Items tagged with academic

Let us consider 

Student[Precalculus]:-LimitTutor(sqrt(x), x = 2);

One expects a nice illustration of the result sqrt(2). But instead of that one reads "f(x) approaches 1.41 as x approaches 2". This is simply ignorant and forms a wrong understanding of limits. It should also be noticed that all the entries (left, 2-sided, and right) produce the same animation. The same issue with other limits I tried, e.g.

Student[Precalculus]:-LimitTutor(sqrt(x), x = 1);

. I think this command should be completely rewritten or excluded from Maple. 

What method of worksheet dissemination maximizes the number of recipients and methods of access?

Our students do have access to Maple on campus. However some students beyond our campus do not. Hence I'm trying maximize the number of recipients. This means that the recipient can accesses my worksheet either via tradition computer OSs such as Mac, Linux, Windows, or via tablet / personal computer OSs: iOS or Android. 

For a static worksheet, I've found PDF is the obvious choice. However, I'm writing more modifiable worksheets. The most obvious options I see are:

* MaplePlayer
* MapleCloud
* Maple WWW, by DigiArea Team

In investigating the options, I have found that:
* Maple Player is useful to those who have no access to Maple, but there appears to be no Android/iOS version of it.
* MapleCloud looks very promising since all OSs read the web, but there appears to be no way a person can join a private group via the MapleCloud website. Am I missing or misreading the help?
* Maple WWW, while pretty, appears to be painfully slow on a tablet (at least the examples I looked at).

Thus the initial purpose of this post is to answer the questions I raise, which I view as far more strategic, with more nuances than most questions that appear on this forum. And the second purpose is to initiate more discussion among the users year, particularly from the educational community, about guiding the direction of Maple.

I look forward to your comments and suggestions.

From October 19-21, the third installment of the Maple T.A. and Möbius User Summit took place. Making the move back to Europe this year, the three-day conference was held at the beautiful Vienna University of Technology in the heart of Vienna, Austria. The scope of this year’s event expanded to include Maplesoft’s newest product, Möbius, which is designed to help academic institutions move their STEM courses online.

This year’s Summit brought together participants from 20 countries, including Australia, the Czech Republic, Poland, China, Norway, India, Egypt, Japan, the Netherlands, and many others. Needless to say, there is great interest in learning more about how Maple T.A. and Möbius can play a role in shaping the educational landscape.

Video recordings of each presentation will be made public soon, so keep an eye out for them!

Conference attendees take in the sights on the veranda at TU Wien

Getting Down to Business

Presentations were divided into 5 overarching themes as they relate to Maple T.A. and Möbius: Shaping Curriculum; Content Creation; Experiences Using Möbius; Integrating with your Technology; and The Future of Online Education. Presentations were given by representatives from schools across Europe, including DTU (Denmark), TH Köln (Germany), Imperial College of London and University of Birmingham (UK), Vienna UT (Austria), KTH Royal Institute of Technology (Sweden), Université de Lausanne (Switzerland), and others.

Many talks showcased the impressive versatility of Maple T.A. and Möbius to have practical applications in all STEM subjects, from Nuclear Engineering to Operations Management and many subjects in between.

Perhaps the discussion that gave Maplesoft the most feedback was led by Steve Furino from the University of Waterloo, who divided attendees up into groups to formulate a wish list of what they’d like to see in a courseware authoring environment. The list had over 40 items.


Linda Simonsen, Country Manager in the Nordic, records a group’s wish list

Notable Quotables

Many thought-provoking statements and questions were posed, but the following few stood out above the rest:

  • “Wouldn’t it be wonderful if you could take the best course from the best instructor anywhere in the world?”
  • “With Maple T.A., we can divert resources away from grading and over to tutoring.”
  • “Möbius rescued us!”

Get the party started!

While each day was full of invigorating conference discussions, evenings provided ample opportunity to ditch the suit jacket and tie, and enjoy the lively Austrian atmosphere. The first evening at the Zwölf Apostelkeller was the perfect venue to break the ice while satisfying those taste buds longing for some traditional Viennese cuisine. Once Schnitzel, Käsespätzle (a delicious German version of Mac and Cheese), Strudel, Kaiserschmarren (shredded pancake), and a glass or two of wine hit the table, people soon forgot about the pouring rain outside.

The evening reception took place 3-4 levels under ground

Michael Pisapia, VP of Europe, serves digestifs to guests

It would have been hard to top the social in the Apostelkeller, but the next evening sure tried.

Day 2 finished with an impressive formal dining experience at the historic Gerstner Beletage in the Palace Todesco, built in 1864 and situated directly across from the Vienna State Opera House. The 500-room palace was home to Eduard Freiherr von Todesco, a well-known Viennese banker.

View from the palace of the Vienna State Opera House

Jonny Zivku, Maple T.A. Product Manager, gives opening remarks at the Gerstner Beletage im Palais Todesco

Jonathan Watkins from the University of Birmingham and Michael Pisapia - both dressed to impress

The skies finally cleared enough to take some photos, but only after most people had gone home. Thankfully Aron Pasieka, Möbius Project Manager, was still around to get some great shots of the city. Enjoy!


Before the skies cleared vs. after the skies cleared

From beginning to end, the entire Summit was very well received by everyone who attended.

We would be remiss if we did not thank our incredible hosts at the Vienna University of Technology. Stefanie Winkler, Professor Andreas Körner, and Professor Felix Breitenecker were beyond helpful in bringing many of the finer details together, as well as helping many people overcome the language barrier.

We can’t wait to do it all again in London, England in 2017, and hope to see just as many new faces as familiar ones.

