MaplePrimes Questions

In answer to a previous question (https://www.mapleprimes.com/questions/228965-How-Can-I-Save-A-Record) acer introduced me to the .mla archives as a possible store location for variables.

I have been working with this and now end up having 4 .mla files (called results1.mla...results4.mla) that each hold output from some lengthy batch running of maple. Each of these has two tables with values called Beams and Beams2 (same name in each of these archives). While named the same the content is different.

My question is: How do I get at each of these? As acer explained, the content (i.e. the tables Beams and Beams2) is mapped into the namespace of the Maple session once libname includes the directory where the .mla files sit. They are actually in the same directory as the associated Maple programs so they mask each other.

I tried to juggle libname prepending the .mla file I want to the rest; but that does not seem to work (I always get the same data). I do need the data all in one worksheet for collating and postprocessing and display.

TIA,

M.D.

Hi everybody:

How can I solve this system of nonlinear equations without use the fsolve command?

eq[1] := -3*c[0]+3*c[1] = c[0]*(1-259*d[0]*(1/192)+43*d[1]*(1/64)-11*d[2]*(1/64)+(1/64)*d[3])-23/6;
eq[2] := -3*d[0]+3*d[1] = d[0]*(-2+23*c[0]*(1/80)+139*c[1]*(1/320)-17*c[2]*(1/160)+3*c[3]*(1/320))-1;
eq[3] := -4*c[0]*(1/3)+c[2]+(1/3)*c[3] = (8*c[0]*(1/27)+4*c[1]*(1/9)+2*c[2]*(1/9)+(1/27)*c[3])*(1-2657*d[0]*(1/5184)-343*d[1]*(1/1728)-185*d[2]*(1/1728)-79*d[3]*(1/5184))-3;
eq[4] := -4*d[0]*(1/3)+d[2]+(1/3)*d[3] = (8*d[0]*(1/27)+4*d[1]*(1/9)+2*d[2]*(1/9)+(1/27)*d[3])*(-2+109*c[0]*(1/8640)+553*c[1]*(1/1440)+559*c[2]*(1/2880)+37*c[3]*(1/1080))-5/6;
eq[5] := -(1/3)*c[0]-c[1]+4*c[3]*(1/3) = ((1/27)*c[0]+2*c[1]*(1/9)+4*c[2]*(1/9)+8*c[3]*(1/27))*(1-673*d[0]*(1/5184)-455*d[1]*(1/1728)-505*d[2]*(1/1728)-767*d[3]*(1/5184))-49/27;
eq[6] := -(1/3)*d[0]-d[1]+4*d[3]*(1/3) = ((1/27)*d[0]+2*d[1]*(1/9)+4*d[2]*(1/9)+8*d[3]*(1/27))*(-2-173*c[0]*(1/4320)+241*c[1]*(1/2880)+59*c[2]*(1/180)+2191*c[3]*(1/8640))-11/36;
eq[7] := c[0] = 1;
eq[8] := d[0] = 0;

tnx...

I'm trying to Graph a Piecewise Defined Function.

j(x) = piecewise(x < 0, x^2 + 1, 0 <= x, x - 1);

It doesn't provide a plot option.

Download piecewisegraph.mw


 

Function Misrepresented

y^2-x = 1"(->)"[[y = (x+1)^(1/2)], [y = -(x+1)^(1/2)]]NULL

NULL

 

Expected output for y

 

y = `&+-`(sqrt(1+x))

 

Yes it's the same meaning but i would like less to look at.


 

Download funct_misrep.mw


data := [[0., 9.1300, 0.931e-1, 0.899e-1, .1000, 0.], [30.0000, 8.9300, .1270, .1230, .2270, 0.49e-2], [60.0000, 8.6000, .1510, .1390, .4920, 0.153e-1], [90.0000, 8.2800, .1540, .1490, .7780, 0.249e-1], [120.0000, 7.9700, .1540, .1570, 1.0700, 0.329e-1], [150.0000, 7.8600, .1540, .1600, 1.1700, 0.348e-1], [180.0000, 7.8100, .1530, .1530, 1.2100, 0.404e-1], [210.0000, 7.7700, .1400, .1420, 1.2800, 0.432e-1]];
des := [diff(y1(t), t) = -k1*y1(t)-k2*y1(t), diff(y2(t), t) = k2*y1(t)-k3*y2(t), diff(y3(t), t) = k1*y1(t)+k3*y2(t)-k4*y3(t), diff(y4(t), t) = k4*y3(t)-k5*y2(t)*y4(t)+k6*y5(t), diff(y5(t), t) = k5*y2(t)*y4(t)-k6*y5(t)];
ics := seq((y || i)(0) = data[1, i+1], i = 1 .. 5):
Error, unable to match delimiters
Typesetting:-mambiguous(Typesetting:-mambiguous(ics Assign seq

