Why does Maple not have texture fill for plotting. Even the old Pascal had it.

Maple Help points out

All Programs>Maple 2017>Classic Worksheet Maple 2017

link that I do not have. Is there a reason for this and how do I fix this?

I don't understand whats wrong with my code
r(x):=proc(x) 
    local y
    if (x=1)
    then    
    y=1;
    else 
        y=0;
    end if; 
    return y
    end proc    

Error, reserved word `if` unexpected

I'm also having trouble getting this code to work:

h(x):=proc(x) 
    if (x= infinity)
    then    
    return 1 ;
    else 
        return 0;
        end if; 
    end proc 
theres no error but h(0) just returns h(0) not either value as I would like

Dear All,

How do I use the piecewise function z[k]:=piecewise(k<>0,0,k=0,1) in the loop 

for k from 0 to 10 do

(sum((m^2-1)/z(z[k]+1),m,0..k))

end:

i want the value of z[k] to be automatically used in the loop for the corresponding value of k

Graph the real roots of the equation x3 + (a − 3)3x2 − a2x + a3=0 for a ∈ [0, 1].

Sol:=[solve(x^3+(a-3)^3*x^2-a^2*x+a^3=0,x)];

for i from 1 to 3 do

R||i:=unapply(Sol[i],a):

print(plot(R||i(a),a=0..1,numpoints=500)); end od:

Is there a simplier way to solve this. If not why did they choose this path?

@Carl Love 

Can you help me to make a 1D plotting by using this data? 

Dear all

Can we use maple to compute integral using the residual theorem

Compute_integral.mw

We can consider the contour a circle that contains one pole (Ipi/2)

Many thanks

Hi! I have been trying to calculate the value of theClenshaw derivative. I can get the regular clenshaw to work but not the derivative. I'm going off of the thread (https://scicomp.stackexchange.com/questions/27865/clenshaw-type-recurrence-for-derivative-of-chebyshev-series). (P.s notice I have halfed the A[z+1] term. I have tried both ways but the overall result is wrong so I am guessing something more important is wrong).

This is my code so far

Clenshaw_Dx_1D:=proc(z,C,Nm,s)
local i,k,A,B: global Clen1D_Dx: 

A := Vector(z..Nm+1+z);
B := Vector(z..Nm+1+z);

for k from Nm-1+z by -1 to 1+z do
A[k] := C[k] + 2*s*A[k+1] - A[k+2]:
od:

for k from Nm-1+z by -1 to 1+z do
B[k] := 2*A[k+1] + 2*s*B[k+1] - B[k+2]:
od:
Clen1D_Dx :=  A[z+1]/2 + s*B[1+z] - B[2+z]:

end:

Where z could be 0 or 1 depending on what index your C array starts at. C is the array of chebyshev coefficients of a Chebyshev series appromating u(x) for example. The Chebycoeff1D procedure calculates the Chebyshev coefficients  and the code below calls the procedure for a specific function. Try it with some values of xM, for example 4-10.
 

Chebycoeff1D:=proc(express,Nn,C2,A,B)
local Cfac,fac1x,fac2x,k,K;

fac1x:=Pi/Nn;
  for k from 1 to Nn do 
  Cfac(k):=eval(subs(x=evalf(cos(fac1x*(k-0.5)))*A+B,evalf(express))); 
  od; unassign('k'):                    
  for K from 1 to Nn do
    fac2x:=Pi*(K-1)/Nn;
    C2[K-1]:=(2/Nn)*add(Cfac(k)*evalf(cos(fac2x*(k-0.5))),k=1..Nn);
  od:
end:

nn := 1.0:
Lc := 0.0: Rc := 1.0:
func:=nn*sin(2*Pi*x)+0.5*nn*sin(Pi*x):
Chebycoeff1D(func,xM+1,C,0.5*(Rc-Lc),0.5*(Rc+Lc)):

Once you have the C array call Clenshaw_Dx_1D. For example
 

Clenshaw_Dx_1D(0,C,xM+1,0.0);  # Evalutes f'(x) in the middle of the domain.

Check with

subs(x=0.5, diff(func,x));

The overall "pattern/shape" of the derivative values is correct, just not the amplitude/values. Any help here? Where is the Clenshaw derivative procedure going wrong.

I have a region that can be graphed by A := plots:-inequal([evalc(abs(z)) < 3, evalc(1 < abs(z - 1))], x = -5 .. 5, y = -5 .. 5)

From there, I need to use the transformation 

M := (3 + sqrt(5))/2*(z - x1)/(z - x2) where x1 := (9 - sqrt(45))/2 and x2 := (9 + sqrt(45))/2 

How do I do this? Do I need to extract data points from the first graph? Is there an easier way to do this? 

