I try to find kernel and image of a application whose i know the matrix.
restart;
with(LinearAlgebra);
A := Matrix([[1, 1, 1, -1], [-1, 1, -1, -1], [1, -1, -1, -1], [-1, -1, 1, 3]]);
k := op(NullSpace(A));#kernel
MatrixVectorMultiply(A, k);#check
C := op(ColumnSpace(A));
X := <x, y, z, t>;
F := MatrixVectorMultiply(A, X) - a*C[1] - b*C[2] - c*C[2];
G := op(convert(F, list));
solve({seq(G[i] = 0, i = 1 .. 4)}, {a, b, c}); why there is no solution ? Thank you.

How to fprintf say a plot into a docx file in maplesoft

kind help

I'm solving a set of non-linear equations:

eqn1 := W__1 + W__2 + W__3 + W__4 = 4;
             eqn1 := W__1 + W__2 + W__3 + W__4 = 4

eqn2 := W__1*zeta__1 + W__2*zeta__2 + W__3*zeta__3 + W__4*zeta__4 = 0;
eqn2 := W__1 zeta__1 + W__2 zeta__2 + W__3 zeta__3 + W__4 zeta__4 = 

  0


eqn3 := W__1*zeta__1^2 + W__2*zeta__2^2 + W__3*zeta__3^2 + W__4*zeta__4^2 = 2/3;
                         2               2               2
     eqn3 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      2   2
        + W__4 zeta__4  = -
                          3


eqn4 := W__1*zeta__1^3 + W__2*zeta__2^3 + W__3*zeta__3^3 + W__4*zeta__4^3 = 0;
                         3               3               3
     eqn4 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      3    
        + W__4 zeta__4  = 0


eqn5 := W__1*zeta__1^4 + W__2*zeta__2^4 + W__3*zeta__3^4 + W__4*zeta__4^4 = 2/5;
                         4               4               4
     eqn5 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      4   2
        + W__4 zeta__4  = -
                          5


eqn6 := W__1*zeta__1^5 + W__2*zeta__2^5 + W__3*zeta__3^5 + W__4*zeta__4^5 = 0;
                         5               5               5
     eqn6 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      5    
        + W__4 zeta__4  = 0


eqn7 := W__1*zeta__1^6 + W__2*zeta__2^6 + W__3*zeta__3^6 + W__4*zeta__4^6 = 2/7;
                         6               6               6
     eqn7 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      6   2
        + W__4 zeta__4  = -
                          7


eqn8 := W__1*zeta__1^7 + W__2*zeta__2^7 + W__3*zeta__3^7 + W__4*zeta__4^7 = 0;
                         7               7               7
     eqn8 := W__1 zeta__1  + W__2 zeta__2  + W__3 zeta__3 

                      7    
        + W__4 zeta__4  = 0


solve({eqn1, eqn2, eqn3, eqn4, eqn5, eqn6, eqn7}, {W__1, W__2, W__3, W__4, zeta__1, zeta__2, zeta__3, zeta__4});

However, the result looks like this:

How should I obtain numerical values for W1,W2,W3,W4,zeta1,zeta2,zeta3,zeta4 ?

I have an array with a list of parameters. Suppose there are three parameters, t1, t2 and t3. I like to evaulate the array for different values of the parameters. I can do it manually by a command such as

eval(array, {t1=1, t2=5, t3=10})

What is a more systematic approach and how can it be done? Suppose I do not know the number of paramters in advance and the number of parameters could possible be large. Once I know the number of parameters, I would like to map a list of values to the parameters.

Thanks in advance!

Is there any equivalent to Excel function arctan2?

This function takes x and y values, and returns values from -Pi to +Pi. In that way one can easily get the x and y values of an angle with correct signs.

Hello,

I have a problem regarding the physics package, which concerns, more specifically, the Define command.

It seems to be the case that it provides the wrong answer, and blatantly so, and I can't see what I am doing wrong for the life of me.

I want to build a fast way of doing coordinate transformations in general relativity, for which the basis vectors are called E[mu,~nu]. XX[~alpha] is the old coordinates expressed in terms of the new (in this case, the spherical coordinates are expressed in terms of cylindrical ones, i.e. [t,r, theta, phi]->[t, rho, phi,z].) As can be seen, only the first component in XX depends on time, but E contains weird elements (look at, for example, E[1,~2].) But when I do the calculation manually, it returns the expected results.

What am I doing wrong? I really appreciate any help you can provide.

Benzema

Are there any examples showing how to plot multiple items in one plot?

• What I have started with is to put a Plot component in my sheet, so that I can use that by its name.
• Next thing on the list was to generate different plot objects, by assigning each of them to a variable.
• To plot the whole bunch, I collected all the items I want to plot in a list
• And plotted them by SetProperty("MyPlotcomponent", value, display(plotlist))

I wonder if this is the best way to do it. Especially when adding more stuff afterwards, would it be possible to just add something to the plot component?

Do I need plot components at all?

What's the correct way of copying the current contents of a mutable object to a variable.

Apparently eval doesn't do the job.

