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MaplePrimes Activity

These are replies submitted by Kitonum

@David Sycamore 

a:= proc(n::posint)
option remember;
   if n=1 then return 1 fi;
   if type(a(n-1)+1,prime) then return a(n-1)+1 else
   if type(a(n-1),even) then return a(n-1)+2 else a(n-1)+3 fi; fi;
end proc:

seq(a(n), n=1..20);


@RaulGutsa  For 3d you can use in the procedure  P  the  plots:-pointplot3d  command:

Sol:=dsolve({Eq, ics}, numeric):
plots:-pointplot3d(eval([[x,y(x),x]], Sol(s)), style=point, symbol=solidcircle, symbolsize=15);
end proc:
plots:-animate(plots:-display,['P'(s)], s=0..Pi, scaling=constrained);


Present here in editable form this matrix so that I can copy it into my Maple. I have a program written many years ago and not published anywhere that allows you to step by step bring the matrix of an arbitrary size to the ReducedRowEchelonForm  (comments to the steps in Russian). 

@mugwort  I ran  your worksheet without any errors (only replaced digits with Digits).

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .


@mmcdara  You wrote: "The main point is in the definition of f3 which should had been f3 := (x, y) -> f(x, y)+f2(x, y)".

But  f3:=f+f2  and  f3 := (x, y) -> f(x, y)+f2(x, y)   are the same, regardless of whether the variables  x  and  y  are predetermined or not.

For example:

f := sin + cos:

@MuhammadSinan  Using the  eval  command you can get a list of values of any of the functions with the necessary step. For example:

[seq(eval([t, B(t)], sol(s)), s = 0 .. 1, 0.01)];


@Carl Love  Since the new 3 teams differ in color, it is not necessary to divide by 3 in the denominator.

@acer  If the graphs should be at different plots, it can be made easier:

A:=plots:-animate(plot,[x^2, x=-2..a, color=red], a=-2..2, frames=60):
B:=plots:-animate(plot,[sin(2*x), x=0..a, color=blue], a=0..2*Pi, frames=60):
plots:-display(<A | B>, size=[700,400]);


@Awesome123  You need to more clearly articulate your task. For starters, I recommend reading this article .

Here's a possible formulation of the problem: find the dimensions of a cylindrical container that has the smallest full surface that could hold all your chocolates. For comparison, the same for the container in the form of a cuboid.

 @Zeineb  My answer is incomplete and only applies to case b = 0 (I just did not notice this number b). k=0 will also be the solution for this case. But in the general case, for an arbitrary b, the answer will be much more complicated and will depend on  . I'm working on it now and soon my answer above will be updated.

@tizozadoxo  This linear system has the unique solution  (0, 0, 0) . But maybe you are confusing the dimension of the solution space with the number of solutions. From a geometric point of view, every solution of the system is a point in the space  R^3 . But a point has dimension  .

@Carl Love  I understood the dots above the variables in the first equation of the system as derivatives.

@Lali_miani   Because  u(1)=sin(u(0)), u(2)=sin(u(1))=sin(sin(u(0)))  and so on.

@Lali_miani  For example  (sin@@3)(x)  means  sin(sin(sin(x))) :



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