 

Photo credits: A. Pasieka, A. French, H. Zunic, J. Cooper

 

Update: The conference presentation recordings are now available here on our website.

I have tried to solve the following ode equation, but I have got error. What is the potencial problem?

http://i65.tinypic.com/xdcl8p.jpg

 

 error

 

The material below was presented in the "Semantic Representation of Mathematical Knowledge Workshop", February 3-5, 2016 at the Fields Institute, University of Toronto. It shows the approach I used for “digitizing mathematical knowledge" regarding Differential Equations, Special Functions and Solutions to Einstein's equations. While for these areas using databases of information helps (for example textbooks frequently contain these sort of databases), these are areas that, at the same time, are very suitable for using algorithmic mathematical approaches, that result in much richer mathematics than what can be hard-coded into a database. The material also focuses on an interesting cherry-picked collection of Maple functionality, that I think is beautiful, not well know, and seldom focused inter-related as here.

 

 

Digitizing of special functions,

differential equations,

and solutions to Einstein’s equations

within a computer algebra system

 

Edgardo S. Cheb-Terrab

Physics, Differential Equations and Mathematical Functions, Maplesoft

Editor, Computer Physics Communications

 

 

Digitizing (old paradigm)

 

• 

Big amounts of knowledge available to everybody in local machines or through the internet

• 

Take advantage of basic computer functionality, like searching and editing

 

 

Digitizing (new paradigm)

• 

By digitizing mathematical knowledge inside appropriate computational contexts that understand about the topics, one can use the digitized knowledge to automatically generate more and higher level knowledge

 

 

Challenges


1) how to identify, test and organize the key blocks of information,

 

2) how to access it: the interface,

 

3) how to mathematically process it to automatically obtain more information on demand

 

 

 

 

                                           Three examples


Mathematical Functions

 

"Mathematical functions, are defined by algebraic expressions. So consider algebraic expressions in general ..."

The FunctionAdvisor (basic)

 

"Supporting information on definitions, identities, possible simplifications, integral forms, different types of series expansions, and mathematical properties in general"

Examples

   

General description

   

References

   

 

Differential equation representation for generic nonlinear algebraic expressions - their use

 

"Compute differential polynomial forms for arbitrary systems of non-polynomial equations ..."

The Differential Equations representing arbitrary algebraic expresssions

   

Deriving knowledge: ODE solving methods

   

Extending the mathematical language to include the inverse functions

   

Solving non-polynomial algebraic equations by solving polynomial differential equations

   

References

   

 

Branch Cuts of algebraic expressions

 

"Algebraically compute, and visualize, the branch cuts of arbitrary mathematical expressions"

Examples

   

References

   

 

Algebraic expresssions in terms of specified functions

 

"A conversion network for arbitrary mathematical expressions, to rewrite them in terms of different functions in flexible ways"

Examples

   

General description

   

References

   

 

Symbolic differentiation of algebraic expressions

 

"Perform symbolic differentiation by combining different algebraic techniques, including functions of symbolic sequences and Faà di Bruno's formula"

Examples

   

References

   

 

Ordinary Differential Equations

 

"Beyond the concept of a database, classify an arbitrary ODE and suggest solution methods for it"

General description

   

Examples

   

References

   

 

Exact Solutions to Einstein's equations

 

 

Lambda*g[mu, nu]+G[mu, nu] = 8*Pi*T[mu, nu]

 

"The authors of "Exact solutions toEinstein's equations" reviewed more than 4,000 papers containing solutions to Einstein’s equations in the general relativity literature, organized the whole material into chapters according to the physical properties of these solutions. These solutions are key in the area of general relativity, are now all digitized and become alive in a worksheet"


The ability to search the database according to the physical properties of the solutions, their classification, or just by parts of keywords (old paradigm) changes the game.

More important, within a computer algebra system this knowledge becomes alive (new paradigm).

• 

The solutions are turned active by a simple call to one commend, called the g_  spacetime metric.

• 

Everything else gets automatically derived and set on the fly ( Christoffel symbols  , Ricci  and Riemann  tensors orthonormal and null tetrads , etc.)

• 

Almost all of the mathematical operations one can perform on these solutions are implemented as commands in the Physics  and DifferentialGeometry  packages.

• 

All the mathematics within the Maple library are instantly ready to work with these solutions and derived mathematical objects.

 

Finally, in the Maple PDEtools package , we have all the mathematical tools to tackle the equivalence problem around these solutions.

Examples

   

References

   

 

Download:  Digitizing_Mathematical_Information.mw,    Digitizing_Mathematical_Information.pdf

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

A string is wound symmetrically around a circular rod. The string goes exactly
4 times around the rod. The circumference of the rod is 4 cm and its length is 12 cm.
Find the length of the string.
Show all your work.

(It was presented at a meeting of the European Mathematical Society in 2001,
"Reference levels in mathematics in Europe at age16").

Can you solve it? You may want to try before seing the solution.
[I sometimes train olympiad students at my university, so I like such problems].

restart;
eq:= 2/Pi*cos(t), 2/Pi*sin(t), 3/2/Pi*t; # The equations of the helix, t in 0 .. 8*Pi:
               
p:=plots[spacecurve]([eq, t=0..8*Pi],scaling=constrained,color=red, thickness=5, axes=none):
plots:-display(plottools:-cylinder([0,0,0], 2/Pi, 12, style=surface, color=yellow),
                         p, scaling=constrained,axes=none);
 

VectorCalculus:-ArcLength(<eq>, t=0..8*Pi);

                           20

 

Let's look at the first loop around the rod.
If we develop the corresponding 1/4 of the cylinder, it results a rectangle  whose sides are 4 and 12/4 = 3.
The diagonal is 5 (ask Pythagora why), so the length of the string is 4*5 = 20.

