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Mathematics for IBA
Claudia Vogel
Europa-Universitat Viadrina
Winter Term 2009/2010
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 1 / 131
Course Details
Course Details
Lecture: Monday 9-13 GD HS 7
Tutorials: Wednesday and Thursday 18-20
Group A: Wednesday: GD 202, Thursday GD 302Group B: GD HS 7
Office Hours: Monday 14-15 HG 241
Email: [email protected]
Course Materials: website of Prof. Dr. Friedel Bolle
Exams:1 30.11.2009 11:00-13:002 22.03.2010 11:00-13:00
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 2 / 131
Course Details
Course Outline
Date Topic12.10.2009 Basic Knowledge
Linear Algebra I19.10.2009 Linear Algebra II
Interest Rates and Present Values26.10.2009 Single-Variable Functions02.11.2009 Integration of Single-Variable Functions
Multi-Variable Functions I09.11.2009 Multi-Variable Functions II
Game Theory I16.11.2009 Game Theory II23.11.2009 Game Theory III27.11.2009 Exam Preparation(11-13)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 3 / 131
Course Details
References
K. Sydsaeter, P. Hammond, Essential Mathematics for EconomicAnalysis, Prentice Hall, 2002
A. Chiang, K. Wainwright, Fundamental Methods of MathematicalEconomics, McGraw-Hill, 2005
Dixit, A., Skeath, S. (1999): Games of Strategy, New York, London:W.W. Norton & Company
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 4 / 131
Basic Knowledge
Outline0 Basic Knowledge1 Linear Algebra
Systems of Linear EquationsLinear equations of two unknownsGaussian Elimination
MatricesMatrix OperationsDeterminantsDefiniteness
2 Interest Rates and Present ValuesIntroductionCalculating InterestPresent Value
3 Single-Variable FunctionsIntroductionDifferentiationSingle-Variable OptimizationProperties of FunctionsIntegration
4 Multi-Variable FunctionsIntroductionDifferentiationMultivariable Optimization
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 5 / 131
Basic Knowledge
Fractions 1/2
a÷ b =a
b
numerator
denominator
ab =
a·/cb·/c = a
b
−a−b = (−a)·(−1)
(−b)·(−1) = ab
− ab = (−1) a
b = −ab
ac ±
bc = a±b
cab ±
cd = a·d±c·b
b·d
2115 = 7·/3
5·/3 = 75
−5−6 = 5
6
−35 = (−1) 3
5 = −35
53 + 13
3 = 183 = 6
35 + 1
6 = 3·6+1·55·6 = 23
30
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 6 / 131
Basic Knowledge
Fractions 2/2
a + bc = a·c+b
c
a · bc = a·bc
ab ·
cd = a·c
b·dab ÷
cd = a
b ·dc = a·d
b·c
5 + 35 = 5·5+3
5 = 285
47 ·
58 = 4·5
7·8 = 514
7 · 35 = 21
538 ÷
614 = 3
8 ·146 = 7
8
WRONG!2/x + 3y
/xy=
2 + 3/y
/y=
2 + 3
1= 5
WRONG!x
x2 + 2x=
/x
x/2+
/x
2/x=
1
x+
1
2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 7 / 131
Basic Knowledge
Rules of Algebra 1/2
a + b = b + a
(a + b) + c = a + (b + c)
a + 0 = a
a + (−a) = 0
ab = ba
(ab)c = a(bc)
1 · a = a
aa−1 = 1 for a 6= 0
(−a)b = a(−b) = −ab(−a)(−b) = ab
a(b + c) = ab + ac
(a + b)c = ac + bc
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 8 / 131
Basic Knowledge
Rules of Algebra 2/2
Combinations of Algebraic Rules:
a(b − c) = a[b + (−c)] = ab + a(−c) = ab − ac
x(a + b − c + d) = xa + xb − xc + xd
(a + b)(c + d) = ac + ad + bc + bd
Quadratic Identities:
(a + b)2 = a2 + 2ab + b2
(a− b)2 = a2 − 2ab + b2
(a + b)(a− b) = a2 − b2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 9 / 131
Basic Knowledge
Integer Powers
am · an = am+n
am
an = am−n
an · bn = (ab)n
an
bn =(ab
)n(am)n = (an)m = am·n
1an = a−n(ab
)−n=(ba
)n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 10 / 131
Basic Knowledge
Fractional Powers
n√a = a
1n
n√am = a
mn
n√a · n√b = n√ab
n√an√b = n
√ab
m√
n√a = n
√m√a = m·n
√a
1n√a = a−
1n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 11 / 131
Basic Knowledge
Logarithms
ln (u · v) = ln u + ln v
ln(uv
)= ln u − ln v
ln ur = r · ln uln n√u = ln u
1n = 1
n ln u
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 12 / 131
Basic Knowledge
Equations 1/2
T1 = T2 ⇔ T1 ± T3 = T2 ± T3
T1 = T2 ⇔ T1 · T3 = T2 · T3 (T3 6= 0)
T1 = T2 ⇔ T1T3
= T2T3
(T3 6= 0)
T1 = T2 ⇔ aT1 = aT2 (a > 0, a 6= 1)
T1 = T2 ⇔ lnT1 = lnT2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 13 / 131
Basic Knowledge
Equations
If and only if, n is an odd number, then
T1 = T2 ⇔ T n1 = T n
2
⇔ n√T1 = n
√T2
If n is an even number, check the solutions as signs may be missing orwrong solutions appear:
T n1 = T n
2 ⇔ T1 = T2
or T1 = −T2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 14 / 131
Basic Knowledge
Quadratic Equations 1/4
Case 1:
ax2 + c = 0
x2 = −c
a
x1,2 = ±√−c
a
Example: x2 − 16 = 9
x2 = 25
x1 = −5 x2 = 5
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 15 / 131
Basic Knowledge
Quadratic Equations 2/4
Case 2:
ax2 + bx = 0
x (ax + b) = 0
x1 = 0 x2 = −b
a
Example: 15x − x2 = 0
x (15− x) = 0
x1 = 0 x2 = 15
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 16 / 131
Basic Knowledge
Quadratic Equations 3/4
Case 3:
ax2 + bx + c = 0
x1,2 =−b ±
√b2 − 4ac
2a
Example: 3x2 + 45x − 48 = 0
x1,2 =−45±
√2025− 4 · 3 · (−48)
2 · 3=−45± 51
6
x1 =−96
6= −16
x2 =6
6= 1
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 17 / 131
Basic Knowledge
Quadratic Equations 4/4
Case 4:
x2 + px + q = 0
x1,2 = −p
2±√(p
2
)2− q
Example: 3x2 + 45x − 48 = 0
x2 + 15x − 16 = 0
x1,2 = −15
2±√
225
4+ 16
= −15
2± 17
2x1 = −16
x2 = 1
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 18 / 131
Linear Algebra
Outline0 Basic Knowledge1 Linear Algebra
Systems of Linear EquationsLinear equations of two unknownsGaussian Elimination
MatricesMatrix OperationsDeterminantsDefiniteness
2 Interest Rates and Present ValuesIntroductionCalculating InterestPresent Value
3 Single-Variable FunctionsIntroductionDifferentiationSingle-Variable OptimizationProperties of FunctionsIntegration
4 Multi-Variable FunctionsIntroductionDifferentiationMultivariable Optimization
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 19 / 131
Linear Algebra Systems of Linear Equations
Solving by substitution
(1) Solve one of the equations (I or II) for one of the variables y (or x) interms of the other.
