MaplePrimes Questions and Posts

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Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

Tried posting responses to a question this evening. Click the question title, hit the "reply" button and enter a phrase like "Clarification" in the Title filed of the reply form, the reply form just disappears. Tried this in Firefox and Chrome - same behaviour. Means I can no longer contribute - has somebody broken something??? Or I have a problem with *all* my browsers

PS not even sure that this post will appear correctly

(1) I keep trying to enter a reply, and it just never appears. What's wrong?

(2) How do I get code from the Standard GUI to paste into a reply?

the calculation is like the following command, the result in the picture

SetCoordinates(spherical[r, theta, phi]);
Fv := rho*VectorField(`<,>`(v[r](r, theta, phi), v[theta](r, theta, phi), v[phi](r, theta, phi)));



1) when the Divergence act on the Fv, then it will be expanded, which is lengthy and not like most book's formulation , especially when I want to continue for a Conversation law like in fluid mechanics, this will be too long and a messy for later check.

could there be a way to not expand this result, just as the eq(3) like.

2) when I want to calculate the Divergence of Fv, I must construct a VectorField at first, but this is in components way, is there a quick way for Vector Field Function


hello guys , i have a metric and i want to find its non-zero components of ricci tensor but i have problem with writing the metric and maple gave me error.

I have the following procedure to do the above. It works but it returns [9,10],[10,9],[12,1] for n=1729(for example). How do I modify this to 

a) to count 9,10 and 10,9 as the same and hence only show one of them

b) get 1,12 to show as a solution?

global listcub:=table();
local k:=0, x:=iroot(iquo(n,3),3),y:=x,x3:=x^3,y3:=y^3;
if 3*x3 <> n then x=x+1; x3:=x^3;y:=x;y3:=x3 end if;
while x3<=n do
y:=iroot(n-x3,3); y3:=y^3;
if(x3+y3 = n) then k:=k+1; listcub[k]:=[x,y]end if;
x:=x+1; x3:=x^3;
end do;
end proc:


I currently have a function quadsum(n) that determines the [x,y] solutions of the above equation for an integer n. :

quadsum:= proc(n::nonnegint)
k:= 0, mylist:= table(),
x:= isqrt(iquo(n,2)), y:= x, x2:= x^2, y2:= y^2;
if 2*x2 <> n then x:= x+1; x2:= x2+2*x-1; y:= x; y2:= x2; end if;
while x2 <= n do
y:= isqrt(n-x2); y2:= y^2;
if x2+y2 = n then k:= k+1; mylist[k]:= [x,y] end if;
x:= x+1; x2:= x2+2*x-1;
end do;
convert(mylist, list)
end proc:

How would I alter this so that I get [x,y] for n= (5^a).(13^b).(17^c)(29^d) for non-negative integers a,b,c,d?

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

I've got the following piecewise function :

(x^2+y^2)^(alpha).arcsin(y/x) if (x,y) are in [-pi/2,pi/2]

0, (x,y)=(0,0)

1. How do I plot this function taking the alpha variable and the piecewise construct into account?

2. How can I check for points of discontinuity, indifferentiability from the plot/function itself?



I need to show what happens to the zero r=20 of f(x)= (x-1)(x-2)..(x-20)-(1/10^8)*(x^19) and the hint given is that the secant method in double precision gives an approximate in [20,21].

At present, I'm calling the secant method on f with a tolerance of 1/(10^12) with an initial x=20, but I'm stuck as to what the second initial value would be. What is the right approach to this question?


I'm trying to get the RHL of exp1:=(2/(1+e^(-1/x)) as x->0+

and have l2:=limit(exp1,x=0,right) but that isn't giving me a value. How do I correct this? 


I've got the following double integral over a region A:

e^(1/x*y)/(y^2)*(x+1)^2 where A={(x,y):1/2<=x*y<=2,1<=x<=3}

to evaluate this:

I've tried :


since the largest lower bound and smallest upper bound for y based on 1/2<=xy<=2 are 1/2 and 2/3 respectively.

This statement however, only evaluates the inner integral; is my approach correct?

I've got a pair of equations :

x^3-4x=y and y^3-4y=x

 which I've defined as eqns:={,}

and 9 solutions as solns1:={,,..}

and being stored as s1,s2,..s9

when I run a command such as testeq(subs(s1,eqns[1])=subs(s1,eqns[2])

I get an error of passing invalid arguments into testeq. What I essentially need to show is that on substituting for x,y from each s1,..s9; both equations get the same result. What am I doing incorrectly?

I've also noticed that just subs(s1,eqns[1]) returns an equality; I don't quite understand why

I'm given the following two equations:

x^3-4x=y, y^3-4y=x

to solve the system, I've just used




and have obtained only four solutions when I should instead get 9. Is there a mistake in my approach?

Hi all,

I google and found a program using C# connect with Maple. The Maple file is mla file - a pakage library type of Maple. I want to review data structure and all interfaces funtions to understand the way to implement this features.

Please help me the way to read the original Maple code. I uploaded the .mla file into mediafire if you want to review it. Link 


Quan Nguyen

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