Hey,

So I have an array of plots [2,1] and I am trying to seperate them into two seperate plots to extract data using the getplot function (which only works on single plots).

I have looked all over the help guide but never found anything relating to my problem. Is there a way to seperate these two plots?

Thank you for the help.

hi

when i copy my results in excell or notepad, they are in the following format

Dear All

Hi

I have a question about Maple 16.

Does Maple 16 is able to use several cores in multi-cores system?

Does Maple 16 is able to execute some common commands such as int, diff, dsolve and solve with capabilities of a multi-cores system?

In the previous question, we discussed a workaround for evaluating a multi-variable integration. 

For the following example 

> g := (e1, e2, e3) -> piecewise(e1+1.200000000 <= e2 and e3-2.400000000 <= e1, evalf(64.00000000*exp((-1)*8.000000000*e1^2)*exp((-1)*8.000000000*e2^2)*exp((-1)*8.000000000*e3^2)/Pi^2), 0);
> 
> evalf(Int((x, y, z) -> g(x, y, z), [-2 .. 2, -2 .. 2, -2 .. 2]), 3);

Maple could start execution and do the computation.

When an integration is successfully evaluated numerically, a float number is returned, otherwise the unevaluated expression is returned. I'm wondering whether Maple has a utility function which simply takes in whatever the integration evaluation gives and return a boolean flag indicating whether it is unevaluated or is a floating point result? 

hi

i write the followinf code in maple

eq1:= f'''(x) - f'(x)

is this posible in maple that i worte the above command like this

Linear:= d3/d(x) - d/d(x)

eq1:= linear (f)

thanks for youe attention in advance

hi

i have 2 equation named eq1 and eq2 with 2 unknown parameter a and b which each equation dependent to a and b. i want to achieve critical a and b to minimize eq1 and eq2.

previously i have eq1 and one parameter a only and i use the NLPSOLVE command. but , for above mention problem i dont know how achieve a and b?

thancks for your attenrion in advance

Hello everyone !

First of all, excuse my English if it's difficult to read, I'm french.

I'd like to draw a surface/shape in 3D in spherical and I used plot3d with coords=spherical to do so. But I'd like to color on the picture to change when R(theta,phi) changes. I could visualize how R is changing and where it is the same. But I have no idea how to do so...

I can't put the picture here, the formula is too long...

Hi,

> assume(y::prime)

> about(y);
Originally y, renamed y~:
  is assumed to be: prime

I don't understand this answer:

> is(>0)

      FAIL

how can I solve memory used limitation problem on MAC OSX LION .After memory indicated on the bowwtom of the sheet reaches 57M it freezes.

Dear Maple Users

I have been testing Maple 16 for some time now, and I am overall very pleased with it. There is however one issue, which is really annoying. In previous version of Maple, images inserted into Maple and plots were printed much bigger than they looked like in the Worksheet. Then me and other users have requested to have the printed output look more like it does in the Worksheet on the computerscreen. Maple has adressed those user complaints in the new...

Not really a review but I think Maple has so much cool stuff one can do with it, that we get sidetracked from one project to the next.   Before you know it, a new release is out and you've put your project on hold to try out and play with the new features - and not really forging ahead on any projects.  One year between new releases isn't enough time to have customers really, and I mean really, dive into Maple.

Here is what a reviewer at computing world said about Maple16

Hi friends

This is strange for me, but Maple could not solve the below integral. I tried with online version

of Mathematica and it solved for me. The solution is Hypergeometry function. 

> f := ((1/2)*r+1/r)^(1/n);

> int(f, r);

What is the problem?

Regards!

This does not make any sense to me :-)

restart:
A := Matrix([[seq(i, i = 1 .. 10)], [seq(i, i = 11 .. 20)], [seq(i, i = 21 .. 30)]]):
A[.. -1,..] ;
A[.. -2,..] ;



You would assume that A[..-1,..] would remove the last row but that is not the case!
Instead you have to do A[..-2,..] he he it seems a bit odd!