MaplePrimes Questions

Respected members!

I downloaded this file because I've to study the geodesics over the cone, the sphere and the cylinder (and in general on a Riemannian manifold). But when I modify the equation, putting one of that I'm interested to, the file doesn't work (for example, it doesn't recognize the "assign" command). Could you help me, please?

 

http://www.maplesoft.com/applications/download.aspx?SF=34940/199536\GeodesicsSurface.pdf

Best,

Marzio

Respected member!

Please help me in finding the solution of this problem....
 

NULL

 

 

NULL

>   

``

NULL

restart

with(RealDomain):

r := .2:

k := 5;

5

(1)

BCSforNum1 := u(0) = 0, (D(u))(0) = 1+beta*(((D@@2)(u))(0)-(D(u))(0)*RealDomain:-`^`(k, -1)), (D(u))(m) = 0, ((D@@2)(u))(m) = 0;

u(0) = 0, (D(u))(0) = 1+.2*((D@@2)(u))(0)-0.4000000000e-1*(D(u))(0), (D(u))(6) = 0, ((D@@2)(u))(6) = 0

 

v(0) = 1, v(6) = 0

(2)

numsol1 := dsolve({BCSforNum1, BCSforNum2, ODEforNum1, ODEforNum2}, numeric, output = listprocedure)

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 

``


 

Download mplprimes.mw

Dear community, 

I'm new to maple and was wondering if you could help me out.

I have this curve where I want to make a line that goes from x=0.5 up to its value on the curve in this case 1.60 and then all the way to the y-axis so there is an area under the curve which I can color if that's even possible?

I have the following in maple:

k := 2.5;
                              2.5
Ca0 := 1;
                               1
v := 20;
                               20
Ca := Ca0*(1-x);
                             1 - x
Fa0 := Ca0*v;
                               20
Cb := Ca0*x;
                               x
ra := k*Ca*Cb;
                         2.5 (1 - x) x
plot(1/ra, x = 0 .. 1);
 

thank you for your help

Best Regards

Saad

I am curious whether anyone here can come up with a way to simplify the expression x1 to 15*Pi/32 in fewer exact, symbolic steps. The following was performed in Maple 2016.2 for Linux.

restart;

x1 := arcsin(1/2*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2));

arcsin((1/2)*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))

x2 := evalc(convert(x1,expln));

arctan((2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))

x3 := convert(x2, expln);

((1/2)*I)*(ln(1-I*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))-ln(1+I*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)))

# non-Pro wolframalpha can simplify x3 to 15*Pi/32 (but not x2 or x1).
#

x4 := combine(x3);

((1/2)*I)*ln(-(I*2^(1/2)*((2+2^(1/2))*(2+(2+2^(1/2))^(1/2)))^(1/2)+I*(2*2^(1/2)+4)^(1/2)+I*2^(1/2)+(-1+I))/(I*2^(1/2)*((2+2^(1/2))*(2+(2+2^(1/2))^(1/2)))^(1/2)+I*(2*2^(1/2)+4)^(1/2)+I*2^(1/2)+1+I))

x5 := simplify(x4);

((1/2)*I)*(-ln(2)+ln(-(2+(2+2^(1/2))^(1/2))^(1/2)*(I*2^(1/2)*(2+2^(1/2))^(1/2)-I*2^(1/2)+1-I)))

x6 := expand(x5);

-((1/2)*I)*ln(2)+((1/4)*I)*ln(2+(2+2^(1/2))^(1/2))+((1/2)*I)*ln(-I*2^(1/2)*(2+2^(1/2))^(1/2)+I*2^(1/2)+(-1+I))

x7 := combine(x6);

-((1/4)*I)*(ln(-(1/2)*(2+2^(1/2))^(1/2)*(I*2^(1/2)+(-1-I)))+(2*I)*Pi)

simplify(x7,constant);

(15/32)*Pi

# It's a pity that last step worked while this next is inadequate.
simplify(x7);

-((1/4)*I)*ln(-(2+2^(1/2))^(1/2)*(I*2^(1/2)+(-1-I)))+((1/4)*I)*ln(2)+(1/2)*Pi

# Another way, using x7
simplify(evalc(x7));

(15/32)*Pi

# Another way, using x7
simplify(combine(expand(x7)));

(15/32)*Pi

 

Download simplify_example.mw

I am trying to solve system linear partial differential equations using command "pdsolve". I am surprised to see that the solution given by this command is not satisfying the system, instead, an additional constraint is obtained for an arbitrary function, is there something about "pdsolve" I am missing? 


