Muhammad Usman

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10 years, 320 days
Beijing, China

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Dear Users!

I want to simple the Gamma function occures in a matrix. I need the simpliest form of this matrix. If there is some thing common in all entries take it common. Thanks

 

Matrix(6, 6, {(1, 1) = 2*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(alpha*(2*alpha+3)*(1+2*alpha)*GAMMA(alpha)*GAMMA(alpha+1/2)^2), (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (2, 1) = (1/2)*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(alpha*(2*alpha+3)*(1+2*alpha)*GAMMA(alpha)*GAMMA(alpha+1/2)^2), (2, 2) = (1/4)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(GAMMA(alpha+1/2)^3*alpha^3*(alpha+1)^3*GAMMA(alpha)^3)*Pi^(1/4)/((2*alpha+3)*(1+2*alpha)*GAMMA(alpha+1/2)^2*alpha^2*(alpha+1)^2*GAMMA(alpha)^2), (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (3, 1) = (1/16)*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(GAMMA(alpha+1/2)^2*alpha*(alpha+1)*(1+2*alpha)*GAMMA(alpha)), (3, 2) = (1/8)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(GAMMA(alpha+1/2)^3*alpha^3*(alpha+1)^3*GAMMA(alpha)^3)*Pi^(1/4)/((2*alpha+3)*(1+2*alpha)*GAMMA(alpha+1/2)^2*alpha^2*(alpha+1)^2*GAMMA(alpha)^2), (3, 3) = (1/32)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(alpha^3*GAMMA(alpha+3/2)^3*(2+alpha)^3*GAMMA(alpha)^3)*Pi^(1/4)/((alpha+1)*(2*alpha+3)*alpha^2*GAMMA(alpha+3/2)^2*(2+alpha)^2*GAMMA(alpha)^2), (3, 4) = 0, (3, 5) = 0, (3, 6) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 2*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(alpha*(2*alpha+3)*(1+2*alpha)*GAMMA(alpha)*GAMMA(alpha+1/2)^2), (4, 5) = 0, (4, 6) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = (3/2)*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(alpha*(2*alpha+3)*(1+2*alpha)*GAMMA(alpha)*GAMMA(alpha+1/2)^2), (5, 5) = (1/4)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(GAMMA(alpha+1/2)^3*alpha^3*(alpha+1)^3*GAMMA(alpha)^3)*Pi^(1/4)/((2*alpha+3)*(1+2*alpha)*GAMMA(alpha+1/2)^2*alpha^2*(alpha+1)^2*GAMMA(alpha)^2), (5, 6) = 0, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = (1/16)*(18*alpha+19)*GAMMA(alpha+5/2)*Pi^(1/4)*sqrt(GAMMA(alpha+1)*GAMMA(alpha+1/2)^3)/(alpha*(alpha+1)*(2*alpha+3)*(1+2*alpha)*GAMMA(alpha)*GAMMA(alpha+1/2)^2), (6, 5) = (3/8)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(GAMMA(alpha+1/2)^3*alpha^3*(alpha+1)^3*GAMMA(alpha)^3)*Pi^(1/4)/((2*alpha+3)*(1+2*alpha)*GAMMA(alpha+1/2)^2*alpha^2*(alpha+1)^2*GAMMA(alpha)^2), (6, 6) = (1/32)*GAMMA(alpha+5/2)*sqrt(2)*sqrt(alpha^3*GAMMA(alpha+3/2)^3*(2+alpha)^3*GAMMA(alpha)^3)*Pi^(1/4)/((alpha+1)*(2*alpha+3)*alpha^2*GAMMA(alpha+3/2)^2*(2+alpha)^2*GAMMA(alpha)^2)})

Dear Users!

