MaplePrimes Questions

fyi, there seems to be a problem here. Maple 2019, Physics version 395 on windows 10.

The solution given to this wave PDE by Maple as sum that starts from zero, has "n" in the denominator. When n=0, this gives division by zero.  Is this a bug?

restart;
L:=3: c:=4: h:=1/10: b:=Pi*c/L:
f:=piecewise(0<=x and x<=L/3,3*h/L*x,L/3<x and x<=L,h):
pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2):
bc  := u(0,t)=0,D[1](u)(L,t)=0:
ic  := D[2](u)(x,0)=0,u(x,0)=f:
sol:=pdsolve([pde,bc,ic],u(x,t));
subs(n=0,sol)

u(x, t) = Sum(-((3/10)*I)*sin((1/6)*(1+2*n)*Pi*x)*(3^(1/2)*sin((1/3)*Pi*n)+cos((1/3)*Pi*n))*((2*I)*exp((4/3)*t*(I*n^(1/2)*(n+1)^(1/2)-1/2)*Pi)*n^(1/2)*(n+1)^(1/2)+(2*I)*exp(-(4/3)*(I*n^(1/2)*(n+1)^(1/2)+1/2)*t*Pi)*n^(1/2)*(n+1)^(1/2)+exp((4/3)*t*(I*n^(1/2)*(n+1)^(1/2)-1/2)*Pi)-exp(-(4/3)*(I*n^(1/2)*(n+1)^(1/2)+1/2)*t*Pi))/(n^(1/2)*(n+1)^(1/2)*Pi^2*(1+2*n)^2), n = 0 .. infinity)

Error, numeric exception: division by zero

 


 

Download bug_july_11_2019.mw

 

 

Let m, n be two monomials with parametric coefficients. How to decide that two monomials are distinct in the variables?

For example, if m= (a-1)x^2y and b= (a-b)z^2 (where a,b are parameters and x,y,z are variables) then m and n are distinct. Is there any command? 

Thank you in advance.

 

Hello,

I need help to approximate Fn(x,y) and Sn(y) for n>=3 to see the convergence. 

Here is the problem:

I have an integral equation Fn(x, y) defined by

F0(x,y) = 0,

F1(x,y) = (9/8)*(y+1) - (21/8)*( y*exp(-2*x) + exp(-x) )

Fn+1(x,y) = (9/8)*(y+1) - (21/8)*( y*exp(-2*x) + exp(-x) ) + (3/4)*int ( Fn( x-z, 2*y*exp(-z) ), z=0..Un(x) ), where the function Un(x) is defined below.

where I suppose that Fn(x,y) >=0 and y>=0 and x>=0. I am iterating the integral equation and find the zeros Sn(y) solutions of Fn(S(y) , y) = 0. The goal is to show that the sequence Fn(x,y) converges to some F(x,y) and Sn(y) converges to some S(y). We can also plot the curves Sn(y) to show the convergence. Notice that Sn(y) >= 0 is increasing in n and decreasing in y.

What I did :

If n = 0, I solve F1(x,y) = 0 and find analytically x = S1(y) = -ln((1/14)*(-7+sqrt(84*y^2+84*y+49))/y). Notice that F1(x,y) = 0 is quadratic in exp(-x) and then F1(x,y) = 0 is easy to solve. S1(y) is the positive function for y>=0.

If n = 1, we need to solve F2(x,y) = (9/8)*(y+1) -(21/8)*( y*exp(-2*x) + exp(-x) ) + (3/4)*int ( F1( x-z, 2*y*exp(-z) ), z, 0, U1(x) ). The function U1(x) is obtained as follows :

i) We define g( U ) = U + S1( 2*y*exp(-U) )

ii) We find the function U1(x,y) solution of g( U1(x,y) ) = x

 

A generalization of Un(x) is given by

i) Define g( U ) = U + Sn( 2*y*exp(-U) )

ii) Find the function Un(x,y) solution of g( Un(x,y) ) = x

The curve S2(y) :

I find the analytic function F2(x, y), but it is impossible to solve F2(x,y) = 0 and find x = S2(y) analytically. So I choose an interval [0, 10] for y and discretize that interval. Now at each point yi in [0, 10], the function F2(x, yi) depends only on x. I apply the bisection method (dichotomic method) and find the value xi such that F2(xi, yi) = 0. The data (yi, xi) gives the curve S2(y). I plot the data (yi, xi) and (y, S1(y)) in the same curve.