I would like to plot a hyperbola using the polarplot command, such as the following:

polarplot(3/(1-1.5*sin(theta)), coordinateview = [0 .. 10, 0 .. 2*Pi])

But the graph includes the asymptotes, which I would not like to be included. I have tried the discont=true command, but it completely changes the shape of the graph and no longer looks like a hyperbola:

polarplot(3/(1-1.5*sin(theta)), coordinateview = [0 .. 10, 0 .. 2*Pi], discont = true)

 How would I get the hyperbola above to display with no asymptotes?

Thanks

The Library:-RedefineTensorComponent in the physics package has got some serious problems, I guess in the update. It is not assigning the components at the right location for a tensor of rank 4 and above. It is working for tensors of rank 1, 2 and 3. Can someone please look into the issue and help out?

I am attaching the code with an example to indiacate the problem. Redefine.mw

Hi,

Defining variables named H₂ or O₂ doesn't pose any problem (H__2 for instance), but could it be possible in Maple 2019 to define a variable named H₂O₂ ?

Thanks in advance

Hello,

I have a question regarding the dsolve function in Maple. I am trying to solve a system of 5 first order ODE's. First I solve the differential equations to arrive at the general solution using the dsolve command. Here Maple already produces very large and slow output while I would actually expect a much more compact output. 

Then I want to evaluate the final solution by using the initial conditons. At this point Maple keeps 'evaluating..' and does not come up with a solution. Only when I fill in all the parameters Maple finds a solution ( still very slow ). However as I want to be able to play with the input parameters after evaluation I want a solution without having to fill in the parameters prior to solving the system.

My question is whether I am presenting the system of ODE's in the right way to Maple. Should I rewrite the system for Maple to be able to produce a more dense solution? Or should I for instance use Laplace Transform to simplify the equations prior to solving?

Please help!

Joa

restart;

eq1 := sig_total-sig2(t)-sig3(t)-sig4(t)-sig5(t)+T1*(-(diff(sig2(t), t))-(diff(sig3(t), t))-(diff(sig4(t), t))-(diff(sig5(t), t))) = u1*(diff(eps(t), t));
eq2 := sig2(t)+T2*(diff(sig2(t), t)) = u2*(diff(eps(t), t));
eq3 := sig3(t)+T3*(diff(sig3(t), t)) = u3*(diff(eps(t), t));
eq4 := sig4(t)+T4*(diff(sig4(t), t)) = u4*(diff(eps(t), t));
eq5 := sig5(t)+T5*(diff(sig5(t), t)) = u5*(diff(eps(t), t));

dsolve({eq1, eq2, eq3, eq4, eq5}, {eps(t), sig2(t), sig3(t), sig4(t), sig5(t)});
sig_total := sig_0;
desys := {eq1, eq2, eq3, eq4, eq5}; ic := {eps(0) = sig_0/(E1+E2+E3+E4+E5), sig2(0) = sig_0/(1+(E1+E3+E4+E5)/E2), sig3(0) = sig_0/(1+(E1+E2+E4+E5)/E3), sig4(0) = sig_0/(1+(E1+E3+E2+E5)/E4), sig5(0) = sig_0/(1+(E1+E3+E4+E2)/E5)};

solution := combine(dsolve(desys union ic, {eps(t), sig2(t), sig3(t), sig4(t), sig5(t)})); assign(solution);

E_total := 33112; a := .1; b := .2; c := .15; d := .2; e := .35;

E1 := a*E_total; E2 := b*E_total; E3 := c*E_total; E4 := d*E_total; E5 := e*E_total; T1 := 1; T2 := 10; T3 := 100; T4 := 1000; T5 := 10000; u1 := T1*E1; u2 := T2*E2; u3 := T3*E3; u4 := T4*E4; u5 := T5*E5; sig_0 := 2;

plot(eps(t), t = 0 .. 672, y = 0 .. 0.12e-3);
y := eval(eps(t), t = 672);
evalf(y);
 

Download Maxwell_solve_new_method_5_units.mwMaxwell_solve_new_method_5_units.mw

How do I extract the real solutions of a polynomial that has solutions both in real and complex domains?

How do I work with polynomials over finite fields?  not just the integers mod p but over  https://www.maplesoft.com/support/help/Maple/view.aspx?path=GF