MaplePrimes Questions

I have the following equation:

.9215999996*z^4+.9999999996*z^4*s^4-2.079999999*z^4*s^2-3.212799999*s*z^3+12.63999999*s^3*z^3-7.999999997*s^5*z^3-24.17694443*s^4*z^2-.3071555554*s^2*z^2+23.99999999*s^6*z^2-31.99999999*s^7*z+6.804911108*s^3*z+11.58777777*s^5*z+15.99999999*s^8+5.692222220*s^6-4.853219442*s^4=0

where s is real and z is complex.

 

How do I plot the graph Im(z) versus s?

 

Thank you very much for your help!

I have the following two equations:

u=1-(8*(10.3968*t^2-5.8368*t*f-.229376*f^2-5.1984))/(4.56*t^2-2.56*t*f+.8192*f^2+2.28)^2;

x = f+t-(8*(-2.28*t+.64*f))/(2.28+2*(-t+.64*f)^2+2.56*t^2);

 

How can I plot

(i) a 2-dimensional graph that u versus x at t=0?

(ii) a 3-dimensional graph that u versus x and t ?

 

Thank you a lot for your help!

 

This is a new error I have not seen before.  

Trying to verify my solution for the diffusion pde in cylinderical coordinates wth no angle theta dependency.

unassign('z,t,r,u');
lap:=diff(u(r,z,t),r$2)+ 1/r*diff(u(r,z,t),r)+diff(u(r,z,t),z$2);
bc:=u(r,0,t)=0,u(r,1,t)=0, u(1,z,t)=0;
ic:=u(r,z,0) = f(r,z);
pdsolve([diff(u(r,z,t),t) = lap,bc,ic],u(r,z,t)) assuming t>0

The error is

 

Error, (in assuming) when calling 'BesselJZeros'. Received: '0th zero of BesselJ(0,x) not defined'

Is this a bug or Am I doing something wrong? 

update

fyi, it also fails with same error when trying boundedseries HINT

restart;
unassign('z,t,r,u');
lap:=VectorCalculus:-Laplacian(u(r, z, t), cylindrical[r, theta,z]);
bc:=u(r,0,t)=0,u(r,1,t)=0, u(1,z,t)=0;
ic:=u(r,z,0) = f(r,z);
pdsolve([diff(u(r,z,t),t) = lap,bc,ic],u(r,z,t),HINT=boundedseries(r=0))

I am using Maple 2019.1 with Physics 366 on windows 10

 

odeplot plots "paths", i. e. keeps the points at every time.

I want to see only the "current" position where the particle is in the frame. I want the animation to show the particle in movement, not the path it is drawing.

Thanks in advance.

Hello,

I converted the results in maple to latex format. I copy them in math type software an error is appeared which shows the volume of latex file is very big to transfer.

If I want to use other software to input latex input and then I copy the equation in  office, please guide me how I can write the results obtained  in maple and transfer in word by other softwares?

Thanks

I want to convert the following expression to form of ‘a+bi’,unfortunately,it can't be successful,Can anyone help me?

 

2*omega^2*B[1]*Zeta*omega[0] - omega^3*B[1]*I + omega*B[1]*omega[0]^2*I - 2*I*B[0]*Zeta*omega*omega[0] - B[0]*omega^2 + B[0]*omega[0]^2

f1:=proc(n) local x,y,i,rx ,ry,dumbindex:  

dumbindex:=1:  

xdumb[1]:=n:  

ydumb[1]:=n:  

for i from 1 to  n do    

d:=0.00197241:  flag:=1:  

while(d>0.0019724 or flag=1) do    

rx:=rand(1..n):  x[i]:=rx():  ry:=rand(x[i]..n):  y[i]:=ry():  flag:=check(x[i],y[i],xdumb,ydumb,dumbindex):  

if flag=1 then  dumbindex:=dumbindex+1  xdumb[dumbindex]:=x[i]:  ydumb[dumbindex]:=y[i]:  fi:  

p,d:=f(x[i],y[i]):  

end :  

xl[i]:=x[i]:yl[i]:=y[i]:  

if  i=1 then   temp:=p:xg:=x[i]:yg:=y[i];  

else  

if p<temp then   temp:=p:xg:=x[i]:yg:=y[i] ;  fi:  

fi:

end do:  

return xg,yg:  

end proc;

 

That's the definition of my procedure. It reported illegal use of an object as a name error. Could anyone help have a look and point out what goes wrong. I just started using Maple. Thanks a million in advance.

