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I am thinking about buying maple 2017 however there are only 4 different categories to choose from when you want to buy: student, commercial, academic and government. I dont belong to any of them! Also the price difference is huge! I am on disability benefits and the academic license cost more or less the same amount that I get in disability benifts each month to cover my food, rent and medicines which is approximatly 1 100 usd. The price for a student licens is completely realistic and is a price that I am willing to pay but I am not a student and I dont feel comfortable claiming that I am even though everyone is a student as long as they live. When you stop learning you life is more or less over anyway. If I am forced to pay around 1 000 usd then I am not going to buy maple 2017. Then I am just going to continue using MathPapa free algbra calculator https://www.mathpapa.com/ because to be honest I dont really need maple that much in my research today but there are a couple of reaons why I want to buy. 1) I would like to support maplesoft because I think you have the best and most userfriendly mathematical software on the market. 2) I want to hedge my bets. My needs might change in the future. 3) I want to be able to run my large number of old worksheets and see if I can improve them. 4) I want to see what changes and improvments have been made to maple 2017 compared to let say 5 years ago and to assess if these changes provide any value to me. 

Everything is simple, until you go underwater – This is what the University of Waterloo Submarine Racing team, or in short ‘WatSub’ coined as their motto. Never mind learning to scuba dive, and dealing with such things as rust, this newly formed team would have to compete against university teams with a decade or more of experience.

But that did not deter the team, and they started work on Ontario’s first submarine racing project. The team approached Maplesoft to be a sponsor and we are proud to have supported this ingenious venture. The team has used Maplesoft technology in the design and testing of the submarine.

“Maple has been our go-to calculations and analysis tool throughout the development of Amy (2015-2016 season), and we will continue using it throughout the development of Bolt (2016-2017 season),” said Gonzalo Espinoza Graham, President of the WatSub Team. “Its familiar interface and computing environment allowed us to set design benchmark targets from early on the design process and follow through with them on the later stage.”

What started as an engineering project in December 2014, becoming officially the first submarine racing team in Ontario. The team soon grew to over 130 general members and a tight core-team, who were eager to tackle new challenges.  The team resides inside the Sedra Student Design Centre, University of Waterloo’s state of the art facility that houses over 25 student teams, the largest of its kind in North America.  

WatSub made its first appearance on the European International Submarine Races (eISR) back in July 2016, with its 1st submarine ‘Amy’, where a single scuba diver piloted the submarine and propelled it through an unforgiving winding course marked by obstacles and turns 10 meters underwater. The team has since then participated in other competitions and is constantly improving the design and performance of the submarine, learning from each competition they participate in.  Next year Amy will participate in the 14th edition of the eISR international competition. “I think the greatest thing we learned is never to give up,” said Ana Krstanovic, a third-year political science student who manages communications for the team. “We’re more motivated now than ever.”

 

Ojaswi Tagore, Gonzalo Espinoza Graham, and Janna Henzl represented WatSub at the European International Submarine Race in Gosport, UK.

 

Another example of an innovative project that Maplesoft supported in 2016 is Waterloop: The Canadian SpaceX Hyperloop Competition Team, Canada's only SpaceX Hyperloop Pod Competition team. This project, which could change the way we travel in the future, is driven by a group of dedicated University of Waterloo students who have taken on the challenge to design and build a functional prototype Hyperloop pod. They will test it on a one-mile test track in Hawthorne, California in January 2017, pitting it against 22 of the 1200+ teams who originally entered the competition.

The Hyperloop is a conceptual next generation high-speed transit system that will take commuters between cities at speeds over 1,000 km/h. The technology will differ from previous rail transit by having pods ride on a cushion of air in a reduced pressure tube in order to reach greater speeds with a smoother ride, and is powered entirely by renewable energy.

 The Hyperloop Pod Competition was launched by Elon Musk, the billionaire engineer and founder of SpaceX and Tesla Motors.  The competition is separated into 3 rounds. The first one was held in late December, where selected teams sent in their initial designs to be reviewed. From there, 180 teams were chosen to compete at Texas A&M University. Each team set up a booth and a panel of judges critiqued them and chose 31 teams to move onto the final, build and test stage.

Waterloop Goose I

Waterloop Goose X

The GOOSE I is Waterloop’s half-scale, functional prototype vehicle pod, which will be the one in the competition.  The GOOSE X pod is a conceptual full size Hyperloop vehicle inspired by the prototype they are building. The full size pod will have a capacity of 26 passengers per pod.

"Our prototype has been designed to be as simple and economical as possible, while still performing all necessary functions for the full size Hyperloop. If it is successful, it has the potential to revolutionize the transit industry in the same manner the train and airplane has before it," said Montgomery de Luna, architectural design lead for Waterloop. “We would like to thank Maplesoft for their generous support.  Without sponsors like Maplesoft supporting our vision and encouraging innovative student projects, we wouldn’t be able to achieve our goal.”

Revolutionizing the transportation industry isn’t easy and is at times frustrating and time consuming for these teams, but having the best tools and resources will ensure that the teams have a good chance at excelling in competitions and creating innovative models that could change our future.

Students using Maple often have different needs than non-students. Students need more than just a final answer; they are looking to gain an understanding of the mathematical concepts behind the problems they are asked to solve and to learn how to solve problems. They need an environment that allows them to explore the concepts and break problems down into smaller steps.

The Student packages in Maple offer focused learning environments in which students can explore and reinforce fundamental concepts for courses in Precalculus, Calculus, Linear Algebra, Statistics, and more. For example, Maple includes step-by-step tutors that allow students to practice integration, differentiation, the finding of limits, and more. The Integration Tutor, shown below, lets a student evaluate an integral by selecting an applicable rule at each step. Maple will also offer hints or show the next step, if asked.  The tutor doesn't only demonstrate how to obtain the result, but is designed for practicing and learning.

For this blog post, I’d like to focus in on an area of great interest to students: showing step-by-step solutions for a variety of problems in Maple.

Several commands in the Student packages can show solution steps as either output or inline in an interactive pop-up window. The first few examples return the solution steps as output.

