Quadratic Approximation at a Stationary Point Let F(X, Y) Be a Given

Multivariable Calculus

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Quadratic approximation at a stationary point

Let f(x, y) be a given function and let (x0, y0) be a point in its domain.

Under proper diﬀerentiability conditions one has

f(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0) 1 2 1 2 + 2 fxx(x0, y0)(x − x0) + fxy(x0, y0)(x − x0)(y − y0) + 2 fyy(x0, y0)(y − y0) + higher−order terms.

Let (x0, y0) be a stationary (critical) point of f: fx(x0, y0) = fy(x0, y0) = 0. Then

2 2 f(x, y) = f(x0, y0) + A(x − x0) + 2B(x − x0)(y − y0) + C(y − y0) + higher−order terms,

1 1 1 1 where A = 2 fxx(x0, y0),B = 2 fxy(x0, y0) = 2 fyx(x0, y0), and C = 2 fyy(x0, y0).

Assume for simplicity that (x0, y0) = (0, 0) and f(0, 0) = 0.

[ This can always be achieved by translation: f˜(x, y) = f(x0 + x, y0 + y) − f(x0, y0). ]

Then f(x, y) = Ax2 + 2Bxy + Cy2 + higher−order terms.

Thus, provided A, B, C are not all zero, the graph of f near (0, 0) resembles the quadric surface z = Ax2 + 2Bxy + Cy2. Generically, this quadric surface is either an elliptic or a hyperbolic paraboloid. We distinguish three scenarios:

* Elliptic paraboloid opening up, (0, 0) is a point of local minimum. * Elliptic paraboloid opening down, (0, 0) is a point of local maximum. * Hyperbolic paraboloid, (0, 0) is a saddle point.

It should certainly be possible to tell which case we are dealing with by looking at the coeﬃcients A, B, and C, and this is the idea behind the Second Partials Test.

The following observation is repeatedly used below. x2 x For y 6= 0, Ax2 + 2Bxy + Cy2 has the same sign as A + 2B + C. y y Therefore everything boils down to a one-variable quadratic, At2 + 2Bt + C. Second Partials Test Elliptic Paraboloid Opening Up. For any (x, y) diﬀerent from (0, 0) we must have Ax2 + 2Bxy + Cy2 > 0. Setting t = x/y, for y 6= 0, we see that At2 + 2Bt + C > 0. This means that A, C > 0 and the discriminant 4(B2 − AC) is negative (no roots). Thus the conditions 2 fxx or fyy > 0, fxxfyy − (fxy) > 0 indicate a local minimum.

Elliptic Paraboloid Opening Down. For any (x, y) diﬀerent from (0, 0) we must have Ax2 + 2Bxy + Cy2 < 0. Setting t = x/y, for y 6= 0, we see that At2 + 2Bt + C < 0. This means that A, C < 0 and the discriminant 4(B2 − AC) is negative (no roots). Thus the conditions 2 fxx or fyy < 0, fxxfyy − (fxy) > 0 indicate a local maximum.

Hyperbolic Paraboloid. The expression Ax2 + 2Bxy + Cy2 is neither nonnegative nor nonpositive. This means that the quadratic At2 + 2Bt + C takes both positive and negative values, i.e., its discriminant 4(B2 − AC) is positive (two roots). Thus the condition 2 fxxfyy − (fxy) < 0 indicates a saddle point. This is the case if, for instance, A or C = 0 and B 6= 0, or if A and C are of opposite sign.

Addendum We have been fairly informal in our critical point analysis and have only examined generic cases.

In degenerate cases, such as B2 − AC = 0 A = B = 0 A = B = C = 0, the higher-order terms of f (and so higher-order partials) become important.

However, even if the Second Partials Test conditions are not met or a quadric does not capture the geometry of f, one may still examine the proﬁles of the graph to classify a critical point.

EXAMPLE. The function f(x, y) = x2 + 1 has a stationary point at the origin: fx(0, 0) = fy(0, 0) = 0. Since fyy(0, 0) = fxy(0, 0) = 0, the test is not applicable. But clearly f(x, y) ≥ f(0, 0), and so (0, 0) is a point of minimum. As well as any point (0, y).

2 EXAMPLE. The function f(x, y) = xy has a stationary point at (0, 0): fx(0, 0) = fy(0, 0) = 0. 3 Since fxx(0, 0) = fyy(0, 0) = fxy(0, 0) = 0, the test does not apply. Note, however, that f(t, t) = t and so (0, 0) cannot be a point of local extremum. Hence (0, 0) is a saddle point.