1. MATH-321 In
One Slide
MATH-321
&
MATLAB Command

2. Final Exam From Coordinator
1. Formulas for the error bounds and RK methods will be provided if needed.
2. The exam consists of written questions on the new material and MCQs on the old material. 
3. Two questions about Matlab .

3. Formula Error Equation
Some Comments on Root finding
Formula
Error Equation
Newton’s
𝑝 𝑛 = 𝑝 𝑛−1 − 𝑓( 𝑝 𝑛−1 ) 𝑓′( 𝑝 𝑛−1 )
Secant
𝑝 𝑛 = 𝑝 𝑛−1 − 𝑓( 𝑝 𝑛−1 )( 𝑝 𝑛−1 − 𝑝 𝑛−2 ) 𝑓 𝑝 𝑛−1 −𝑓( 𝑝 𝑛−2 )
𝑝 𝑛 = 𝑎 𝑛−1 + 𝑏 𝑛−1 2
𝑝 ∗ − 𝑝 𝑛 ≤ 𝑏−𝑎 2 𝑛
Bisection
Fixed-point
𝑝 𝑛 =𝑓( 𝑝 𝑛−1 )
𝑔′(𝑥) ≤𝑘 ∀𝑥

4. Sec:4.1 Numerical Differentiation
Backward-difference Formula
𝑓 ′ 𝒙 𝟎 = 𝑓 𝒙 𝟎 −𝑓( 𝒙 𝟎 −𝒉) ℎ + ℎ 2 𝑓′′(𝜉)
Forward-difference Formula
𝑓 ′ 𝒙 𝟎 = 𝑓 𝒙 𝟎 +𝒉 −𝑓( 𝒙 𝟎 ) ℎ − ℎ 2 𝑓′′(𝜉)
where 𝜉 between 𝒙 𝟎 and 𝒙 𝟎 −𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
where 𝜉 between 𝒙 𝟎 and 𝒙 𝟎 +𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
Central Difference for f’(x)
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
Central Difference for f’’(x)
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡

5. Sec:4.3 Elements of Numerical Integration
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
ℎ=𝑏−𝑎
Simpson’s Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
ℎ=(𝑏−𝑎)/2
The midpoint Rule 𝑎 𝑏
𝑥 0
𝒚=𝒇(𝒙)
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =(𝒃−𝒂)𝒇 𝒂+𝒃 𝟐 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
𝑥 0 =(𝑏+𝑎)/2

6. Sec:5.2 EULER’S METHOD
Euler’s Method
𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝑖 , 𝑤 𝑖
𝑤 =𝛼

7. Sec:3.3 Divided Differences
Example
𝒙 𝒊 𝒇(𝒙 𝒊 )
Calculate f (4) using Newton’s interpolating polynomials of order 3.
Given the data
𝒙 𝒊
𝒇[𝒙 𝒊 ]
First
Second
Third 1 3 2 6 19 5 99 𝟑 𝟓 𝟏𝟑 𝟏 𝟗 𝟒𝟎
𝒑 𝟑 𝒙 =𝟑+𝟑 𝒙−𝟏 +𝟓 𝒙−𝟏 𝒙−𝟐 +𝟏(𝒙−𝟏)(𝒙−𝟐)(𝒙−𝟑)
The coefficients of the Newton forward divided-difference form of the interpolating polynomial are along the diagonal in the table.
𝒑 𝟑 𝟒 =𝟒𝟖

8. MATH-321 and MATLAB Command
Root-finding
(Ch.2)
Numerical Differentiation
Initial Value Prob (IVP)
(Ch.4)
(Ch.5)
The Bisection Method
Euler
Mod. Euler
Huen’s
RK4 
Forward Difference
Ode45, ode23, ode113, ode15s, ode23s, ode23t, ode23tb, ode15i,
Fixed- Point Iteration
Newton's Method
Backward Difference
Secant Method
Central Difference y’
Central Difference y’’
x^3+x+1;
fzero(f,1)
Linear System
(Ch.6,7)
h=0.01; x=0:h:pi/2;
diff(sin(x))/h \
linsolve
pcg
minres
Gmres lu
Gaussian elim
Jacobi
Gauss-Seidel
LU fact 
roots([ ])
Numerical integration
(Ch.4)
Interpolation
(Ch.3)
Midpoint
Trapozidal
Simpson’s
Composite 
Lagrange Poly
integral,
trapz,
quad
Divided Differences
Least Squares
(Ch.8)
Cubic Spline
\ linsolve
spline ,
interp1 ,
csape
x.^3+x+1;
quad(f,0,1)
Finite-Difference Methods
integral(f,0,1)
(Ch.11)