 

This presentation is on an undergrad intermediate Quantum Mechanics topic. Tackling the problem within a computer algebra worksheet in the way shown below is actually the novelty, using the Physics package to formulate the problem with quantum operators and related algebra rules in tensor notation.

 

Quantization of the Lorentz Force

 

Pascal Szriftgiser1 and Edgardo S. Cheb-Terrab2 

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

 

We consider the case of a quantum, non-relativistic, particle with mass m and charge q evolving under the action of an arbitrary time-independent magnetic field "B=Curl(A(x,y,z)), "where `#mover(mi("A",mathcolor = "olive"),mo("&rarr;"))` is the vector potential. The Hamiltonian for this system is

H = (`#mover(mi("p",mathcolor = "olive"),mo("&rarr;"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("&rarr;"))`(X))^2/(2*m)

where `#mover(mi("p",mathcolor = "olive"),mo("&rarr;"))` is the momentum of the particle, and the force acting in this particle, also called the Lorentz force, is given by

 

`#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = m*(diff(v(t), t))

 

where `#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))` is the quantized velocity of the particle, and all of  H, `#mover(mi("p",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("A",mathcolor = "olive"),mo("&rarr;"))` and `#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` are Hermitian quantum operators representing observable quantities.

 

In the classic (non-quantum) case, `#mover(mi("F"),mo("&rarr;"))` for such a particle in the absence of electrical field is given by

 

`#mover(mi("F"),mo("&rarr;"))` = `&x`(q*`#mover(mi("v"),mo("&rarr;"))`, `#mover(mi("B"),mo("&rarr;"))`) ,

 

Problem: Departing from the Hamiltonian, show that in the quantum case the Lorentz force is given by [1]

 

`#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`))

 

[1] Photons et atomes, Introduction à l'électrodynamique quantique, p. 179, Claude Cohen-Tannoudji, Jacques Dupont-Roc et Gilbert Grynberg - EDP Sciences janvier 1987.

 

Solution

 

We choose to tackle the problem in Heisenberg's picture of quantum mechanices, where the state of a system is static and only the quantum operators evolve in time according to

``

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`&hbar;`

 

Also, the algebraic manipulations are simpler using tensor abstract notation instead of the standard 3D vector notation. We then start setting the framework for the problem, a system of coordinates X, indicating the dimension of the tensor space to be 3 and the metric Euclidean, and that we will use lowercaselatin letters to represent tensor indices. In addition, not necessary but for convenience, we set the lowercase latin i to represent the imaginary unit and we request automaticsimplification so that the output of everything comes automatically simplified in size.

 

restart; with(Physics); interface(imaginaryunit = i)

Setup(mathematicalnotation = true, automaticsimplification = true, coordinates = X, dimension = 3, metric = Euclidean, spacetimeindices = lowercaselatin, quiet)

[automaticsimplification = true, coordinatesystems = {X}, dimension = 3, mathematicalnotation = true, metric = {(1, 1) = 1, (2, 2) = 1, (3, 3) = 1}, spacetimeindices = lowercaselatin]

(1)

 

Next we indicate the letters we will use to represent the quantum operators with which we will work, and also the standard commutation rules between position and momentum, always the starting point when dealing with quantum mechanics problems

 

Setup(quantumoperators = {F}, hermitianoperators = {A, B, H, p, r, v, x}, realobjects = {`&hbar;`, m, q}, algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`&hbar;`*KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0}, hermitianoperators = {A, B, H, p, r, v, x}, quantumoperators = {A, B, F, H, p, r, v, x}, realobjects = {`&hbar;`, m, q, x1, x2, x3, %dAlembertian, Physics:-dAlembertian}]

(2)

 

Note that we start not indicating F as Hermitian, in order to arrive at that result. The quantum operators A, B, and F are explicit functions of X, so to avoid redundant display of this functionality on the screen we use

 

CompactDisplay((A, B, F)(X))

A(x1, x2, x3)*`will now be displayed as`*A

 

B(x1, x2, x3)*`will now be displayed as`*B

 

F(x1, x2, x3)*`will now be displayed as`*F

(3)

Define now as tensors the quantum operators that we will use with tensorial notation (recalling: for these, Einstein's sum rule for repeated indices will be automatically applied when simplifying)

 

Define(x, p, v, A, B, F, quiet)

{A, B, F, p, v, x, Physics:-Dgamma[a], Physics:-Psigma[a], Physics:-d_[a], Physics:-g_[a, b], Physics:-KroneckerDelta[a, b], Physics:-LeviCivita[a, b, c], Physics:-SpaceTimeVector[a](X)}

(4)

The Hamiltonian,

H = (`#mover(mi("p",mathcolor = "olive"),mo("&rarr;"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("&rarr;"))`(X))^2/(2*m)

in tensorial notation, is given by

H = (p[n]-q*A[n](X))^2/(2*m)

H = (1/2)*Physics:-`^`(p[n]-q*A[n](X), 2)/m

(5)

Generally speaking to arrive at  ```#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`)) what we now need to do is

1) Express this Hamiltonian (5) in terms of the velocity

 

And, recalling that, in Heisenberg's picture, quantum operators evolve in time according to

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`&hbar;`

 

2) Take the commutator of H with the velocity itself to obtain its time derivative and, from `#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = m*(diff(v(t), t)) , that commutator is already the force up to some constant factors.