Precalculus problems:

The Student:-Basics sub-package provides a collection of commands to help students and teachers explore fundamental mathematical concepts that are core to many disciplines. It features two commands, both of which return step-by-step solutions as output.

The ExpandSteps command accepts a product of polynomials and displays the steps required to expand the expression:

with(Student:-Basics):
ExpandSteps( (a^2-1)/(a/3+1/3) );

The LinearSolveSteps command accepts an equation in one variable and displays the steps required to solve for that variable.

with(Student:-Basics):
LinearSolveSteps( (x+1)/y = 4*y^2 + 3*x, x );

This command also accepts some nonlinear equations that can be reduced down to linear equations.

Calculus problems:

The Student:-Calculus1 sub-package is designed to cover the basic material of a standard first course in single-variable calculus. Several commands in this package provide interactive tutors where you can step through computations and step-by-step solutions can be returned as standard worksheet output.

Tools like the integration, differentiation, and limit method tutors are interactive interfaces that allow for exploration. For example, similar to the integration-methods tutor above, the differentiation-methods tutor lets a student obtain a derivative by selecting the appropriate rule that applies at each step or by requesting a complete solution all at once. When done, pressing “Close” prints out to the Maple worksheet an annotated solution containing all of the steps.

For example, try entering the following into Maple:

with(Student:-Calculus1):
x*sin(x);

Next, right click on the Matrix and choose “Student Calculus1 -> Tutors -> Differentiation Methods…

The Student:-Calculus1 sub-package is not alone in offering this kind of step-by-step solution finding. Other commands in other Student packages are also capable of returning solutions.

Linear Algebra Problems:

The Student:-LinearAlgebra sub-package is designed to cover the basic material of a standard first course in linear algebra. This sub-package features similar tutors to those found in the Calculus1 sub-package. Commands such as the Gaussian EliminationGauss-Jordan Elimination, Matrix Inverse, Eigenvalues or Eigenvectors tutors show step-by-step solutions for linear algebra problems in interactive pop-up tutor windows. Of these tutors, a personal favourite has to be the Gauss-Jordan Elimination tutor, which were I still a student, would have saved me a lot of time and effort searching for simple arithmetic errors while row-reducing matrices.

For example, try entering the following into Maple:

with(Student:-LinearAlgebra):
M:=<<77,9,31>|<-50,-80,43>|<25,94,12>|<20,-61,-48>>;

Next, right click on the Matrix and choose “Student Linear Algebra -> Tutors -> Gauss-Jordan Elimination Tutor

This tutor makes it possible to step through row-reducing a matrix by using the controls on the right side of the pop-up window. If you are unsure where to go next, the “Next Step” button can be used to move forward one-step. Pressing “All Steps” returns all of the steps required to row reduce this matrix.

When this tutor is closed, it does not return results to the Maple worksheet, however it is still possible to use the Maple interface to step through performing elementary row operations and to capture the output in the Maple worksheet. By loading the Student:-LinearAlgebra package, you can simply use the right-click context menu to apply elementary row operations to a Matrix in order to step through the operations, capturing all of your steps along the way!

An interactive application for showing steps for some problems:

While working on this blog post, it struck me that we did not have any online interactive applications that could show solution steps, so using the commands that I’ve discussed above, I authored an application that can expand, solve linear problems, integrate, differentiate, or find limits. You can interact with this application here, but note that this application is a work in progress, so feel free to email me (maplepm (at) Maplesoft.com) any strange bugs that you may encounter with it.

More detail on each of these commands can be found in Maple’s help pages.

 

I downloaded a maplesoft apps, and wat to activate, an activeation form displayed requiring me to put in a purchase code of which i don't know or have. How can i get it or where can i find it? please some one help.

Walking into the big blue Maplesoft office on August 3rd was a bit nerve wracking. I had no idea who anyone was, what to expect, or even what I would be doing. As I sat in the front hall waiting for someone to receive me, I remember thinking, “What have I gotten myself into?”. Despite my worries on that first day, interning at Maplesoft has been a great experience! I never knew that I would be able to learn so much about programming and working in a company in such a short amount of time. Although Maple was a programming language that was foreign to me a couple weeks ago, I feel like I’m relatively well versed in it now. Trying to learn a new language in this short timespan hasn’t been easy, but I think that I picked it up quickly, even if I’ve had my fair share of frustrations.

Chaos Game example on Rosetta Code

At Maplesoft, I’ve been contributing to the Rosetta Code project by writing short programs using Maple. The Rosetta Code project is dedicated to creating programming examples for many different tasks in different programming languages. My summer project has been to create solutions using Maple for as many tasks as possible and to post these to Rosetta Code; the goal being to have the list of tasks without Maple implementation shrink with each passing day. It’s nice to feel like I’m leaving a mark in this world, even if it is in such a small corner of the internet.

Flipping Bits example on Rosetta Code/MapleCloud

This internship, of course, came with its share of challenges. During my work on the Rosetta Code project, I posted solutions for a total of 38 tasks. Some of them were easy, but some of them took days to complete. On some days, I felt like I was on top of the world. Everything I made turned out great and I knew exactly how to tackle each problem. Other days were slower. I’ve spent ages just staring at a computer monitor trying to figure out just how on earth I was going to make this machine do what I wanted it to do! The 24 Game task was particularly hard, but also very educational. Through this task, I learned about modules, a concept previously unknown to me. I’m fairly sure that the 24 Game also took me the longest, whereas the Increment a numerical string task took me no time at all. Despite it being easy, the Increment a numerical string task wasn’t particularly fun; a bit of a challenge is required for something to be entertaining, after all. My personal favourite was the Fibonacci n-step number sequences task. It was the first really challenging task I encountered, and for after which the feeling of finally completing a task that I spent so long on, of finally overcoming that mountain, was extremely satisfying. Not all challenges end in satisfaction, however. I often found myself accidentally doing something that made the window freeze. I would close the program, then cry a bit on the inside when I realized I just lost the past half an hour’s worth of unsaved work. Nevertheless, I’m glad I got to face all these obstacles because they have made me more resilient and a better programmer.