(2) Substitute the resulting term for y (or x) into the other equation (II orI) and solve for the only existing variable.
(3) Insert the solution of (2) into the other equation.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 20 / 131
Linear Algebra Systems of Linear Equations
Solving by equalization
(1) Solve both equations (I and II) for the same variable (x or y).
(2) Equalize both terms and solve for the only variable.
(3) Insert the solution in one of the two equations and solve for the othervariable.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 21 / 131
Linear Algebra Systems of Linear Equations
Solving by Addition
(1) Add (or subtract) a multiple of one equation to (from) the otherequation, so that one variable is eliminated.
(2) Solve the resulting equation.
(3) Insert the solution of (2) into one of the original equations.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 22 / 131
Linear Algebra Systems of Linear Equations
Gaussian Elimination
Idea: Rearrange a system of equations into the following form byeliminating variables.
Elimination of variables by:
1 Adding a row or a multiple of a row to another row
2 interchanging rows
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 23 / 131
Linear Algebra Systems of Linear Equations
Systems of Equations in Matrix Form
a11X1 + a12X2 + ...+ a1nXn = Y1
a21X1 + a22X2 + ...+ a2nXn = Y2
...
am1X1 + am2X2 + ...+ amnXn = Ym
Define: A =
a11 .. a1j .. a1n
......
...a12 .. a22 .. a2n
......
...am1 .. amj .. amn
; x =
x1
x2...xm
; b =
b1
b2...bm
Ax = b
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 24 / 131
Linear Algebra Systems of Linear Equations
Matrices
A matrix is a rectangular array of numbers considered as an entity.
(m × n)-Matrix:
A =
a11 a12 ... a1n
......
...a12 a22 ... a2n
......
...am1 am2 ... amn
row vector:(1× n)-Matrix
b = (b1...bj ...bn)
column vector:(m × 1)-Matrix
c =
c1...ci...cm
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 25 / 131
Linear Algebra Systems of Linear Equations
Square-Matrix
A matrix with equal numbers of rows and columns, m=n is calledsquare-matrix.
a11 ... a1n...
. . ....
an1 ... ann
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 26 / 131
Linear Algebra Systems of Linear Equations
Identity Matrix
A diagonal matrix where all elements on the main diagonal equal 1 iscalled Identity Matrix.
I =
1 0 00 1 00 0 1
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 27 / 131
Linear Algebra Systems of Linear Equations
Triangular Matrix
A square matrix, where all elements on one side of the main diagonal arezero is called triangular matrix.
upper triangular matrix lower triangular matrixa11 a12 a13 ... a1n
0 a22 a23 ... a2n
0 0 a33 ... a3n...
......
......
0 0 0 ... ann
a11 0 0 ... 0a21 a22 0 ... 0a31 a32 a33 ... 0
......
......
...an1 an2 an3 ... ann
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 28 / 131
Linear Algebra Matrices
Equality
Two matrices A and B are equal if
(1) they have the same order and
(2) if their corresponding entries are equal
(1) A = (aij)m×n and B = (bij)m×n
(2) aij = bij for all i = 1, 2, ...,m and j = 1, 2, ..., n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 29 / 131
Linear Algebra Matrices
The Transpose
A =
a11 .. a1j .. a1n
......
...ai1 .. aij .. ain...
......
am1 .. amj .. amn
→ A′ =
a11 .. ai1 .. am1
......
...a1j .. aij .. amj...
......
a1n .. ain .. amn
A matrix is called symmetric if A = A’.
Rules for Transposition:
(A′)′ = A
(A + b)′ = A′ + B ′(αA)′ = αA′
(AB)′ = B ′A′
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 30 / 131
Linear Algebra Matrices
Addition/Subtraction
Addition and Substraction are only possible for matrices of the samedimension.
a11 .. a1j .. a1n...
......
ai1 .. aij .. ain...
......
am1 .. amj .. amn
±
b11 .. b1j .. b1n...
......
bi1 .. bij .. bin...
......
bm1 .. bmj .. bmn
=
a11 ± b11 .. a1j ± b1j .. a1n ± b1n
......
...ai1 ± bi1 .. aij ± bij .. ain ± bin
......
...am1 ± bm1 .. amj ± bmj .. amn ± bmn
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 31 / 131
Linear Algebra Matrices
Multiplication with a scalar
α · A = α ·
a11 .. a1j .. a1n
......
...ai1 .. aij .. ain...
......
am1 .. amj .. amn
=
α · a11 .. α · a1j .. α · a1n
......
...α · ai1 .. α · aij .. α · ain
......
