@Carl Love I made the examples up, it doesn't matter what exactly p1 looks like.

In your latest example, the expanded polynomial does not have any terms of degree 3 or less in t. That's why it shows just O(t^4). If you want to include terms of degree 4 in t, use

Ahh, that's what I get for making up new examples...

Now it works. Many thanks again.

@Carl Love It showed me only O(t^4) and nothing else. But you have the idea of what I want.

Here is an even easier example

example.mw

@Carl Love I just meant we could, for clarifying the labelling we could do.

plot([x,x^2,x^3], x=-1..1, thickness=2, legend=['f1' = x,'f2' = x^2,'f3'= x^3])

@Carl Love Found it, it was the cup of ace under the star I better do something about my old questions now.

@Thomas Richard One thing I experimented is that it may be better to write 'f1' = x

I didn't know this existed after many experiments.

@Kitonum Is it possible to do this in the command "animate"?

@Carl Love Thank you! I gave you a thumbs-up, is that the same as "accept answer"?

@tomleslie thanks for your reply. You have the rough idea. But basically I want to solve the system of coefficients up to a certain power.

Your code outputs whether it is consistent or not, if it is consistent i want Maple to tell me what the results are.

 EDIT: actually I figured a way around the problem. This method is not necesaarily.

@Rouben Rostamian  I know what you are saying, but my system has some singularities that can't be easily removed by Maple using dsolve.

Here is an excerpt of what I mean, (this willbe  hard to read without LaTeX) for example, ONE equation out a system may look like

x4'(t) = x4(t)* (x1'(t)/x1(t)) + 1

One of my inital condition say, tells me that x1(0) = 0, but if I tell Maple this, it will not return anything because it will htink that it is dividing by 0.

But one algebraicially notices,

x1(0) = 0 => if x1(t) = c[0]+c[1]u + c[2]u^2 + ... => c0 = 0, but the other coefficients are not determined to be 0, so x1(u) = c[1]u + c[2]u^2 + ....

Now plugging in the above if x4(u) = 2a[2] + 3a[3]u + .... (this is because in my system, I set x4 to be the derivative of another function, say x3(u)), then

(2a[2] + 3a[3]u +...) = ( 2a[2]u + 3a[3]u^2 + .. )[ (2c[2]u + 3c[3]u^2 + ...)/(c[1]u + c[2]u^2) ] + 1

(2a[2] + 3a[3]u +...)(c[1]u + c[2]u^2) = ( 2a[2]u + 3a[3]u^2 + .. ) (2c[2]u + 3c[3]u^2 + ...)(c[1]u + c[2]u^2)

Grouping now one can solve the coffiecients with initial conditions.

I know this example isn't fully solved (because i jsut made it up on the spot). The problem is knowing that my polynomial can get around the singuarlity using the method I illustrated.

(I will return tomorrow to see if this makes sense or not)

@Rouben Rostamian  

I know that method exists, but the actual system I am working with is much larger and is nonlinear and can't be handled using dsolve (I have tried and it does not work).

That's why I have to use an arbitrary polynomial first.

@Rouben Rostamian  

I meant to say that the solution is x(t) = 2 + t + t^2 + O(t^3) and y(t) = 1 + 2t + 0.5t^2 + O(t^3), that appear to be the root of my error. But the nature of my question remains.

I was just trying to approximate it under a terminated polynomial, which in this case is a 3rd-degree polynomial.

@JohnS 

 Can you explain

"map(coeffs,lhs~({e1,e2}),{x,y}) =~ 0:"

I can't figure out what the x,y do

Just fixed the code in the OP; you can now copy and paste my code. The current issue is still at hand however

@Carl Love 

Yes, I will in the future type out the codes.

Unfortunately the result isn't what I wanted. The problem right now is that all the $a_ij$ should return a number and I don't want them in terms of $t$ because the initial conditions determine the other a_ij.

@Carl Love thanks, but I don't really  understand the part

" because you want i to be evaluated at the time the function is defined rather than the time it is called"

To me, both my and your code is trying to achieve the same goal and i is being iterated first.