The following is the code for the Fibonacci n-step number sequences task

numSequence := proc(initValues :: Array)
	local n, i, values;
n := numelems(initValues);
values := copy(initValues);
for i from (n+1) to 15 do
values(i) := add(values[i-n..i-1]);
end do;
return values;
end proc:
 
initValues := Array([1]):
for i from 2 to 10 do
initValues(i) := add(initValues):
printf ("nacci(%d): %a\n", i, convert(numSequence(initValues), list));
end do:
printf ("lucas: %a\n", convert(numSequence(Array([2, 1])), list));

Maple was a great software to program with and a fairly straightforward language to learn. Having previously programmed in Java, I found Maple similar enough that transitioning wasn’t too difficult. In fact, every once in a while when I didn`t know what to do for a task, I would take a look at the Java example in Rosetta Code and it would point me in a direction or give me some hints. While the two languages are similar, there are still many differences. For example, I liked the fact that in Maple, lists started at an index of 1 rather than 0 and arrays could an arbitrary starting index. Although it was different from what I was used to, I found that it made many things much less confusing. Another thing I liked was that the for loop syntax was very simple. I never once had to run through in my head how many times something would loop for. There were such a wide variety of commands in Maple. There was a command for practically anything, and if you knew that it existed and how to use it, then so much power could be at your fingertips. This is where the help system came in extremely handy. With a single search you might find that the solution to the exact problem you were trying to solve already existed as a Maple command. I always had a help window open when I was using Maple.

Multiplication Tables example on Rosetta Code

Spending my summer coding at Maplesoft has been fun, sometimes challenging, but an overall rewarding experience. Through contributing to the Rosetta Code project, I’ve learned so much about computer programming, and it certainly made the 45 minute drive out to Waterloo worth it!

Yili Xu,
Maplesoft SHAD Intern

I am having 26th degree polynomial univariate equation , I used Isolate to get the roots. but I am getting some extra roots which are not true they I even tried to substitute those roots in original equation then I got non zero answer instead of getting nearly zero answer.How is it possible??

 

equation looks like:

-12116320194738194778134937600000000*t^26+167589596741213731838990745600000000*t^24+1058345691529498270472972795904000000*t^22-4276605572538658673086219419648000000*t^20-23240154739806540070988490473472000000*t^18-5442849111209103187871341215744000000*t^16+49009931453396028716875310432256000000*t^14+74247033158233643322704589225984000000*t^12-2762178990802317464801412907008000000*t^10-25947900993773120244883450232832000000*t^8-7468990043547273070742668836864000000*t^6-567730116675454293925108383744000000*t^4+3703566799705707258760396800000000*t^2-4742330812072533924249600000000

Solutions i got:

[t = -4.162501845, t = -2.295186769, t = -1.300314688, t = -.8048430445, t = -0.6596008501e-1, t = -0.4212510777e-1, t = 0.4212510777e-1, t = 0.6596008501e-1, t = .8048430445, t = 1.300314688, t = 2.295186769, t = 4.162501845]

t=4.162501845 give me non zero answer when I substitute it in the equation given above:

I got this answer: 4.750212083*10^39

 

Dear all

I have a simple question about geometric series

What's the difference between g(2) and gg(2)

 Many thanks

 

 

Geometric_series.mw

 

Hi,

I did some hypothesis testing exercises and I cross checked the result with Maple. I just used following vectors for an unpaired test

a := [88, 89, 92, 90, 90];
b := [92, 90, 91, 89, 91];

I ended up with the following solution:

HFloat(1.5225682336585966)
HFloat(-3.122568233658591)
for a 0.95 confidence interval.

 

Using

TwoSampleTTest(a, b, 0, confidence = .95, summarize = embed)

and

TwoSampleTTest(a, b, 0, confidence = .975, summarize = embed)

I get following results:

-2.75177 .. 1.15177

-3.13633 .. 1.53633

respectively. I can not explain the discrepancy.

 

Best regards,

Oliver

 

PS:

Maple Code in case files won´t be attached.

 

 

Unpaired t Test
restart;
Unpaired test-test dataset
a := [88, 89, 92, 90, 90];
b := [92, 90, 91, 89, 91];
The se² estimate is given by:
se²=var(a)+var(b)+2*cov(a*b)=var(a)+var(b)
se²=
sigma[a]^2/Na+sigma[b]^2/Nb;
with Na, Nb being the length of vector a and b respectively.
                             2                              2
  sigma[[88, 89, 92, 90, 90]]    sigma[[92, 90, 91, 89, 91]]
  ---------------------------- + ----------------------------
               Na                             Nb             
sigma[a]^2;
 and
sigma[b]^2;
 are approximated by
S[a]^2;
 and
S[b]^2;
                                             2
                  sigma[[88, 89, 92, 90, 90]]
                                             2
                  sigma[[92, 90, 91, 89, 91]]
                                           2
                    S[[88, 89, 92, 90, 90]]
                                           2
                    S[[92, 90, 91, 89, 91]]
with
S[X]^2;
 defined as
S[X]*`²` = (sum(X[i]-(sum(X[j], j = 1 .. N))/N, i = 1 .. N))^2/N;
                                 2
                             S[X]
                                                 2
                      /      /         N       \\
                      |      |       -----     ||
                      |  N   |        \        ||
                      |----- |         )       ||
                      | \    |        /    X[j]||
                      |  )   |       -----     ||
                      | /    |       j = 1     ||
                      |----- |X[i] - ----------||
                      \i = 1 \           N     //
             S[X] ᅡᄇ = ----------------------------
                                   N              
with(Statistics);
Sa := Variance(a);
                   HFloat(2.1999999999999993)
Sb := Variance(b);
                   HFloat(1.3000000000000003)
Now we are ready to do hypothesis testing (0.95).
We have (with k=min(Na,Nb)=5):
C = mean(a)-mean(b); Deviation := t_(alpha/a, k-1)*se(Sa/k-Sb/k);
c := Mean(a)-Mean(b); deviation := 2.776*sqrt((1/5)*Variance(a)+(1/5)*Variance(b));
                  HFloat(-0.7999999999999972)
                   HFloat(2.3225682336585938)
upperlimit := c+deviation; lowerlimit := c-deviation;
                   HFloat(1.5225682336585966)
                   HFloat(-3.122568233658591)

