Preliminary Course on Mathematics winter term 2014/2015 Veronika - - PowerPoint PPT Presentation

Preliminary Course on Mathematics – winter term 2014/2015

Veronika Penner vpenner@economics.uni-kiel.de CAU Kiel

Literature

• Parts of the course content are based on

‘A Cook-Book Of Mathematics’ by V. Vinogradov (Publisher: CERGE-EI 1999)

Download link: http://www.cerge-ei.cz/pdf/lecture_notes/LN01.pdf

▫ recipes for solving problems students might face in their studies of economics ▫ target audience is supposed to have some mathematical background (BA level) ▫ the main goal is to refresh students' knowledge of mathematics ▫ supplementary reading list is provided

• Further literature

▫ K. Sydsaeter, P. J. Hammond and A. Strom - Essential Mathematics for Economic Analysis, Prentice Hall International, 2012

▫ C.P. Simon and L. Blume - Mathematics for Economists, Norton, 1994

→ C.P Simon and L. Blume - Answers Pamphlet for Mathematics for Economists available online

Part 1 – Linear Algebra

Veronika Penner vpenner@economics.uni-kiel.de CAU Kiel

Outline Part 1 – Linear algebra

I. Vector and matrix algebra

▫ Operations ▫ Scalar product ▫ Special matrices ▫ Hessian matrix ▫ Determinant ▫ Matrix inverse

II. Systems of equations

▫ Linear combinations ▫ Rank and basis ▫ Solution methods ▫ Nonlinear equations

• III. Eigenvalue problems

▫ Linear transformations ▫ Eigenvalues

Vector and matrix algebra

Definition

• Row/column order of matrices

▫ rows in the first index, columns in the second

• Notation
• Dimensions (or order)

▫ square ▫ special cases of matrices → vectors

 column and row vectors 𝐵𝑛𝑜 = 𝑏11 ⋯ 𝑏1𝑜 ⋮ ⋱ ⋮ 𝑏𝑛1 … 𝑏𝑛𝑜 𝐵𝑛𝑜 = 𝑏𝑗𝑘 [𝑛𝑦𝑜] 𝐵𝑜 = 𝑏𝑗𝑗 [𝑜𝑦𝑜] 𝐵𝑛1 = 𝑏𝑗1 [𝑛𝑦1] = 𝑏11 ⋮ 𝑏𝑛1 𝐵1𝑜 = 𝑏1𝑗 [1𝑦𝑜] = (𝑏11, ⋯ , 𝑏1𝑜)

Matrix operations

• Addition/Substraction

▫ only if the matrices have the same dimension

• Scalar multiplication
• Multiplication

▫ note the dimensions! ▫ inner dimensions must be the same

𝐵𝑛𝑜±𝐶𝑛𝑜= 𝑏𝑗𝑘 ± 𝑐𝑗𝑘 [𝑛𝑦𝑜] 𝜇𝐵𝑛𝑜 = 𝜇𝑏𝑗𝑘 [𝑛𝑦𝑜] 𝐵𝑛𝑜 ∙ 𝐶𝑜𝑙 = 𝑑𝑗𝑘

𝑛𝑦𝑙

𝑥ℎ𝑓𝑠𝑓 𝐷𝑛𝑙 = 𝑑𝑗𝑘 [𝑛𝑦𝑙] and 𝑑𝑗𝑘 = 𝑏𝑗𝑚𝑐𝑚𝑘

𝑜 𝑚=1

Examples

• Example 1
• Example 2

⇒ 𝐵23 ∙ 𝐶32= 3 ∙ 9 + 2 ∙ 12 + 1 ∙ 14 3 ∙ 10 + 2 ∙ 11 + 1 ∙ 13 4 ∙ 9 + 5 ∙ 12 + 6 ∙ 14 4 ∙ 10 + 5 ∙ 11 + 6 ∙ 13 𝐵23 = 3 2 1 4 5 6 , 𝐶32 = 9 10 12 11 14 13 = 27 + 24 + 14 30 + 22 + 13 36 + 60 + 84 40 + 55 + 78 = 65 65 180 173 = 𝐷22, 𝐷 ∈ 𝕅2×2 𝑣 = 2,1,0 , 𝑣 ∈ 𝕅1×3 𝑤 = 1 2 1 , 𝑣 ∈ 𝕅3×1

• 1. 𝑣 ∙ 𝑤 = 2,1,0 ∙

1 2 1 = 4

• 2. 𝑤 ∙ 𝑣 =

1 2 1 ∙ 2,1,0 = 2 1 4 2 2 1 ∈ 𝕅3×3

Laws of operations

• Commutative/associative law
• f addition
• Associative law of

multiplication

• Distributiv law:
• NO commutative law of

multiplication

𝐵 + 𝐶 = 𝐶 + 𝐵 𝐵 + 𝐶 + 𝐷 = 𝐵 + (𝐶 + 𝐷) 𝐵(𝐶𝐷) = (𝐵𝐶)𝐷 𝐵 𝐶 + 𝐷 = 𝐵𝐶 + 𝐵𝐷 𝐶 + 𝐷 𝐵 = 𝐶𝐵 + 𝐷𝐵 𝐵𝐶 ≠ 𝐶𝐵 𝐵 = 2 3 8 , 𝐶 = 7 2 6 3 𝐵𝐶 = 14 4 69 30 ≠ 𝐶𝐵 = 20 16 21 24

Scalar product

• Geometric interpretation:

Product of the magnitudes of the two vectors and the cosine of the angle between them

• If cos 𝜄 = 90° then 𝑏 ∙ 𝑐 = 0, a and b are orthogonal
• Algebraic interpretation:

Sum of the products of the corresponding entries of the two vectors

𝑏 ∙ 𝑐 = 𝑏 𝑐 cos (𝜄) 𝑏 = 𝑏1 ⋮ 𝑏𝑜 , 𝑐 = 𝑐1 ⋮ 𝑐𝑜 → 𝑏 ∙ 𝑐 = 𝑏𝑗𝑐𝑗

𝑜 𝑗=1

𝑏 = 1 3 5 , 𝑐 = 2 4 6 → 𝑏 ∙ 𝑐 = 1 ∙ 2 + 3 ∙ 4 + 5 ∙ 6 = 44

Special matrices

• Identity matrices are always

square matrices

• Null matrices can have any dimension
• Diagonal matrices are always square matrices
• Triangular matrices are always square matrices

𝐵I = IA = A 𝐵IB = AB 𝟏𝑜 = ⋯ ⋮ ⋱ ⋮ ⋯ 𝐵 + 𝟏 = A 𝐵 + (−A) = 𝟏 𝐽𝑜 = 1 ⋯ ⋮ ⋱ ⋮ ⋯ 1𝑜 𝐵𝑜 = 𝑏11 ⋯ ⋮ ⋱ ⋮ ⋯ 𝑏𝑜𝑜 𝐵𝑜 = 𝑏11 ⋯ ⋮ ⋱ ⋮ 𝑏𝑜1 ⋯ 𝑏𝑜𝑜

Transpose matrices I

• Definition
• Example
• Properties

▫ (𝐵𝑈)𝑈 = 𝐵 ▫ (𝛽𝐵)𝑈 = 𝛽𝐵𝑈 ▫ (𝐵 + 𝐶)𝑈= 𝐵𝑈 + 𝐶𝑈 ▫ (𝐵𝐶)𝑈= 𝐶𝑈𝐵𝑈 𝐵 = 𝑏𝑗𝑘 [𝑛𝑦𝑜] 𝐵𝑈 = 𝑐𝑗𝑘 [𝑜𝑦𝑛] 𝑏𝑘𝑗 = 𝑐𝑗𝑘 for 𝑗 = 1 … 𝑜 and 𝑘 = 1 … 𝑛 𝐵 = 3 2 1 4 5 6 𝐵𝑈 = 3 4 2 5 1 6

Transpose matrices II

• Transpose matrices are

▫ symmetric if ▫ anti-symmetric if ▫ orthogonal if ▫ idempotent if and 𝐵𝑈 = 𝐵 𝐵𝑈 = −𝐵 𝐵𝑈𝐵 = 𝐽 𝐵𝑈 = 𝐵 𝐵𝐵 = 𝐵

• The symmetric Hessian matrix
• f second partial derivatives(see

calculus[Part 2]) is defined as

𝐼 = 𝜖2𝑔 𝜖𝑦1𝜖𝑦1 … 𝜖2𝑔 𝜖𝑦1𝜖𝑦𝑜 ⋮ ⋱ ⋮ 𝜖2𝑔 𝜖𝑦𝑜𝜖𝑦1 … 𝜖2𝑔 𝜖𝑦𝑜𝜖𝑦𝑜

• Example 𝑨(𝑦, 𝑧) = 𝑦2𝑧 + 6𝑦 − 𝑧3

▫ 𝑨𝑦 = 2𝑦𝑧 + 6

⇒ 𝑨𝑦𝑦 = 2𝑧 ⇒ 𝑨𝑦𝑧 = 2𝑦

▫ 𝑨𝑧 = 𝑦2 − 3𝑧2

⇒ 𝑨𝑧𝑧 = −6𝑧 ⇒ 𝑨𝑧𝑦 = 2𝑦

• 𝐼 =

𝑨𝑦𝑦 𝑨𝑦𝑧 𝑨𝑧𝑦 𝑨𝑧𝑧 = 2𝑧 2𝑦 2𝑦 −6𝑧

Determinant I

• Properties

▫ det 𝐵 ∙ 𝐶 = det 𝐵 ∙ det 𝐶 ▫ in general: det 𝐵 + 𝐶 ≠ det 𝐵 + det 𝐶

• Method 1 :

Determinant of a 2 × 2 -matrix:

• Method 2: SARRUS

Determinant of a 3 × 3 -matrix

det (𝐵2) = 𝑏11 𝑏12 𝑏21 𝑏22 = 𝑏11𝑏22 − 𝑏12𝑏21 det (𝐵3) = 𝑏11 𝑏12 𝑏13 𝑏21 𝑏22 𝑏23 𝑏31 𝑏32 𝑏33 𝑏11 𝑏12 𝑏21 𝑏22 𝑏31 𝑏32 = 𝑏11𝑏22𝑏33 + 𝑏12𝑏23𝑏31 + 𝑏13𝑏21𝑏32 − 𝑏13𝑏22𝑏31 − 𝑏11𝑏23𝑏32 − 𝑏12𝑏21𝑏33