Hi,

Is there any way to define a specific part of an equation to be integrated in respect to time T[1]?

In my attached example I try to integrate the right hand side after isolating for C1D__1 (removing the effect of C1D__1) and maybe isolation is not the right thing to do. Any idea?


coeff1 := 5*L__1*C*exp(I*T[1]*e*eta)*e*eta*C2/omega__n-(2*I)*D__1*C1/omega__n+L__1*C*exp(I*T[1]*e*eta)*e^2*eta^2*C2/omega__n^2-(3*I)*R__n*C*C1^2*C2/omega__n+3*gamma__1*C*C1^2*C2-I*R*C*C1/omega__n+13*L__1*C*exp(I*T[1]*e*eta)*C2*(1/2):
isolate(coeff1, D__1)*C1:

integrate the right hand side of the equation with T[1]

Thanks,

Bahar

 

 

Hi,

I have an expression like x:= C*exp(I*T[1]*e*eta) and want to define e*T[1] as T[2]. How can I do this with coding?

Regards,

Bahar

 

 

Maple 2019.1 with Physics version 362  gives this strange error on this pde

restart;
pde := x*diff(u(x, y), x) + y*diff(u(x, y), y) = -4*x*y*u(x, y);
ic := u(x, 0) = exp(-x);
sol:=pdsolve([pde,ic],u(x,y));

Error is

Error, (in PDEtools:-casesplit) equation of unknown type integer : 1

Is this a bug? It shows only when using exp(-x). Changing it to exp(-x^2) or exp(x) do not show the error, even though Maple can't solve it.

On windows 10.

Hi, I want to transform the formula as following,how can I do?

PS: code 

(2*I*Zeta*omega*omega[0] - omega^2 + omega[0]^2)*fourier(omega[r](t), t, omega) = (omega*B[1]*I + B[0])*fourier(delta(t), t, omega)


 

restart:
with(LinearAlgebra):
with(plots):
with(geometry):
with(plottools): # On appelle alpha la moitié de l'angle de rotation de la roue menée par tour de roue menante. alpha=Pi/n en raduans soit Pi/5=36° pour 5 rainures.. On a alors les relations suivantes entre l'entaxe E, le rayon de la roue ùenante R1 et le rayon de la roue menée R2 : R1=E.sin(alpha), R2=E*cos(alpha) Intersection du cercle (O,R2) avec la droite tan(phi)x-r/cos(phi), on obtient les coordonnées de P3
sol:=allvalues(solve([tan(phi)*x-r/cos(phi)=y,y^2+x^2=R2^2],[x,y])): # Intersection de 2 cercles
sol1:=allvalues(solve([(x-E)^2+y^2=(R-a)^2,y^2+x^2=R2^2],[x,y])): # Coordonnées des points du pourtour de l'élément de croix #point(O,0,0):
phi:=Pi/5:
R2:=5:
r:=1/4:
E:=R2/cos(phi):
R:=R2*tan(phi):
a:=0.5:
[(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)];
geometry[point](P1,(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)):
geometry[point](P2,(R2/2)*cos(phi)+r*sin(phi),(R2/2)*sin(phi)-r*cos(phi)):

[(9/4)*cos((1/5)*Pi), (9/4)*sin((1/5)*Pi)]

xP2:=(R2/2)*cos(phi)+r*sin(phi):
yP2:=(R2/2)*sin(phi)-r*cos(phi):
xP1:=(R2/2-r)*cos(phi):
yP1:=(R2/2-r)*sin(phi): # Equation paramétrique du segment OP1 (t varie de 0 à 1) ; non pris en compte
x1:=t*(0-xP1)+xP1:
y1:=t*(0-yP1)+yP1:
n1:=5:
dt:=1/(n1-1): #t varie entre 0 et 1

