Mathematics A/3 - Ildikó Perjési-Hámori

Page 1

Mathematics A/3 Engineering Mathematics 3 Function series

1. Find the Fourier series generated by ݂ ሺ‫ݔ‬ሻǤ ͳ ݂݅ Ͳ ൑ ‫ ݔ‬൏ ߨ ݂ ሺ‫ ݔ‬ሻ ൌ ቊ ǡ ݂ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ ሺ‫ݔ‬ሻǡ െͳ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ Ͳ

݇‫א‬Ժ

‫ݔ‬ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ ߨǡ ݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ ʹ

ߨ ൅ ‫ Ͳ ݂݅ ݔ‬൑ ‫ ݔ‬൏ ߨ ʹ ǡ ݂ ሺ‫ ݔ‬ሻ ൌ ൞ ߨ െ ‫ ݂݅ ݔ‬െ ߨ ൑ ‫ ݔ‬൏ Ͳ ʹ

݇‫א‬Ժ

݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ

݇‫א‬Ժ

݂ሺ‫ݔ‬ሻ ൌ ȁ‫ݔ‬ȁ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ ߨǡ ݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ ‫ ݔ‬െ ͳ ݂݅ Ͳ ൑ ‫ ݔ‬൏ ߨ ǡ െͳ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ Ͳ

݂ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ

݇‫א‬Ժ

݇‫א‬Ժ

݂ ሺ‫ݔ‬ሻ ൌ ‫ ݔ‬ଶ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ ߨǡ ݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ

݇‫א‬Ժ

݂ ሺ‫ ݔ‬ሻ ൌ

݂ ሺ‫ ݔ‬ሻ ൌ ቊ

ߨ ߨ ‫ۓ‬െ ʹ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ െ ʹ ۖ ߨ ߨ ሺ ሻ ‫ ݂݅ ݔ‬െ ൑ ‫ ݔ‬൏ ǡ ݂ ‫ ݔ‬ൌ ʹ ʹ ‫۔‬ ۖ ߨ ݂݅ ߨ ൑ ‫ ݔ‬൏ ߨ ‫ʹ ە‬ ʹ ݂ ሺ‫ ݔ‬ሻ ൌ ቊ

െʹ‫ ݔ‬൅ ͳ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ Ͳ ǡ ʹ‫ ݔ‬൅ ͳ ݂݅ Ͳ ൑ ‫ ݔ‬൏ ߨ

‫ݔ‬ ൅ ͳ ݂݅ െ ߨ ൑ ‫ ݔ‬൏ Ͳ ǡ ݂ ሺ‫ ݔ‬ሻ ൌ ቐʹ ͳ ݂݅ Ͳ ൑ ‫ ݔ‬൏ ߨ

݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ ሺ‫ݔ‬ሻǡ

݂ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ ݂ ሺ‫ ݔ‬൅ ݇ʹߨሻ ൌ ݂ሺ‫ݔ‬ሻǡ

݇‫א‬Ժ

݇‫א‬Ժ ݇‫א‬Ժ

Ildikó Perjési-Hámori

Mathematics A/3

Pécs 2019

The Mathematics A/3 course material was developed under the project EFOP 3.4.316-2016-00005 "Innovative university in a modern city: open-minded, value-driven and inclusive approach in a 21st century higher education model".

EFOP 3.4.3-16-2016-00005 "Innovative university in a modern city: open-minded, value-driven and inclusive approach in a 21st century higher education model".

Ildikó Perjési-Hámori

Mathematics A/3

Pécs 2019

A Mathematics A/3 tananyag az EFOP-3.4.3-16-2016-00005 azonosító számú, „Korszerű egyetem a modern városban: Értékközpontúság, nyitottság és befogadó szemlélet egy 21. századi felsőoktatási modellben” című projekt keretében valósul meg.

EFOP-3.4.3-16-2016-00005 Korszerű egyetem a modern városban: Értékközpontúság, nyitottság és befogadó szemlélet egy 21. századi felsőoktatási modellben

Engineering Mathematics 3 Differential equations

PTE MIK Perjésiné dr. Hámori Ildikó

Introduction Thomas Robert Malthus: An Essay on the Principle of Population (1798) Problem: The simplest mathematical model of population growth is obtained by assuming that the rate of increase of the population at any time is proportional to the size of the population at that time. đ?’…đ?‘ľ đ?’…đ?’•

differential equation

= đ?’“đ?‘ľ,

initial condition

đ?‘ 0 = đ?‘ 0

where đ?‘ (đ?‘Ą) is the size of population (dependent variable), đ?‘Ą is the time (dependent variable), r is a positive constant. unknown: function Question: đ?‘ľ đ?’• =? 1 ′ đ?‘ đ?‘

Solution: ln( đ?‘ đ?‘Ą

= đ?‘&#x;đ?‘Ą + đ?‘™đ?‘› đ??ś

= đ?‘&#x; if đ?‘ ≠0

đ??śâ‰ 0

ln( đ?‘ đ?‘Ą

[đ?‘™đ?‘› đ?‘ (đ?‘Ą) ′ = đ?‘&#x; = đ?‘™đ?‘›đ?‘’ đ?‘&#x;đ?‘Ą + đ?‘™đ?‘› đ??ś = đ?‘™đ?‘› đ??śđ?‘’ đ?‘&#x;đ?‘Ą

đ?‘ľ(đ?’•) = đ?‘Şđ?’†đ?’“đ?’• If đ?‘ = 0 then đ?‘ ′ = 0, then đ?‘ ′ = đ?‘&#x;đ?‘ is true. So đ?‘ľ=đ?&#x;Ž Substituting the initial condition:

đ?‘ 0 = đ??śđ?‘’ đ?‘&#x;0 = đ??ś đ?‘ľ(đ?’•) = đ?‘ľđ?&#x;Ž đ?’†đ?’“đ?’• PTE MIK PerjĂŠsinĂŠ dr. HĂĄmori IldikĂł

general solution singular solution

particular solution

Introduction

The population of Earth between 400-2000

PTE MIK Perjésiné dr. Hámori Ildikó

Concept and classification of ODE Definítion: A differential equation is an equation that relates some unknown function with its derivatives. Comment: In applications, the functions usually represent physical quantities, the derivatives represent their rates of change, and the equation defines a relationship between the two. Definítion: An ordinary differential equation (ODE) is an equation containing a function of one independent variable and its derivatives. Classification: Order the highest derivative order that appears in the equation. first order: f  y, y , x   0 e.g. yx  2 sin x  cos y x second order: f  y, y, y , x   0 e.g. y   y  2e Linearity:

linear if the unknown function’s and its derivatives degree is only 1 and products of the unknown function and its derivatives are not allowed. y  y  tg x  3 y  y  2  x non linear: ln y  2 y  3 Homogeneity: homogeneous if each term consists of the unkown function or its derivatives y  y  2 yx  0 y  y  2 yx  cos x

inhomogeneous otherwise PTE MIK Perjésiné dr. Hámori Ildikó

Types of solutions of ODE Def: A function is a solution of the differential equation if the equation is satisfied for all values of the independent variable in the domain of the function. Def: Initial Condition(s) are a condition, or a set of conditions, on the solution that will allow us to determine which solution we are after.

Def: General solution: is a solution that contains all possible solutions Def: Particular solution: a solution that satisfies the initial condition Def: Singular solution, one that cannot be obtained from the general solution. It appears in differential equations when there is a need to divide a term that might be equal to zero

PTE MIK Perjésiné dr. Hámori Ildikó

Direction field of ODE The method of directional fields is a graphical method for displaying the general shape and behavior of solution to � ′ = �(�, �) . It soes not require solving differential equation. A direction field line segment is represented by an arrow, to show the direction of the tangent. The tangent vector for � = �� + � � � are drawn from �′ = �� + �(�, �)�. Each time we specify an initial condition � �0 = �0 for the solution � ′ = � �, � , the solution curve (graph of the solution) is required to pass through the point (�0 , �0 ) and to have slope �(�0 , �0 ) there.

direction field

direction field and particular solutions

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

First order ODE F x , y , y  0 implicit form

y  f x , y  explicit form

Separable differential equation

y  f x   g  y  Solution: If g  y   0

y  f x gy

Integration of both sides respect to x (independent variable) y

 g  y dx   f x dx Knowing that l y  yx

ydx  dy

1  g  y dy   f x dx

G y   F  x   c If y(x) can be expressed, we get the solution in explicit form, if not, the solution is in implicit form.

PTE MIK Perjésiné dr. Hámori Ildikó

Application Verhulst equation: when the population growth is constrained by limited resources, a heuristic modification of the Malthusian growth model results in the Verhulst equation (If the the population increases, several factors will begin to affect the growth rate. For example, there will be increased competition for the limited resources that are available, increases in disease, and overcrowding of the limited available space, all of which would serve to slow the growth rate.)

đ?‘‘đ?‘ đ?‘ = đ?‘&#x;đ?‘ 1 − đ?‘‘đ?‘Ą đ?‘˜ where đ?‘&#x; is the growth rate, đ?‘˜ is the carrying capacity of the environment.

Let

đ?‘ đ?‘˜

= đ?‘Ś,

đ?‘&#x;đ?‘Ą = đ?‘Ľ. Then

General solution:

đ?‘Ś(đ?‘Ľ) =

đ?‘Śâ€˛ = đ?‘Ś 1 − đ?‘Ś

1 1+đ??śđ?‘’ −đ?‘Ľ

đ?‘Ś đ?‘Ľ =?

đ??śâ‰ 0

Singular solutions: đ?‘Ś = 0 (no population) đ?‘Ś = 1 (k=N) Initial condition: đ?‘Ś 0 = đ?‘Ś0 , Particular solution: đ?‘Ś đ?‘Ľ =

đ?‘Ś0 đ?‘Ś0 +(1−đ?‘Ś0 )đ?‘’ −đ?‘Ľ

lim đ?‘Ś đ?‘Ľ = 1

đ?‘Ľâ†’∞

The population, therefore, grows in size until it reaches the carrying capacity of its environment. PTE MIK PerjĂŠsinĂŠ dr. HĂĄmori IldikĂł

:

First order ODE Linear, inhomogeneous differential equation

y  px y  qx

If qx   0

y  px  y  0

equation is homogenous

Solution: : Variable coefficient method

First solve the homogeneous ODE. This is a separable ODE

Y   px Y  0

Y   p x  Y ln Y    p x  dx  ln C C0

(to distinguish Y was written.).

