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as screen shot shown i just was wondering if the mechanic is available my engine dispatched again so um i guess i can pay in reputation points or food stamps, i just dont see why engine would desire either tho being an artificial entity and all. i mean its not like you need to eat.

I was asked if I would put together a list of top resources to help students who are using Maple for the first time.  An awful lot of students will be cracking Maple open in the next few weeks (the ones who are keeping up with their assignments, at least – for others, it sometimes takes little longer :-), so it seemed like a good idea.

So then I had to decide what to do. I know Top N lists are very popular (Ten Things that Will Shock You about Your Math Software!), and there are tons of Maple training resources available to fill such a list without any difficulties.  But personally, I don’t always like Top N lists. What are the chances that there are exactly N things you need to know, for nice values of N? And how often you are really interested in all N items? I just want to get straight to the points I care about.

I decided I’d try a matrix. So here you go: a mini “choose your own adventure” guide for getting to know Maple.  Pick the row that corresponds to what you want to do, and the column for how you want to do it.  All on a single, page, and ad-free!

And best of luck for the new school year.

 

 

I like words

I like videos

Just let me try it

Product Overview

Inside Maple, from the Help menu, select Take a Tour of Maple then click on the Ten Minute Tour button.

 

(Okay, even though I like words, too, you might also want to watch the video in the next column. The whole “picture is worth a thousand words” does have some truth to it, much as I don’t always like to admit it. J)

Watch Clickable Math

 

Keep in mind that if you prefer to use commands instead of these Clickable Math tools, you can do that too.  Personally, I mix and match.

You’ll figure it out.

Getting Started Info

Read the Maple Quick Start Tutorial Guide, as a PDF, or from the Help system. To access this guide from within Maple, start Maple, click on the Getting Started icon the left, then select the Quick Start Guide (first icon in the second row).

Watch the Maple Quick Start Tutorial Video.

The most important things to remember are

  1. Right click on your math expression to bring up a menu of things you can do, like plotting or integrating or solving your expression
  2. If you have just entered an exponent or the denominator of a fraction, use the right arrow key to get out of it.

How do I? Essentials

Look at the “How do I” section of the Maple Portal (Start Maple, click on the Getting Started icon, click on the Maple Portal icon; or search for “MaplePortal” in the help system).  Also look at the Maple Portal for Students, using the button from the Maple Portal.

Check out the dozens of videos in the Maple Training Video collection.

You can do a lot with the context menus and the various tools you’ll find on the Tools menu. But when in doubt, look at the list of “How do I” tasks from the Maple Portal described in the “words” column and pull out what you need from there.

What now?

The help system is your friend. Not only does it have help pages for every feature and every command, but it includes both the Maple User Manual and the Maple Programming Guide (also available as PDFs).

Check out the collection of videos on the Maplesoft YouTube channel.  (And the help system is your friend, too. We can’t make videos to cover every last thing, and if we did, you wouldn’t have time to watch them all!)

Maple comes with many examples and applications you can look at and modify.  You can browse through the Start page resources, or search for “examples,index” in the help system to see the full list.

 

And yes, the help system is your friend, too.  But don’t worry, no one is going to make you read the manual.

 

 

 

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. 

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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