for i to n1 do tau:=(i-1)*dt:
xx[i]:=evalf(subs(t=tau,x1)):
yy[i]:=evalf(subs(t=tau,y1)): #print(i,xx[i],yy[i]);
od: # Equation paramétrique du quart de cercle P1P2 de la rainure (t varie de 0 à 1)
x2:=R2/2*cos(phi)+r*cos(t): #attention au sens de rotation du parcours de l'objet
y2:=R2/2*sin(phi)+r*sin(t):
n2:=6:
dt:=Pi/2/(n2-1): #arc de Pi/2
for i to n2 do
tau:=phi-Pi+(i-1)*dt:
xx[i]:=evalf(subs(t=tau,x2)):
yy[i]:=evalf(subs(t=tau,y2)):
od:
for i to n2 do
Vector[row]([i,xx[i],yy[i]])
od:

droite:=plot((tan(phi)*x-r/cos(phi),x=0..3),linestyle=dot,color=blue):

xP3:=evalf(subs(op(1,sol[1]),x)):
yP3:=evalf(subs(op(1,sol[1]),y)):
xP2:=evalf(subs(op(1,sol[2]),x)):
yP2:=evalf(subs(op(1,sol[2]),y)):
xP4:=evalf(subs(op(1,sol1[1]),x)):
yP4:=evalf(subs(op(1,sol1[1]),y)):
xP5:=E-(R-a):
yP5:=0:
x3:=t*(xP3-xP2)+xP2:
y3:=t*(yP3-yP2)+yP2:
n3:=10:
dt:=1/(n3-1): #t varie entre 0 et 1

for i to n3 do
tau:=(i-1)*dt:
xx[i+n2]:=evalf(subs(t=tau,x3)):
yy[i+n2]:=evalf(subs(t=tau,y3)):
od:

for i to n3 do Vector[row]([i,xx[i],yy[i]]) od:

x4:=xP5+R-a+(R-a)*cos(t):#attention au sens de rotation du parcours de l'objet
y4:=(R-a)*sin(t):
n4:=11:
eta:=arcsin(yP4/(R-a)):
dt:=(-eta)/(n2-1)/2:#arc de Pi/2

for i to n4 do
tau:=(Pi+eta)+(i-1)*dt: #recherche de tau ?
xx[i+n2+n3]:=evalf(subs(t=-tau,x4)):
yy[i+n2+n3]:=evalf(subs(t=-tau,y4)):
od:

for i to n4 do
Vector[row]([i,xx[i],yy[i]])
od:

n:=n2+n3+n4;
for i to n do
Vector[row]([i,xx[i],yy[i]])
od:

27

figure:=NULL:
for i from 0 to n do
xx[0]:=0:
yy[0]:=0:
figure:=figure,[xx[i],yy[i]]:
od:

figure:=[figure]:

polygonplot(figure,scaling=constrained,color=yellow);#,view=[-0.1..5,-0.1..3]

for i to n do
X[i]:=xx[i]:
Y[i]:=yy[i]
od:

d1:=plottools[disk]([xP1,yP1],0.05,color=blue):############ non définis
d2:=plottools[disk]([xP2,yP2],0.05,color=red):############ non définis

n := 27; ##### ci-dessus, il faut arrêter le trait à l'axe Ox ############### il suffit de prendre le symétrique puis 4 rotations de Pi/5 #############
symOX:= pt -> [pt[1],-pt[2]]:
rot:=proc(t,pt) [pt[1]*cos(t)-pt[2]*sin(t),pt[1]*sin(t)+pt[2]*cos(t)]; end:
figure1:=[ op(figure),op(map(symOX,figure))]:
croix:=op(figure1):

27

for i to 4 do croix:=croix,op(map( pt -> rot(i*2*Pi/5,pt),figure1) ): od:
croix:=[croix]:
polygonplot(croix,scaling=constrained,color=yellow); #How arrange this drawning and animate it. Thank you.

