Political Science Math Camp:
Problem Set 5
Answer Key
1. Calculate the following derivatives:
(a)
d
1/3
dx 3x
2
d 1/3 1
3x = × 3x− 3
dx
3
2
= x− 3
(b)
d
2
dx (x
+ 1)(x3 − 1)
d 2
d 5
(x + 1)(x3 − 1) =
(x − x2 + x3 − 1)
dx
dx
= 5x4 − 2x + 3x2
(c)
d x2 −3x+2
dx e
d x2 −3x+2
2
e
= ex −3x+2 (2x − 3)
dx
(d)
h
d 4x−12x2
dx x3 −4x2
i2
h 4x − 12x2 i d h 4x − 12x2 i
d h 4x − 12x2 i2
=
2
dx x3 − 4x2
x3 − 4x2 dx x3 − 4x2
h 4x − 12x2 ih (4 − 24x)(x3 − 4x2 ) − (4x − 12x2 )(3x2 − 8x) i
=2 3
x − 4x2
(x3 − 4x2 )2
h 4x − 12x2 ih (4x3 − 16x2 − 24x4 + 96x3 ) − (12x3 − 36x4 − 32x2 + 96x3 ) i
=2 3
x − 4x2
(x3 − 4x2 )(x3 − 4x2 )
h 4x − 12x2 ih 12x4 − 8x3 + 16x2 i
=2 3
x − 4x2
x6 − 8x5 + 16x4
2. Calculate the following integrals:
1
(a)
R
(2x + 12x2 )dx
Z
(b)
R
(2x + 12x2 )dx = x2 + 4x3 + c
360t6 dt
Z
(c)
R
xe3x
2 +1
360t6 dt =
360 7
t +c
7
dx
Let u = 3x2 + 1. Then, using the chain rule, du = (3x2 + 1)0 dx=6xdx. Therefore,
xdx = 61 du.
Z
e
(d)
R
3x2 +1
eu
du
6
Z
1
=
eu du
6
eu
=
6
2
e3x +1
+c
=
6
Z
xdx =
x2
dx
1−x2
Z
−1(x2 )
dx
−1(1 − x2 )
Z
−x2
=
dx
−1 + x2
Z 2
x −1+1
=−
dx
−1 + x2
Z 2
x −1
1 =−
+
dx
x2 − 1 x2 − 1
Z 1 =−
1+ 2
dx
x −1
Z
Z
1
= − 1dx −
dx
x2 − 1
Z
1
= −x −
dx
x2 − 1
Z
1
1
= −x −
dx
x−1x+1
x2
dx =
1 − x2
Z
2
3. The utility that a legislator obtains from a policy in a one-dimensional policy space is U (x) =
−a(x − i)2 where i is the legislator’s ideal point and a is the salience of the issue. This utility
function formalizes the idea that the farther the policy x is from the legislator’s ideal policy,
i, the worse off the legislator is. The larger a is, the larger the effect that a given change in
x has on the legislator’s utility. Suppose a = 20 and i = 60.
(a) Prove that the legislator’s utility maximizing policy is x = 60.
U (x) = −a(x − i)2 with a = 20 and i = 60, so U (x) = −20(x − 60)2
First, we take the first order condition in order to find the utility maximizing policy.
U 0 (x) = 2 × −20(x − 60) = 0
x − 60 = 0
x∗ = 60
Now, we need to show that this critical point is a maximum and not a minimum. Taking
the second derivative:
U 0 (x) = −20 × 2(x − 60)
= −40x − 120
00
U (x) = −40
Because −40 < 0, we know that the function is concave down, and therefore, the critical
point at x=60 is a maximum.
(b) Suppose that the constitution requires x to be in the range x ∈ [70, 80]. What policy in
this range maximizes the legislator’s utility? Be sure to justify your answer.
x=70 maximizes the legislator’s utility when x ∈ [70, 80] because we know that 1) the
graph is strictly concave downward and 2) that there is a unique maximum at x=60.
Therefore, as x increases or decreases from x=60, the value of U(x) will decrease. Therefore, the utility maximizing point is always the point that is closest to x=60, which in
this case is x=70.
4. The balance-of-power school in international relations argues that peace is most likely when
there is a balance of power between states. By contrast, the preponderance-of-power school
argues that that peace is least likely when there is a balance of power and is most likely when
one side has a preponderance of power. Suppose that the analysis of a formal model yields
the result that the probablilty of war in the model is
π=
1 h
1 i2
− p−
4
2
where π is the probability of war and p is the distribution of power between states 1 and 2.
That is, p is the probability that 1 prevails in a war against 2. There is a balance of power
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when p = 1 / 2 . By contrast, 1 has a preponderance of power (in the extreme) when p = 1
and 2 has a preponderance of power when p = 0. Is the result of the formal analysis consistent
with one of the two schools or neither? Be sure to explain your answer.
First, we want to find the distribution of power, p, that is most likely to result in peace (i.e.
minimize π). To do this, we first take the first order condition of the model.
1
π 0 (p) = −2(p − )
2
0 = −2p + 1
1
p∗ =
2
To determine if this is a maximum or minimum, we take the second derivative.
π 00 (p) = −2
Because the second order condition is less than 0, we know that the function is strictly concave downward, and p = 12 is a unique maximum. Therefore, the values of p that minimize
π 0 (p) must be as far from 12 as possible. However, because we are dealing with probabilities,
the function is bounded by p=0 and p=1. Therefore, there must be a minimum at either
p=0, p=1, or both. To check, we can plug in these values to see what yields the smallest π(p).
1
− (0 −
4
1
π(0) = − (1 −
4
π(0) =
1 2
) =0
2
1 2
) =0
2
Therefore, both p=0 and p=1 are minima. This suggests that the preponderance of power
theory is correct.
5. For the following functions, determine the interval over which the function is convex or concave
(if at all), and identify any inflection points:
(a) f (x) =
1
x
f 0 (x) = −(x)−2
f 00 (x) = 2x−3
The function is concave downward when x < 0 because f 00 (x) < 0 and concave upward
when x > 0 because f 00 (x) > 0.
(b) f (x) = x2 + 4x + 8
f 0 (x) = 2x + 4
f 00 (x) = 2
4
Because the second derivative is equal to a constant greater than 0, the function is
strictly concave upward.
(c) f (x) =
1
1+x2
f 0 (x) =
=
f 00 (x) =
=
=
=
(1 + x2 )(0) − (2x)
(1 + x2 )2
−2x
(1 + x2 )2
(1 + x2 )2 (−2) − (−2x)(2)(2x)
(1 + x2 )4
(1 + x2 )(−2) − (−2x)(2)(2x)
(1 + x2 )3
−2 − 2x2 + 8x2
(1 + x2 )3
6x2 − 2
(1 + x2 )3
Then, we set the second order condition equal to zero.
f 00 (x) = 0 =
x2 =
6x2 − 2
(1 + x2 )3
1
3
1
x = ±√
3
Thus, the function is concave down between the inflection points (± √13 ) and concave up
everywhere else.
(d) f (x) = log(x)
f 0 (x) =
1
x
f 00 (x) = −
1
x2
Because of x2 , all values of x will yield a positive value in the denominator of − x12 .
Because the sign on the fraction is negative, however, the value of the second order condition will always be negative, which means the function is strictly concave downward.
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