Numerical methods เนื้อหามิดเทอม 2023/1
ไม่อยากอ่านแล้ว! เอาสูตร มาเลยน้อง
Table of Contents
Type of error
ϵ \epsilon ϵ 0.000001 0.000001 0.000001 0.00001 % 0.00001\% 0.00001%
∣ f ( x m ) ∣ < ϵ |f(x_m)| < \epsilon ∣ f ( x m ) ∣ < ϵ
สำหรับ Root of equation เพื่อ Check ว่า f ( x ) f(x) f ( x ) 0 0 0
∣ x m n e w − x m o l d x m o l d ∣ × 100 < ϵ |\dfrac{x_{m_{new}}-x_{m_{old}}}{x_{m_{old}}}| \times 100 < \epsilon ∣ x m o l d x m n e w − x m o l d ∣ × 100 < ϵ
ยิ่งค่าเข้าใกล้ x x x
Root of equation
การหาว่า x x x 0 0 0
Graphical method (Sequential search)
Plot graph x x x f ( x ) f(x) f ( x ) f ( x ) = 0 f(x)=0 f ( x ) = 0 x x x
ex. หาตั้งแต่ [ 0 , 7 ] [0,7] [ 0 , 7 ]
Solution ก็คือหาจนกว่าจะเจอการกลับเครื่องหมายจึงค่อย search ละเอียด
Bisection search (Binary search)
ปัญหาของ Graphical methods คือการ Search ที่เปลือง
จึงใช้วิธีหาทีละครึ่งและดูว่าเครื่องหมายเปลี่ยนที่ด้านซ้ายหรือด้านขวา
สูตร: x M = x L + x R 2 x_M = \dfrac{x_L + x_R}{2} x M = 2 x L + x R
False-position method
คล้ายๆ กับ Bisection Method เพิ่ม การให้น้ำหนักของ x L x_L x L x R x_R x R
จากสูตรสามเหลี่ยมคล้าย
f ( x R ) x R − x 1 = f ( x L ) x L − x 1 \dfrac{f(x_R)}{x_R-x_1}=\dfrac{f(x_L)}{x_L-x_1} x R − x 1 f ( x R ) = x L − x 1 f ( x L ) x L f ( x R ) − x 1 f ( x R ) = x R f ( x L ) − x 1 f ( x L ) x_Lf(x_R)-x_1f(x_R)=x_Rf(x_L)-x_1f(x_L) x L f ( x R ) − x 1 f ( x R ) = x R f ( x L ) − x 1 f ( x L ) x 1 = x L f ( x R ) − x R f ( x L ) f ( x R ) − f ( x L ) x_1=\dfrac{x_Lf(x_R)-x_Rf(x_L)}{f(x_R)-f(x_L)} x 1 = f ( x R ) − f ( x L ) x L f ( x R ) − x R f ( x L ) One-Point Iteration method
Open method (only one initial value needed)
Simple but may not converge
Bonus: Taylor Series
f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + ( x − x 0 2 ) 2 ! f ′ ′ ( x 0 ) + . . . + ( x − x 0 ) n n ! f ( n ) ( x 0 ) f(x)=f(x_0)+(x-x_0)f'(x_0)+\dfrac{(x-x_0^2)}{2!}f''(x_0)+...+\dfrac{(x-x_0)^n}{n!}f^{(n)}(x_0) f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + 2 ! ( x − x 0 2 ) f ′′ ( x 0 ) + ... + n ! ( x − x 0 ) n f ( n ) ( x 0 )
ฟังค์ชั่นใดๆ ในโลกนี้สามารถหาด้วย Taylor series ได้
ต้องมี x 0 x_0 x 0 f n ( x ) f^n(x) f n ( x ) d y d x \dfrac{dy}{dx} d x d y f ( x ) f(x) f ( x )
As the degree of the Taylor polynomial rises, it approaches the correct function. This image shows sin x and its Taylor approximations by polynomials of degree 1, 3, 5, 7, 9, 11, and 13 at x = 0.