Execution of built in student test
TwoSampleTTest(a, b, 0, confidence = .95, summarize = embed);

 

 

Hello,

 

I tried to plot the problem presented below:

restart; with(plots); C := setcolors(); with(LinearAlgebra);

formula1 := 2.6*BodyWeight*abs(sin(4*Pi*t));
2.6 BodyWeight |sin(4 Pi t)|
BodyWeight := 80*9.81;
plot(formula1, t = 0 .. 2);


eq2 := formula1-SpringConstant*y(t)-DampConstant*(diff(y(t), t)) = Mass*(diff(y(t), `$`(t, 2)));
2040.480 |sin(4 Pi t)| - SpringConstant y(t)

/ d \ / d / d \\
- DampConstant |--- y(t)| = Mass |--- |--- y(t)||
\ dt / \ dt \ dt //
DampConstant := 50;
50
Mass := .200;
Springt := 200;
200
SpringConstant := Youngsmodulus*Surface/DeltaLength;
DeltaLength := 0.2e-1-y(t);
Surface := .15;
Youngsmodulus := 6.5*10^6/(t+1)+6.5*10^6;
plot(Youngsmodulus, t = 0 .. 10000);

eq2;
2040.480 |sin(4 Pi t)|

/ 6 \
|6.5000000 10 6|
0.15 |------------- + 6.5000000 10 | y(t)
\ t + 1 / / d \
- ----------------------------------------- - 50 |--- y(t)| =
0.02 - y(t) \ dt /

/ d / d \\
0.200 |--- |--- y(t)||
\ dt \ dt //

incs := y(0) = 0, (D(y))(0) = 0;
eq4 := dsolve({eq2, incs}, y(t), type = numeric, method = lsode[backfull], maxfun = 0);
proc(x_lsode) ... end;

plots:-odeplot(eq4, [t, y(t)], 0 .. 5);

 When I try to plot it beyond t=5, Maple gives the following error:

Warning, could not obtain numerical solution at all points, plot may be incomplete

Does anyone know how to plot it even further?

 

 

Hi everyone,

I'm kinda new here, and I really hope you guys can help me through this. In my new case study, after some revision, i thought i might be trying to implement a shooting method. I tried my best to make it work/understand but i couldn't get to any result.

So, as attached (i re-do PV Satya Naraya's paper first to be more understand but .....)

 

Here is my questions and the worksheet:

1) really stuck in mind - what is the purpose of shooting method for some related study?

2) what is the meaning of error .............'use midpoint method intead" 

3) Worksheet - 1MASS_JEFF_SATYA_on_Beta.mw

Thanks in advanced. Really hope that someone can help/teach me how to solve the boundary value problem by shooting method. 

 

 

restart; with(plots); lambda := 1.0; m := 2.0; M := 2; R := .1; Pr := .75; G := .1; Sc := .6; Kr := .2; blt := 5

Eq1 := diff(f(eta), eta, eta, eta)+(1+lambda)*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2)-(1+lambda)*M*(diff(f(eta), eta))+beta*((diff(f(eta), eta, eta))^2-f(eta)*(diff(f(eta), eta, eta, eta, eta))) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+2.0*f(eta)*(diff(diff(f(eta), eta), eta))-2.0*(diff(f(eta), eta))^2-4.0*(diff(f(eta), eta))+beta*((diff(diff(f(eta), eta), eta))^2-f(eta)*(diff(diff(diff(diff(f(eta), eta), eta), eta), eta))) = 0

(1)

``

Eq2 := (1+(4/3)*R)*(diff(theta(eta), eta, eta))+Pr*(f(eta)*(diff(theta(eta), eta))-m*(diff(f(eta), eta))*theta(eta)+G*theta(eta)) = 0;
NULL``

1.133333333*(diff(diff(theta(eta), eta), eta))+.75*f(eta)*(diff(theta(eta), eta))-1.500*(diff(f(eta), eta))*theta(eta)+0.75e-1*theta(eta) = 0

(2)

Eq3 := diff(phi(eta), eta, eta)+Sc*(f(eta)*(diff(phi(eta), eta))-m*(diff(f(eta), eta))*phi(eta)-Kr*phi(eta)) = 0;

diff(diff(phi(eta), eta), eta)+.6*f(eta)*(diff(phi(eta), eta))-1.20*(diff(f(eta), eta))*phi(eta)-.12*phi(eta) = 0

(3)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(blt) = 0, (D(D(f)))(blt) = 0, theta(0) = 1, theta(blt) = 0, phi(0) = 1, phi(blt) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(5) = 0, ((D@@2)(f))(5) = 0, theta(0) = 1, theta(5) = 0, phi(0) = 1, phi(5) = 0

(4)

L := [1.0, 1.5, 2.0, 2.5];

[1.0, 1.5, 2.0, 2.5]

(5)

for k to 4 do R := dsolve(eval({Eq1, Eq2, Eq3, bcs1}, beta = L[k]), [f(eta), theta(eta), phi(eta)], numeric, output = listprocedure); Y || k := rhs(R[3]); YA || k := rhs(R[6]); YB || k := rhs(R[5]); YC || k := -rhs(R[8]) end do

Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

 

R

 

``

 

NULL

 

Download 1MASS_JEFF_SATYA_on_Beta.mw

Hi, i am trying to solve my PDEs with HPM method ,but i get strange errors.

first one is :Error, (in trig/reduce/reduce) Maple was unable to allocate enough memory to complete this computation.  Please see ?alloc,

but when i run my last function again,the error chages,let me show you.