𝐵 = 2 3 1 4 5 7 2 1

Determinant II

• Method 3: Laplace Expansion

▫ important concepts  the minor 𝑁𝑗𝑘 of element 𝑏𝑗𝑘 is the resulting matrix after removing the ith row and the jth col  the co-factor 𝐷𝑗𝑘 of element 𝑏𝑗𝑘 :

(−1)𝑚+𝑙∙ det (𝑁𝑚𝑙)

det 𝐵 = (−1)𝑚+𝑙∙ 𝑏𝑚𝑙∙ det (𝑁𝑚𝑙)

𝑜 𝑙=1

for some integer 𝑚 𝑗𝑜 1 ≤ 𝑚 ≤ 𝑜

𝑁23 = 2 3 7 2 𝐷23 = (−1)2+3 ∙ 5 ∙ −17 = 85 det 𝐵 = 𝐷13 + 𝐷23 + 𝐷33

Determinant III

• Laplace Expansion

▫ examples

𝐷22 = (−1)2+2 ∙ 4 ∙ 2 = 8 det 𝐵 = (−1)𝑚+𝑙∙ 𝑏𝑚𝑙∙ det (𝑁𝑚𝑙)

𝑜 𝑙=1

for some integer 𝑚 𝑗𝑜 1 ≤ 𝑚 ≤ 𝑜

𝐵 = 2 3 1 4 5 7 2 1 𝐵 = 2 3 7 2 𝐷21 = (−1)2+1 ∙ 7 ∙ 3 = −21 det 𝐵 = 𝐷21 + 𝐷22 + 𝐷23 = 90 det 𝐵 = 𝐷21 + 𝐷22 = − 17 𝐷21 = (−1)2+1 ∙ 1 ∙ 3 = −3 𝐷23 = (−1)2+3 ∙ 5 ∙ (−17) = 85 𝐷22 = (−1)2+2 ∙ 2 ∙ 2 = 4

Determinant IV

• Simplification rules for calculation

1 – the multiplication of any row/column by a scalar changes the determinant according to the scalar 2 – the interchange of two rows/columns changes the sign of the determinant 3 – the determinant is not affected by linear transformations of rows/columns 4 – if two rows/columns are identical, the determinant equals zero 5 – the determinant of a diagonal/triangular matrix is the product of its diagonal elements

Determinant V

• Example

1) divide the first row by 2 (rule 1) 2) subtract the first row from the third row (rule 3) 3) rule 5

𝐵 = 4 −2 −1 6 2 5 −3 2 −1 0 8 −2 2 det 𝐵 = 2 ∙ 𝑒𝑓𝑢 2 −1 −1 3 1 5 −3 2 −1 0 8 −2 2 det 𝐵 = 2 ∙ 𝑒𝑓𝑢 2 −1 −1 3 1 5 −3 0 5 −3 2 det 𝐵 = 2 ∙ 2 ∙ −1 ∙ 5 ∙ 2 = −40

Matrix inverse I

• Only non-singular matrices are invertible
• Equivalent statements

matrix A is invertible ⇔ det 𝐵 ≠ 0 ⇔ rank(A) = order(A) (definition later) ⇔ matrix A is non-singular Several ways to check for singularity

Matrix inverse II

• Method 1: Adjoint matrix

▫ first check for singularity ↑!

𝑒𝑗𝑘 = 1 det (𝐵) (−1)𝑗+𝑘det (𝑁

𝑘𝑗)

𝐵−1 = 𝑒𝑗𝑘 , 𝐵 = 2 3 1 4 5 7 2 1 → det 𝐵 = 8 + 105 − 20 − 3 = 90 ≠ 0 𝑒11 = 1 90 −1 1+1 4 − 10 = −6/90 𝑒12 = 1 90 −1 1+2 3 − 0 = −3/90 𝑒21 = 1 90 −1 2+1 1 − 35 = 34/90 𝑒31 = 1 90 −1 3+1 2 − 28 = −26/90 𝑒22 = 1 90 −1 2+2 2 − 0 = 2/90 𝑒32 = 1 90 −1 3+2 4 − 21 = 17/90 𝑒13 = 1 90 −1 1+3 15 − 0 = 15/90 𝑒23 = 1 90 −1 2+3 10 − 0 = −10/90 𝑒33 = 1 90 −1 3+3 8 − 3 = 5/90 𝐵−1 = −6/90 −3/90 15/90 34/90 2/90 −10/90 −26/90 17/90 5/90

Matrix inverse III

• Method of adjoint for (2 × 2) − matrices

A = 𝑏 𝑐 𝑑 𝑒 ⟹ 𝐵−1 = 1 𝑏𝑒 − 𝑐𝑑 𝑒 −𝑐 −𝑑 𝑏 compare with 𝑒𝑗𝑘 = 1 det (𝐵) (−1)𝑗+𝑘det (𝑁

𝑘𝑗)

Matrix inverse IV

• Method 2: Gauss elimination

▫ first check for singularity ▫ place an identity matrix alongside and perform basic

• perations

 1st row divided by 2  2nd row minus 4 times 1st row  2nd row divided by -4  1st row minus 3/2 times second row 𝐵 = 2 3 4 2 2 3 4 2 1 1 ⇔ 1 3/2 4 2 1/2 1 ⇔ 1 3/2 −4 1/2 −2 1 ⇔ 1 3/2 1 1/2 1/2 −1/4 ⇔ 1 1 −1/4 3/8 1/2 −1/4 ⟹ 𝐵−1 = −1/4 3/8 1/2 −1/4 → det 𝐵 = 4 − 12 = −8 ≠ 0

Systems of equations

Linear combinations

• Linear combination of vectors

▫ 𝑟𝑗 - real numbers ▫ 𝑏𝑗 - vectors, 𝑗 ∈ 1, … , 𝑙

• Linear dependency if
• Example:

as system of equations: Solve 1) for x: Use 2):

𝑟1𝑏1 + ⋯ + 𝑟𝑙𝑏𝑙 𝑟1𝑏1 + ⋯ + 𝑟𝑙𝑏𝑙 = 0, with 𝑟𝑗 ≠ 0 3 2 4 + (−2) 3 6 = 0 1) 2𝑦 + 3𝑧 = 0 2) 4𝑦 + 6𝑧 = 0 2𝑦 = −3𝑧, 𝑦 = − 3𝑧 2 4 − 3𝑧 2 + 6𝑧 = 0, 𝑧 = 𝑧 → 0 = 0

Rank

• Definition of rank:

▫ max number of linearly independent rows/columns

• If there are no linearly dependent rows/columns

▫ 𝑠𝑏𝑜𝑙 𝐵 = 𝑝𝑠𝑒𝑓𝑠 𝐵 𝑏𝑜𝑒 det 𝐵 ≠ 0

• Rank and determinant

▫ if 𝑠𝑏𝑜𝑙 𝐵 = 𝑝𝑠𝑒𝑓𝑠 𝐵 then det (𝐵) ≠ 0 ▫ if det 𝐵 ≠ 0 then 𝑠𝑏𝑜𝑙 𝐵 = 𝑝𝑠𝑒𝑓𝑠 𝐵 ▫ in these cases, matrix is called non-singular

• Example

▫ first two rows are linearly dependent ▫ 𝑠𝑏𝑜𝑙 𝐵 = 2 < 𝑝𝑠𝑒𝑓𝑠 𝐵 = 3 ▫ det 𝐵 = 0

𝐵 = 1 3 2 2 6 4 −5 7 1

Basis

• Definition of basis:

▫ set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space → define a coordinate system → linearly independent spanning set

• All bases of a vector space have the same number of elements

→ dimension of the vector space

• Linear independence property:
• Example

▫ ℝ2 with basis 𝑓1 = 1 0 , 𝑓2 = 0 1 ▫ 𝑤 = 𝑏 𝑐 vector in ℝ2 then 𝑤 = 𝑏 1 0 + 𝑐 0 1

if 𝑟1𝑏1 + ⋯ + 𝑟𝑙𝑏𝑙 = 0 → 𝑟1 = ⋯ = 𝑟𝑙 = 0

System of equations

• System of equations can be represented in matrix form
• Example:

𝑏11𝑦1 + ⋯ + 𝑏1𝑜𝑦𝑜 = 𝑐1 ⋮ 𝑏𝑛1𝑦1 + ⋯ + 𝑏𝑛𝑜𝑦𝑜 = 𝑐𝑛 𝑦 = 𝑦1 ⋮ 𝑦𝑜 , 𝐵𝑦 = 𝑐 ⇒ 𝐵 = 𝑏11 ⋯ 𝑏1𝑜 ⋮ ⋱ ⋮ 𝑏𝑛1 … 𝑏𝑛𝑜 𝑐 = 𝑐1 ⋯ 𝑐𝑛 , 6𝑨 + 3𝑧 = 4 12𝑨 + 2𝑧 = 5 𝑦 = 𝑨 𝑧 , 𝑐 = 4 5 , 𝐵 = 6 3 12 2

System of linear equations I

• Example 1

▫ homogeneous case: 𝑐 = 0 ▫ det 𝐵 = 3 ≠ 0 ▫ unique trivial solution

2𝑨 + 𝑧 = 0 3𝑨 + 3𝑧 = 0 → 𝐵 = 2 1 3 3 , 𝑐 = 0 0 , 𝑦 = 𝑨 𝑧 , 𝐵𝑦 = 𝑐 → 𝑨 = −𝑧 2 → 3 −𝑧 2 + 3𝑧 = 0 → − 3𝑧 2 + 6𝑧 2 = 0 𝑧 = 0 𝑨 = 0 2𝑨 + 𝑧 = 0 3𝑨 + 3𝑧 = 0 →

System of linear equations II

• Example 2

▫ homogeneous case: 𝑐 = 0 ▫ det 𝐵 = 0 ▫ infinite number of solutions

2𝑨 + 2𝑧 = 0 3𝑨 + 3𝑧 = 0 𝐵 = 2 2 3 3 , 𝑐 = 0 0 , 𝑦 = 𝑨 𝑧 , 𝐵𝑦 = 𝑐 → 𝑨 = −𝑧 → 3 −𝑧 + 3𝑧 = 0 → 0 = 0 2𝑨 + 2𝑧 = 0 3𝑨 + 3𝑧 = 0 →

System of linear equations III

• Example 3

▫ non-homogeneous case: 𝑐 ≠ 𝟏 ▫ det 𝐵 = 0 ▫ two possibilities

 infinite solutions  system is inconsistent

• How to tell?