9. After
EXAM-2

10. Sec:11.3 Finite-Difference Methods for Linear Problems
cq19
Final171
𝒚(𝒙 𝒊+𝟏 )−𝟐𝒚( 𝒙 𝒊 )+ 𝒚(𝒙 𝒊−𝟏 ) 𝒉 𝟐
+𝟑 𝒚(𝒙 𝒊+𝟏 )− 𝒚(𝒙 𝒊−𝟏 ) 𝟐𝒉
2 𝒚 (𝒙 𝒊 ) ≈ −
2 𝒙 𝒊 +𝟑
𝒘 𝒊+𝟏 −𝟐 𝒘 𝒊 + 𝒘 𝒊−𝟏 𝒉 𝟐 +𝟑 𝒘 𝒊+𝟏 − 𝒘 𝒊−𝟏 𝟐𝒉 −𝟐 𝒘 𝒊 =𝟐 𝒙 𝒊 +𝟑

11. Sec:7.3 The Jacobi and Gauss-Siedel Iterative Techniques
Example
Keep the diagonal on the left hand side
Use GS to find the second iteraton
GS Method
Change the coeff of LHS to be one by division
𝒙 (𝟎) =(𝟎,𝟎,𝟎,𝟎 ) 𝑻 find 𝒙 (𝟐)

12. Sec:7.3 The Jacobi and Gauss-Siedel Iterative Techniques
Convergence Criterion for the Gauss-Seidel Method
Definition
A is strictly diagonally dominant
That is, the diagonal coefficient in each of the equations must be larger than the sum of the absolute values of the other coefficients in the equation
𝑎 𝑖𝑖 > 𝑗=1 𝑗≠𝑖 𝑛 𝑎 𝑖𝑗 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖
Example
Example1
Use GS to find the second iteraton
THM:
A is strictly diagonally dominant

13. 𝑨=𝑳𝑼 𝑳= 𝑼= Sec:6.5 Matrix Factorization LU factorization Find
-2R1+R2R2
-3R2+R3R3
-4R2+R4R4
-4R1+R3R3
-3R1+R4R4
-R3+R4R4 𝑳= 𝑼=
Solve
𝑳𝑼𝒙 =[𝟏,𝟏,𝟏,𝟏] 𝑻
𝑨𝒙=𝒃 1
𝑳𝒚=𝒃
𝑳𝑼𝒙=𝒃 2
𝑼𝒙=𝒚

14. 𝑼= Sec:6.1 Linear systems of Equations 𝑷= (Pivoting) 𝑷𝑨= -2R1+R2R2
permutation matrix 𝑷=
-(4/3)R2+R4R4
𝑷𝑨=
(2/3)R3+R4R4 𝑼=

15. 3 Cases: Sec:6.1 Linear systems of Equations unique solution
infinite number of solutions
no solution

16. Sec:8.1 Discrete Least Squares Approximation
Fit the data in Table with the discrete least squares polynomial of degree at most 2.
Derivation for linear or polynomial of degree two
𝒑 𝟐 (𝒙) = 𝒂 𝟎 + 𝒂 𝟏 𝒙+ 𝒂 𝟐 𝒙 𝟐
Matrix Form
= 𝒊=𝟏 𝒎 𝒚 𝒊 − 𝒂 𝟎 − 𝒂 𝟏 𝒙 𝒊 𝟐
= 𝒂 𝟎 𝒂 𝟏 𝒂 𝟐
𝑬 𝒂 𝟎 , 𝒂 𝟏
𝝏𝑬 𝝏 𝒂 𝟎 =𝟎
𝝏𝑬 𝝏 𝒂 𝟏 =𝟎
8.7680
𝒂 𝟎 𝒂 𝟏 𝒂 𝟐 =
𝑚 𝒙 𝒊 𝑥 𝑖 𝒙 𝒊 𝑥 𝑖 𝑥 𝑖 𝑥 𝑖 𝑥 𝑖 𝑥 𝑖 𝒂 𝟎 𝒂 𝟏 𝒂 𝟐 = 𝒚 𝒊 𝒙 𝒊 𝒚 𝒊 𝑥 𝑖 2 𝒚 𝒊

17. EXAM-1
and
EXAM-2

18. first divided difference second divided difference
Sec:3.3 Divided Differences
Example 𝒙
𝑓[𝑥]
𝑥 0 =1
𝑓[ 𝑥 0 ] =1
𝑥 1 =2
𝑓[ 𝑥 1 ] =8
𝑥 2 =4
𝑓 𝑥 2 = 14
Fit a second-order polynomial to the three points 𝒙
𝑓[𝑥]
first divided difference
second divided difference
𝑥 0 =1
𝑓[ 𝑥 0 ] =1
𝑓 𝑥 0 , 𝑥 1 = 8−1 2−1 =7
𝑥 1 =2
𝑓[ 𝑥 1 ] =8
𝑓 𝑥 1 , 𝑥 2 = 14−8 4−2 =3
𝑥 2 =4
𝑓 𝑥 2 = 14
= 𝒇 𝒙 𝟏 , 𝒙 𝟐 −𝒇 𝒙 𝟎 , 𝒙 𝟏 𝒙 𝟐 − 𝒙 𝟎
= 𝟑−𝟕 𝟒−𝟏
=− 𝟒 𝟑
𝒇 𝒙 𝟎 , 𝒙 𝟏 , 𝒙 𝟐
𝑝 2 𝑥 =𝒇[ 𝒙 𝟎 ]+𝒇[ 𝒙 𝟎 , 𝒙 𝟏 ] 𝒙− 𝑥 0 +𝒇[ 𝒙 𝟎 , 𝒙 𝟏 , 𝒙 𝟐 ] 𝒙− 𝑥 0 𝒙− 𝑥 1
𝑝 2 𝑥 =𝟏+𝟕 𝒙−1 − 𝟒 𝟑 𝒙−1 𝒙−2

