This differential equation has an analytical solution.

However, I am looking for a numerical solution for it.

A single boundary condition with respect to t would be sufficient to solve the problem, but the command does not accept this.

In the end, I try to insert boundary conditions in relation to the variables x and t, but again this does not work.

Where am I going wrong?

Hello my friends

I have some problems with maple 18. I try to consider and extract some things about tensor such as contraction.

for instance, suppose we have metric=-exp(alpha(r))*(dt^2)+exp(beta(r))*(dr^2)+r^2*(dtheta^2)+r^2*(sin(theta)^2)*(dphi^2). how we can find all Riemann tensor and corresponding contraction, Ricci tensor and its contraction and even Weyl tensor and its contraction. unfortunately, I attempt to find them by using some other examples on the net but they don't help me to calculate them when time is the first element in coordinate, not last ( t,r,theta,phi) not (r,theta,phi)

thanks with the best regard

In Maple 2017, a simple one-liner can crash the current worksheet (in any mode with any text type it seems).

z[x]:=z

z[x]:=z(x)

Note that z and x can be any two letters, and that you can replace z[x] with the equivalent z+(ctrl shift _)+x (which displays as z_x)

I was just wondering if someone could explain why kernelopts(maxdigits) = 38654705646, as in is this different for a differing computer, or is there a design aspect of maple that requires it to be this number, or is there a mathematical reason?

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 ?