Homework 1, Optimization 550.362, Spring 2017
Problem 1: Use a Taylor expansion with remainder term to show that for all x : 0 ≤ x <
π
2
it
holds that tan(x) ≥ x. (Hint: Expand about the value 0, and consider the possible values that
the secant function can take on.)
Solution: Expanding tan(x) about the value 0, with remainder term at the first derivative, we
get tan(x) =
tan(0)
(x
0!
− 0)0 +
tan0 (ξ)
(x
1!
2
− 0)1 for some ξ between 0 and x. Noting that, for all t,
| cos(t)| ≤ 1, we get tan0 (t) = sec (t) =
1
cos(t)
2
≥ 1 hence, by the above Taylor expansion we
have tan(x) = 0 + sec2 (ξ) · x ≥ x for all x : 0 ≤ x < π2 .
Problem 2: Construct a polynomial p(x) of degree 2 such that p(3) = −1, p0 (3) = 5, p00 (3) = −4.
Simplify this polynomial to the form ax2 + bx + c.
Solution: The polynomial is p(x) = −1 + 5(x − 3) +
−4
(x
2
− 3)2 = −2x2 + 17x − 34.
Problem 3: Construct a polynomial p(x) of degree 2 such that p(3) = −1, p(5) = 6, p(8) = 2.
Simplify this polynomial to the form ax2 +bx+c. (Hint: It might be helpful to consider, for exam(x−8)(x−9)
ple, what happens when you evaluate the polynomial 16 (x−9)(x−10)
− 13 (x−8)(x−10)
+ 62 (10−8)(10−9)
(8−9)(8−10)
(9−8)(9−10)
at the points x = 8, 9, 10.)
29 2
Solution: The polynomial is p(x) = −1 (x−5)(x−8)
+ 6 (x−3)(x−8)
+ 2 (x−3)(x−5)
= − 30
x + 337
x − 26.
(3−5)(3−8)
(5−3)(5−8)
(8−3)(8−5)
30
Problem 4: a) Show that for any A ∈ Rm×n it holds that AT A is positive semidefinite. b) Show
that for any symmetric B ∈ Rn×n there is a real number c such that cI + B is positive definite.
Solution: a) Clearly (AT A)T = AT AT
T
= AT A is symmetric and, for any x ∈ Rn , we have
xT AT Ax = kAxk22 ≥ 0, hence AT A is positive semidefinite by the definition. b) Suppose the least
eigenvalue of B is λ0 . Taking c := −λ0 + 1, we have each eigenvalue λ of B shifted in cI + B to
become λ − λ0 + 1 > 0, and thus cI + B is positive definite since its eigenvalues are all positive.
Problem 5: a) Derive a first order approximation and also a second order approximation for
the function f (x) =
x1
x2
(where x = [x1 , x2 ]T ) about the point x = [6, 2]T . (Hint: The answer
will be as follows. The first order approximation is g(x) = 3 + 21 x1 − 32 x2 and the second order
approximation is g(x) = 3 + x1 − 3x2 − 14 x1 x2 + 34 x22 . Your job is to do the derivation.)
1
Solution: It is straightforward to compute

∇f (x) = 
1
x2
− xx12
2

,

∇f (x) = 
1
2
− 32


,
∇2 f (x) = 




∇2 f (x) = 
0 − x12 
2
− x12
2
2x1
x32
0 − 14
− 14
3
2
Thus, the first order approximation is
g(x) = f (x) + ∇f (x)T (x − x) = 3 + 12 (x1 − 6) − 32 (x2 − 2) = 3 + 12 x1 − 23 x2 .
The second order approximation is g(x) = f (x) + ∇f (x)T (x − x) + 21 (x − x)T ∇2 f (x)(x − x)
= 3 + 12 (x1 − 6) − 23 (x2 − 2) − 14 (x1 − 6)(x2 − 2) + 43 (x2 − 2)2 = 3 + x1 − 3x2 − 14 x1 x2 + 43 x22 .
2