 Y dY   px  dx 1

 p ( x ) dx Y  Ce 

Using this solution to find the solution of the inhomogeneous ODE, substitute C constant with  p  x dx C(x) unknown function. y  C x e   p  x dx  p  x dx y   C x e   C x  px e 

Our goal is to determine C(x). Substitue y and y  back to the original ODE.

 p  x dx  p  x dx  p  x dx Cx e   Cx   px   e   px   Cx e   qx 

 p  x dx Cx e   qx 

Cx   qx  e 

p  x dx

y   qx e 

Cx    qx  e  p  x dx

p  x dx

 p  x dx dx  e 

PTE MIK Perjésiné dr. Hámori Ildikó

dx

:

Application RL circuit The figure represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Close: The equation accounting for both resistance and inductance is đ??ż

đ?‘‘đ?‘– đ?‘‘đ?‘Ą

+ �� = �, � 0 = 0

Open:

PTE MIK PerjĂŠsinĂŠ dr. HĂĄmori IldikĂł

:

Numerical solution of first order ODE – Euler method If we do not require or cannot find an exact solution giving an explicit formula for an initial value problem � ′ = � �, � , � �0 = �0 , we can generate a table of approximate numerical values of y for values x in an approximate interval. This type of the solution is called numerical solution of the problem. � ′ � = � �, � � ,

y đ?‘Ľ0 = đ?‘Ś0

approximate the derivative with the difference quotient đ?‘Ś đ?‘Ľ +â„Ž −đ?‘Ś(đ?‘Ľ )

0 đ?‘Śâ€˛ đ?‘Ľ ≈ 0 so â„Ž đ?‘Ś đ?‘Ľ0 + â„Ž − đ?‘Ś(đ?‘Ľ0 ) ≈ đ?‘“ đ?‘Ľ0 , đ?‘Ś đ?‘Ľ0 â„Ž đ?‘Ś đ?‘Ľ0 + â„Ž ≈ đ?‘Ś đ?‘Ľ0 + â„Žđ?‘“ đ?‘Ľ0 , đ?‘Ś đ?‘Ľ0 đ?‘Ś đ?‘Ľ0 + 2â„Ž ≈ đ?‘Ś đ?‘Ľ0 + â„Ž + â„Žđ?‘“ đ?‘Ľ0 + â„Ž, đ?‘Ś đ?‘Ľ0 + â„Ž â‹Ž

đ?‘Ś đ?‘Ľ0 + đ?‘›â„Ž ≈ đ?‘Ś đ?‘Ľ0 + (đ?‘› − 1)â„Ž + â„Žđ?‘“ đ?‘Ľ0 + (đ?‘› − 1)â„Ž, đ?‘Ś đ?‘Ľ0 + (đ?‘› − 1)â„Ž

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

Second order ODE Linear, constant coefficients, homogeneous differential equation b, c  R y  by  cy  0 Def: y1(x) and y2(x) functions are lineary independent of each other, if c1 y1 x  c2 y2 x  0  c1  c2  0 c1 ,c2 R Theoreme: If y1 and y2 are two independent particular solutions of a linear, constant coefficient(s), homogenous differential equation, then the general solution is :

y  c1 y1  c2 y2 Solution: Find two independent particular solutions

y  e x 2 e x  be x  cex  0

y   e x

y   2 e x

e x 2  b  c  0

2  b  c  0 characteristic equation of ODE (quadratic equation to -ra) Depending on the discriminant D = b2-4c there are 3 different cases: PTE MIK Perjésiné dr. Hámori Ildikó

:

Second order ODE I. D  b 2  4c  0

y1  e 1 x

two different real roots 1, 2

y 2  e 2 x

II. D  b 2  4c  0 1  2   It can be proved, that of each other.

y  c1e 1 x  c2 e 2 x

general solution b 2

y 2  xe

y1  e

one double real root b  x 2

is a solution as well, and y1, y2 are linearly independent

general solution III. D  b2  4c  0

b  x 2

y  c1e

b  x 2

 c2 xe

b  x 2

there is no real solution b 4c  b 2 b b 2  4c 1 1,2       2 2 2 2 i    

complex unit real number It can be proved, that the two linearly independent solutions are: y1  e

b  x 2

 cosx

y2  e

b  x 2

 sin x

general solution

PTE MIK Perjésiné dr. Hámori Ildikó

ye

b  x 2

c1  cos x  c2 sin x 

Application When a spring is slightly deformed, it creates a force proportional and in opposite direction to the deformation. Consider a spring-plus-body system with a spring constant D, body mass m. From Newton’s law: đ?‘šđ?‘Ž = −đ??ˇđ?‘Ś đ?‘‘2đ?‘Ś đ?‘š 2 = −đ??ˇđ?‘Ś đ?‘‘đ?‘Ą Let

đ??ˇ đ?‘š

=

đ?œ”2 ,

and

đ?‘‘2 đ?‘Ś đ?‘‘đ?‘Ą 2

= đ?‘Śáˆˇ then đ?‘Śáˆˇ + đ?œ”2 đ?‘Ś = 0

General solution: Initial conditions: đ?‘Ś 0 = 0, đ?‘Śáˆś 0 = đ?‘Łđ?‘šđ?‘Žđ?‘Ľ, Particular solution:

đ?‘Łđ?‘šđ?‘Žđ?‘Ľ đ?œ”

= đ??´.

= đ??´đ?‘ đ?‘–đ?‘›(đ?œ”đ?‘Ą)

PTE MIK PerjĂŠsinĂŠ dr. HĂĄmori IldikĂł

:

Second order ODE Linear, constant coefficients, inhomogenous differential equation

y  by  cy  r (x)

b, c ďƒŽ R

Where đ?‘&#x; đ?‘Ľ is a linear combination of đ?‘’ đ?›źđ?‘Ľ , đ?‘ đ?‘–đ?‘›đ?›˝đ?‘Ľ, đ?‘?đ?‘œđ?‘ đ?›˝đ?‘Ľ and a power function. Solution: Unknown coefficient method Steps. 1. Solve the homogeneous equation. 2. Write a function similar to đ?‘&#x; đ?‘Ľ with unknown coefficients 3. Substitute this function with its derivatives in the original differential equation. 4. Write an equation system, knowing that the coefficients of the same functions are the same on both sides. 5. Solving the equation system get the unknown coefficients and the particular solution of the ODE. 6. The general solution of the original ODE is the sum of the general solution of the homogeneous ODE and the particular solution of the inhomogeneous ODE.

PTE MIK PerjĂŠsinĂŠ dr. HĂĄmori IldikĂł

:

Second order ODE – numerical method �′′ � = � �, � � , � ′ (�) ,

y �0 = �0 , � ′ �0 = �0

Reduce the ODE into two first order ODE Notation: Initial conditions: Using the Euler method:

đ?‘Ł đ?‘Ľ = đ?‘Ś ′ (đ?‘Ľ), đ?‘Ł ′ (đ?‘Ľ)=đ?‘“ đ?‘Ľ, đ?‘Ś đ?‘Ľ , đ?‘Ł(đ?‘Ľ) y đ?‘Ľ0 = đ?‘Ś0 , đ?‘Ł đ?‘Ľ0 = đ?‘Ł0 y đ?‘Ľ0 + â„Ž ≈ đ?‘Ś0 + h đ?‘Ł đ?‘Ľ0 v đ?‘Ľ0 + â„Ž ≈ đ?‘Ł0 + hđ?‘“ đ?‘Ľ, đ?‘Ś đ?‘Ľ , đ?‘Ł(đ?‘Ľ) â‹Ž

Example: Pendulum motion, when the initial deflection is greater then 5o

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

Laplace transformation The Laplace transform is most useful for solving linear, constant-coefficient ODE’s when the inhomogeneous term or its derivative is discontinuous. The main idea is to Laplace transform the constant-coefficient differential equation for y(đ?‘Ą) into a simpler algebraic equation for the Laplace-transformed functionY(đ?‘ ) solve this algebraic equation, and then transform Y(đ?‘ ) back into y(đ?‘Ą). Def. Laplace transformation of đ?‘“(đ?‘Ą), denoted by đ??š đ?‘ = đ??ż đ?‘“ đ?‘Ą transform

is defined by the integral

∞

đ??š đ?‘ = ŕśą đ?‘’ −đ?‘ đ?‘Ą đ?‘“ đ?‘Ą đ?‘‘đ?‘Ą 0

đ?‘“(đ?‘Ą)

đ??š(đ?‘ )

1

1 đ?‘ 1 đ?‘ −đ?‘Ž đ?‘›! đ?‘ đ?‘›+1 đ?‘›! (đ?‘ − đ?‘Ž)đ?‘›+1

đ?‘’ đ?‘Žđ?‘Ą đ?‘Ąđ?‘› đ?‘Ą đ?‘› đ?‘’ đ?‘Žđ?‘Ą

đ?‘“(đ?‘Ą)

đ??š(đ?‘ )

đ?‘? đ?‘ 2 + đ?‘?2 đ?‘ cos(đ?‘?đ?‘Ą) đ?‘ 2 + đ?‘?2 đ?‘? đ?‘’ đ?‘Žđ?‘Ą sin(đ?‘?đ?‘Ą) (đ?‘ − đ?‘Ž)2 +đ?‘? 2 đ?‘ −đ?‘Ž đ?‘’ đ?‘Žđ?‘Ą cos(đ?‘?đ?‘Ą) (đ?‘ − đ?‘Ž)2 +đ?‘? 2 sin(đ?‘?đ?‘Ą)

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

Laplace transformation Heaviside function: đ??ť đ?‘Ą = ቊ

0 ℎ� � < 0 1 ℎ� � ≼ 0

Laplace transform: đ??š đ?‘ =

1 đ?‘

0 đ?‘–đ?‘“ đ?‘Ą < đ?‘Ž Discontinuous function: đ?‘“ đ?‘Ą = á‰?1 â„Žđ?‘Ž đ?‘Ž ≤ đ?‘Ą < đ?‘? 0 đ?‘–đ?‘“ đ?‘Ą ≼ đ?‘? Laplace transform:

đ??š đ?‘ =

đ?‘’ −đ?‘?đ?‘ −đ?‘’ −đ?‘Žđ?‘ đ?‘

Using Heaviside function: đ?‘“ đ?‘Ą = đ??ť đ?‘Ą − đ?‘Ž − đ??ť(đ?‘Ą − đ?‘?) The Dirac delta function denoted as đ?›ż đ?‘Ą is defined by requiring that for any function đ?‘“ đ?‘Ą ∞ â€Ť×Źâ€Źâˆ’âˆž đ?‘“ đ?‘Ą đ?›ż đ?‘Ą đ?‘‘đ?‘Ą = đ?‘“(0) with other words

0 â„Žđ?‘Ž đ?‘Ą ≠0 đ?›ż đ?‘Ą =ቊ ∞ â„Žđ?‘Ž đ?‘Ą = 0

Remark:

đ?‘‘đ??ť(đ?‘Ą) đ?‘‘đ?‘Ą

=� �

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

Laplace transforms Notation:

đ?‘‘ đ?‘“ đ?‘‘đ?‘Ą

đ?‘Ą = đ?‘“áˆś đ?‘Ą ,

đ?‘‘2 đ?‘“ đ?‘‘đ?‘Ą 2

đ?‘Ą = đ?‘“áˆˇ đ?‘Ą

(If t is the time in physical applications)

áˆś is piecewise Theorem1 Suppose lim đ?‘’ −đ?‘ đ?‘Ą đ?‘“ đ?‘Ą = 0 and that đ?‘“(đ?‘Ą) is continuous and đ?‘“(đ?‘Ą) đ?‘Ąâ†’∞

continuous on any interval 0 ≤ đ?‘Ą ≤ đ??´ , then đ??ż đ?‘“áˆś đ?‘Ą Proof

= đ?‘ đ??ż đ?‘“ t

∞

đ??ż đ?‘“áˆś đ?‘Ą

− đ?‘“(0)

∞

−đ?‘ đ?‘Ą = ŕśą đ?‘“áˆś đ?‘Ą đ?‘’ −đ?‘ đ?‘Ą đ?‘‘đ?‘Ą = [đ?‘’ −đ?‘ đ?‘Ą đ?‘“ đ?‘Ą ]∞ đ?‘“ đ?‘Ą đ?‘‘đ?‘Ą = −đ?‘“ 0 + đ?‘ đ??ż đ?‘“ t 0 − ŕśą −đ?‘ đ?‘’ 0

0

áˆś áˆˇ is piecewise Theorem2 Suppose lim đ?‘’ −đ?‘ đ?‘Ą đ?‘“ đ?‘Ą = 0 and that đ?‘“(đ?‘Ą) is continuous and đ?‘“đ?‘Ą) đ?‘Ąâ†’∞

continuous on any interval 0 ≤ đ?‘Ą ≤ đ??´ , then

đ??ż đ?‘“áˆśáˆˇ đ?‘Ą

= đ?‘ 2đ??ż đ?‘“ t

áˆś − sđ?‘“ 0 − đ?‘“(0)

Def. The inverse Laplace transform, denoted đ??żâˆ’1 , of a function F is đ??żâˆ’1 đ??š đ?‘

= đ?‘“ đ?‘Ą â&#x;ş F s = L{đ?‘“ đ?‘Ą }

PTE MIK IldikĂł PerjĂŠsi-HĂĄmori PhD

:

Solution linear ODEs using Laplace transformation Let the ODEРђЎ s shape a­ЮЉдрѕи ­ЮЉА + ­ЮЉЈ­ЮЉд ­ЮЉА + ­ЮЉд ­ЮЉА = ­ЮЉЪ ­ЮЉА ,

­ЮЉдрѕХ 0 = ­ЮЉд0 ,

рѕХ ­ЮЉд 0 = ­ЮЉБ0

The steps of the solution: 1. Take the Laplace transforms of both sides, using the Theorem 1 and 2 a ­ЮЉа 2 ­Юљ╣ ­ЮЉа Рѕњ s­ЮЉд 0 Рѕњ ­ЮЉдрѕХ 0

+ b sF s Рѕњ y 0

+F s =­Юљ┐ ­ЮЉЪ ­ЮЉА

2. Solve this equation to ­Юљ╣ ­ЮЉа 3. Take the inverse Laplace transform to ­Юљ╣ ­ЮЉа . It is the particular soltion of the original ODE.