restart;
lambda:=0.5;K[r]:=0.5;Sc:=0.5;Nb:=0.1;Nt:=0.1;Pr:=10;
                              0.5
                              0.5
                              0.5
                              0.1
                              0.1
                               10
> EQUATIONS;


equ1:=diff(f(eta),eta$4)-R*(diff(f(eta),eta)*diff(f(eta),eta$2)-f(eta)*diff(f(eta),eta$2))-2*K[r]*diff(g(eta),eta)=0;

equ2:=diff(g(eta),eta$2)-R*(diff(f(eta),eta)*g(eta)-f(eta)*diff(g(eta),eta))+2*K[r]*diff(f(eta),eta)=0;

equ3:=diff(theta(eta),eta$2)+Pr*R*f(eta)*diff(theta(eta),eta)+Nb*diff(phi(eta),eta)*diff(theta(eta),eta)+Nt*diff(theta(eta),eta)^2=0;

equ4:=diff(phi(eta),eta$2)+R*Sc*f(eta)*diff(phi(eta),eta)+diff(theta(eta),eta$2)*(Nt/Nb)=0;
/  d   /  d   /  d   /  d         \\\\     //  d         \ /  d  
|----- |----- |----- |----- f(eta)|||| - R ||----- f(eta)| |-----
\ deta \ deta \ deta \ deta       ////     \\ deta       / \ deta

   /  d         \\          /  d   /  d         \\\
   |----- f(eta)|| - f(eta) |----- |----- f(eta)|||
   \ deta       //          \ deta \ deta       ///

         /  d         \    
   - 1.0 |----- g(eta)| = 0
         \ deta       /    
     /  d   /  d         \\
     |----- |----- g(eta)||
     \ deta \ deta       //

            //  d         \                 /  d         \\
        - R ||----- f(eta)| g(eta) - f(eta) |----- g(eta)||
            \\ deta       /                 \ deta       //

              /  d         \    
        + 1.0 |----- f(eta)| = 0
              \ deta       /    
  /  d   /  d             \\               /  d             \
  |----- |----- theta(eta)|| + 10 R f(eta) |----- theta(eta)|
  \ deta \ deta           //               \ deta           /

           /  d           \ /  d             \
     + 0.1 |----- phi(eta)| |----- theta(eta)|
           \ deta         / \ deta           /

                             2    
           /  d             \     
     + 0.1 |----- theta(eta)|  = 0
           \ deta           /     
    /  d   /  d           \\                /  d           \
    |----- |----- phi(eta)|| + 0.5 R f(eta) |----- phi(eta)|
    \ deta \ deta         //                \ deta         /

                     /  d   /  d             \\    
       + 1.000000000 |----- |----- theta(eta)|| = 0
                     \ deta \ deta           //    
> BOUNDARY*CONDITIONS;


ics:=
f(0)=0,D(f)(0)=1,g(0)=0,theta(0)=1,phi(0)=1;
f(1)=lambda,D(f)(1)=0,g(1)=0,theta(1)=0,phi(1)=0;
   f(0) = 0, D(f)(0) = 1, g(0) = 0, theta(0) = 1, phi(0) = 1
  f(1) = 0.5, D(f)(1) = 0, g(1) = 0, theta(1) = 0, phi(1) = 0
> HPMs;


hpm1:=(1-p)*(diff(f(eta),eta$4)-2*K[r]*diff(g(eta),eta))+p*(diff(f(eta),eta$4)-R*(diff(f(eta),eta)*diff(f(eta),eta$2)-f(eta)*diff(f(eta),eta$2))-2*K[r]*diff(g(eta),eta))=0;

hpm2:=(1-p)*(diff(g(eta),eta$2)+2*K[r]*diff(f(eta),eta))+p*(diff(g(eta),eta$2)-R*(diff(f(eta),eta)*g(eta)-f(eta)*diff(g(eta),eta))+2*K[r]*diff(f(eta),eta))=0;

hpm3:=(1-p)*(diff(theta(eta),eta$2))+p*(diff(theta(eta),eta$2)+Pr*R*f(eta)*diff(theta(eta),eta)+Nb*diff(phi(eta),eta)*diff(theta(eta),eta)+Nt*diff(theta(eta),eta)^2)=0;

hpm4:=(1-p)*(diff(phi(eta),eta$2)+diff(theta(eta),eta$2)*(Nt/Nb))+p*(diff(phi(eta),eta$2)+R*Sc*f(eta)*diff(phi(eta),eta)+diff(theta(eta),eta$2)*(Nt/Nb))=0;

        //  d   /  d   /  d   /  d         \\\\
(1 - p) ||----- |----- |----- |----- f(eta)||||
        \\ deta \ deta \ deta \ deta       ////

         /  d         \\     //  d   /  d   /  d   /  d         \
   - 1.0 |----- g(eta)|| + p ||----- |----- |----- |----- f(eta)|
         \ deta       //     \\ deta \ deta \ deta \ deta       /

  \\\     //  d         \ /  d   /  d         \\
  ||| - R ||----- f(eta)| |----- |----- f(eta)||
  ///     \\ deta       / \ deta \ deta       //

            /  d   /  d         \\\       /  d         \\    
   - f(eta) |----- |----- f(eta)||| - 1.0 |----- g(eta)|| = 0
            \ deta \ deta       ///       \ deta       //    
        //  d   /  d         \\       /  d         \\     //  d  
(1 - p) ||----- |----- g(eta)|| + 1.0 |----- f(eta)|| + p ||-----
        \\ deta \ deta       //       \ deta       //     \\ deta

   /  d         \\
   |----- g(eta)||
   \ deta       //

       //  d         \                 /  d         \\
   - R ||----- f(eta)| g(eta) - f(eta) |----- g(eta)||
       \\ deta       /                 \ deta       //

         /  d         \\    
   + 1.0 |----- f(eta)|| = 0
         \ deta       //    
                                       /                         
        /  d   /  d             \\     |/  d   /  d             \
(1 - p) |----- |----- theta(eta)|| + p ||----- |----- theta(eta)|
        \ deta \ deta           //     \\ deta \ deta           /

  \               /  d             \
  | + 10 R f(eta) |----- theta(eta)|
  /               \ deta           /

         /  d           \ /  d             \
   + 0.1 |----- phi(eta)| |----- theta(eta)|
         \ deta         / \ deta           /

                           2\    
         /  d             \ |    
   + 0.1 |----- theta(eta)| | = 0
         \ deta           / /    
        //  d   /  d           \\
(1 - p) ||----- |----- phi(eta)||
        \\ deta \ deta         //