 

with(PDEtools):

DepVars := [f(x, y, t, u)]

[f(x, y, t, u)]

(1)

Sys := {diff(f(x, y, t, u), u, t)-(diff(f(x, y, t, u), x, y)) = 0, diff(f(x, y, t, u), u, u) = 0, diff(f(x, y, t, u), u, y) = 0, diff(f(x, y, t, u), x, u) = 0, diff(f(x, y, t, u), x, x) = 0, diff(f(x, y, t, u), y, y, y) = 0}

{diff(diff(f(x, y, t, u), t), u)-(diff(diff(f(x, y, t, u), x), y)) = 0, diff(diff(diff(f(x, y, t, u), y), y), y) = 0, diff(diff(f(x, y, t, u), u), u) = 0, diff(diff(f(x, y, t, u), u), x) = 0, diff(diff(f(x, y, t, u), u), y) = 0, diff(diff(f(x, y, t, u), x), x) = 0}

(2)

pdsolve(Sys)

{f(x, y, t, u) = (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t)}

(3)

f := proc (x, y, t, u) options operator, arrow; (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t) end proc

proc (x, y, t, u) options operator, arrow; (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t) end proc

(4)

Sys

{0 = 0, diff(_F3(t), t)-_F3(t) = 0}

(5)

``


 

Download pdsolve_command.mw

Good day sirs,

I am trying to plot graphs on stream function but its given me a lot of multiple error using my codes. Anyone with useful information should please share. 

Below is the code.

Thanks

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/Test.mw .
 

Download Test.mw

 

HI Maple primes.  We try to make sense of 'rsolve' in the Maple world.  What is command for source code?

 

Regards,

Matt

 

Can't understand what's wrond. This text from my academic books but it doesn't work on practice. Help please

 

plot3d(polygons([[0, 0, 0], [1, 0, 0], [1, 0, 0], [0, 1, 0]], [[0, 0, 0], [0, 1, 0], [0, 1, 1], [0, 0, 1]], [[1, 0, 0], [1, 1, 0], [1, 1, 1], [1, 0, 1]], [[0, 0, 0], [1, 0, 0], [1, 0, 1], [0, 0, 1]], [[0, 1, 0], [1, 1, 0], [1, 1, 1], [0, 1, 1]], [[0, 0, 1], [1, 0, 1], [1, 1, 1], [0, 1, 1]]), light(0, 0, 0.0, 0.7, 0.0), light(100, 45, 0.7, 0.0, 0.0), light(100, -45, 0.0, 0.0, 0.7), ambientlight(.4, .4, .4), title(cube), style(patch), color(zhue));

Error, (in plot3d) bad range arguments: light(0, 0, 0., .7, 0.), light(100, 45, .7, 0., 0.)
 

Hi Mapleprimes,

We know that '' rsolve '' is a recurrence equation solver.  It is more than an expression simplifier.

Congratulations to the Maple computer algebra team for creating such a great computer tool.  simply want to know more.

rsolve_on_May_16_2017.pdf

Surely there are many steps to determine the values to place.

Regards,

Matt

 

I recently encontered a very strange result.

