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Real symmetric matrices                    131
                       Example 10.2. Application of the Jacobi algorithm in celestial mechanics
                       It is appropriate to illustrate the use of algorithm 14 by Jacobi’s (1846) own
                       example. This arises in the study of orbital perturbations of the planets to
                       compute corrections to some of the parameters of the solar system. Unfortunately
                       at the time Jacobi was writing his paper, Neptune had not been discovered.
                      Leverrier reported calculations suggesting the existence of this planet on 31
                      August 1846, to l’Académie des Sciences in Paris, and Galle in Berlin confirmed
                       this hypothesis by observation less than three weeks later on 18 September. The
                       derivation of the eigenproblem in this particular case is lengthy and irrelevant to
                       the present illustration, so we will begin with Jacobi’s equations V. These give a
                       non-symmetric matrix à which can be symmetrised by a diagonal transformation
                       resulting in Jacobi’s equations VIII, where the off-diagonal elements are expres-
                       sed in terms of their common logarithms to which 10 has been added. I decided to
                       work with the non-symmetric form and symmetrised it by means of
                                                                   ½
                                                 A ij  = A ji  = (Ã Ã ) .
                                                                 ji
                                                              ij
                       The output from a Hewlett-Packard 9830 (machine precision = 1E–11) is given
                       below, and includes the logarithmic elements which in every case approximated
                       very closely Jacobi’s equations VIII. For comparison, he computed eigenvalues
                       –2·2584562,    –3·7151584,   –5·2986987,    –7·5740431,    –17·1524687,
                       –17·8632192 and –22·4267712 after 10 rotations. At this point the largest
                       off-diagonal element (which is marked as being negative) had a logarithm
                       (8·8528628–10), which has the approximate antilog –7·124E–2. Jacobi used as a
                       computing system his student Ludwig Seidel, apparently operating in eight-digit
                       decimal arithmetic!

                       ENHJCB JACOBI WITH ORDERING MAR 5 75
                       ORDER= 7
                       INPUT JACOBI'S MATRIX
                       ROW 1   :
                       -5.509882    1.870086   0.422908    8.81400E-03    0.148711
                        3.90800E-03   4.50000E-05
                       ROW 2   :
                        0.287865   -11.811654  5.7119   0.058717    0.728088    0.018788
                        2.24000E-04
                       ROW 3   :
                        0.049099    4.308033  -12.970687   0.229326    1.689087    0.04258
                        5.04OOOE-04
                       ROW 4   :
                        6.23500E-03   0.269851    1.397369   -17.596207   5.304038    0.125346
                        1.45100E-03
                       ROW 5   :
                        2.23100E-05    7.09480E-04   2.18227E-03    l.l2462E-03   -7.489041
                        4.815454    0.035319
                       ROW 6   :
                        1.45000E-06    4.52200E-05   1.35880E-04    6.56500E-05    11.893979
                       -18.58541    0.232241
                       ROW 7   :
                        6.00000E-08    1.94000E-06   5.79000E-06    2.73000E-06    0.313829
                        0.835482   -2.325935
                       SYMMETRIZE A(I,J)=A(J,I):=SQR(A(I,J)*A(J,I))
                       LOG(S) GIVEN FOR COMPARISON WITH JACOBI'S TABLE VIII
                       S=A( 1    ,2     )= 0.733711324    LOG10(S)+l0= 9.865525222
                       S=A( 1    ,3     )= 0.144098438    LOG10(S)+l0= 9.158659274
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