 

To get in contact with the basic commutation rules between position and momentum behind quantum phenomena, the quantized velocity itself can be computed as the time derivative of the position operator, i.e as the commutator of x[k] with H

I*Commutator(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, x[k])/`&hbar;`

I*Physics:-Commutator(H, x[k])/`&hbar;` = (1/2)*(I*q^2*Physics:-AntiCommutator(A[n](X), Physics:-Commutator(A[n](X), x[k]))-I*q*Physics:-AntiCommutator(p[n], Physics:-Commutator(A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics:-KroneckerDelta[k, n]*`&hbar;`)/(`&hbar;`*m)

(6)

This expression for the velocity, that involves commutators between the potential A[n](X), the position x[k] and the momentum p[n], can be simplified taking into account the basic quantum algebra rules between position and momentum. We assume that A[n](X)(X) can be decomposed into a formal power series (possibly infinite) of the x[k], hence all the A[n](X) commute between themselves as well as with all the x[k]

 

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

(7)

(Note: in some cases, this is not true, but those cases are beyond the scope of this worksheet.)

 

Add these rules to the algebra rules already set so that they are all taken into account when simplifying things

 

Setup(algebrarules = {%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(8)

Simplify(I*Physics[Commutator](H, x[k])/`&hbar;` = (1/2)*(I*q^2*Physics[AntiCommutator](A[n](X), Physics[Commutator](A[n](X), x[k]))-I*q*Physics[AntiCommutator](p[n], Physics[Commutator](A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics[KroneckerDelta][k, n]*`&hbar;`)/(`&hbar;`*m))

I*Physics:-Commutator(H, x[k])/`&hbar;` = (-A[k](X)*q+p[k])/m

(9)

The right-hand side of (9) is then the kth component of the velocity tensor quantum operator, the relationship is the same as in the classical case

v[k] = rhs(I*Physics[Commutator](H, x[k])/`&hbar;` = (-A[k](X)*q+p[k])/m)

v[k] = (-A[k](X)*q+p[k])/m

(10)

and with this the Hamiltonian (5) can now be rewritten in term of the velocity completing step 1)

simplify(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, {SubstituteTensorIndices(k = n, (rhs = lhs)(v[k] = (-A[k](X)*q+p[k])/m))})

H = (1/2)*m*Physics:-`^`(v[n], 2)

(11)

For step 2), to compute

 `#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = m*(diff(v(t), t)) and m*(diff(v(t), t)) = I*m*Physics:-Commutator(H, v(t)[k])/`&hbar;` 

 

we need the commutator between the different components of the quantized velocity which, contrary to what happens in the classical case, do not commute. For this purpose, take the commutator between (10) with itself after replacing the free index

Commutator(v[k] = (-A[k](X)*q+p[k])/m, SubstituteTensorIndices(k = n, v[k] = (-A[k](X)*q+p[k])/m))

Physics:-Commutator(v[k], v[n]) = -q*(Physics:-Commutator(A[k](X), p[n])+Physics:-Commutator(p[k], A[n](X)))/m^2

(12)

To simplify (12), we use the fact that if f  is a commutative mapping that can be decomposed into a formal power series in all the complex plan (which is assumed to be the case for all A[n](X)(X)), then

Physics:-Commutator(p[k], f(x, y, z)) = -I*`&hbar;`*`&PartialD;`[k](f(x, y, z))

where p[k]"=-i `&hbar;` `&PartialD;`[k] " is the momentum operator along the x[k] axis. This relation reads in tensor notation:

Commutator(p[k], A[n](X)) = -I*`&hbar;`*d_[k](A[n](X))

Physics:-Commutator(p[k], A[n](X)) = -I*`&hbar;`*Physics:-d_[k](A[n](X), [X])

(13)

Add this rule to the rules previously set in order to automatically take it into account in (12)

Setup(Physics[Commutator](p[k], A[n](X)) = -I*`&hbar;`*Physics[d_][k](A[n](X), [X]))

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`&hbar;`*Physics:-d_[k](A[n](X), [X]), %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(14)

Physics[Commutator](v[k], v[n]) = -q*(Physics[Commutator](A[k](X), p[n])+Physics[Commutator](p[k], A[n](X)))/m^2

Physics:-Commutator(v[k], v[n]) = -I*q*`&hbar;`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2

(15)

Also add this other rule so that it is taken into account automatically

Setup(Physics[Commutator](v[k], v[n]) = -I*q*`&hbar;`*(Physics[d_][n](A[k](X), [X])-Physics[d_][k](A[n](X), [X]))/m^2)

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`&hbar;`*Physics:-d_[k](A[n](X), [X]), %Commutator(v[k], v[n]) = -I*q*`&hbar;`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2, %Commutator(x[k], p[l]) = I*`&hbar;`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(16)

Recalling now the expression of the Hamiltonian (11) as a function of the velocity, one can compute the components of the force operator  "()Component(v*B,k)=m (v[k])=(i m [H,v[k]][-])/`&hbar;`"

F[k](X) = I*m*%Commutator(rhs(H = (1/2)*m*Physics[`^`](v[n], 2)), v[k])/`&hbar;`

F[k](X) = I*m*%Commutator((1/2)*m*Physics:-`^`(v[n], 2), v[k])/`&hbar;`

(17)

Simplify this expression for the quantized force taking the quantum algebra rules (16) into account

Simplify(F[k](X) = I*m*%Commutator((1/2)*m*Physics[`^`](v[n], 2), v[k])/`&hbar;`)

F[k](X) = (1/2)*q*(-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])+Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))+Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X])))

(18)