                 /  d   /  d             \\\     //  d   /  d   
   + 1.000000000 |----- |----- theta(eta)||| + p ||----- |-----
                 \ deta \ deta           ///     \\ deta \ deta

          \\                /  d           \
  phi(eta)|| + 0.5 R f(eta) |----- phi(eta)|
          //                \ deta         /

                 /  d   /  d             \\\    
   + 1.000000000 |----- |----- theta(eta)||| = 0
                 \ deta \ deta           ///    
f(eta)=sum(f[i](eta)*p^i,i=0..1);
                f(eta) = f[0](eta) + f[1](eta) p
g(eta)=sum(g[i](eta)*p^i,i=0..1);
                g(eta) = g[0](eta) + g[1](eta) p
theta(eta)=sum(theta[i](eta)*p^i,i=0..1);
          theta(eta) = theta[0](eta) + theta[1](eta) p
phi(eta)=sum(phi[i](eta)*p^i,i=0..1);
             phi(eta) = phi[0](eta) + phi[1](eta) p
> FORequ1;


A:=collect(expand(subs(f(eta)=f[0](eta)+f[1](eta)*p,g(eta)=g[0](eta)+g[1](eta)*p,hpm1)),p);
/      /  d            \ /  d   /  d            \\
|-1. R |----- f[1](eta)| |----- |----- f[1](eta)||
\      \ deta          / \ deta \ deta          //

                 /  d   /  d            \\\  3   /
   + R f[1](eta) |----- |----- f[1](eta)||| p  + |
                 \ deta \ deta          ///      \
      /  d            \ /  d   /  d            \\
-1. R |----- f[0](eta)| |----- |----- f[1](eta)||
      \ deta          / \ deta \ deta          //

          /  d            \ /  d   /  d            \\
   - 1. R |----- f[1](eta)| |----- |----- f[0](eta)||
          \ deta          / \ deta \ deta          //

                 /  d   /  d            \\
   + R f[0](eta) |----- |----- f[1](eta)||
                 \ deta \ deta          //

                 /  d   /  d            \\\  2   //  d   /  d   /
   + R f[1](eta) |----- |----- f[0](eta)||| p  + ||----- |----- |
                 \ deta \ deta          ///      \\ deta \ deta \

    d   /  d            \\\\       /  d            \
  ----- |----- f[1](eta)|||| - 1.0 |----- g[1](eta)|
   deta \ deta          ////       \ deta          /

          /  d            \ /  d   /  d            \\
   - 1. R |----- f[0](eta)| |----- |----- f[0](eta)||
          \ deta          / \ deta \ deta          //

                 /  d   /  d            \\\  
   + R f[0](eta) |----- |----- f[0](eta)||| p
                 \ deta \ deta          ///  

     /  d   /  d   /  d   /  d            \\\\
   + |----- |----- |----- |----- f[0](eta)||||
     \ deta \ deta \ deta \ deta          ////

         /  d            \    
   - 1.0 |----- g[0](eta)| = 0
         \ deta          /    
A1:=diff(f[0](eta),eta$4)-2*K[r]*(diff(g[0](eta),eta))=0;
A2:=diff(f[1](eta),eta$4)-2*K[r]*(diff(g[1](eta),eta))-R*(diff(f[0](eta),eta))*(diff(f[0](eta),eta$2))+R*f[0](eta)*(diff(f[0](eta),eta$2))=0;
/  d   /  d   /  d   /  d            \\\\       /  d            \   
|----- |----- |----- |----- f[0](eta)|||| - 1.0 |----- g[0](eta)| =
\ deta \ deta \ deta \ deta          ////       \ deta          /   

  0
/  d   /  d   /  d   /  d            \\\\       /  d            \
|----- |----- |----- |----- f[1](eta)|||| - 1.0 |----- g[1](eta)|
\ deta \ deta \ deta \ deta          ////       \ deta          /

       /  d            \ /  d   /  d            \\
   - R |----- f[0](eta)| |----- |----- f[0](eta)||
       \ deta          / \ deta \ deta          //

                 /  d   /  d            \\    
   + R f[0](eta) |----- |----- f[0](eta)|| = 0
                 \ deta \ deta          //    
icsA1:=f[0](0)=0,D(f[0])(0)=1,g[0](0)=0,f[0](1)=lambda,D(f[0])(1)=0,g[0](1)=0;
icsA2:=f[1](0)=0,D(f[1])(0)=0,g[1](0)=0,f[1](1)=0,D(f[1])(1)=0,g[1](1)=0;
   f[0](0) = 0, D(f[0])(0) = 1, g[0](0) = 0, f[0](1) = 0.5,

     D(f[0])(1) = 0, g[0](1) = 0
    f[1](0) = 0, D(f[1])(0) = 0, g[1](0) = 0, f[1](1) = 0,

      D(f[1])(1) = 0, g[1](1) = 0
>
FORequ2;


B:=collect(expand(subs(f(eta)=f[0](eta)+f[1](eta)*p,g(eta)=g[0](eta)+g[1](eta)*p,hpm2)),p);
/      /  d            \          
|-1. R |----- f[1](eta)| g[1](eta)
\      \ deta          /          

                 /  d            \\  3   /
   + R f[1](eta) |----- g[1](eta)|| p  + |
                 \ deta          //      \
      /  d            \          
-1. R |----- f[0](eta)| g[1](eta)
      \ deta          /          

          /  d            \          
   - 1. R |----- f[1](eta)| g[0](eta)
          \ deta          /          

                 /  d            \
   + R f[0](eta) |----- g[1](eta)|
                 \ deta          /

                 /  d            \\  2   //  d   /  d            
   + R f[1](eta) |----- g[0](eta)|| p  + ||----- |----- g[1](eta)
                 \ deta          //      \\ deta \ deta          

  \\       /  d            \        /  d            \          
  || + 1.0 |----- f[1](eta)| - 1. R |----- f[0](eta)| g[0](eta)
  //       \ deta          /        \ deta          /          

                 /  d            \\     /  d   /  d            \\
   + R f[0](eta) |----- g[0](eta)|| p + |----- |----- g[0](eta)||
                 \ deta          //     \ deta \ deta          //

