MaplePrimes Questions

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≔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.

I'm trying to be able to add bullet points to this Plot as well as where they cross. Download provided below. Thanks in Advance.

Example 7: Equations of Perpendicular Lines

 

NULLy = -(3/2)*x+2

 

2*x-3*y = 5; "_noterminate"

2, -1; "_noterminate"and m = 2/3; "_noterminate"

y = -5/3+(2/3)*x

(1)

 

``


Download mp.mw

Why the performance of Maple GUI is bad. It is really hard to type anything in thew gui. 

I have a relatively involved modeling application that tracks  a particle beam through a beam line for multiple passes. The beam is described by a Maple Record with a number of different entries; one of them being a table with a 6-vector of numbers for each particle in the beam.

Each pass this beam gets updated. I would like to be able to save the whole record into a filel (i.e. append the file each pass) and then be able to read it back in; regenerating the Record structures. Is this possible?

(I'd normally upload a mwe; except this code uses a large package that would have to be installed. This is something I do not consider reasonable to ask for.)

If there is an explicit answer like: call this function, give it your Record and it should work; and call this reading function to recreate the Record, that would be swell. Or maybe some pointers in the right direction...Maple has so many read, write, save and restore functions that a systematic test is daunting.

TIA,

Mac Dude


 

NULLrestart

Digits := 10:

with(plots):

with(CurveFitting):

with(plottools):

v := .7:

Disp := 20:

esp := 800000:

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) 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) end proc

(1)

f := proc (x) options operator, arrow; 20*exp(-(1/2000000)*(x-10000)^2)+15*exp(-(1/2000000)*(x-13800)^2) end proc:

 

 

u := proc (x, t) options operator, arrow; int(E(x-xi, t)*f(xi), xi = 0. .. 20000) end proc;

proc (x, t) options operator, arrow; int(E(x-xi, t)*f(xi), xi = 0. .. 20000) end proc

(2)

``

plot(u(x, t), x = 1500, t = 0 .. 60000, numpoints = 100)

Error, (in plot) unexpected options: [x = 1500, t = 0 .. 60000]

 

 

NULL

 

``

NULL


 

Download antegral.mw

I'm reading this book : Algebra and Trigonometry - Real Mathematics, Real People, 7th Edition

Chapter 1 Functions and Their Graphs, Example 7, Monthly Wage , Page 80.

I'm trying to plot for a monthly wage.

The inital equation is:

y = 2000 + 0.1*x

Then other changes:

a. Sales are $1480 in August. What are your wages for that month?
b. You receive $2225 for September. What are your sales for that month?

I'm lost. Which is better Numerical Solution or Graphical Solution?

Thanks in Advance,

Paul

 

Hi,

I want to have the fractal of von koch as shown in the photo, and then animate it? the command Kochcurve does not give exactly the objective. some ideas?

Thanks

VonKochFractal.mw

f := proc (a, b) options operator, arrow; (1/2)*b+(1/2)*arccos(sin(2*a-b)/tan(b)) end proc; dis := proc (A, B) options operator, arrow; sqrt(inner(A-B, A-B)) end proc; bisA := proc (A, B, C) local b, c, M; b, c := dis(A, C), dis(A, B); M := (b*B+C*c)/(b+c); x*(A[2]-M[2])+y*(M[1]-A[1])+A[1]*M[2]-A[2]*M[1] end proc; P := proc (a0, b0) local a, b, c, p1, p2, p3, p4, r, II; a := evalf(a0); b := evalf(b0); c := f(a, b); if b0-0.1e-2

Hi,

The VectorCalculus package offers a convenient way to compute multivariate integral over a non-square geometry. See help("VectorCalculus/int"). Here is an example taken from this help page:

 

with(VectorCalculus):

int( x*y, [x,y] = Triangle( <0,0>, <1,0>, <0,1> ) );

 

The question is: is there a way to draw the area “Triangle( <0,0>, <1,0>, <0,1> )” in order to quickly check that it matches the expected region of integration that one needs?

I am trying to plot the solution of the DE in the interval [0.3,4]

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