Lecture 1A: Sums and product (and Chapters 2.8, 2.9 and 2.11)
Summation operator Σ (the uppercase of sigma, 𝜎):
𝑛
∑ 𝑥𝑖 = 𝑥𝑚 + 𝑥𝑚+1 + ⋯ + 𝑥𝑛−1 + 𝑥𝑛
𝑖=𝑚
where





Σ is the summation sign;
𝑥𝑖 is its typical elements;
𝑖 the index of summation, i.e. the variable which is being summed;
𝑚 the starting point (or the lower limit of summation);
𝑛 the stopping point (or upper limit of summation).
This reads as “the sum from 𝑖 = 𝑚 to 𝑖 = 𝑛, of 𝑥𝑖 ”.
Properties of sums:
∑(𝑥𝑖 + 𝑦𝑖 ) = ∑ 𝑥𝑖 + ∑ 𝑦𝑖
Additivity:
Homogeneity: ∑ 𝑐𝑥𝑖 = 𝑐 ∑ 𝑥𝑖
𝑛+𝑚
Extending terms: ∑𝑛𝑖=1 𝑥𝑖 + ∑𝑛+𝑚
𝑖=𝑛+1 𝑥𝑖 = ∑𝑖=1 𝑥𝑖
𝑛
∑
Constant term:
𝑖=1 𝑐 = 𝑛𝑐
∑𝑛𝑖=1(𝑥𝑖 − 𝑥𝑖−1 ) = 𝑥𝑛 − 𝑥0
Telescopic series:
Double summation ∑ ∑ :
𝑛
𝑚
∑ ∑ 𝑥𝑖𝑗 = ⏟
𝑥11 + ⋯ + 𝑥1𝑚 + ⏟
𝑥21 + ⋯ + 𝑥2𝑚 + ⋯ + ⏟
𝑥𝑛1 + ⋯ + 𝑥𝑛𝑚
𝑖=1 𝑗=1
𝑖=1
𝑖=2
Product operator ∏ :
𝑛
∏ 𝑥𝑖 = 𝑥𝑚 × 𝑥𝑚+1 × ⋯ × 𝑥𝑛−1 × 𝑥𝑛
𝑖=𝑚
Properties:
𝑛+𝑚
Extending terms: ∏𝑛𝑖=1 𝑥𝑖 × ∑𝑛+𝑚
𝑖=𝑛+1 𝑥𝑖 = ∑𝑖=1 𝑥𝑖
𝑛
𝑛
𝑛
∑
∑
Constant terms:
𝑖=1 𝑐𝑥𝑖 = 𝑐
𝑖=1 𝑥𝑖
𝑖=𝑛
Lecture 5A: Indefinite integrals
The antiderivative (also called the indefinite integral) is the inverse operation of the derivative.
Example:
𝐹(𝑥) = 4𝑥 2 − 2𝑥 + 𝐶
𝑑𝑦
𝑑𝑥
(4𝑥 2 − 2𝑥 + 𝐶)
∫(8𝑥 − 2) 𝑑𝑥 + 𝐶
𝑑𝑦
𝑑𝑥
= 8𝑥 − 2
Definition:
We write 𝐹(𝑥) for the antiderivative of 𝑓(𝑥)
In formula
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐
where

𝑓(𝑥) is the integrand

𝑥 is the variable of integration

𝐶 is the constant of integration
Basic indefinite integrals:

𝑓(𝑥) = 𝑎: ∫ 𝑎𝑑𝑥 = 𝑎𝑥 + 𝐶
o example: 𝑓(𝑥) = 10 ⇒ ∫ 10𝑑𝑥 = 10𝑥 + 𝐶

𝑓(𝑥) = 𝑥 𝑎 : ∫ 𝑥 𝑎 𝑑𝑥 =
1
𝑎+1
𝑥 𝑎+1 + 𝐶 (with 𝑎 ≠ −1)
1
o example: 𝑓(𝑥) = 𝑥 4 ⇒ ∫ 𝑥 4 𝑑𝑥 = 𝑥 5 + 𝐶
5

