Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hi

I want to employ FDM to solve nonlinear ODE. Since large expressions in terms of s[1] are generated, the program lose its efficiency for N>6 (It requires long run-time). Please amend the program, if it is possible to reach more precision for N>10. Moreover, is there a Maple command to dsolve this ode?

Thanks alot

restart;

sy := 2400.:

Hp := 0.1e9:

P := -900.:

a := 20.:

c := 21.:

N := 6:

s[0] := P:

h := (c-a)/N:

Et := (r*(diff(sr(r), r))-sy)/Hp:

Er := (sy-r*(diff(sr(r), r)))/Hp:

ode := simplify(Er-Et-r*(diff(Et, r))*(1+(diff(Et, r))*r/(2+4*Et))):

ode:=subs(diff(sr(r), r, r) = (s[k+1]-2*s[k]+s[k-1])/h^2, diff(sr(r), r) = (s[k+1]-s[k-1])/(2*h), r = h*k+a, ode):

for k to N-1 do

s[k+1] := solve(ode, s[k+1])[1]

end do:

s[1] := fsolve(s[N] = -200., s[1] = -900 .. -100);

plots[pointplot]([seq([h*k+a, s[k]], k = 0 .. N)]);

Hi, i have accelerometer readings saved as a CVS file., these file contains time data, then the accelerometer data in g's. (recorded at 700hz)

I need to analyse the data for hand-arm vibration and be able to plot the frequency/time. 

i have managed to fft the data but im struggling to plot this.

thanks for the help!!!

I am trying to calculate the divergence of a tensor field (in this specific case of a cauchy stress tensor (tensor of second order)). I am trying to use the DifferentialGeometry as well as the tensor package. I tried doing it somewhat similar to calculating the divergence of a tensor of first order, which worked fine with this code:

>with(DifferentialGeometry): with(Tensor): with(Tools):
>DGsetup([x,y,z],T);
>v := DGzip([v__x(x,y,z,t),v__y(x,y,z,t),v__z(x,y,z,t)],[D_x,D_y,D_z],"plus");
>Divergence(v);


However when I tried something similar with a tensor of second order it returened an error:

>sigma := evalDG(sigma__11 * D_x &t D_x + sigma__12 * D_x &t D_y + sigma__13 * D_x &t D_z
                 + sigma__21 * D_y &t D_x + sigma__22 * D_y &t D_y + sigma__23 * D_y &t D_z 
                 + sigma__31 * D_z &t D_x + sigma__32 * D_z &t D_y + sigma__33 * D_z &t D_z);
>Divergence(sigma);
Error, (in DifferentialGeometry:-Tools:-Divergence) expected argument to be a vector field. Received _DG([["tensor", T, [["con_bas", "con_bas"], []]], [`...`]])


Is there a different command in order to compute the divergence of a tensor of second order compared to the command for a tensor of first order or am I making an enitrely different mistake?
Would really appreciate some help 

Hello!

I have a question about a calculation of a differential of a functional.

I think dW(Y(t), t)/dt equals W  +W'Y .

There is some problem when I try to calculate it in Maple as below:

diff(w(y(t), t), t);
                          / d      \                   
         D[1](w)(y(t), t) |--- y(t)| + D[2](w)(y(t), t)
                          \ dt     /                   
I don't know it right or wrong; and  what does the D[1],D[1] mean?

I hope i can get your help!

Thanks!

 

please help me export data of for(do) to excel


 

restart

Digits := 8

with(plots)

with(CurveFitting)

with(plottools)

with(ExcelTools)

v := .7

Disp := 20

esp := 1000000

k := 0

E := proc (x, t) options operator, arrow; Int(exp((-esp*w^4+Disp*w^2+k)*t)*cos(w*(x+v*t))/Pi, w = 0 .. infinity, epsilon = 0.1e-6) end proc

proc (x, t) options operator, arrow; Int(exp((-esp*w^4+Disp*w^2+k)*t)*cos(w*(x+v*t))/Pi, w = 0 .. infinity, epsilon = 0.1e-6) end proc

