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.

non_zero_components_of_ricci_tensor.mw

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?

cubesum:=proc(n::nonnegint)
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;
convert(listcub,list);
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)
local
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 :

int6:=int(int(e^(1/x*y)/((y^2)*(x+1)^2),x=1..3),y=(1/2)..(2/3));

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

eqns:={x^3-4x=y,y^3-4y=x};

vars:={x,y};

solns:=solve(eqns,vars);

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 http://www.mediafire.com/download/abd9kpk3q6oq8lb/CHEMISTRY.mla 

Regards,

Quan Nguyen

Hello everyone, I am having problem while trying to evaluate the double integral written in the file attached double_inegral.mw. Please help me out.
Many thanks in advance.

>restart:
>with(LinearAlgebra):
>with(Student[NumericalAnalysis]):
s := {E[1], E[2], E[3]};
v := {x[1], x[2], x[3]};
A := GenerateMatrix(s, v);
B := augment(A)
Then what i do that for any matric i can use same program.

It appears that email notifications have gone on Christmas vacation.
When will they be back?

I have a statement:

print(plots[display](seq(colr2[i],i=1..numplayers, xtickmarks=0,axes=NONE,view=[0..5*numplayers,0..1])):

which outputs a horizontal display of the various "colors" of the players, and it does so.

    However, the height of the output is quite big and makes the output look "chunky".  Inserting the option "scaling=constrained" makes the width of the output much narrower - but there is a bounding box of the output (which is not actually displayed.)  It is displayed when the mouse cursor is clicked on the display.  I was hoping the view= [..] option would provide a solution, but this does not alter the exterior bounding rectangle.

  Is there any means of modifying the dimensions of the rectangular display through Maple code?

Thanks,

   David