         /  d            \    
   + 1.0 |----- f[0](eta)| = 0
         \ deta          /    
B1:=diff(g[0](eta),eta$2)+2*K[r]*(diff(f[0](eta),eta))=0;
B2:=diff(g[1](eta),eta$2)+2*K[r]*(diff(f[1](eta),eta))-R*(diff(f[0](eta),eta))*g[0](eta)+R*f[0](eta)*(diff(g[0](eta),eta))=0;
     /  d   /  d            \\       /  d            \    
     |----- |----- g[0](eta)|| + 1.0 |----- f[0](eta)| = 0
     \ deta \ deta          //       \ deta          /    
       /  d   /  d            \\       /  d            \
       |----- |----- g[1](eta)|| + 1.0 |----- f[1](eta)|
       \ deta \ deta          //       \ deta          /

              /  d            \          
          - R |----- f[0](eta)| g[0](eta)
              \ deta          /          

                        /  d            \    
          + R f[0](eta) |----- g[0](eta)| = 0
                        \ deta          /    
icsB1:=f[0](0)=0,D(f[0])(0)=1,g[0](0)=0,f[0](1)=lambda,D(f[0])(1)=0,g[0](1)=0;
icsB2:=f[1](0)=0,D(f[1])(0)=0,g[1](0)=0,f[1](1)=0,D(f[1])(1)=0,g[1](1)=0;
   f[0](0) = 0, D(f[0])(0) = 1, g[0](0) = 0, f[0](1) = 0.5,

     D(f[0])(1) = 0, g[0](1) = 0
    f[1](0) = 0, D(f[1])(0) = 0, g[1](0) = 0, f[1](1) = 0,

      D(f[1])(1) = 0, g[1](1) = 0
> FORequ3;


C:=collect(expand(subs(theta(eta)=theta[0](eta)+theta[1](eta)*p,phi(eta)=phi[0](eta)+phi[1](eta)*p,f(eta)=f[0](eta)+f[1](eta)*p,hpm3)),p);
 /                                     
 |                /  d                \
 |10. R f[1](eta) |----- theta[1](eta)|
 \                \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[1](eta)| |----- theta[1](eta)|
          \ deta            / \ deta              /

                               2\                              
          /  d                \ |  3   /                /  d   
    + 0.1 |----- theta[1](eta)| | p  + |10. R f[0](eta) |-----
          \ deta              / /      \                \ deta

                \                   /  d                \
   theta[1](eta)| + 10. R f[1](eta) |----- theta[0](eta)|
                /                   \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[0](eta)| |----- theta[1](eta)|
          \ deta            / \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[1](eta)| |----- theta[0](eta)|
          \ deta            / \ deta              /

                                                            /
          /  d                \ /  d                \\  2   |/
    + 0.2 |----- theta[0](eta)| |----- theta[1](eta)|| p  + ||
          \ deta              / \ deta              //      \\

     d   /  d                \\
   ----- |----- theta[1](eta)||
    deta \ deta              //

                      /  d                \
    + 10. R f[0](eta) |----- theta[0](eta)|
                      \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[0](eta)| |----- theta[0](eta)|
          \ deta            / \ deta              /

                               2\  
          /  d                \ |  
    + 0.1 |----- theta[0](eta)| | p
          \ deta              / /  

      /  d   /  d                \\    
    + |----- |----- theta[0](eta)|| = 0
      \ deta \ deta              //    
C1:=diff(theta[0](eta),eta$2)=0;
C2:=diff(theta[1](eta), eta, eta)+Pr*R*f[0](eta)*(diff(theta[0](eta), eta))+Nb*(diff(phi[0](eta), eta))*(diff(theta[0](eta), eta))+Nt*(diff(theta[0](eta), eta))^2=0;
                  d   /  d                \    
                ----- |----- theta[0](eta)| = 0
                 deta \ deta              /    
       /  d   /  d                \\
       |----- |----- theta[1](eta)||
       \ deta \ deta              //

                           /  d                \
          + 10 R f[0](eta) |----- theta[0](eta)|
                           \ deta              /

                /  d              \ /  d                \
          + 0.1 |----- phi[0](eta)| |----- theta[0](eta)|
                \ deta            / \ deta              /

                                     2    
                /  d                \     
          + 0.1 |----- theta[0](eta)|  = 0
                \ deta              /     
icsC1:=theta[0](0)=1,theta[0](1)=0;
icsC2:=theta[1](0)=0,theta[1](1)=0,phi[0](0)=0,phi[0](1)=0;
                theta[0](0) = 1, theta[0](1) = 0
 theta[1](0) = 0, theta[1](1) = 0, phi[0](0) = 0, phi[0](1) = 0
> FORequ4;


E:=collect(expand(subs(theta(eta)=theta[0](eta)+theta[1](eta)*p,phi(eta)=phi[0](eta)+phi[1](eta)*p,f(eta)=f[0](eta)+f[1](eta)*p,hpm4)),p);
                 3 /  d              \   /                /  d   
0.5 R f[1](eta) p  |----- phi[1](eta)| + |0.5 R f[0](eta) |-----
                   \ deta            /   \                \ deta

             \                   /  d              \\  2   //
  phi[1](eta)| + 0.5 R f[1](eta) |----- phi[0](eta)|| p  + ||
             /                   \ deta            //      \\

    d   /  d              \\
  ----- |----- phi[1](eta)||
   deta \ deta            //

                 /  d   /  d                \\
   + 1.000000000 |----- |----- theta[1](eta)||
                 \ deta \ deta              //

                     /  d              \\  
   + 0.5 R f[0](eta) |----- phi[0](eta)|| p
                     \ deta            //  

     /  d   /  d              \\
   + |----- |----- phi[0](eta)||
     \ deta \ deta            //

                 /  d   /  d                \\    
   + 1.000000000 |----- |----- theta[0](eta)|| = 0
                 \ deta \ deta              //    
E1:=diff(phi[0](eta),eta$2)+Nt*(diff(theta[0](eta),eta$2))/Nb=0;
E2:=diff(phi[1](eta),eta$2)+Nt*(diff(theta[1](eta),eta$2))/Nb+R*Sc*f[0](eta)*(diff(phi[0](eta),eta))=0;
       /  d   /  d              \\
       |----- |----- phi[0](eta)||
       \ deta \ deta            //