(1)

"f(x):=15.5*(e)^((-(x-12590)^(2))/(3710000))+14.55*(e)^((-(x-16100)^(2))/(3000000)):"

NULL

u := proc (x, t) options operator, arrow; Int(E(x-xi, t)*f(xi), xi = 0 .. 0.2e5, epsilon = 0.1e-4) end proc

proc (x, t) options operator, arrow; Int(E(x-xi, t)*f(xi), xi = 0 .. 0.2e5, epsilon = 0.1e-4) end proc

(2)

NULL

NULL

uu := evalf(Int(E(0-xi, i)*f(xi), xi = 0 .. 20000, method = _NCrule, epsilon = 10^(-6)))

Int((Int(.31830988*exp((-1000000.*w^4+20.*w^2)*i)*cos(w*(-1.*xi+.7*i)), w = 0. .. Float(infinity)))*(15.5*exp(-0.26954178e-6*(xi-12590.)^2)+14.55*exp(-0.33333333e-6*(xi-16100.)^2)), xi = 0. .. 20000.)

(3)

for i from 0 by 300 to 39000 do evalf(Int(E(0-xi, 39000-i)*f(xi), xi = 0 .. 20000, method = _NCrule, epsilon = 10^(-6))) end do

-0.40453181e-3

 

0.34992127e-3

 

0.12637278e-2

 

0.16816398e-2

 

0.98280208e-3

 

-0.90049939e-3

 

-0.31323275e-2

 

-0.41557391e-2

 

-0.25326936e-2

 

0.18889271e-2

 

0.72213081e-2

 

0.99627384e-2

 

0.68245493e-2

 

-0.27701468e-2

 

-0.15052225e-1

 

-0.22679194e-1

 

-0.18343538e-1

 

0.20091767e-3

 

0.26631663e-1

 

0.47005735e-1

 

0.46070677e-1

 

0.16544944e-1

 

-0.33724783e-1

 

-0.81923601e-1

 

-0.99668403e-1

 

-0.68223938e-1

 

0.83269218e-2

 

.10029034

 

.16367514

 

.16160510

 

0.85214080e-1

 

-0.37861317e-1

 

-.15374440

 

-.20787576

 

-.17365568

 

-0.69367811e-1

 

0.46138347e-1

 

0.98175763e-1

 

0.30797102e-1

 

-.16333068

 

-.42882780

 

-.65291065

 

-.68519320

 

-.36491699

 

.45065352

 

1.8601309

 

3.8933757

 

6.4921833

 

9.4993084

 

12.664412

 

15.671497

 

18.185563

 

19.912262

 

20.654429

 

20.352632

 

19.099001

 

17.122716

 

14.750504

 

12.351527

 

10.277969

 

8.8110726

 

8.1121013

 

8.2595148

 

9.1484300

 

10.625106

 

12.470000

 

14.442900

 

16.334977

 

17.814817

 

18.760786

 

19.181587

 

18.732976

 

17.780388

 

16.318411

 

14.486252

 

12.440398

 

10.334385

 

8.3022216

 

6.4471886

 

4.8365207

 

3.5019327

 

2.4446257

 

1.6429774

 

1.0611606

 

.65712596

 

.38895586

 

.21914009

 

.11682497

 

0.58405315e-1

 

0.26982858e-1

 

0.11209327e-1

 

0.39351902e-2

 

0.94525962e-3

 

-0.74540895e-4

 

-0.29495088e-3

 

-0.25204186e-3

 

-0.16004845e-3

 

-0.86532870e-4

 

-0.41463181e-4

 

-0.17899580e-4

 

-0.69278927e-5

 

-0.23491366e-5

 

-0.66829402e-6

 

-0.13686514e-6

 

-0.76837315e-8

 

0.49022551e-8

 

-0.69948640e-9

 

0.21531116e-8

 

0.94383197e-8

 

0.13935450e-7

 

0.74857886e-8

 

-0.17655857e-7

 

-0.20268315e-7

 

0.19079956e-6

 

0.10964273e-6

 

