User:EGM6341.S11.team5.cavalcanti/Mtg39

NM1 Mtg 39: Fri, 8 Apr 11 [[media: Nm1.s11.mtg39.djvu | Page 39-1]] (5) & (6)p.26-2: $$F(z_{i+1})=0$$ $$z^{(k+1)}_{i+1}=z^{(k)}_{i+1}- \left [ \frac{dF}{dz}\left ( z^{(k)}_{i+1} \right ) \right ]^{\color{red}{-1}}F\left ( z^{(k)}_{i+1} \right )$$ Note: matrix ODES: $$\underline{\dot{z}}=\underline{f}\left ( \underline{z}, t \right )$$
 * (p. 36-4)

(2) - (3)p.38-4: $$\underline{F}\left ( \underline{z}_{i+1} \right )=\underline{0}$$ (e.g., p.36-4, $$ \underline{F}_{\color{red}{4 \times 1}}$$) $$\underline{z}^{(k+1)}_{i+1}=\underline{z}^{(k)}_{i+1}-\left [ \frac{\delta \underline{F}}{\delta \underline{z}}\left ( \underline{z}^{(k)}_{i+1} \right ) \right ]^{\color{red}{-1}} \underline{F}\left ( \underline{z}^{(k)}_{i+1} \right )$$ Jacobian matrix: $$\frac{\delta \underline{F}}{\delta \underline{z}}=\left [ \frac{\delta F_{\color{blue}i \leftarrow row}}{\delta z_{\color{red}j \leftarrow col}} \right ]$$ Initial guess: $$z^{(0)}_{i+1}=z_{i}$$ [[media:Nm1.s11.mtg39.djvu | Page 39-2]] Convergence: $$\left | z^{(k+1)}_{i+1}-z^{(k)}_{i+1} \right | \leqslant \color{red}{AbsTol} \ \mathit{(Absolute \ Tolerance)}$$ or $$\frac{\left | z^{(k+1)}_{i+1}-z^{(k)}_{i+1} \right |}{z^{(k)}_{i+1}}\leq \color{red}{RelTol \mathit{(Relative Tolerance)}}$$ Note: Comments on HW5, Cauchy distribution.