How can one the last value of variable

when I have variables Like C1;C2,C3....., F1,F2,F32,....so on

always number would be at last of the variable.

Depending upon the number in the variable I need to call values from other list. It would be really helpful to know how one can get the number present in variable?

Hi, I am working on an assignment and have some doubts about my solution to plotting the position of an anharmonic oscilliator with the Runge-Kutta method.

My attempt on the problem:

Handin_2.mw

A picture of the problem is attached below. In addition to this, we are to repeat the problem with k = 0.1 (f = 0 still), and k = 0.1 and f = 0.5. 

For example the elemetary charge e constant has the derive descriptor.  Having the descriptor in it, I might have thought there is a builtin way to pull out the derive equation there isn't so one has to do it this way

with(ScientificConstants):
rhs(GetConstant(e)[3])
                                    

However you wouldn't know the 3rd indexed position in the constant was the derive descriptor - it could be in a different location for another constant.

Also, what I wonder is, where is the derive for the Fine Structure Constant ? 

I am carrying out a research in dynamical system to which end I need to do optimal control. I have coded the control equations but its not displaying any result. I need guidance. The code is shown below....

restart;
with(plots);
r := 3; r[1] := 3; k := 10; a := 0.2e-1; b := 0.1e-1; c := 0.1e-1; beta := 0.3e-1; alpha := 0.3e-1; m := 0.5e-1;
z := 40; q := 5; p := 100; T := 3;
sigma := 0.1e-1; k[1] := 10; rho := 0.5e-1;

u[1] := min(max(0, z), 1); z := (a*m*k*lambda[2](t)*x(t)*y(t)-lambda[1](t)*r*(1+b*x(t)+c*y(t))*x(t)*x(t))/(z*k*(1+b*x(t)+c*y(t))); u[2] := min(max(0, q), 1); q := -lambda[1](t)*beta*x(t)*s(t)/q; u[3] := min(max(0, p), 1); p := -(r[1]*lambda[3](t)*s(t)*s(t))/(p*k[1]);
NULL;
sys := diff(x(t), t) = r*x(t)*(1-(1-u[1])*x(t)/k)-a*m*x(t)*y(t)/(1+b*x(t)+c*y(t))-beta*(1-u[2])*x(t)*s(t), diff(y(t), t) = -alpha*y(t)+a*m*x(t)*y(t)/(1+b*x(t)+c*y(t)), diff(s(t), t) = sigma*s(t)+r[1]*s(t)*(1-(1-u[3])*s(t)/k[1])-rho*s(t)*y(t), diff(lambda[1](t), t) = -lambda[1](t)*(r-2*r*(1-u[1])*x(t)/k-a*y(t)*(1+c*y(t))/((1+b*x(t)+c*y(t)) . (1+b*x(t)+c*y(t)))-beta*(1-u[2])*s(t))-lambda[2](t)*a*m*(1-u[1])*(1+c*y(t))*y(t)/((1+b*x(t)+c*y(t)) . (1+b*x(t)+c*y(t))), diff(lambda[2](t), t) = -lambda[1](t)*a*x(t)*(1+b*x(t))/((1+b*x(t)+c*y(t))*(1+b*x(t)+c*y(t)))+lambda[2](t) . (-alpha+(a*m*(1-u[1]) . (1+b*x(t)))*x(t)/((1+b*x(t)+c*y(t))*(1+b*x(t)+c*y(t))))+lambda[3](t)*rho*s(t), diff(lambda[3](t), t) = lambda[1](t)*beta*(1-u[2])*x(t)-lambda[1](t)*(r[1]-2*r[1]*(1-u[3])*s(t)/k[1]-sigma-rho*y(t)), x(0) = 100, y(0) = 200, s(0) = 100, lambda[1](T) = 0, lambda[2](T) = 0, lambda[3](T) = 0;
p1 := dsolve({sys}, type = numeric, method = bvp[midrich], abserr = .1);
 

Hi!

I have the following problem: to create a special elimination ordering which is a weighted degree ordering suitable for elimination + pure lexicographic ordering. Let me provide an example. Assume my variables are x,y,z,t,u,v and I want to eliminate x,y,z. For this purpose, I define the weight vector [1,1,1,0,0,0] and compare monomials wrt corresponding weighted degree. If such degrees are equal, I compare monomials by pure lexicographic ordering. This is exactly what I need. How to obtain such monomial ordering by the package Groebner? I have tried monomial orderings defined by matrices using command  'matrix'(M,vars) but it seems it does not work for 30 variables.

Any suggestion? Thanks in advance.

I am unable to the get the output  in ans2 , error is comming

restart:
with(DETools):
with(PDEtools):
u[o](r,z):=(-1/4)*diff(p[o](z),z)*(1-r^2):
ode:=gamma1*diff(u[o](r,z),z)+(1/r)*diff(v[1](r)*r,r)=0:
#dsolve( (ode), { v[1](r) } ):
IC1 := {v[1](0) = 0}:
ans2 := combine(dsolve(`union`(ode, IC1),{v[1](r)}));
 

Hi, I have a long expression (differential polynomial).