-0.46401214e-7

 

0.65111854e-8

 

0.19030541e-9

 

-0.32052754e-6

 

-0.33434254e-7

 

0.21130779e-7

 

-0.60700260e-9

 

-0.15371908e-9

 

0.56652633e-10

 

-0.29781841e-10

 

-0.17334684e-11

 

0.20761743e-11

 

0.98264048e-9

 

0.43165138e-11

 

-0.28235638e-7

(4)

NULL

NULL

NULL


 

Download getD1.mw

Second Try:

f:= c/(c - 1) - c*(Pi^2 - 12*ln(c))*(1 + c)/(12*(c - 1)^3*n) + (((144*c^3 + 1584*c^2 + 1584*c + 144)*ln(c)^2 - 24*Pi^2*(1 + c)*(c^2 + 10*c + 1)*ln(c) + (-96*c^3 - 288*c^2 + 288*c + 96)*Zeta(3) + Pi^2*((Pi^2 + 24)*c^3 + (11*Pi^2 + 72)*c^2 + (11*Pi^2 - 72)*c + Pi^2 - 24))*c)/(288*(c - 1)^5*n^2)

This expression is already arranged with respect to n. However, the nominator of each term is not collected with respect to c. So naturally I thought

collect(f,[n,c]) or collect(f,[n,c],simplify)

would work. But now he messes up the nominator i.e. he can not factor and simplify. Note that the single variable case collect(f,n,simplify) works in not messing up, but this is not what I want, since the nominator of each n-term is not in c-collected form.

collect(f,[n],u->collect(u,[c],...))

also does not work, since he messes up again.

By c-collected form I mean the following:

The n^0 and n^{-1} term are actually fine. The factorization in the second term is ok. But for the third term

f2:=(144*c^3 + 1584*c^2 + 1584*c + 144)*ln(c)^2 - 24*Pi^2*(1 + c)*(c^2 + 10*c + 1)*ln(c) + (-96*c^3 - 288*c^2 + 288*c + 96)*Zeta(3) + Pi^2*((Pi^2 + 24)*c^3 + (11*Pi^2 + 72)*c^2 + (11*Pi^2 - 72)*c + Pi^2 - 24)

what I mean by c-collected is

collect(f2,c,simplify)

This is the nominator of the n^{-2} term.

I sort of managed by the following to procedings:

of:=[op(f)];
add(`~`[`/`](collect~(numer~(of), c, factor), ` $`, denom(of)));
add(collect~(of, c, simplify));

but the first one seems cumbersome for such a trivial thing that should be handled by collect. It also does not factor out the e.g. 12*ln(c)-Pi^2. This seems to be a general behaviour

g:=a*x*(x+1);
collect(g,x) does not factor out the a as in a*(x^2+x).

With the second method I'm more or less happy, but I didn't manage to completely collect the c terms i.e. the n^{-1} term is still c*(c+1) and not c^2+c.

Instead of getting a pop message telling me I have been cut off from the maple kernel can I please know how to recieve that error in the output of my worksheet instead?

Why does it insist I have to close and reopen the worksheet?

I was just thinking that if it were in my worksheets output I could have the current procedure in a try catch statement and write the occurance of the loss of server connection to a text file that is being checked by a script that will then send the necessary keystrokes to maple to save the worksheet close it then reopen it

Maple2020 was installed in windows 10 home. When installing maplesim 2019, it was looking for maple2019. Do I have to install maple 2019 before maplesim 2019? 

with(plots);
P1 := plot([-sin(t), t, t = 0 .. 2*Pi], coords = polar, color = red);
P2 := plot([cos(t), t, t = 0 .. 2*Pi], coords = polar, color = blue);
display(P1, P2, scaling = constrained);
 

I have two polar equation in the same graph but how do i shade the region between those two polar curve?

s1 := RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S)

s2 := -(D1*RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S)*D6-S*D6+RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))/(D2*D6*RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))
algsubs(s = RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S), s2)

I try to use s1 to represent s2, and I get the following result:

-(D1*__SELECTION(RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))*D6-S*D6+RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))/(D2*D6*RootOf(D1*D2*D6*_Z^2+(-D1*D4*D6-D2*D6*S-D4)*_Z+D4*D6*S))
what is mean of __SELECTION, If algsubs cannot work well, what should I do?

