vv

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These are replies submitted by vv

@C_R Actually, in both situations, solve only finds a few surfaces contained in the interior of the torus (the main components of the solution are missing).

@Preben Alsholm The result of

sol2:=SolveTools:-SemiAlgebraic({torus1<0}, [x,y,z]);

seems to be correct but more complicated than necessary.

After completing my last post I was said that there are no tags; but they were introduced! I accepted a proposed tag (a stupid one) and the post was accepted.

@janhardo As seen, Maple knows the symbolic value of Zeta(2).
It is possible to deduce it e.g. from the infinite product expansion of sin(x) (see below), but it would be much more useful for you to read a few known mathematical proofs in the given references.

 

f:=Product(1-x^2/(k^2*Pi^2), k=1..infinity); # sin(x)/x

Product(1-x^2/(k^2*Pi^2), k = 1 .. infinity)

(1)

eval(diff(f,x,x)/2, x=0);

(1/2)*(Sum(-2*(Product(1, i1 = 1 .. k-1))*(Product(1, i1 = k+1 .. infinity))/(k^2*Pi^2), k = 1 .. infinity))

(2)

C2:=simplify(%); # coeff of x^2 in f

-(Sum(1/k^2, k = 1 .. infinity))/Pi^2

(3)

S2:=coeff(series(sin(x)/x, x), x^2);

-1/6

(4)

isolate(C2=S2, indets(C2, function)[]);

Sum(1/k^2, k = 1 .. infinity) = (1/6)*Pi^2

(5)

 

Download zeta2.mw

@janhardo Euler product formula can be written in Maple this way:

but note that Maple cannot use ithprime symbolically.

@Jesús Guillera You have to be careful because e.g.  1 + 4 + 4^2 + ...  = -1/3.
Also, F(z) converges only for |z| < 1 and for some points with |z|=1.

@NIMA112  y[2] still present! Check with indets

lprint(indets(zz, name));

# {Delta, Q__01, Q__02, T__1, T__2, a, alpha, b, beta, lambda, v, x, x__01, x__02, y, y__01, y__02, y[2]}

 

@NIMA112 Of course it does not work because you also have y[2].

@sand15 Yes, but try to compare the timings for e.g.

d := rand(10^20 .. 10^30)():
n := d^rand(100 .. 1000)()*rand(10^10..10^20)():


 

@nm Try  n:=90; d:=3;

@dharr Just a remark. The eigenvalues are not correctly sorted; the option key=evalf should be added.

@mehdi jafari 
 

with(LinearAlgebra):

A := <<-1,-3,-6>|<3,5,6>|<-3,-3,-4>>;

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

(1)

(V,Q):=Eigenvectors(A);

V, Q := Vector(3, {(1) = -4, (2) = 2, (3) = 2}), Matrix(3, 3, {(1, 1) = 1/2, (1, 2) = -1, (1, 3) = 1, (2, 1) = 1/2, (2, 2) = 0, (2, 3) = 1, (3, 1) = 1, (3, 2) = 1, (3, 3) = 0})

(2)

Q^(-1).A.Q = DiagonalMatrix(V);

(Matrix(3, 3, {(1, 1) = -4, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 2})) = (Matrix(3, 3, {(1, 1) = -4, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 2}))

(3)

(Vt,Qt):=Eigenvectors(A^+); # left eigenvectors; the eigenvalues are always the same

Vt, Qt := Vector(3, {(1) = 2, (2) = 2, (3) = -4}), Matrix(3, 3, {(1, 1) = -2, (1, 2) = -1, (1, 3) = 1, (2, 1) = 0, (2, 2) = 1, (2, 3) = -1, (3, 1) = 1, (3, 2) = 0, (3, 3) = 1})

(4)

A . Q[..,1] = V[1] * Q[..,1];         # A.v = lambda*v

(Vector(3, {(1) = -2, (2) = -2, (3) = -4})) = (Vector(3, {(1) = -2, (2) = -2, (3) = -4}))

(5)

Qt[..,1]^+ . A = Qt[..,1]^+ * Vt[1];  # w.A = lambda*w  

(Vector[row](3, {(1) = -4, (2) = 0, (3) = 2})) = (Vector[row](3, {(1) = -4, (2) = 0, (3) = 2}))

(6)

 

 

 

@acer However,

restart;
local a; 	
a := 17:
:-a := 18:
p := proc()
  local a;
  a := 25;
  return parse("a",statement);
end proc:

p();  #17

 

@mmcdara Sorry, probably I have seen an older version of your answer, or maybe I was not careful enough.

@mmcdara You have ommited useassumptions. So, solve has another task: a single inequality instead of a system.

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