                        /  d   /  d                \\    
          + 1.000000000 |----- |----- theta[0](eta)|| = 0
                        \ deta \ deta              //    
         /  d   /  d              \\
         |----- |----- phi[1](eta)||
         \ deta \ deta            //

                          /  d   /  d                \\
            + 1.000000000 |----- |----- theta[1](eta)||
                          \ deta \ deta              //

                              /  d              \    
            + 0.5 R f[0](eta) |----- phi[0](eta)| = 0
                              \ deta            /    
icsE1:=theta[0](0)=1,theta[0](1)=0,phi[0](0)=1,phi[0](1)=0;
icsE2:=theta[1](0)=0,theta[1](1)=0,phi[1](0)=0,phi[1](1)=0;
 theta[0](0) = 1, theta[0](1) = 0, phi[0](0) = 1, phi[0](1) = 0
 theta[1](0) = 0, theta[1](1) = 0, phi[1](0) = 0, phi[1](1) = 0
       
theta[0](eta) = -(152675527/100000000)*eta+1;
                                152675527        
              theta[0](eta) = - --------- eta + 1
                                100000000        
U:=f[1](eta)=0;
                         f[1](eta) = 0
Dsolve(A1,B1,icsA1,icsB1);
                  Dsolve(A1, B1, icsA1, icsB1)


sys:={ diff(g[0](eta), eta, eta)+1.0*(diff(f[0](eta), eta)) = 0, diff(f[0](eta), eta, eta, eta, eta)-1.0*(diff(g[0](eta), eta)) = 0};
    //  d   /  d   /  d   /  d            \\\\
   { |----- |----- |----- |----- f[0](eta)||||
    \\ deta \ deta \ deta \ deta          ////

            /  d            \      
      - 1.0 |----- g[0](eta)| = 0,
            \ deta          /      

     /  d   /  d            \\       /  d            \    \
     |----- |----- g[0](eta)|| + 1.0 |----- f[0](eta)| = 0 }
     \ deta \ deta          //       \ deta          /    /
IC_1:={ f[0](0) = 0, (D(f[0]))(0) = 1, g[0](0) = 0, f[0](1) = .5, (D(f[0]))(1) = 0, g[0](1) = 0,f[0](0) = 0, (D(f[0]))(0) = 1, g[0](0) = 0, f[0](1) = .5, (D(f[0]))(1) = 0, g[0](1) = 0};
    {f[0](0) = 0, f[0](1) = 0.5, g[0](0) = 0, g[0](1) = 0,

      D(f[0])(0) = 1, D(f[0])(1) = 0}
ans1 := combine(dsolve(sys union IC_1,{f[0](eta),g[0](eta)}),trig);
Error, (in dsolve) expecting an ODE or a set or list of ODEs. Received `union`(IC_1, sys)
>

Hello

I have a subscripts error, or it seems to be an error.

As you can see on the picture, then I have defined the varible I__K, but when I need it again I get another result or It seems to be another result that looks like this I[K]. I use the esc buttom to recall the varible.

Are there anybody that has a solution to this? I have been looking at other treads, but there seems not to be a solution that works or maybe I'm looking the wrong places.

Regards

Heide

 

 

How difficult is it to simulate gravitational influences and perturbing effects on celestial orbits with Maple? Could this syntax http://www.maplesoft.com/applications/view.aspx?SID=4484&view=html be altered without excessive changes to consider these aspects?

Are there somewhere worksheets to take a look at as an introduction and to see how such goals would be approached and implemented?


with(PDEtools, casesplit, declare)
``

L := 1651.12; m := 3205.12; r1 := .1875; r2 := 2; z1 := 0; z2 := 12; ld := 4.5

NULL

declare(u(r, z), w(r, z))``

with(DEtools, gensys)

rr := (L+2*m)*(diff(u(r, z), r))+L*(diff(w(r, z), z))+L*u(r, z)/r

zz := L*(diff(u(r, z), r))+(L+2*m)*(diff(w(r, z), z))+L*u(r, z)/r

rz := m*(diff(u(r, z), z))+m*(diff(w(r, z), r))

BCS := {rr(r1, ld) = 0, rz(r1, z) = T, w(r, 0) = 0, zz(r, z2) = 0}

{3205.12*(diff(u(r, z), z))(.1875, z)+3205.12*(diff(w(r, z), r))(.1875, z) = T, 8061.36*(diff(u(r, z), r))(.1875, 4.5)+1651.12*(diff(w(r, z), z))(.1875, 4.5)+1651.12*(u(r, z))(.1875, 4.5)/r(.1875, 4.5) = 0, 1651.12*(diff(u(r, z), r))(r, 12)+8061.36*(diff(w(r, z), z))(r, 12)+1651.12*(u(r, z))(r, 12)/r(r, 12) = 0, w(r, 0) = 0}

(1)

``

NULL

sys3 := [(L+2*m)*(diff(u(r, z), r, r))+(L+m)*(diff(w(r, z), r, z))+(L+2*m)*(diff(u(r, z), r))/r-(L+2*m)*u(r, z)/r^2+m*(diff(u(r, z), z, z)) = 0, (L+m)*(diff(u(r, z), r, z))+m*(diff(w(r, z), r, r))+(L+2*m)*(diff(w(r, z), z, z))+(L+m)*(diff(u(r, z), z))/r+m*(diff(w(r, z), r))/r = 0]

pdsolve(sys3, BCS, numeric)

 

 

``

``


Download PDE_equation2.mw

Hi all,

I have the following PDE, is it solveable by Maple or not. Do I need a boundary condition and how many or I can get a general solution? I am new to Maple. Any help will be appreciated.

Thank you.

 

 

 

Hi all, 

 

I was wondering if it is possible to add a colour gradient to a point plot to represent another perameter in the data.

 

 

For example, each point corresponds to a value of 'f' ranging from 0.1 to 1, and I was wondering how to display this by means of a colour gradient.

 

 

Thanks,

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