Question:

It is impossible to iterate the process below and find F3(x, y), F4(x,y), etc... because we cannot find U2(x), U3(x), etc... analytically. Any suggestions or helps to plot the curves (y, Sn(y)) for n = 3, 4, etc. to see the convergence ?

Thanks !

 I'm confused as to why Maple can't perform a certain integral for me. Please see my attached code - it get's stuck on the last step.

 

070919_ThetaIntegral.mw

        I  want save a set as shown below:
       
 E:={{1,2,3,4,5,6},{4,5,6,7,8,9,10,11},{10,11,12,13,14}};
E := {{10, 11, 12, 13, 14}, {1, 2, 3, 4, 5, 6}, {4, 5, 6, 7, 8, 9, 10, 11}}
     
    The order of set is changed; I have known the list can do it following my order,
       E:=[{1,2,3,4,5,6},{4,5,6,7,8,9,10,11},{10,11,12,13,14}];
but I want to know what can I do in set?

Does anyone know how collect zero epsilon coefficient from follow expression?

coeff_test.mw

Is it possible?

Thank you!

How I can calculate integral?

Thanks


 

"restart;    f[1,1](r,theta,phi):=r^4 sin(6 theta) sin(3 phi):    L(r, theta,phi):=(2.784615385 10^10 ((&PartialD;)^2)/(&PartialD;r^2) `f__11`(r,theta,phi)+(2.784615385 10^10 (2+2 r cos(theta)) ((&PartialD;)/(&PartialD;r) `f__11`(r,theta,phi)))/(r (2+r cos(theta)))-(0.1175000000 (((&PartialD;)^4)/(&PartialD;theta^4) `f__11`(r,theta,phi)))/(r^4)-(0.1175000000 (((&PartialD;)^4)/(&PartialD;phi^4) `f__11`(r,theta,phi)))/((2+r cos(theta))^4)-(0.1175000000 (((&PartialD;)^4)/(&PartialD;phi^2&PartialD;r^2) `f__11`(r,theta,phi)))/((2+r cos(theta))^2)-(0.1175000000 (((&PartialD;)^4)/(&PartialD;r^2&PartialD;theta^2) `f__11`(r,theta,phi)))/(r^2)-(0.2350000000 (((&PartialD;)^4)/(&PartialD;phi^2&PartialD;theta^2) `f__11`(r,theta,phi)))/(r^2 (2+r cos(theta))^2)+(0.1175000000 ((cos(theta))^2 r^2+4 (cos(theta))^2 r^4+16 cos(theta) r^3-4+17 r^2) (((&PartialD;)^2)/(&PartialD;theta^2) `f__11`(r,theta,phi)))/((2+r cos(theta))^2 r^4)+(0.1175000000 (2 (cos(theta))^2 r^2+4 (cos(theta))^2 r^4+16 cos(theta) r^3+4+12 r^2) (((&PartialD;)^2)/(&PartialD;phi^2) `f__11`(r,theta,phi)))/(r^2 (2+r cos(theta))^4)-(2.784615385 10^10 (2 (cos(theta))^2 r^2+4 r cos(theta)+4) `f__11`(r,theta,phi))/(r^2 (2+r cos(theta))^2)+(0.2350000000 (((&PartialD;)^3)/(&PartialD;r&PartialD;theta^2) `f__11`(r,theta,phi)))/(r^3 (2+r cos(theta)))-(0.2350000000 (((&PartialD;)^3)/(&PartialD;phi^2&PartialD;r) `f__11`(r,theta,phi)))/(r (2+r cos(theta))^3)+(0.2350000000 sin(theta) (((&PartialD;)^3)/(&PartialD;theta^3) `f__11`(r,theta,phi)))/(r^3 (2+r cos(theta)))-(0.2350000000 sin(theta) (((&PartialD;)^3)/(&PartialD;phi^2&PartialD;theta) `f__11`(r,theta,phi)))/(r (2+r cos(theta))^3)+(0.1175000000 sin(theta) (((&PartialD;)^3)/(&PartialD;r^2&PartialD;theta) `f__11`(r,theta,phi)))/(r (2+r cos(theta)))-(0.1175000000 (2 r cos(theta)+3) sin(theta) (2 r cos(theta)+5) ((&PartialD;)/(&PartialD;theta) `f__11`(r,theta,phi)))/(r (2+r cos(theta))^3)) r^4 sin(6 theta) sin(3 phi):"

with(Student[Calculus1]); K[rr, s] := evalf(ApproximateInt(L(r, theta, phi), r = .2 .. 1, method = simpson)); KK[rr, s] := evalf(ApproximateInt(K[rr, s], theta = 0 .. 2*Pi, method = simpson)); k2 := evalf(ApproximateInt(KK[rr, s], phi = 0 .. 2*Pi, method = simpson))

``


 

Download

Hello,

Real beginner, never used Maplw before. I want to analyse PDE's through Lie symmetry analysis -- get the infiniteeimals and then generate the invariants. I am going through the standard examples on the help within Maple, but I am stuck on a basic issue which prompts me to think that there may be something different in my version of Maple (although I downloaded it from TTU which I presume should be the latest version).

