@Alejandro Jakubi It's like you say. In Maple 17.00/32/Windows 8, I killed it after 90 seconds. In Maple 16.02/64/Windows 8, it finished in 9-10 seconds (obtaining essentially the same answer).

seq(evalf(1/(1+50*(x[0]+i*h)^2)), i= 1..5);

seq(evalf(1/(1+50*(x[0]+i*h)^2)), i= 1..5);

@Alejandro Jakubi wrote:

On the other hand the computation of this integral in Maple 17.01 by method elliptic, or equivalently leaving the default, does not finish here (on Linux 32-bit) in any reasonable time. What version have you used?

I got 39 seconds on 17.01 / 64 / Windows 8:

restart:
J:= Int(sin(x)/cos(x)/sqrt(1+sin(x)^3), x= 0..Pi/4):

IntegrationTools:-Change(J, t= sin(x)):
CodeTools:-Usage(value(%)):
memory used=3.72GiB, alloc change=390.01MiB, cpu time=38.88s, real time=37.92s
kernelopts(version);
   Maple 17.01, X86 64 WINDOWS, Jun 25 2013, Build ID 849430

I've done it multiple times without failure.

@Alejandro Jakubi wrote:

On the other hand the computation of this integral in Maple 17.01 by method elliptic, or equivalently leaving the default, does not finish here (on Linux 32-bit) in any reasonable time. What version have you used?

I got 39 seconds on 17.01 / 64 / Windows 8:

restart:
J:= Int(sin(x)/cos(x)/sqrt(1+sin(x)^3), x= 0..Pi/4):

IntegrationTools:-Change(J, t= sin(x)):
CodeTools:-Usage(value(%)):
memory used=3.72GiB, alloc change=390.01MiB, cpu time=38.88s, real time=37.92s
kernelopts(version);
   Maple 17.01, X86 64 WINDOWS, Jun 25 2013, Build ID 849430

I've done it multiple times without failure.

@Markiyan Hirnyk Addressing a minor point from earlier in this thread: Function application does indeed distribute over the list-building operator [] and the set-building operator {}. So, in particular,

{seq}(expr, k= 1..n) = {seq(expr, k= 1..n)},

and there was no problem with Mehdi's syntax on this point.

The undistributed syntax is especially useful when there are multiple functions in the list or set:

[f,g](x,y) = [f(x,y), g(x,y)]

@Markiyan Hirnyk Addressing a minor point from earlier in this thread: Function application does indeed distribute over the list-building operator [] and the set-building operator {}. So, in particular,

{seq}(expr, k= 1..n) = {seq(expr, k= 1..n)},

and there was no problem with Mehdi's syntax on this point.

The undistributed syntax is especially useful when there are multiple functions in the list or set:

[f,g](x,y) = [f(x,y), g(x,y)]

@spradlig asked: Can you please tell us more about who you are and why you are asking?

The Poster is an employee of Maplesoft, as is indicated by the blue background on the name and maple leaf icon in the lower right of the gravatar.

@Mathematix

Can you describe in words what you were intending by subs(exp= 3, Y)? That does not do anything mathematically valid; rather it changes the exponential function into a constant function, the constant being 3. This works:

plot(Y, x= -1..1);

@Mathematix

Can you describe in words what you were intending by subs(exp= 3, Y)? That does not do anything mathematically valid; rather it changes the exponential function into a constant function, the constant being 3. This works:

plot(Y, x= -1..1);

@Mathematix I don't know. Those don't look like difference equations or differential equations. There's no independent variable and no derivatives. Also, it is obvious that the system does not satisfy its own initial conditions.

@Mathematix I don't know. Those don't look like difference equations or differential equations. There's no independent variable and no derivatives. Also, it is obvious that the system does not satisfy its own initial conditions.

I haven't looked at your worksheet yet, but generally you want to raise epsilon and decrease the precision to get rid of holes in a plot. Four significant digits should be enough for any plot, smaller differences not being perceivable. The higher precision can lead to holes for complicated integrands because there may be some points for which the integral returns unevaluated because it cannot achieve the requested precision.

You would need to have an A and a B to use the procedure. What do A and B represent? Obviously, they are n x n matrices of nonnegative integers. Are they zero-one? Where did you find procedure rOrt?

By powers of Y_1*Y_2 do you mean (Y_1*Y_2)^2, (Y_1*Y_2)^3, etc., but not Y_1^3*Y_2^2? And should the latter be considered as Y_1*(Y_1*Y_2)^2, i.e., as a "power" of Y_1*Y_2 with a coefficient of Y_1?