Adri van der Meer

Adri vanderMeer

1420 Reputation

19 Badges

21 years, 154 days
University of Twente (retired)
Enschede, Netherlands

MaplePrimes Activity


These are answers submitted by Adri van der Meer

...it is automatically filled with zeros on the empty places:

restart;
a1:=[1,2]:
a2:=[5,3,4,5]:
a3:=[4,8,4,5,3,2,3,4,5]:
a4:=[2,2,5,4,7,8]:
b := Matrix([seq(a||i,i=1..4)]);
B1 := b[..,1]; #first column
# etc. If you want to remove the zeros:
convert(b[..,5],list); B5 := remove(x->x=0, %);

 

restart;

Next line: choose 2D input, "×" symbol from palette; select the whole expression; Format→Convert to→Atomic Identifier (or Ctrl-Shift A)

`#mrow(mi(

`#mrow(mi(

(1)

In the next line: make an alias so you can type "axb" whenever you want "a×b" displayed

alias( %=axb ):

avector:=Vector(3,symbol=a):

bvector:=Vector(3,symbol=b):

axb := LinearAlgebra:-CrossProduct(avector,bvector);

 

`&x`(a, b) := Vector(3, {(1) = a[2]*b[3]-a[3]*b[2], (2) = a[3]*b[1]-a[1]*b[3], (3) = a[1]*b[2]-a[2]*b[1]})

(2)

 

 

Download AtomicIdentifier.mw

Alternative (if you really don't want to use the menu or the palette):

alias( `#mrow(mi("a"),mo("×"),mi("b"))`=axb);

...to make g such that g(sigma^2) yields the expression in the question?

f := 1/sqrt(1/3*q^2*sigma^2/Pi^2);
g := unapply( eval(f,sigma=sqrt(x)), x ): g(sigma^2);

This means that the limit doesn't exist (it is +∞). Bacuase you are using floats, maple makes a numerical approxximation. To make an exact calculation:

f := convert( LommelS1(4.04, 3.04, z)/z^5.04, rational );
limit( f, z=0 );

equ := 0 = -32*sigmas^2*ups*rp/(R*T)^2*(2/ln(S(x))^3)*(diff(ln(S(x)), x)) + 
   (1/2)*x*(sqrt(K(x)/S(x))*(diff(S(x), x))+sqrt(S(x)/K(x))*(D(K))(x))+sqrt(S(x)*K(x));

(thanks to Markiyan for the idea of the logplot)

restart;
A := [[0.5e-1, 2.7], [.1, 6.8], [.15, 12.7], [.2, 1100]]:
B := [[0.5e-1, 3.2], [.1, 7.4], [.15, 13.7], [.2, 1110.5]]:
p1 := plot([A, B], style=point, color=[red,blue], symbol = [diamond,cross], symbolsize = 30):
p2 := plot([A,B], color=[red,blue] ):
plots:-display({p1,p2}, axes=boxed, axis[1, 2] = [mode = log] );

n:=readstat("Input n:"):
A:=Matrix(n,n,0): # use Matrix, not matrix
for i from 1 to n do
for j from 1 to n do
A[i,j]:=readstat('A'[i,j] );
od;
od;
A;

What happened with your worksheet? I see semicolons in the first line, but no output.

With this worksheet: LaTeX.mw I get LaTeX.txt (I changed the *tex file to a *.txt file bacause I cannot upload tex-files). Note that I used Maple input in the worksheet.
With 2D input I get LaTeX2D.txt, which is also perfect.

Perhaps in older versions of Maple, the 2D input is not supported?

I observe the following:
(1)  p(max(m))=p(max(n)): you don't give a value for this, so I suppose it is random.
(2)  the list n has more elements than the list m; I make both lists random.

Than a table that satisfies all the requirements can be made as follows:

restart;
m := convert(LinearAlgebra:-RandomVector(6),list);
n := convert(LinearAlgebra:-RandomVector(10),list);
p[0]:=0:
p[max(m)] := rand();
p[max(n)] := p[max(m)];
M := [seq( sort(m)[i], i=1..nops(m)-1 )];
N := sort(n);
for i to nops(M) do p[M[i]]:= N[i] end do:
eval(p);

(1) This is the adjacency matrix of your network (graph). See ?AdjacencyMatrix in the GraphTheory package

(2) To get the vector of row totals you can take the matrix-vectorproduct of this adjacency matrix with a vector consisting of ones:

A := LinearAlgebra:-RandomMatrix(10, generator=0..1 );
v := A.Vector(10,fill=1);

To get the ranking vector:

u := Statistics:-Rank(v,order=descending);

The "and" construction isn't necessary in a piecewise function, because x<1, y1, x<x2, y2,x<x3,y3, y4 means:
if x<x1 then y1
elif x<x2 then y2
elif x<x3 then y3
else y4
end if

(you missed the else-clause)

Now we can construct the sequence tu use in a piecewise definition: (I generate random X and Y lists)

restart;
N := 5:
Y := [seq( randpoly(x,degree=2), i=1..N )]:
X := sort( [seq(RandomTools:-Generate(float(range=1..10.0), digits=2), i=1..N)] ):
H := ListTools:-Flatten( [seq( [x<=X[i],Y[i]], i=1..N )] ):
yp := piecewise( x<=0, 0, op(H), 0 );

You have (converted to Maple input):

f(x[]):=Re( add(Ei(z(n)*ln(x)), n=1..10)):

This has to be:

f := x -> Re( add(Ei(z[n]*ln(x)), n=1..10)):

Now you can calculate

 f(0.5);
                         0.01259288459
limit(f(x),x=1);
                        -Float(infinity)
plot(f, 0..0.1 );

What is Q? How can - h124 y  - h124 appear in the output ( it should automatically be simplified to
-2* h124*y). Please privide us with the output of lprint(Q);

Q := <3, 1 - h121 - h124*y  - h124*y - h125*x + h125*y>:
lprint(Q);
Vector[column](2, {1 = 3, 2 = -2*h124*y-h125*x+h125*y-h121+1}, datatype = anything,
storage = rectangular, order = Fortran_order, shape = [])
rtable_num_elems(Q);
                               2

Your cc is a list of vectors, so not a matrix. If you want a Matrix with given vectors as columns, you can do

g1 := Vector([0, y, x]):
g2 := Vector([0, y^2-x-y, 0]):
g3 := Vector([x, x+y, 0]):
g4 := Vector([y, -y, 0]):
g5 := Vector([0, x*y+x/2+y/2, 0]):
g6 := Vector([0, x^2-x/4-y/4, 0]):
cc := <g1|g2|g3|g4|g5|g6>;
cc[1,1..-1]; # First row

Maple cannot find a solotion to the IVP (but I get an other errormessage)

restart;
de := (1/4)*U*(diff(S(y)^4, `$`(y, 2))) =
   C*P1*((O*exp(G*y)-R*exp(J*y))/(P1*(exp(J*L)-exp(G*L)))+P2/P1):
sol := dsolve({de, S(0) = .1, S(v) = 0, (D(S))(0) = 0, (D(S))(L) = 0}, S(L));
sol :=
Y := rhs(sol);
Error, invalid input: rhs expects 1 argument, but received 0


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