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Gradient Vector Proof

  • Thread starter bubblesewa
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  • #1

Homework Statement



Suppose that the function f: Rn --> R has first-order partial derivatives and that the point x in Rn is a local minimizer for f: Rn --> R, meaning that there is a positive number r such that
f(x+h) > f(x) if dist(x,x+h) < r.
Prove that Df(x)=0.

Homework Equations



Df(x)=(df/dx1,df/dx2,...,df/dxn)

The Attempt at a Solution



We know that the function has first-order partial derivatives, which makes finding the gradient vector possible. And the definition for local minimizer is already given in the problem. I just need to prove that all partial derivatives are equal to zero. But how does knowing the local minimizer help me figure out the gradient vector?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Suppose that the function f: Rn --> R has first-order partial derivatives and that the point x in Rn is a local minimizer for f: Rn --> R, meaning that there is a positive number r such that
f(x+h) > f(x) if dist(x,x+h) < r.
Prove that Df(x)=0.

I just need to prove that all partial derivatives are equal to zero. But how does knowing the local minimizer help me figure out the gradient vector?
Hi bubblesewa! Welcome to PF! :smile:

Find the directional derivative of f along each coordinate axis (keeping al the other coordinates constant) :wink:
 
  • #3
Is this about what my proof should look like then? And thanks for welcoming me btw.

Since x is an interior point of Rn, we can choose a positive number r such that the open ball Br(x) is contained in Rn. Fix an index i with 1 < i < n. Then, do I suppose that I have some function, let's say q(t). Where q(t) = f(x+th) for |t| < r. Then the point 0 is an extreme point of the function q: (-r,r) --> R, so q'(0) = (df/dxi)(x) = 0.
 
  • #4
tiny-tim
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Homework Helper
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Then, do I suppose that I have some function, let's say q(t). Where q(t) = f(x+th) for |t| < r. Then the point 0 is an extreme point of the function q: (-r,r) --> R, so q'(0) = (df/dxi)(x) = 0.
Hi bubblesewa! :smile:

Yes, in principle that's right …

you define a line with that parameter t, and since, as you say, 0 is an extreme point of the line, the derivative along the line must be zero. :smile:

However, you haven't yet defined h so as to get the line that gives you ∂f/∂xi, have you? :wink:
 

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