Newton-Raphson method
วิธีนี้จะใช้การประมาณค่า x x x x n x_n x n x n + 1 x_{n+1} x n + 1 f ( x ) f(x) f ( x ) x x x
สูตร x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n )
มาจาก Taylor Series ที่ n = 1 n = 1 n = 1
0 = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) 0 = f(x_0) + (x - x_0)f'(x_0) 0 = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) ( x − x 0 ) f ′ ( x 0 ) = − f ( x 0 ) (x - x_0)f'(x_0) = -f(x_0) ( x − x 0 ) f ′ ( x 0 ) = − f ( x 0 ) x − x 0 = − f ( x 0 ) f ′ ( x 0 ) x - x_0 = - \dfrac{f(x_0)}{f'(x_0)} x − x 0 = − f ′ ( x 0 ) f ( x 0 ) x = x 0 − f ( x 0 ) f ′ ( x 0 ) x = x_0 - \dfrac{f(x_0)}{f'(x_0)} x = x 0 − f ′ ( x 0 ) f ( x 0 ) Secant method
ปัญหาของ Newton-Raphson methon คือการที่จำเป็นต้องรู้ x 0 , f ( x 0 ) , f ′ ( x 0 ) , f n ( x 0 ) x_0, f(x_0), f'(x_0), f^{n}(x_0) x 0 , f ( x 0 ) , f ′ ( x 0 ) , f n ( x 0 ) d y d x \dfrac{dy}{dx} d x d y
ใช้เส้นตรงที่ผ่านทางสองจุดบนกราฟของฟังก์ชันเพื่อประมาณค่าของ x x x
วิธีนี้เหมาะสำหรับฟังก์ชันที่ไม่สามารถหาอนุพันธ์ของฟังก์ชันได้
สูตร x n = x n − 2 − f ( x n − 2 ) ( x n − 1 − x n − 2 ) f ( x n − 1 ) − f ( x n − 2 ) x_{n} = x_{n-2} - \dfrac{f(x_{n-2})(x_{n-1} - x_{n-2})}{f(x_{n-1}) - f(x_{n-2})} x n = x n − 2 − f ( x n − 1 ) − f ( x n − 2 ) f ( x n − 2 ) ( x n − 1 − x n − 2 ) x n + 1 = x n − f ( x n ) ( x n − x n − 1 ) f x ( x n ) − f x ( x n − 1 ) x_{n+1}=x_n - \dfrac{f(x_n)(x_n - x_{n-1})}{fx(x_n) - fx(x_{n-1})} x n + 1 = x n − f x ( x n ) − f x ( x n − 1 ) f ( x n ) ( x n − x n − 1 )
จาก Newton-Raphson
x 2 = x 0 − f ( x 0 ) f ′ ( x 0 ) ; f ′ ( x 0 ) ≈ s l o p e = f ( x 0 ) − f ( x 1 ) x 0 − x 1 = f ( x 1 ) − f ( x 0 ) x 1 − x 0 x_2 = x_0 - \dfrac{f(x_0)}{\color{red}f'(x_0)} \space\space\space\space ; \space\space\space\space f'(x_0) \approx slope = {\color{red}\dfrac{f(x_0) - f(x_1)}{x_0 - x_1} = \dfrac{f(x_1) - f(x_0)}{x_1 - x_0}} x 2 = x 0 − f ′ ( x 0 ) f ( x 0 ) ; f ′ ( x 0 ) ≈ s l o p e = x 0 − x 1 f ( x 0 ) − f ( x 1 ) = x 1 − x 0 f ( x 1 ) − f ( x 0 ) x 2 = x 0 − f ( x 0 ) f ( x 0 ) − f ( x 1 ) x 0 − x 1 = x 0 − f ( x 0 ) f ( x 1 ) − f ( x 0 ) x 1 − x 0 x_2 = x_0 - \dfrac{f(x_0)}{\color{red}\dfrac{f(x_0) - f(x_1)}{x_0 - x_1}} = x_0 - \dfrac{f(x_0)}{\color{red}\dfrac{f(x_1) - f(x_0)}{x_1 - x_0}} x 2 = x 0 − x 0 − x 1 f ( x 0 ) − f ( x 1 ) f ( x 0 ) = x 0 − x 1 − x 0 f ( x 1 ) − f ( x 0 ) f ( x 0 ) x 2 = x 0 − f ( x 0 ) ( x 0 − x 1 ) f ( x 0 ) − f ( x 1 ) = x 0 − f ( x 0 ) ( x 1 − x 0 ) f ( x 1 ) − f ( x 0 ) x_2 = x_0 - \dfrac{f(x_0)(x_0 - x_1)}{f(x_0) - f(x_1)} = x_0 - \dfrac{f(x_0)(x_1 - x_0)}{f(x_1) - f(x_0)} x 2 = x 0 − f ( x 0 ) − f ( x 1 ) f ( x 0 ) ( x 0 − x 1 ) = x 0 − f ( x 1 ) − f ( x 0 ) f ( x 0 ) ( x 1 − x 0 ) x n = x n − 2 − f ( x n − 2 ) ( x n − 2 − x n − 1 ) f ( x n − 2 ) − f ( x n − 1 ) = x n − 2 − f ( x n − 2 ) ( x n − 1 − x n − 2 ) f ( x n − 1 ) − f ( x n − 2 ) x_n = x_{n-2} - \dfrac{f(x_{n-2})(x_{n-2} - x_{n-1})}{f(x_{n-2}) - f(x_{n-1})} = x_{n-2} - \dfrac{f(x_{n-2})(x_{n-1} - x_{n-2})}{f(x_{n-1}) - f(x_{n-2})} x n = x n − 2 − f ( x n − 2 ) − f ( x n − 1 ) f ( x n − 2 ) ( x n − 2 − x n − 1 ) = x n − 2 − f ( x n − 1 ) − f ( x n − 2 ) f ( x n − 2 ) ( x n − 1 − x n − 2 ) Linear algebra equation
หรือพีชคณิตเชิงเส้น เป็นการแก้และแทนระบบสมการเชิงเส้นด้วย Matrix
a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 ⋮ A x = B [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] { x 1 x 2 x 3 } = { b 1 b 2 b 3 } a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = b_1 \\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = b_2 \\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = b_3 \\
\vdots \\
Ax=B \\
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{bmatrix}
\begin{Bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{Bmatrix}=
\begin{Bmatrix}
b_1 \\
b_2 \\
b_3 \\
\end{Bmatrix} a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 ⋮ A x = B a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎩ ⎨ ⎧ x 1 x 2 x 3 ⎭ ⎬ ⎫ = ⎩ ⎨ ⎧ b 1 b 2 b 3 ⎭ ⎬ ⎫ Cramer's rule
x i = d e t ( A i ) d e t ( A ) ; A 1 = [ b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 ] , A 2 = [ a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 ] , A 3 = [ a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 ] x_i = \dfrac{det(A_i)}{det(A)} \space ; \space A_{\color{red}1} = \begin{bmatrix}
{\color{red}b_1} & a_{12} & a_{13} \\
{\color{red}b_2} & a_{22} & a_{23} \\
{\color{red}b_3} & a_{32} & a_{33} \\
\end{bmatrix},A_{\color{red}2} = \begin{bmatrix}
a_{11} & {\color{red}b_1} & a_{13} \\
a_{21} & {\color{red}b_2} & a_{23} \\
a_{31} & {\color{red}b_3} & a_{33} \\
\end{bmatrix},A_{\color{red}3} = \begin{bmatrix}
a_{11} & a_{12} & {\color{red}b_1} \\
a_{21} & a_{22} & {\color{red}b_2} \\
a_{31} & a_{32} & {\color{red}b_3} \\
\end{bmatrix} x i = d e t ( A ) d e t ( A i ) ; A 1 = b 1 b 2 b 3 a 12 a 22 a 32 a 13 a 23 a 33 , A 2 = a 11 a 21 a 31 b 1 b 2 b 3 a 13 a 23 a 33 , A 3 = a 11 a 21 a 31 a 12 a 22 a 32 b 1 b 2 b 3
A i A_i A i B B B A A A i i i
Bonus: Row Operation
5 x 1 + 7 x 2 = 10 2 x 1 + 5 x 2 = 5 {\begin{align*}
\begin{equation}
5x_1 + 7x_2 = 10 \space\space\space\space
\end{equation} \\
\begin{equation}
2x_1 + 5x_2 = 5 \space\space\space\space
\end{equation}
\end{align*}} 5 x 1 + 7 x 2 = 10 2 x 1 + 5 x 2 = 5
Multiplication
เช่น 5 × ( 2 ) ⇒ 10 x 1 + 25 x 2 = 25 5 \times (2) \rArr 10x_1+25x_2 = 25 5 × ( 2 ) ⇒ 10 x 1 + 25 x 2 = 25