▫ augmented matrix 𝐵 = 𝐵 𝑐 ▫ if 𝑠𝑏𝑜𝑙 𝐵 = 𝑠𝑏𝑜𝑙 Ã

 infinite solutions

▫ if 𝑠𝑏𝑜𝑙 𝐵 ≠ 𝑠𝑏𝑜𝑙 Ã

 system is inconsistent 2𝑨 + 3𝑧 = 1 4𝑨 + 6𝑧 = 2 2𝑨 + 2𝑧 = 1 3𝑨 + 3𝑧 = 0 → 𝑨 = 1 − 2𝑧 2 → 3 1 − 2𝑧 2 + 3𝑧 = 0 →

3 2 = 0, false statement

→ 𝑨 = 1 − 3𝑧 2 → 4 1 − 3𝑧 2 + 6𝑧 = 2 → 0 = 0, true statement 𝐵 = 2 3 4 6 , Ã = 2 3 1 4 6 2 𝑐 = 1 2 → 𝐵 = 2 2 3 3 , Ã = 2 2 1 3 3 𝑐 = 1 0 → → 𝑠𝑏𝑜𝑙 𝐵 = 1 = 𝑠𝑏𝑜𝑙 Ã → 𝑠𝑏𝑜𝑙 𝐵 = 1 ≠ 2 = 𝑠𝑏𝑜𝑙 Ã

System of linear equations IV

• Example 4

▫ non-homogeneous case: 𝑐 ≠ 𝟏 ▫ det 𝐵 = 3 ≠ 0 ▫ unique solution

2𝑨 + 𝑧 = 1 3𝑨 + 3𝑧 = 0 𝐵 = 2 1 3 3 , 𝑐 = 1 0 , 𝑦 = 𝑨 𝑧 , 𝐵𝑦 = 𝑐 → 𝑨 = 1 − 𝑧 2 → 3 1 − 𝑧 2 + 3𝑧 = 0 → − 3𝑧 2 + 6𝑧 2 = − 3 2 𝑧 = −1 𝑨 = 1 → 2𝑨 + 𝑧 = 1 3𝑨 + 3𝑧 = 0 →

Solution methods

• If 𝑐 ≠ 0 and det 𝐵 ≠ 0

⟹ unique solution

• 3 methods to solve

▫ Inverse matrix method ▫ Gauss method

 Same idea as before, but placing b alongside with A

▫ Cramer’s rule

 We can find all elements of x by applying the following formula 𝐵 = 2 1 3 3 , 𝑐 = 1 0 , 𝑦 = 𝑨 𝑧 , 𝐵𝑦 = 𝑐 𝑦 = 𝐵−1𝑐 𝑦𝑘 = det (𝐵𝑘) det (𝐵) 𝐵𝑘 = 𝑏11 … 𝑏1𝑘−1 ⋮ ⋱ ⋮ 𝑏𝑜1 … 𝑏𝑜𝑘−1 𝑐1 ⋮ 𝑐

𝑘

𝑏1𝑘+1 … 𝑏1𝑜 ⋮ ⋱ ⋮ 𝑏𝑜𝑘+1 … 𝑏𝑜𝑜

Example I

𝐵 = 2 3 4 −1 𝑐 = 12 10 𝑦 = 𝑨 𝑧 𝐵𝑦 = 𝑐 𝑦 = 𝐵−1𝑐 det 𝐵 = 2 −1 − 12 = −14 𝑒11 = 1 −14 −1 1+1(−1) = 1/14 𝑩−𝟐 = 𝟐/𝟐𝟓 𝟒/𝟐𝟓 𝟓/𝟐𝟓 −𝟑/𝟐𝟓 𝑦 = 1/14 3/14 4/14 −2/14 12 10 𝑦 = 12/14 + 30/14 48/14 − 20/14 𝒚 = 𝟒 𝟑 CRAMER’S RULE det 𝐵 = 2 −1 − 12 = −14 𝒜 = det (𝐵1) det (𝐵) = det ( 12 3 10 −1 ) −14 = −42 −14 = 𝟒 𝒛 = det (𝐵2) det (𝐵) = det ( 2 12 4 10 ) −14 = −28 −14 = 𝟑 𝟑 𝟒 𝟓 −𝟐 𝟐𝟑 𝟐𝟏 → 1 1.5 −7 6 −14 → 1 1.5 1 6 2 → 1 1 3 2 𝒜 = 𝟒, 𝒛 = 𝟑 𝑒21 = 1 −14 −1 2+14 = 4/14 𝑒12 = 1 −14 −1 1+23 = 3/14 𝑒22 = 1 −14 −1 2+22 = −2/14 MATRIX INVERSE METHOD GAUSS METHOD

Example II – Part 1

• Consider a market for three goods

▫ solve for price by using:

 Cramer’s rule  Inverse matrix method 𝐸1 = 5 − 2𝑄

1 + 𝑄2 + 𝑄3

𝑇1 = −4 + 3𝑄

1 + 2𝑄2

𝐸2 = 6 + 2𝑄

1 − 3𝑄2 + 𝑄3

𝑇2 = 3 + 2𝑄2 𝐸3 = 20 + 𝑄

1 + 2𝑄2 − 4𝑄3

𝑇3 = 3 + 𝑄2 + 3𝑄3 5𝑄

1 + 𝑄2 − 𝑄3 = 9

⇒ −2𝑄

1 + 5𝑄2 − 𝑄3 = 3

−𝑄

1 − 𝑄2 + 7𝑄3 = 17

⇒ 𝐵 = 5 1 −1 −2 5 −1 −1 −1 7 , 𝑐 = 9 3 17 , 𝑦 = 𝑄

1

𝑄2 𝑄3 → det 𝐵 = 𝐷21 + 𝐷22 + 𝐷23 = 178 𝐷21 = −1 2+1 −2 ∙ 6 = 12 𝐷22 = −1 2+2 ∙ 5 ∙ 34 = 170 𝐷23 = −1 2+3 −1 ∙ −4 = −4

Example II – Part 2

• Using Cramer’s rule

𝐵 = 5 1 −1 −2 5 −1 −1 −1 7 𝑐 = 9 3 17 𝑦 = 𝑄

1

𝑄2 𝑄3 𝑄

1 =

𝑒𝑓𝑢 9 1 −1 3 5 −1 17 −1 7 𝑒𝑓𝑢 𝐵 𝑄2 = 𝑒𝑓𝑢 5 9 −1 −2 3 −1 −1 17 7 𝑒𝑓𝑢 𝐵 𝑄3 = 𝑒𝑓𝑢 5 1 9 −2 5 3 −1 −1 17 𝑒𝑓𝑢 𝐵 𝐷21(𝐵1) = −1 2+1 ∙ 3 ∙ 6 = −18 𝐷22(𝐵1) = −1 2+2 ∙ 5 ∙ 80 = 400 𝐷23(𝐵1) = −1 2+3 −1 ∙ −4 = −26 𝑄

1 = 356

178 = 2 𝐷21(𝐵2) = −1 2+1 ∙ (−2) ∙ 80 = 160 𝐷22(𝐵2) = −1 2+2 ∙ 3 ∙ 34 = 102 𝐷23(𝐵2) = −1 2+3 −1 ∙ 94 = 94 𝑄2 = 356 178 = 2 𝐷21(𝐵3) = −1 2+1 ∙ (−2) ∙ 26 = 52 𝐷22(𝐵3) = −1 2+2 ∙ 5 ∙ 94 = 470 𝐷23(𝐵3) = −1 2+3 ∙ 3 ∙ −4 = 12 𝑄3 = 534 178 = 3

Example II – Part 3

• Using the inverse matrix method

𝐵 = 5 1 −1 −2 5 −1 −1 −1 7 , 𝑐 = 9 3 17 , 𝑦 = 𝑄

1

𝑄2 𝑄3 𝑒11 = 1 178 −1 1+1 35 − 1 = 34/178 𝑒12 = 1 178 −1 1+2 7 − 1 = −6/178 𝑒13 = 1 178 −1 1+3 −1 + 5 = 4/178 𝑒21 = 1 178 −1 2+1 −14 − 1 = 15/178 𝑒31 = 1 178 −1 3+1 2 + 5 = 7/178 𝑒22 = 1 178 −1 2+2 35 − 1 = 34/178 𝑒32 = 1 178 −1 3+2 −5 + 1 = 4/178 𝑒23 = 1 178 −1 2+3 −5 − 2 = 7/178 𝑒33 = 1 178 −1 3+3 25 + 2 = 27/178 𝐵−1 = 1 178 34 −6 4 15 34 7 7 4 27 𝐵𝑦 = 𝑐 ⇔ 𝑦 = 𝐵−1𝑐 𝑦 = 𝐵−1𝑐 = 1 178 34 −6 4 15 34 7 7 4 27 9 3 17 𝑦 = 1 178 34 ∙ 9 + −6 ∙ 3 + 4 ∙ 17 15 ∙ 9 + 34 ∙ 3 + 7 ∙ 17 7 ∙ 9 + 4 ∙ 3 + 27 ∙ 17 = 2 2 3

