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k variables, k functions

One may also use Newton's method to solve systems of k equations, which amounts to finding the (simultaneous) zeroes of k continuously differentiable functions f: R k → R. {\displaystyle f:\mathbb {R} ^{k}\to \mathbb {R}.} This is equivalent to finding the zeroes of a single vector-valued function F: R k → R k. {\displaystyle F:\mathbb {R} ^{k}\to \mathbb {R} ^{k}.} In the formulation given above, the scalars x are replaced by vectors x and instead of dividing the function f (x) by its derivative f ′ (x) one instead has to left multiply the function F (x) by the inverse of its k × k Jacobian matrix J (x). This results in the expression

x n + 1 = x n − J F (x n) − 1 F (x n). {\displaystyle \mathbf {x} _{n+1}=\mathbf {x} _{n}-J_{F}(\mathbf {x} _{n})^{-1}F(\mathbf {x} _{n}).}

or, by solving the system of linear equations

J F (x n) (x n + 1 − x n) = − F (x n) {\displaystyle J_{F}(\mathbf {x} _{n})(\mathbf {x} _{n+1}-\mathbf {x} _{n})=-F(\mathbf {x} _{n})}

for the unknown x − x.