hi Barton
In this case,I think
Calculating residue is not necessary,if you use Jordan form.
My diag package is almost fine in this case.
Because it uses the residue not in CAS but in our brain.
matrix A has Jordan form,such that A= X diag(J1,J2,..,Jk)X^-1
Ji is jordan cell mi x mi,m1+m2+...mk=n
By Cauchy's theorem ,when f is analytic function and $B&C(B is a closed path
within the eigenvalue,
f(Ji)=1/(2$B&P(Bi) $B"i(B[$B&C(B] f(z)(zI-Ji)^^-1 dz,
for example Ji=matrix([$B&K(B,1,0],[0,$B&K(B,1],[0,0,$B&K(B])
(zI-Ji)^^-1=matrix([1/(z-$B&K(B),1/(z-$B&K(B)^2,1/(z-$B&K(B)^3],[0,1/(z-$B&K(B),1/(z-
$B&K(B)^2],[0,0,1/(z-$B&K(B)])
So f(ji)=matrix([f($B&K(B),f'($B&K(B),f''($B&K(B)/2!],[0,f($B&K(B),f'($B&K(B)],[0,0,f($B&K(B)])
f(A)=X diag(f(J1),f(J2),..,f(Jk))X^-1
(C1) m : matrix([2,1,0],[1,2,1],[0,1,10])$
(C2) load("diag.mac");
Warning - you are redefining the MACSYMA function EIGENVALUES
Warning - you are redefining the MACSYMA function EIGENVECTORS
(D2) diag.mac
(C3) mat_function(exp,m);
OK
another example
(C5) a6:matrix([1,-1,0,-1],[0,2,0,1],[-2,1,-1,1],[2,-1,2,0])$
(C6) fpow(x):=block([k],declare(k,integer),x^k)$ //k>=1
(C7) mat_function(fpow,a6);
(D7) MATRIX([-(-1)^k/2-k+3/2,-k,-(-1)^k/2-k+1/2,-k],
[(-1)^k/2+k-1/2,k+1,(-1)^k/2+k-1/2,k],[3*(-1)^k/2+k-3/2,k,3*(-1)^k/2+k-1/2,k],
[-3*(-1)^k/2-k+3/2,-k,-3*(-1)^k/2-k+3/2,1-k])
but in maxima we can't substitute some expression such that
(C8) jordan(m);
(D8)
[[(-SQRT(3)*%I/2-1/2)*(14*SQRT(22)*%I/(3*SQRT(3))+.)..,1],[..,1],[..,1]]
we will change this long expression to [[$B&K(B1,1],[$B&K(B2,1],[$B&K(B3,1]],and
mat_function will return the answer with $B&K(B1,$B&K(B2,$B&K(B3 .but I cannot.
I think this trouble with the substitution is serious in the useability.
GF
(B.
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