The Hessian, Taylor’s Theorem and ExtremaJonathan LeeNovember 27, 2007These handouts are crudely extracted from my ongoing set of notes posted on my websiteat http://math.stanford.edu/~jlee/math51/.Differentials and Taylor’s Theorem• given a k-times differentiable function f : R → R, we can define its k-th order Taylorpolynomial at the point p to bepk(x) =kXi=0f(i)(p)i!· (x − p)i• Taylor’s theorem provides us an error estimate: if f : R → R is a (k + 1)-timesdifferentiable function, then there exists some number ξ between p and x such thatRk(x, a) = f(x) − pk(x) =f(k+1)(ξ)(k + 1)!· (x − a)k+1• given a differentiable function f : Rn→ R, we can define its first-order Taylor polyno-mial at the point p = (p1, . . . , pn) to bep1(x) = f(p) + Df(p) · (x − p)= f(p) +nXi=1∂f∂xi(p) · (xi− pi) ;similarly, if f is twice-differentiable, we can define its s econd-order Taylor polynomialat the point p to bep2(x) = f(p) + Df(p) · (x − p) +12(x − p)T· Hf(p) · (x − p) = f(p) +nXi=1∂f∂xi(p) · (xi− pi) +12nXi,j=1∂2f∂xixj(p) · (xi− pi)(xj− pj) ,where for notational sanity, we define the Hessian matrix to be the n× n matrix whose(i, j)-th entry is∂2f∂xixj1– find the first- and second-order Taylor polynomials forthe function at the pointf(x, y) = 1/(x2+ y2+ 1) a = (0, 0)f(x, y) = 1/(x2+ y2+ 1) a = (1, −1)f(x, y) = e2xcos(3y) a = (0, π)• as before, Taylor’s theorem provides an error estimate; the same formula holds, withthe change that ξ is taken to be a point on the line segment connecting p and xExtrema of Functions• given a function f : Rn→ R, know what its extrema, both global and local, are definedto be; this (and the following) works analogously to the single-variable case• define a point p to be a critical point of a differentiable function f : Rn→ R ifDf(p) = 0• Theorem: lo cal extrema of differentiable functions must be critical points• Theorem: let p be a f : Rn→ R be a twice-differentiable function; thenif Hf(p) is then p is a . . . of fpositive definite local minimumnegative definite local maximumneither of the above but still invertible saddle point– find the point on the plane 3x − 4y − z = 24 closest to the origin– determine the absolute extrema off(x, y) = x2+ xy + y2− 6yon the rectangle given by x ∈ [−3, 3] and y ∈ [0, 5]– determine the absolute extrema off(x, y, z) = exp(1 − x2− y2+ 2y − z2− 4z)on the ball{(x, y, z) ∈ R3: x2+ y2− 2y + z2+ 4z ≤ 0}• by Heine- Borel, say that a subset X of Rnis compact if it is closed and bounded• Extreme Value Theorem: any continuous R-valued function on a compact topologicalspace attains global minima and
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