...α · am1 .. α · amj .. α · amn
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 32 / 131
Linear Algebra Matrices
Rules for Matrix Operations:
A + B = B + A
(A + B) + C = A + (B + C )
A + 0 = A
A + (−A) = 0
(λ1 + λ2)A = λ1A + λ2A
λ (A + B) = λA + λB
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 33 / 131
Linear Algebra Matrices
Matrix Multiplication 1/3
(1) two linear equation systems:
(i) z1 = a11y1 + a12y2 + a13y3 (ii) y1 = b11x1 + b12x2
z2 = a21y1 + a22y2 + a23y3 y2 = b21x1 + b22x2
y3 = b31x1 + b32x2
(2) take expressions for y1, y2 and y3 from (II) and insert in (I):
z1 = a11 (b11x1 + b12x2) + a12 (b21x1 + b22x2) + a13 (b31x1 + b32x2)
z2 = a21 (b11x1 + b12x2) + a22 (b21x1 + b22x2) + a23 (b31x1 + b32x2)
(3) rearranging terms:
z1 = (a11b11 + a12b21 + a13b31) x1 + (a11b12 + a12b22 + a13b23) x2
z2 = (a21b11 + a22b21 + a23b31) x1 + (a21b12 + a22b22 + a23b32) x2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 34 / 131
Linear Algebra Matrices
Matrix Multiplication 2/3
(1) in terms of matrices:(i)
(z1
z2
)=
(a11 a12 a13
a21 a22 a23
) Y1
Y2
Y3
(ii) Y1
Y2
Y3
=
b11 b12
b21 b22
b31 b32
X1
X2
X3
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 35 / 131
Linear Algebra Matrices
Matrix Multiplication 3/3
(2) insert the expression for Y from (ii) into (i):(z1
z2
)=
(a11 a12 a13
a21 a22 a23
) b11 b12
b21 b22
b31 b32
X1
X2
X3
(3) coefficient matrix C as the matrix product of A and B:
C =
(a11b11 + a12b21 + a13b31 a11b12 + a12b22 + a13b32
a21b11 + a22b21 + a23b31 a21b12 + a22b22 + a23b32
)If A is a (m × n)-Matrix and B is a (n × p)-Matrix then their productC = AB is a (m × p)-Matrix where each component cij is defined asfollows:
cij =n∑
r=1
airbrj
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 36 / 131
Linear Algebra Matrices
Falk-Schema
b11 . . . b1j . . . b1p...
... b11dotsbn1 . . . bnj . . . bnp
a11 . . . a1n c11 . . . . . . c1p...
... · · ·...
...ai1 . . . ain . . . cij . . ....
... · · ·...
...am1 . . . amn cm1 . . . . . . cmp
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 37 / 131
Linear Algebra Matrices
Rules for Multiplication
AB is defined only if the number of columns in A is equal to thenumber of rows in B. The resulting matrix has the same number ofrows like A and the same number of columns like B.
A · B 6= B · A(A · B) · C = A · (B · C )
A2 = A · A(A + B)C = AC + BC
λ (AB) = (λA)B = A (λB)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 38 / 131
Linear Algebra Matrices
Determinants of 1× 1- and 2× 2-Matrices
A determinant detA or |A| assigns a real number to each square matrix.
How to find determinants for (n × n)-matrices:
n=1:
|A| = |a11| = a11
n=2: ∣∣∣∣ a11 a12
a21 a22
∣∣∣∣ = a11 · a22 − a12 · a21
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 39 / 131
Linear Algebra Matrices
Determinants of 3× 3-Matrices
n=3: Rule of Sarrus:
∣∣∣∣∣∣a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣ −→a11 a12 a13 a11 a12
a21 a22 a23 a21 a22
a31 a32 a33 a31 a32
= a11 · a22 · a33 + a12 · a23 · a31 + a13 · a21 · a32
−a11 · a23 · a32 − a12 · a21 · a33 − a13 · a22 · a31
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 40 / 131
Linear Algebra Matrices
Determinants of larger Matrices
n ≥ 3: Rule of Laplace: The determinant can be found either bydevelopment for row i
|A| =n∑
j=1
(−1)i+j · aij · |Aij |
or by development for column j
|A| =n∑
i=1
(−1)i+j · aij · |Aij |
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 41 / 131
Linear Algebra Matrices
Characteristics of Determinants
The determinant |A| of a matrix A equals the determinant |A′| of thetranspose matrix A′.
For two matrices A and B of the same order holds: |A| · |B| = |AB|A determinant has the value of 0 if two rows (columns) are the same,or if one row (column) is a multiple of another row (column).
The determinant of a triangular matrix is the product of the elementsin the main diagonal.
∣∣∣∣∣∣∣∣∣a11 a12 ... a1n
0 a22 ... a2n...
......
...0 0 ... ann
∣∣∣∣∣∣∣∣∣ = a11 · a22 · ... · ann
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 42 / 131
Linear Algebra Matrices
Positive/ Negative Definiteness
Each Matrix consists of several submatrices. The structure of thedeterminants of these submatrices helps to identify the definiteness of amatrix:
The matrix is
negative definite, if the determinants have alternating signs, startingwith minus.
positive definite, if all determinants have positive signs.
semi-definite, if in one of the above structures some of thedeterminants are zero.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 43 / 131
Interest Rates and Present Values
Outline0 Basic Knowledge1 Linear Algebra
Systems of Linear EquationsLinear equations of two unknownsGaussian Elimination
MatricesMatrix OperationsDeterminantsDefiniteness
2 Interest Rates and Present ValuesIntroductionCalculating InterestPresent Value
3 Single-Variable FunctionsIntroductionDifferentiationSingle-Variable OptimizationProperties of FunctionsIntegration
4 Multi-Variable FunctionsIntroductionDifferentiationMultivariable Optimization
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 44 / 131
Interest Rates and Present Values Introduction
Introduction
If you deposit money in a bank account, you are paid for leaving thecontrol over the money to the bank.
If you raise a credit with a bank, you get control over the bank’s moneyand the bank asks financial compensation.
The price for money, that is temporarily left to someone else is calledinterest. Three factors influence the amount of interest:
the amount of the capital borrowed or deposited
time for which the money is left to someone else
the interest rate
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 45 / 131
Interest Rates and Present Values Introduction
Interest Rates
The interest loan denotes that percentage of the capital, that has to bepaid as interest and is denoted by p.
