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Math Help - Local extrema in multiple variables

  1. #1
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    Local extrema in multiple variables

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

    Proof so far.

    Now, \nabla f(x) = ( \frac { \partial f}{ \partial x_1 } , \frac { \partial f}{ \partial x_2 } , ... , \frac { \partial f}{ \partial x_n }
    = ( \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 dist(x,x+h)<r, we will have f(x+h) \geq f(x)

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

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

    so:

     <br />
\nabla f(x)=0<br />

    CB
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