Here's what I get >>>>>

Loading PDEtools

with(PDEtools, InfinitesimalGenerator, declare);
               [InfinitesimalGenerator, declare]

declare(u(x, t));
               u(x, t) will now be displayed as u
U: diff_table(u(x, t)):
PDE := U_x,x - U_t = 0;
                    PDE := U_x,x - U_t = 0
show;
                       U_x,x - U_t = 0
Infinitesimals(PDE)
Error, (in PDEtools:-Infinitesimals) missing dependent variables

>>>>>

 

I have also tried:

Infinitesimals(PDE, u)
Error, (in PDEtools:-Infinitesimals) not a PDE system w.r.t u

and:

Infinitesimals(PDE, U)
Error, (in PDEtools:-Infinitesimals) not a PDE system w.r.t U
 

There are many other issues, but let's deal with this one first.

 

Thanks

Best wishes

Nadeem

 

 

 

Hello all, 

Would you teach me how to cancel a denominator term by multiplying the same term?

For example, I got this expression:

-L2*diff(i2(t), t) + L1*diff(i1(t), t)/n12

Then, I want to come to this expression by multiplying the expression with 'n12'

-L2*n12*diff(i2(t), t) + L1*diff(i1(t), t), 

But the result from my simple attempt was this:

(-L2*diff(i2(t), t) + L1*diff(i1(t), t)/n12)*n12

Thank you in advance!

 

Hey there, I have a pretty general question.

I'm trying to fill out the entries of a matrix. Each matrix element is the result of a very complicated calculation. Currently, I am using nested do loops (two loops total) to scan through each row and column respectively. I find that this process racks up a ton of memory, and that I in fact run out of memory before the matrix has been completely specified.

First, why is this happening? And second, what is a smarter way to perform the calculation?

 

Cheers.

Hi,

Due to an unexpected maintenance operation, I had to uninstall Maple 18 from my Windows 7-64 bit PC.

Later on I installed it without any problems but, to my surprise now I can't configure it properly. This is:

* Enabling Maple Text as Input (classical input method)
* Removing numbers from equations
* Setting Maple language to English (it took Spanish by default because of Windows)
* Hiding left panel

As usual, this is performed under Tools/Options/Interface and so on...

After I modify my preferred settings, apply globally and close-open again, the program is again in its original form.

How can I do in this case? I have installed it several times. It is also worth noting that the maintenance that I performed was related to deep Windows registry modifications. 

Thanks and regards.

 

X := [0, -3]; Y := [3, 1]; Z := [5, -2]; PP := proc (M::list, N::list, K::list) local A, B, P, distPH, H, lambda, mdist; A := M; B := N; P := K; H := simplify(expand((1-lambda)*A+lambda*B)); distPH := sqrt((H[1]-P[1])^2+(H[2]-P[2])^2); mdist := diff(distPH, lambda); lambda := solve(mdist, lambda); simplify(distPH); H end proc; debug(PP); PP(X, Y, Z);#the calculation isn't finished

Would you teach me how to collect and isolate specific terms in an equation?

For example, this is a given equation:

v1(t) - R1*(-i3(t)/n13 - i2(t)/n12) - L1*diff(-i3(t)/n13 - i2(t)/n12, t) = n12*(v2(t) - R2*i2(t) - L2*diff(i2(t), t))

From this equation, I wonder if there is a way to collect and isolate the derivative terms at the RHS, like this:

(- L1/n12+(n12 * L2))*diff(i2(t), t) + (L1/n13)*diff(i3(t), t) = -v1(t) + R1*(-i3(t)/n13 - i2(t)/n12) + n12*v2(t) - n12*R2*i2(t)

Thank you in advance!

 

Hi 

Is there a way to solve it numericall

evalf(int(exp((5+I*6)*x)*sin(1+erf(x)),x=-to..to));

evalf(int(exp((5+I*6)*x)*cos(1+erf(x)),x=-to..to));

 

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