PTE MIK Ildik├│ Perj├Еsi-H├Аmori PhD

Engineering Mathematics 3 Linear Algebra

PTE MIK Ildikó Perjésiné Hámori PhD

What is it about? The fundamental problem of linear algebra is to solve n linear equations in m unknowns Example from mechanics

PTE MIK Ildikó Perjésiné Hámori PhD

Forms of a linear equation system 2đ?‘Ľ − 3đ?‘Ś = 0 −đ?‘Ľ + 2đ?‘Ś = 2

đ?‘Ľ

2 −3 0 +đ?‘Ś = −1 2 2

2 −3 0 ,đ??? = , b= −1 2 2 are vectors, x, y are scalars. đ?‘Ľđ??œ + đ?‘Śđ??? is the linear combination of vectors c and d đ??œ=

xc+yd=b is the vector form The intersection of the plots (if they do intersect) represent the solution to the system of equations. đ?‘Ľ = 6, y = 4

Matrix form:

Vector form:

Row form:

If đ?‘Ľ = 6, y = 4, then 6

2 −3 0 +4 = −1 2 2

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

−3 đ?‘Ľ 0 = 2 đ?‘Ś 2

2 −1 đ??´=

2 −1

−3 2

is called a coefficient matrix đ?‘Ľ đ??ą = đ?‘Ś is an unknown vector, đ??´đ??ą = đ??› is the matrix form If đ??ą = 2 −1

6 then 4 −3 6 0 = 2 4 2

Matrices and matrix operations DefinĂ­tion: A matrix is a rectangular array of numbers. The numbers in the array are called the entries in the matrix

đ??´ = đ??´đ?‘šđ?‘Ľđ?‘›

đ?‘Ž11 = â‹Ž đ?‘Žđ?‘š1

â‹Ż đ?‘Ž1đ?‘› â‹ą â‹Ž Size is m by n (written đ?‘š Ă— đ?‘› or mn). In a size description, â‹Ż đ?‘Žđ?‘šđ?‘›

the first number denotes the number of rows, and the second denotes the number of columns. A matrix with only one column is called a column vector or a column matrix, and a matrix with only one row is called a row vector or a row matrix

đ??š = đ?‘Žŕ´¤ = đ?‘Ž1 , đ?‘Ž2 ‌ ‌ . . , đ?‘Žđ?‘› đ?‘?1 đ??› = đ?‘?ŕ´¤ = â‹Ž đ?‘?đ?‘›

A matrix A with n rows and n columns is called a square matrix of order n, the đ?‘Ž11 , đ?‘Ž22 ‌ ‌ . . , đ?‘Žđ?‘›đ?‘› entries are the main diagonal of A. A square matrix is called identity matrix, if the entries of the main diagonal are 1, the other entries are 0. ďƒŠ1 0 0ďƒš ďƒŞ ďƒş I 3  ďƒŞ0 1 0 ďƒş ďƒŞďƒŤ0 0 1ďƒşďƒť

If A is any matrix, then the transpose of A, denoted by AT, is defined to be the matrix that results by interchanging the rows and columns of A. PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Matrices and matrix operations Definítion: Two matrices are equal if they have the same size and their corresponding entries are equal. Amxn=Bmxn if aik=bik  i,, k (i=1…m, k=1…n)

Definition: If A and B are matrices of the same size, then the sum A+B (difference A-B) a ik  bik  a ik  bik m,n m,n m,n

 

Properties:

A+B=B+A

 

commutative

(A+B)+C=A+(B+C) associative A+0=A

zero element (a matrix for which every entry is 0)

A+X=0 (X=-A) Definítion: If A is any matrix and l is any scalar, then the product lA is the matrix obtained by multiplying each entry of the matrix A by l. The matrix lA is said to be a scalar multiple of A

l aik   laik 

Properties::

(l1l2)A=l1(l2A)

associative

(l1+l2)A=l1A+l2A

distributive

l(A+B)=lA+lB

distributive

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Multiplying Matrices 2 −1

−3 6 0 = 2 4 2

đ?‘Ž11 Definition: đ??´đ??ľ = â‹Ž đ?‘Žđ?‘š1

because 2 ∙ 6 + −3 ∙ 4 = 0, and (−1) ∙ 6 + 2 ∙ 4 = 2 â‹Ż â‹ą â‹Ż

đ?‘Ž1đ?‘&#x; â‹Ž đ?‘Žđ?‘šđ?‘&#x;

đ?‘?11 â‹Ž đ?‘?đ?‘&#x;1

â‹Ż â‹ą â‹Ż

đ?‘?1đ?‘› â‹Ž the entry in row i and column j of AB is given đ?‘?đ?‘&#x;đ?‘›

by (đ??´đ??ľ)đ?‘–đ?‘— = đ?‘Žđ?‘–1 đ?‘?1đ?‘— + đ?‘Žđ?‘–2 đ?‘?2đ?‘— + â‹Ż đ?‘Žđ?‘–đ?‘&#x; đ?‘?đ?‘&#x;đ?‘— . Remark: The definition of matrix multiplication requires that the number of columns of the first factor A be the same as the number of rows of the second factor B in order to form the product AB. Properties:

AB≠BA

non commutative

(AB)C=A (BC)

associative

(A+B)C=A B+AC distributive

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Determinant Definition: The determinant of the matrix A2x2 is the following number: a a det A  11 12  a11a22  a12a21 a21 a22 Definition: If A is a square matrix then the minor of entry aij is denoted Aij and is defined to be the determinant of the submatrix that remains after the ith row and jth column are deleted from A. The number đ??śđ?‘–đ?‘— = (−1)đ?‘–+đ?‘— đ??´đ?‘–đ?‘— is called the cofactor of entry aij Definition: If A is a nxn matrix, then the number obtained by multiplying the entries in any row or column of A by the corresponding cofactor and adding the resulting product is called the determinant of A, and the sums themselves are called cofactor expansions of A. det đ??´ = đ?‘Ž1đ?‘— đ??ś1đ?‘— + đ?‘Ž2đ?‘— đ??ś2đ?‘— + â‹Ż + đ?‘Žđ?‘›đ?‘— đ??śđ?‘›đ?‘— (cofactor expansion along the jth column) det đ??´ = đ?‘Žđ?‘–1 đ??śđ?‘–1 + đ?‘Žđ?‘–2 đ??śđ?‘–2 + â‹Ż + đ?‘Žđ?‘–đ?‘› đ??śđ?‘–đ?‘›

(cofactor expansion along the ith row)

Remark: Note that a minor and its corresponding cofactor are either the same or negatives of each other and that the relating sign (−1)đ?‘–+đ?‘— is either +1 or −1 in accordance with the pattern in the “checkerboardâ€? array

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Determinant Definition: If det A ≠0 then A is called regular matrix, if det A =0 then a is called singular matrix. Rule of Sarrus: Determinants of 3x3 matrices can be evaluated very efficiently using the pattern suggested in the next figure: a11 a12 a13 a11 a12 a13 det A  a21 a22 a23 a21 a22 a23  a11a22a33 + a12a23a31 + a13a21a32  (a11a23a32 + a12a21a33 + a12a21a33 )

a31 a32 a33 a31 a32 a33 Properties: Let A be a square matrix, 1. if A has a row of zeros or a column of zeros, then detA=0. 2. det A=det AT 3. if B is the matrix that results when a single row or single column of A is multiplied by a scalar k, then det B=kdet A. 4. If B is the matrix that results when two rows or two columns of A are interchanged then det B= −det A. 5. If B is the matrix that results when a multiple of one row of A is added to another row or when a multiple of one column is added to another column then det B=det A.

PTE MIK Ildikó Perjésiné Hámori PhD

Adjoint and inverse of a matrix Definition If A is any đ?‘› Ă— đ?‘› matrix and Cij is the cofactor of aij then the matrix ďƒŠC11 C21 ďƒŞ C C22 adjA  ďƒŞ 12 ďƒŞ ďƒŞ ďƒŞďƒŤC1n

Cn1 ďƒš ďƒş Cn 2 ďƒş ďƒş ďƒş Cnn ďƒşďƒť

is called the adjoint of A. Remark: adjA is the transpose of the cofactor matrix. Theoreme: If A is any nxn matrix (adj A)A=A(adj A)=(detA)I, where I is the identity matrix.

Corrolary: If det A ≠0, then

A(adjA) I det A

Definition: If A is a regular square matrix (det A ≠0), đ??´âˆ’1 is called the inverse of the matrix, if đ??´âˆ’1 đ??´ = đ??´đ??´âˆ’1 = đ??ź , where I is the identity matrix. Corollary:

A1 

adjA det A

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Solution of a regular linear equation system Defintion The form of a linear equation system with as many equations as unknowns (n) is đ?‘Ž11 đ?‘Ľ1 + đ?‘Ž12 đ?‘Ľ2 + â‹Ż + đ?‘Ž1đ?‘› đ?‘Ľđ?‘› = đ?‘?1 đ?‘Ž21 đ?‘Ľ1 + đ?‘Ž22 đ?‘Ľ2 + â‹Ż + đ?‘Ž2đ?‘› đ?‘Ľđ?‘› = đ?‘?2 â‹Ž đ?‘Žđ?‘›1 đ?‘Ľ1 + đ?‘Žđ?‘›2 đ?‘Ľ2 + â‹Ż + đ?‘Žđ?‘›đ?‘› đ?‘Ľđ?‘› = đ?‘?đ?‘›1 where the xk (k = 1,2,...,n ) is the kth unknown, the aik (i = 1,2, . . . ,n; k = 1,2, . . .,n) , bi are real numbers. Define a matrix from the coefficients, from the unknown and from the constants from the right hand side. ďƒŠ a11 a12 ďƒŞ ďƒŞa21 a22 A  A nxn  ďƒŞ ďƒŞ ďƒŞ ďƒŞa ďƒŤ n1 an 2

ďƒŠ x1 ďƒš a1n ďƒš ďƒŞ ďƒş ďƒş a2n ďƒş ďƒŞ x2 ďƒş ďƒş xď€˝ďƒŞ . ďƒş ďƒŞ ďƒş ďƒş ďƒŞ . ďƒş ďƒş ďƒŞx ďƒş ďƒŤ nďƒť ann ďƒşďƒť

ďƒŠ b1 ďƒš ďƒŞ ďƒş ďƒŞb2 ďƒş bď€˝ďƒŞ . ďƒş ďƒŞ ďƒş ďƒŞ.ďƒş ďƒŞb ďƒş ďƒŤ nďƒť

The short form of the system is Ax=b.

Definition: A linear equation system is called a regular one, if it has as many equations as unknowns and det A ≠0.

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Cramer’s rule Cramer’s rule: The solution of a regular equation system is x=A-1 b

Theorem: If Ax=b is a system of n linear equations in n unknown such det A ≠0, then the system •

is solvable

•

has a unique solution. det đ??´

det đ??´

1 2 The solution is đ?‘Ľ1 = , đ?‘Ľ2 = ,‌, đ?‘Ľđ?‘› = det đ??´ det đ??´ replacing the entries in jth column of A in matrix b.

det đ??´đ?‘› det đ??´

where Aj is the matrix obtained by

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Matrix form of a general linear equation system Definition: A system of m linear equations in n unknown is a11x1 + a12 x2 + ... + a1k xk + ... + a1n xn  b1 a21x1 + a22 x2 + ... + a2k xk + ... + a2n xn  b2 . ai1x1 + ai 2 x2 + ... + aik xk + ... + ain xn  bi . am1x1 + am 2 x2 + ... + amk xk + ... + amn xn  bm

where the xk (k = 1,2,...,n ) is the kth unknown, the aik (i = 1,2, . . . ,m; k = 1,2, . . .,n) , bi are real numbers.  a11 a12 . . a1n b1   b1   x1  a1n   a11 a12   b  x     a a . . a b 21 22 2n 2 a2n   2  a21 a22  2    x .  b .  B  Bmx( n +1)   . A  Amxn   . . . . . Ax=b           . . . . .  .  .   .  a    amn  x n  b m   m1 am 2   am1 am 2 . . amn bn    coefficient matrix

unknown vector

constant vector

PTE MIK Ildikó Perjésiné Hámori PhD

augmented matrix

Existence and uniqueness of the solution, elementary row operations A solution of the system is a list {s1 , s2, . . . , sn } of numbers that makes each equation a true statement when the values s1 , s2, . . . , sn are substituted for x1 , x2, . . . , xn , respectively. A system of linear equations has • no solution or • exactly one solution or • infinitely many solutions

A system of linear equations is said to be consistent if it has either one solution or infinitely many solutions; a system is inconsistent if it has no solution.

