517 Midterm 3
1. Let F : Rn → Rm be linear. Show that F is differentiable and dFa = F for every a ∈ Rn .
Solution. By linearity,
F (a + h) − F (a) − F (h)
F (a) + F (h) − F (a) − F (h)
0
= lim
= lim
= 0.
h→0
h→0
h→0 |h|
|h|
|h|
lim
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2. Let f : Rn → R be differentiable. Use the single variable MVT to show that for all a, b ∈ Rn ,
there is a point c on the line segment from a to b such that
f (b) − f (a) = f 0 (c)(b − a).
Solution. Fix a, b ∈ Rn and define φ(t) = a + t(b − a). Note that φ0 (t) = b − a, so by the chain
rule and single variable MVT,
f (b) − f (a) = f (φ(1)) − f (φ(0)) = (f ◦ φ)0 (s) = f 0 (φ(s))(b − a)
for some s ∈ (0, 1). Letting c = φ(s), we are done.
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3. Let f (x, y) = e−x
2 −y 2
. Use Taylor’s theorem1 to show that f has a local maximum at 0.
Solution. The degree 2 Taylor polynomial of f at 0 is P2 (h) = 1−|h|2 , and f (h) = P2 (h)+R2 (h)
where R2 (h)/|h|2 → 0 as h → 0. For h sufficiently small, |R2 (h)| < |h|2 and so f (h) =
1 − |h|2 + R2 (h) < 1.
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Of course, we know that f (x, y) < 1 = f (0, 0) for all (x, y) 6= (0, 0) so Taylor’s theorem is not really needed. In
general, for classifying a local optimum, Taylor’s theorem is useful whenever the second order partial derivatives
of f at the optimum can be explicitly computed.
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4. Suppose F : Rn → Rn is class C 1 , with F (0) = 0 and F 0 (0) = I. Show that φ(x) := F (x) − x
is a contraction mapping on a sufficiently small ball around 0. What is its fixed point?
Solution. Choose δ > 0 such that ||F 0 (z) − I|| ≤ 1/2 for all z ∈ Bδ (0), and let x, y ∈ Bδ (0).
With L the line segment between x and y, the multivariate MVT implies
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|φ(x) − φ(y)| ≤ |x − y| max ||φ0 (z)|| ≤ |x − y|.
z∈L
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This shows φ is a Lipschitz continuous with constant 1/2. Setting y = 0 into the last display
shows that φ(Bδ (0)) ⊂ Bδ (0). Thus, φ is a contraction mapping on Bδ (0), and its fixed point
must be 0 (since φ(0) = 0 and the fixed point is unique).
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5. Use the definition of derivative to show that f (x, y) = x2 y is differentiable and
df(a,b) (x, y) = 2abx + a2 y.
Solution. Note that
|f (a + x, b + y) − f (a, b) − (2abx + a2 y)|
|bx2 + 2axy + x2 y|
=
|(x, y)|
|(x, y)|
2
|bx | + 2|axy| + |x2 y|
≤
|(x, y)|
≤ |bx| + 2|ay| + |xy| → 0 as (x, y) → 0.
(The last line uses the fact that |(x, y)| ≥ |x|.)
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