Problem Set: Mathematical Foundations SOLUTIONS
for some a, b ∈ R Normality condition:
Orthogonality condition:
v·v=1 β2(9 + 16) = 1
Copyright By PowCoder代写 加微信 powcoder
β = ± 15 a
a2 + b2 = 1 v·w=0
β(3a + 4b) = 0
Rearrange 2nd condition and sub in 1st:
=⇒ a2 + 9 a2 = 1 16
=⇒ 25a2 =1 16
=⇒ a2 = 16 25
=⇒ b = −3 5
w=14 5 −3
(c) No, because −w is also orthonormal to v
(e) If A is invertible then detA ̸= 0 by the Invertible Matrix Theorem.
detA = (4.9) − α.α = 36 − α2
=⇒ 36 − α2 ̸= 0 =⇒ 36̸=α2
=⇒ α ̸= ±6
(f) If A is positive definite then its eigenvalues,{λi}2i=1, are positive. Eigenvalues are satisfied by the following characteristic polynomial:
Soforλi >0:
det(A − λiI) = 0 4−λi α =0
α 9−λi (4−λi)(9−λi)−α2 = 0 (λi)2 −13λi +36−α2 =0
=⇒ λi = 13±132 −4(36−α2) 2
13−132 −4(36−α2)>0
132 > 132 − 4(36 − α2)
36 > α2 =⇒ −6 < α < 6
(g) If A represents a covariance matrix then the diagonal elements give the vari- ances, σ12, σ2:
σ12 = 4 =⇒ σ1 = 2 σ22 = 9 =⇒ σ1 = 3
The correlation, ρ, is given by the off-diagonal elements, so that: ρσ1σ2 = α
(h) The shape will be an ellipse
The ellipse will be axis-aligned
The ratio of the semi-major to the semi-minor axis will be: 3:2
Note that the form of f(x) is:
f(x)=2x2 −2(a1 +a2)x+(a21 +a2)
So, forming the Hessian:
Thus the Hessian is positive definite. So f(x) is strictly convex
So the solution is globally optimal and unique (by properties of strictly convex functions)
(b) Note that g(x) has a characteristic ‘V’ shape, with change of slope at b (a sketch would also be acceptable in order to demonstrate this)
State that g(x) is convex, but not strictly convex
So the solution is globally optimal (by properties of convex functions). But the solution is unique (by geometry).
(c) Note that h(x) has a characteristic ‘\ /’ shape, with changes of slope at b1 and b2 (a sketch would also be acceptable in order to demonstrate this)
State that h(x) is convex, but not strictly convex
So the solution is globally optimal (by properties of convex functions). And the solution is not unique (by geometry).
(d) Note that k(x) = f(x) + h(x), and that the sum of a convex function and a strictly convex function is strictly convex.
So the solution is globally optimal (by properties of convex functions) And the solution is unique (by properties of strictly convex functions)
3. Consider x ≥ 1:
(1 − x) ≤ 0 < e−x
Consider 0 < x < 1:
Binomial expansion, valid for |x| < 1:
Taylor expansion of ex:
1 =1+x+x2 +x3 +... 1−x
ex =1+x+1x2 ++1x3 +... 2 3!
For positive x this implies:
Consider x = 0:
Consider x < 0: Let y = −x > 0:
So, since y > 0:
=⇒ (1−x)