1
1
𝑥
𝑥
𝑓(𝑥) = : ∫ 𝑑𝑥 = ln(|𝑥|) + 𝐶
o example: 𝑓(𝑥) =
1
10
⇒∫
1
10
𝑑𝑥 = ln(|10|) + 𝐶
1

𝑓(𝑥) = 𝑒 𝑎𝑥 : ∫ 𝑒 𝑎𝑥 𝑑𝑥 = 𝑒 𝑎𝑥 + 𝐶 (with 𝑎 ≠ 0)
𝑎
o example: 𝑓(𝑥) = 𝑒 𝑥 ⇒ ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶)

𝑓(𝑥) = 𝑎 𝑥 : ∫ 𝑎 𝑥 𝑑𝑥 =
1
ln(𝑎)
𝑎 𝑥 + 𝐶 (with 𝑎 > 0 𝑎𝑛𝑑 𝑎 ≠ 1)
o example: 𝑓(𝑥) = 10𝑥 ⇒ ∫ 10𝑥 𝑑𝑥 =
1
ln 10
10𝑥 + 𝐶)
Properties of indefinite integrals
𝑎 ⋅ 𝑓(𝑥):
∫(𝑎𝑓 (𝑥))𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥)𝑑𝑥
𝑓(𝑥) + 𝑎:
∫(𝑓(𝑥) + 𝑎)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + 𝑎𝑥
𝑓(𝑥) + 𝑔(𝑥): ∫(𝑓(𝑥) + 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥
Playing with constants:
Suppose that
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶
and
∫ 𝑔(𝑥)𝑑𝑥 = 𝐺(𝑥) + 𝐷
then:
∫(𝑓(𝑥) + 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥 =
= 𝐹(𝑥) + 𝐶 + 𝐺(𝑥) + 𝐷 =
= 𝐹(𝑥) + 𝐺(𝑥) + 𝐸
Lecture 9B: Numbers and units
An integer is any number in the set {… , −3, −2, −1, 0, 1, 2, 3, … }

A positive integer is any number in the set {1, 2, 3, … }

A negative integer is any number in the set {−1, −2, −3, … }

A nonnegative integers is {0}
A natural number is any integer greater or equal to than 0. Natural numbers begin at 1 and increment to infinity: {1, 2, 3, 4, 5, … }
One of the numbers 0, 1, 2, 3,…
Fractions:
2/3, 3 ¼, …
Decimal numbers:
1.82, 0.034, …

Mind the confusion between decimal point and decimal comma

Mind the optional use of a thousand separator (as in 1.234.567,8)

So what is 1.234.567? It is ambiguous
Leading and trailing zeros:

Are often skipped: 00123.45600=123.456

Not always in monetary amounts: $12.50, not $12.5

What is 3. and what is .3?
Negative numbers:
-3, -4 ½, -0.012, …

Mind that -4 ½ < 4

Mind that the minus sign is long (not -4 but −4)
Non-computed numbers
Scientific notation:
Decimal notation
0.001
0.01
0.1
1
10
100
1000
-10
Scientific notation
1 × 10−3
1 × 10−2
1 × 10−1
1 × 100
1 × 101
1 × 102
1 × 103
−1 × 101
Software
1E-3
1E-2
1E-1
1E0
1E1
1E2
1E3
-1E1
Words:
Value in
scientifi
c
notation
10^0
10^1
10^2
10^3
10^6
10^9
Value in propositional notation
1
10
100
1,000
1,000,000
Short scale (US, modern
UK)
Name
Logic
Long scale (continental, older UK)
one
ten
hundred
thousand
million
one
ten
hundred
thousand
million
1,000,000,000
billion
10^12
1,000,000,000,000
trillion
10^15
1,000,000,000,000,000
10^18
1,000,000,000,000,000,000
10^21
1,000,000,000,000,000,000,000
sextillion
10^24
1,000,000,000,000,000,000,000,0
00
septillion
Unit:
quadrillio
n
quintillion
1,000×1000^
1
1,000×1000^
2
1,000×1000^
3
1,000×1000^
4
1,000×1000^
5
1,000×1000^
6
1,000×1000^
7
Name
thousand
million
billion
Alternativ
e name
1,000,000^1
milliard
billiard
thousand
billion
trillion
thousand
trillion
quadrillio
n
Logic
1,000×1,000,000^
1
1,000,000^2
1,000×1,000,000^
2
1,000,000^3
trilliard
1,000×1,000,000^
3
1,000,000^4
Lecture 10A: Constrained optimization
Problem formulation
max
𝑓(𝑥, 𝑦)
{
s. t. 𝑔(𝑥, 𝑦) = 𝑐
Constrained optimization problem

𝑓(𝑥, 𝑦) is the objective function

𝑔(𝑥, 𝑦) = 𝑐 is the constraint

𝑥 and 𝑦 are the decision variables
Lagrange method
Introduce extra variable (Lagrange multiplier) 𝜆
Define new function Lagrangain) ℒ(𝑥, 𝑦, 𝜆) as follows
ℒ(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) − 𝜆(𝑔(𝑥, 𝑦) − 𝑐)
Find stationary points of the Lagrangian
Motivation:
Solutions of the constrained optimization are among the stationary points of ℒ. In other words: all stationary points
of ℒ are candidate solutions of the original constrained maximization problem
max
So, {
s. t.
𝑓(𝑥, 𝑦)
is in some sense equivalent to max ℒ(𝑥, 𝑦, 𝜆)
𝑔(𝑥, 𝑦) = 𝑐
Where we have defined ℒ(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) − 𝜆(𝑔(𝑥, 𝑦) − 𝑐)
We have written the constraint optimization with 2 decision variables as an unconstrained optimization problem
with 3 decision variables
1. Rewrite constrained optimization problem in 2D as an unconstrained optimization problem in 3D (from
max
𝑓(𝑥, 𝑦)
{
to max ℒ(𝑥, 𝑦, 𝜆))
s. t. 𝑔(𝑥, 𝑦) = 𝑐
𝜕ℒ
=0
2. Find 𝑥 and 𝑦 such that
𝜕𝑥
𝜕ℒ
𝜕𝑦
𝜕ℒ
=0
{𝜕𝜆 = 0
Constrained optimization:
Suppose:
The price of 1 unit of 𝐾 is 𝑘
The price of 1 unit of 𝐿 is 𝑙
The available budget is 𝑚
Budget line: 𝑘𝐾 + 𝑙𝐿 = 𝑚
Iso-production
line for 𝑞1
Iso-production
line for 𝑞2
𝐿
Optimum point on
line for optimum
production 𝑞 ∗
Budget line
𝐿∗
𝑞∗
𝑞2
𝑞1
𝐾∗
𝐾
Problem formulation
+local
min
local max
+
local min
+
Example
Given is the profit function
𝜋(𝑎, 𝑏) = 5𝑎2 + 12𝑏 2
and the capacity constraint
8𝑎 + 4𝑏 = 20
min/max
Write as {
subject to
2
2
Sol {max 5𝑎 + 12𝑏
s. t. 8𝑎 + 4𝑏 = 20
+local
max
max
𝑓(𝑥, 𝑦)
{
s. t. 𝑔(𝑥, 𝑦) = 𝑐
Constrained optimization problem

𝑓(𝑥, 𝑦) is the objective function

𝑔(𝑥, 𝑦) = 𝑐 is the constraint

𝑥 and 𝑦 are the decision variables
Visualization of 𝑧 = 𝑓(𝑥, 𝑦)