It is not difficult to verify that this is the antisymmetrized vector product `&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`). Departing from `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))` = `&x`(VectorCalculus[Nabla], `#mover(mi("A",mathcolor = "olive"),mo("&rarr;"))`) expressed using tensor notation,

B[c](X) = LeviCivita[c, n, m]*d_[n](A[m](X))

B[c](X) = -Physics:-LeviCivita[c, m, n]*Physics:-d_[n](A[m](X), [X])

(19)

and taking into acount that

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`), k) = `&epsilon;`[b, c, k]*v[b]*B[c](X) 

multiply both sides of (19) by `&epsilon;`[b, c, k]*v[b], getting

LeviCivita[k, b, c]*v[b]*(B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = -Physics:-LeviCivita[b, c, k]*Physics:-LeviCivita[c, m, n]*Physics:-`*`(v[b], Physics:-d_[n](A[m](X), [X]))

(20)

Simplify(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = -Physics[LeviCivita][b, c, k]*Physics[LeviCivita][c, m, n]*Physics[`*`](v[b], Physics[d_][n](A[m](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[m], Physics:-d_[k](A[m](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(21)

Finally, replacing the repeated index m by n 

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[m], Physics[d_][k](A[m](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(22)

Likewise, for

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`), k) = `&epsilon;`[b, c, k]*B[b]*B[c](X) 

multiplying (19), this time from the right instead of from the left, we get

Simplify(((B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))*LeviCivita[k, b, c])*v[b])

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[m](X), [X]), v[m])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(23)

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[m](X), [X]), v[m])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(24)

Simplifying now the expression (18) for the quantized force taking into account (22) and (24) we get

simplify(F[k](X) = (1/2)*q*(-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n])+Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))+Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))), {(rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))), (rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))})

F[k](X) = (1/2)*q*Physics:-LeviCivita[b, c, k]*(Physics:-`*`(v[b], B[c](X))+Physics:-`*`(B[c](X), v[b]))

(25)

i.e.

`#mover(mi("F",mathcolor = "olive"),mo("&rarr;"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("&rarr;"))`, `#mover(mi("v",mathcolor = "olive"),mo("&rarr;"))`))

in tensor notation. Finally, we note that this operator is Hermitian as expected

(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))-Dagger(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))

F[k](X)-Physics:-Dagger(F[k](X)) = 0

(26)



Download:  Quantization_of_the_Lorentz_force.mw,   Quantization_of_the_Lorentz_force.pdf


Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Editor, Computer Physics Communications

The presentation below is on undergrad Quantum Mechanics. Tackling this topic within a computer algebra worksheet in the way it's done below, however, is an exciting novelty and illustrates well the level of abstraction that is now possible using the Physics package.

 

Quantum Mechanics: Schrödinger vs Heisenberg picture

Pascal Szriftgiser1 and Edgardo S. Cheb-Terrab2 

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

 

Within the Schrödinger picture of Quantum Mechanics, the time evolution of the state of a system, represented by a Ket "| psi(t) >", is determined by Schrödinger's equation:

I*`&hbar;`*(diff(Ket(psi, t), t)) = H*Ket(psi, t)

where H, the Hamiltonian, as well as the quantum operators O__S representing observable quantities, are all time-independent.

 

Within the Heisenberg picture, a Ket Ket(psi, 0) representing the state of the system does not evolve with time, but the operators O__H(t)representing observable quantities, and through them the Hamiltonian H, do.

 

Problem: Departing from Schrödinger's equation,

  

a) Show that the expected value of a physical observable in Schrödinger's and Heisenberg's representations is the same, i.e. that

Bra(psi, t)*O__S*Ket(psi, t) = Bra(psi, 0)*O__H(t)*Ket(psi, 0)

  

b) Show that the evolution equation of an observable O__H in Heisenberg's picture, equivalent to Schrödinger's equation,  is given by:

diff(O__H(t), t) = (-I*Physics:-Commutator(O__H(t), H))*(1/`&hbar;`)

where in the right-hand-side we see the commutator of O__H with the Hamiltonian of the system.

Solution: Let O__S and O__H respectively be operators representing one and the same observable quantity in Schrödinger's and Heisenberg's pictures, and H be the operator representing the Hamiltonian of a physical system. All of these operators are Hermitian. So we start by setting up the framework for this problem accordingly, including that the time t and Planck's constant are real. To automatically combine powers of the same base (happening frequently in what follows) we also set combinepowersofsamebase = true. The following input/output was obtained using the latest Physics update (Aug/31/2016) distributed on the Maplesoft R&D Physics webpage.

with(Physics):

Physics:-Setup(hermitianoperators = {H, O__H, O__S}, realobjects = {`&hbar;`, t}, combinepowersofsamebase = true, mathematicalnotation = true)

[combinepowersofsamebase = true, hermitianoperators = {H, O__H, O__S}, mathematicalnotation = true, realobjects = {`&hbar;`, t}]

(1)

Let's consider Schrödinger's equation

I*`&hbar;`*(diff(Ket(psi, t), t)) = H*Ket(psi, t)

I*`&hbar;`*(diff(Physics:-Ket(psi, t), t)) = Physics:-`*`(H, Physics:-Ket(psi, t))

(2)

Now, H is time-independent, so (2) can be formally solved: psi(t) is obtained from the solution psi(0) at time t = 0, as follows:

T := exp(-I*H*t/`&hbar;`)

exp(-I*t*H/`&hbar;`)

(3)

Ket(psi, t) = T*Ket(psi, 0)

Physics:-Ket(psi, t) = Physics:-`*`(exp(-I*t*H/`&hbar;`), Physics:-Ket(psi, 0))