Example III – Part 1

• Solve 𝐽 − 𝐵 𝑦 = 𝑐

𝐵 = 0.2 0.2 0.1 0.3 0.3 0.1 0.1 0.2 0.4 , 𝑐 = 8 4 5 , 𝑦 = 𝑦1 𝑦2 𝑦3 𝐽 − 𝐵 = 1 − 0.2 0 − 0.2 0 − 0.1 0 − 0.3 1 − 0.3 0 − 0.1 0 − 0.1 0 − 0.2 1 − 0.4 det 𝐽 − 𝐵 = 𝐷21 + 𝐷22 + 𝐷23 = 0.269 = 0.8 −0.2 −0.1 −0.3 0.7 −0.1 −0.1 −0.2 0.6 𝐷21 = (−1)2+1 ∙ (−0.3) ∙ (−0.14) = −0.042 𝐷22 = (−1)2+2 ∙ (0.7) ∙ (0.47) = 0.329 𝐷23 = (−1)2+3 ∙ −0.1 ∙ −0.18 = −0.018

Example III – Part 2

• solve 𝐽 − 𝐵 𝑦 = 𝑐

𝑒11 =

1 0.269 −1 1+1 0.4 = 0.4 0.269

𝑒21 =

1 0.269 −1 2+1 −0.19 = 0.19 0.269

𝑒31 =

1 0.269 −1 3+1 0.13 = 0.13 0.269

𝑒12 =

1 0.269 −1 1+2 −0.14 = 0.14 0.269

𝑒22 =

1 0.269 −1 2+2 0.47 = 0.47 0.269

𝑒32 =

1 0.269 −1 3+2 −0.18 = 0.18 0.269

𝑒13 =

1 0.269 −1 1+3 0.09 = 0.09 0.269

𝑒23 =

1 0.269 −1 2+3 −0.11 = 0.11 0.269

𝑒33 =

1 0.269 −1 3+3 0.5 = 0.5 0.269

𝑦 = (𝐽 − 𝐵)−1𝑐 = 0.4 0.269 ∙ 8 + 0.14 0.269 ∙ 4 + 0.09 0.269 ∙ 5 0.19 0.269 ∙ 6 + 0.47 0.269 ∙ 4 + 0.11 0.269 ∙ 5 0.13 0.269 ∙ 8 + 0.18 0.269 ∙ 4 + 0.5 0.269 ∙ 5 = 4210 269 3950 269 4260 269 =

0.4 0.269 0.14 0.269 0.09 0.269 0.19 0.269 0.47 0.269 0.11 0.269 0.13 0.269 0.18 0.269 0.5 0.269

∙ 8 4 5

Nonlinear equations

• Quadratic form 𝑅 = 𝑤′𝐵 𝑤 with 𝑤 =

𝑦 𝑧 𝑨 , 𝑤′ = (𝑦, 𝑧, 𝑨)

• Corresponding matrix form
• Important concept of leading principal minors

Q 𝑦, 𝑧, 𝑨 = 3𝑦2 + 3𝑧2+ 5𝑨2 − 2𝑦𝑧

𝐵 = 3 −1 −1 3 5 𝐸1 = det 3 = 3 > 0, 𝐸𝑙 = det 𝑏11 … 𝑏1𝑙 ⋮ ⋱ ⋮ 𝑏𝑙1 … 𝑏𝑙𝑙 𝐸2 = det 3 −1 −1 3 = 8 > 0, 𝐸3 = det 3 −1 −1 3 5 = 40 > 0 𝐸3 = 𝐷31 + 𝐷32 + 𝐷33 = 0 + 0 + (−1)3+3∙ 5 ∙ 8 = 40

Quadratic forms

• A quadratic form is said to be:

▫ positive definite (PD)

 if 𝑅 > 0 ∀ 𝑦 ≠ 0 ⇔ 𝑅 𝑗𝑡 𝑄𝐸 ⇔ 𝐸𝑙 > 0 ∀𝑙 = 1, … , 𝑜

▫ negative definite (ND)

 if 𝑅 < 0 ∀ 𝑦 ≠ 0 ⇔ 𝑅 𝑗𝑡 𝑂𝐸 ⇔ −1 𝑙𝐸𝑙 > 0 ∀ 𝑙 = 1, … , 𝑜

▫ positive semidefinite (PSD)

 if 𝑅 ≥ 0 ∀ 𝑦 ⇔ 𝑅 𝑗𝑡 𝑄𝑇𝐸 ⇔ 𝐸𝑙 ≥ 0 ∀ 𝑙 = 1, … , 𝑜

▫ negative semidefinite (NSD)

 if 𝑅 ≤ 0 ∀ 𝑦 ⇔ 𝑅 𝑗𝑡 𝑄𝑇𝐸 ⇔ 𝐸𝑙 ≤ 0 ∀ 𝑙 = 1, … , 𝑜

all minors are greater than zero:

→ Q(x, y, z) is positive definite [example previous slide]

𝐸1 = 3 > 0, 𝐸2 = 8 > 0, 𝐸3 = 40 > 0

Eigenvalue problems

Linear transformations I

• Consider a linear map 𝑔: ℝ2 → ℝ2, 𝑔 𝑦 = 𝐵 ∙ 𝑦

▫ 4 7 3 2 1 2 = 18 7 ▫ 4 7 3 2 2 1 = 15 8 ▫ 4 7 3 2 1 1 = 11 5 ▫ identity transformation: 1 1 1 2 = 1 2 , 1 1 2 1 = 2 1 ▫ dilation: 2 2 1 2 =

1 2

1 , 2 2 2 1 = 1

1 2

▫ compression:

1 2 1 2

1 2 =

1 2

1 ,

1 2 1 2

2 1 = 1

1 2

Linear transformations II

▫ reflection against the x-axis: 1 −1 1 2 = 1 −2 , 1 −1 2 1 = 2 −1 ▫ reflection against the y-axis: −1 1 1 2 = −1 2 , −1 1 2 1 = −2 1 ▫ point reflection: −1 −1 1 2 = −1 −2 , −1 −1 2 1 = −2 −1 ▫ Rotation by 90°, clockwise: 1 −1 1 2 = 2 −1 , 1 −1 2 1 = 1 −2

Eigenvalues I

• Any number λ such that the equation

𝐵𝑦 = 𝜇𝑦

has a non-zero vector solution (𝑦 ≠ 0) is called an eigenvalue (or characteristic root) of the equation

• The resulting vector x is called the eigenvector (or characterisitc vector)
• f A
• Analytic geometry:

eigenvector is a vector whose direction is either preserved or exactly reversed after multiplication by 𝐵 → corresponding eigenvalue determines change of the length of the vector → direction maybe is reversed, dependent on the sign of the eigenvalue

Eigenvalues II

• Calculation

𝐵𝑦 = 𝜇𝑦 ⇔ 𝐵𝑦 − 𝜇𝑦 = 0 ⇔ 𝐵 − 𝜇𝐽 𝑦 = 0

• Since 𝑦 ≠ 0

→ 𝐵 − 𝜇𝐽 singular/ not invertible → det 𝐵 − 𝜇𝐽 = 0 characteristic equation

𝐵 = 3 −1 −1 3 5 det 𝐵 − 𝜇𝐽 = 𝐷21 + 𝐷22 + 𝐷23 = 0 → 𝐵 − 𝜇𝐽 = 3 − 𝜇 −1 −1 3 − 𝜇 5 − 𝜇 𝐷21 = (−1)2+1 ∙ −1 ∙ (𝜇 − 5) 𝐷22 = (−1)2+2 ∙ (3 − 𝜇) ∙ (3 − 𝜇) ∙ (5 − 𝜇) 𝐷23 = 0 ⇔ − 5 − 𝜇 + 3 − 𝜇 ∙ 3 − 𝜇 ∙ 5 − 𝜇 = 0 ⇔ 5 − 𝜇 (𝜇2 − 6𝜇 + 9 − 1) = 0 ⇔ 5 − 𝜇 𝜇 − 2 𝜇 − 4 = 0 ⇒ 𝜇 = 2, 𝜇 = 4, 𝜇 = 5 det 𝐵 − 𝜇𝐽 = 0

Eigenvalues and quadratic forms

• Characteristic root test for sign definiteness
• A quadratic form is said to be:

▫ positive definite (PD)

 𝑅 𝑗𝑡 𝑄𝐸 ⇔ 𝜇𝑗 > 0 for all 𝑗 = 1 … 𝑜

▫ negative definite (ND)

 𝑅 𝑗𝑡 𝑂𝐸 ⇔ 𝜇𝑗 < 0 for all 𝑗 = 1 … 𝑜

▫ positive semidefinite (PSD)

 𝑅 𝑗𝑡 𝑄𝑇𝐸 ⇔ 𝜇𝑗 ≥ 0 for all 𝑗 = 1 … 𝑜

▫ negative semidefinite (NSD)

 𝑅 𝑗𝑡 𝑄𝑇𝐸 ⇔ 𝜇𝑗 ≤ 0 for all 𝑗 = 1 … 𝑜

Properties

• det A = λ1 ∙ ⋯ ∙ λn
• if λ1, … , λn are eigenvalues of A

→ then f(λ1), … , f(λn) are the eigenvalues of f(A)

• if λ1, … , λn are eigenvalues of A

→ then

1 𝜇1 , … , 1 𝜇𝑜 are the eigenvalues of 𝐵−1 with the same eigenvector

• trace A =

aii

n i=1

= λ1 + ⋯ + λn ▫ trace A + B = trace A + trace B if A and B of same order ▫ trace λA = λ ∙ trace A ▫ trace AB =trace BA (if A and B are square) ▫ trace A =trace AT

Part 2 – Calculus

Veronika Penner vpenner@economics.uni-kiel.de CAU Kiel

Outline Part 2 – Calculus

I. Differential calculus

▫ Continuity and differentiability ▫ Necessity and sufficiency ▫ Differentiation rules ▫ Partial derivative

II. Optimization

▫ Maximum and minimum ▫ Concavity and convexity ▫ Optimization with constraints ▫ Lagrange multipliers ▫ Kuhn-Tucker conditions

III. Approximation

▫ Total differential ▫ Taylor series

IV. Integral calculus

▫ Definite integral ▫ Primitive ▫ Integration by parts ▫ Substitution formula

Differential calculus

Continuity and differentiability

• A function 𝑔 𝑦 is continuous at 𝑦 = 𝑏 if lim

𝑦→𝑏 𝑔(𝑦) = 𝑔(𝑏)

• Example: 𝑔 𝑦 = 𝑦 − 2 is continuous at 𝑦 = 2 because

lim

𝑦→2 𝑦 − 2 = 0 = 𝑔(2)

• A function 𝑔 𝑦 is differentiable at 𝑦 = 𝑏 if the derivative

𝑔′ 𝑦 = lim

𝑦→𝑏

𝑔 𝑏 + Δ𝑦 − 𝑔(𝑏) Δ𝑦 exists at 𝑦 = 𝑏

• Example: 𝑔 𝑦 = 𝑦 − 2 is not differentiable at 𝑦 = 2 because

𝑔′ 𝑦 = lim

𝑦→𝑏 𝑔 𝑏+Δ𝑦 −𝑔(𝑏) Δ𝑦

does not exist!

