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9 years, 318 days
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Symbolic...

vv 13837
January 08 2019
1 0

The integrand has the form

F := a*exp(b*v^2)*v^c;
So,
int(F, v) :  simplify(%);


i.e. it can be expressed via incomplete GAMMA ( = the  2-arguments GAMMA).

 

1D...

vv 13837
January 06 2019
1 0

If you select that line and convert it to 1D Math input (even if it appears to be 1D), the asterisk will be visible.

Why don't you use worksheet mode with 1D input?

union & intersect...

vv 13837
January 04 2019
0 0 
L1 := [{3,5},{4,3},{4,6},{3,2},{2,6}]:
L2 := [1,5,4]:
evalb( `union`(L1[]) intersect {L2[]} <> {} );

    true

semi-bug...

vv 13837
January 04 2019
1 2
> 

pde0:=diff(u(x,t),t)+diff(u(x,t),x)=0;
sol0:=pdsolve(pde0,u(x,t)) assuming x>0;
pdetest(sol0,pde0);

diff(u(x, t), t)+diff(u(x, t), x) = 0

  

u(x, t) = _F1(-x+t)

  

0

 (1)
> 

pde:=diff(u(x,t),t)+diff(u(x,t),x)=0;
bci:= u(0,t)=0, u(x,0)=sin(x);
sol:=pdsolve([pde,bci],u(x,t)) assuming x>0;
pdetest(sol,pde);

diff(u(x, t), t)+diff(u(x, t), x) = 0

  

u(0, t) = 0, u(x, 0) = sin(x)

  

u(x, t) = -sin(-x+t)*Heaviside(x-t)

  

0

 (2)
> 

restart;
> 

# Executing all (with (!!!)) ==> 0
# Executing then only the 2nd execution group ==> <>0
> 

#    Dirac(t-x)*sin(t)*cos(x)-Dirac(t-x)*cos(t)*sin(x)-Dirac(x-t)*sin(t)*cos(x)+Dirac(x-t)*cos(t)*sin(x)

 

This is strange but not a real bug. Let's call it "semi-bug".

Note first that the solution of the pde (with bci) is not a classic one. In fact, the solution (containing Heaviside) is continuous but not C^1; actually the partial derivatives do not exist on the line x=t.

I don't know what pdetest does to test the solution but obviously sol does not satisfy the pde everywhere. On the other side, the nonzero result of pdetest  simplifies to 0 with assuming x>t; also for assuming x<t

So, both results of pdetest(sol,pde)  (zero or not zero)  are correct; the problem is that this depends on the order of execution!

 

I mention that I used Maple 2018.2 without any update.

 

> 

 


Download pde!!!.mw

seq+eval...

vv 13837
January 03 2019
1 0

I'd use

A:=[x1,x2,x3]:
B:=[f(x),g(x)]:
[seq([seq(eval(b,x=a),a=A)],b=B)];

        [[f(x1), f(x2), f(x3)], [g(x1), g(x2), g(x3)]]

parse...

vv 13837
January 03 2019
2 1

The command  Get('xxx1')   returns a string, but you need an exprseq. Use parse(Get('xxx1')).
Or, simply add the line
xxx := parse(xxx);  o := parse(o);  co := parse(co);

Conventions...

vv 13837
January 02 2019
1 1

These are mathematical conventions. For degree(0) alternative conventions are -1  and "undefined".
- infinity is preferable because e.g. the equality  degree(f*g) = degree(f) + degree(g)  remains valid.
These conventions are related (and compatible) with other two standard ones:
max({}) = - infinity  and  min({}) = infinity.

readline...

vv 13837
January 02 2019
1 0

Try the old-fashioned solution:

f:=x->x^2:
printf("a=?");
a:=parse(readline());
'f(a)'=f(a);

 

Manny solutions...

vv 13837
January 02 2019
0 3

Using a translation, we may suppose that the center of the sphere is (0,0,0).
Even so, the number of the solutions is large and not very easy to obtain (the brute force is out of the question).
E.g.  for the radius of the sphere in the interval  2.10 .. 2.11  here are 32 solutions (actually almost 6 times greater because the permutations are not included; probably other symmetries should have been excluded too):