Usually the interest rate is used for calculations instead of the interest loan.
r =p
100= 0.01p
An interest loan of 4.7 percent means that for 100 Euro of capital 4.70Euro have to be paid as interest. So the interest rate
r =4.7
100= 0.047
denotes the price you get (have to pay) for depositing (borrowing) 1 Euro.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 46 / 131
Interest Rates and Present Values Introduction
Symbols
The following symbols are usually when talking about interest problems:
p interest loan (in percent)
r interest rate(r = p
100
)K0 initial capital
Kn capital at the end of period n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 47 / 131
Interest Rates and Present Values Calculating Interest
Simple Interest
With simple interest, interest is disbursed after every period. The initialcapital in the account, for which interest is payed, stays the same all thetime.
K1 = K0 + rK0 = K0 (1 + r)
K2 = K1 + rK0 = K0 (1 + r) + rK0 = K0 (1 + 2r)
...
Kn = K0 (1 + nr)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 48 / 131
Interest Rates and Present Values Calculating Interest
Compound yearly Interest
If interest payments from one period are added to the initial capital andthen interest is charged for the whole sum, it is called compound interest.
K1 = K0 + rK0 = K0 (1 + r)
K2 = K1 + rK1 = K1 (1 + r) = K0 (1 + r)2
...
Kn = K0 (1 + r)n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 49 / 131
Interest Rates and Present Values Calculating Interest
Compound periodical interest
Compound interest can also be payed for periods that are shorter than oneyear. If interest is credited to an account already after 1
m years, then theinital capital K0 increases after n years at an interest rate r to:
Kn = K0
(1 +
r
m
)nm
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 50 / 131
Interest Rates and Present Values Calculating Interest
Continuous Compounding
If the number of periods m becomes larger, the time intervals betweeninterest payments become smaller. For m→∞ interest is payed everymoment and is then interest-bearing as well. This is called continuouscompounding.
Kn = K0ern
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 51 / 131
Interest Rates and Present Values Present Value
Present Value
In order to compare payments in different time periods one has to comparetheir values in the same time period.
Often one chooses the time period n = 0 and therefore calculates thePresent Value of the payments. But also every other point in time can bechosen.
If the interest or discount rate is p % per year and r = p/100, an amountK that is payable in t years has the Present Value (or Present DiscountedValue, PDV):
K (1 + r)−t with annual interest payments
Ke−rt with continuous compounding of interest
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 52 / 131
Single-Variable Functions
Outline0 Basic Knowledge1 Linear Algebra
Systems of Linear EquationsLinear equations of two unknownsGaussian Elimination
MatricesMatrix OperationsDeterminantsDefiniteness
2 Interest Rates and Present ValuesIntroductionCalculating InterestPresent Value
3 Single-Variable FunctionsIntroductionDifferentiationSingle-Variable OptimizationProperties of FunctionsIntegration
4 Multi-Variable FunctionsIntroductionDifferentiationMultivariable Optimization
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 53 / 131
Single-Variable Functions Introduction
Basics
One variable is a function of another if the first depends upon the second.
A function of a real variable x with domain D is a rule that assigns aunique real number to each number x in D. As x varies over the wholedomain, the set of all possible resulting values f (x) is called the range of f.
y = f (x), y = y(x)f : x 7→ y , f : x 7→ f (x)
x is called argument or independent variable.
y is called functional value or dependent variable.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 54 / 131
Single-Variable Functions Introduction
Describing functions 1/2
(1) General Formula
f (x) = x2 − 4x + 3
f (x) = sign(x) =
1 for x > 00 for x = 0−1 for x > 0
(2) Value Table
x 0 1 2 3 4
f (x) = x2 − 4x + 3 3 0 −1 0 3
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 55 / 131
Single-Variable Functions Introduction
Describing functions 2/2
(3) Graph
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 56 / 131
Single-Variable Functions Introduction
Linear Functions
General Formula: f (x) = mx + n
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 57 / 131
Single-Variable Functions Introduction
Power Functions
General Formula: f (x) = xn
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 58 / 131
Single-Variable Functions Introduction
Root Functions
General Formula: f (x) = xn where n = pq
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 59 / 131
Single-Variable Functions Introduction
Exponential Functions
General Formula: f (x) = ax where a ∈ R, a > 0, a 6= 1
Special Case: f (x) = ex
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 60 / 131
Single-Variable Functions Introduction
Logarithmic Functions
General Formula: f (x) = loga x where a ∈ R, a > 0, a 6= 1
Special Case: f (x) = ln x
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 61 / 131
Single-Variable Functions Introduction
Shifting the graph of y = f (x)
(1) If y = f (x) is replaced by y = f (x) + c , the graph is moved upwardsby c units if c > 0 (downwards if c < 0)
(2) If y = f (x) is replaced by y = f (x + c), the graph is moved c units tothe left if c > 0 (to the right if c < 0)
(3) If y = f (x) is replaced by y = cf (x), the graph is stretched verticallyif c > 0 (stretched vertically and reflected about the x-axis if c < 0)
(4) If y = f (x) is replaced by y = f (−x), the graph is reflected about they-axis.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 62 / 131
Single-Variable Functions Introduction
Composite Functions
If y is a function of u, and u is a function of x , then y can be regarded asa function of x . y is called a composite function of x .
y = f (u) and u = g(x)
y = f (g(x))
g(x) is called interior function, f exterior function.
(f ◦ g)(x) = f (g(x)) (g ◦ f )(x) = g(f (x))
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 63 / 131
Single-Variable Functions Introduction
Inverse Functions
Let f be a function with domain A and range B. If and only if f isone-to-one, it has an inverse function g with domain B and range A. Thefunction g is given by the following rule :For each y ∈ B the value g(y) is the unique number x in A such thatf (x) = y .
Then
g(y) = x ⇔ y = f (x) (x ∈ A, y ∈ B)
When two functions f and g are inverses of each other, then the graphs ofy = f (x) and y = g(x) are symmetric about the line y = x
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 64 / 131
Single-Variable Functions Introduction
Limits 1/2
If the functional values of the function f (x) move towards a certain valuea, while the arguments x move closer to a value x0 than a is called thelimit of the function f (x) for point x0. One writes lim
x→x0
f (x) = a.
If a functions f (x) has unlimitedly rising (falling) functional values if xmoves towards x0, than the function has the improper limit +∞ (−∞).