Questions: 1. Is the system consistent that is does at least one solution exists? 2. If a solution exists, is it the only one; that is, is the solution unique? Algebraic operations that do not alter the solution set: E1. Multiply an equation through by a nonzero constant. E2. Interchange two equations. E3. Add a constant times one equation to another.

Row operations with an augmented matrix correspond to the equation: M1. Multiply a row through by a nonzero constant. M2. Interchange two rows. M3. Add a constant times one row to another.

PTE MIK Ildikó Perjésiné Hámori PhD

Reduced row echelon form: Gauss-Jordan elimination Forward phase: A matrix is in row echelon form if it has the following properties: 1. All nonzero rows are above any rows of all zeros. 2. Each leading entry of a row is in a column to the right of the leading entry of the row above it, 3. All entries in a column below a leading entry are zeros. Backward phase: In reduced echelon form there are two additional conditions: 4. The leading entry in each nonzero row is 1. 5. Each leading 1 is the only nonzero entry in its column.

row echelon form REF

reduced row echelon form RREF

A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position. A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column – that is, if and only if an echelon form of the augmented matrix has no row of the form [0 ‌ 0 đ?‘?] with b nonzero. PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Solving a linear system using Gauss-Jordan elimination 1. Write the augmented matrix of the system. 2. Use the reduction algorithm to obtain an equivalent augmented matrix in echelon form. Decide whether the system is consistent. If there is no solution, stop, otherwise go to the next step. 3. Continue row reduction to obtain the reduced echelon form. 4. Write the system of equation corresponding to the matrix obtained in step 3. 5. Rewrite each nonzero equation from step 4 so that its one basic variable is expressed in terms of any free variables appearing in the equation. x1 + 2 x2 + 3x3  1 x1 + 3x2 + 5 x3  1 3x1  x2  4 x3  1 9 x1 + 2 x2  x3  1 5 x1 + 2 x2 + x3  1

x1  8x 2 + 9x 3  32 2x1  x 2 + 3x 3  1 x1 + 2x 2  x 3  12

x1  1

x1 + x 2 + x 3  x 4  4

x2  4

x1  x 2 + x 3 + x 3  8

x3  2

3x1 + x 2 + 3x 3  x 4  16

one solution

x1  6  x3 x2  2 + x4

no solution

infinite number of solutions with two free variables

PTE MIK Ildikó Perjésiné Hámori PhD

Vector space Definition: Let V be an arbitrary nonempty set of objects, on which the addition and multiplication by scalars are defined. If the following axioms are satisfied by all object u, v, w in V and all scalars l and m, then we call V a vector space and we call the objects in V vectors. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

If u and w are objects in V, then u+v is in V. u+v= v+u u+(v+w)=(u+v)+w There is an object 0 in V, called a zero vector for V, such that 0+u= u+0= u . For each u in V, there is an object –u in V, called a negative of u, such that u+(–u) = (–u) +u=0. If l is any scalar and u is an object in V, then lu is in V. l(u+v)=lu+lv (l+m)u= lu+mu l(mu)= (lm)u 1u=u

Remark: The definition of a vector space does not specify the nature of vectors or the operations. Any kind of object can be a vector . The only requirement is that the ten vector space axioms be satisfied. Examples: 1. Geometrical vectors: � = �2 (vectors in plane is called 2-space vector), � = �3 (vectors in space, 3-space ) 2. � = �3�3 (3 by 3 matrices) 3. V : convergent infinite sequences of real numbers 4. V : real functions with one variable PTE MIK Ildikó PerjÊsinÊ Håmori PhD

Subspace, linear combination, linearly independent and dependent set Definition: A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V. Subspace examples 1. Planes through the origin are subspaces of â„?3 0 đ?‘Ž 2. đ?‘€ = matrices, đ?‘Ž, đ?‘? ∈ â„? are subspaces of đ?‘€2đ?‘Ľ2 đ?‘? 0 3. Number sequence approaches to 0 is a subspace convergent infinite sequences of real numbers

Definition: If w is a vector in a vector space V , then w is said to be a linear combination of the vectors v1, v2, ‌., vn in V, if w can be expressed in the form đ??° = đ?œ†1 đ??Ż1 + đ?œ†2 đ??Ż2 +‌+đ?‘Ľđ?œ†đ?‘› đ??Żđ?‘› Definition: Let S={v1, v2, ‌., vr }be a nonempty set of vectors in a vector space V. If the vector equation đ?œ†1 đ??Ż1 + đ?œ†2 đ??Ż2 +‌+đ?œ†đ?‘› đ??Żđ?‘› = đ?&#x;Ž has only one solution (trivial solution) namely đ?œ†1 = 0, đ?œ†2 = 0, ‌ . , đ?œ†đ?‘› = 0, then we call S a linearly independent set. If there are solutions in addition to the trivial solution, then S is said to be a linearly dependent set. Linearly independent examples 1. In â„?3 space đ??˘ =(1,0,0), đ??Ł =(0,1,0), đ??¤ =(0,0,1) 2. In â„?đ?‘› space the set of standard unit vectors đ??žđ?&#x;? =(1,0,0,‌.,0), đ??žđ?&#x;? =(0,1,0,‌.,0),‌ , đ??žđ?’? =(0,0,0,‌.,1) PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Linearly independent and dependent vectors

PTE MIK Ildikó Perjésiné Hámori PhD

Span, basis, coordinates, coordinate mapping Definition: If any w vector in a vector space V is expressed as a linear combination đ??° = đ?‘?1 đ??Ż1 + đ?‘?2 đ??Ż2 +‌+đ?‘?đ?‘› đ??Żđ?‘› for some coefficients ci, then the vectors đ??Ż1 , đ??Ż2 , ‌ , đ??Żđ?‘› are said to span the space. Notation: span(V) Definition: If V is any vector space and v1, v2, ‌., vn is a finite set of vectors in V, then it is called a basis for V, if • v1, v2, ‌., vn set is linearly independent • v1, v2, ‌., vn set spans V. Definition: If S={v1, v2, ‌., vn } is a basis for a vector space V and đ??° = đ?‘?1 đ??Ż1 + đ?‘?2 đ??Ż2 +‌+đ?‘?đ?‘› đ??Żđ?‘› is the expression for a vector w in terms of the basis S, then scalars đ?‘?1 , đ?‘?2 , ‌ , đ?‘?đ?‘› are called the coordinates of w relative to the basis S, the đ??°đ?‘† =( đ?‘?1 , đ?‘?2 , ‌ , đ?‘?đ?‘› ) vector is called coordinate vector of w. Definition: If S={v1, v2, ‌., vn } is a basis for a finite-dimensional vectorspace V and if đ??°đ?‘† =( đ?‘?1 , đ?‘?2 , ‌ , đ?‘?đ?‘› ) is the coordinate vector of w relative S, then, the mapping đ??° → (đ??°)đ?‘† creates a one-to-one correspondence between vectors in the generated space V and vectors in the familiar vector space â„?đ?‘› . It is called a coordinate map from V to â„?đ?‘› . The coordinate vector in matrix form: đ?‘?1 đ?‘?2 [đ??°]đ?‘† = â‹Ž đ?‘?đ?‘› PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Coordinates in â„?2 and â„?3 Trivial bases 1. In â„?3 space đ??˘ = (1,0,0), đ??Ł = (0,1,0), đ??¤ = (0,0,1) 2. In â„?đ?‘› space the set of standard unit vectors đ??žđ?&#x;? = (1,0,0,‌.,0), đ??žđ?&#x;? = (0,1,0,‌.,0), â‹Ž đ??žđ?’? = (0,0,0,‌.,1)

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Dimension of a vector space, rank of a matrix Definition: The dimension of a finite-dimensional vector space V is denoted by dim(V) and is defined to be the number of vectors in a basis for V. Dimension in other words: degrees of freedom.

For an nxm matrix

ďƒŠ a11 a12 ďƒŞ ďƒŞ a21 a22 Aď€˝ďƒŞ ďƒŞ ďƒŞ ďƒŞa ďƒŤ m1 am 2

a1n ďƒš ďƒş a2n ďƒş ďƒş ďƒş ďƒş amn ďƒşďƒť

Definition: The vectors đ??Ż1 = đ?‘Ž11 đ?‘Ž12 ‌ đ?‘Ž1đ?‘› đ??Ż2 = đ?‘Ž21 đ?‘Ž22 ‌ đ?‘Ž2đ?‘› â‹Ž đ??Żđ?‘š = đ?‘Žđ?‘š1 đ?‘Žđ?‘š2 ‌ đ?‘Žđ?‘šđ?‘› đ?‘› in â„? that are formed from the rows of A are called the row vectors of A.

Definition:

đ?‘Ž1đ?‘› đ?‘Ž11 đ?‘Ž12 đ?‘Ž21 đ?‘Ž22 đ?‘Ž2đ?‘› The vectors đ??°1 = â‹Ž , đ??°2 = , ‌ , đ??° = 1 â‹Ž â‹Ž đ?‘Žđ?‘š1 đ?‘Žđ?‘š21 đ?‘Žđ?‘šđ?‘› đ?‘š in â„? that are formed from the column of A are called the column vectors of A.

Theoreme: The row space and column space of a matrix A have the same dimensions Definition: The common dimension of the row space and column space of a matrix A is called the rank of A and is denoted by rank(A); PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Change of basis Problem: If w is a vector in a finite-dimensional vector space V and if we change the basis from basis B to a basis B’, how are the coordinate vectors đ??° đ??ľ and đ??° đ?‘Šâ€˛ .

Definition: Let B={b1, b2, ‌ bk,‌, bn } be a basis of V, đ??œ = đ?‘?1 đ??›1 + đ?‘?2 đ??›2 +‌đ?‘?đ?‘˜ đ??›đ?‘˜ +‌+đ?‘?đ?‘› đ??›đ?‘› , đ?‘?đ?‘˜ ≠0 be a vector in V. We call it an elementary change of basis, if we change bk into c, the new basis is B’={b1, b2, ‌ c,‌, bn }

Procedure: đ??œ = đ?‘?1 đ??›1 + đ?‘?2 đ??›2 +‌+đ?‘?đ?‘˜ đ??›đ?‘˜ +‌+đ?‘?đ?‘› đ??›đ?‘› , đ?‘?đ?‘˜ ≠0 đ??ą = đ?‘Ľ1 đ??›1 + đ?‘Ľ2 đ??›2 +‌+đ?‘Ľđ?‘˜ đ??›đ?‘˜ +‌+đ?‘Ľđ?‘› đ??›đ?‘› Express đ??›đ?‘˜ from c and substitute it into x. Then we get the coordinates of x in the new basis. đ?‘? đ?‘? 1 đ?‘? đ??›đ?‘˜ = − 1 đ??›1 − 2 đ??›2 âˆ’â€Ś+ c+â€Śâˆ’ đ?‘› đ??›đ?‘› đ?‘?đ?‘˜

đ?‘?đ?‘˜

đ?‘?đ?‘˜

đ?‘?đ?‘˜

đ??ą = (đ?‘Ľ1 − đ?›żđ?‘?1 )đ??›1 + (đ?‘Ľ2 −đ?›żđ?‘?2 )đ??›2+‌+đ?›żđ??œ+‌+(đ?‘Ľđ?‘› −đ?›żđ?‘?đ?‘˜ )đ??›đ?‘› đ?‘Ľ where đ?›ż = đ?‘˜ đ?‘?đ?‘˜

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Change of basis - example Example: Let đ??›1 = −1,1, −1 , đ??›2 = 2,1,0 , đ??›3 = 1, −1,1 be vectors in â„?3 . 1. Are these vectors linearly dependent or independent vectors? 2. What is the dimension of the space generated by đ??›1 , đ??›2 , đ??›3 ? 3. Choose as many basis vectors from đ??›1 , đ??›2 , đ??›3 , as you can. 4. If đ??œ = 4,3,2 , đ??? = 4, −1, 2 , are these vectors in the space generated by đ??›1 , đ??›2 , đ??›3 ? 5. If one of c or d is in the generated space, give its coordinates in the new basis. Solution: Create a table from the vectors, where đ??ž1 , đ??ž2 , đ??ž3 are vectors of standard basis and use elementary change of basis, as many as possible. We need the coordinate vectors of the old basis relative to the new basis.

