Summarize for Numerical Analysis (Math 517) . it's Lectures for students of Computer Science and especially students of graduate studies of the Institute of Statistical Studies and Research - Cairo University .

1. Numerical Analysis
1- Taylor Series =
𝑓( 𝑐)
0!
(𝑥 − 𝑐)0 +
′𝑓( 𝑐)
1!
(𝑥 − 𝑐)1 +
′′𝑓( 𝑐)
2!
(𝑥 − 𝑐)2 …..
2- Maclaurin series =
𝑓(0)
0!
+
′𝑓(0)
1!
(𝑥)1 +
′′𝑓( 𝑐)
2!
(𝑥)2 …..
3- Relative Error (RE) = |
𝑋−𝑥
𝑋
|∗ 100 4- Absolute Error (AE) = |𝑋 − 𝑥| Ɛ=.0001
- Solving non Linear Equation , using iterationMethods :
1- Bisection Method : Given : 𝑎 , 𝑏 - 𝑓(𝑐)
Sol : 𝑎 , 𝑏 , 𝑐 - 𝑓(𝑐) 𝑐 = 𝑎 +/2
2- Newton Method :
Given : 𝑥 - 𝑓(𝑥)
Sol : 𝑓( 𝑥) , ′𝑓( 𝑥) - 𝑥2 𝑥2 = 𝑥1 -
𝑓(𝑥)
′𝑓(𝑥)
Given : 𝑥0 , 𝑦0 - 𝑓1(𝑥, 𝑦) - 𝑓2(𝑥, 𝑦)
𝐽𝑜 = [
′𝑥 ′𝑦
′𝑥 ′𝑦
] 𝐽𝑜−1=
1
|𝐽𝑜|
[′𝑦 ′𝑦
′𝑥 ′𝑥
] = [ 𝐽𝑜−1]
[
𝑥1
𝑦1
] = [
𝑥0
𝑦0
] - [ 𝐽𝑜−1][
𝑣1
𝑣2
]
3- Secant Method : Given : 𝑥0 , 𝑥1 - 𝑓(𝑥)
Sol : 𝑓(𝑥0) , 𝑓(𝑥1) , 𝑥2 𝑥2 = 𝑥1 -
𝑓( 𝑥1)(𝑥0−𝑥1)
𝑓(𝑥0)− 𝑓(𝑥1)
: 𝑓(𝑥1) , 𝑓(𝑥2) , 𝑥3
4- Fixed-Point : Given : 𝑥0 , root 𝑓(𝑥) < [root]
′𝑓(𝑥) < 1
Sol : 𝑥1 = 𝑓( 𝑥0)
: 𝑥2 = 𝑓( 𝑥1)
- + ±
- Solving 1 non Linear Equation
𝑣2𝑣1
𝑓1
𝑓2
- Solving 2 non Linear Equation
Matrix Inverse

4. Eigen Values & Eigen Vectors :
Given : A = [
𝑛 𝑛
𝑛 𝑛
]
Sol : A − λT
: A − λT = 0
: (A − λT) 𝑥 = 0
𝐀 − 𝛌𝐓 :
[ 𝐴] - [
𝜆 0
0 𝜆
]
𝐀 − 𝛌𝐓 = 𝟎 :
[
𝐴 − 𝜆 1
1 𝐴 − 𝜆
] = 0 𝜆1 = 𝑣 𝜆2 = 𝑣
𝐀 − 𝛌𝐓 = 𝟎 :
{[ 𝐴] − [
𝜆1 0
0 𝜆1
] } [
𝑥1
𝑥2
] = [
0
0
] 𝑥1 = 𝑣1 𝑥2= 𝑣1
{[ 𝐴] − [
𝜆2 0
0 𝜆2
] }[
𝑥1
𝑥2
] = [
0
0
] 𝑥1 = 𝑣2 𝑥2 = 𝑣2
- Power Method :
Given : [
𝑛 𝑛 𝑛
𝑛 𝑛 𝑛
] 𝑥0 = [1,1,1] error% || iteration
Sol : [
𝑛 𝑛 𝑛
𝑛 𝑛 𝑛
] [
1
1
1
] = [
𝑣1
𝑣2
𝑣3
] = 𝑣1 [
1
𝑣2
𝑣3
]largest element
𝑐1 𝑥1