Necessity and sufficiency I

• Logic digression:

A condition B is said to be necessary for a condition A, if the falsity

• f B guarantees the falsity of A B ⇒ A

A condition A is said to be sufficient for a condition B, if the truth

• f A guarantees the truth of B A ⇒ B

▫ A ⇒ B implies B ⇒ A → A is a sufficient condition for B / B is a necessary condition for A ▫ A ⇔ B → equivalency: A is necessary and sufficient for B / B is necessary and sufficient for A

Necessity and sufficiency II

• Continuity is a NECESSARY but NOT SUFFICIENT condition for

differentiability differentiability ⇒ continuity [NC for cont./SC for diff.] continuity ⇏ differentiability !

• Differentiability is a sufficient condition for continuity
• In addition the following implication holds:

continuity ⇒ differentiability ! BUT if continuity ⇒ ???; differentiability ⇒ ???

Necessity and sufficiency III

• Examples:

▫ air is necessary for human life: human life ⇒ air air ⇏ human life, but air ⇒ human life ▫ having four sides is necessary to be a square: square ⇒ four sides four sides ⇏ square [see rhomboid] ▫ glass of water is sufficient for having a drink: water ⇒ have a drink have a drink ⇏ water ▫ if it rains the street is wet ⇒rain is sufficient for a wet street: rain⇒wet street wet street ⇏ rain

Differentiation rules I

• Suppose we have two differentiable functions

𝑔 𝑦 𝑏𝑜𝑒 𝑕(𝑦)

• Sum / difference rule: 𝑔 𝑦 ± 𝑕 𝑦

′ = 𝑔 ′ 𝑦 ± 𝑕 ′(𝑦)

• Product rule: 𝑔 𝑦 ∙ 𝑕 𝑦

′ = 𝑔 ′ 𝑦 ∙ 𝑕 𝑦 + 𝑔 𝑦 ∙ 𝑕 ′( 𝑦)

• Quotient rule:

𝑔(𝑦) 𝑕(𝑦) ′

=

𝑔 ′ 𝑦 ∙ 𝑕 𝑦 − 𝑔 𝑦 ∙ 𝑕 ′( 𝑦) [𝑕 𝑦 ]²

Differentiation rules II

Differentiation rules III

• Suppose we have two differentiable functions 𝑔 = 𝑔 𝑧 and 𝑧 = 𝑕 𝑦

then the chain rule can be applied:

𝑔′ 𝑦 = 𝑔′ 𝑧 ∙ 𝑕′ 𝑦

• Example: z = (3𝑦 + 1)2
• Here 𝑔 𝑧 = 𝑧2 𝑕 𝑦 = 3𝑦 + 1 = 𝑧

𝑔′ 𝑦 = 2 ∙ 𝑧 ∙ 3 ⇔ 𝑔′ 𝑦 = 2 ∙ 3𝑦 + 1 ∙ 3 ⇔ 𝒈′ 𝒚 = 𝟐𝟗𝒚 + 𝟕

Example

• Find the derivative of

𝑧 = 2𝑦 − 1 𝑦 − 1 2 Set 𝑧 =

𝑔(𝑦) 𝑕(𝑦) with 𝑔 𝑦 = 2𝑦 − 1 𝑏𝑜𝑒 𝑕 𝑦 = 𝑦 − 1 2

𝑧′ = 𝑔′ 𝑦 𝑕 𝑦 − 𝑔 𝑦 𝑕′(𝑦) (𝑕(𝑦))2 = 𝑕 ℎ 𝑦 = ℎ(𝑦) 2; ℎ 𝑦 = 𝑦 − 1 CHAIN RULE: 𝑕′ 𝑦 = 𝑕′ ℎ 𝑦 ℎ′(𝑦) 𝑕′ 𝑦 = 2ℎ(𝑦) = 2𝑦 − 2 and 𝑔′(𝑦) = 2 2 𝑦 − 1 2 − (2𝑦 − 1)(2𝑦 − 2) (𝑦 − 1)4 = (2𝑦2 − 4𝑦 + 2) − (4𝑦2 − 6𝑦 + 2) (𝑦 − 1)4 = (−2𝑦2 + 2𝑦) (𝑦 − 1)4 = −2𝑦(𝑦 − 1) (𝑦 − 1)4 𝑧′ = −2𝑦 (𝑦 − 1)3

Partial derivative

• Now consider a function 𝑧 = 𝑔 𝑦1, 𝑦2, … , 𝑦𝑜 where all variables

𝑦𝑗, 𝑗 = 1, … , 𝑜 are independent of one another

• The partial derivative of 𝑧 with respect to 𝑦𝑗 is defined as

𝜖𝑧 𝜖𝑦𝑗 = lim

∆𝑦→0

∆𝑧 ∆𝑦𝑗

• Example:

𝑧 = 4𝑦1

3 + 𝑦1𝑦2 + 𝑦2

→

𝜖𝑧 𝜖𝑦1 = 𝑔 𝑦1 = 12𝑦12+𝑦2

→ 𝜖𝑧 𝜖𝑦2 = 𝑔

𝑦2 = 𝑦1 + 1

Optimization

Maxima and minima

• If 𝑔 𝑦 is continuous in a

neighborhood 𝑉 of a point 𝑦0

• local maximum/minimum at 𝑦0

if for all 𝑦 ∈ 𝑉, 𝑦 ≠ 𝑦0 𝑔 𝑦 < 𝑔(𝑦0) {𝑔 𝑦 > 𝑔(𝑦0)}

𝑔 𝑦 = 𝑦3 − 3𝑦 𝑝𝑜 [−2,3]

• Check the following points

▫ stationary interior points ▫ extrema at the boundaries

• First, find stationary points

▫ first derivative test

 𝑔′ 𝑦 = 0

• Then, decide which one is the

maximum/minimum

▫ second derivative test

 𝑔′′ 𝑦 > 0, local minimum  𝑔′′ 𝑦 < 0, local maximum

• Finaly, check for boundary

solutions

𝑔 𝑦 = 𝑦3 − 3𝑦 𝑝𝑜 [−2,3] 𝑔′ 𝑦 = 3𝑦2 − 3 = 0 ↔ 3𝑦2 = 3 𝑦 = 1 𝑝𝑠 𝑦 = −1 𝑔′′ 𝑦 = 6𝑦 𝑔′′ −1 = −6 < 0 → 𝑦 = −1 𝑗𝑡 𝑚𝑝𝑑𝑏𝑚 𝑛𝑏𝑦𝑗𝑣𝑛 𝑔′′ 1 = 6 > 0 → 𝑦 = 1 𝑗𝑡 𝑚𝑝𝑑𝑏𝑚 𝑛𝑗𝑜𝑗𝑛𝑣𝑛 𝑔 −1 = 2; 𝑔 1 = −2 / 𝑔 −2 = −2; 𝑔 3 = 18 𝑦 = 3 𝑗𝑡 𝑕𝑚𝑝𝑐𝑏𝑚 𝑛𝑏𝑦𝑗𝑛𝑣𝑛 𝑐𝑝𝑢ℎ 𝑦 = 1 𝑏𝑜𝑒 𝑦 = −2 𝑏𝑠𝑓 𝑕𝑚𝑝𝑐𝑏𝑚 𝑛𝑗𝑜𝑗𝑛𝑏

Concavity and convexity I

• A function 𝑔 is called

convex /concave at 𝑦0 if

𝑔′′ 𝑦0 ≥ 0 /𝑔′′(𝑦0) ≤ 0

and strictly convex /concave if the inequalities are strict(>, <).

• Consider 𝑔 𝑦 = 𝑦³ − 3𝑦
• n [−2,3]

𝑔′ 𝑦 = 3𝑦2 −3 𝑔′′ 𝑦 = 6𝑦 𝑔′′ 1 = 6; 𝑔′′ 3 = 18 → strict convexity [1] 𝑔′′ −2 = −12 → strict concavity [2] 𝑔′′ 0 = 0 → saddle point [3] [3] [2] [1]

Concavity and convexity II

• A twice continuously differentiable function 𝑨 = 𝑔 𝑦1, … , 𝑦𝑜 is

concave/convex if and only if

𝑒2𝑨

is negative/positive semidefinite everywhere

• It is strictly concave/convex if but not only if 𝑒2𝑨 is negative/positive

definite everywhere

• Example:

𝑨 = 𝑦2 + 𝑦𝑧 + 𝑧2 ▫ 𝐼 = 2 1 1 2 ▫ 𝐸1 = det 2 = 2 > 0, 𝐸2 = det 𝐼 = 3 > 0 → all leading minors are greater than zero → H is positive definite → z is strictly convex

• 𝑨 = 𝑔 𝑦1, … , 𝑦𝑜
• there is a stationary point at

𝑦∗ = 𝑦1∗, … , 𝑦𝑜∗ if 𝑒𝑨 = 0 at 𝑦∗

• that is

▫ 𝑔

𝑦𝑗 = 0 ∀ 𝑗 = 1, … , 𝑜

which are called the first order necessary conditions (FONC)

• Example

 𝑨 = 𝑦3 − 8𝑧3 + 6𝑦𝑧 + 1

• FONC

▫ 𝑨𝑦 = 3(𝑦2 + 2𝑧) = 0

⇒ 𝑧 = −𝑦2/2

▫ 𝑨𝑧 = 6(−4𝑧2 + 𝑦) = 0

⇒ −4(−𝑦2/2)2+𝑦 = 0 ⇔ −4

𝑦4 4

+ 𝑦 = 0 ⇔ −𝑦4 + 𝑦 = 0 ⇔ 𝑦(1 − 𝑦3) = 0 𝑦 = 0 𝑝𝑠 𝑦 = 1 𝑧 = 0 𝑝𝑠 𝑧 = −1/2

• Second Order Necessary Conditions (SONC)