đ??ž1 â&#x;š đ??›1

1. 2. 3. 4.

5.

đ??ž3 â&#x;š đ??›2

đ??›3 = −đ??›1 , so e.g. 1đ??›1 + 0đ??›2 − 1 đ??›3 = đ?&#x;Ž dependent vector system. dim(đ??›1 , đ??›2 , đ??›3 )=2 B{đ??›1 , đ??›3 } c is not in the space generated by đ??›1 , đ??›2 , đ??›3 (đ??œ = −đ?&#x;?đ??›1 + 4đ??ž2 +1đ??›3 ) d is in the space generated by đ??›1 , đ??›2 , đ??›3 (đ??? = −đ?&#x;?đ??›1 +1đ??›3 ) đ??? = (−2,1)đ??ľ

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Vector form of a linear equation system Definition: A vector form of a linear equation system of m linear equations in n unknown is đ?‘Ľ1 đ??š1 + đ?‘Ľ2 đ??š2 +‌+đ?‘Ľđ?‘› đ??šđ?‘› = đ??› đ?‘Žđ?‘›1 đ?‘?1 đ?‘Ž21 đ?‘Ž11 đ?‘Žđ?‘›2 đ?‘?2 đ?‘Ž22 đ?‘Ž12 đ??šđ?‘› = â‹Ž đ??› = â‹Ž are vectors. đ??š2 = â‹Ž ‌. đ??š1 = â‹Ž where â‹Ž â‹Ž â‹Ž â‹Ž đ?‘Ž đ?‘›đ?‘š đ?‘Ž2đ?‘š đ?‘?đ?‘š đ?‘Ž1đ?‘š Alternative questions: 1. Can vector b be expressed as a linear combination of a1, a2, ‌an vectors? (Is the system consistent, that is, does at least one solution exist? ) 2. Are vectors a1, a2, ‌an linearly independent of each other? (If a solution exists, is it the only one; that is, is the solution unique?)  x1 + 2 x2 + x3  4 x1 + x2  x3  3  x1 +

+ x3  2

 x1 + 2 x2 + x3  4 x1 + x2  x3  1  x1 +

+ x3  2

1 ďƒ— x1 + 0 ďƒ— x2 + (1) ďƒ— x3  2 0 ďƒ— x1 + 0 ďƒ— x2 + 0 ďƒ— x3  4 0 ďƒ— x1 + 1 ďƒ— x2 + 0 ďƒ— x3  1

1 ďƒ— x1 + 0 ďƒ— x2 + (1) ďƒ— x3  2 0 ďƒ— x1 + 0 ďƒ— x2 + 0 ďƒ— x3  0 0 ďƒ— x1 + 1 ďƒ— x2 + 0 ďƒ— x3  1

No solution, because of the second row

đ?‘Ľ1 = −2 + đ?‘Ľ3 đ?‘Ľ2 = 1 Infinite number of solutions with one free parameter

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Eigenvalue, eigenvector Definition: If A is an đ?‘›đ?‘Ľđ?‘› matrix, then a nonzero v in â„?đ?‘› is called an eigenvector of A đ??´đ??Ż = Îťđ??Ż (1) for some scalar of l. l is called an eigenvalue of A. and v is said to be an eigenvector corresponding to l. Equivalently :

(đ??´ − Îťđ??ź)đ??Ż = đ?&#x;Ž where I is the identity matrix. If the coefficient đ??´ − Îťđ??ź were nonsingular, the unique solution of equation (1) would be đ??Ż = đ?&#x;Ž , which is of no interest. To obtain a nonzero solution of the eigenvalue problem, đ??´ − Îťđ??ź must be singular. det(đ??´ − Îťđ??ź) = 0 This equation is called a characteristic equation of A. Example: If đ??´ =

đ?‘Ľ1 If đ??Ż = đ?‘Ľ 2

−5 2 −5 − đ?œ† 2 , then the characteristic equation is = 0 or 2 −2 − đ?œ† 2 −2 the characteristic polynomial is (−5 − đ?œ†)(−2 − đ?œ†) − 4 = đ?œ†2 + 7đ?œ† + 6 = (đ?œ† + 6)(đ?œ† + 1) = 0 So the eigenvalues are đ?œ†1 = −6 , đ?œ†2 = −1.

for đ?œ†1 = −6

1 2 đ?‘Ľ1 0 = 2 4 đ?‘Ľ2 0

for đ?œ†1 = −1

−4 2 đ?‘Ľ1 0 = 2 −1 đ?‘Ľ2 0

đ?‘Ľ1 + 2đ?‘Ľ2 = 0

đ?‘Ľ1 = −2đ?‘Ľ2

e.g. đ??Żđ?&#x;? =

−2 1

2đ?‘Ľ1 = đ?‘Ľ2

e.g. đ??Żđ?&#x;? =

1 2

2đ?‘Ľ1 +4đ?‘Ľ2 = 0 −4đ?‘Ľ1 + 2đ?‘Ľ2 = 0 2đ?‘Ľ1 − đ?‘Ľ2 = 0

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Diagonization Properties: 1. The eigenvalues of an upper or lower trigonal matrix are the diagonal entries. 2. The eigenvalues of a diagonal matrix (all the entries except for the main diagonal are zero) are the diagonal entries, and the eigenvectors are the trivial basis of the column vector space. 3. The eigenvectors of a real, symmetric matrix with distinct eigenvalues are mutually orthogonal. 4. The product of eigenvalues equals to the determinant of the matrix. Definition: A square matrix A is said to be diagonalizable if it is similar to some diagonal matrix, that is, if there exists an invertible matrix P such that đ??ˇ = đ?‘ƒâˆ’1 đ??´đ?‘ƒis diagonal.

Remark: If đ??ˇ = đ?‘ƒâˆ’1 đ??´đ?‘ƒ then đ??´ = đ?‘ƒđ??ˇđ?‘ƒâˆ’1 . Theorem: An A nxn square matrix is diagonalizable if and only if A has n linearly independent eigenvectors and P is the matrix constructed from the linearly independent eigenvectors. Example: Show that đ??´ =

6 −1 2 1 is a diagonizable, đ??ľ = is not diagonizable matrix. 2 3 0 2

6 −1 1 1 5 0 1 1 −1 1 1 1 1 5 0 2 −1 Solution: For A đ?œ†1 = 5, đ?œ†2 = 4, đ??Ż1 = , đ??Ż2 = . So = = 1 2 2 3 1 2 0 4 1 2 1 2 0 4 −1 1 For B there is only one eigenvalue đ?œ†1 = 2 and for all eigenvectors the second coordinate is equal to 0. It has no two independent eigenvectors, so it is not a diagonizable matrix.

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Principal axes theorem Definition: If A is a symmetric đ?‘›đ?‘Ľđ?‘› matrix, and x is a nonzero column vector in â„?đ?‘› then the number đ?‘„ đ??´ đ??ą = đ??ą đ?‘‡ đ??´đ??ą is called the quadratic form associated with A. đ?‘Ľ1 đ?‘Ž11 If n=2, đ??ą = đ?‘Ľ , đ??´ = đ?‘Ž 2 12

đ?‘Ž12 2 2 đ?‘Ž22 , then đ?‘„ = đ?‘Ž11 đ?‘Ľ1 + đ?‘Ž22 đ?‘Ľ2 + 2đ?‘Ž12 đ?‘Ľ1 đ?‘Ľ2 (in geometry conic sections)

đ?‘Ž11 đ?‘Ž12 đ?‘Ž13 đ?‘Ľ1 If n=3, đ??ą = đ?‘Ľ2 , đ??´ = đ?‘Ž12 đ?‘Ž22 đ?‘Ž23 , then đ?‘Ľ3 đ?‘Ž13 đ?‘Ž23 đ?‘Ž33 2 2 đ?‘„ = đ?‘Ž11 đ?‘Ľ1 + đ?‘Ž22 đ?‘Ľ2 + đ?‘Ž33 đ?‘Ľ32 + 2đ?‘Ž12 đ?‘Ľ1 đ?‘Ľ2 + 2đ?‘Ž13 đ?‘Ľ1 đ?‘Ľ3 + 2đ?‘Ž23 đ?‘Ľ2 đ?‘Ľ3 (in geometry second ordered surfaces) The numbers đ?‘Ľ1 , đ?‘Ľ2 , đ?‘Ľ3 are the coordinates in the đ??ž1 , đ??ž2 , đ??ž3 trivial basis. Let the eigenvalues of đ??´3đ?‘Ľ3 symmetrical matrix be đ?œ†1 , đ?œ†2 , đ?œ†3 and let the appropriate eigenvectors be đ??Ż1 , đ??Ż2 , đ??Ż3 . Make a basis transformation, and let the new basis be the eigenvectors be đ??Ż1 , đ??Ż2 , đ??Ż3 . In the new coordinate system đ??ą đ?‘‡ = đ?‘Ś1 , đ?‘Ś2 , đ?‘Ś3 = đ?‘Ś1 đ??Ż1 + đ?‘Ś2 đ??Ż2 + đ?‘Ś3 đ??Ż3 , đ??´đ??Ż1 = đ?œ†1 đ??Ż1 , đ??´đ??Ż2 = đ?œ†2 đ??Ż2 , đ??´đ??Ż3 = đ?œ†3 đ??Żđ?&#x;‘. Then đ??ą đ?‘‡ đ??´đ??ą = đ?‘Ś1 đ??Ż1 + đ?‘Ś2 đ??Ż2 + đ?‘Ś3 đ??Ż3 đ?‘Ś1 đ?œ†1 đ??Ż1 + đ?‘Ś2 đ?œ†2 đ??Ż2 + đ?‘Ś3 đ?œ†31 đ??Ż3 = đ?œ†1 đ?‘Ś12 + đ?œ†2 đ?‘Ś22 + đ?œ†3 đ?‘Ś32 . 0 if đ?‘– ≠đ?‘— We used đ??Żđ?‘– đ??Żđ?‘— = ቊ 1 if đ?‘– = đ?‘—

Definition: We call a basis transformation a principal axes transformation, if we transform a quadratic form into the basis of eigenvectors be đ??Ż1 , đ??Ż2 , đ??Ż3 ‌., đ??Żđ?‘› . PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Principal axes theorem Central conics in standard position: đ?‘Ľ

0 đ?‘Ľ 0 đ?‘? đ?‘Ś

đ?‘Ś đ?‘Ž

Central quadratics in standard position:

đ?‘Ľ

A central conic in not standard form

8đ?‘Ľ 2 + 5đ?‘Ś 2 − 4đ?‘Ľđ?‘Ś = 1

đ?‘Ś

�

A central quadratic in not standard form

2� 2 + 6� 2 + 2� 2 + 8�� = 1

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

đ?‘Ž 0 0 đ?‘Ľ 0 đ?‘? 0 đ?‘Ś 0 0 đ?‘? đ?‘§

First-order, linear, homogenous differential equation systems Definition: Equation

đ??˛ ′ = đ??´đ??˛

đ?‘Ś1 đ?‘Ś1′ đ?‘Ž â‹Ż đ?‘Ž 11 1đ?‘› đ?‘Ś đ?‘Śâ€˛ â‹ą â‹Ž ,đ??˛= 2 where đ??˛ ′ = 2 , A= â‹Ž â‹Ž â‹Ž đ?‘Ž â‹Ż đ?‘Ž đ?‘›1 đ?‘›đ?‘› đ?‘Śđ?‘› đ?‘Śđ?‘›â€˛ is called a first order, linear, homogenous differential equation system. Solution: • Find a matrix P that diagonalizes A. • Make a substitution đ??˛ = đ?‘ƒđ??Ž, đ??˛ ′ = đ?‘ƒđ??Žâ€˛ to obtain a new ‚diagonal systemâ€? đ??Žâ€˛ = đ??ˇu , where đ??ˇ = đ?‘ƒâˆ’1 đ??´đ?‘ƒ • Solve đ??Žâ€˛ = đ??ˇu • Determine y from the equation đ??˛ = đ?‘ƒđ??Ž. Example: Solve the system đ?‘Ś1′ = đ?‘Ś1 + đ?‘Ś2 đ?‘Ś2′ = 4đ?‘Ś1 − 2đ?‘Ś2 . Find the solution that satisfies the initial conditions đ?‘Ś1 0 = 1, đ?‘Ś2 0 =6.