Hope you would be fine. In the following maple code, I want to write the derivative of psi in term of psi like it did manually in red portion. For higher M and k it very hard to do it manully. It there any command to fix my problem for any value of k and M.

restart; k := 2; M := 4;

with(linalg); with(LinearAlgebra);

printlevel := 2;

for i while i <= 2^(k-1) do

for j from 0 while j <= M-1 do

psi[M*i+j-M+1] := simplify(2^((1/2)*k)*sqrt(GAMMA(j+1)*(j+alpha)*GAMMA(alpha)^2/(Pi*2^(1-2*alpha)*GAMMA(j+2*alpha)))*(sum((-1)^i1*GAMMA(j-i1+alpha)*(2*(2^k*x-2*i1+1))^(j-2*i1)/(GAMMA(alpha)*factorial(i1)*factorial(j-2*i1)), i1 = 0 .. floor((1/2)*j))));

`&psi;&psi;`[M*i+j-M+1] := simplify(diff(psi[M*i+j-M+1], x))

end do

end do; r := 2^(k-1)*M;

VV := Vector[column](r, proc (i) options operator, arrow; psi[i] end proc);

DV := Vector[column](r, proc (i) options operator, arrow; `&psi;&psi;`[i] end proc);

``&psi;&psi;`[2] := 8*sqrt((alpha+1)*(1/2))*sqrt(2)*sqrt(alpha*GAMMA(alpha)^2*4^alpha/GAMMA(2*alpha))/sqrt(Pi) = 8*sqrt((alpha+1)*(1/2))*psi[1];

`&psi;&psi;`[3] := 16*sqrt((2+alpha)*(alpha+1)/(1+2*alpha))/sqrt(2)*(2*sqrt(2)*sqrt((alpha+1)*GAMMA(alpha)^2*4^alpha/GAMMA(1+2*alpha))*alpha*(4*x+1)/sqrt(Pi)) = 16*sqrt((2+alpha)*(alpha+1)/(1+2*alpha))*psi[2]/sqrt(2)

I am waiting your response. Thanks

Dear Users! 

Hope you would be fine with everything. I want to define a matrix F of M+1 by M+1 order having element of the following form:

I derived the F[r,s] but confuse who to generate matrix now.

restart; M := 5; printlevel := 3; for r from 2 while r <= M+1 do for s while s <= r-1 do if type(r+s, odd) then F[r, s] := 2^(k+1)*sqrt((2*r-1)*(2*s-1)) end if end do end do

Dear Users!

Hoped everyone fine with everything. I the following maple expression, I need a matrix A for each n. Like if I take k =1 I want A[1]; if I take k=2, I want A[1], A[2]; for k=3 I want A[1], A[2], A[3] and so on. A[i]'s is square matrix having order M-1 by M-1.

Further I want to generate a block matrix for k. Like for k=1 I want a block matrix as Vector(1, {(1) = A[1]}), for k=2 I want a block matrix as Matrix(2, 2, {(1, 1) = A[1], (1, 2) = 0, (2, 1) = 0, (2, 2) = A[2]}), for k =3 I want a block matrix as Matrix(3, 3, {(1, 1) = A[1], (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = A[2], (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = A[3]}) and so on.

restart; alpha := 1;
k := 2; M := 3;
printlevel := 3;

for n while n <= 2^(k-1) do

for m from 0 while m <= M-1 do

for j from 0 while j <= M-1 do

Omega[m, j] := 2^((1/2)*k)*sqrt(GAMMA(j+1)*(j+alpha)*GAMMA(alpha)^2/(Pi*2^(1-2*alpha)*GAMMA(j+2*alpha)))*(sum((-1)^i*GAMMA(j-i+alpha)*2^(j-2*i)*(sum((1/2)*binomial(m, l)*(2*n-1)^(m-l)*(1+(-1)^(j-2*i+l))*GAMMA((1/2)*j-i+(1/2)*l+1/2)*GAMMA(alpha+1/2)/GAMMA(alpha+1+(1/2)*j-i+(1/2)*l), l = 0 .. m))/(GAMMA(alpha)*factorial(i)*factorial(j-2*i)), i = 0 .. floor((1/2)*j)))/2^(k*(m+1))

end do

end do;

A[n]:=???

end do;

I am waiting for your positive response.

Thanks
 

Dear Users!

Hoped everyone is good. I am facing problem to write the following sigma notation for any m.

Please help me to fix this problem. Thanks

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