𝑨 = 𝑔 𝑦1, … , 𝑦𝑜

• has a maximum(minimum) at

𝑦∗ = 𝑦1∗, … , 𝑦𝑜∗

• if 𝑔 𝑦∗ − 𝑔 𝑦 ≥ 0(≤ 0) ∀𝑦 in the neighborhood of 𝑦∗
• check for the sign semidefiniteness of the second order total

differential

• Use the Hessian Matrix of

second partial derivatives to check for conditions

▫ the sign definiteness of 𝑒2𝑨 is equivalent to the sign definiteness of the quadratic form 𝐼 evaluated at 𝑦∗

• Can apply either the

▫ principal minors test ▫ or the eigenvalue test

• Example 𝑨 = 𝑦3 − 8𝑧3 + 6𝑦𝑧 + 1
• SONC

▫ 𝑨𝑦 = 3 𝑦2 + 2𝑧

⇒ 𝑨𝑦𝑦 = 6𝑦, 𝑨𝑦𝑧 = 6

▫ 𝑨𝑧 = 6(−4𝑧2 + 𝑦)

⇒ 𝑨𝑧𝑧 = −48𝑧, 𝑨𝑧𝑦 = 6

▫ 𝐼 = 𝑨𝑦𝑦 𝑨𝑦𝑧 𝑨𝑧𝑦 𝑨𝑧𝑧 = 6𝑦 6 6 −48𝑧

 𝐼(0,0) = 0 6 6  all leading minors are not greater than zero  negative semidefinite  𝐼(1,−1/2) = 6 6 6 24  all leading minors are greater than zero  positive definite

• Second order sufficient condition (SOC)

𝑦∗ is a maximum/minimum if 𝑒2𝑨 at 𝑦∗ is negative/positive definite OR 1) If (−1)𝑙 𝐸𝑙(𝑦∗) > 0, 𝑙 = 1, … , 𝑜, then (𝑦∗) is a local maximum 2) If 𝐸𝑙(𝑦∗) > 0, 𝑙 = 1, … , 𝑜, then (𝑦∗) is a local minimum 3) If 𝐸𝑜(𝑦∗) ≠ 0, and neither of the two conditions above is satisfied, then (𝑦∗) is a saddle point

• Example: 𝑨 = 𝑦3 − 8𝑧3 + 6𝑦𝑧 + 1; 𝐼 = 6𝑦

6 6 −48𝑧

 𝐼(0,0) = 0 6 6 0 → negative semidefinite ▫ saddle point  𝐼(1,−1/2) = 6 6 6 24 → positive definite ▫ local minimum

Optimization with constraints

• Consider cost function

𝐿 𝑦, 𝑧 = 2𝑦 + 3𝑧 with respect to production function g 𝑦, 𝑧 = 10 𝑦𝑧 = 100, 𝑦 ≥ 0, y ≥ 0 1. Rewrite g 𝑦, 𝑧 : 2. Plug into 𝐿 𝑦, 𝑧 : 3. Take the derivative of 𝐿 𝑦 :

𝑧 = 100 𝑦 𝐿 𝑦 = 2𝑦 + 3 ∙ 100 𝑦 𝐿′(𝑦) = 2 − 300 𝑦2 → 𝐿′ 𝑦 = 0 ⇒ 𝑦0 = 150 ≈ 12.2 ⇒ 𝑧0 = 100 150 ≈ 8.2

Method of Lagrange multipliers

• Problem: maximize/minimize

𝑔 𝑦1, … , 𝑦𝑜 subject to (s.t.) 𝑕𝑘 𝑦1, … , 𝑦𝑜 = 𝑐𝑘, 𝑘 = 1, … , 𝑛 < 𝑜

• Conditions

▫ Jacobian matrix is full-ranked

 𝑠𝑏𝑜𝑙 𝐾 = 𝑛  𝐾 =

𝜖𝑕1 𝜖𝑦1

…

𝜖𝑕1 𝜖𝑦𝑜

⋮ ⋱ ⋮

𝜖𝑕𝑛 𝜖𝑦1

…

𝜖𝑕𝑛 𝜖𝑦𝑜

• all functions are continuously

differentiable

• First step:

▫ build the Lagrangian function 𝑀 = 𝑔 𝑦1, … , 𝑦𝑜 + 𝜇𝑘 𝑐

𝑘 − 𝑕𝑘 𝑦1, … , 𝑦𝑜

= 0

𝑛 𝑘=1

• Second step:

▫ equate all partials of 𝑀 with respect to 𝑦1, … , 𝑦𝑜, 𝜇1, … , 𝜇𝑛 to zero

• Third step:

▫ solve these equations for 𝑦1, … , 𝑦𝑜, 𝜇1, … , 𝜇𝑛 to get a set of stationary points of the Lagrangian

Example I – Part 1

• Maximize

▫ 𝑦𝑧 + 𝑦 + 𝑧 + 1

• subject to

▫ 𝑦 + 2𝑧 = 30

• Lagrangian function

𝑀 = 𝑦𝑧 + 𝑦 + 𝑧 + 1 + 𝜇 30 − (𝑦 + 2𝑧)

• First order necessary conditions

1.

𝜖𝑀 𝜖𝑦 = 𝑧 + 1 − 𝜇 = 0

2.

𝜖𝑀 𝜖𝑧 = 𝑦 + 1 − 2𝜇 = 0

3.

𝜖𝑀 𝜖𝜇 = 30 − 𝑦 − 2𝑧 = 0

• Solving

1. 𝜇 = 𝑧 + 1 2. 𝜇 = (𝑦 + 1) 2

⇒ 𝑦 = 2 y + 1 − 1

▫ Then

• 3. 30 − 𝑦 − 2𝑧 = 0

⇔ 30 − 2 y + 1 + 1 − 2𝑧 = 0 ⇔ 30 − 2𝑧 − 2 + 1 − 2𝑧 = 0

▫ 𝑧 =

29 4 = 7,25; 𝑦 = 31 2 ; 𝜇 = 15,5

these are stationary points

Example I – Part 2

• Second order sufficient conditions

▫ check whether a stationary point is a maximum/minimum ▫ check the sign definiteness of the bordered Hessian

• Bordered Hessian

𝐼 𝑠 = … ⋮ ⋱ ⋮ … 𝜖𝑕1 𝜖𝑦1 … 𝜖𝑕1 𝜖𝑦𝑠 ⋮ ⋱ ⋮ 𝜖𝑕𝑛 𝜖𝑦1 … 𝜖𝑕𝑛 𝜖𝑦𝑠 𝜖𝑕1 𝜖𝑦1 … 𝜖𝑕𝑛 𝜖𝑦1 ⋮ ⋱ ⋮ 𝜖𝑕1 𝜖𝑦𝑠 … 𝜖𝑕𝑛 𝜖𝑦𝑠 𝜖2𝑀 𝜖𝑦1𝜖𝑦1 … 𝜖2𝑀 𝜖𝑦1𝜖𝑦𝑠 ⋮ ⋱ ⋮ 𝜖2𝑀 𝜖𝑦𝑠𝜖𝑦1 … 𝜖2𝑀 𝜖𝑦𝑠𝜖𝑦𝑠

• Maximize 𝑦𝑧 + 𝑦 + 𝑧 + 1

s.t. 𝑦 + 2𝑧 = 30

• Recall: Lagrangian function

𝑀 = 𝑦𝑧 + 𝑦 + 𝑧 + 1 + 𝜇 30 − (𝑦 + 2𝑧)

𝐼 𝑠 = 1 2 1 2 1 1 det (𝐼 𝑠) = 𝐷1 + 𝐷2 + 𝐷3 = 4 𝐷1 = (−1)1+1∙ 0 ∙ (−1) = 0 𝐷2 = (−1)2+1∙ 1 ∙ (−2) = 2 𝐷3 = (−1)3+1∙ 2 ∙ 1 = 2 det (𝐼 𝑠) > 0, solution is local maximum

Example II – Part 1

• Minimize

▫ 2𝑦2 + 𝑧2 + 3𝑨2

• subject to

▫ 2𝑦 − 3𝑧 − 4𝑨 = 49

• Lagrangian function

𝑀 = 2𝑦2 + 𝑧2 + 3𝑨2 + 𝜇 49 − 2𝑦 + 3𝑧 + 4𝑨

• First order necessary conditions

1.

𝜖𝑀 𝜖𝑦 = 4𝑦 − 2𝜇 = 0

2.

𝜖𝑀 𝜖𝑧 = 2𝑧 + 3𝜇 = 0

3.

𝜖𝑀 𝜖𝑨 = 6𝑨 + 4𝜇 = 0

4.