1 1 Solution: The coefficient matrix đ??´ = , 4 −2 2 0 1 1 đ?œ†1 = 2, đ?œ†2 = −3, đ?‘ƒ = ,đ??ˇ = . 1 −4 0 −3 đ?‘˘1′ = 2đ?‘˘1 , đ?‘˘2′ = −3đ?‘˘2 đ?‘?1 đ?‘’ 2đ?‘Ľ đ??Ž= , đ??˛ = đ?‘ƒđ??Ž then đ?‘?2 đ?‘’ −3đ?‘Ľ đ?‘?1 đ?‘’ 2đ?‘Ľ đ?‘?1 đ?‘’ 2đ?‘Ľ + đ?‘?2 đ?‘’ −3đ?‘Ľ 1 1 đ??˛= = . 1 −4 đ?‘?2 đ?‘’ −3đ?‘Ľ đ?‘?1 đ?‘’ 2đ?‘Ľ − 4đ?‘?2 đ?‘’ −3đ?‘Ľ From the initial conditions: đ?‘?1 + đ?‘?2 = 1 đ?‘?1 −4đ?‘?2 = 6 đ?‘?1 = 2, đ?‘?2 = −1. So đ?‘Ś1 = 2đ?‘’ 2đ?‘Ľ − đ?‘’ −3đ?‘Ľ đ?‘Ś2 = 2đ?‘’ 2đ?‘Ľ + 4đ?‘’ −3đ?‘Ľ

PTE MIK IldikĂł PerjĂŠsinĂŠ HĂĄmori PhD

Mathematics 3

Differential equations1 First order- Analitical and numerical solutions First order separable ODE 1. The number of bacteria in a certain culture grows at a rate that is proportional to the number present. If the number increased from 500 to 2000 in 2 hours, determine a. the number present after 12 hours b. the doubling time. 2. In some cases it is reasonable to assume that the resistance encountered by a moving object, such as a car coasting to a stop, is proportional to the object’s velocity. The faster the object moves, the more its forward progress is resisted by the air through which it passes. đ?‘‘đ?‘Ł đ?‘š = −đ?‘˜đ?‘Ł đ?‘Ł(0) = đ?‘Ł0 đ?‘˜ > 0 đ?‘‘đ?‘Ą

What can we learn from the solution of the ODE? 3. A body with mass đ?‘š free falling under gravity but with air resistance. We assume that the force of air resistance is proportional to the speed of the mass and opposes the direction of motion.

đ?‘š

đ?‘‘đ?‘Ł = −đ?‘šđ?‘” − đ?‘˜đ?‘Ł đ?‘‘đ?‘Ą

đ?‘Ł(0) = 200

đ?‘˜đ?‘š â„Ž

As an example, a skydiver of mass đ?‘š = 100 kg with his parachute closed may have a terminal velocity of đ?‘š đ?‘˜đ?‘” 200 km/h, g = 9.81 2 , đ?‘˜ = 63.54 . What is the speed after đ?‘Ą = 4 sec? đ?‘ â„Ž 4. Give the general solution a. (2đ?‘Ľ + 1)đ?‘Ś ′ − 3đ?‘Ś = 0 b. (2đ?‘Ľđ?‘Ś + 3đ?‘Ś)đ?‘Ś ′ − đ?‘Ś 2 − 2 = 0 c. 2(đ?‘Ľđ?‘Ś + đ?‘Ľ − đ?‘Ś − 1) = (đ?‘Ľ 2 − 2đ?‘Ľ)đ?‘Ś ′ d. (đ?‘Ľ 2 − 1)đ?‘Ś ′ = đ?‘Ś 5. Find the particular solution a. đ?‘Ľ(1 + đ?‘Ś) = đ?‘Ś ′ (1 + đ?‘Ľ) đ?‘Ś(0) = 2 1 1 b. đ?‘Ľâˆš1 − đ?‘Ś 2 = đ?‘Śđ?‘Ś ′ √1 − đ?‘Ľ 2 đ?‘Ś ( ) = 2 2 c. (2đ?‘Ľ + 1)đ?‘Ś ′ = 2đ?‘Ś(đ?‘™đ?‘›đ?‘Ś + 1) đ?‘Ś(1) = đ?‘’

First order linear ODE 6. Give the general solution a. đ?‘Ś ′ − 2đ?‘Ś = 5 b. đ?‘Ś ′ + 3đ?‘Ś = đ?‘’ 2đ?‘Ľ c. đ?‘Ś ′ đ?‘?đ?‘œđ?‘ đ?‘Ľ + đ?‘Śđ?‘ đ?‘–đ?‘›đ?‘Ľ = 1 d. đ?‘Ľđ?‘Ś ′ − 2đ?‘Ś = đ?‘Ľ 3 + 1 e. đ?‘Ľđ?‘Ś ′ + đ?‘Ś = đ?‘Ľđ?‘™đ?‘›đ?‘Ľ 7. Find the particular solution

a. đ?‘Ś ′ − 2đ?‘Ś = 5 b. đ?‘Ś ′ + 3đ?‘Ľ 2 đ?‘Ś = đ?‘Ľ 2

đ?‘Ś(1) = 3 4 đ?‘Ś(1) = 3

c. đ?‘Ś ′ + đ?‘Ś = đ?‘’ −đ?‘Ľ

� (�� 2) = 2

1

2

1

d. đ?‘Ś ′ + 2đ?‘Ľđ?‘Ś = 2đ?‘Ľđ?‘’ −đ?‘Ľ đ?‘Ś(√đ?‘™đ?‘›2) = (1 + đ?‘™đ?‘›2) 2 8. Find the analitical and numerical solutions

a. đ?‘Ś ′ = 2đ?‘Ľđ?‘Ś b. đ?‘Ś ′ = đ?‘Ś − đ?‘Ľ 2 c. đ?‘Ś ′ = 2đ?‘Ś − đ?‘’ đ?‘Ľ d. đ?‘Ś ′ = đ?‘Ľ − đ?‘Ś

đ?‘Ś(0) = 1 đ?‘Ś(0) = −1 đ?‘Ś(0) = 0 đ?‘Ś(0) = 1

đ?‘Ž=0 đ?‘?=1 đ?‘Ž=0 đ?‘?=1 đ?‘Ž = 0 đ?‘? = 1.2 đ?‘Ž=0 đ?‘?=1

đ?‘›=5 đ?‘› = 10 đ?‘›=4 đ?‘›=4

1. a. m=2 048 000 𝑘𝑚 3. v=1.54 ℎ

b. t=1h 3

4. a. 𝑦 = 𝐶(2𝑥+1)2

b. 𝑦 = ±√2𝐶𝑥 + 3𝐶 − 2

c. 𝑦 = −1 + 𝐶 𝑥(𝑥 − 2)

√1−𝑥 2 d. 𝑦 = 𝐶 𝑥+1 3𝑒 𝑥 −1−𝑥 5. a. 𝑦 = 1+𝑥 5 6. a. 𝑦 = 𝐶𝑒 2𝑥 − 2 2 3

b. 𝑦 = 𝑥

c. 𝑦 = 𝑒 3𝑥−3 1

1

d. 𝑦 = 𝐶𝑥 + 𝑥 − 2 5 2 2

7. a. 𝑦 = − +

11 2𝑥−2 𝑒 2 −𝑥 2

d. 𝑦 = (𝑥 + 1)𝑒

8. a.

8.b.

8.c.

8.d.

4

b. 𝑦 = 𝐶𝑒 −3𝑥 + 5 𝑒 2𝑥 𝐶

1

1

e. 𝑦 = 𝑥 + 2 𝑥 ln 𝑥 − 4 𝑥 1 3

b. 𝑦 = + 𝑒 1−𝑥

3

1

1

c. 𝑦 = 𝐶𝑒 −3𝑥 + 5 𝑒 2𝑥 c. 𝑦 = (𝑥 + 1 + ln 2)𝑒 −𝑥

Engineering Mathematics 3 Differential equations2 Incomplete second order ODE 1. Find the general solution a. 𝑦 ′′ = 𝑥 + sin 𝑥

f. 𝑦 ′′ =

𝑥 ′ ′′

′′

+𝑥

g, 2𝑥𝑦 𝑦 = (𝑦 ′ )2 + 1 h. 𝑦 ′′ (1 − 𝑦) + 2(𝑦 ′ )2 = 0 i. (𝑦 ′ )2 + 2𝑦𝑦 ′′ = 0 j. 𝑦𝑦 ′′ = 1 + (𝑦 ′ )2

b. 𝑦 = ln 𝑥 c. 𝑥𝑦 ′′ = 𝑦 ′ d. 𝑦 ′′ = 𝑦 ′ + 𝑥 e. 𝑦 ′′ + 2𝑦 ′ = 𝑒 𝑥 2. Find the particular solution a. 𝑦 ′′ = 𝑦 ′ + 𝑒 𝑥 b. 𝑦 ′′ 𝑦 ln 𝑦 = −(𝑦 ′ )2 c. 𝑦 ′′ (𝑥 2 + 1) = 2𝑥𝑦 ′ 1

d. 𝑦 ′′ = √1−𝑥 2

e. 𝑦𝑦 ′′ = 2(𝑦 ′ )2 Linear, second order ODE with constant coefficients 3. Find the general solution a. b. c. d. e. f. g. h.

𝑦′

𝑦 ′′ − 𝑦 ′ − 2𝑦 = 0 2𝑦 ′′ + 5𝑦 ′ − 3𝑦 = 0 𝑦 ′′ − 16𝑦 ′ + 64𝑦 = 0 3𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0 9𝑦 ′′ − 6𝑦 ′ + 𝑦 = 0 𝑦 ′′ − 4𝑦 ′ + 13𝑦 = 0 𝑦 ′′ − 𝑦 ′ − 6𝑦 = 0 𝑦 ′′ − 2𝑦 ′ + 𝑦 = 0

𝑦(0) = −1 𝑦(0) = 𝑒 𝑦(0) = 1

𝑦(1) = 𝑒 𝑦 ′ (0) = 2 𝑦 ′ (0) = 3

𝑦(0) = 2

𝑦(1) =

𝑦(0) = 1

𝑦(1) = 2

𝑦(0) = 0, 𝑦 ′ (0) = 1 𝑦(0) = 1, 𝑦(2) = 𝑒 𝑦(0) = 0, 𝑦 ′ (0) = −1 𝑦(0) = 0, 𝑦 ′ (0) = −1 𝑦(0) = 0, 𝑦 ′ (0) = −1 𝑦(0) = 1, 𝑦 ′ (0) = −2 𝑦(0) = 2, 𝑦(1) = 1 1 𝑦(0) = 𝑒 , 𝑦(1) = 1 + 𝑒

4. Find the particular solution a. b. c. d. e.

𝑦 ′′ + 𝑦 ′ − 2𝑦 = 2𝑥 2 − 3 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 17𝑐𝑜𝑠3𝑥 𝑦 ′′ − 3𝑦 ′ − 4𝑦 = 6𝑒 2𝑥 𝑦 ′′ − 𝑦 ′ − 2𝑦 = 2𝑒 3𝑥 − 4𝑥 2 + 8𝑥 𝑦 ′′ + 9𝑦 = 13𝑥𝑒 2𝑥

𝑦𝑝 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶 𝑦𝑝 = 𝐴𝑠𝑖𝑛3𝑥 + 𝐵𝑐𝑜𝑠3𝑥 𝑦𝑝 = 𝐴𝑒 2𝑥 𝑦𝑝 = 𝐴𝑒 3𝑥 + 𝐵𝑥 2 + 𝐶𝑥 + 𝐷 𝑦𝑝 = 𝐴𝑒 2𝑥 (𝐵𝑥 + 𝐶)

5. Find the general solution (resonance case) 1

a. b.

