numerical method basic concept exercise

7/23/2019 Numerical Method Basic Concept Exercise

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I N S T I T U T E O F S P A C E T E C H N O L O G Y

MAT-721

Advanced Numerical TechniquesOHT-02-A

Fall 2015 December 22, 2015

ns!ruc!"r# Fai$an Ahmed Ma% Mar&s# 20

'ues!i"n 1# (5 mar&s)*

Starting from the initial guess (1,2) , use Newton’s method to solve the system (perform three

iterations):

f 1 :=3 x1

2− x2

2=0 f 2≔3 x1 x2

2− x1

3−1=0

The formula for Newton;s method is:

X i= X i−1−J −1 ( X i−1 ) F ( X i−1)

Here,

3 x1

2

− x2

2

F ( X )=¿ 3 x1 x22− x1

3−1¿

And

J =[ ∂ f

1

∂ x1

∂ f 1

∂ x2

∂ f 2

∂ x1

∂ f 2

∂ x1

] ¿[ 6 x1

−2 x2

3 x2

2−3 x1

26 x1 x2

]And

J −1=

1

36 x1

2 x2+2 x

2(3 x

2

2−3 x1

2)

[

6 x1 x

2 +2 x

2

−3 x2

2+3 x1

26 x1

]N"+, "u us! need !" &ee. .lu/in/ values and "b!ain ans+ers*

Here

X 0=(1,2 )T

",

X 1= X 0−J −1 ( X 0 ) F ( X 0 )

¿[1

2]− 1

36 (1 )2 (2 )+2 (2 ) (3 (2 )2−3 (1 )2 ) [

6 (1 ) (2 ) +2 (2 )

−3 (2 )2+3 (1 )

26 (1 ) ][ 3 (1 )

2−(2 )

2

3 (1 ) (2 )2−(1 )

3−1]

Rest is computations .

O Prove that a polynomial function of degree n can have at most n−1 etremas over (−∞ , ∞ ) .

Sol:

The etreme values lies either on the critical point or on the point where the function is not

differentia!le" Since the polynomial function is differentia!le everywhere so the only candidates formaimi#ers are critical points" Since the derivative of degree n polynomial is a polynomial of degreen−1 " The n−1 degree polynomial can have at most n−1 roots according to fundamental

theorem of linear alge!ra"

'ues!i"n 2(5 mar&s)#

$ind the first three iterations of the Power method applied to the following matri




[ 1 −1 0

−2 4 −2

0 −1 1 ]

%se x0=(1,−1,2 )t

Sol:

&ou 'ust need to find x1

= A x0

, x2

= A x1

, x3

= A x3

,O

$ind the etreme values of:

f ( x , y , z )=2 xy+2 xz+2 yz Su!'ect to x2+ y

2+ z2=1

Sol:

The agrange function is:

L ( x , y , λ )=f ( x , y , z )+ λ ( x2+ y2+ z

2−1)The critical points are solution of the system:

∂ L

∂ x=2 y+2 z+2 λx=0

∂ L

∂ y=2 x+2 z+2 λy=0

∂ L

∂ z =2 x+2 y+2 λz=0

∂ L∂ λ = x2

+ y2

+ z2

−1=0

Solve the a!ove system to find critical points and then use essian matri to determine if they are

point of maima or minima,

'ues!i"n # (3 mar&s)* *etermine the singular value decomposition of the matri:

(2 1

0 1)Sol:

$ind the eigenvalues of the matri A T A =

[4 22 2] , the eigenvalues are: "

So S matri is :

&ou can alsofind the eigenvalues to find matri +"

nd now use it to find the matri %"

'ues!i"n 3# a) ( mar&s)* %se -uler method to approimate the solution to the initial value pro!lem:

y' =cos2 t +sin3 t , 0≤ t ≤ 1, y (0 )=1, withh=0.25

Sol:

.ts 'ust computations nothing special"

b)( mar&s) Show that the following initial value pro!lems have uni/ue solution

y' = ycos t , 0≤ t ≤ 1, y (0 )=1

"l# 4"u need !" chec& i !he unc!i"n f ( t , y )= ycost is#




1. Is continuous2. Satisfy Lipschit con!itions in th" #a$ia%&" y.'.




MAT-721

Advanced Numerical TechniquesOHT-02-6

Fall 2015 December 22, 2015

ns!ruc!"r# Fai$an Ahmed Ma% Mar&s# 20

'ues!i"n 1# (5 mar&s)*

%se Newton’s method to solve the following system (perform three iterations with initial guess (0,1))"

x1

2

+ x2

2

− x1=0 x1

2− x2

2− x2=0

O

*etermine the value of a such that the matri:

(1 a

a 4 ).s: a) positive definite b) negative definite"

'ues!i"n 2(5 mar&s)#

$ind the first three iterations of the Power method applied to the following matri

[ 1 −1 0

−2 4 −2

0 −1 2 ]

%se x0=(−1,2,1 )t

O

$ind etreme value of:

f ( x , y , z )=3 x2+2 y

2+3 z2

2 xz su!'ect to x2+ y

2+ z2=1

'ues!i"n # (3 mar&s)*

*etermine the singular value decomposition of the matri:

(−1 1

1 1)'ues!i"n 3# a) ( mar&s)* %se -uler method to approimate the solution to the initial value pro!lem:

y' =1+(t − y)2,

0 ≤t ≤ 3, y (2 )=1, withh=0.25

b)( mar&s) Show that the following initial value pro!lems have uni/ue solution

y' =( 2

t ) y+t 2

et , 0 ≤t ≤ 2, y (1 )=0

numerical method basic concept exercise

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