𝜖𝑀 𝜖𝜇 = 49 − 2𝑦 + 3𝑧 + 4𝑨 = 0

• Solving

1. 𝜇 = 2𝑦 2. 𝜇 = − 2 3 𝑧 ⇒ 𝑦 = − 𝑧 3 3. 𝜇 = − 3 2 𝑨

⇒ 𝑨 = 4 9 𝑧

▫ Then

• 4. 2 − 𝑧 3

− 3𝑧 − 4 4 9 𝑧 = 49 ⇔ −6𝑧 9 − 27𝑧/9 − 16𝑧/9 = 49

▫ 𝑧 = −9; 𝑦 = 3; 𝑨 = −4; 𝜇 = 6

these are stationary points

Example II – Part 2

• Second order sufficient conditions

▫ check whether a stationary point is a maximum/minimum ▫ check the sign definiteness of the bordered Hessian

• Bordered Hessian (case of one

constraint)

𝐼 𝑠 = 𝜖𝑕 𝜖𝑦1 … 𝜖𝑕 𝜖𝑦𝑠 𝜖𝑕 𝜖𝑦1 ⋮ 𝜖𝑕 𝜖𝑦𝑠 𝜖2𝑀 𝜖𝑦1𝜖𝑦1 … 𝜖2𝑀 𝜖𝑦1𝜖𝑦𝑠 ⋮ ⋱ ⋮ 𝜖2𝑀 𝜖𝑦𝑠𝜖𝑦1 … 𝜖2𝑀 𝜖𝑦𝑠𝜖𝑦𝑠

• Minimize 2𝑦2 + 𝑧2 + 3𝑨2

s.t. 2𝑦 − 3𝑧 − 4𝑨 = 49

𝐼 𝑠 = 2 −3 −4 2 −3 −4 4 2 6 det (𝐼 𝑠) = 𝐷1 + 𝐷2 + 𝐷3 + 𝐷4 = −392 𝐷1 = (−1)1+1∙ 0 ∙ 48 = 0 𝐷2 = (−1)2+1∙ 2 ∙ 24 = −48 𝐷3 = (−1)3+1∙ −3 ∙ 72 = −216 𝐷4 = (−1)4+1∙ −4 ∙ −32 = −128 det (𝐼 𝑠) < 0, solution is local minimum

Necessary Kuhn Tucker conditions

• As in the equality case

▫ start by setting up the Lagrangian 𝑀 = 𝑔 𝑦1, . . , 𝑦𝑜 + 𝜇𝑘 𝑐𝑘 − 𝑕𝑘 𝑦1, . . , 𝑦𝑜

𝑛 𝑘=1

• The necessary (but not sufficient) Kuhn-Tucker conditions for a point

to be a maximum are:

▫

𝜖𝑀 𝜖𝑦𝑗 ≤ 0, 𝑦𝑗 ≥ 0, 𝑦𝑗 𝜖𝑀 𝜖𝑦𝑗 = 0 for all 𝑗 = 1, … , 𝑜

▫ 𝑕𝑘 𝑦1, … , 𝑦𝑜 ≤ 𝑐𝑘, 𝜇𝑘 ≥ 0, 𝜇𝑘 𝑐𝑘 − 𝑕𝑘 𝑦1, … , 𝑦𝑜 = 0 for all 𝑘 = 1, … , 𝑛

Example

• Example

max 4𝑦1 + 3𝑦2

𝑡. 𝑢. 2𝑦1 + 𝑦2 ≤ 10 𝑏𝑜𝑒 𝑦1, 𝑦2 ≥ 0

• Lagrangian

𝑀 = 4𝑦1 + 3𝑦2 + 𝜇 10 − 2𝑦1 − 𝑦2

• KT necessary conditions

1.

𝜖𝑀 𝜖𝑦1 = 4 − 2 𝜇 ≤ 0

2.

𝜖𝑀 𝜖𝑦2 = 3 − 𝜇 ≤ 0

3.

𝜖𝑀 𝜖𝜇 = 10 − 2𝑦1 − 𝑦2 ≤ 0

4. 𝑦1 ≥ 0, 𝑦2 ≥ 0, 𝜇 ≥ 0 5. 𝑦1 4 − 2 𝜇 = 0 6. 𝑦2 3 − 𝜇 = 0 7. 𝜇 10 − 2𝑦1 − 𝑦2 = 0

• Find possible maxima

▫ from 2.:

a. 𝜇 ≥ 3

▫ from 5 and a:

b. 𝑦1 = 0

▫ from 6, b, and 3:

c. 𝜇 = 3

▫ from b, c, and 7:

d. 𝑦2 = 10

The candidate point for maximum is

𝑦1 = 0; 𝑦2 = 10; 𝜇 = 3 check for the Kuhn-Tucker sufficient conditions

Sufficient Kuhn Tucker conditions

• Kuhn Tucker Sufficient Conditions (max)

1. 𝑔 𝑦1, . . , 𝑦𝑜 is differentiable and concave in the non negative orthant (𝑦𝑗 ≥ 0, 2D - quadrant) 2. each constraint function

𝑕𝑘 𝑦1, . . , 𝑦𝑜 ≥ 𝑐𝑘 is differentiable and convex in the non negative orthant

3. a point 𝑦∗ = 𝑦1, . . , 𝑦𝑜 satisfies the KT necessary conditions

• Example

▫ max 𝑔 𝑦1, 𝑦2 = 4𝑦1 + 3𝑦2 𝑡. 𝑢. 𝑕 𝑦1, 𝑦2 = 2𝑦1 + 𝑦2 ≤ 10 𝑏𝑜𝑒 𝑦1, 𝑦2 ≥ 0

• All linear funtions are both concave and convex

▫ condition 1 and 2 are also fulfilled

Example – Part 1

• max 𝑔 𝑦1, 𝑦2 = 2𝑦 + 3𝑧

𝑡. 𝑢. 𝑕 𝑦1, 𝑦2 = 𝑦2 + 𝑧2 ≤ 2

• Lagrangian

𝑀 = 2𝑦 + 3𝑧 + 𝜇 2 − 𝑦2 − 𝑧2

• KT necessary conditions

1.

𝜖𝑀 𝜖𝑦 = 2 − 2𝑦𝜇 ≤ 0

2.

𝜖𝑀 𝜖𝑧 = 3 − 2𝑧𝜇 ≤ 0

3.

𝜖𝑀 𝜖𝜇 = 2 − 𝑦2 − 𝑧2 ≤ 0

4. 𝑦 ≥ 0, 𝑧 ≥ 0, 𝜇 ≥ 0 5. 𝑦 2 − 2𝑦𝜇 = 0 6. 𝑧 3 − 2𝑧𝜇 = 0 7. 𝜇 2 − 𝑦2 − 𝑧2 = 0

• 5 implies that

▫ 𝑦 = 0 𝑝𝑠 𝑦 = 1/𝜇

• If 𝑦 = 0

▫ 1 cannot be statisfied. Thus

 𝑦 = 1/𝜇

• 6 implies that

▫ 𝑧 = 0 𝑝𝑠 𝑧 = 3/2𝜇

• If 𝑧 = 0

▫ 2 can not be satisfied. Thus

 𝑧 = 3/2𝜇

• Plugging these in 7

▫ 2 − 1/𝜇 2− 3/2𝜇 2= 0 ▫ 𝜇 =

13 8 (recall 𝜇 ≠ 0)

• Solution 𝑦∗ =

8 13 , 𝑧∗ = 18 13 , 𝜇∗ = 13 8

Example – Part 2

• max 𝑔 𝑦1, 𝑦2 = 2𝑦 + 3𝑧 𝑡. 𝑢. 𝑕 𝑦1, 𝑦2 = 𝑦2 + 𝑧2 ≤ 2
• Show that 𝑔 and 𝑕 satisfy the KT sufficient conditions and find the

maxima

• 𝑔 is linear, thus concave
• 𝑕

→ recall that the Hessian matrix of a convex function is positive definite

• 𝐼 =

𝑕𝑦𝑦 𝑕𝑧𝑦 𝑕𝑦𝑧 𝑕𝑧𝑧 = 2 2

▫ 𝐸1 = det 2 = 2 > 0, 𝐸2 = det 𝐼 = 4 > 0

• All leading minors of the Hessian matrix are greater than zero

→ 𝑕 is convex → (𝑦∗, 𝑧∗, 𝜇∗) is maximum

Approximation

Total differential I

• The total differential of 𝑧 is defined as

𝑒𝑧 = 𝜖𝑧 𝜖𝑦𝑗 𝑒𝑦𝑗

𝑜 𝑗=1

• Example

𝑧 = 𝑦1

2𝑦2 + 6𝑦1 − 𝑦2 3

𝜖𝑧 𝜖𝑦1 = 𝑔

𝑦1 = 2𝑦1𝑦2 + 6

𝜖𝑧 𝜖𝑦2 = 𝑔

𝑦2 = 𝑦1 2 − 3𝑦2 2

𝑒𝑧 = 𝜖𝑧 𝜖𝑦1 𝑒𝑦1 + 𝜖𝑧 𝜖𝑦2 𝑒𝑦2 𝑒𝑧 = (2𝑦1𝑦2+6)𝑒𝑦1 + (𝑦1

2 − 3𝑦2 2)𝑒𝑦2

Total differential II

• The total differential of 𝑧 = 𝑔(𝑦1, 𝑦2) closely approximates the

functional change ∆y for small ∆𝑦1 and ∆𝑦2. → How can ∆𝑧 = 𝑔 𝑦1 + ∆𝑦1, 𝑦2+∆𝑦2 − 𝑔(𝑦1, 𝑦2) be approximated?

• Using the approximation 𝑒𝑔(𝑦1, 𝑦2) ≈ ∆ 𝑔(𝑦1, 𝑦2), 𝑒𝑧 ≈ ∆𝑧

⟹ 𝑒𝑧 = 𝑒𝑔 = 𝜖𝑔 𝜖𝑦1 𝑒𝑦1 + 𝜖𝑔 𝜖𝑦2 𝑒𝑦2 ⟹ 𝑔 𝑦1 + ∆𝑦1, 𝑦2+∆𝑦2 ≈ 𝑔(𝑦1, 𝑦2) + 𝜖𝑔 𝜖𝑦1 𝑒𝑦1 + 𝜖𝑔 𝜖𝑦2 𝑒𝑦2

Total differential III

• Example: Consider 𝑧 = ln 𝑦1𝑦2 + 𝑦1

2 + 𝑦2

• Assume:

𝑦1 = 1, 𝑒𝑦1 = 0.05; 𝑦2 = 2, 𝑒𝑦2 = −0.02 ⟹ 𝑒𝑧 = 𝑦2 𝑦1𝑦2 + 2𝑦1 𝑒𝑦1 + 𝑦1 𝑦1𝑦2 + 1 𝑒𝑦2 = ( 1 𝑦1 + 2𝑦1)𝑒𝑦1 + ( 1 𝑦2 + 1)𝑒𝑦2 ⟹ 𝑒𝑧 = 1 1 + 2 ∙ 1 0.05 + 1 2 + 1 −0.02 = 0.12 ⟹ Δ𝑧 = 𝑚𝑜 1.05 ∙ 1.98 + 1.052 + 1.98 − [ln 1 ∙ 2 + 12 + 2] = 0.1212

Taylor series

• General notation

𝑔 𝑜 (𝑏) 𝑜! ∞ 𝑜=0

(𝑦 − 𝑏)𝑜= 𝑔 𝑏 +

𝑔′ 𝑏 1!