{{[-20, -3, -3], [-20, 3, 3], [20, -3, 3], [20, 3, -3]}, {[-20, -3, 3], [-20, 3, -3], [20, -3, -3], [20, 3, 3]}, {[-19, -3, -3], [-19, 3, 3], [19, -3, 3], [19, 3, -3]}, {[-19, -3, 3], [-19, 3, -3], [19, -3, -3], [19, 3, 3]}, {[-18, -3, -3], [-18, 3, 3], [18, -3, 3], [18, 3, -3]}, {[-18, -3, 3], [-18, 3, -3], [18, -3, -3], [18, 3, 3]}, {[-17, -3, -3], [-17, 3, 3], [17, -3, 3], [17, 3, -3]}, {[-17, -3, 3], [-17, 3, -3], [17, -3, -3], [17, 3, 3]}, {[-16, -3, -3], [-16, 3, 3], [16, -3, 3], [16, 3, -3]}, {[-16, -3, 3], [-16, 3, -3], [16, -3, -3], [16, 3, 3]}, {[-15, -7, -1], [-15, 7, 1], [15, -5, -5], [15, 5, 5]}, {[-15, -7, -1], [-5, -13, -9], [5, 15, 5], [15, 5, 5]}, {[-15, -7, 1], [-15, 7, -1], [15, -5, 5], [15, 5, -5]}, {[-15, -7, 1], [-5, -13, 9], [5, 15, -5], [15, 5, -5]}, {[-15, -5, -5], [-15, 5, 5], [15, -7, -1], [15, 7, 1]}, {[-15, -5, -5], [-5, -15, -5], [5, 13, 9], [15, 7, 1]}, {[-15, -5, 5], [-15, 5, -5], [15, -7, 1], [15, 7, -1]}, {[-15, -5, 5], [-5, -15, 5], [5, 13, -9], [15, 7, -1]}, {[-15, -3, -3], [-15, 3, 3], [15, -3, 3], [15, 3, -3]}, {[-15, -3, 3], [-15, 3, -3], [15, -3, -3], [15, 3, 3]}, {[-15, 5, -5], [-5, 15, -5], [5, -13, 9], [15, -7, 1]}, {[-15, 5, 5], [-5, 15, 5], [5, -13, -9], [15, -7, -1]}, {[-15, 7, -1], [-5, 13, -9], [5, -15, 5], [15, -5, 5]}, {[-15, 7, 1], [-5, 13, 9], [5, -15, -5], [15, -5, -5]}, {[-13, -6, -1], [0, -14, -11], [0, 11, 4], [13, 4, 5]}, {[-13, -6, 1], [0, -14, 11], [0, 11, -4], [13, 4, -5]}, {[-13, -4, -5], [0, -11, -4], [0, 14, 11], [13, 6, 1]}, {[-13, -4, 5], [0, -11, 4], [0, 14, -11], [13, 6, -1]}, {[-13, 4, -5], [0, -14, 11], [0, 11, -4], [13, -6, 1]}, {[-13, 4, 5], [0, -14, -11], [0, 11, 4], [13, -6, -1]}, {[-13, 6, -1], [0, -11, 4], [0, 14, -11], [13, -4, 5]}, {[-13, 6, 1], [0, -11, -4], [0, 14, 11], [13, -4, -5]}}

The radii above are:
2.100420126, 2.102918181, 2.104995276, 2.106740650, 2.108221156, 2.108878475, 2.109487662

I don't post the code now because it is not clean & tested.

 

Digits...

vv 13837
December 30 2018
2 3

Increasing Digits (e.g. to 25) will solve the problem.
(There are also options such as 'abserr', 'relerr').
BTW, why are you so sure that rkf45 is the best method for your ODE? If you are not, let dsolve choose the method.

Newton...

vv 13837
December 30 2018
0 1
> 

F:=<x[1]^(2)*sin(x[1]*x[2])+x[2]^(3)-5,    
    exp(x[1]^(2))*x[2]^(2)+3* x[1]^(2)*x[2]+x[2]^(2)-6>;

Vector(2, {(1) = x[1]^2*sin(x[1]*x[2])+x[2]^3-5, (2) = exp(x[1]^2)*x[2]^2+3*x[1]^2*x[2]+x[2]^2-6})

 (1)
> 

V:=<x[1],x[2]>;  # Variables

Vector(2, {(1) = x[1], (2) = x[2]})

 (2)
> 

G:=Matrix(2, (i,j) -> diff(F[i],V[j])):  # Jacobian
> 

G1:=G^(-1):
> 

X[0] := <0.1, 2.>;  # Initial iteration

Vector(2, {(1) = .1, (2) = 2.})

 (3)
> 

for k from 0 to 6 do
  X[k+1] := eval(V - G1.F, Equate(V, X[k]));
od;

Vector(2, {(1) = 0.58600855e-1, (2) = 1.750059558})

  

Vector(2, {(1) = .1847971795, (2) = 1.710602709})

  

Vector(2, {(1) = .1444078366, (2) = 1.709551622})

  

Vector(2, {(1) = .1388413340, (2) = 1.709461608})

  

Vector(2, {(1) = .1387315960, (2) = 1.709460320})

  

Vector(2, {(1) = .1387315523, (2) = 1.709460320})

  

Vector[column](%id = 18446744074402171950)

 (4)
> 

fsolve([F[1],F[2]], {x[1]=0..1, x[2]=1..3}); # compare with fsolve

{x[1] = .1387315528, x[2] = 1.709460319}

 (5)
> 

 


Download Newton2.mw

fsolve...

vv 13837
December 26 2018
1 0

For functions of a single variable such as fsolve(f(x), x=a..b)  the initial iteration is usually indeed (a+b)/2.
It is not easy to grasp the fsolve code for such a complex command. But you can do some tests to see the evaluation points of the function.
For example:

f:=proc(x)  print(x) ; cos(x)-x end:
fsolve(f, 0.5 .. 10);
                          5.250000000
                          5.250005250
                          5.249947500
                          2.875000000
                          2.875002875
                          2.874971250
                          7.625000000
                          7.625007625
                          7.624923750
                        3.87708345542864
                          1.687500000
                          1.687501688
                          1.687483125
                       0.782569730439221
                       0.742447770776506
                       0.739097995069612
                       0.739085133328356
                  0.7390851333290950851333284
                  0.7390851332151606416581575
                          0.7390851332

 

 

declare...

vv 13837
December 25 2018
4 2

After declare(...), the derivatives of any function (declared or not) are printed indexed.

A non-derivated function f(x,y) is printed as f  iff  f(x,y) appears in declare().

 

> 

PDEtools[declare](y(t));

` y`(t)*`will now be displayed as`*y

 (1)
> 

diff(y(t),t) + y(t) + y(x);  
diff(p(s),s) + p(s) + p(x)

diff(y(t), t)+y(t)+y(x)

  

diff(p(s), s)+p(s)+p(x)

 (2)
> 

 

randomize()...

vv 13837
December 25 2018
4 0

The restart command resets the "seed" of the random number generator to a default value.
You can remove the restart  or  insert a  randomize();  after it.

width...

vv 13837
December 24 2018
0 2

By default the arrow width depends on length. So, use e.g.:

PlotVector( [...], width=0.1);

 

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