It is limx→∞
f (x) = a
(lim
x→−∞f (x) = a
), if for unlimited rising (falling) x the
values of f (x) moves towards the value a.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 65 / 131
Single-Variable Functions Introduction
Limits 2/2
f (x) has the left (right) limit a at point x0, if the functional value f (x)move towards a while the arguments move towards x0 from left (right).
One writes limx→x0−
f (x) = a
(lim
x→x0+f (x) = a
)Example: f (x) = 1
x
limx→x0−
f (x) = −∞
limx→x0+
f (x) = +∞
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 66 / 131
Single-Variable Functions Differentiation
An Economic Example
A firm produces a good at total costs K , that depend on the producedquantities x . The cost function K = K (x) describes the relationshipbetween K and x . How do the costs change, if the quantities change?
In general: If the quantity changes from x0 by ∆x to x0 + ∆x , the costschange from K (x0) to K (x0 + ∆x), i.e. the costs change byK (x0 + ∆x)− K (x0) = ∆K
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 67 / 131
Single-Variable Functions Differentiation
Example: K = 10√x + 100
x K = 10√x + 100
0 100
1 110
2 114,14
3 117,32
4 120
9 130
16 140
25 150
49 170
64 180
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 68 / 131
Single-Variable Functions Differentiation
The change in costs ∆K not only depends on the change in quantities∆x , but also on the initial quantity:
x K = 10√x + 100
0 100
1 110
2 114,14
3 117,32
4 120
9 130
16 140
25 150
49 170
64 180
Example:
change from to by ∆x = 1x = 1 x = 2 → ∆K = 4, 14x = 2 x = 3 → ∆K = 3, 18
change from to by ∆x = 15x = 1 x = 16 → ∆K = 30x = 49 x = 64 → ∆K = 10
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 69 / 131
Single-Variable Functions Differentiation
initialquantity
increasedto
∆x ∆K ∆K∆x
1 2 1 4,14 4,14
1 3 2 7,32 3,66
1 4 3 10 3,33
1 9 8 20 2,5
1 16 15 30 2
1 64 63 70 1,11
0 9 9 30 3,33
16 25 9 10 1,11
49 64 15 10 0,67
costs of one additional unit:
∆K
∆x=
K (x + ∆x)− K (x)
∆x
The difference quotient ∆K∆x
shows the average rise of thecost function related to achange in the quantity x.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 70 / 131
Single-Variable Functions Differentiation
The costs of an additionalproduction unit independent ofthe size of the productionchange is found for ∆x → 0.
The limit
lim∆x→0
∆K
∆x=
dK
dx
describes the relationshipbetween the change in costsand the infinitesimal smallquantity change at point x.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 71 / 131
Single-Variable Functions Differentiation
The first derivative of a function
There is a continuous function y=f(x) with two points x1 and x2 andf (x1) = y1 and f (x2) = y2.
Difference of the independent variables:
∆x = x2 − x1
Difference of the function:
∆y = y2 − y1 = f (x2)− f (x1) = f (x1 + ∆x)− f (x1)
Changing the independent variable from x1 by ∆x to x2 changes thedependent variable by ∆y .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 72 / 131
Single-Variable Functions Differentiation
The average slope of the functionbetween x and x + ∆x is describedby the difference quotient
∆y
∆x=
f (x + ∆x)− f (x)
∆x
The limit
lim∆x→0
∆y
∆x= lim
∆x→0
f (x + ∆x)− f (x)
∆x
is called first order derivative and isindicated by f ′(x) or df (x)
dx
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 73 / 131
Single-Variable Functions Differentiation
Rules for Differentiation 1/3
f (x) = c
f (x) = xn
f (x) = c · g(x)
y(x) = g(x)± f (x)
dfdx = 0dfdx = nxn−1
dfdx = c · dgdxdydx = dg
dx ±dfdx
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 74 / 131
Single-Variable Functions Differentiation
Rules for Differentiation 2/3
Product Rule
y(x) = f (x) · g(x) dydx = df
dx · g(x) + dgdx · f (x)
short: [uv ] = u′v + uv ′
Quotient Rule
y(x) = f (x)g(x)
dydx =
dfdx·g(x)−f (x)· dg
dx(g(x))2
short:[uv
]= u′v−uv ′
v2
Chain Rule
y(x) = f (g(x)) dydx = df
dx ·dgdx
short: [u (v(x))] = u′(v) · v ′
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 75 / 131
Single-Variable Functions Differentiation
Special Functions
f (x) = ex
f (x) = ax = e ln ax = ex ln a
f (x) = ln x
dfdx = ex
dfdx = ex ln a · ln a = ax ln adfdx = 1
x
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 76 / 131
Single-Variable Functions Differentiation
Higher-order Derivatives
The first-order derivative of the function y = f (x), y ′ = f ′(x) is adifferentiable function.Then the first-order derivative of y ′ = f ′(x) is called second-orderderivative of y = f (x) and is described by
f ′′(x) or d2f (x)dx2 .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 77 / 131
Single-Variable Functions Differentiation
Implicit Differentiation
y = f (x) and x = g(y) are the explicit form of a certain function.f (x , y) ≡ 0 is the implicit form.From the implicit form one cannot necessarily identify the dependent andindependent variable.
If two variables x and y are related by and equation, to find y ′:
(1) Differentiate each side of the equation w.r.t. x , considering y as afunction of x .
(2) Solve the resulting equation for y ′.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 78 / 131
Single-Variable Functions Differentiation
The Differential of a Function
Consider a differentiable function f (x), and let dx denote an arbitrarychange in the variable x . The expression f ′(x)dx is called the differentialof y = f (x), and it is denoted by dy (or df ), so that:
dy = f ′(x)dx
If x changes by dx , then the corresponding change in y = f (x) is :
∆y = f (x + dx)− f (x)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 79 / 131
Single-Variable Functions Single-Variable Optimization
Single-Variable Optimization
A main task in economics is to find solutions for optimization problems.
Some examples of such problems:
Maximization of profits
Production of goods at minimum average costs
Maximization of sales, turnover or profitability of equity
Maximization of utility when analyzing household behavior
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 80 / 131
Single-Variable Functions Single-Variable Optimization
Definition of Optimium
Formally, a maximum (minimum) is a point, where the function takes hishighest (lowest) value.
If this is only true for a part of the domain it is called local maximum(minimum).