𝑦 − 𝑏 +

𝑔′′ 𝑏 2!

𝑦 − 𝑏 2 + ⋯

• Approximation of functions by using a finite number of terms of its

Taylor series

• Examples

1. Expansion around 𝑏 = 0: 2. Expansion around 𝑏 = 1:

𝑓𝑦 = 𝑦𝑜 𝑜!

∞ 𝑜=0

= 1 + 𝑦 + 𝑦2 2! + 𝑦3 3! + ⋯ ln 𝑦 = −1 𝑜+1 𝑜 𝑦 − 1 𝑜

∞ 𝑜=1

= 𝑦 − 1 − 𝑦 − 1 2 2 + 𝑦 − 1 3 3 − ⋯

Integral calculus

Integration theory

• The fundamental theorem of

calculus

differentiation and integration

are inverse processes

• 𝑔 𝑦 is a continuous function, its

indefinite integral is defined by

𝑔 𝑦 𝑒𝑦 = 𝐺 𝑦 + 𝐷 → set of all primitives of f(x)

• where

▫ 𝐺′ 𝑦 = 𝑔 𝑦 ▫ 𝐷 𝑗𝑡 𝑏𝑜 𝑏𝑠𝑐𝑗𝑢𝑠𝑏𝑠𝑧 𝑑𝑝𝑜𝑡𝑢𝑏𝑜𝑢

• Rules for integration

▫ 𝑑𝑔 𝑦 𝑒𝑦 = 𝑑 𝑔 𝑦 𝑒𝑦 ▫ [𝑔 𝑦 + 𝑕 𝑦 ]𝑒𝑦 = 𝑔 𝑦 𝑒𝑦 + 𝑕 𝑦 𝑒𝑦 ▫ 𝑦𝑜𝑒𝑦 =

𝑦𝑜+1 𝑜+1 + 𝐷

▫ 𝑏𝑔 𝑦 + 𝑐𝑕(𝑦) 𝑒𝑦 = 𝑏 𝑔 𝑦 𝑒𝑦 + 𝑐 𝑕 𝑦 𝑒𝑦

Definite integral

• The indefinite integral is a function
• The definite integral is a number

𝑔(𝑦)𝑒𝑦

𝑐 𝑏

= 𝐺 𝑐 − 𝐺(𝑏)

• Example:

▫ (𝑦2 + 1)𝑒𝑦

2

= [

𝑦3 3 + 𝑦]0 2 = 23 3 + 2 − 03 3 + 0 = 14 3

Primitive

• 𝐺′ 𝑦 = 𝑔 𝑦 , F is the primitive of f
• Examples

▫ 𝐺 𝑦 = 𝑦2, 𝑔 𝑦 = 2𝑦 ▫ 𝐺 𝑦 = 𝑓𝑦, 𝑔 𝑦 = 𝑓𝑦 ▫ 𝐺 𝑦 = ln 𝑔 𝑦 , 𝑔 𝑦 =

𝑔′ 𝑦 𝑔 𝑦

▫ 𝐺 𝑦 =

𝑏𝑦 ln (𝑏) , 𝑔 𝑦 = 𝑏𝑦

▫ 𝐺 𝑦 = sin (𝑦), 𝑔 𝑦 = cos (𝑦)

Integration

• Examples

▫ 𝑦7𝑒𝑦 =

𝑦8 8 + 𝐷

▫ 5𝑒𝑦 = 5𝑦 + 𝐷 ▫ (2𝑦2 + 3𝑦 + 2)𝑒𝑦 = 2𝑦2𝑒𝑦 + 3𝑦𝑒𝑦 + 2𝑒𝑦 =

2 3 𝑦3 + 3 2 𝑦2 + 2𝑦 + 𝐷

▫ sin (𝑦)𝑒𝑦 = − cos 𝑦 + 𝐷 ▫ cos (𝑦)𝑒𝑦 = sin (𝑦) + 𝐷

Integration by parts I

• Recall the product rule in differentiation

𝑔 𝑦 𝑕(𝑦) ′ = 𝑔 𝑦 𝑕′ 𝑦 + 𝑔′ 𝑦 𝑕 𝑦

• Integrate both sides → 𝑔 𝑦 𝑕(𝑦) = 𝑔 𝑦 𝑕′ 𝑦 + 𝑔′ 𝑦 𝑕 𝑦
• Solve for the first integral → 𝑔 𝑦 𝑕′ 𝑦 = 𝑔 𝑦 𝑕 𝑦 − 𝑔′ 𝑦 𝑕 𝑦
• Let

𝑣 = 𝑔 𝑦 𝑒𝑣 = 𝑔′ 𝑦 ; 𝑤 = 𝑕 𝑦 𝑒𝑤 = 𝑕′ 𝑦 ⇒ 𝑣𝑒𝑤 = 𝑣𝑤 − 𝑤𝑒𝑣

Integration by parts II

• Example

𝑦 𝑦 + 1𝑒𝑦

• Let

𝑣 = 𝑦 𝑏𝑜𝑒 𝑒𝑤 = 𝑦 + 1𝑒𝑦

• Then

𝑣𝑒𝑤 = 𝑦 𝑦 + 1𝑒𝑦 , 𝑤 = 2 3 (𝑦 + 1)3/2

• Using the formula

𝑦 𝑦 + 1𝑒𝑦 = 𝑦 2 3 (𝑦 + 1)3/2− 2 3 𝑦 + 1

3 2𝑒𝑦

= 𝑦

2 3 (𝑦 + 1)3/2− 2 3 2 5 𝑦 + 1

5 2 + 𝐷

= 𝑦

2 3 (𝑦 + 1)3/2− 4 15 𝑦 + 1

5 2 + 𝐷

𝑣𝑒𝑤 = 𝑣𝑤 − 𝑤𝑒𝑣

Substitution formula

• Complicated integration due to the composition of different functions

⟹ appropriately chosen substitution can simplify an integral problem → effect of changing the variable, integrand and the bounds of integration in case of a definite integral

𝑔 𝜒 𝑦 ∙ 𝜒′ 𝑦 𝑒𝑦 = 𝑔 𝑢 𝑒𝑢

𝜒(𝑐) 𝜒(𝑏) 𝑐 𝑏

• Example:

sin 2𝑦 𝑒𝑦

𝑑

= sin 𝑢 ∙ 1 2 𝑒𝑢

2𝑑

= 1 2 sin 𝑢 𝑒𝑢

2𝑑

= 1 2 [− cos 𝑢 ]0

2𝑑

= 1 2 − cos 2𝑑 + cos 0 = 1 2 1 − cos 2𝑑 2x = φ 𝑦 = 𝑢

⇒

𝑒𝑢 𝑒𝑦 = 2 ↔ 𝑒𝑢 = 2 ∙ 𝑒𝑦 = 𝜒′ 𝑦 ∙ 𝑒𝑦

↔ 𝑒𝑦 =

1 2 𝑒𝑢

φ 𝑐 = φ 𝑑 = 2𝑑

↓

φ 𝑏 = φ 0 = 0

↳

Appendix

Exercises I

1. Calculate : a) 𝐵 ∙ 𝐶 and 𝐶 ∙ 𝐵 with A = 1 2 3 , 𝐶 = 4, 5, 6 b) (𝐵 ∙ 𝐶)𝑈 with A = 3 2 1 −1 3 , 𝐶 = 1 1 −2 1 c) 𝐵 ∙ 𝐶 with 𝐵 = 1 −1 4 4 8 9 7 −2 1 , B = 1 2 3 + −2 1 4

Exercises II

2. Calculate the Hessian for the following functions: a) 𝑔 𝑦1, 𝑦2 = 440 + 4𝑦1 + 10𝑦2 − 𝑦1

2 + 3𝑦1𝑦2 − 2.5𝑦2 2

b) 𝑕 𝑦, 𝑧 = 𝑏0𝑦𝑏1𝑧𝑏2 − 𝑑 c) ℎ 𝑧1, 𝑧2 = (𝑧1 − 1)2+(𝑧2 − 1)2 d) 𝑙1 𝑏, 𝑐 = sin 𝑓𝑏𝑦 + 𝑐𝑦 , 𝑙2 𝑏, 𝑐 = cos ( 𝑐𝑦 +

𝑦 𝑏)

Exercises III

3. Calculate the determinant and the inverse matrix of the following matrices using different methods:

a) 𝐵 =

2 2 2 1 1 3 3 2 2 b) 𝐶 = 1 3 2 2 6 4 −5 7 1 c) C = 1 2 3 0 and D = 8𝑏 2 2 𝑏

Exercises IV

4. Solve the following systems using different methods: a) 3𝑦1 − 𝑦2 = 7 6𝑦1 + 2𝑦2 = −14 b) x − 4y + 2z = −2 x + 2y − 2z = −3 x − y = 4 5. For which values of 𝑏, 𝑑 ∈ ℝ does the following inhomogeneous system

• f linear equations exhibit no solution, exactly one solution or infinitely

many solutions? 𝑦1 − 𝑦4 = 2 𝑦2 − 𝑦3 + 𝑦4 = 3 𝑏𝑦3 = 1 𝑦2 + 𝑑𝑦4 = 0

Exercises V

6. Compute all eigenvalues of the following matrices: a) 𝐵1 = 2 −4 −1 −1 , 𝐵2 = 1 1 −1 1 b) 𝐶1 = 1 2 3 , 𝐶2 = 2 −2 3 3 −2 −1 2 c) 𝐷 = 𝑏 1 𝑏

Exercises VI

7. Differentiate the following functions: a) 𝑔: ℝ → ℝ, 𝑔 𝑦 = sin 𝑦 ∙ 𝑦2 b) 𝑕: ℝ\ −1,1 → ℝ, 𝑕 𝑦 =

2𝑦−1 𝑦2−1

c) ℎ: ℝ+ → ℝ+, ℎ 𝑦 = 𝑦3 + 2𝑦

3

8. Compute the following integrals: a) 𝑦 𝑓𝑦𝑒𝑦 b) sin 𝑦 cos 𝑦 𝑒𝑦