3𝑦 ′′ + 5𝑦 ′ − 2𝑦 = 7𝑒 3𝑥 − 4𝑥 𝑦 ′′ − 3𝑦 ′ = 12𝑥 + 8

c.

9𝑦 ′′ − 6𝑦 ′ + 𝑦 = 36𝑒 3𝑥

1

1

𝑦𝑝 = 𝐴𝑥 + 𝐵 + 𝐶𝑒 3𝑥 𝑥 𝑦𝑝 = (𝐴𝑥 + 𝐵)𝑥 1

𝑦𝑝 = 𝐴𝑒 3𝑥 𝑥 2

𝜋 2

1.a. b. c. d. e. f. g. h. i. j.

2. a. b. 𝑦(𝑥)(ln(𝑦(𝑥)) − 1) = 2𝑥 c. d. e. 1

1

1

b. 𝑦 = 𝑒 2𝑥

3.a. 𝑦 = − 3 𝑒 −𝑥 + 3 𝑒 2𝑥

1

c. 𝑦 = 𝑒 8𝑥 − 9𝑥𝑒 8𝑥 4

e. 𝑦 = 𝑒 3𝑥 (1 − 3 𝑥)

d.

(1−2𝑒 3 )𝑒 −2𝑥 +(2𝑒 −2 −1)𝑒 3𝑥

4

f. 𝑦 = − 3 𝑒 2𝑥 sin 3𝑥 + 𝑒 2𝑥 cos 3𝑥 g. 𝑦 = 𝑒 −2 −𝑒 3 𝑥−1 𝑥 h. 𝑦 = 𝑒 + 𝑥𝑒 6 7 4. a. 𝑦 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −2𝑥 − 𝑥 − 𝑥 2 b. 𝑦 = 𝐶1 𝑒 −𝑥 sin 𝑥 + 𝐶2 𝑒 −𝑥 cos 𝑥 + 5 sin 3𝑥 − 5 cos 3𝑥 1

c. 𝑦 = 𝐶1 𝑒 4𝑥 + 𝐶2 𝑒 −𝑥 − 𝑒 2𝑥 1

3

d. 𝑦 = 𝐶1 𝑒 2𝑥 + 𝐶2 𝑒 −𝑥 + 2 𝑒 3𝑥 + 2𝑥 2 − 6𝑥 + 5

1

1

5.a. 𝑦 = 𝐶1 𝑒 −2𝑥 + 𝐶2 𝑒 3𝑥 − 7 𝑒 3𝑥 + 2𝑥 + 5 + 𝑥𝑒 3𝑥 1

1

1

c. 𝑦 = 𝐶1 𝑒 3𝑥 + 𝐶2 𝑥𝑒 3𝑥 + 2𝑥 2 𝑒 3𝑥

1

b. . 𝑦 = 𝐶1 3 𝑒 3𝑥 + 𝐶2 − 2𝑥 2 − 4𝑥

Engineering Mathematics 3. Differential equations3 Second order ODE Euler method 1. Find the numerical solution on the given interval a. 3đ?‘Śđ?‘Ś ′′ + (đ?‘Ś ′ )2 = 0 đ?‘Ś(0) = 1 đ?‘Ś ′ (0) = 3 ′′ ′ b. đ?‘Ś = −đ?‘Ś + đ?‘Ś đ?‘Ś(0) = 1 đ?‘Ś ′ (0) = 0 c. đ?‘Ś ′′ + đ?‘Ľđ?‘Ś = 0 đ?‘Ś(1) = 1 đ?‘Ś ′ (1) = 3

[đ?‘Ž, đ?‘?] = [0,2] [đ?‘Ž, đ?‘?] = [0,1] [đ?‘Ž, đ?‘?] = [1,3]

đ?‘›=5 đ?‘›=4 đ?‘›=6

Laplace transformation

2. Find the Laplace transform a; đ?‘“(đ?‘Ą) = 1 b; đ?‘“(đ?‘Ą) = đ?‘Ą 2

c; đ?‘“(đ?‘Ą) = đ?‘’ −3đ?‘Ą d. đ?‘“(đ?‘Ą) = sin 2đ?‘Ą 0 đ?‘–đ?‘“ đ?‘Ą < 0 đ?‘Ą e; đ?‘“(đ?‘Ą) = cos f. đ??ť(đ?‘Ą) = { (Heaviside or unit step function) 3 1 đ?‘–đ?‘“ đ?‘Ą ≼ 0 0 đ?‘–đ?‘“ đ?‘Ą < 4 g; đ??ť4 (đ?‘Ą) = đ?‘“(đ?‘Ą) = { express it using Heaviside function 1 đ?‘–đ?‘“ đ?‘Ą ≼ 4 0 đ?‘–đ?‘“ đ?‘Ą < 2 1 đ?‘–đ?‘“ 2 ≤ đ?‘Ą < 4 (step up step down function) express it using h; đ??ť2 (đ?‘Ą) − đ??ť4 (đ?‘Ą) = đ?‘“(đ?‘Ą) = { 0 đ?‘–đ?‘“ đ?‘Ą ≼ 4 Heaviside function 3. Find the inverse Laplace transform a. đ??š(đ?‘ ) = d. đ??š(đ?‘ ) =

đ?‘ −1 (đ?‘ −2)(đ?‘ +1) −3đ?‘ 2 −12đ?‘ −8 (đ?‘ +2)3

b. đ??š(đ?‘ ) = e. đ??š(đ?‘ ) =

2 (đ?‘ 2 +1)(đ?‘ 2 +4) 2đ?‘ −5 đ?‘ 2 +4đ?‘ +8

c. đ??š(đ?‘ ) = f. đ??š(đ?‘ ) =

đ?‘ −9 đ?‘ 2 −6đ?‘ +5 đ?‘ 2 −6đ?‘ +10 (đ?‘ −2)(đ?‘ −1)(đ?‘ −3)

4. Using the function show that đ??ż{đ?‘“̇ (đ?‘Ą)} = đ?‘ đ??ż{đ?‘“(t)} − đ?‘“(0) and đ??ż {đ?‘“Ě‡Ěˆ (đ?‘Ą)} = đ?‘ 2 đ??ż{đ?‘“(t)} − sđ?‘“(0) − đ?‘“̇ (0) a; đ?‘“(đ?‘Ą) = 1 b; đ?‘“(đ?‘Ą) = đ?‘’ −3đ?‘Ą b; đ?‘“(đ?‘Ą) = 2đ?‘Ą 5. Solve the ODE-s using Laplace transformation a. đ?‘Ś ′′ − 6đ?‘Ś ′ + 5đ?‘Ś = 0 đ?‘Ś(0) = 1, đ?‘Ś ′ (0) = −3 b. đ?‘Ś ′ + 2đ?‘Ś = 4đ?‘Ąđ?‘’ −2đ?‘Ą đ?‘Ś(0) = −3 ′′ ′ 3đ?‘Ą c. đ?‘Ś − 3đ?‘Ś + 2đ?‘Ś = đ?‘’ đ?‘Ś(0) = 1, đ?‘Ś ′ (0) = 0

1 đ?‘–đ?‘“ 0 ≤ đ?‘Ą < 4 đ?‘“(đ?‘Ą) = { 0 đ?‘–đ?‘“ đ?‘Ą ≼ 4 đ?‘œđ?‘&#x; đ?‘Ą < 0 e. 2đ?‘Ś ′′ + đ?‘Ś ′ + 2đ?‘Ś = đ??ť5 (đ?‘Ą) − đ??ť20 (đ?‘Ą) đ?‘Ś(0) = 0, đ?‘Ś ′ (0) = 0 f. 2đ?‘Ś ′′ + đ?‘Ś ′ + 2đ?‘Ś = đ?›ż(đ?‘Ą − 5) đ?‘Ś(0) = 0, đ?‘Ś ′ (0) = 0 d. đ?‘Ś ′′ + 4đ?‘Ś = đ?‘“(đ?‘Ą) đ?‘Ś(0) = 3, đ?‘Ś ′ (0) = −2

Solution 1.a.

b. c.

1

2

2. a. 𝐹(𝑠) = 𝑠

2

d. 𝐹(𝑠) = 𝑠3 +4 h. 𝐹(𝑠) =

3a

𝑒 −4𝑠 −𝑒 −2𝑠 𝑠

1

b. 𝐹(𝑠) = 𝑠3

c. 𝐹(𝑠) = 3+𝑠

e. 𝐹(𝑠) = 9𝑠3 +1

g. 𝐹(𝑠) =

9𝑠

𝑒 −4𝑠 , 𝑓(𝑡) 𝑠

, 𝑓(𝑡) = 𝐻(𝑡 − 2) − 𝐻(𝑡 − 4)

b.

c.

d.

e.

f.

5.a.

b.

d.

c.

= 𝐻(𝑡 − 4)

Engineering Mathematics 3 Determinant, adjoint and inverse of matrices, Cramer’s rule, Gauss-Jordan elimination CAS practices Download Determinant, gauss elimination.mw file, solve the exercises. with(LinearAlgebra): Determinant, ColumnOperation, RowOperation, Adjoint, GenerateMatrix, GaussianElimination, ReducedRowEchelonForm, BackwardSubstitution

Paper work 1. Evaluate the determinants of the matrices: 1 −2 3 3 5 đ??´=[ ] đ??ľ=[ 6 7 −1] −2 4 −3 1 4

đ?‘Žâˆ’3 5 đ??ś=[ ] −3 đ?‘Žâˆ’2

đ?‘? −4 đ??¸ = [2 1 4 đ?‘?−1

3 đ?‘?2] 2

2. Find all values of ď Ź for which đ?‘‘đ?‘’đ?‘Ąđ??´ = 0 đ?œ†âˆ’4 4 0 đ?œ†âˆ’2 1 đ??´=[ ] đ??´ = [ −1 đ?œ† 0 ] −5 đ?œ† + 4 0 0 đ?œ†âˆ’5 3. Decide whether the given matrix is invertible, and if so, use the adjoint method to find its inverse. 2 5 5 2 −3 5 đ??´ = [−1 −1 0] đ??´ = [0 1 −3] 2 4 3 0 0 2 4. Solve the next linear equation systems (LES) by Cramer's rule, where it applies. 7đ?‘Ľ1 − 2đ?‘Ľ2 = 3 -4đ?‘Ľ + 5đ?‘Ś =2 đ?‘Ľ1 − 3đ?‘Ľ2 + đ?‘Ľ3 = 4 3đ?‘Ľ1 + đ?‘Ľ2 = 5 2đ?‘Ľ1 − đ?‘Ľ2 = −2 11đ?‘Ľ + đ?‘Ś + 2đ?‘§ = 3 4đ?‘Ľ − 3đ?‘Ľ đ?‘Ľ + 5đ?‘Ś + 2đ?‘§ = 1 1 3 =0 5. By examining the determinant of the coefficient matrix, show that the following system has a nontrivial solution if and only if ď Ąď€˝ď ˘. đ?‘Ľ + đ?‘Ś + đ?›źđ?‘§ = 0 đ?‘Ľ + đ?‘Ś + đ?›˝đ?‘§ = 0 đ?›ź đ?‘Ľ + đ?›˝đ?‘Ś + đ?‘§ = 0 6. Solve by Gauss-Jordan elimination đ?‘Ľ1 + đ?‘Ľ2 + 2đ?‘Ľ3 = 8 đ?‘Ľ + đ?‘Ś + 2đ?‘§ = 1 đ?‘Ľ1 + 3đ?‘Ľ2 − 2đ?‘Ľ3 + 2đ?‘Ľ5 =0 −đ?‘Ľ1 − 2đ?‘Ľ2 + 3đ?‘Ľ3 = 1 4đ?‘Ľ + 4 đ?‘Ś + 5đ?‘§ = 6 2đ?‘Ľ1 + 6đ?‘Ľ2 − 5đ?‘Ľ3 − 2đ?‘Ľ4 + 4đ?‘Ľ5 − 3đ?‘Ľ6 = −1 3đ?‘Ľ1 − 7đ?‘Ľ2 + 4đ?‘Ľ3 = 10 7đ?‘Ľ + 7đ?‘Ś + 8đ?‘§ = 10 5đ?‘Ľ3 + 10đ?‘Ľ4 + 15đ?‘Ľ6 = 5 2đ?‘Ľ1 + 6đ?‘Ľ2 + 8đ?‘Ľ4 + 4đ?‘Ľ5 + 18đ?‘Ľ6 = 6 7. Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions. đ?‘Ľ+ 2đ?‘Ś = 1 đ?‘Ľ + 2đ?‘Ś − 3đ?‘§ = 4 3đ?‘Ľ − đ?‘Ś + 5đ?‘§ = 2 2đ?‘Ľ + (đ?‘Ž2 − 14)đ?‘Ś = đ?‘Ž − 1 4đ?‘Ľ + đ?‘Ś + (đ?‘Ž2 − 14)đ?‘§ = đ?‘Ž + 2

