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For a 2-variable function $f(x,y)$, the Hessian matrix is $$\mathcal{H}(f) = \left[\begin{array}{cc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial {xy}} \\ \frac{\partial^2 f}{\partial {xy}} & \frac{\partial^2 f}{\partial y^2}\end{array}\right]$$, how does it describe the local curvature of $f$?

avocado
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1 Answers1

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Are you familiar with the Taylor expansion of a single variable function?$$f(x)=f(x_0)+f'(x_0)\Delta x+\frac12f''(x_0)\Delta x^2+\dots$$We can also find Taylor expansions for multivariable functions; the first two terms should be familiar:$$f(\mathbf{x})=f(\mathbf{x}_0)+J(\mathbf{x}_0)\Delta\mathbf{x}+\frac12\Delta\mathbf{x}^TH(\mathbf{x}_0)\Delta\mathbf{x}+\dots$$... so we see that while the Jacobian corresponds to our original idea of the derivative, the Hessian instead serves a purpose like our second derivative -- it indeed captures local curvature information. In fact, if you look at its elements, this should become very clear.

obataku
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  • I know Taylor expansion, and I know that for a single variable function, its curvature is $\kappa=\frac{|f''|}{(1+f'^2)^{3/2}}$, but I don't know what does the curvature for 2-variable functions look like. – avocado Sep 01 '13 at 23:40
  • http://web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node32.html#eqn:surf_explicit_K – obataku Sep 02 '13 at 02:36
  • I still don't understand why Hessian captures the local curvature info, just because it encodes the second derivative? Moreover, to compute the curvature, we need both first and second derivatives, right? – avocado Sep 03 '13 at 00:23
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    The second derivative itself doesn't exactly encode 'curvature' - as loganecolss points out, you need the first and second derivative for that. Rather, the second derivative should be better thought of as a measure of the concavity/convexity of the function. (Think back to the classic second derivative test for maxima/minima). – Baron Mingus Sep 04 '13 at 10:53
  • @BaronMingus, each entry in Hessian matrix is just a 2nd order derivative, which indicates how fast the 1st order derivative changes in x or y axis, right? So I can understand the 2nd order derivatives show the concavity/convexity. – avocado Sep 06 '13 at 13:47
  • well just as the curvature $\kappa$ relies on $f',f''$ so does the extrinsic curvature surface rely on $f_x,f_y,f_{xx},f_{xy},f_{yy}$ – obataku Sep 06 '13 at 20:26
  • @loganecolss yes, in the sense that concavity tells you the way the curve bends; a curve bending 'upwards' has positive curvature whereas a curve bending 'downwards' has negative curvature. The partial derivatives give you similar information along different axes. – obataku Sep 06 '13 at 20:29