  lparlpary verbarverbar irpar(0) equals data(1comma i + 1)comma 

  i equals 1 periodperiod 5rparcolon, 

  Typesetting:-merror("unable to match delimiters")))
res := dsolve({ics, des[]}, numeric, parameters = [k1, k2, k3, k4, k5, k6]);
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations
timeList := [0, 30, 60, 90, 120, 150, 180, 210];
              [0, 30, 60, 90, 120, 150, 180, 210]
sse := proc (k1, k2, k3, k4, k5, k6) res(parameters = [k1, k2, k3, k4, k5, k6]); add((rhs(select(has, res(timeList[i]), y1)[])-data[i, 2])^2+(rhs(select(has, res(timeList[i]), y2)[])-data[i, 3])^2+(rhs(select(has, res(timeList[i]), y3)[])-data[i, 4])^2+(rhs(select(has, res(timeList[i]), y4)[])-data[i, 5])^2+(rhs(select(has, res(timeList[i]), y5)[])-data[i, 6])^2, i = 2 .. 8) end proc;
proc(k1, k2, k3, k4, k5, k6)  ...  end;
c := GlobalOptimization:-GlobalSolve(('sse')(k1, k2, k3, k4, k5, k6), k1 = 0 .. 1, k2 = 0 .. 1, k3 = 0 .. 1, k4 = 0 .. 1, k5 = 0 .. 1, k6 = 0 .. 1, timelimit = 10); [.219132447080011505, [k1 = 0.852482740113834e-3, k2 = 0.52683998680924474e-4, k3 = 0., k4 = 0.5113239298267808e-1, k5 = 0.4363021255887466e-2, k6 = 0.]];
Error, `GlobalOptimization` does not evaluate to a module
[0.219132447080011505, [k1 = 0.000852482740113834, 

  k2 = 0.000052683998680924474, k3 = 0., 

  k4 = 0.05113239298267808, k5 = 0.004363021255887466, k6 = 0.]]

res(parameters = c[2]):  p:=Array(1..5):  for n from 1 to 5 do     p[n]:=plots:-display(plots:-odeplot(res, [t, (y || n)(t)], t = 0 .. 210),plots:-pointplot([seq([data[i, 1], data[i, n+1]], i = 1 .. 8)]));  end do; plots:-display(p) ;
Error, unable to match delimiters
Typesetting:-mambiguous(Typesetting:-mambiguous(res(parameters 

  equals c(2))colon  pAssignArray(1periodperiod5)colon  for n 

  from 1 to 5 do     p(n)Assignplotscolon - displaylparplotscolon

   - odeplotlparrescomma lsqbtcomma lpary verbarverbar nrpar(t)

  rsqbcomma t equals 0 periodperiod 210rparcommaplotscolon - 

  pointplot((seq((data(icomma 1)comma data(icomma n + 1))comma i 

  equals 1 periodperiod 8)))rparsemi  end dosemi plotscolon - 

  display(p) , Typesetting:-merror("unable to match delimiters")))

 I cannot solve this equation. Please help me

How can I use maple to get the numerical solution of this non-autonomous system

sys_ode := diff(x(t), t) = ax(t)-bx(t)y(t)+2sin(t), diff(y(t), t) = -cy(t)+dx(t)y(t)

where a,b,c,d is parameters, they can take any value.

Hi! 

 

I'm having a weird issue. :( 

DEplot command worked when I initially ran the executed it, but then when I executed the entire worksheet, it didn't plot. Instead I just get this output. 

 

This is the entire worksheet so far without the output. 

 

If someone could please help me, I would greatly appreciate it!!!!!

 


restart;
with(DEtools);

eq1 := diff(y(x), x) = 3*x*y(x);

dsolve(eq1, y(x));

ini1 := y(0) = 3;

dsolve({eq1, ini1}, y(x));

ini2 := y(1) = -3;

dsolve({eq1, ini2}, y(x));

eq2 := y(x)*diff(y(x), x) = -x;

dsolve(eq2, y(x));

dsolve({eq2, ini1}, y(x));

restart;
with(plots);

eq := diff(y(x), x) = 3*x*y(x);

ini := y(0) = 5;


dsolve({eq, ini}, y(x));




sol := rhs(%);

plot(sol, x = -2 .. 2, y = 0 .. 10);
DEplot(eq, y(x), x = -2 .. 2, y = 0 .. 10, [[ini]]);





 

Does anyone use MTM toobox and why?

 

I discovered this through the eig help search and found the MATLAB format of [V, D] = eig(Matrix) for eigenvectors and eigenvalues as a single function call with output a nice shortcut vs two calls to LinearAlgebra[Eigenvalues], LinearAlegebra[Eigenvectors]

 

But, is the overhead and overmapping of standard maple functions worth it?   I find very little descriptions about the Maple Toolbox, specifically the MTM package.

Hi everyone:

how can I obtain the B from A?

A:=u[1, 0](t)*v[1, 3](t)+u[1, 1](t)*v[1, 2](t)+u[1, 2](t)*v[1, 1](t)+u[1, 3](t)*v[1, 0](t)
B:=u[1, 0](t)*v[1, 3](tau)+u[1, 1](t)*v[1, 2](tau)+u[1, 2](t)*v[1, 1](tau)+u[1, 3](t)*v[1, 0](tau)

tnx... 

