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One option for choosing g(x) is to select a hyperplane passing through X0 that is
orthogonal to the vector vi:
So, the Newton iteration becomes:
Xk+1 = Xk - H-1x(Xk)H(Xk) |
| (4) |
H(X) = |
æ ç
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ö ÷
÷ ø
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, Hx(X) = |
æ ç
ç è
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ö ÷
÷ ø
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. |
| (5) |
Then one can prove that the Newton iteration for (2) will converge to a point xi+1
on the curve from X0 provided that the stepsize h is sufficiently small and that the
curve is regular (rank Fx(x)=n).
Having found the new point xi+1 on the curve we need to compute the tangent vector at
that point:
Furthermore the direction along the curve must be preserved: ávi,vi+1 ñ = 1, so we get the (n+1)-dimensional appended system
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vi+1 = |
æ ç
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ö ÷
÷ ø
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. |
| (7) |
Upon solving this system, vi+1 must be normalized.