MaplePrimes - answers and comments on Question, New to this forum need help with Maple for general equations

Not sure whether I've understood your problem

Joerg Picard — Sat, 22 Sep 2012 07:01:59 Z

Not sure whether I've understood your problem correctly, but if xi represents a real (or complex) number then try:

w := (x)->a[0]+sum(a[i]*x^i, i=1..4); # define the polynom

solve([w(0), w(1), D(D(w))(0), D(D(w))(1)]); # solve system of equations, D() is the differential operator

To expand a little on Jörg's answer: In

Mac Dude — Sat, 22 Sep 2012 19:12:09 Z

To expand a little on Jörg's answer:

In Maple, assignments are done by ":=". The "=" is a logical expression, not an assignment.

You define a function by writing f:=(x)-> a*x+b*x^2+c i.e. using the arrow operator (there are other ways but this is the quickest way for a statement function). Assigning an expression to f(x) makes the whole f(x) a name (which is not f(y)); usually not what you want.

As to Maple vs Matlab: it really depends what you want to do. Maple is primarily a Computer Algebra System; Matlab is primarily a tool for numeric evaluations esp. involving matrices. So if you are looking for analytic solutions, Maple is your friend. If you routinely deal with large numerical problems; Matlab is. (And, yes, I am simplifying greatly here.)

Note however, that both systems have aspects of the other side as well (certainly Maple can do a lot of numerics; I believe late versions of Matlab have a certain amount of symbolic capability added) and in fact can access each other at a certain level (the details of which I have not explored at all).

HTH,

Mac Dude

This is a good approach

Nicolo — Sat, 22 Sep 2012 21:32:12 Z

> solve([w(0), w(1), (D(D(w)))(0), (D(D(w)))(1)]);
 {a[0] = 0, a[1] = a[4], a[2] = 0, a[3] = -2 a[4], a[4] = a[4]}
> solve([w(1), (D(D(w)))(1)]);
      {a[0] = -a[1] + 2 a[3] + 5 a[4], a[1] = a[1], 

        a[2] = -3 a[3] - 6 a[4], a[3] = a[3], a[4] = a[4]}

This is what i got. My question is how can I make maple remember that a[0] = 0 and a[2] = 0. so i can keep       solving my equation to get an answer in terms of w(x) and x with the constants being replaced.

Like Mac Dude mention before: you can use a[0

Joerg Picard — Sun, 23 Sep 2012 05:46:49 Z

Like Mac Dude mention before: you can use

a[0] := 0; a[2] := 0;

or use the assign command like this:

w := (x)->a[0]+sum(a[i]*x^i, i=1..4); # define the polynom

sol := solve([w(0), w(1), D(D(w))(0), D(D(w))(1)]); # solve system of equations, D() is the differential operator

assign(sol); # assign solution to parameters a[0]-a[4]