Problem session solutions
1. For the function f (x) = exp(−x2 /2), defined for x ∈ R,
(a) On which intervals is f non-negative? On which is it non-positive?
(b) On which intervals is f increasing? On which is it decreasing?
(c) On which intervals is f convex? On which is it concave?
(d) Does f have a maximum, and if so where? Does f have a minimum, and if so
where?
Solution
(a) f is postive, and hence non-negative, on all of R, because the exponential is
always positive.
(b) By the chain rule, f 0 (x) = −x exp(−x2 /2), so f 0 is non-negative, and hence
f is increasing, in (−∞, 0]. f 0 is non-positive, and hence f is decreasing, in
[0, ∞)
(c) f 00 (x) = (x2 − 1) exp(−x2 /2), so f 00 is non-negative, and hence f is convex, in
(−∞, −1] and [1, ∞). f 00 is non-positive, and hence f is concave, in [−1, 1].
(d) If x 6= 0 then −x2 /2 < 0 and hence exp(−x2 /2) < exp(0) = f (0), so f has a
maximum, and in fact a strict maximum, at 0. It has no minimum, since for
any x we can find a y with f (y) < f (x). If x = 0 then any y 6= 0 will suffice,
and y = 2x works in all other cases.
2. The matrix exponential function is defined by
exp(A) = I + A +
Ak
A2
+ ... +
+ ...
2
k!
Compute exp(Jm (λ)) (the exponential of the Jordan block of size m with λ on the
diagonal).
3. (a) Is {0, −1, −2, −3, −4, . . . } a subgroup of (Z, +)?
(b) Is 6Z = {6n : n ∈ Z} a subgroup of (Z, +)?
(c) Find four different subgroups H of (Z, +) such that 6 ∈ H.
Solution (a) No, since it is not closed under inversion. Indeed, −1 is in this set,
but its inverse in (Z, +) is 1, which is not in the set.
(b) Yes. We clearly have 6Z ⊆ Z, and
• 6Z 6= ∅ since, for example, 6 ∈ 6Z
• if x, y ∈ 6Z then x = 6n and y = 6m for some n, m ∈ 6Z, so x+y = 6n+6m =
6(n + m) and n + m ∈ Z, so x + y = 6(n + m) ∈ 6Z
• if x ∈ 6Z then x = 6n for some n ∈ Z, so x−1 = −x = 6 · (−n) and −n ∈ Z,
so x−1 = 6 · (−n) ∈ 6Z.
(c) Consider the four sets Z, 6Z, 3Z = {3n : n ∈ Z} and 2Z = {2n : n ∈ Z}.
These sets all contain 6 and are subsets of Z. Clearly, Z is a subgroup of (Z, +),
and the proof that 2Z and 3Z are subgroups is very similar to (b). So these are all
subgroups of (Z, +).
4. (a) Is {(1), (1 5)} a subgroup of (S5 , ◦)?
(b) Is {(1), (2 3 4), (2 4 3), (2 4)} a subgroup of (S5 , ◦)?
(c) Is {α ∈ S3 : α ◦ α = (1)} a subgroup of (S3 , ◦)?
[Note: (1) is a cycle of length 1, so it maps 1 7→ 1 and x 7→ x if x 6= 1.
So (1) is identity mapping in Sn .]
Solution (a) Yes, H = {(1), (1 5)} is a subgroup of (S5 , ◦). Indeed,
• H ⊆ S5 and H 6= ∅
• if α, β ∈ H then there are four possibilities:
α
(1)
(1)
(1 5)
(1 5)
β
(1)
(1 5)
(1)
(1 5)
α◦β
(1)
(1 5)
(1 5)
(1)
In each case, α ◦ β ∈ H. So α, β ∈ H =⇒ α ◦ β ∈ H.
• We have (1)−1 = (1) ∈ H and (1 5)−1 = (1 5) ∈ H. So α ∈ H =⇒ α−1 ∈ H.
(b) No, H = {(1), (2 3 4), (2 4 3), (2 4)} is not a subgroup of (S5 , ◦) since it not
closed under ◦. For example, (2 3 4) ◦ (2 4) = (3 4) 6∈ H.
(c) No, H = {α ∈ S3 : α ◦ α = (1)} is not a subgroup of (S3 , ◦), since it is not closed
under ◦. For example, if α = (1 2) and β = (2 3) then α ◦ α = (1) = β ◦ β so
α, β ∈ H, but if γ = α◦β then γ = (1 2)◦(2 3) = (1 2 3) so γ ◦γ = (1 2 3)◦(1 2 3) =
(1 3 2) 6= (1), so γ = α ◦ β 6∈ H.
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