Why not use output=['determinant']?  That way the external NAG routine does all the work.  Thus:

determinant := LUDecomposition(M, inplace=true,output=['determinant']);

Why not use output=['determinant']?  That way the external NAG routine does all the work.  Thus:

determinant := LUDecomposition(M, inplace=true,output=['determinant']);

Yes, if time is run backwards, starting points become ending points.

Yes, if time is run backwards, starting points become ending points.

Here's what works for me, in copying 1D Maple input from Classic to Standard under Windows XP.

Rather than paste directly into Standard, I open a Wordpad window and paste into that.  The result has one black ">" at the beginning.  I highlight the Wordpad text, not including that ">", and copy, then paste it into Standard.

Here's what works for me, in copying 1D Maple input from Classic to Standard under Windows XP.

Rather than paste directly into Standard, I open a Wordpad window and paste into that.  The result has one black ">" at the beginning.  I highlight the Wordpad text, not including that ">", and copy, then paste it into Standard.

The expression you gave involving pochhammer didn't have any k in it.

For all integer values of i and j from 1 to 10, the expression simplifies to a rational function of mu with all coefficients in numerator and denominator positive.

The expression you gave involving pochhammer didn't have any k in it.

For all integer values of i and j from 1 to 10, the expression simplifies to a rational function of mu with all coefficients in numerator and denominator positive.

We can remove a few parameters.

Assuming f <> 0, we can divide numerator and denominator by f, so we can take f=1.

Next, by a translation in the x and y directions, we can assume c = 0.

With those reductions, checking the Abel classes seems to be considerably less time-consuming.  But since Maple doesn't find closed-form solutions in the particular case I mentioned, it's very unlikely to do so in the general case.  I don't think it's realistic to expect closed-form solutions, whether in Maple 13 or Maple 113.  It's quite likely that they simply don't exist.  Instead, you might look at numerical solutions, series solutions, or phase-plane analysis.

We can remove a few parameters.

Assuming f <> 0, we can divide numerator and denominator by f, so we can take f=1.

Next, by a translation in the x and y directions, we can assume c = 0.

With those reductions, checking the Abel classes seems to be considerably less time-consuming.  But since Maple doesn't find closed-form solutions in the particular case I mentioned, it's very unlikely to do so in the general case.  I don't think it's realistic to expect closed-form solutions, whether in Maple 13 or Maple 113.  It's quite likely that they simply don't exist.  Instead, you might look at numerical solutions, series solutions, or phase-plane analysis.

We don't even know what the real life situation is.  You do.  You should think about what's going on there, and what that -1 does.  As I said, it may have to do with running time backwards, but there may be other symmetries involved.  Why not go back to the advisor who told you to multiply by -1 and ask him/her for some clarification?

We don't even know what the real life situation is.  You do.  You should think about what's going on there, and what that -1 does.  As I said, it may have to do with running time backwards, but there may be other symmetries involved.  Why not go back to the advisor who told you to multiply by -1 and ask him/her for some clarification?

In the usual picture, I con't think the trajectories correspond to contour lines. 
Instead, you're looking at trajectories for the system of differential equations corresponding to the gradient of the z function.  However, not every vector field is a gradient, so not every system of differential equations, even if it has a saddle-point equilibrium, has a corresponding "saddle function".

In the usual picture, I con't think the trajectories correspond to contour lines. 
Instead, you're looking at trajectories for the system of differential equations corresponding to the gradient of the z function.  However, not every vector field is a gradient, so not every system of differential equations, even if it has a saddle-point equilibrium, has a corresponding "saddle function".

What exactly do you mean by that?  Do you mean trajectories corresponding to the eigenvectors of the system? What do you want the 3D picture to look like?