MaplePrimes Questions and Posts

These are Posts and Questions associated with the product, MaplePrimes

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.

Hi everyone,

at the suggestion of Carl I am making my question a post.


Newbies often get fascinated with the power tower: x^x^x^...

The generalized power tower z^z^z^... is a special case of the Euler sequence:

z_{n+1}=c^z_n, for c\in C.

Like the Mandelbrot set: z_{n+1}=z^2-c, the power tower, often called infinite exponential, also has a general periodic map. It is included below and is taken from Daniel Geisler's site:

Shel and Thron proved that the infinite exponential conveges whenever c belongs to the red region, called today Shell-Thron region.

Definition of Julia Sets for the iterated exponential map:

Also like the Mandelbrot set, the infinite exponential admits Juila Sets. The Julia Sets of the infinite exponential however, are defined differently from the Julia Sets of the Mandelbrot set. They are defined to be  the closure of the periodic repellers of the Euler sequence . They are Cantor Bouquets.  Geisler's colored map then is a general map of how the corresponding Julia set behaves roughly, with c taken from the map. 

We can then introduce small cuts which go from the interior of the Shell-Thron region to the exterior, crossing at various angles, and this will tells us how the infinite exponential evolves. Generally speaking, each time one crosses the Shell-Thron boundary, one wittnesses what's called a Knaster explosion, wherein the exponential explodes into p subregions, where p is the pre-period of the multiplier.

When the parameter c exits the Shell-Thron region at angles of 2*Pi and Pi from the real axis (cuts right and left, p=1, 2), the infinite exponential either transitions from converging to a single feature to exploding into multiple indecomposable contiua (p=1), or it breaks into a period 2 bifurcation (p=2), which itself, also may explode into continua.

When it exits at angles 2*Pi/p, where p>2 is the preperiod of the multiplier, then the infinite exponential evolves from converging to a single feature, to exploding into  p major regions, called Fatou regions, each one having its own attractor, displaying a p-furcation.

In all cases,  the Knaster explosions may introduce the presence of indecomposable continua, as some Fatou regions end up covering entire parts of the complex plane after each transition. In the animations, Knaster explosions occur whenever the background is red. There may be more than one explosion present in the evolution of the power tower. 

Cantor Bouquets are strange creatures. They are essentially quantum sets, and no point of them is actually visible. The probability a point is visible varies directly with the area of the corresponding bouquet "finger" which is rendered in the area of interest. Devaney uses these "fingers"  to obtain "iteneraries" of the iterated exponential map.

Points "close"  to the fingers-hairs o the Cantor Bouquets escape towards complex infinity at (final) angles 2*Pi/p.

The "hairs" of a Cantor Bouquet are C^\infty curves, hence they can be termed easthetically pretty. Only hairs from the main Cantor bouquet for c=e^{1/e} are globally convex/concave. Every other bouquet may contain hairs which change curvature "unpredictably".

Inlcuded files:





Code for static fractal with given parameters in 1). Parameters can be changed in the constant section (Try p=2,4 or Pi)

Code for morphing animations through cuts in the Shell-Thron region in 2). Parameters d1-d2, N

Code for zooming animations in 3). Parameters M, M1-M2,


Will show attractors of broken Fatou basins only up to pre-period p=5. No big deal though. If you want to increase the preperiod past p=5, you need to add the relavnt tests in procedure ppc(c), otherwise the new attractors are not calculated and the plot ends up red.

Colors are assigned a bit more dispersed, using ln(m). You can also use m in that place. It depends on which area you are in.

Basins of attraction and Fatou regions change appearance under different epsilons. If you want different shapes for the basins, you can try using the Manhattan metric in proc Jhf, instead of |z-lim|.

Included below are the main map of tetration by Daniel Geisler, a short excerpt of the Shell-Thron region which shows pre-periodic cuts and  6 morhing transitions, for p\in {1,2,3,4,5} and one which exits at the angle of 2 radians (p=Pi).


More than 300. To be published with my thesis. Patches, problems and code corrections in the original question page:

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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