 


 

Download GenovaDrive.mw

[Just in case the above is not a 100%-correct presentation of the original code, I've left it below in its origianally posted form.--Carl Love as Moderator]

restart:with(LinearAlgebra):with(plots):with(geometry):with(plottools): On appelle alpha la moitié de l'angle de rotation de la roue menée par tour de roue menante. alpha=Pi/n en raduans soit Pi/5=36° pour 5 rainures.. On a alors les relations suivantes entre l'entaxe E, le rayon de la roue ùenante R1 et le rayon de la roue menée R2 : R1=E.sin(alpha), R2=E*cos(alpha) Intersection du cercle (O,R2) avec la droite tan(phi)x-r/cos(phi), on obtient les coordonnées de P3 sol:=allvalues(solve([tan(phi)*x-r/cos(phi)=y,y^2+x^2=R2^2],[x,y])): Intersection de 2 cercles sol1:=allvalues(solve([(x-E)^2+y^2=(R-a)^2,y^2+x^2=R2^2],[x,y])): Coordonnées des points du pourtour de l'élément de croix #point(O,0,0): phi:=Pi/5:R2:=5:r:=1/4:E:=R2/cos(phi):R:=R2*tan(phi):a:=0.5: [(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)]: geometry[point](P1,(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)): geometry[point](P2,(R2/2)*cos(phi)+r*sin(phi),(R2/2)*sin(phi)-r*cos(phi)): xP2:=(R2/2)*cos(phi)+r*sin(phi):yP2:=(R2/2)*sin(phi)-r*cos(phi): xP1:=(R2/2-r)*cos(phi):yP1:=(R2/2-r)*sin(phi): Equation paramétrique du segment OP1 (t varie de 0 à 1) ; non pris en compte x1:=t*(0-xP1)+xP1: y1:=t*(0-yP1)+yP1: n1:=5: dt:=1/(n1-1):#t varie entre 0 et 1 for i to n1 do tau:=(i-1)*dt: xx[i]:=evalf(subs(t=tau,x1)): yy[i]:=evalf(subs(t=tau,y1)): #print(i,xx[i],yy[i]); od: Equation paramétrique du quart de cercle P1P2 de la rainure (t varie de 0 à 1) x2:=R2/2*cos(phi)+r*cos(t):#attention au sens de rotation du parcours de l'objet y2:=R2/2*sin(phi)+r*sin(t): n2:=6: dt:=Pi/2/(n2-1):#arc de Pi/2 for i to n2 do tau:=phi-Pi+(i-1)*dt: xx[i]:=evalf(subs(t=tau,x2)): yy[i]:=evalf(subs(t=tau,y2)): od: for i to n2 do Vector[row]([i,xx[i],yy[i]]) od: droite:=plot((tan(phi)*x-r/cos(phi),x=0..3),linestyle=dot,color=blue): sol[1]: xP3:=evalf(subs(op(1,sol[1]),x)):yP3:=evalf(subs(op(1,sol[1]),y)): xP2:=evalf(subs(op(1,sol[2]),x)):yP2:=evalf(subs(op(1,sol[2]),y)): xP4:=evalf(subs(op(1,sol1[1]),x)):yP4:=evalf(subs(op(1,sol1[1]),y)): xP5:=E-(R-a):yP5:=0: x3:=t*(xP3-xP2)+xP2: y3:=t*(yP3-yP2)+yP2: n3:=10: dt:=1/(n3-1):#t varie entre 0 et 1 for i to n3 do tau:=(i-1)*dt: xx[i+n2]:=evalf(subs(t=tau,x3)): yy[i+n2]:=evalf(subs(t=tau,y3)): od: for i to n3 do Vector[row]([i,xx[i],yy[i]]) od: x4:=xP5+R-a+(R-a)*cos(t):#attention au sens de rotation du parcours de l'objet y4:=(R-a)*sin(t): n4:=11: eta:=arcsin(yP4/(R-a)): dt:=(-eta)/(n2-1)/2:#arc de Pi/2 for i to n4 do tau:=(Pi+eta)+(i-1)*dt:#recherche de tau ? xx[i+n2+n3]:=evalf(subs(t=-tau,x4)): yy[i+n2+n3]:=evalf(subs(t=-tau,y4)): od: for i to n4 do Vector[row]([i,xx[i],yy[i]]) od: n:=n2+n3+n4; for i to n do Vector[row]([i,xx[i],yy[i]]) od: figure:=NULL: for i from 0 to n do xx[0]:=0:yy[0]:=0: figure:=figure,[xx[i],yy[i]]: od: figure:=[figure]: polygonplot(figure,scaling=constrained,color=yellow):#,view=[-0.1..5,-0.1..3] for i to n do X[i]:=xx[i]: Y[i]:=yy[i] od: d1:=plottools[disk]([xP1,yP1],0.05,color=blue):############ non définis d2:=plottools[disk]([xP2,yP2],0.05,color=red):############ non définis n := 27 ##### ci-dessus, il faut arrêter le trait à l'axe Ox ############### il suffit de prendre le symétrique puis 4 rotations de Pi/5 ############# symOX:= pt -> [pt[1],-pt[2]]: rot:=proc(t,pt) [pt[1]*cos(t)-pt[2]*sin(t),pt[1]*sin(t)+pt[2]*cos(t)]; end: figure1:=[ op(figure),op(map(symOX,figure))]: croix:=op(figure1): for i to 4 do croix:=croix,op(map( pt -> rot(i*2*Pi/5,pt),figure1) ): od: croix:=[croix]: polygonplot(croix,scaling=constrained,color=yellow); How arrange this drawning and animate it. Thank you.