If there is no higher (lower) point in the domain, we have an absolutemaximum (minimum).
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 81 / 131
Single-Variable Functions Single-Variable Optimization
Conditions for Optimum
Necessary condition for extreme points:
f ′(x) = 0
Sufficient condition:f ′′(x0) < 0→ Maximum
f ′′(x0) > 0→ Minimum
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 82 / 131
Single-Variable Functions Single-Variable Optimization
Boundaries 1/2
Economic functions usually have some constraints:
Production capacity limits the produced quantities.
It is not possible to produce negative amounts of goods.
Investments are limited by available financial resources.
Such constraints result in bounded domains for the respective functions. Afunctions is said to have a lower boundary if there is a number m, suchthat f (x) ≥ m, and an upper boundary if f (x) ≤ m
Extreme points lying at boundaries are called boundary maximum orboundary minimum.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 83 / 131
Single-Variable Functions Single-Variable Optimization
Boundaries 2/2
x1, x4 boundaries of the domain
x1 local boundary minimumx2 local maximumx3 absolute local minimumx4 absolute boundary maximum
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 84 / 131
Single-Variable Functions Single-Variable Optimization
Search for Maxima/Minima
Problem: Find the maximum and minimum values of a differentiablefunction f defined on a closed, bounded interval [a,b]
Solution:
(1) Find all stationary points of f in (a,b) - that is, find all points x in(a,b) that satisfy the equation f ′(x) = 0
(2) Evaluate f at the end points a and b on the interval and also at allstationary points.
(3) The largest function value found in (2) is the maximum value, andthe smallest function value is the minimum value of f in [a,b]
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 85 / 131
Single-Variable Functions Properties of Functions
Zero of the function
The argument x0 for which f (x0) = 0 holds is called zero of a function.
The zero of the function is the intersection of the function with the x-axis.
To find the zero, one has to solve the equation f(x)=0 for x.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 86 / 131
Single-Variable Functions Properties of Functions
Continuity
f (x) is called continuous at point x = x0 if limx→x0
f (x) = f (x0) holds.
f (x) is continuous in the interval (a,b), if f(x) is continuous at every pointof the interval.
Graphically a function is continuous if its graph is connected and does nothave any breaks.
Example: f (x) = x f (x) = 1x
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 87 / 131
Single-Variable Functions Properties of Functions
Limits
For some functions, that can be seen as a quotient of two other functionsit is possible that the limit is an indefinite term like
0
0,∞∞, 0 · ∞, 0 · −∞
In such cases we apply L’Hospital’s Rule:
limx→∞
f (x)
g(x)=
f ′(x)
g ′(x)= L
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 88 / 131
Single-Variable Functions Properties of Functions
Monotony 1/2
A function is monotonically increasing [decreasing] if for any twox1, x2 ∈ D the following holds:
x1 < x2 ⇒ f (x1) ≤ f (x2) [f (x1) ≥ f (x2)]
If x1 < x2 ⇒ f (x1) < f (x2) [f (x1) > f (x2)] holds, the function strictlymonoton increasing [decreasing].
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 89 / 131
Single-Variable Functions Properties of Functions
Monotony 2/2
Monotony changes in extreme points.
Criteria for monotony:f ′(x) ≥ 0 monotonically increasingf ′(x) > 0 strictly monotonically increasing
f ′(x) ≤ 0 monotonically decreasingf ′(x) < 0 strictly monotonically decreasing
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 90 / 131
Single-Variable Functions Properties of Functions
Curvature 1/2
f (x) is convex (concave) in an interval if for any two arguments x1, x2 theline between the points (x1, f (x1)), (x2, f (x2)) within the interval (x1, x2)is always above (below the graph of f(x).
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 91 / 131
Single-Variable Functions Properties of Functions
Curvature 2/2
Curvature changes in inflection points.
Conditions for an inflection point:f ′′(x) = 0f ′′′(x) 6= 0
Criteria for curvaturef ′′(x) ≥ 0 convexf ′′(x) > 0 strictly convex
f ′′(x) ≤ 0 concavef ′′(x) < 0 strictly concave
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 92 / 131
Single-Variable Functions Properties of Functions
Summary of Monotony and Curvature
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 93 / 131
Single-Variable Functions Properties of Functions
Curve Sketching
The curve sketching process offers the possibility of a more detailedanalysis of a function. Therefor you follow the below steps.
(1) Domain
(2) Zeros of the function
(3) Points of discontinuity
(4) Relative extreme points
(5) Inflection points
(6) Monotony
(7) Curvature
(8) Behavior towards infinity
(9) Graph of the function
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 94 / 131
Single-Variable Functions Integration
Definite Integrals
A major task of integration is to determine the area under a graph of acontinuous and non-negative function.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 95 / 131
Single-Variable Functions Integration
Lower Sum
The sum A1 + A2 + A3 is called lower sum of area A, because always thelowest value of the function determines the height of the rectangle. Thearea determined by this lower sum is always smaller than the are betweenthe graph and the x-axis.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 96 / 131
Single-Variable Functions Integration
Upper Sum
The sum B1 + B2 + B3 is then called upper sum. This area is alwaysgreater than the respective are between the graph and the x-axis.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 97 / 131
Single-Variable Functions Integration
Summary
lower sum ≤ area under the graph ≤ upper sum
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 98 / 131
Single-Variable Functions Integration
Approximation 1/2
The difference between upper and lower sum decreases if ∆x becomessmaller.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 99 / 131
Single-Variable Functions Integration
Approximation 2/2
If ∆x becomes infinitesimal small, the difference between upper and lowersum also becomes infinitesimal small, so that:
lower sum = upper sum for ∆x → 0
The inequation
lower sum ≤ area A ≤ upper sum
then becomes the equation
lower sum = area A = upper sum for ∆x → 0
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 100 / 131
Single-Variable Functions Integration
The Integrl of a function
The common limit between upper and lower sum for an infinitesimal smallsegmentation of the interval [a, b] is called definite interval over [a, b]. Itcan also be written as:
A = limn→∞
n∑i=1
f (x i ) ∆x = limn→∞
n∑i=1
f (xi ) ∆x for ∆x =b − a
n
The limit can also be written by using the integral sign:
A =
∫ b
af (x)dx
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 101 / 131
Single-Variable Functions Integration
Properties of certain integrals
∫ ba f (x)dx = −
∫ ab f (x)dx∫ a
a f (x)dx = 0∫ ba αf (x)dx = α
∫ ba f (x)dx∫ b
a f (x)dx =∫ ca f (x)dx +
∫ bc f (x)dx
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 102 / 131
Single-Variable Functions Integration
Indefinite Integrals
Another major task of integration is to find information about a functionfrom its derivatives.