19. Sec:5.2 The Bisection Method
Example:
Use Bisection method to find the root of the function
𝑓 (𝑥) = 10 𝑥 6 −149 𝑥 5 +10𝑥−149 10( 𝑥 4 +1)
in [12, 16] 12 16
Change of sign
-34.8
17.6
𝒙 𝟏 =
True root: 𝑥 𝑟 ∗ =14.9 12 14
Change of sign 16
Iter1 𝒏
𝒙 𝒏
-34.8
-12.6
17.6
𝒙 𝟐 = 14 15 16
Iter2
Change of sign
-12.6
1.5
17.6
𝒙 𝟑 =
Change of sign
14.5 14 15
Iter3
-5.8
1.5
-12.6

20. Sec:6.1 Fixed- Point Iteration
Step 1
Example1
Use simple fixed-point iteration to locate the root of
𝒇 𝒙 = 𝒆 −𝒙 −𝒙
rearranging the function
𝒙= 𝒆 −𝒙  
𝒙=𝒈(𝒙) 
𝒙 𝒏
Step 2 𝒏
𝒙 𝒊+𝟏 =𝒈( 𝒙 𝒊 )
𝒙 𝒊+𝟏 = 𝒆 − 𝒙 𝒊  
Starting with an initial guess of x0 = 0
𝒙 𝟏 =𝒈 𝒙 𝟎 = 𝒆 −𝟎 =𝟏
𝒙 𝟐 =𝒈 𝒙 𝟏 = 𝒆 −𝟏 =
𝒙 𝟑 =𝒈 𝒙 𝟐 = 𝒆 −𝟎.𝟑𝟔𝟕𝟖𝟕𝟗 = ⋮ ⋮ ⋮ ⋮
Thus, each iteration brings the estimate closer to the true value of the root:

21. Sec:2.3 Newton’s Method and Its Extensions
Backward difference
𝑓 ′ 𝑥 𝑛 ≈ 𝑓 𝑥 𝑛 −𝑓( 𝑥 𝑛−1 ) 𝑥 𝑛 − 𝑥 𝑛−1
To approximate the roots of
𝑓 𝑥 =0
Given initial guess 𝑥 1
The Secant Method
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 ) 𝑓′( 𝑥 𝑛 )
𝑥 𝑛+1 = 𝑥 𝑛 − 𝑓( 𝑥 𝑛 )( 𝑥 𝑛 − 𝑥 𝑛−1 ) 𝑓 𝑥 𝑛 −𝑓( 𝑥 𝑛−1 )

22. Sec:5.2 EULER’S METHOD 𝑡 𝑖 𝑤 𝑖 𝑦(𝑡 𝑖 ) Example
𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑤 𝑖 − 𝑡 𝑖 2 +1
𝑤 0 = 0.5
Use Euler’s method with h = 0.4 to approximate the solutions for the following initial-value problems
𝑤 1 = 𝑤 0 +ℎ 𝑤 0 − 𝑡
= −0+1
𝑦 ′ =𝑦− 𝑡 2 +1
0≤ 𝑡 ≤2,
𝑦(0) = 0.5
=1.1
→𝑦(0.4)≈1.1
𝑤 2 = 𝑤 1 +ℎ 𝑤 1 − 𝑡
= −
𝒘 𝟎
𝒘 𝟏
𝒘 𝟐
𝒘 𝟑
𝒘 𝟒
𝒘 𝟓
=1.876
→𝑦(0.8)≈1.876
𝒕 𝟎
𝒕 𝟏
𝒕 𝟐
𝒕 𝟑
𝒕 𝟒
𝒕 𝟓
0.4
0.8
1.2
1.6 2
𝑡 𝑖
𝑤 𝑖
𝑦(𝑡 𝑖 )
Euler’s Method
𝑦 ′ =𝑓 𝑡,𝑦
𝑎≤ 𝑡 ≤𝑏,
𝑦(𝑎) =𝛼
𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝑖 , 𝑤 𝑖
𝑤 =𝛼

23. Sec:5.2 EULER’S METHOD 𝑡 𝑖 𝑤 𝑖 𝑦(𝑡 𝑖 ) Example Euler’s Method
𝑦 ′ =𝑦− 𝑡 2 +1
0≤ 𝑡 ≤4,
𝑦(0) = 0.5
Euler’s Method
𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑤 𝑖 − 𝑡 𝑖 2 +1
𝑤 0 = 0.5
𝑡 𝑖
𝑤 𝑖
𝑦(𝑡 𝑖 )
𝒚(𝒕)= (𝒕+ 𝟏) 𝟐 − 𝟎.𝟓 𝒆 𝒕
𝑻𝒓𝒖𝒆
𝑬𝒖𝒍𝒆𝒓