Hello,

Could you please help me with the following system of equations:

> restart; with(student);
> Z[0] = 377;
> solve({Z = 1/(j*w*C[1]+1/(j*w*L[1])+1/R)+1/(j*w*C[2]+1/(j*w*L[2])), C[1] = w[c]*(w[0]^2-w[c]^2)*(w[2]^2-w[1]^2)/(2*Z[0]*(w[1]^2-w[c]^2)*(w[0]^2-w[1]^2)), C[2] = w[c]*(w[0]^2-w[c]^2)*(w[2]^2-w[1]^2)/(2*Z[0]*(w[1]^2-w[c]^2)*(w[2]^2-w[c]^2)*(w[2]^2-w[0]^2)), L[1] = 1/(w[1]^2*C[1]), L[2] = 1/(w[2]^2*C[2]), R[1] = subs(w = w[1], 2*Z[0]*(1-S[21...

I'm trying to write a program that solves sudoku's using a Groebner basis. I introduced 81 variables x₁ to x₈₁, this is a linearisation of the sudoku board.

The space of valid sudokus is defined by:

for i=1,…,81 : F_i=(x_i−1)(x_i−2)⋯(x_i−9) This represents the fact that all squares have integer values between 1 and 9.

for all x_i and x_j which...

circle_line_intersec.mw

i'm trying to write some code that will output an angle phi when a null event is reached.

see worksheet

Hello,

is there a way to activate the "undo" command after executing the entire worksheet? 

Before executing every change can be undone, but since I press the execute button 

often, I loose my chance to undo.


thanks

Does anyone know the release date of Maple 16?

A common example to emphasize that it is not OK to bring absolute values inside the integral compares

abs( int( cos(n*x), x=0..Pi ) ) asuming n::integer

and

int( abs( cos(n*x) ), x=0..Pi ) assuming n::integer

Maple correctly formulates the first to 0. But the second expression gives it more trouble, returning two messages:

Warning, unable to determine if (1/2)*Pi*(1+2*_Z7)/n is between 0 and Pi; try to use assumptions or use the AllSolutions option
Warning, unable to determine if (1/2)*Pi*(1+2*_Z8)/n is between 0 and Pi; try to use assumptions or use the AllSolutions option

I want to assign the output of a function that has been passed to timelimit. Currently I have to pass, say, X in by reference (as a single element list) and do  timelimit( 0.5, F(X,args) ); where F := proc( X::list ); X[1] := f(args); end proc; --- which is not really desireable. 

More generally I want my proceudre to skip loop iterations that are taking too long. 

Dear all,

I am trying to simulate the penetration of the wavefunction into a potential barrier consisting of two barrier steps of different heights, V1 and V2.

My code looks like that:

> restart;
> m := 0.91e-30;h := 0.1055e-33; V1 := 10*0.16e-18; V2 := .8*V1;a := 0.10e-9;En := .6*V1; w := En/h;
                      ...

Hey guys,

How do I find a max in a 3d plot? I've tried plottools[getdatat] but keep getting the error of

"Error, index must evaluate to a name when indexing a module"

Thanks!

Hello,

I am trying to solve the following system of quasilinear PDEs for dependent variables pA(z,t), pB(z,t), pC(z,t) using Maple 14:

> system:= [

.15e-3*diff(pA(z,t),t)+.83e-1*diff(pA(z,t),z)+.47e-3/(1+.11e-6*pA(z,t)+.14e-4*pB(z,t)+.73e-4*pC(z,t))^2*((1+.14e-4*pB(z,t)+.73e-4*pC(z,t))*diff(pA(z,t),t)-.14e-4*pA(z,t)*diff(pB(z,t),t)-.73e-4*pA(z,t)*diff(pC(z,t),t)) = 0,

.15e-3*diff(pB(z,t),t)+.83e-1*diff(pB(z,t),z)+.90e-1/(1+.11e-6*pA(z,t)+.14e-4*pB(z,t...

Hello everyone,

I have a problem with the Brownian Motion (I'm using Maple 15).

Here is the code:

X:=BrownianMotion(0,0,1,t);

X1:=SamplePath(X(t),t=0..T,timesteps =T/d):

Mean(X1);

The last line give me a number different from zero, which is not true for a Brownian Motion as defined.

Could you please tell me where is the problem?

PS: I have the same problem with WienerProcess(sigma).

1.how to plot a equation include two complex variables?

e.g.,0.1501071588582623569e-19 * s1 ^ 2 - 0.1770422954406499377e-11 * s1 ^ 2 * s2 ^ 2 + 0.1526369514180028298e-19 * s2 ^ 2 - 0.8979700309848614490e-12 * s1 ^ 4 - 0.9507487955845430138e-12 * s2 ^ 4 + 0.4941234607542336257e-4 * s1 ^ 4 * s2 ^ 2 + 0.5029096170369867504e-4 * s1 ^ 2 * s2 ^ 4 + 0.1631266049971375518e-4 * s1 ^ 6 + 0.1557291987414385136e-4 * s2 ^ 6 - 0.3925287034197085366e3 * s1 ^ 6 * s2 ^ 2 - 0.5855115617172747465e3...

If you use all the convergents of the simple continued fraction of the MRB constant as the terms of a generalized continued fraction, then likewise use the new convergents in another generalized continued fraction, and so on... you arrive at 0.5557531....  For more on this process see https://oeis.org/wiki/Convergents_constant .

Hi,

I would be thankful if someone help me to generate primitive polynomials (of deferent degree if possible) defined over GF(2^16), i.e., the coefficients of the polynomials are in GF(2^16). Thanks in advance.

Hello!

I have to  solve following ODE set as a part of my MsC thesis