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140 Compact numerical methods for computers
we can substitute
j - 1 = 2 (q -1) + w (11.31a)
i- 1 = 2 (p-1) + w (11.31b)
where p and q will be the new indices for the matrices A and B running from 1 to
n'= n/2 (assuming n even). Thus the matrix elements are
2
à p q =-(j – 1)(j – 2)I + 2 a(2j – 1)I + (k – 4a )I s+4 + k I (11.32)
2
s
s+2
4 s + 6
and
(11.33)
where
s = i + j – 4 =2(p + q – 3+ w)
and j is given by (11.31a). The tilde is used to indicate the re-numeration of A
and B.
The integrals (11.30) are easily computed recursively.
STEP DESCRIPTION
0 Enter s, a. Note s is even.
1 Let v = 1.
2 If s<0, stop. I is in v. For s<0 this is always multiplied by 0.
s
3 For k = 1 to s/2.
Let v = v * (2 * k- 1) * 0·25/ a.
End loop on k.
4 End integral. I is returned in v.
s
As an example, consider the exactly solvable problem using n' = 2, and a= 0·5
for w = 0 (even parity). Then the eigenproblem has
with solutions
e =1 c = (1, 0) T
and
-½ T
e = 5 c = 2 (-1, 2) .
The same oscillator (a = 0·5) with w = 1 and n' = 10 should also have exact
solutions. However, the matrix elements range from 0·5 to 3·2E+17 and the
solutions are almost all poor approximations when found by algorithm 15.
Likewise, while the problem defined by n' = 5, w = 0, a = 2, k = 0, k = 1 is
4
2
solved quite easily to give the smallest eigenvalue e= 1·06051 with eigenvector
1
T
c = (0·747087, 1·07358, 0·866449, 0·086206, 0·195257)
the similar problem with n' = 10 proves to have a B matrix which is computation-
ally singular (step 4 of algorithm 15). Inverse iteration saves the day giving, for
n' = 5, e = 1·0651 and for n' = 10, e = 1·06027 with eigenvectors having small
residuals. These results were found using an inverse iteration program based on