Dear All,

This is probably an easy issue to solve.

I have imported an .xls file using the 1D lookup table.

I need the signal to be executed for 4 seconds which is also how long my simulation duration is.

So I set the Ramp height and rise time to 4s. For some reason when I display the signal using a probe it gives me the constant value of the first row of my Excel sheet. I need help to solve this! I have provided some visuals below.

Thank you for your time!

Can

Hello, thanks for read me

I don't know why my code don't work, I'm trying to calculate the magnitude of a complex vector but I get a error in the next image you can view it, 

Can someone help me? thank you

My program have more than 100 subsections. It is tedious to close them after run.

How is it possible the maple do not open those subsections I specified, during running the program?

Is it possible to close subsections abruptly?

Greetings People!

I am modelling a dynamical system using variational formulation. The final step is to plug in, the energy values, in the Euler-Lagrange Equation. The functional has both explicit dependence on time, and implicit dependence (in the form of the generalised coordinate). Since motive of the exercise is to obtain the governing equations for the generalized coordinates, this implicit dependence is unknown. How can Maple be used to derive the differential equations in this situation?

As an example, consider,

Here, ld and l are variables, which are function of time. In order to calculate the term,

How do I proceed? Thanks in advance.

I have a piecewise defined linear function, at one point it has a set value. How can i plot it in maple??

How is it possible to stop Maple from running, in specific line of program when using execute the entire worksheet???

Is it possible to capture and convert the output of dismantle to one (long) string?

For example, if I do

r:=dismantle(x+sin(x));

SUM(5)
   NAME(4): x
   INTPOS(2): 1
   FUNCTION(3)
      NAME(4): sin #[protected, _syslib]
      EXPSEQ(2)
         NAME(4): x
   INTPOS(2): 1

I'd like to first capture the output in "r", which now it does not, it just goes to screen and "r" is left empty, and then convert the result to long string. (using convert()) So the result of the above will becomes

r:="SUM(5) NAME(4): x INTPOS(2): 1 FUNCTION(3) NAME(4): sin #[protected, _syslib]..."

The reason I need to capture the output, is so I can more easily look for some specific name/content inside the expression. which is now in a string, using string tools later on, which I think will be easier. For example, if I want to look for "sin" I can now search the above string and see if "sin" is there. Context is not important, just need to see if a name happens to be in the string.  The problem I do not know how to convert output of dismantle to string.

Maple 2016.2 on windows

TQ.mw

Can any one help for finding the solution of these differntial equations and then plotting the graph for differnt values of M

(FILE ATTACHED)
 

Download TQ.mw

A lemniscate is a polar curve of the form r^2=a^2*cos(2*theta) or r^2=a^2*sin(2*theta). I have just started using Maple and I wrote the following commands: 

> with(plots):
> polarplot(2*sqrt(cos(2*t)), axiscoordinates = cartesian, angularunit = radian, color = "Black");
 

But I am getting the following graph 

which is not satisfactory since some points are missing. I know that using the square root may have caused this, but I am not sure as to how should I resolve this issue. I used plus/minus symbol before the expression 2(cos(2t))^(0.5) but there was an error and the discontinuity still persisits. Kindly help me in plotting this curve. 

Thank you.

How can i copy paste some differential equations with boundary conditions from maple work sheet to here

I have a big square matrix(1000*1000) having parameter x within some of its entries.

How do you suggest to find close form eigenvalue of such matrix with maple? (eigenvalue will be function of x)

Hello

Can Maple compute the weak derivative of piecewise function like

f:=x->piecewise(0<x and x<0.5,x,0.5<x and x<1,1-x,  elsewhere 0);

Many thanks

hello agian i have the following problem. i need to fit data using the following model:

ode_sub := diff(S(t), t) = -k1*S(t)-S(t)/T1_s;

ode_P1 := diff(P1(t), t) = 2*k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1;

ode_P2 := diff(P2(t), t) = -k2*(-keq*P1(t)+P2(t))/keq-k4*P2(t)-P2(t)/T1_p2;

ode_P2e := diff(P2_e(t), t) = k4*P2(t)-P2_e(t)/T1_p2_e;

ode_system := ode_sub, ode_P1, ode_P2, ode_P2e

known paramters: s0 := 10000; k2 := 1000; T1_s := 14; T1_p2_e := 35; T1_p2 := T1_p1

initial conditions: init := S(0) = s0, P1(0) = 0, P2(0) = 0, P2_e(0) = 0

using the following data the fitting is fine:

T := [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100]

data_s := [10000.00001377746, 7880.718836217512, 6210.572917625112, 4894.377814704496, 3857.121616806618, 3039.689036293312, 2395.493468824973, 1887.821030424934, 1487.738721378254, 1172.445015238064, 923.970970533911, 728.1555148262845, 573.8389058920359, 452.2263410725434, 356.3868133709972, 280.8584641247961, 221.3366863178145, 174.4291648589732, 137.4627967029932, 108.3305246342751, 85.37230370803576, 67.2794974831867, 53.0210755540487, 41.78436556408739, 32.92915589156605, 25.95049906181449, 20.45092590987601, 16.11678944747619, 12.70115104646253, 10.009439492356, 7.888058666357939, 6.216464754698855, 4.899036441885205, 3.860793948052506, 3.042584031379331, 2.397778731385364, 1.889601342133927, 1.489145259940784, 1.173591417634974, .9248694992255094, .7288443588090404, .5743879613705721, .4526024635132348, .3567687570029392, .2810508404011898, .2215650452813882, .174603181767829, .1376243372528356, .1084232889842753, 0.8542884707952822e-1, 0.6738282660463157e-1]

data_p1 := [0.1194335334401124e-4, 244.8746046039949, 374.8721199398692, 430.5392805383767, 439.6598364813143, 421.0353424914179, 387.1842556288343, 346.2646897222593, 303.4377508746471, 261.8283447091155, 223.1996547160051, 188.4213144493491, 157.7924449350029, 131.2622073983344, 108.5771635112278, 89.37951009190863, 73.26979150957087, 59.84578653950572, 48.72563358658898, 39.56010490461378, 32.03855466968536, 25.88922670933322, 20.87834763772145, 16.80708274458702, 13.50774122768974, 10.84014148654258, 8.687656394505874, 6.954093898245485, 5.560224107929433, 4.441209458524726, 3.544128529596104, 2.825755811619965, 2.251247757181308, 1.792233305651086, 1.425861347838012, 1.133566009019768, .90081361320016, .7153496336919163, .5678861241754847, .4505952916932289, .3572989037753538, .2832489239941939, .2244289868248577, .1778450590752305, .1408633578784151, .1114667192753896, 0.8826814044702111e-1, 0.6979315954603076e-1, 0.5526502783606788e-1, 0.4370354298880999e-1, 0.3456334307662573e-1]

data_p2_p2e := [-0.1821397630630296e-4, 1000.40572909871, 1568.064904416198, 1848.900129881268, 1944.147939710225, 1923.352973299286, 1833.705314342611, 1706.726235937363, 1563.036042115902, 1415.741121363331, 1272.825952816517, 1138.833091575137, 1016.03557293539, 905.2470623752856, 806.3754051707843, 718.7979384094563, 641.6091822001032, 573.7825966275556, 514.2718125966452, 462.0710416682647, 416.2491499591012, 375.9656328260581, 340.4748518743264, 309.124348579787, 281.3484612079911, 256.6599441255977, 234.6416876403397, 214.9378461256197, 197.2453333305823, 181.3059798685686, 166.90059218783, 153.8422174795751, 141.9717783871606, 131.1529759058513, 121.2692959115991, 112.2199678247825, 103.9187616370328, 96.29007054510998, 89.26877137303566, 82.79743967408479, 76.82586273439793, 71.30944943617081, 66.2085909515396, 61.48838150744805, 57.11714763242225, 53.06666224006544, 49.31130738219119, 45.82807853990728, 42.59597194910467, 39.59575450632008, 36.81013335261527]

the fitting is done the following way: 

P1fu, P2fu, P2e_fu, Sfu := op(subs(res, [P1(t), P2(t), P2_e(t), S(t)]))

making residuals:

Q := proc (T1_p1, k1, keq, k4) local P1v, P2v, P2e_v, Sv, resid; option remember;

res(parameters = [T1_p1, k1, keq, k4]);

try P1v := `~`[P1fu](T); P2v := `~`[P2fu](T); P2e_v := `~`[P2e_fu](T); Sv := `~`[Sfu](T); resid := [P1v-data_p1, P2v+P2e_v-data_p2_p2e, Sv-data_s]; return [seq(seq(resid[i][j], i = 1 .. 3), j = 1 .. nops(T))] catch: return [1000000$3*nops(T)] end try end proc;
q := [seq(subs(_nn = n, proc (T1_p1, k1, keq, k4) Q(args)[_nn] end proc), n = 1 .. 3*nops(T))];

finding inital point for the LSsolve:

L := fsolve(q[2 .. 5], [10, 0.2e-1, 4, 4])

fitting the data with the intial point: 

solLS := Optimization:-LSSolve(q, initialpoint = L)

this is all good however, when i used the following data it did not turn out so well (using the same approch as above):

data_s := [96304.74567, 77385.03700, 62621.83067, 51239.94333, 42663.82367, 35084.74100, 28480.28367, 23066.01467, 18774.73700, 15179.13700, 12278.50767, 9937.652000, 8046.848333, 6521.242000, 5287.811667, 4277.779000, 3466.518333, 2835.467000, 2297.796333, 1861.249667, 1529.654000, 1235.353000, 999.6626667, 826.2343333, 667.9480000, 559.9230000, 449.2790000, 376.4860000, 289.1203333, 236.1483333]

data_p1 := [0.86e-1, 3.904, 26.975, 31.719, 41.067, 46.779, 52.115, 43.101, 44.344, 41.094, 36.523, 27.742, 26.543, 28.062, 22.178, 21.303, 14.951, 17.871, 11.422, 12.051, 9.232, 6.817, 6.1, .717, 1.215, 6.146, .772, .375, 2.595, .518]

data_p2_p2e := [-3.024, 22.238, 61.731, 103.816, 132.695, 159.069, 167.302, 160.188, 158.398, 152.943, 146.745, 135.22, 132.145, 120.413, 107.864, 95.339, 90.775, 81.828, 71.065, 70.475, 62.872, 49.955, 40.858, 42.938, 41.311, 35.583, 31.573, 29.841, 29.558, 21.762]

the known parameters is the case are:

s0 := 96304.74567; k2 := 10^5; T1_s := 14; T1_p2_e := 35; T1_p2 := T1_p1

additionally the following fuction affects the solution: 

K:=t->cos((1/180)*beta*Pi)^(t/Tr)

i included this by doing this:

P1fu_K := proc (t) options operator, arrow; P1fu(t)*K(t) end proc;

P2fu_K := proc (t) options operator, arrow; P2fu(t)*K(t) end proc;

P2e_fu_K := proc (t) options operator, arrow;

P2e_fu(t)*K(t) end proc;

Sfu_K := proc (t) options operator, arrow; Sfu(t)*K(t) end proc

resulting in the following residuals:

Q := proc (T1_p1, k1, keq, k4) local P1v, P2v, P2e_v, Sv, resid; option remember;

res(parameters = [T1_p1, k1, keq, k4]);

try P1v := `~`[P1fu_K](T); P2v := `~`[P2fu_K](T); P2e_v := `~`[P2e_fu_K](T); Sv := `~`[Sfu_K](T); resid := [P1v-data_p1, P2v+P2e_v-data_p2_p2e, Sv-data_s]; return [seq(seq(resid[i][j], i = 1 .. 3), j = 1 .. nops(T))] catch: return [1000000$3*nops(T)] end try end proc;
q := [seq(subs(_nn = n, proc (T1_p1, k1, keq, k4) Q(args)[_nn] end proc), n = 1 .. 3*nops(T))];

i think my problem is that the inital point in this case is not known. all i know is that all the fitted parameters should be positive and that k1<1, k4>10, keq>1 and T1_p1>100 (more i do not know) - is there a way to determin the inital point without guessing?

i also know the results, which should be close to these values:

k1=0.000438, k4=0.0385, keq=2.7385 and T1_p1=36.8 the output fit should look something like this

where the red curve is (PP2(t)+PP2_e(t))*K(t)

the blue cure is PP1(t)*K(t) 

anyone able to help - i've tried for 2 days now. it might be that  ode_P1 := diff(P1(t), t) = 2*k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1 should be changed into ode_P1 := diff(P1(t), t) = k1*S(t)-k2*(P1(t)-P2(t)/keq)-P1(t)/T1_p1;

i've tried this but it didnt seem to do much 

anyone able to help?:)

NB stiff=true can be used within the dsolve to speed up the process if needed:)

Hi, here a maple newbie... I'm trying to obtain de coordinates of the "double points" of a planar Lissajous curve, but I have no idea how =S.

I have two symmetric real matrices as below (Mt,Kk).

I multilpy them in Mt.Kk.Mt.

But the result is not symmetric.!!!!why????

Download question.mw

I couldnt upload Kk.mlp and Mt.mlp so changed their extensions to mw. for using please change them to .mlp

Download Kk.mw

Download Mt.mw