Given the first-order derivative y ′ = f (x) of the unknown functiony = F (x), we come from F to f by taking the derivative.
The reverse process F is called an indefinite integral of f and written
∫f (x)dx = F (x) + C when F ′(x) = f (x)
F (x) + C is not a single function, but a class of functions, all with thesame derivative f .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 103 / 131
Single-Variable Functions Integration
The Relationship between Integration and Differentiation
Integration and Differentiation cancel out each other:
d∫f (x)dx
dx= f (x)
∫F ′(x)dx = F (x) + C
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 104 / 131
Single-Variable Functions Integration
Rules of Integration 1/2
∫0dx = C∫adx = ax + C∫xndx = 1
n+1xn+1 + C∫
1ndx = ln |x |+ C∫exdx = ex + C
Constant Factor∫af (x)dx = a
∫f (x)dx + C
Sums∫f (x)± g(x)dx =
∫f (x)dx ±
∫g(x)dx + C
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 105 / 131
Single-Variable Functions Integration
Rules of Integration 2/2
Integration by Substitution∫f [g(x)] · g ′(x)dx =
∫f (u)du
where: u = g(x) and du = g ′(x)dx
Integration by Parts∫u′ (x) · v (x) dx = u (x) v (x)−
∫u (x) v ′ (x) dx
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 106 / 131
Single-Variable Functions Integration
Calculation of Areas
If for a function all f (x) ≥ 0 for all x ∈ [a, b], then the area of the function
is determined by the integral∣∣∣∫ b
a f (x)dx∣∣∣
If the sign changes, the integral has to be separated at the zeros of thefunction. The area is then the sum of the absolute values of allsubintegrals.
∫ ba f (x)dx =
∣∣∫ c1
a f (x)dx∣∣+∣∣∣∫ c2
c1f (x)dx
∣∣∣+∣∣∣∫ c3
c2f (x)dx
∣∣∣+∣∣∣∫ b
c3f (x)dx
∣∣∣Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 107 / 131
Single-Variable Functions Integration
Infinite Intervals of Integration
∫ +∞
af (x) dx = lim
c→+∞
∫ c
af (x) dx
∫ c
−∞f (x) dx = lim
a→−∞
∫ c
af (x) dx
∫ +∞
−∞f (x) dx = lim
a→−∞
∫ 0
af (x) dx + lim
c→+∞
∫ c
0f (x) dx
if the limit exists
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 108 / 131
Multi-Variable Functions
Outline0 Basic Knowledge1 Linear Algebra
Systems of Linear EquationsLinear equations of two unknownsGaussian Elimination
MatricesMatrix OperationsDeterminantsDefiniteness
2 Interest Rates and Present ValuesIntroductionCalculating InterestPresent Value
3 Single-Variable FunctionsIntroductionDifferentiationSingle-Variable OptimizationProperties of FunctionsIntegration
4 Multi-Variable FunctionsIntroductionDifferentiationMultivariable Optimization
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 109 / 131
Multi-Variable Functions Introduction
Introduction
In many cases economic variables depend on more than one parameter,e.g. consumer’s demand for a good depends on the price, the income ofthe consumers and the prices of other goods, that could be bought instead.
Functions with one variable cannot describe those relationships adequately,we therefor need multivariable functions.
y = f (x1, x2, ..., xn)
y → dependent variable
x1, ..., xn → independent variable
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 110 / 131
Multi-Variable Functions Introduction
Description of Functions
(1) Table: only for a few variables
(2) General Formula
(3) Graph: only for two independent variables
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 111 / 131
Multi-Variable Functions Differentiation
Differentiation
The slope of a two-variable function z = f (x , y) not only depends on xand y , but also on the direction of the movement:
movement in direction a: slope = 0movement in direction b: slope maximum slopemovement in direction c: some value between a and b
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 112 / 131
Multi-Variable Functions Differentiation
To conclude about the slope of the function’s plain depending on the pointand the direction, one can proceed as follows:
Replace y in the function z = f (x , y) by the constant y1:⇒ The emerging function z = f (x , y1) = z(x) only depends on x .
Replace x in the function z = f (x , y) by the constant x1:⇒ The emerging function z = f (x1, y) = z(y) only depends on x .
As the functions z(x) and z(y) are now functions of only one variable theycan now be differentiated for x and y .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 113 / 131
Multi-Variable Functions Differentiation
The slope of the tangent inx-direction equals thederivative of z(x) for x in x1.
The slope of the tangent iny -direction equals thederivative of z(y) for y in y1.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 114 / 131
Multi-Variable Functions Differentiation
The partial derivative of the function f (x , y1) with y1 = constant at pointx1 can be defined as:
lim∆x→0
f (x1 + ∆x , y1)− f (x1, y1)
∆x=∂f (x1, y)
∂x= f ′x(x1, y)
and is called first-order partial derivative of f (x , y) for x at point (x1, y1).
Accordingly
lim∆y→0
f (x1, y1 + ∆y)− f (x1, y1)
∆y=∂f (x , y1)
∂y= f ′y (x , y1)
and is called first-order partial derivative of f (x , y) for y at point (x1, y1).
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 115 / 131
Multi-Variable Functions Differentiation
Given the function z = f (x , y). If for all (x , y) ∈ D(f ) the following limitsexist:
lim∆x→0
f (x + ∆x , y)− f (x , y)
∆x=∂f (x , y)
∂x
lim∆y→0
f (x , y + ∆y)− f (x , y)
∆y=∂f (x , y)
∂y
then the function is partially differentiable for x and for y .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 116 / 131
Multi-Variable Functions Differentiation
The partial derivatives can be written as:
∂f (x , y)
∂xor f ′x(x , y)
and
∂f (x , y)
∂yor f ′y (x , y)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 117 / 131
Multi-Variable Functions Differentiation
A function of two variables has therefor two first-order derivatives. Whendetermining the derivative for one variable the other is kept constant.The partial differentiation equals therefor the differentiation of asingle-variable function and the same rules can be applied.