Hi,

Two weeks ago, I started loading data on the CoVid19 outbreak in order to understand, out of any official communication from any country, what is really going on.

From february 29 to march 9 these data come from https://bnonews.com/index.php/2020/02/the-latest-coronavirus-cases/ and from 10 march until now from https://www.worldometers.info/coronavirus/#repro.In all cases the loading is done manually (copy-paste onto a LibreOffice spreadsheet plus correction and save into a xls file) for I wasn't capable to find csv data (csv data do exist here https://github.com/CSSEGISandData/COVID-19, by they end febreuary 15th).
So I copied-pasted the results from the two sources above into a LibreOffice spreadsheet, adjusted the names of some countries for they appeared differently (for instance "United States" instead of "USA"), removed the unnessary commas and saved the result in a xls file.

I also used data from https://www.worldometers.info/world-population/population-by-country/ to get the populations of more than 260 countries around the world and, finally, csv data from https://ourworldindata.org/coronavirus#covid-19-tests to get synthetic histories of confirmed and death cases (I have discovered this site only yesterday evening and I think it could replace all the data I initially loaded).

The two worksheet here are aimed to exploratory and visualization only.
An other one is in progress whose goal is to infer the true death rate (also known as CFR, Case Fatality Rate).

No analysis is presented, if for no other reason than that the available data (except the numbers of deaths) are extremely dependent on the testing policies in place. But some features can be drawn from the data used here.
For instance, if you select country = "China" in file Covid19_Evolution_bis.mw, you will observe very well known behaviour which is that the "Apparent Death Rate", I defined as the ratio of the cumulated number of death at time t by the cumulatibe number of confirmed cases at the same time, is always an underestimation of the death rate one can only known once the outbreak has ended. With this in mind, changing the country in this worksheet from China to Italy seems to lead to frightening  scary interpolations... But here again, without knowing the test policy no solid conclusion can be drawn: maybe Italy tests mainly elder people with accute symptoms, thus the huge "Apparent Death Rate" Italy seems to have?


The work has been done with Maple 2015 and some graphics can be improved if a newer version is used (for instance, as Maple 2015 doesn't allow to change the direction of tickmarks, I overcome this limitation by assigning the date to the vertical axis on some plots).
The second Explore plot could probably be improved by using newer versions or Maplets or Embeded components.

Explore data from https://bnonews.com/index.php/2020/02/the-latest-coronavirus-cases/ and https://www.worldometers.info/coronavirus/#repro
Files to use
Covid19_Evolution.mw
Covid19_Data.m.zip
Population.xls

Explore data from  https://ourworldindata.org/coronavirus#covid-19-tests
Files to use
Covid19_Evolution_bis.mw
daily-deaths-covid-19-who.xls
total-cases-covid-19-who.xls
Population.xls


I would be interested by any open collaboration with people interested by this post (it's not in my intention to write papers on the subject, my only motivation is scientific curiosity).

 

If I execute a program like:

restart;
pathname := "C:\\Users\\Gregory McColm\\Desktop\\Scratch2";
filename := "WhatGives6.txt";
stream := cat(pathname, "\\", filename);
Y := seq(rand(1..10)(), j = 1..10);
writeto(stream);
for i from 1 to 10 do
 printf("X[ %d ]:= %d;\n", i, Y[i]);
end do;
writeto(terminal);
printf("DONE\n");

 

Maple will print exactly what I told it to print.  But when I enter

restart;
pathname := "C:\\Users\\Gregory McColm\\Desktop\\Scratch2";
filename := "WhatGives44.txt";
stream := cat(pathname, "\\", filename);
writeto(stream);
for i from 1 to 10 do
 Y := rand(1..10)();
 printf("X[ %d ]:= %d;\n", i, Y);
end do;
writeto(terminal);
printf("DONE\n");

