Advanced Search

vv

14187 Reputation

20 Badges

10 years, 281 days
Contact vv

MaplePrimes Activity


These are answers submitted by vv

rank...

vv 14187
December 18 2016
0 10

In your example, you can do this only if the rank of the system is r=19 and the rank corresponding to the unknowns Z1,...,Z19 is also r=19.
In this case you can solve wrt Z1,...,Z19, e.g.
solve( sys, {seq(Z||i,i=1..r)} );

`.`...

vv 14187
December 17 2016
0 5

At top level `.` is used as a noncommutative (but associative) operator not related to any algebraic structure.
It was designated mainly for rtables.
You can see this executing

> 

showstat(`.`);

 

Note that generally, (a.b)^(-1)  is not equal necessarily to  b^(-1) . a^(-1), because e.g. a^(-1) may not exist.
For example:

 

> 

a:=<1,2,3;4,5,6>;
b:=<1,2;3,4;5,6>;

a := Matrix(2, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (2, 1) = 4, (2, 2) = 5, (2, 3) = 6})

  

b := Matrix(3, 2, {(1, 1) = 1, (1, 2) = 2, (2, 1) = 3, (2, 2) = 4, (3, 1) = 5, (3, 2) = 6})

 (1)
> 

(a.b)^(-1);

Matrix([[16/9, -7/9], [-49/36, 11/18]])

 (2)
> 

a^(-1);

Error, (in rtable/Power) power -1 not defined for non-square Matrices; try LinearAlgebra[MatrixInverse] to obtain a pseudo-inverse

  
> 

 

Of course, as you have noticed, some packages (e.g. GroupTheory) may implement special simplification rules concerning `.` in some contexts.

AFAIK in Physics package, a noncommutative multiplication is implemented
(actually, here `.` and `*` are redefined), but I don't use this package and I cannot say more.

With a little help...

vv 14187
December 16 2016
3 6

The problem reduces to compute

 

> 

J:=Int( ln(1-y)*y^a*(1-y)^b, y=0..1);

Int(ln(1-y)*y^a*(1-y)^b, y = 0 .. 1)

 (1)

(a>-1, b>-1)

 

Maple is not able to find it without a little help.

 

> 

Jn:=-(1/n)*Int( y^(a+n)*(1-y)^b,y=0..1);

-(Int(y^(a+n)*(1-y)^b, y = 0 .. 1))/n

 (2)
> 

jn:=value(Jn);

-GAMMA(b+1)*GAMMA(a+n+1)/(n*GAMMA(a+b+n+2))

 (3)
> 

simplify( sum(jn, n=1..infinity) ) assuming a>-1,b>-1;

-(Psi(a+b+2)*b^2-Psi(3+b)*b^2+3*b*Psi(a+b+2)-3*Psi(3+b)*b+2*Psi(a+b+2)-2*Psi(3+b)+2*b+3)*GAMMA(a+1)*GAMMA(b+1)/((b^2+3*b+2)*GAMMA(a+b+2))

 (4)


 

area...

vv 14187
December 16 2016
1 4


 

> 

M := Matrix(14, 2, [[0, 0], [0, 4170], [1, 3966], [1, 3466],
[3, 3058], [3, 3058], [4, 1854], [4, 1354], [7, -2258],
[7, -2758], [8, -3962], [8, -3962], [10, -4370], [10, 0]]):
> 

pcw:=proc(M::Matrix,x)
local m:=upperbound(M,1),i,p:=NULL;
for i to m-1 do
  if M[i,1]<>M[i+1,1] then
    p:=p, x<M[i+1,1], M[i,2]+(x-M[i,1])/(M[i+1,1]-M[i,1])*(M[i+1,2]-M[i,2]);
  fi
od;
piecewise(p,undefined)
end:
> 

f:=pcw(M,x):
#plot(f,x=0..10);
a:=solve(f,x); evalf(a);
'area'=int(f,x=0..a);  evalf(%);

3085/602

  

5.124584718

  

area = 8313225/602

  

area = 13809.34385

 (1)
> 

restart;

Edit:
I have interpreted your matrix as representing a piecewise linear (discontinuous) function (as your plot shows).