(4)

To check that (4) is a solution of (2), substitute it in (2):

eval(I*`&hbar;`*(diff(Physics[Ket](psi, t), t)) = Physics[`*`](H, Physics[Ket](psi, t)), Physics[Ket](psi, t) = Physics[`*`](exp(-I*H*t/`&hbar;`), Physics[Ket](psi, 0)))

Physics:-`*`(H, exp(-I*t*H/`&hbar;`), Physics:-Ket(psi, 0)) = Physics:-`*`(H, exp(-I*t*H/`&hbar;`), Physics:-Ket(psi, 0))

(5)

Next, to relate the Schrödinger and Heisenberg representations of an Hermitian operator O representing an observable physical quantity, recall that the value expected for this quantity at time t during a measurement is given by the mean value of the corresponding operator (i.e., bracketing it with the state of the system Ket(psi, t)).

So let O__S be an observable in the Schrödinger picture: its mean value is obtained by bracketing the operator with equation (4):

Dagger(Ket(psi, t) = Physics[`*`](exp(-I*H*t/`&hbar;`), Ket(psi, 0)))*O__S*(Ket(psi, t) = Physics[`*`](exp(-I*H*t/`&hbar;`), Ket(psi, 0)))

Physics:-`*`(Physics:-Bra(psi, t), O__S, Physics:-Ket(psi, t)) = Physics:-`*`(Physics:-Bra(psi, 0), exp(I*t*H/`&hbar;`), O__S, exp(-I*t*H/`&hbar;`), Physics:-Ket(psi, 0))

(6)

The composed operator within the bracket on the right-hand-side is the operator O in Heisenberg's picture, O__H(t)

Dagger(T)*O__S*T = O__H(t)

Physics:-`*`(exp(I*t*H/`&hbar;`), O__S, exp(-I*t*H/`&hbar;`)) = O__H(t)

(7)

Analogously, inverting this equation,

(T*(Physics[`*`](exp(I*H*t/`&hbar;`), O__S, exp(-I*H*t/`&hbar;`)) = O__H(t)))*Dagger(T)

O__S = Physics:-`*`(exp(-I*t*H/`&hbar;`), O__H(t), exp(I*t*H/`&hbar;`))

(8)

As an aside to the problem, we note from these two equations, and since the operator T = exp((-I*H*t)*(1/`&hbar;`)) is unitary (because H is Hermitian), that the switch between Schrödinger's and Heisenberg's pictures is accomplished through a unitary transformation.

 

Inserting now this value of O__S from (8) in the right-hand-side of (6), we get the answer to item a)

lhs(Physics[`*`](Bra(psi, t), O__S, Ket(psi, t)) = Physics[`*`](Bra(psi, 0), exp(I*H*t/`&hbar;`), O__S, exp(-I*H*t/`&hbar;`), Ket(psi, 0))) = eval(rhs(Physics[`*`](Bra(psi, t), O__S, Ket(psi, t)) = Physics[`*`](Bra(psi, 0), exp(I*H*t/`&hbar;`), O__S, exp(-I*H*t/`&hbar;`), Ket(psi, 0))), O__S = Physics[`*`](exp(-I*H*t/`&hbar;`), O__H(t), exp(I*H*t/`&hbar;`)))

Physics:-`*`(Physics:-Bra(psi, t), O__S, Physics:-Ket(psi, t)) = Physics:-`*`(Physics:-Bra(psi, 0), O__H(t), Physics:-Ket(psi, 0))

(9)

where, on the left-hand-side, the Ket representing the state of the system is evolving with time (Schrödinger's picture), while on the the right-hand-side the Ket `&psi;__0`is constant and it is O__H(t), the operator representing an observable physical quantity, that evolves with time (Heisenberg picture). As expected, both pictures result in the same expected value for the physical quantity represented by O.

 

To complete item b), the derivation of the evolution equation for O__H(t), we take the time derivative of the equation (7):

diff((rhs = lhs)(Physics[`*`](exp(I*H*t/`&hbar;`), O__S, exp(-I*H*t/`&hbar;`)) = O__H(t)), t)

diff(O__H(t), t) = I*Physics:-`*`(H, exp(I*t*H/`&hbar;`), O__S, exp(-I*t*H/`&hbar;`))/`&hbar;`-I*Physics:-`*`(exp(I*t*H/`&hbar;`), O__S, H, exp(-I*t*H/`&hbar;`))/`&hbar;`

(10)

To rewrite this equation in terms of the commutator  Physics:-Commutator(O__S, H), it suffices to re-order the product  H  exp(I*H*t/`&hbar;`) placing the exponential first:

Library:-SortProducts(diff(O__H(t), t) = I*Physics[`*`](H, exp(I*H*t/`&hbar;`), O__S, exp(-I*H*t/`&hbar;`))/`&hbar;`-I*Physics[`*`](exp(I*H*t/`&hbar;`), O__S, H, exp(-I*H*t/`&hbar;`))/`&hbar;`, [exp(I*H*t/`&hbar;`), H], usecommutator)

diff(O__H(t), t) = I*Physics:-`*`(exp(I*t*H/`&hbar;`), H, O__S, exp(-I*t*H/`&hbar;`))/`&hbar;`-I*Physics:-`*`(exp(I*t*H/`&hbar;`), Physics:-`*`(H, O__S)+Physics:-Commutator(O__S, H), exp(-I*t*H/`&hbar;`))/`&hbar;`

(11)

