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Esta correcto solo dice jacobina en vez de Jacobiana xDEditar

Método de NewtonEditar

$ f (x); $ $ x_0 $

$ x_i + 1 = x_i - {f(x_i)\over f'(x_i)} $

$ f(x) + f'(x_0)(x_1 - x_0) $

$ l(x) = f'(x_i)(x - x_i) $

$ l(x) = 0 $

$ f(x_i) + f'(x_(i+1))(x_1 - x_i) = 0 $

$ f'(x_i) =(x_(i+1) - x_i)= 0 $

RecordandoEditar

$ ln(ab) = ln(a) + ln(b) $

$ f(x) = a\exp^{bx} $

$ y = a\exp^{bx} $ / $ ln() $

$ ln(y) = ln(a) + ln(xp^{bx}) $

$ ln(y) = ln(a) + bx $


$ f(x) = a\exp^{bx + cx^{2} } $

$ y = a\exp^{bx + cx^{2} } $

$ ln(y) = ln(a\exp^{bx + cx^{2}}) $

$ ln(y) = ln(a) + ln(\exp^{bx + cx^{2}}) $

$ ln(y) = ln(a) bx + cx^{2} $


$ ln(y) = ln(a) (bx + cx^{2})x $

Metodo de NewtonEditar

$ f(x) = f(x_i) + f'(x_i)(x- x_i) $

$ f(x) = 0 $

$ x_i+1 = x_i - {f(x_i)\over f'(x_i)} $


Metodo de Gauss-NewtonEditar

$ l(\vec{x}) = \vec{f}'(\vec{x_i}) + j(\vec{x} - \vec{x_i}) $

$ l(\vec{x}) = 0 $

$ = \binom{n}{k} \quad $

$ \vec{f}'(\vec{x_i}) $ $ \mathbb{} = \; \begin{pmatrix} f_1 = ( x_1, x_2,...., x_n ) \\ f_2 = ( x_1, x_2,...., x_n ) \\ f_3 = ( x_1, x_2,...., x_n ) \\ .\\ .\\ .\\ f_n = ( x_1, x_2,...., x_n ) \\ \end{pmatrix} $

$ \vec{x_i} $ $ \mathbb{} = \; \begin{pmatrix} x_1i \\ x_2i \\ x_3i \\ .\\ .\\ x_ni \\ \end{pmatrix} $ , $ \vec{x} $ $ \mathbb{} = \; \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ .\\ .\\ x_n \\ \end{pmatrix} $


Matriz JacobinaEditar

 $ \mathbb{j} = \;         \begin{pmatrix}      \frac{\partial f_1(x_1, x_2)}{\partial x_1\,} & \frac{\partial f_1(x_1, x_2)}{\partial x_2\,}\\      \frac{\partial f_2(x_1, x_2)}{\partial x_1\,} & \frac{\partial f_2(x_1, x_2)}{\partial x_2\,}\\         \end{pmatrix} $

Ejemplo del metodo de Gauss-Newton, ejercicio 4Editar

$ \vec{f}: R^{2} \rightarrow \ R^{3} $

$ \vec{f} $ $ \mathbb{} = \; \begin{pmatrix} f(x,y) \\ g(x,y) \\ h(x,y) \\ \end{pmatrix} $

$ f(x_i,y_i) + \frac{\partial f(x_i, y_i)}{\partial x\,} (x_(i+1) - x_i) + \frac{\partial f(x_i, y_i)}{\partial y\,} (y_(i+1) - x_i) = 0 $

$ g(x_i,y_i) + \frac{\partial g(x_i, y_i)}{\partial x\,} (x_(i+1) - x_i) + \frac{\partial g(x_i, y_i)}{\partial y\,} (y_(i+1) - x_i) = 0 $

$ f(x_i) + f'(x)(x - x_i) = 0 $

 $ \mathbb{j} = \;         \begin{pmatrix}      \frac{\partial f(x_i, x_i)}{\partial x_\,} & \frac{\partial f(x_i, x_i)}{\partial y\,}\\      \frac{\partial g(x_i, x_i)}{\partial x\,} & \frac{\partial g(x_i, x_i)}{\partial y\,}\\     \frac{\partial h(x_i, x_i)}{\partial x\,} & \frac{\partial h(x_i, x_i)}{\partial y\,}\\         \end{pmatrix} $