Calculus - Day 3
October 25, 2007
Topics: Main topics: Lagrange multipliers, Implicit Function Theorem. Other
topics: Implicit differentiation, normals/tangents to a surface/line, finding extrema, Taylor’s theorem, Taylor’s remainder, Mean Value Theorem, Fubini’s
theorem.
Problems:
J92#2.
J95#4, S03#4, S03#5, J95#3, S91#2, S94#4, S96#5, J01#1,
———————————————————————————————–
Lagrange Multiplier Suppose we would like to minimize/maximize f (~x)
subject to constraints g1 (~x) = 0, g2 (~x) = 0, . . . gm (~x) = 0. If x∗ is a local
minimizer/maximizer of the problem, and {∇gi (x∗ )} are linearly independent,
then x∗ is a critical point of the function
L(~x, λ1 , λ2 , . . . λm ) = f (~x) − λ1 g1 (~x) − . . . − λm gm (~x).
In other words, to find a minimizer we must solve the equations ∇x L = 0,
∇λ L = 0 for x∗ , {λi }. We must also check points on the boundaries of the set,
if such boundaries exist, as well as points where the constraints are degenerate.
Implicit Function Theorem Let F = F (x, y, z) ∈ C 1 on an open set D.
Let p0 = (x0 , y0 , z0 ) ∈ D satisfy F (p0 ) = 0. Suppose Fz (p0 ) 6= 0. Then there
exists a function φ ∈ C 1 defined in a neighbourhood N of (x0 , y0 ) such that
z = φ(x, y) solves F (x, y, z) = 0 for (x, y) ∈ N , and φ(x0 , y0 ) = z0 .
In general, if we have a system of m equations in n unknowns,
Φ(x) = 0
Φ : Rn → Rm ∈ C 1 ,
and we know one solution Φ(p) = 0, p ∈ Rn , then we can solve for m of
the variables in terms of the others provided the Jacobian of the transformation
doesn’t vanish at p:
(x1 , x2 , . . . , xm ) = f (xm+1 , . . . , xn )
1
⇔
∂(φ1 , . . . φm ) 6= 0.
∂(x1 , . . . xm ) p
Taylor’s Remainder Let f ∈ C n+1 on an interval I about x = c, and let
P (x) be the Taylor polynomial of degree n at c. Then f (x) = P (x) + Rn (x) for
x ∈ I, with
Z
1 x (n+1)
Rn (x) =
f
(t)(t − c)n dt.
n! c
Corollary:
Rn (x) = f (n+1) (τ )
(x − c)n+1
(n + 1)!
for some τ ∈ (c, x).
R
Fubini’s Theorem If A×B |f (x, y)|d(x, y) < ∞, (integral with respect to
R R
product measure),
then
we
can
do
the
integration
in
any
order:
f
(x,
y)dy
dx =
A
B
R R
f (x, y)dx dy.
B
A
2