Manuscript XVI-J@1 · XVI-J.tex · section · 6c1f306d

Closed formula for the Hessian

Independent polynomial proofs of the numerical tables › Closed formula for the Hessian · lines 872–906

XVI-J@1#eq:Fexpanded, XVI-J@1#eq:Ltable

Expanding (eq:F) gives the full quadratic polynomial

F = - \frac{1}{2} (z_{0}^{2} + z_{7}^{2} + z_{9}^{2}) - \frac{9}{2} z_{4}^{2} - z_{0} z_{4} - z_{0} z_{7} - z_{0} z_{9} + z_{4} z_{7} + z_{4} z_{9} + z_{7} z_{9} + 6 z_{1}^{2} + 2 z_{2}^{2} + 2 z_{3}^{2} - 6 z_{5}^{2} - 6 z_{6}^{2} - 2 z_{8}^{2} .

Consequently the complete Hessian is

L = - 1 000 - 1 00 - 1 0 - 1 0120000000000400000000004000000 - 1 000 - 9 00101 00000 - 12 0000 000000 - 12 000 - 1 000100 - 1 01 00000000 - 4 0 - 1 00010010 - 1 .

This proves every entry, including every zero, without a script or a numerical approximation.

Displayed formulas in this passage

XVI-J@1#eq:Fexpanded · line 875
$F = - \frac{1}{2} (z_{0}^{2} + z_{7}^{2} + z_{9}^{2}) - \frac{9}{2} z_{4}^{2} - z_{0} z_{4} - z_{0} z_{7} - z_{0} z_{9} + z_{4} z_{7} + z_{4} z_{9} + z_{7} z_{9} + 6 z_{1}^{2} + 2 z_{2}^{2} + 2 z_{3}^{2} - 6 z_{5}^{2} - 6 z_{6}^{2} - 2 z_{8}^{2} .$
XVI-J@1#eq:Ltable · line 887
$L = - 1 000 - 1 00 - 1 0 - 1 0120000000000400000000004000000 - 1 000 - 9 00101 00000 - 12 0000 000000 - 12 000 - 1 000100 - 1 01 00000000 - 4 0 - 1 00010010 - 1 .$