I am working with the Physics package in Maple 2018. I have a spacetime with a metric g and the vector field l is the tangent to a null curve. I have two problems:

(1) I want to check in maple that indeed g(l,l)=0. For this I wrote:

SumOverRepeatedIndices(g_[mu,nu]l[~mu]l[~nu])

Firstly, when I try to execute this command, the program keeps on evaluating for a long time. So, I have to `interrupt the current operation', undo this command, save and reopen the file, execute the entire worksheet again and then somehow it works. However, as you can see in step(13) of the attached worksheet, Maple returns an expression for this command but doesn't cancel out terms and show that it is indeed zero. How can I do this?

(2) I wish to find $(\nabla_l l)^\nu$. Is there a way to directly do this? Or should I do something like:

Define(L[mu,~nu]=D_mu l[~nu])

SumOverRepeatedIndices(l[~mu]L[mu,~nu])

Thanks a lot.

Edit: I actually carried out these steps too. First, I defined $L_\mu^\nu$ in step (14). Then, when I ask for the non-zero components of L, I find that the L\mu,\nu components are given. How can I get the non-zero L\mu\nu components? Also, in step (16), I did the `SumOverRepeatedindices' but it returned only a symbolic result without evaluationg.

try2.mw

Hi I have encounted this error:

Error, (in diffalg/DAparse_polynomial) unknown indeterminate or coefficient outside base field, -Cp+Cf/(1+Kw*(-sigma*`&Delta;Pi`+`&Delta;P`)*(1-(1/2)*Fp/Ff)/Ks)

The text highlighted in bold is an equation that I typed in.

 

Can I know what it means?

I was trying to verify some solution to pde in textbook using Maple. The book gives the Cauchy data for this first order PDE in the form that Maple does not like when I used it as input. Here is an example

When I typed

 

pde:=u(x,y)*(x+y)*diff(u(x,y),x)+u(x,y)*(x-y)*diff(u(x,y),y)=x^2+y^2;
ic:=u(x,2*x)=0;
pdsolve([pde,ic],u(x,y))

Maple complained

Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unexpected 
occurrence of the variables {x} in the 2nd operand of u(x, 2*x) in the given initial conditions

In Mathematica it accepts such form of Cauchy data:

pde=u[x,y]*(x+y)*D[u[x,y],x]+u[x,y]*(x-y)*D[u[x,y],y]==x^2+y^2;
ic=u[x,2*x]==0;
DSolve[{pde,ic},u[x,y],{x,y}]

(I have not verified the above answer is correct or not).

Am I doing something wrong in Maple?

Or is there a trick or option or method to allow Maple to accepts such initial conditions? The book I am looking at has many problems where Cauch data is given on such form (i.e. u=0 on specific curve or in 3D on some specific surface). Here is another example

 

Which I'd like write its initial conditions as u(x,1/x)=0 but can not.

I could ofcourse solve the pde without these initial conditions, and then post process the answer to find the constants of integration from the Cauchy data given. But it will be nice if Maple would accept the IC as is.

 

 

 

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