Question: Generalization of a matrix known for some values

Preliminaries: 

and 

I wrote a code for above. You can Download the code for Above:  

 

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Now, we can calculate the Matrix  as follows  for k=2, M=3,

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Question:  How can we find the Matrix  for ?

(If it is not possible to find the Matrix for any k and M,

can we find it for k=3, M=4? )

Before finding Matrix,   we must find to 

in terms of any of Psi_i,j    like

and others.

 

But how? :)
Update:

I found elements of Matrix 
NOW, How to create  Matrix A:=  by the elements in the_code.mw

J:=proc(j1,j2,j3,m1,m2,m3)
local i,f:
i:=max(0,j2-j3-m1,j1-j3+m2):
f:=min(j1+j2-j3,j1-m1,j2+m2):
if m1+m2+m3<>0 then:
0
elif j3>j1+j2 then:
# printf("Error. Does not satisfy the triangle relation");
0
elif j3<abs(j1-j2) then:
# printf("Error. Does not satisfy the triangle relation");
0
elif abs(m1)>j1 or abs(m2)>j2 or abs(m3)>j3 then:
0
else:
simplify(((-1)^(j1-j2-m3))*((((j1+j2-j3)!*(j1-j2+j3)!*(-j1+j2+j3)!*(j1+m1)!*(j1-m1)!*(j2+m2)!*(j2-m2)!*(j3+m3)!*(j3-m3)!)/(j1+j2+j3+1)!))^(1/2)*sum(((-1)^t)/((j1+j2-j3-t)!*(j1-m1-t)!*(j2+m2-t)!*(j3-j2+m1+t)!*(j3-j1-m2+t)!*t!),t=i..f));
end if;
end proc:

###### That is matrix which we want to find ###############
#psi_1aa.psi_1bb:=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(1,c),c=abs(aa-bb)..(aa+bb));
#psi_2aa.psi_2bb:=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(2,c),c=abs(aa-bb)..(aa+bb));
#Let A:= Psi.Transpose(Psi). So, it is matrix which we want to find.


aa:=0:bb:=0: 
A(1,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(1,c),c=abs(aa-bb)..(aa+bb)); 
# it is element of A in first row and column or psi_10.psi_10
                               
aa:=1:bb:=0: 
A(2,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(c),c=abs(aa-bb)..(aa+bb)); 
# it is element of A in second row and column or psi_11.psi_10 
#We can proceed so so for aa=0..M-1 and bb=0..M-1

#Similarly; Let's find psi_20.psi_20
A(M+1,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(2,c),c=abs(aa-bb)..(aa+bb));
 # it is element of A in M+1 row and column or psi_20.psi_10

 

Best regards.

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