, `&omega;IC`) local a, Mh, Msat, G, e, beta, M, Eqns, ICs; global `&omega;H`, Th, soln; a := 1.501*10^9; Mh := 5.5855*10^18; Msat := 5.6832*10^26; G := 6.67259/10^11; e := .232; beta := .89; M := Mh+Msat; Eqns := diff(theta(t), t) = omega(t), diff(omega(t), t) = -G*Msat*beta^2*(xH(t)*sin(theta(t))-yH(t)*cos(theta(t)))*(xH(t)*cos(theta(t))+yH(t)*sin(theta(t)))/(xH(t)^2+yH(t)^2)^2.5, diff(xH(t), t) = vxH(t), diff(vxH(t), t) = -G*M*xH(t)/(xH(t)^2+yH(t)^2)^(3/2), diff(yH(t), t) = vyH(t), diff(vyH(t), t) = -G*M*yH(t)/(xH(t)^2+yH(t)^2)^(3/2); ICs := xH(0) = a*(1+e), yH(0) = 0, vxH(0) = 0, vyH(0) = sqrt(G*M*(1-e)/(a*(1+e))), theta(0) = `&theta;IC`, omega(0) = `&omega;IC`; soln := dsolve({Eqns, ICs}, numeric); odeplot(soln, [theta(t), omega(t)/`&omega;H`], 0 .. 5*Th, numpoints = 2000, labels = ["&theta;(t)","&omega;(t)/&omega;H"], axes = boxed, size = [.25, .75]) end proc

``

 

Download New_Poincare_section.mw

http://www.maplesoft.com/support/help/Maple/view.aspx?path=Physics/.

i see bra and ket expression are so beautiful,

however,

how do real valued eigenvectors involve in calculation of bra and ket style computation?

 

equ1 := -l*cos(xi)^2*(1-cos(`&beta;__f`))/(alpha^2.sin(sigma))-`&lambda;__2`*w*(v^2.sin(sigma))/(g*l*cos(xi)^2) = 0

-l*cos(xi)^2*(1-cos(`&beta;__f`))/(alpha^2.sin(sigma))-`&lambda;__2`*w*(v^2.sin(sigma))/(g*l*cos(xi)^2) = 0

(1)

equ2 := -l*cos(xi)^2*(1-cos(beta[f]))/(alpha*sin(sigma)*tan(sigma))+Typesetting:-delayDotProduct(l, cos(xi)^2)*z__0*sin(`&beta;__f`)/(alpha*sin(sigma)*(2*l*cos(sigma)^2))-`&lambda;__1`*`#mi("L")`*`#mi("sin",fontstyle = "normal")`(sigma)*cos(xi)+`&lambda;__2`*L*cos(sigma)*cos(xi)-`&lambda;__2`*w*alpha*v^2*sin(sigma)/(g*l*tan(sigma)*cos(xi)^2) = 0

-l*cos(xi)^2*(1-cos(beta[f]))/(alpha*sin(sigma)*tan(sigma))+(1/2)*(l.(cos(xi)^2))*z__0*sin(`&beta;__f`)/(alpha*sin(sigma)*l*cos(sigma)^2)-`&lambda;__1`*`#mi("L")`*`#mi("sin",fontstyle = "normal")`(sigma)*cos(xi)+`&lambda;__2`*L*cos(sigma)*cos(xi)-`&lambda;__2`*w*alpha*v^2*sin(sigma)/(g*l*tan(sigma)*cos(xi)^2) = 0

(2)

equ3 := l*cos(xi)^2*sin(`&beta;__f`)*tan(sigma)/(alpha*sin(sigma)*(2*l)) = 0

(1/2)*cos(xi)^2*sin(`&beta;__f`)*tan(sigma)/(alpha*sin(sigma)) = 0

(3)

equ4 := -`&lambda;__1`*`#mi("L")`*`#mi("cos",fontstyle = "normal")`(sigma)*sin(xi)+`&lambda;__2`*L*sin(sigma)*sin(xi)-2*`&lambda;__2`*tan(xi)*w*alpha*v^2*sin(sigma)/(g*l*cos(xi)^2)-l*sin(2*xi)*(1-cos(beta[f]))/(alpha*sin(sigma)) = 0