A function with more than two independent variables y = f (x1, x2, ..., xn)can be partially differentiated for each variable x1, x2, ..., xn. All othervariables are then kept constant. For a function of n independent variables,there are n first-order partial derivatives.
The gradient vector is a vector of all partial derivatives.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 118 / 131
Multi-Variable Functions Differentiation
Absolute Differential 1/2
For a single-variable function y = f (x) the differential can be interpretedas the change dy = df (x) when the independent variable is changed by dx .For multivariable functions partial differentials in relation to eachindependent variable can be found:
Given the function z = f (x , y) with the first-order partial derivativesf ′x(x , y) and f ′y (x , y), then is
dzx = f ′x(x , y)dx the partial differential for x
and
dzy = f ′y (x , y)dy the partial differential for y .
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 119 / 131
Multi-Variable Functions Differentiation
Absolute Differential 2/2
Given the function z = f (x , y) and the two partial differentialsdzx = f ′x(x , y)dx and dzy = f ′y (x , y)dy . If x changes by dx and y changesby dy at the same time, the total change in the function dz is calledabsolute differential and is the sum of the partial differentials:
dz = dzx + dzy = f ′xdx + f ′ydy
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 120 / 131
Multi-Variable Functions Differentiation
Second-order Partial Derivatives 1/2
The first-order partial derivatives are again functions of multipleindependent variables.
They can then again be differentiated for each of the n independentvariables. ⇒ There are n2 second-order partial derivatives.
The second-order derivatives can be shown in the form of the HessianMatrix.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 121 / 131
Multi-Variable Functions Differentiation
Second-order Partial Derivatives 2/2
y = f (x1, x2)
First-order partial derivatives:
yx1 =∂f
∂x1
yx2 =∂f
∂x2
Second-order partial derivatives (Hessian Matrix):
H(y) =
∂2f∂x2
1
∂2f∂x1∂x2
∂2f∂x2∂x1
∂2f∂x2
2
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 122 / 131
Multi-Variable Functions Differentiation
Implicit Differentiation 1/2
Let F be a function of two variables
F (x , y) = c (c is a constant)
where y is implicitly defined as a function y = f (x) of x in some Interval I.The auxiliary function u defined for all x in I is
u(x) = F (x , f (x))
According to the chain rule
u′(x) = F ′1(x , f (x)) ∗ 1 + F ′2(x , f (x)) ∗ f ′(x)
As u(x) = F (x , y) = c and the derivative of a constant is 0, we have
u′(x) = F ′1(x , f (x)) ∗ 1 + F ′2(x , f (x)) ∗ f ′(x) = 0
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 123 / 131
Multi-Variable Functions Differentiation
Implicit Differentiation 2/2
Replacing f (x) by y and f ′(x) by y ′:
F (x , y) = c ⇒ y ′ = −F ′1(x , y)
F ′2(x , y)(F ′2(x , y 6= 0))
If y is an implicit function of x in F (x , y) = 0 then the above formulagives the derivative of y for x .Therefore:
F (x , y) = c ⇒ dy
dx= −∂F/∂x
∂F/∂y
(∂F
∂y6= 0
)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 124 / 131
Multi-Variable Functions Multivariable Optimization
Functions with two variables 1/2
Given the function z = f (x , y) and its first-order partial derivatives:
Necessary Condition:
(1) f ′x (x0, y0) = 0 and f ′y (x0, y0) = 0
Sufficient Condition:
(2) f ′′xx (x0, y0) f ′′yy (x0, y0) >(f ′′xy (x0, y0)
)2
(3a) Maximum: f ′′xx (x0, y0) < 0 and therefore f ′′yy (x0, y0) < 0
(3b) Minimum: f ′′xx (x0, y0) > 0 and therefore f ′′yy (x0, y0) > 0
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 125 / 131
Multi-Variable Functions Multivariable Optimization
Functions with two variables 2/2
If instead of condition (2)
f ′′xx (x0, y0) f ′′yy (x0, y0) <(f ′′xy (x0, y0)
)2
holds, the function has a saddle point.
For f ′′xx (x0, y0) f ′′yy (x0, y0) =(f ′′xy (x0, y0)
)2no statement can be made.
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 126 / 131
Multi-Variable Functions Multivariable Optimization
Extreme values of functions with two variables
Maximum Minimum
Sattelpunkt
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 127 / 131
Multi-Variable Functions Multivariable Optimization
Functions with more than two variables
Necessary Condition:
gradf =
∂f∂x1
... = 0∂f∂xn
= 0
Sufficient Condition:
Hessian Matrix:
∂2f∂x2
1· · · ∂2f
∂x1∂xn...
. . ....
∂2f∂xn∂x1
· · · ∂2f∂x2
n
If the Hessian Matrix is positive (negative) definite, the function has aminimum (maximum).
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 128 / 131
Multi-Variable Functions Multivariable Optimization
Constrained Optimization
Structure of the problem:
max (min) y = f (x1, ..., xn)
subject to: g (x1, ..., xn) ≡ 0 and j = 1, ...,m
All constraints are multiplied by the Lagrange multiplier and added to theobjective function:
L = f (x1, ..., xn) +m∑j=1
λjgj (x1, ..., xn)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 129 / 131
Multi-Variable Functions Multivariable Optimization
Solving the problem by the Lagrange multiplier method:
(1) Identify the objective function and the constraints:
objective function:f (x1, x2)→ min or max
constraint:g(x1, x2) = 0
(2) Lagrange function:
L(x1, x2, λ) = f (x1, x2) + λ · g(x1, x2)
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 130 / 131
Multi-Variable Functions Multivariable Optimization
(3) Find all first-order derivatives:
∂L
∂x1=
∂f
∂x1+ λ · ∂g
∂x1= 0
∂L
∂x2=
∂f
∂x2+ λ · ∂g
∂x2= 0
∂L
∂λ=∂f
∂λ+ λ · ∂g
∂λ= 0
(4) solve for all variables and λ
Claudia Vogel (EUV) Mathematics for IBA Winter Term 2009/2010 131 / 131