I found the following inconsitencies when studying the 2nd branch of the LambertW function, and i was hoping someone can help me understand what i am doing wrong here:

#Forming an equation based on what the simplify command produces as output with the lhs on the rhs:

exp(-(1/2)*LambertW(1, 2*exp(2))+1) = simplify(exp(-(1/2)*LambertW(1, 2*exp(2))+1));

#A float approximation made directly for the first branch of LambertW directly:
exp(-(1/2)*LambertW(1, 2*exp(2))+1)=evalf[10](convert(evalf[10](LambertW(1, 2*exp(2))), 'rational'));
# A float approximation made for the value implied by equation stated in the first line gives a slightly differing value:
evalf[10](rhs(isolate((1/2)*sqrt(2)/sqrt(1/x) = convert(evalf[10](exp(-(1/2)*LambertW(1, 2*exp(2))+1)), 'rational'), x)));
#The symbolic simplify command gives the following output, which is contradictory to the evalf command's output for that argument:
exp(-(1/2)*LambertW(1, 2*exp(2))+1)=simplify(exp(-(1/2)*LambertW(1, 2*exp(2))+1),'symbolic');
#So i investigated this a little further in observing that this seemed like a few of the functions i have found very interesting, have a Number theoretic proportionality to the number of significant digits we have restricted the evalf command to:
F:=Re(exp(-(1/2)*LambertW(1, 2*exp(2))+1))+Re((1/2)*sqrt(2)/sqrt(1/LambertW(1, 2*exp(2)))):
fLOATY:=proc (k) options operator, arrow; evalf[k](F) end proc;
DATA := [seq([k, convert(fLOATY(k)*10^(k-1), 'rational')], k = 1 .. 40)];
plots[:-pointplot](DATA, color = "Black");

I'm working on a problem requiring non-commutative variables. I have this functionality working properly using the code below:

Interestingly solves runtimes vary by orders of magnitude when new equations are added to it. Interestingly in one of the cases shown here when  a larger more complex equation is added it runs orders of magnitude faster.

This graph shows the runtimes of solve on my laptop, as i pass it more members of a family of equations. Notably runtime jumps up by an order of magnitude at 4 but then reduces when i add the next equation, despite it being a much longer and more complex one


The worksheet this is taken from ,where you can see all the arguments passed to solve is here.

Hi all. I have two questions about polynomials over Q.

i) Let f,g be polynomials over Q. How to show (with Maple) that the decomposition field of f is included in the decomposition field of g?

ii) Let f be a polynomial of degree n  over Q with Galois group C_n, the cyclic group of order n. Then, there is a polynomial P with coefficients in Q, of degree n-1, s.t. the iteration: u0=some root of f, u_{k+1}=P(u_k) gives all the roots of f; how to find P (with Maple)?

Thanks in advance.

Hi everiboty

This is a notional example derived from my real application.
I have different quantities named Object1, Object2, ... ObjectN (N being a nonnegative integer) and a boolean sequence A of length N.
By construction A contains only one "true" values (and thus N=-1 "false").
I wrote a procedure Choice that takes as input such a sequence and returns the single object among Object1, ... ObjectN for which [A][n]=true.

I know that it would have been clever to name the quantities Object1, ... ObjectN differently (for instance by indexing them), but I'm constrained by what already exists.
Here is my procedure

restart:
Choice := proc(a)
  local l, n, object:
  l := [seq(args(k), k=1..nargs)]:
  n := ListTools:-Search(true, l):
  object := Object||n:
  print(object):
end proc:

# a few examples of Object(s)
Object1 := {x1, y1, z1}: 
Object2 := {x2, y2}:
Object3 := {x3}:

# application:
A := false, true, false:
Choice(A);
    {x2, y2}

Now I want to insert this procedure within a another one named “f”:
restart:

f := proc():
local Object1, Object2, Object3:
local Choice, A:
Choice := proc(a)
  local l, n, object:
  l := [seq(args(k), k=1..nargs)]:
  n := ListTools:-Search(true, l):
  object := Object||n:
  print(object):
end proc:

Object1 := {x1, y1, z1}: 
Object2 := {x2, y2}:
Object3 := {x3}:

# here I give A some values, in the true application A is the sequence
# returned by a Maplet through a Shutdown command
# More precisely A correspond to “N” buttons of the same group … then
# only one is “true”.

A := false, true, false:
Choice(A);
end proc;

f();
    Object2

Replacing print(object): by print(eval(object)): within procedure  Choice produces  the same ‘uneval’ answer.

I cannot figure out why procedure f doesn’t return the expected {x2, y2} answer
The only workaround I have found is to replace
  object := Object||n:
  print(object):

by this awfull sequence
    if n=1 then print(Object1)
  elif n=2 then print(Object1)
  …
  end if

Could you please help me to understand where is my error and how could I fix it?