 

Maple prints:

             YAssign7, [Typesetting:-mprintslash([Y := 7], [7])]

X[ 1 ]:= 7;
           YAssign10, [Typesetting:-mprintslash([Y := 10], [10])]

X[ 2 ]:= 10;
             YAssign6, [Typesetting:-mprintslash([Y := 6], [6])]

X[ 3 ]:= 6;
             YAssign2, [Typesetting:-mprintslash([Y := 2], [2])]

X[ 4 ]:= 2;
             YAssign4, [Typesetting:-mprintslash([Y := 4], [4])]

X[ 5 ]:= 4;
             YAssign6, [Typesetting:-mprintslash([Y := 6], [6])]

X[ 6 ]:= 6;
             YAssign5, [Typesetting:-mprintslash([Y := 5], [5])]

X[ 7 ]:= 5;
             YAssign1, [Typesetting:-mprintslash([Y := 1], [1])]

X[ 8 ]:= 1;
             YAssign8, [Typesetting:-mprintslash([Y := 8], [8])]

X[ 9 ]:= 8;
             YAssign5, [Typesetting:-mprintslash([Y := 5], [5])]

X[ 10 ]:= 5;
 

This must be more than a matter of calling rand when I am writing to terminal, for if I enter:

restart;
pathname := "C:\\Users\\Gregory McColm\\Desktop\\Scratch2";
filename := "WhatGives8.txt";
stream := cat(pathname, "\\", filename);
writeto(stream);
X := rand(0..1)():
printf("Mary had a little lamb\n");
writeto(terminal);
printf("DONE\n");

 

then all Maple prints is "Mary is a little lamb".  I have tried this on Maple 2018 and Maple 2019, with the same results.

 

Any ideas or suggestions?

Hi,

I'm having trouble converting a static plot to animated plot:

Also I've been considering using functional operators instead of expressions so that there's no reuse of variable s when drawing different curves, though I'm not sure if this will be harder to differentiate since diff(expr, s) does not work on a functional operator meaning I'd have to do unapply(diff(f(s),s),s) which seems a long route and I'm not sure if it's what I'm looking for (in terms of simplification).

 

Thanks guys

agentpath.mw

 Hello everyone!

 I want to find all solutions of  following equations :

I used Maple 2019: 

solutions:=solve([abs(1+1/3*lambda+1/18*lambda^2-1/324*lambda^3+1/1944*lambda^4)-1=0],[lambda]);
evalf(solutions)

The output is:

So we have two solutions. But when I use Matlab 2018, 
four solutions are returned.

syms lambda
eqn =abs(1+1/3*lambda+1/18*lambda^2-1/324*lambda^3+1/1944*lambda^4)-1==0;
solx1=solve(eqn, lambda) 
%%
solx1 =
 root (z1 ^ 4-6 * z1 ^ 3 + 108 * z1 ^ 2 + 648 * z1, z1, 1)
 root (z1 ^ 4-6 * z1 ^ 3 + 108 * z1 ^ 2 + 648 * z1, z1, 2)
 roots (z1 ^ 4-6 * z1 ^ 3 + 108 * z1 ^ 2 + 648 * z1, z1, 3)
 roots (z1 ^ 4-6 * z1 ^ 3 + 108 * z1 ^ 2 + 648 * z1, z1, 4)


 

solx3=solve(eqn, lambda,   'MaxDegree', 4)
double(solx3)
%%The solution is:
 -4.2681 + 0.0000i
 0.0000 + 0.0000i
 5.1340 -11.2012i
5.1340
+ 11.2012i

It is easy to check that  first two in Maple and Matlab are  same.

Who is right? Does Maple miss complex solutions?

 

 

 

 

 

 

 

 

Dear,

I need to attach CPU time to my iterations/Computations, please how do I obtain that in Maple. I tried 

a:=time () 

Iterations code

cputime:=time() -a; 

 

But the issue is that once I rerun the code, It produces different cpu time, kindly help me out here. 

 

Thank you.

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