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Prove that if $p^T▽f(x_k)<0$, then $f(x_k+εp)<f(x_k)$ for $ε>0$ sufficiently small.

I think we can expand $f(x_k+εp)$ in a Taylor series about the point $x_k$ and look at $f(x_k+εp)-f(x_k)$, but what's then?

(Taylor series: $f(x_0+p)=f(x_0)+p^T▽f(x_0)+(1/2)p^T▽^2f(x_0)p+...$
=> here is what's $p$)

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You may want a Taylor's series with remainder, of the form $f(x_0+p)=f(x_0)+p^T▽f(x_c)$ where $x_c$ is between (in some sense) $x_0$ and $x_0+p$. –  marty cohen Feb 18 '13 at 6:04
    
@marty cohen: but we can't evaluate the value of $p^T▽f(x_c)$ –  i_a_n Feb 18 '13 at 6:33
    
@marty cohen: Or we may write $f(x_k+\varepsilon p)=f(x_K)+(\varepsilon p)^T▽f(x_K)+(1/2)(\varepsilon p)^T▽^2f(x_C)(\varepsilon p)$ but we still can't evaluate $(1/2)(\varepsilon p)^T▽^2f(x_C)(\varepsilon p)$ –  i_a_n Feb 18 '13 at 6:38

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