Thanks in advance

hi guys,

i having trouble solving 3 equations with 3 unknowns. in fact i have 3 independent unknowns p,v,w and 1 dependent unknown t, i want to plot p(t)! actually in a simple system of equations one can solve p(t) analytically and then plots it but in a complicated system of equations it is impossible to solve p(t) analytically. i consider a simple system of equations in attachment to understand the procedure of ploting the solutions of a system of equations numerically.

numerically.mw

Could someone pleease clarify this error message below.

sol1 := dsolve({syst, theta1(0) = (1/2)*Pi, theta2(0) = (1/4)*Pi, (D(theta1))(0) = 0, (D(theta2)(0) = 0}, theta1(t), theta2(t), type = numeric, output = listprocedure);
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations
 

singular(ln(y^2+1),y);

        {y = -I}, {y = I}, {y = infinity}, {y = -infinity}.

But if "y" was real, then there are no finite singularities, since y^2 is always positive and hence y^2+1>0 always. Adding assumptions did not help

singular(ln(y^2+1),y) assuming y::real;

gives same result. RealDomain does not support singular.  But I am no longer using RealDomain as it seems bugy.

I know I could filter out these complex results using remove(), but it would be nice if there was a way to singular supports assumptions. Is there a way to do it?

Why are the following 2 commands produce different result?

restart;
RealDomain:-solve({y=y,y<>0},y);
solve({y=y,y<>0},y);

                            [y = 0]
                            {y <> 0}
 

Should not the result be the same?

solve({y=y,y<>0},y) assuming y::real;
                           {y <> 0}

Maple 2018.1. 

Dear friends,

I noticed that the function numbcomb from the package combinat fails to calculate a value properly.

Let's try to calculate the example from the function's help page:

combinat[numbcomb](7, 3);
combinat[numbcomb](5, 4);

returns 35, 5 correspondingly, whereas should be 15 and 4.

Please, check this in your systems. Maybe, this is a bag in my installation?

Suppose I have a subset of R^2, say (x,y) € [0,1] x [0,1]

Now I map

x=r*exp(t)

y=r*exp(-t)

Is it possible to do a 2d contourplot to just see where [0,1]x[0,1] is mapped onto ?

Hi everyone,

I'm having some difficulty replicating a vector diagram. 

So I have learnt how to plot the vector diagram and I have been getting the correct results however the graph I am trying to replicate has some sort of limit within the range they are using which changes the direction of the vector. I was hoping to get this result, but I don't know how to achieve it. Could anyone point me to the right direction please.

Graph I want to replicate (i've highlighted the section that I want to achieve), what I get.

I look forward to your replies.

Thank you in advance

Are these errors to be expected? Why do they happen?

restart;
solve({x<>10, -infinity<x , x<infinity, -infinity<y , y<infinity},{x,y});

Error, (in solver) invalid input: SolveTools:-Inequality:-LinearUnivariateSystem expects its 1st argument, eqns, to be of type ({list, set})({`<`, `<=`, `=`}), but received {x <> -infinity, x < 10}

But it works when replacing x<>10 by y<>10

restart;
solve({y<>10,-infinity<x , x<infinity, -infinity<y , y<infinity},{x,y});
            {10 < y, x < infinity, y < infinity, -infinity < x}, {y <> -infinity, x < infinity, y < 10, -infinity < x}

What is the difference in the above two?

It also work when replacing x<>10 by x=10

solve({x=10, -infinity<x, x<infinity, -infinity<y , y<infinity},{x,y});
               {x = 10, y < infinity, -infinity < y}

It also works when removing the y parts by keeping x<>10

solve({x<>10, -infinity<x , x<infinity},{x,y});
           {y = y, 10 < x, x < infinity}, {y = y, x < 10, -infinity < x}

it also works when removing x<>10 and putting back the y stuff

solve({-infinity<x , x<infinity, -infinity<y , y<infinity},{x,y});
              {x < infinity, y < infinity, -infinity < x, -infinity < y}

Why Maple gives an error for some cases and not the others?

Maple 2018.1

Hi

I need to find a relation between delta [m,k] in terms of m and k

delta[m,k]=f(m,k), where k=0,1,2,...,m

A code is written (delta.mw) to find delta[m,k] for a certain amoun of m.

Is there a way or a code to find a general form of f(m,k)?

Thanks

Hello

I've written a completely symbolic expression involving sum function. When I hit enter, Maple 2016.2 evaluates the expression although it's completely symbolic and not evaluable.

How to stop Maple from falsely evaluating a symbolic expression? I dis use the restart command to make sure all variables are empty. When evaluating Maple generate a very large open form expression that makes no mathematical sense.

Maple file in the following link

https://www.dropbox.com/s/13ciwothh1pqijk/sumBug.mw?dl=0

Please advice

Thanks