On purpose?...

vv 14187
December 15 2016
0 0

I think that this is intentional. After all, we can see 10 as a binary number.

b:=convert(0.01,binary);
    .1010001111 10^-6
op(b);
    1010001111, -16
``(op(1,b))*2^(``(op(2,b)));

 

 

 

 

 

 

plot...

vv 14187
December 14 2016
1 0

Use one of the following:

plot3d(p1, -1 .. 1, -1 .. 1);

expr:=piecewise(x^2+y^2 <= 1, x*y-y^2, 0);
plot3d(expr, x = -1 .. 1, y = -1 .. 1);

plot3d('p1(x,y)', x = -1 .. 1, y = -1 .. 1);

 

 

eval...

vv 14187
December 14 2016
1 0

The command to plug n=1 into the expression S is:
eval(S, n=1);

The command solve is to find the root(s) of the equation S=0, and the syntax solve(S, n=1) is wrong.
You should read (or at least browse) the Maple User Manual.

No need for assumptions, but instead: -...

vv 14187
December 14 2016
0 0

No need for assumptions, but instead:

- Don't use square brackets [ ... ],  they are list delimiters.
- Don't use inert Int , or use value(%) after that.
- (optional) use Maple (1D) input.

int( (x/L)^3 - 1, x=0..L );

           
                              

 

workaround...

vv 14187
December 14 2016
2 3

Yes, arrays of plots refuse to be exported (by context menu or by plottools:-exportplot).
Workaround:

restart;
p:=seq(plot(x^k,x=-1..1),k=1..4):
plotsetup(png,plotoutput="d:/temp/ex-array.png");
plots:-display(<p[1],p[2];p[3],p[4]>);
plotsetup(default);

Omit tilde...

vv 14187
December 12 2016
2 1


subs(_Z9 = n, rhs(symbolic));
(of course the numeric suffix may differ).

 

int...

vv 14187
December 12 2016
2 2


 

Just describe the body using inequalities:

 

c(x) <= y <= d(x),   a <= x <= b

and use the simple and well-known formulae.

This is in two dimensions. Similarly for 3 or more.

(It may be needed to decompose the body into several such regions.)

 

Example for a upper half disk:

> 

A:=  [y = 0 .. sqrt(1-x^2), x = -1..1] ;

[y = 0 .. (-x^2+1)^(1/2), x = -1 .. 1]

 (1)
> 

d:=1; # density
M := int(d,A);
xM:=int(x*d,A)/M;
yM:=int(y*d,A)/M;

1

  

(1/2)*Pi

  

0

  

(4/3)/Pi

 (2)

r(theta,z)...

vv 14187
December 12 2016
1 2

When coords=cylindrical,  r  is ploted as a function of theta and z.
Even in your worksheet the notation is misleading; you use plot3d( f(r), theta=..., r=...)
but actually r should be z; it does not matter, of course, but your intentions are then not clear.

evalf...

vv 14187
December 12 2016
1 0

evalb(evalf[3](Pi=3.14));
                              true
evalb(evalf[4](Pi=3.14));
                             false

 

x0,y0,z0:=1,2,3: eq:= u*~[cos(t),sin(t)...

vv 14187
December 10 2016
2 0 
x0,y0,z0:=1,2,3:
eq:= u*~[cos(t),sin(t),0] + (1-u)*~[x0,y0,z0]:
plot3d( eq, u=0..1,t=0..2*Pi, scaling=constrained );

syntax...

vv 14187
December 10 2016
1 7

For the 2nd order derivative you must use

fdiff(dydata, [1,1], 0);

[Note that to compute this it would be much better to use the RHS of the ODE].

 

First 98 99 100 101 102 103 104 Last Page 100 of 122