5. - Interpolation and polynomial approximation :
Given : 𝑥0 , 𝑥1 , 𝑥2
𝑓( 𝑥0) , 𝑓( 𝑥1) , 𝑓( 𝑥2)
Sol : 𝒑( 𝒙) = 𝑓( 𝑥0) 𝐿0( 𝑥) + 𝑓( 𝑥1) 𝐿1( 𝑥) + 𝑓( 𝑥2) 𝐿2( 𝑥)
𝐿0( 𝑥) =
( 𝑥−𝑥1)(𝑥−𝑥2)
( 𝑥0−𝑥1)(𝑥0−𝑥1)
𝐿1( 𝑥) =
( 𝑥−𝑥0)(𝑥−𝑥2)
( 𝑥1−𝑥0)(𝑥1−𝑥2)
𝐿2( 𝑥) =
( 𝑥−𝑥0)(𝑥−𝑥1)
( 𝑥2−𝑥0)(𝑥2−𝑥1)
- Newton’s Interpolating and Polynomial (Divided difference ) :
Given : 𝑥0 , 𝑥1 , 𝑥2
𝑓( 𝑥0) , 𝑓( 𝑥1) , 𝑓( 𝑥2)
Sol : 𝒑( 𝒙) = 𝑓[ 𝑥0] + 𝑓[ 𝑥0 ,𝑥1]( 𝑥 − 𝑥0)
+ 𝑓[ 𝑥0 , 𝑥1 , 𝑥2]( 𝑥 − 𝑥0)( 𝑥 − 𝑥1)
+ 𝑓[ 𝑥0 ,𝑥1 , 𝑥2 , 𝑥3]( 𝑥 − 𝑥0)( 𝑥 − 𝑥1)( 𝑥 − 𝑥3)
𝒙 𝒇[ 𝒙𝒊] 𝑓[𝑥 𝑖 , 𝑥 𝑖+1] 𝑓[ 𝑥 𝑖 , 𝑥 𝑖+1 , 𝑥 𝑖+2] 𝑓[ 𝑥 𝑖 , 𝑥 𝑖+1 , 𝑥 𝑖+2 , 𝑥 𝑖+3]
𝑥0 𝑓[ 𝑥0]
𝑓[ 𝑥0 , 𝑥1]
𝑥1 𝑓[ 𝑥1] 𝑓[ 𝑥0 ,𝑥1 , 𝑥2]
𝑓[ 𝑥1 , 𝑥2] 𝑓[ 𝑥0 ,𝑥1 , 𝑥2 ,𝑥3]
𝑥2 𝑓[ 𝑥2] 𝑓[ 𝑥1 , 𝑥2 , 𝑥3]
𝑓[ 𝑥2 ,𝑥3]
𝑥3 𝑓[ 𝑥3]
𝑓[ 𝑥0 , 𝑥1]= 𝑓[ 𝑥1] - 𝑓[ 𝑥0] / 𝑥1 - 𝑥0
𝑓[ 𝑥0 , 𝑥1 ,𝑥2]= 𝑓[ 𝑥1 ,𝑥2] - 𝑓[ 𝑥0 , 𝑥1] / 𝑥2 - 𝑥0
𝑓[ 𝑥0 , 𝑥1 , 𝑥2 , 𝑥3]= 𝑓[ 𝑥1 , 𝑥2 , 𝑥3] - 𝑓[ 𝑥0 ,𝑥1 , 𝑥2] / 𝑥3 - 𝑥0
- Solving 3 non Linear Equation

6. Numerical Differentiation
Two Points :
Given : 𝑥 - 𝑓(𝑥) ′𝑓(𝑥)
𝑥1 , 𝑥2 , 𝑥3
𝑓( 𝑥1) , 𝑓( 𝑥2) , 𝑓( 𝑥3)
Forward : ′𝑓( 𝑥3) =
𝑓( 𝑥3)− 𝑓( 𝑥2)
𝑥3− 𝑥2
Backward : ′𝑓( 𝑥1) =
𝑓( 𝑥2)− 𝑓( 𝑥1)
𝑥2− 𝑥1
Central : ′𝑓( 𝑥2) =
𝑓( 𝑥3)− 𝑓( 𝑥1)
𝑥3− 𝑥1
Three Point :
Given : 𝑥 - 𝑓(𝑥) ′𝑓(𝑥)
𝑥1 , 𝑥2 , 𝑥3
𝑓( 𝑥1) , 𝑓( 𝑥2) , 𝑓( 𝑥3)
′𝒇( 𝒙 𝒊) =
−3𝑓( 𝑥1)+4𝑓( 𝑥2)−𝑓( 𝑥3)
2.ℎ
DifferentiationFormulasusingLagrange Polynomials :
Given : 𝑥 - 𝑓(𝑥) ′𝑓(𝑥)
𝑥1 , 𝑥2 , 𝑥3
𝑓( 𝑥1) , 𝑓( 𝑥2) , 𝑓( 𝑥3)
′𝒇( 𝒙 𝟐) = 𝑓( 𝑥1)
𝑥2 − 𝑥3
( 𝑥1− 𝑥2)(𝑥1− 𝑥3)
+
𝑓( 𝑥2)
2𝑥2 − 𝑥1− 𝑥3
( 𝑥2− 𝑥1)(𝑥2 − 𝑥1− 𝑥3)
+
𝑓( 𝑥3)
𝑥2 − 𝑥1
( 𝑥3− 𝑥2)(𝑥3− 𝑥1)
+
2
Ɛ =
′𝑓( 𝑥 𝑖)− ′𝑓( 𝑥)
′𝑓( 𝑥)
Ɛ =
′𝑓( 𝑥 𝑖)− ′𝑓( 𝑥)
′𝑓( 𝑥)
Ɛ =
′𝑓( 𝑥 𝑖)− ′𝑓( 𝑥)
′𝑓( 𝑥)