-`&lambda;__1`*`#mi("L")`*`#mi("cos",fontstyle = "normal")`(sigma)*sin(xi)+`&lambda;__2`*L*sin(sigma)*sin(xi)-2*`&lambda;__2`*tan(xi)*w*alpha*v^2*sin(sigma)/(g*l*cos(xi)^2)-l*sin(2*xi)*(1-cos(beta[f]))/(alpha*sin(sigma)) = 0

(4)

equ5 := L*cos(sigma)*cos(xi)-w = 0

L*cos(sigma)*cos(xi)-w = 0

(5)

`#mi("equ6")` := `#mi("L")`*`#mi("sin",fontstyle = "normal")`(sigma)*cos(xi)-w*alpha*v^2*sin(sigma)/(g*l*cos(xi)^2)

`#mi("L")`*`#mi("sin",fontstyle = "normal")`(sigma)*cos(xi)-w*alpha*v^2*sin(sigma)/(g*l*cos(xi)^2)

(6)

answer := solve({equ1, equ2, equ3, equ4, equ5, equ6}, {alpha, sigma, xi, `&lambda;__1`, `&lambda;__2`, beta[f]})

``

(7)

``

(8)

NULL

 

Download Bryson_sesson1_p6.mw

Maple's isprime is not a definitive primality test. The input has to pass a "strong pseudo-primality test" and "one Lucas test". This is well documented. I thought I remembered that there is also a way to get Maple to perform a true primality test, but I don't remember how and don't see anything about this in the Maple help system.

Is my memory faulty, or is there no definitive primality test in Maple?

Thanks in advance,

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu

Hello,

How to change the length underline in latex ?

For example in this environment :

$\underline{\mathcal{G}}$

 

Thank you,

Gérard.

So I am trying to apply the Optimization function to a rather complicated problem I am currently working on, and having some trouble getting maple to cooporate. This is the part of the code that is giving me the error, with error included. All of the variables imputtet when calling OptimizeSpring are constants. If there is any other details I should add, please say so and I will add them promptly :) Thanks alot!
Maple problem

I suspect that it might be related to this suggestion, but I am not sure how to apply it http://www.mapleprimes.com/posts/95113-Optimization-Workaround-For-Operator-Form

I'm trying to solve a beam vibration problem. I am relatively new to Maple and trying to solve PDE's. I feel like I'm pretty close, but I don't understand what's wrong here.

This is my maple file.

Thesis_Pde2_attempt.mw

as a note, I made some simplifying assumptions. I am just using part of this equation since it is less complex. And my boundary conditions are simply supported on both ends so no displacement and no 2nd derivative at the ends.

 

Any help would be greatly appreciated!!

Hello everyone, 

I have some problems with the "isolve" command on Maple. I am trying to solve for integer a very easy system of equations. When I type the commands

 

restart; 

n := 2;
isolve({sum(a[k], k = 1 .. n)-1 = 1}, d)


I get the expected {a[1] = 2-d, a[2] = d}. However, if I add conditions a[1],a[2] >= 0, that is the commands



restart;
n := 2;
isolve({ge(a[1], 0), ge(a[2], 0), sum(a[k], k = 1 .. n)-1 = 1}, d)


I get the warning "Warning, solutions may have been lost". What am I doing wrong? Is there a way to get Maple to give me the possible values?

 

Thank you in advance,

David

I'm trying to solve a system of equations thats expressed as a summation (the original has the summation symbol as opposed to 'sum'):

d_actual := solve(W_actual = sum(W_guess(def-asp_rad_inverse[i], E, asp_rad[i]), i = 1 .. n), def);

When n<5, I get an answer after a few seconds, but when n is higher, the program sits and 'evaluates' forever...I've waited up to 30 min. 

Background:

I have a plate with a number of hemispheres on the surface (# of impacted hemispheres given a force = n). Each has a unique radius and they're listed from largest to smallest in 'asp_rad[]'. I have an equation for the deflection of a single hemisphere as a function of Force and material properties that I have rearranged with respect to Force (W_guess(deflection, E, r)). There's an opposing plate that stays parallel to the original plate while pushing down on the asperities with Force W_actual. The total deflection of the opposing plate is the sum of: 1) the difference between the tallest radius and the impacted radius in question (asp_rad_inverse[]), and 2) the deflection of the impacted radius in question. 

I'm attempting to solve for the total deflection of the opposing plate via solving for the 'def' in the summation above, but when I run it, the program is not able to compute a solution.

Thanks in advance.

First 206 207 208 209 210 211 212 Last Page 208 of 361