In 3D

As level curve
Lagrange method
Introduce extra variable (Lagrange multiplier) 𝜆
Define new function Lagrangain) ℒ(𝑥, 𝑦, 𝜆) as follows
ℒ(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) − 𝜆(𝑔(𝑥, 𝑦) − 𝑐)
Find stationary points of the Lagrangian
Example
Given is the profit function
𝜋(𝑎, 𝑏) = 5𝑎2 + 12𝑏 2
and the capacity constraint
8𝑎 + 4𝑏 = 20
Write the Lagrangian
Sol ℒ(𝑎, 𝑏, 𝜆) = 5𝑎2 + 12𝑏 2 − 𝜆(8𝑎 + 4𝑏 − 20)
Motivation:
Solutions of the constrained optimization are among
the stationary points of ℒ. In other words: all stationary
points of ℒ are candidate solutions of the original
constrained maximization problem
max
𝑓(𝑥, 𝑦)
So, {
is in some sense equivalent to
s. t. 𝑔(𝑥, 𝑦) = 𝑐
max ℒ(𝑥, 𝑦, 𝜆)
Where we have defined ℒ(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) −
𝜆(𝑔(𝑥, 𝑦) − 𝑐)
We have written the constraint optimization with 2
decision variables as an unconstrained optimization
problem with 3 decision variables
Full procedure
1. Rewrite constrained optimization problem in 2D as
an unconstrained optimization problem in 3D (from
max
𝑓(𝑥, 𝑦)
{
to max ℒ(𝑥, 𝑦, 𝜆))
s. t. 𝑔(𝑥, 𝑦) = 𝑐
𝜕ℒ
=0
2. Find 𝑥 and 𝑦 such that
𝜕𝑥
𝜕ℒ
𝜕𝑦
𝜕ℒ
=0
{𝜕𝜆 = 0
3. Check if the stationary points are indeed a
maximum
Note that you should not use (
(
𝜕2 ℒ
𝜕𝑥𝜕𝑦
2
) for
𝜕2 ℒ
𝜕𝑥 2
)(
𝜕2 ℒ
𝜕𝑦 2
)−
Example 1
Constrained problem:
max
𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2
{
s. t. 𝑔(𝑥, 𝑦) = 𝑥 2 + 𝑥𝑦 + 𝑦 2 = 3
Lagrangian:
ℒ(𝑥, 𝑦, 𝜆) = 𝑥 2 + 𝑦 2 − 𝜆(𝑥 2 + 𝑥𝑦 + 𝑦 2 − 3)
First-order conditions for stationary points:
𝜕ℒ
= 2𝑥 − 𝜆(2𝑥 + 𝑦) = 0
𝜕𝑥
𝜕ℒ
= 2𝑦 − 𝜆(𝑥 + 2𝑦) = 0
𝜕𝑦
𝜕ℒ
2
2
{ 𝜕𝜆 = −𝑥 − 𝑥𝑦 − 𝑦 + 3 = 0
(1)
(2)
(3)
the constrained case
Phase 1:
Solve (1) and (2) for 𝜆 and equate
2𝑥
2𝑥 + 𝑦
2𝑦
2𝑦 − 𝜆(𝑥 + 2𝑦) ⇒ 2𝑦 = 𝜆(𝑥 + 2𝑦) ⇒ 𝜆 =
2𝑦 + 𝑥
2𝑥
2𝑦
⋅
= 2𝑥(2𝑦 + 𝑥) = 2𝑦(2𝑥 + 4)
2𝑥 + 𝑦 2𝑦 + 𝑥
𝑥 2 = 𝑦2
𝑥 = ±𝑦
2𝑥 − 𝜆(2𝑥 + 𝑦) ⇒ 2𝑥 = 𝜆(2𝑥 + 𝑦) ⇒ 𝜆 =
Phase 2 (fill in constraints)
−𝑥 2 − 𝑥𝑦 − 𝑦 2 + 3 = 0
Let 𝑥 = 𝑦 in (3)
−𝑥 2 − 𝑥 2 − 𝑥 2 + 3 = 0
−3𝑥 2 = −3
𝑥2 = 1
𝑥 = ±1
(1,1), (−1,1)
Let 𝑥 = −1 ⇔ −𝑥 = 𝑦
−𝑥 2 − 𝑥(−𝑥)— 𝑥2 + 3 = 0
−𝑥 2 = −3
𝑥2 = 3
𝑥 = ±√3
(√3, −√3), (−√3, √3)
Phase 3
Candidates:
(1,1), (−1,1)
(√3, −√3), (−√3, √3)
Determine the nature of these points
(min/max/saddle) and then determine the maximum
𝑓(1,1) = 12 + 12 = 2
𝑓(−1, −1) = (−1)2 + (−1)2 = 2
2
2
𝑓(√3, −√3) = (√3) + (−√3) = 6
2
2
√−√3, √3 = −(√3) + (√3) = 6
So:

(√3, −√3) and (−√3, √3) are maximum points

And (1,1) and (−1, −1) are minimum points
Example 2:
Constrained problem:
max 𝑞(𝐾, 𝐿) = 𝐴𝐾 0.4 𝐿0.7 (𝐴 > 0)
{
s. t. 25𝐾 + 10𝐿 = 𝑚
(𝑚 > 0)
Lagrangian:
ℒ(𝐾, 𝐿, 𝜆) = 𝐴𝐾 0.4 𝐿0.7 − 𝜆(25𝐾 + 10𝐿 − 𝑚)
First-order conditions for stationary points:
𝜕ℒ
= 0.4𝐴𝐾 −0.6 𝐿0.7 − 25𝜆
𝜕𝐾
𝜕ℒ
= 0.7𝐴𝐾 0.4 𝐿0.3 − 10𝜆
𝜕𝐿
𝜕ℒ
{ 𝜕𝜆 = −(25𝐾 + 10𝐿 − 𝑚)
0.4𝐴𝐾 −0.6 𝐿0.7
(1)
(2)
(3)
0.7𝐴𝐾 0.4𝐿−0.3
(𝜆) =
=
25
10
Multiply both sides with 𝑘 0.6 (to get rid of 𝐾 −0.6 ) and with 𝐿0.3 (to get rid of 𝐿−0.3 )
0.4𝐴𝐿 0.7𝐴𝐾
10 × 0.4
=
⇒ 25𝐾 =
𝐿
25
10
0.7
10×0.4
0.7 𝑚
0.4 𝑚
𝐿 + 10𝐿 = 𝑚 ⇒
and 𝐾 =
0,7
1.1 10
1.1 25
𝜆 = ⋯ (not interesting)
Candidate maximum point: (𝐾, 𝐿) = (
0.4 𝑚 0.7 𝑚
,
1.1 25 1.1 10
)
Determine nature of the stationary point:
𝑚
Take 𝐾 = 0 ⇒ 𝐿 = and see that 𝑞 = 0
10
Likewise take 𝐿 = 0 ⇒ 𝑘 =
So for 0 <
0.4 𝑚
1.1 25
<
𝑚
25
𝑚
25
and 0 <
Therefore (𝐾, 𝐿) = ((
and see that 𝑞 = 0
0,7 𝑚
1.1 10
0.4 𝑚 0.7 𝑚
,
1.1 25 1.1 10
<
𝑚
10
, also 𝑞 > 0
) is a maximum
Different point on
the same budget
line with 𝑞 < 𝑞∗
𝐿
𝐿
Optimal CobbDouglas line
𝑞∗ = 𝐴𝐾 0.4𝐿0.7
Optimum point on
line for optimum
production 𝑞∗
0 .7 𝑚
Budget line
25 𝐾 + 10 𝐿 = 𝑚
Different point on
the same budget
line with 𝑞 < 𝑞∗
0 .4 𝑚
𝐾∗ = 1 .
𝐾