WA573

80 Reputation

5 Badges

2 years, 235 days

MaplePrimes Activity


These are replies submitted by WA573

@dharr @Carl Love How to use simple notation in soldis.mw? I want 'n~, m~' should be 'n, m' and 'psi(n~,m~), phi(n~,m~)' should be 'psi,phi'. When I use 'declare', then get 'Error, (in PDEtools:-declare) invalid input: conjugate expects 1 argument, but received 2' 

@dharr I am trying to plot the solution for the discrete values of 'n' and continuous values of 't' by using pointplot3d. But encountered an error. How to fix it?

restart

expression := 1-I*((2*I)*I^(n+1)*exp(-2*t)*(2+I)^(n+1)*exp(2*t)/(I^(n+1)*(exp(-2*t))^2*(-I)^(n+1)+(2+I)^(n+1)*(exp(2*t))^2*(2-I)^(n+1))-(2*I)*I^n*exp(-2*t)*(2+I)^n*exp(2*t)/(I^n*(exp(-2*t))^2*(-I)^n+(2+I)^n*(exp(2*t))^2*(2-I)^n)); n_values := [1, 2, 3, 4, 5]; t_values := [-2 .. 2]; with(plots); pointplot3d([seq([n_val, t_val, evalf(subs([n, t], [n_val, t_val], expression))], `in`(n_val, n_values), `in`(t_val, t_values))], style = point, symbol = cross)

1-I*((2*I)*I^(n+1)*exp(-2*t)*(2+I)^(n+1)*exp(2*t)/(I^(n+1)*(exp(-2*t))^2*(-I)^(n+1)+(2+I)^(n+1)*(exp(2*t))^2*(2-I)^(n+1))-(2*I)*I^n*exp(-2*t)*(2+I)^n*exp(2*t)/(I^n*(exp(-2*t))^2*(-I)^n+(2+I)^n*(exp(2*t))^2*(2-I)^n))

 

[1, 2, 3, 4, 5]

 

[-2 .. 2]

 

Error, invalid input: seq expects its 3rd argument, step, to be of type numeric, but received t_val in [-2 .. 2]

 

Download plot_dis.mw

@Carl Love As in (5), the bar is over eta but in (6)-(8) some terms have bar on 'n' and 'm'.

@mmcdara How to apply difference operator 'Delta' on eq. (7)? Further, when b -> 0 we get exponential fun for continuous variable 't'. How to do it in Maple and then plot the solution for discrete variable 'n' and continuous variable 't'? soldis.mw

@mmcdara Many thanks. Yes, I have a continuous PDE and want its discrete version.

@Preben Alsholm How can we write the solution like u=v*exp(w*t)?

Is there anyone who have ONEOptimal package?

@dharr In the next step, I am trying to solve polynomial equations to determine the values of parameters a's and lambda's by setting all the coefficients of the different powers of csc(eta) to zero. But I got only "{0 = 0, 15*cot(eta)*lambda__1*a[-2]^2*a[-1] = 0}".CM_TW.mw

@Thomas Richard Thanks, it works. Secondly,  is it possible that Maple gives the step by step solution?

@acer Can we put three different symbols on these peaks? i.e, solid circle, asteric, box, etc

@mmcdara How can we fix it? By changing the range of h?

@Rouben Rostamian  You wrote h+2 instead of h on RHS of the first equation, why? 

 
 

@mmcdara I did it by taking the terms up to 6th order in summation. But the problem is that I did it step by step and it takes time almost the same as what we did by hand. Can we do it in maple in short steps? Further, how do solve (20) to get A[i]'s?

PA.mw 

@mmcdara A[0] =6 (given), it should not be zero. And system of algebraic equations:

Note: zeta is eta, and z[0] is x[0]. Can we obtain these equations in maple?

@Kitonum Since from determinant(A)=0, we need to find the relation between w and k. How to find it?

1 2 3 Page 2 of 3