Download Mutable.mw

the error in summation , How can this be corrected?

restart;

U[0] := x^2;

for k from 0 to 1 do U[k+1] := sum(U[s]*(diff(U[k-s], x)), s = 0 .. k) end do;
U[0]*(diff(U[0], x));
                                 3
                              2 x 
U[1];
                               0

simplified_MPH_to_RPM_to_engineering_units.mw

Good Morning,

I don't know why I have such difficulties with rotational units, esp the 2*pi conversions with radians and revolutions. 

   We are often back solving from MPH of the vehicle back to RPM of the wheel (revolutions per minute [or seconds]) to Rad/sec for engineering units such as torque*RPM calcs for power - kW.   

Almost everytime I try to use units to convert the units the worksheet will revert to "atomic" units.   I quote "atomic" as this is what I appears to be.:  See graphic (also simplifiied worksheet attached)

The unit length(radius) seems to be a factor that is not well documented.   It also seems the earlier rplies include extra work to use convert and possibly "symbolic = true") as an extra step per Acer in an earlier posting several years ago.   

When running calcs the convert, or simplify (to cancel length units m and inches out) the cancelations and the unit combinations seem to work awkwardly in this particular area.

Let  

f := beta/(2*sigma*GAMMA(1/beta))*exp(-(abs(x-mu)/sigma)^beta);

where mu::real, beta > 0, sigma > 0.
How can we obtain the expression of F?

F := int(f, x=-infinity..s)

Reverse problem (a priori simpler):
It's known that 

F := 1/2+signum(x-mu)/(2*GAMMA(1/beta))*(GAMMA(1/beta)-GAMMA(1/beta, abs((x-mu)/sigma)^beta))

How can we check that diff(F, x)=f?
Even with the assumptions on beta, mu and sigma and additional assumptions x>0 or x<0, I can't verify that diff(F, x)=f.

 

Hello, dear members! I try to solve the 1-D Wave Equation with initial and boundary conditions. And I have a problem when I'm trying to find the coeffs of Fourier series, that they depend on initial conditions. This is my code for Maple:
 

restart;
with(PDETools);

# set the PDE, ic(initial conds) and bc (boundary conds)
pde := diff(u(x, t), t, t) = a^2*diff(u(x, t), x, x);
ic := u(x, 0) = 0, D[2](u)(x, 0) = psi(x);
bc := u(0, t) = 0, u(l, t) = 0;
psi := piecewise(0 <= x and x <= l/2, x, l/2 <= x and x <= l, l - x);

# using Fourier method for solving (I require use this methods)
res := pdsolve(pde, HINT = T(t)*X(x));

ode1 := op(1, op(1, op(2, res)));

ode2 := op(2, op(1, op(2, res)));               

ode2 := subs(_c[1] = -lambda, ode2);
              

# solve 2 ODE
# equation rely X(x)
dsolve({ode2, X(0) = 0}, X(x));

sin(sqrt(lambda)*l) = 0;

solve(%, lambda, allsolutions);

lambda := (Pi*n/l)^2;

X := (x, n) -> sin(Pi*n*x/l);


# equation rely T(t)
ode1 := subs(_c[1] = -lambda, ode1);

dsolve({%, T(0) = 0}, T(t));

T := (t, n) -> C1[n]*sin(Pi*n*a*t/l);

# write the general solutions
Un := (x, t, n) -> T(t, n)*X(x, n);

# find the coeffs series
simplify(subs(t = 0, diff(u(x, t), t)) = psi);

((2/l)*l/(Pi*n*a)*int(psi*X(x, n), x = 0 .. l) assuming (x < l))
simplify(%) assuming n::posint;
combine(%);
C1[n] := factor(%)

But I got it:



How to find this C1 coefficient and solve this equation? Maple doesn't work now. This command does not give the result.
I attached mws files, so pdeLR_11.mw is another code for solving, but it is not working so((! 

Maybe I use this command wrong, or step-function is not convenient for maple?

Any helpful advice would be very cool. I am grateful in advance!

https://drive.google.com/file/d/1VznAvxcpF3ME9MMBjfKKifsbrbOb9uSY/view?usp=sharing

https://drive.google.com/file/d/15D7BCRwHliYb2OseH81i81RjQP7NRFgS/view?usp=sharing
 

I know that the "rotate" can rotate image of a function, but we see that the coordinate axes don't change as much. For example:

Although someone seems to have asked a similar question a long time ago, it was far from what I needed.

https://www.mapleprimes.com/questions/97812-Flip-Plot-Exchange-X-And-Yaxes

What I want to do is rotate the axes and the image at the same time rather than just flip axes.  The schematic diagram is as follows.

How can the value of uu(.1, .1, .1)  as a number, and what is the error when drawing?

restart;
uu := proc (x, y, t) options operator, arrow; e^(.2*t)*(e^(-x)+e^(-y)) end proc;

evalf(uu(.1, .1, .1));
                              2.  
                             -----
                              0.08
                             e    
plot3d(uu(x, y, .1), x = 1 .. 2, y = 1 .. 2);
Warning, expecting only range variables [x, y] in expression e^.2e-1*(e^(-x)+e^(-y)) to be plotted but found name e

I have a pair of vectors, one containing x values, and the other containing the corresponding y values.

I plot them using: plot(x_vector, y_vector)

If I have another pair of vectors, x2_vector and y2_vector.

How to I plot them on the same graph so that I can compare them?