ECON30020 Mathematical Economics Tutorial 4. Functions and Sequences
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ECON30020 Mathematical Economics
Tutorial 4. Functions and Sequences
Question 1. Functions
Provide a plot of the following functions and their ranges.
(a) f : R → R defined as f (x) = x2
(b) f : R+ → R defined as f (x) = ^x
(c) f : Z+ → Q defined as f (x) = 1/x
(d) f : R → R defined as f (x1 , x2 ) = x1 max{1 - x1 - x2 , 0}
Answer: (a): The range is R+
f
100
80
60
40
20
x
-10 -5 5 10
(b): The range is R+
f3.0
2.5
2.0
1.5
0.5
x
2 4 6 8 10
(c): The range is {1, 1/2, 1/3, . . . }
f
1.0
0.8
0.6
0.4
0.2
x
2 4 6 8 10
(d): The range is [0, 1/4] (since f is maximized at x1 = 1/2 and x2 = 0, in which case it is 1/4, and it can’t be less than 0)
Question 2. Quasiconcave and concave functions
Derive an expression for a typical level set (an indifference curve) corresponding to some con- stant c > 0 for the following Cobb-Douglas utility functions. (Express x2 as a function of x1 and c.)
Sketch a few of the indifference curves and determine whether the functions are (strictly) qua- siconcave or (strictly) quasiconvex. Then say whether they are concave, convex or neither.
(a) y = x10.8x20.2 , when (x1 , x2 ) e R .
(b) y = x1 3x2 3 , when (x1 , x2 ) e R .
Answer: In order to sketch level curves of the function, let’s derive their equations.
(a) Holding y constant at some fixed level c gives us an equation of the corresponding level
curve:
c = x1(0) .8x2(0) .2 =÷ x2(0) .2 = =÷ x2 =
(b) We have
c = x1(3)x2(3) =÷ x2(3) = =÷ x2 =
Below are some sketches of the level curves (indifference curves) for both utility functions, for different values of c.
(a) Level curves of y = z1 0 .8 z2 0 .2
(b) Level curves of y = z1 3 z2 3
Given the shape of indifference curves, we can see that the better sets (the upper contour sets, the sets of points on and above a level curve) are strictly convex. Therefore, both functions are strictly quasiconcave. They are not quasiconvex as their worse sets (the lower contour sets, the sets of points on and below an indifference curve) are not convex.
The function y = x10.8x20.2 in part (a) is concave. At this stage, we currently don’t have a way to prove this easily, but this property should become clear if you look at a three-dimensional graph. ”If it looks like a cave, then it is concave.”
Link: https://www.geogebra.org/3d/qqvrt2nc
At wolframalpha.com: Plot3D [Re[xˆ0.8 yˆ0.2], {x, -0, 10}, {y, -0, 10}]
The function y = x1 3x2 3 is neither concave nor convex. Link: https://www.geogebra.org/3d/sxsrj8rx
How do we show this? Let’s prove it directly, using the definitions of concavity and convexity.
First, let us show that this function is not concave. To do this, let’s pick the points (1,1) and (3,3). Take their convex combination (2,2) which corresponds to λ = 0.5. Then
f(λ ) = (2)3 (2)3 = 64 < λf (1, 1)+(1 - λ)f(3, 3) = 0.5f(1, 1)+0.5f(3, 3) = 0.5+0.5*729 = 365,
contradicting the definition of concavity. (The straight line segment connecting points (1,1) and (3,3) lies above the graph of the function.)
To show that function is not convex, take (3,1) and (1,3). Take their convex combination (2,2) which corresponds to λ = 0.5. Then
f(λ ) = (2)3 (2)3 = 64 > λf (3, 1)+(1 -λ)f(1, 3) = 0.5f(3, 1)+0.5f(1, 3) = 0.5*27+0.5*27 = 27,
contradicting the definition of convexity. (The straight line segment connecting points (3,1) and (1,3) lies below the graph of the function).
Note that we can also use the following result:
Convex functions are quasiconvex, which is equivalent to ”if a function is not quasiconvex, then it is not convex” .
We have already determined that neither function is quasiconvex, therefore, none of them is convex.
Question 3. One-to-one functions
Let B = {B1, B2 } and S = {S1, S2 } be two sets with cardinality 2 each, where B has the interpretation of buyers and S the interpretation of sellers. Consider functions f : B → S with the property that f(b) = s implies that there is no b\ ( b) e B such that f(b\ ) = s. Hence, f is a one-to-one function, that is, f(b) = s implies f -1 (s) = b, and can be interpreted as a
matching as defined in L1. Suppose that each seller s e S has an object, the sale of which costs her cs . Let cS1 = 5 and cS2 = 3. Each buyer has a valuation for each of the sellers’ objects summarized by a vector vb = (vS(b)1 , vS(b)2 ), where vS(b)i is buyer b’s valuation or willingness to pay for seller Si’s object. The surplus that is created if buyer b is matched to seller Si is max{vS(b)i - cSi , 0}. Assuming
vB1 = (11, 8) and vB2 = (9, 5),
find the matching (or one-to-one function) that maximizes the sum of the surpluses.
Answer: Note that we have only two matchings available, denoted fI and fII , with fI defined by fI (B1) = S1 and fI (B2) = S2 and fII (B1) = S2 and fII (B2) = S1 . The sum of the surpluses created by fI is (11 - 5) + (5 - 3) = 8 while the one created by fII is (8 - 3) + (9 - 5) = 9. Hence, fII is the surplus-maximizing matching: we should match B1 with S2, and B2 with S1 .
Question 4. The Limit of a Sequence
Show that the sequence {an} with an = δn and δ e [0, 1) converges to the limit 0. Please use ε - N definition given in Lecture 3, slide 5.
Answer: Note that an+1 = δan < an . Hence, this is a monotonically decreasing sequence.
To show that limn二& an = 0, we need to show that for any ε > 0 there is a number N such that for all n > N , |an - 0| = |an| = δn < ε .
For given ε we can choose N to be the smallest integer such that δ N < ε .
Let n* satisfy δn* = ε, then n* ln δ = ln ε and n* = .
Note that if ε > 1, then < 0. In that case, take N = 1.
Thus, we can choose
N = max {┌ ┐ , 1 } .
Here「.] is a ceiling function:「x] is equal to a smallest integer that is greater than or equal to x. For example,「2.3] = 3 and「2] = 2.
Indeed, let ε > 0.
If ε > 1, then N = 1 and δn < ε for any n > N .
If e < 1, then take N = ┌ ┐ . For any n > N we have
n > ┌ ┐ > =÷ n ln 6 < ln e =÷ 6n < e,
as required. Thus, the sequence converges to 0.
Note that N = max{「│ , 1}, where「.| is a floor function, also works.
2022-06-18