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Test of Second Partials - Proof

  1. Nov 4, 2004 #1
    Can anyone give me a proof or some kind of general explaination about why there is a local max at (a,b) if D>0 and fxx<0, a min if D>0 and fxx>0, etc. My text book doesn't give any kind of explaination at all and I'm just a little curious as to why it works.
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  3. Nov 5, 2004 #2


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    Second Derivatives Test:
    Suppose the second partial derivatives of [itex]f[/itex] are continuous on a disk with center [itex](a,b)[/itex], and suppose that [itex]f_x(a,b)=0[/itex] and [itex]f_y(a,b)=0[/itex] [that is, [itex](a,b)[/itex] is a critical point of [itex]f[/itex]]. Let


    (a) If D>0 and [itex]f_{xx}(a,b)>0[/itex], then f(a,b) is a local minimum.
    (b) If D>0 and [itex]f_{xx}(a,b)<0[/itex], then f(a,b) is a local maximum.
    (c) If D<0 and [itex]f_{xx}(a,b)>0[/itex], then f(a,b) is not a local maximum or minimum.

    Proof of part (a):

    We compute the second-order directional derivative of f in the direction of [itex]\vec u = \langle h, k \rangle[/itex]. The first-order derivative is given by:
    [tex]D_uf=f_xh+f_yk \quad \mbox{(from a different theorem)}[/tex]
    Applying this theorem a second time, we have:
    D^2_uf & = & D_u(D_uf)=\frac{\partial}{\partial x}(D_uf)h+\frac{\partial}{\partial y}(D_uf)k \nonumber \\
    & = & (f_{xx}h+f_{yx}k)h+(f_{xy}h+f_{yy}k)k \nonumber\\
    & = & f_{xx}h^2+2f_{xy}hk+f_{yy}k^2 \mbox{(by Clairaut's theorem)}\nonumber

    If we complete the square in this expression, we obtain:
    [tex]D_u^2f=f_{xx}\left(h+\frac{f_{xy}}{f_{xx}}k\right)^2+\frac{k^2}{f_{xx}}(f_{xx}f_{yy}-f^2_{xy}) \quad \mbox(<- Equation 1)[/tex]
    We are given that [itex]f_{xx}(a,b)>0[/itex] and [itex]D(a,b)>0[/itex]. But [itex]f_{xx}[/itex] and [itex]D=f_{xx}f_{yy}-f^2_{xy}[/itex] are continuous functions, so there is a disk B with center (a,b) and radius [itex]\delta>0[/itex] such that [itex]f_{xx}>0[/itex] and [itex]D>0[/itex] whenever (x,y) is in B. Therefore, by looking at Equation 1, we see that [itex]D_u^2f(x,y)>0[/itex] whenever (x,y) is in B. This means that if C is the curve obtained by intersecting the graph of f with the vertical plane through P(a,b,f(a,b)) in the direction of [itex]\vec u[/itex], then C is concave upward on an interval of lenght [itex]2\delta[/itex]. This is true in the direction of every vector [itex]\vec u[/itex], so if we restrict (x,y) to lie in B, the graph lies above its horizontal tangent plane at P. Thus [itex]f(x,y)\geq f(a,b)[/itex] whenever (x,y) is in B. This shows that f(a,b) is a local minimum.

    Parts (b) and (c) have similar proofs.
    Last edited: Nov 5, 2004
  4. Nov 5, 2004 #3


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    Notice that what Galileo is really doing is expanding f(x,y) in a Taylor's series with two variables. The derivatives at a critical point are 0 so there are no first power terms. For x, y very close to the critical point, we can ignore higher powers so we have only the constant term (the value of f AT the critical point) and the quadratic terms. We can always change the coordinate system to eliminate any "xy" term and have left ax2+ by2 (with a, b, positive, negative, or 0). If both a and b are positive, that's a minimum, if negative, maximum. If either is 0, we need to look at higher powers.

    What's REALLY happening is that, with a real valued function of of R2 to R, the second derivative is a linear function from R2 to R2 which can be represented as a 2 by 2 matrix (the entries are: first row fxx, fxy, second row fyx, fyy).

    Since that is a symmetric matrix we can always change the coordinate system to make that a diagonal matrix resulting in the "ax2+ b2" above (a, b are the diagonal elements). Of course the determinant of such a matrix is ab so everything depends upon whether that determinant is positive or negative. But the determinant is independent of changing the coordinate system so we can just look at the determinant of the original matrix: D= (fxxfyy-fxy2).
  5. Nov 5, 2004 #4


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    The second derivative test can seem rather tricky and abstract to begin with, but, as HallsofIvy said, what you really get out of it, is either:
    1) Locally, your graph looks like a paraboloid expanding upwards (i.e, you've got positive curvatures and a minimum point)
    2) Locally, your graph looks like a paraboloid expanding downwards (i.e, you've got negative curvatures and a maximum point)
    3) Locally, your graph looks like a "saddle", neither maximum nor minimum.
    4) Second derivatives test is insufficient in elucidating local behaviour (i.e, you must look on higher derivatives, (assuming those exist))
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