It may contain different functions a(x,y), b(x,y), c(x,y) and its derrivatives.

Visually i do not see variable a (and its derivative) in expression.

But I want to be sure. How can I check it automatically?

Also I don't know maximal order of derivative that could appear in this expression.

Hello!

Assume we have the first N positive integres, 1,..,N, and we assing to these numbers a (discrete) probability distribution p1,...,pN. Of course, p1+...+pN=1.

Then, How can we select a number in {1,..,N} according to the given probability distribution? That is, the number 1 can be chosen with probability p1, 2 with a probability p2, etc.

Many thanks in advance for your comments.

Hi,

I am collecting the coefficients of funciton terms(like sin, cos, log,exp,abs) form the expression. I was able to collect using the function 'coeff'. Initially am getting all functions in the expression using Indets[flat(expression, funciton)] then using seq and coeff trying to get all funcitons

expression := a*sin((a+b)/(a-b))*log(a/b)/c+a*b/c+2*sin(a+b);
numOfFuncs := numelems(indets[flat](expression, function));
Funcs := convert(indets[flat](Expression, function), list);

funcCoeffList := [seq(coeff(Expression, Funcs[i]), i = 1 .. numOfFuncs )];

funcCoeffList := [a*sin((a+b)/(a-b))/c, a*ln(a/b)/c, 2]

When there are terms of form funtion*function I would like to collect the coefficient for function*function as one term rather than two terms. simply I want to write a code which reads the functions having product between them as one term gives me back the coefficient.

Looking for the output as : [a/c,2] or [a/c,1,2];

Hello

I have an expression which invokes the LambertW function.

LambertW(-ln(1+i)*EP*p*(1+i)^(-(365*EP*hr*kw*p+SC*i)/(365*FIT*hr*i*kw*(-1+p)))/(FIT*i*(-1+p)))

I was trying to import this expression into Excel, but my version doesn't have LambertW.

Does someone know an analagous function in a form Excel can compute?

According to wiki The Lambert W relation cannot be expressed in terms of elementary functions.

I have gotten around the problem using Newton-Raphson method, but it takes a few cells to converge....

how i can remove root of from result.

I want to plot function.

Thnaks

root_of.mw
 

Download root_of.mw

Hi, I'm using Maple 2018 and I tried to run coding from https://www.maplesoft.com/applications/view.aspx?sid=4194&view=html

however, it said : unable to parse. I figured out that the problem maybe is in the if loop. though it seems perfectly fine, but it has some goto commands that i cannot search on maple website. does this mean that the goto cannot be used here and should be replaced? if yes, then how? 

i am still learning on how to use maple. any help would be much appreciated. thank you!

this is the coding for if loop:

label_7;

rv:=vector([p1(x1pt,x2pt),p2(x1pt,x2pt)]):

numgeval:=numgeval+1;

printf("%5d (%8.4f,%8.4f)",numIter,rv[1],rv[2]);

max:=n;

mg:=convert(sqrt(dotprod(rv,rv)),float);

printf("%12.4f",mg);

if(mg<tol or numIter>=max) then

goto(label_6);

else

numIter:=numIter+1;

fi;

v1:=x1pt+t*rv[1];

v2:=x2pt+t*rv[2];

newt:=evalf(subs({x1=v1,x2=v2},f1));

numfeval:=numfeval+1;

lam:=fsolve(diff(newt,t)=0,t,maxsols=1);

nv1:=evalf(subs({t=lam},v1));

nv2:=evalf(subs({t=lam},v2));

printf(" (%8.4f,%8.4f)%13.4f\n",x1pt,x2pt,lam);

x1pt:=nv1;

x2pt:=nv2;

goto(label_7);

label_6;

printf("\n\n-----------------------------------------");

printf("---------------------------------------------");

printf("\n\n Approximate Solution: ");

printf(" (%8.4f,%8.4f)\n",x1pt,x2pt);

Fvalue:=evalf(subs(x1=x1pt,x2=x2pt,f));

printf(" Maximum Functional Value: ");

printf("%21.4f",Fvalue);

printf("\n Number gradient evaluations:");

printf("%22d",numgeval);

printf("\n Number function evaluations:");

printf("%22d",numfeval);

printf("\n\n-----------------------------------------");

printf("---------------------------------------------");

end:

If q := [q1(t),q2(t),q3(t)];

and L=cos(q1(t))+sin(q2(t))+5*dq1 + 4*dq3

now I want to get the result of the following expression

d(dL/d dq1)/dt=?,

how can i write the expression?

In above expression dq1 is the derivative of q1(t), and dq3 is that of q3(t),

How do I solve (1/2)*n^2 +2n-1=O(n^2) in maple2018?Thank you

curry(`?[]`,f)([1]);