Good day house.

Please I don't know why the solve command does not display any results in the following code. Kindly assist. Thank you in anticipation.

restart;
omega := v/h;
t := sum(a[j]*x^j, j = 0 .. 6)+a[7]*cos(omega*x)+a[8]*sin(omega*x);
r1 := diff(t, x$2);
r2 := diff(t, x$4);
c1 := eval(t, x = q+2*h) = y[n+2];
c2 := eval(r1, x = q) = f[n];
c3 := eval(r1, x = q+h) = f[n+1];
c4 := eval(r1, x = q+2*h) = f[n+2];
c5 := eval(r1, x = q+3*h) = f[n+3];
c6 := eval(r2, x = q) = g[n];
c7 := eval(r2, x = q+h) = g[n+1];
c8 := eval(r2, x = q+2*h) = g[n+2];
c9 := eval(r2, x = q+3*h) = g[n+3];
b1 := seq(a[i], i = 0 .. 8);
`k&Assign;solve`({c1, c2, c3, c4, c5, c6, c7, c8, c9}, {a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8]});

 

Please I found out that the MatrixInverse on the assignment statement P3 does not run for about three days now. Please kindly help to simplify the code. Thank you and kind regards.

restart; omega := v/h;
r := a[0]+a[1]*x+a[2]*sinh(omega*x)+a[3]*cosh(omega*x)+a[4]*cos(omega*x)+a[5]*sin(omega*x);
b := diff(r, x);

c := eval(b, x = q) = f[n];
d := eval(r, x = q+3*h) = y[n+3]; e := eval(b, x = q+3*h) = f[n+3];
g := eval(b, x = q+2*h) = f[n+2];
i := eval(b, x = q+h) = f[n+1];
j := eval(b, x = q+4*h) = f[n+4];
k := solve({c, d, e, g, i, j}, {a[0], a[1], a[2], a[3], a[4], a[5]});
Warning,  computation interrupted
assign(k);
cf := r;
s4 := y[n+4] = simplify(eval(cf, x = q+4*h));
s3 := y[n+2] = simplify(eval(cf, x = q+2*h));
s2 := y[n+1] = simplify(eval(cf, x = q+h));
s1 := y[n] = simplify(eval(cf, x = q));

with(LinearAlgebra);
with(plots);
h := 1;
YN_1 := seq(y[n+k], k = 1 .. 4);
A1, a0 := GenerateMatrix([s1, s2, s3, s4], [YN_1]);
eval(A1);
YN := seq(y[n-k], k = 3 .. 0, -1);
A0, b1 := GenerateMatrix([s1, s2, s3, s4], [YN]);
eval(A0);
FN_1 := seq(f[n+k], k = 1 .. 4);
B1, b2 := GenerateMatrix([s1, s2, s3, s4], [FN_1]);
eval(B1);
FN := seq(f[n-k], k = 3 .. 0, -1);
B0, b3 := GenerateMatrix([s1, s2, s3, s4], [FN]);
eval(B0);
ScalarMultiply(R, A1)-A0;
det := Determinant(ScalarMultiply(R, A1)-A0);
P1 := A1-ScalarMultiply(B1, z);
P2 := combine(simplify(P1, size), trig);
P3 := MatrixInverse(P2);
P4 := A0-ScalarMultiply(B0, z);
P5 := MatrixMatrixMultiply(P3, P4);
P6 := Eigenvalues(P5);
f := P6[4];
T := unapply(f, z);
implicitplot(f, z = -5 .. 5, v = -5 .. 5, filled = true, grid = [5, 5], gridrefine = 8, labels = [z, v], coloring = [blue, white]);

 

The equation 

(x3−13x2+55x−73)/(x−1)=0

has at least one real solution. Give an interval on which the Intermediate Value Theorem assures that there is a solution to this equation.

Note: Enter your interval in the form [a,b].

can I get help please? I cant seem to do this.

Dear Helpers

I have a RootOf such as below. My question is how can I obtain a simple answer for it, or how can I simplify it? Also, what's the difference between the case "index=1" and the cases "index=2" and "index=3"?

" RootOf(6*_Z^3+(27+3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z^2+(3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2-9*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+90*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-18*l^4+6*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2-81+45*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z-324-3*l^8+l^10*RootOf(_Z^2*l^2+3*_Z^4-3)^2+108*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)-3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^6+sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^8*RootOf(_Z^2*l^2+3*_Z^4-3)^2-63*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+30*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2+45*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2+351*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-108*l^4, index = 1) "

Thank you so much!

 

We are in the very beginning of implementation of Maple in our office, and are trying to write the first programs.

For the moment we are building a material database, starting from Excel, and generating a Maple library out of that. Afterwards we are going to pull the data from the library, and show it in a nice output.

We've so far managed to write the library, but we are running into problems with variable names.

In Maple 2D notation a variable with indexes including a comma is no problem. My specific problem is, how can I use those variables in a program in Maple 1D notation.

Here are some examples of variables which are relevant.

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