vv

12453 Reputation

19 Badges

9 years, 283 days

MaplePrimes Activity


These are replies submitted by vv

@ReactionUra 

Some variables are mathematically negative; they are small in absolute value, so probably they can be considered 0.

Note that dharr's solution is similar; actually if you start his worksheet with Digits:=25, fsolve fails just because of those variables.

Explore(plot([[[0,0],[1,0],[1/3,2],[0,0]],[(12*s*(t-1)*(s-1)*xi^2+t)/(3+12*(t-1)*(s-1)*xi^2+12*(t-1)*(s-1)*xi),2*t/(1+4*(t-1)*(s-1)*xi^2+4*(t-1)*(s-1)*xi), xi=-infinity..infinity]]), parameters=[s=0. .. 1, t=0. .. 1]);

 

@mmcdara 

1. Not for permutations. E.g. it is known the number of invertible nxn matrices over Zp (i.e. the order of the group GL(n, Zp)).

2. Yes, I have mysef a post here with a Sudoku solver using a Groebner basis: 
https://www.mapleprimes.com/posts/207910-A-Compact-Sudoku-Solver-Via-Groebner

@one man 

The parametric equation of an inellipse is here. Probably in Maple 17 this will work:

restart;
Explore(plot([[[0,0],[1,0],[1/3,2],[0,0]],[(12*s*(t-1)*(s-1)*xi^2+t)/(3+12*(t-1)*(s-1)*xi^2+12*(t-1)*(s-1)*xi),2*t/(1+4*(t-1)*(s-1)*xi^2+4*(t-1)*(s-1)*xi), xi=-infinity..infinity]]), s=0. .. 1, t=0. .. 1);


 

restart;
A:=[1,0]: B:=[1/3, 2]:
K:= 1/(s-1)/(t-1): U:=s*~A: V:=t*~B:
ell := (4*xi^2*~U+K*~V ) /~ (4*xi^2+4*xi+K):
Explore(plot([ [[0,0],A,B,[0,0]], [ell[], xi=-infinity..infinity]]),s=0.0 .. 1, t=0.0..1);


 

This reminds me of the simplest bijection f : N^2 --> N.
f(n,k) = 2^(k-1) * (2*n-1).

 

@dharr 

Nice, but unfortunately for n=4 the  reduction is almost the same and n>4 is out of the question.
Also, rhe diagonal dominance could just eliminate a "few" nonsingular matrices.

@dharr 

It is possible a reduction by a factor of n!*n.
I was curious about a compiled version. I have used a custom "nextperm" and "det" and the runtime was 0.015 sec for n=3.

But it is not very useful. Trying n=4, the estimated runtime is about a week!!! Without compilation it would be years!
It would be interestiong a theoretical approach, but I have no idea about that.
 

 

@janhardo 

If you cannot find the book, you can at least download (from the author's site) and work the exercises and their solutions. They are for Maple 8 but are valid for Maple 2020 too.

@janhardo 

The age of a book is not always an issue. For example, Heck's book, https://staff.fnwi.uva.nl/a.j.p.heck/Maplebook/
is excellent after almost 20 years. Almost all the code still works.
I recommend it (but it contains more advanced material; of course, newer topics are not covered). It has many examples with full solutions.

@mmcdara 

In simpler words, if f(...) is an unevaluated function:
[1]. diff(f(u), x)  is computed in Maple by calling  `diff/f`(u, x),
provided that the procedure `diff/f`  is defined.
[2]. If f is indexed i.e. diff(f[i](u), x)  is to be computed, then inside the procedure
`diff/f`, i is retrived as op(procname).
 
[1] is documented, [2] was a guess as Carl said.

@ecterrab 

My point was just that debugging is more difficult than it used to be long time ago. I know that the regular commands are documented. I was referring to some implementation notes to be used by advanced users and the fact that part of the general Maple functionality is now located in the Physics package.

@gawati2611 

In this case you can simply plot  Re(z)>0, |z| > 1.
But note that inequal also accepts boolean operators, such that the region can be as complicated as you want (see the help pages).

@Preben Alsholm 

Nice catch! I remember that long time ago, such bugs were much easier to detect. Now, there are Maple fundamental functions which are implemented in a strange and undocumented manner. Of course, most of these functions are not designed to be used directly by the user, but it would be nice to know something about them. For example, some of such functions (including a part of the assume facility)  are in the Physics package, where almost everything is redefined.

@mmcdara 

restart

The parametrization of the intersection, including the ranges, is contained in the result of solve (see below).
It can be extracted programatically, but I did not have the patience to do it in general.

with(plots):

cylx := y^2 + z^2 - 1:

cylz := x^2 + (y+1)^2 - 1/2:

cyl:=implicitplot3d([cylx,cylz], x=-2..2, y=-2..2, z=-2..2,
scaling=constrained, style=surface, grid=[40, 40, 40], labels=[x, y, z], color=blue, transparency=0.5):
display(cyl):

opt:=thickness=4, scaling=constrained, style=surface:

s:=solve([cylx,cylz, x^2>=0,y^2>=0,z^2>=0],[x,y,z], explicit);
 

[[x = -(1/2)*2^(1/2), y = -1, z = 0], [x < 0, -(1/2)*2^(1/2) < x, y = -1+(1/2)*(-4*x^2+2)^(1/2), z = -(1/2)*(4*x^2-2+4*(-4*x^2+2)^(1/2))^(1/2)], [x < 0, -(1/2)*2^(1/2) < x, y = -1+(1/2)*(-4*x^2+2)^(1/2), z = (1/2)*(4*x^2-2+4*(-4*x^2+2)^(1/2))^(1/2)], [x = 0, y = -1+(1/2)*2^(1/2), z = -(1/2)*(-2+4*2^(1/2))^(1/2)], [x = 0, y = -1+(1/2)*2^(1/2), z = (1/2)*(-2+4*2^(1/2))^(1/2)], [x < (1/2)*2^(1/2), 0 < x, y = -1+(1/2)*(-4*x^2+2)^(1/2), z = -(1/2)*(4*x^2-2+4*(-4*x^2+2)^(1/2))^(1/2)], [x < (1/2)*2^(1/2), 0 < x, y = -1+(1/2)*(-4*x^2+2)^(1/2), z = (1/2)*(4*x^2-2+4*(-4*x^2+2)^(1/2))^(1/2)], [x = (1/2)*2^(1/2), y = -1, z = 0]]

(1)

s2:=tubeplot([x,-1 + sqrt(-4*x^2 + 2)/2, -sqrt(4*x^2 - 2 + 4*sqrt(-4*x^2 + 2))/2], x=-sqrt(2)/2..0, radius=0.05, opt, color=red):
s3:=tubeplot([x,-1 + sqrt(-4*x^2 + 2)/2, sqrt(4*x^2 - 2 + 4*sqrt(-4*x^2 + 2))/2], x=-sqrt(2)/2..0, radius=0.05, opt, color=red):

s6:=tubeplot([x, -1 + sqrt(-4*x^2 + 2)/2,-sqrt(4*x^2 - 2 + 4*sqrt(-4*x^2 + 2))/2], x=0..sqrt(2)/2,radius=0.05, opt, color=red):
s7:=tubeplot([x, -1 + sqrt(-4*x^2 + 2)/2,sqrt(4*x^2 - 2 + 4*sqrt(-4*x^2 + 2))/2], x=0..sqrt(2)/2,radius=0.05, opt, color=red):

display(cyl, s2,s3,s6,s7);

 

 

 

First 33 34 35 36 37 38 39 Last Page 35 of 166