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Thread: Local extrema in multiple variables

  1. #1
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    Local extrema in multiple variables

    Suppose that the function $\displaystyle f: \mathbb {R} ^n \rightarrow \mathbb {R} $ has first order partial derivatives and that the point $\displaystyle x \in \mathbb {R} ^n $ is a local minimizer for f.
    Prove that $\displaystyle \nabla f(x) = 0 $.

    Proof so far.

    Now, $\displaystyle \nabla f(x) = ( \frac { \partial f}{ \partial x_1 } , \frac { \partial f}{ \partial x_2 } , ... , \frac { \partial f}{ \partial x_n } $
    $\displaystyle = ( \lim _{t \rightarrow 0 } \frac {f(x+te_1)-f(x)}{t}, \lim _{t \rightarrow 0 } \frac {f(x+te_2)-f(x)}{t}, ...,\lim _{t \rightarrow 0 } \frac {f(x+te_n)-f(x)}{t} ) $

    Now, x is a local min implies that if $\displaystyle dist(x,x+h)<r$, we will have $\displaystyle f(x+h) \geq f(x) $

    So how should I incorporate them together? Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Suppose that the function $\displaystyle f: \mathbb {R} ^n \rightarrow \mathbb {R} $ has first order partial derivatives and that the point $\displaystyle x \in \mathbb {R} ^n $ is a local minimizer for f.
    Prove that $\displaystyle \nabla f(x) = 0 $.
    As $\displaystyle \bf{x}$ is a local minimiser of $\displaystyle f$, then $\displaystyle \bf{x}$ is a local minimum of the 1-D minimisation of $\displaystyle f({\bf{x}}+\lambda {\bf{e}}_i)$ where $\displaystyle {\bf{e}}_i$ is the unit vector in the direction of the $\displaystyle i$-th axis. Hence:

    $\displaystyle \left.\frac{d}{d\lambda}f({\bf{x}}+\lambda {\bf{e}}_i)\right|_{\lambda=0}=[\nabla f(x)]_i=0$

    so:

    $\displaystyle
    \nabla f(x)=0
    $

    CB
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