Lets define the procedure:

Fg := proc(x0,y0)
if (x0>=0)and(x0<=3) and (y0<=x0+2) and (y0>=x0-1) and (y0>=0) and (y0 <=3) then
return y0*(3-y0)*x0*(3-x0)*(x0+2-y0)*(y0-x0+1);
else
return 0;
end if:
end proc:

The plot looks like needed:

plot3d('Fg'(x,y), x=0..3, y=0..3);

But integration returns weird result:

evalf(Int('Fg'(x,y), [x=0..1, y=0..2.1]));

7.888753239

evalf(Int('Fg'(x,y), [x=0..1, y=0..2.2]));

Error, (in evalf/int) when calling 'Fg'. Received: 'cannot determine if this expression is true or false: 0 <= x and x <= 3 and y <= x+2 and x-1 <= y and 0 <= y and y <= 3'

hi...i have a problem with pdsolve this equations?

please help me.

thanks

mt.mw

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/mt.mw .
 

Download mt.mw

 

I have a expression p:=C*A+A, when i type this to maple, maple display into p:=A*C+A because commutative property of multiplication

but I dont want to do like that. How can I display into p:=C*A+A?

I have an important question regarding the symbolic toolbox from maple for matlab

I have declared three symbolic variables:

syms t;

x = sym('x(t)', 'real');

y = sym('y(t)', 'real');

z = sym('z(t)', 'real'); 

f = x + y + z;

Now i want to  calculate the derivation of f regarding the variable t:

df = diff (f, t);

The result should be 

df = dx(t) / dt + dy(t) / dt + dz(t) / dt

but  instead i get the result df = 0. If i use the command diff (f) i get the result

df = dx(t) / dt 

Does anyone know how to fix this problem?

 

Hi Guys,

 

got my last Problem solved :) Now i do have a new one...

restart; with(RealDomain); with(CodeGeneration); with(ExcelTools); with(plots);
dx1:=133;
dy1:=132;

n := 1; 
for i to 256 do 
for j to 256 do 
r := evalf(sqrt((i-dx1)^2+(j-dy1)^2)); 
Ints := R0[i, j]; 
IntsR[n] := [r, Ints]; 
n := n+1 
end do; 
end do; 
IntsR := [seq(IntsR[i], i = 1 .. n-1)]

The list IntsR consists of unsorted values doublets. Now it would be nice to get some sort of mean value of my Ints over r.  Didnt really find a solution for it until now... do you have any hint?

Hi, i'm working with some spectral acceleration data and i'm having troubles with a 'local' command on a Newmark method code i found on the internet. I'm not a maple expert so i wish someone could help me. i'm ussing maple 2015. Greetings

Newmark=proc(z0,u0,F0);  local beta,  dt, c, kg,a,b, N, i, dF, dz, du, dw, T, m, k ;        dt:=0.01;    T:=4;    N:=T/(dt);    gam:=0.5;    beta:=0.25;      m:=0.320;    k:=435;    c:=2;   z(0):=z0;   u(0):=u0;   F(0):=F0;   w(0):=(1/(m))*(F0-k*z0-c*u0);        t=vector(i,1);    t[1,1]=0;      for i from 2 to N do :  t(i,1)=t(i-1,1)+dt:  end do:           kg:=k+(gam*c)/(beta*dt)+m/(beta*dt*dt);    a:=m/(beta*dt)+(gam*c)/(beta);    b:=  (0.5*m)/(beta)+dt*((0.5*gam)/(beta)-1)*c;       for i from 2 to N do  dF(i):=diff(F,t)+a*u(i)+b*w(i);    dz(i):=(dF(i))/(kg);    du(i):=((gam*dz(i))/(beta*dt))-((gam*u(i-1))/(beta))+dt*(1-gam/(2*beta))*w(i-1);    dw(i):=((du(i))/(beta*dt*dt))-((u(i-1))/(beta*dt))-((w(i-1))/(2*beta));     z(i+1)=dz(i)+z(i);    u(i+1):=du(i)+u(i);    w(i+1):=dw(i)+w(i);     od;       plot([z(i), u(i), w(i)],x=0..50,y=-50..50);       end;

First 976 977 978 979 980 981 982 Last Page 978 of 2428