Normal(diff(O__H(t), t) = I*Physics[`*`](exp(I*H*t/`&hbar;`), H, O__S, exp(-I*H*t/`&hbar;`))/`&hbar;`-I*Physics[`*`](exp(I*H*t/`&hbar;`), Physics[`*`](H, O__S)+Physics[Commutator](O__S, H), exp(-I*H*t/`&hbar;`))/`&hbar;`)

diff(O__H(t), t) = -I*Physics:-`*`(exp(I*t*H/`&hbar;`), Physics:-Commutator(O__S, H), exp(-I*t*H/`&hbar;`))/`&hbar;`

(12)

Finally, to express the right-hand-side in terms of  Physics:-Commutator(O__H(t), H) instead of Physics:-Commutator(O__S, H), we take the commutator of the equation (8) with the Hamiltonian

Commutator(O__S = Physics[`*`](exp(-I*H*t/`&hbar;`), O__H(t), exp(I*H*t/`&hbar;`)), H)

Physics:-Commutator(O__S, H) = Physics:-`*`(exp(-I*t*H/`&hbar;`), Physics:-Commutator(O__H(t), H), exp(I*t*H/`&hbar;`))

(13)

Combining these two expressions, we arrive at the expected result for b), the evolution equation of a given observable O__H in Heisenberg's picture

eval(diff(O__H(t), t) = -I*Physics[`*`](exp(I*H*t/`&hbar;`), Physics[Commutator](O__S, H), exp(-I*H*t/`&hbar;`))/`&hbar;`, Physics[Commutator](O__S, H) = Physics[`*`](exp(-I*H*t/`&hbar;`), Physics[Commutator](O__H(t), H), exp(I*H*t/`&hbar;`)))

diff(O__H(t), t) = -I*Physics:-Commutator(O__H(t), H)/`&hbar;`

(14)


Download:    Schrodinger_vs_Heisenberg_picture.mw     Schrodinger_vs_Heisenberg_picture.pdf

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Editor, Computer Physics Communications

I am solving "Fx=0" for geting "roots:x" using "solve(Fx,x)". Solution is in the form of "a+sqrt(b)", "a-sqrt(b)". One solution "f1" is given below.

f1:=1/2*(-8*R*d1^2*r^2*C+10*d1*r^2*C*R+5*d1*r^3*C+2*r*L*d1^2-2*C*r^3+2*R*L*d1^2-R*L*d1^3-r*L*d1^3-4*C*r^2*R+2*R*d1^3*r^2*C-4*r^3*d1^2*C+r^3*d1^3*C+sqrt(26*r^6*d1^4*C^2+41*r^6*d1^2*C^2-44*r^6*d1^3*C^2-20*C^2*r^6*d1+16*C^2*r^5*R-16*C*r^4*L-176*r^5*d1^3*C^2*R+164*r^5*d1^2*C^2*R-74*r^4*d1^4*C*L+136*r^4*d1^3*C*L-136*r^4*d1^2*C*L-80*C^2*r^5*d1*R+72*C*r^4*L*d1-64*C*r^3*R*L+104*R^2*d1^4*r^4*C^2-176*R^2*d1^3*r^4*C^2+164*R^2*d1^2*r^4*C^2-8*r^6*d1^5*C^2+r^2*L^2*d1^6-4*R^2*L^2*d1^5+104*r^5*d1^4*C^2*R+40*r*L*R^3*d1^5*C-72*r*L*R^3*d1^4*C+56*r*L*R^3*d1^3*C-16*r*L*R^3*d1^2*C+R^2*L^2*d1^6+20*r^4*L*C*d1^5-32*r^4*d1^5*C^2*R^2+2*R*L^2*d1^6*r-2*r^4*L*d1^6*C+4*R^2*d1^6*r^4*C^2+4*R*d1^6*r^5*C^2-306*r^3*d1^4*C*R*L+548*r^3*d1^3*C*L*R-544*r^3*d1^2*C*R*L+288*C*r^3*L*d1*R+16*C^2*r^4*R^2+4*R^2*L^2*d1^4-16*R^2*L*d1^6*r^2*C-10*R*L*d1^6*r^3*C+r^6*d1^6*C^2-32*r^5*d1^5*C^2*R-4*r^2*d1^5*L^2-352*R^2*d1^4*r^2*C*L+580*R^2*d1^3*r^2*C*L-552*R^2*d1^2*r^2*C*L-80*d1*r^4*C^2*R^2-8*R^3*d1^6*L*C*r+88*r^3*L*C*d1^5*R+116*r^2*L*R^2*d1^5*C+4*C^2*r^6-8*r*R*L^2*d1^5+288*d1*r^2*C*R^2*L-64*C*r^2*R^2*L+8*r*L^2*d1^4*R+4*r^2*L^2*d1^4)^(1/2))/(-3*r^2*d1*L*C-6*R*d1*L*C*r+2*L*C*r^2+r^2*d1^2*L*C+4*L*C*r*R+2*R*d1^2*L*C*r);

I used the following Maple syntax

patmatch(f1,XT::algebraic+sqrt(YT::algebraic),'q1');

the answer is "false"

Is there any modification in the syntax "patmatch" is required.

Here, my question is how to separate "a" and "b" in "a+sqrt(b)" (a, b are big expressions involving many variables).

Thanking you advance for your help.

 

What is the funcional difference between individual, academic, and professional editions of Maple?

Hi There,

Im currently on using a trial version of maple for engineering students. I am looking to see how you can view a step by step method of solving equations. I can't seem to find a way of doing so and I keep just getting the final answer. Can you assist?

 

Also I can't find the quadratic equation option.

 

thanks,

 

stuart

Dear All,

I have a problem solving the attached nonlinear system of equations using shooting method.
I will be grateful if you could help me finding the solutions out.