Engineering Mathematics 3

Vector space, elementary change of basis 1. Let V be the set of all ordered pairs of real numbers, and consider the following „additionâ€? and „scalar multiplicationâ€? operations on u=(u1,u2) and v=(v1,v2): u+v=(u1+v1,u2+v2), ku=(0,ku2). 1.1. Compute u+v and ku for u=(-1,2), v=(3,4) and k=3. 1.2. In words, explain why V is closed under addition and scalar multiplication. 1.3. Show that Axiom 10 fails and hence V is not a vector space under the given operations. đ?‘Ž 0 2. Show that the set of all 2x2 matrices of the form [ ]with the standard matrix addition and scalar 0 đ?‘? multiplication is a vectorspace. 3. Show that lines through the origin are subspaces of â„?2 and of â„?3 . 4. Show that the set of all points (x, y) in for which đ?‘Ľ ≼ 0 and đ?‘Ś ≼ 0 is not the subspace of â„?2 . 5. Show that the set of function with a continuous derivative is a subspace of continous functions. 6. Determine whether the following vectors are linearly independent or linearly dependent in â„?3 . Find the dimension of S={a,b,c}. Construct matrixes where the vectors are in the rows or in the column. Find the rank of the matrixes. 6.1. a=(1,2,3), b=(5,6,-1), c=(3,2,1) in â„?3 6.2. a=(1,2,3), b=(5,6,-1), d=(-3,-2,7) in â„?3 6.3. a=(1,2), b=(5,6), c=(3,2) in â„?2 7. Find the coordinate vector of w relative to the basis S={a,b} 7.1. a=(1,0), b=(0,1), w=(3,-7) 7.2. a=(2,-4), b=(3,8), w=(1,1) 7.3. a=(1,1), b=(0,2), w=(x,y) 8. Show that a=(1,2,1), b=(2,9,0), c=(3,3,4) could be a basis for â„?3 . 8.1. Find the coordinate vector of v=(5,-1,9) relative to the basis S={a,b,c}. 8.2. Find the w vector in â„?3 ={e1,e2,e3} standard basis whose coordinate relative to S is wS=(-1,3,2). 9. Find the coordinate vector of w relative to the basis S={a,b,c} 9.1. a=(3,3,3), b=(1,0,0), c=(2,2,0), w=(2,-1,3), 9.2. a=(1,2,3), b=(-4,5,6), c=(7,-8,9), w=(5,-12,3) 10. Solve Exercise 9 using elementary change of basis. 11. Let a=(-1,1,3), b=(-4,-2,0), c=(-7,-5,-3) be vectors. 11.1. Find the dimension of S={a,b,c}. 11.2. If v=(-10,-8,-6), w=(-10,-8,-6) , is the vektor v or w in the space generated by S? 12. Find the matrix inverse using elementary basis transformation. 2 5 5 2 −3 5 đ??´ = [−1 −1 0] đ??´ = [0 1 −3] 2 4 3 0 0 2

Engineering Mathematics 3 Application of change of basis, eigenvalue, eigenvector

1. Solve the next linear equation systems (LES) by change of basis. x 1  3x 2  x 3  x 4  7; x 1  x 2  x 3  4; x 1  x 2  x 3  3x 4  2; 2x 1  5x 2  x 3  2x 4  22; 3x 1  8x 2  x 3  x 4  24.

x  3y  4z  19; x  4 y  3z  18

x 1  2x 2  3x 3  4x 4  6;

x 1  2x 2  x 3  13;

3x 1  4 x 2  x 3  2x 4  12;

2x 1  4x 2  2x 3  26;

 2x 1  3x 2  2x 3  x 4  2. x 1  3x 2  x 3  x 4  7;

4x 1  5x 2  4x 3  43.

2x 1  5x 2  x 3  2x 4  22;

x 1  2x 2  3x 3  x 4  2x 5  2; 3x 1  x 2  5x 3  3x 4  x 5  6;

2x 1  x 2  2x 3  2x 4  3x 5  8. x  3y  3z  16 3x 1  8x 2  x 3  x 4  24. 2. Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions. đ?‘Ľ + 2đ?‘Ś − 3đ?‘§ = 4 đ?‘Ľ+ 2đ?‘Ś = 1 3đ?‘Ľ − đ?‘Ś + 5đ?‘§ = 2 2đ?‘Ľ + (đ?‘Ž2 − 14)đ?‘Ś = đ?‘Ž − 1 4đ?‘Ľ + đ?‘Ś + (đ?‘Ž2 − 14)đ?‘§ = đ?‘Ž + 2

3. Show that the eigenvalues of the diagonal matrices are the diagonal entries, if 3 2 −1 3 0 0 đ??ľ = [0 8 −1], đ??ś = [2 8 0]. 0 0 2 1 0 2 4. Show that the product of eigenvalues equals to the determinant of the matrix, if 2 0 −3 1 0 2 đ??´ = [ 0 2 3 ] , đ??ľ = [0 1 0] −3 3 −2 0 0 2 5. Show that the eigenvectors of a real, symmetric matrix with distinct eigenvalues are mutually 3 2 2 orthogonal if đ??´ = [2 2 0] 2 0 4

Engineering Mathematics 3 Applications of Eigenvectors and Eigenvalues. 1. Show that the eigenvectors of symmetrical matrixes are orthogonal to each other. 1 −2 0 1 2 −1 2 3 2 −2 đ??´ = [−2 0 ] đ??ľ = [ ] đ??ˇ=[ ] 2 2 7 4 ] đ??ś=[ 3 −5 −2 −1 0 2 −1 −1 4 −3 2. Are the next matrixes diagonalizable or not. If the answer is yes, give the diagonal matrix D, and the P, for which đ?‘šđ?‘Žđ?‘Ąđ?‘&#x;đ?‘–đ?‘Ľ = đ?‘ƒđ??ˇđ?‘ƒâˆ’1 3 0 0 5 0 0 1 1 2 −2 đ??´=[ ] đ??ľ=[ ] đ??ś = [0 2 0] đ??ˇ = [1 5 0] 1 1 2 −2 0 1 2 0 1 5 3. Solve the next systems: đ?‘Ś1′ = đ?‘Ś1 + 4đ?‘Ś2 đ?‘Ś2′ = 2đ?‘Ś1 + 3đ?‘Ś2 đ?‘Ś1 (0) = đ?‘Ś2 (0) = 0

�1′ = 2 �1 + 9�2 �2′ = �1 + 2�2 �1 (0) = 1, �2 (0) = 1

đ?‘Ś1′ = 4đ?‘Ś1 + đ?‘Ś3 đ?‘Ś2′ = −2đ?‘Ś1 + đ?‘Ś2 đ?‘Ś3′ = −2đ?‘Ś1 + đ?‘Ś3 đ?‘Ś1 (0) = −1, đ?‘Ś2 (0) = 1, đ?‘Ś3 (0) = 0 đ?‘Ś1′ = đ?‘Ś1 + 2đ?‘Ś2 − đ?‘Ś3 đ?‘Ś2′ = đ?‘Ś1 + đ?‘Ś3 đ?‘Ś3′ = 4đ?‘Ś1 − 4đ?‘Ś2 + 5đ?‘Ś3 đ?‘Ś1 (0) = 1, đ?‘Ś2 (0) = 1, đ?‘Ś3 (0) = 1

4. Express the quadratic form in matrix notation đ?’™đ?‘‡ đ??´đ?’™, where A is symmetric: 2đ?‘Ľ 2 + 6đ?‘Ľđ?‘Ś − 5đ?‘Ś 2 ;

đ?‘Ľ12 + 7đ?‘Ľ22 − 3đ?‘Ľ32 + 4đ?‘Ľ1 đ?‘Ľ2 − 2đ?‘Ľ1 đ?‘Ľ3 + 8đ?‘Ľ2 đ?‘Ľ3 ;

5. Find the orthogonal change of variable that eliminates the cross product terms in the quadratic form, and express Q in terms of the new variable: đ?‘„ = đ?‘Ľ12 − đ?‘Ľ32 − 4đ?‘Ľ1 đ?‘Ľ2 + 4đ?‘Ľ2 đ?‘Ľ3 ;

đ?‘„ = 3đ?‘Ľ12 + 2đ?‘Ľ1 đ?‘Ľ2 + 2đ?‘Ľ1 đ?‘Ľ3 + 4đ?‘Ľ2 đ?‘Ľ3

6. Identify the conic by rotating the xy-axes to put the conic in standard position. Find the angle Θ through which you rotated the xy-axes. 5đ?‘Ľ 2 − 4đ?‘Ľđ?‘Ś + 8đ?‘Ś 2 − 36 = 0

2đ?‘Ľ 2 − 4đ?‘Ľđ?‘Ś − đ?‘Ś 2 + 8 = 0

7. Identify the quadratic rotating the xyz-axes to put the qudratic in standard position. 4� 2 + 4� 2 + 4� 2 + 4�� + 4�� + 4�� = 1

2� 2 + � 2 + � 2 + 2�� + 2�� = 1

Engineering Mathematics 3

Function series

1. Find the Fourier series generated by 𝑓(𝑥). 1 𝑖𝑓 0 ≤ 𝑥 < 𝜋 𝑓(𝑥) = { , 𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥), −1 𝑖𝑓 − 𝜋 ≤ 𝑥 < 0 𝑓(𝑥) =

𝑥 𝑖𝑓 − 𝜋 ≤ 𝑥 < 𝜋, 2

𝑘∈ℤ

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝜋 + 𝑥 𝑖𝑓 0 ≤ 𝑥 < 𝜋 2 𝑓(𝑥) = { 𝜋 , − 𝑥 𝑖𝑓 − 𝜋 ≤ 𝑥 < 0 2

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝑓(𝑥) = |𝑥| 𝑖𝑓 − 𝜋 ≤ 𝑥 < 𝜋,

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝑥 − 1 𝑖𝑓 0 ≤ 𝑥 < 𝜋 , −1 𝑖𝑓 − 𝜋 ≤ 𝑥 < 0

𝑓(𝑥) = {

𝑓(𝑥) = 𝑥 2 𝑖𝑓 − 𝜋 ≤ 𝑥 < 𝜋, − 𝑓(𝑥) = { 𝑓(𝑥) = {

𝜋 2 𝑥 𝜋 2

𝜋 𝑖𝑓 − 𝜋 ≤ 𝑥 < − 2 𝜋 𝜋 𝑖𝑓 − ≤ 𝑥 < , 2 2 𝜋 𝑖𝑓 ≤𝑥<𝜋 2

−2𝑥 + 1 𝑖𝑓 2𝑥 + 1 𝑖𝑓

𝑥 + 1 𝑖𝑓 𝑓(𝑥) = {2 1 𝑖𝑓

−𝜋 ≤𝑥 <0 , 0≤𝑥<𝜋

−𝜋 ≤𝑥 <0 0≤𝑥<𝜋

,

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑓(𝑥 + 𝑘2𝜋) = 𝑓(𝑥),

𝑘∈ℤ

𝑘∈ℤ

𝑘∈ℤ

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Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.