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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 $p$)

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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