 

restart; Shootlib := "C:/Shoot9"; libname := Shootlib, libname; with(Shoot);
with(plots);
N1 := 1.0; N2 := 2.0; N3 := .5; Bt := 6; Re_m := N1*Bt; gamma1 := 1;
FNS := {f(eta), fp(eta), fpp(eta), g(eta), gp(eta), m(eta), mp(eta), n(eta), np(eta), fppp(eta)};
ODE := {diff(f(eta), eta) = fp(eta), diff(fp(eta), eta) = fpp(eta), diff(fpp(eta), eta) = fppp(eta), diff(g(eta), eta) = gp(eta), diff(gp(eta), eta) = N1*(2.*g(eta)+(eta-2.*f(eta)).gp(eta)+2.*g(eta)*fp(eta)+2.*N2.N3.(m(eta).np(eta)-n(eta).mp(eta))), diff(m(eta), eta) = mp(eta), diff(mp(eta), eta) = Re_m.(m(eta)+(eta-2.*f(eta)).mp(eta)+2.*m(eta)*fp(eta)), diff(n(eta), eta) = np(eta), diff(np(eta), eta) = Re_m.(2.*n(eta)+(eta-2.*f(eta)).np(eta)+2.*N2/N3.m(eta).gp(eta)), diff(fppp(eta), eta) = N1*(3.*fpp(eta)+(eta-2.*f(eta)).fppp(eta)-2.*N2.N2.m(eta).(diff(mp(eta), eta)))};
blt := 1.0; IC := {f(0) = 0, fp(0) = 0, fpp(0) = alpha1, g(0) = 1, gp(0) = beta1, m(0) = 0, mp(0) = beta2, n(0) = 0, np(0) = beta3, fppp(0) = alpha2};
BC := {f(blt) = .5, fp(blt) = 0, g(blt) = 0, m(blt) = 1, n(blt) = 1};
infolevel[shoot] := 1;
S := shoot(ODE, IC, BC, FNS, [alpha1 = 1.425, alpha2 = .425, beta1 = -1.31, beta2 = 1.00, beta3 = 1.29]);
Error, (in isolate) cannot isolate for a function when it appears with different arguments
p := odeplot(S, [eta, fp(eta)], 0 .. 15);
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution
display(p);
Error, (in plots:-display) expecting plot structure but received: p
p2 := odeplot(S, [eta, theta(eta)], 0 .. 10);
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution
display(p2);
Error, (in plots:-display) expecting plot structure but received: p2

 

 

      General description of the method of solving underdetermined systems of equations. As a particular application of the idea proposed a universal method  kinematic analysis for all kinds of linkage (lever) mechanisms. With the description and examples.
      The method can be used for powerful CAD linkages.

Description: Calculation_method_of_linkages.pdf

Attachment:
figure_1.mw
figure_2.mw

Or all in one
Calculation_method_of_linkages_(with_attach.).pdf


        Some examples of a much larger number calculated by the proposed method. Examples gathered here not to look for them on the forum and opportunity to demonstrate the method.  Among the examples, I think, there are very complicated.

https://vk.com/doc242471809_408704758
https://vk.com/doc242471809_408704572
https://vk.com/doc242471809_376439263
https://vk.com/doc242471809_402619761
https://vk.com/doc242471809_402610228
https://vk.com/doc242471809_401188803
https://vk.com/doc242471809_400465891
https://vk.com/doc242471809_400711315
https://vk.com/doc242471809_387358164
https://vk.com/doc242471809_380837279
https://vk.com/doc242471809_379935473
https://vk.com/doc242471809_380217387
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 Hello,every one,i want to solve system of equations but i recieve an error ,how can i find the coeffecients c1,c2,c3,c4?thank.

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restart

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A := 45*x*c4+72*c3 = 0:

 

B := 56*c2*c4+28*c3^2 = 0:

C := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c4+42*beta*c1*c4+42*beta*c2*c3 = 0:

E := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c3+30*beta*c1*c3+15*beta*c2 = 0:

F := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c2+20*beta*c1*c2-20*beta*c1*c4-20*beta*c2*c4-20*beta*c3*c4-20*beta*c4^2-20*beta*c4-20*c4 = 0:

G := M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n*c1+6*beta*c1^2-12*beta*c1*c3-12*beta*c2*c3-12*beta*c3^2-12*beta*c3*c4-12*beta*c3-12*c3 = 0:

``

beta*c1+beta*c2^2+beta*c2*c3+beta*c2*c4+beta*c2+c2 = 0:

M^2*(-x^5*c4-x^4*c3-x^3*c2-x^2*c1+c1+c2+c3+c4+1)^n = 0:

with(SolveTools):

``

PolynomialSystem({{45*c4*x+72*c3 = 0}, {30*beta*c1*c3+15*beta*c2 = 0}, {42*beta*c1*c4+42*beta*c2*c3 = 0}, {20*beta*c1*c2-20*beta*c1*c4-20*beta*c2*c4-20*beta*c3*c4-20*beta*c4^2-20*beta*c4-20*c4 = 0}, {6*beta*c1^2-12*beta*c1*c3-12*beta*c2*c3-12*beta*c3^2-12*beta*c3*c4-12*beta*c3-12*c3 = 0}}, {c1, c2, c3, c4}, {beta = 2, x = 1/5})

Error, invalid input: too many and/or wrong type of arguments passed to SolveTools:-PolynomialSystem; first unused argument is {beta = 2, x = 1/5}

 

NULL

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Hello,how can i find the lambda in this equation? and x=0..2 , t=0..2

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