Add / Subtract
เช่น ( 2 ) + ( 1 ) ⇒ 7 x 1 + 12 x 2 = 15 (2) + (1) \rArr 7x_1 + 12x_2 = 15 ( 2 ) + ( 1 ) ⇒ 7 x 1 + 12 x 2 = 15
Guass elimination
Step: วิธีการทำ
Forward elimination
[ a 11 a 12 a 13 ∣ b 1 a 21 a 22 a 23 ∣ b 2 a 31 a 32 a 33 ∣ b 3 ] ⇒ r o w o p e r a t i o n [ a 11 a 12 a 13 ∣ b 1 0 a 22 ′ a 23 ′ ∣ b 2 ′ 0 0 a 33 ′ ′ ∣ b 3 ′ ′ ] \begin{bmatrix}
a_{11} & a_{12} & a_{13} & | & b_{1} \\
a_{21} & a_{22} & a_{23} & | & b_{2} \\
a_{31} & a_{32} & a_{33} & | & b_{3} \\
\end{bmatrix} \xRightarrow[]{row\space operation} \begin{bmatrix}
a_{11} & a_{12} & a_{13} & | & b_{1} \\
0 & a'_{22} & a'_{23} & | & b'_{2} \\
0 & 0 & a''_{33} & | & b''_{3} \\
\end{bmatrix} a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ∣ ∣ ∣ b 1 b 2 b 3 ro w o p er a t i o n a 11 0 0 a 12 a 22 ′ 0 a 13 a 23 ′ a 33 ′′ ∣ ∣ ∣ b 1 b 2 ′ b 3 ′′
Back substitution
x 3 = b 3 ′ ′ / a 33 x 2 = ( b 2 ′ ′ − a 23 ′ x 3 ) / a 22 x 1 = ( b 3 ′ ′ − a 12 x 2 − a 13 x 3 ) / a 11 o r x i = b i − ∑ j = i + 1 n a i j x j a i i \begin{align*}
& x_3 = b''_3 / a_{33} \\
& x_2 = (b''_2 - a'_{23}x_3) / a_{22} \\
& x_1 = (b''_3 - a_{12}x_2 - a_{13}x_3) / a_{11}
\end{align*} \\
\bold{or} \\
x_i = \dfrac{b_{i}- \sum\limits_{j=i + 1}^{n} {a_{ij}x_j} }{a_{ii}} x 3 = b 3 ′′ / a 33 x 2 = ( b 2 ′′ − a 23 ′ x 3 ) / a 22 x 1 = ( b 3 ′′ − a 12 x 2 − a 13 x 3 ) / a 11 or x i = a ii b i − j = i + 1 ∑ n a ij x j
Guass Jordan elimination
คล้ายๆ กับ Guass elimination
Step: วิธีทำ
Forward elimination
Back elimination
F r o m G u a s s e l i m i n a t i o n ; [ a 11 a 12 a 13 ∣ b 1 0 a 22 ′ a 23 ′ ∣ b 2 ′ 0 0 a 33 ′ ′ ∣ b 3 ′ ′ ] ⇒ r o w o p e r a t i o n [ a 11 ′ ′ 0 0 ∣ b 1 ′ ′ 0 a 22 ′ ′ 0 ∣ b 2 ′ ′ 0 0 a 33 ′ ′ ∣ b 3 ′ ′ ] From\space Guass\space elimination; \\
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & | & b_{1} \\
0 & a'_{22} & a'_{23} & | & b'_{2} \\
0 & 0 & a''_{33} & | & b''_{3} \\
\end{bmatrix} \xRightarrow[]{row\space operation} \begin{bmatrix}
a''_{11} & 0 & 0 & | & b''_{1} \\
0 & a''_{22} & 0 & | & b''_{2} \\
0 & 0 & a''_{33} & | & b''_{3} \\
\end{bmatrix} F ro m G u a ss e l imina t i o n ; a 11 0 0 a 12 a 22 ′ 0 a 13 a 23 ′ a 33 ′′ ∣ ∣ ∣ b 1 b 2 ′ b 3 ′′ ro w o p er a t i o n a 11 ′′ 0 0 0 a 22 ′′ 0 0 0 a 33 ′′ ∣ ∣ ∣ b 1 ′′ b 2 ′′ b 3 ′′
( r n ) = ( r n ) / ( r n ) (r_n) = (r_n) / (r_n) ( r n ) = ( r n ) / ( r n )
[ a 11 ′ ′ 0 0 ∣ b 1 ′ ′ 0 a 22 ′ ′ 0 ∣ b 2 ′ ′ 0 0 a 33 ′ ′ ∣ b 3 ′ ′ ] ⇒ r o w o p e r a t i o n [ 1 0 0 ∣ b 1 ∗ 0 1 0 ∣ b 2 ∗ 0 0 1 ∣ b 3 ∗ ] \begin{bmatrix}
a''_{11} & 0 & 0 & | & b''_{1} \\
0 & a''_{22} & 0 & | & b''_{2} \\
0 & 0 & a''_{33} & | & b''_{3} \\
\end{bmatrix} \xRightarrow[]{row\space operation} \begin{bmatrix}
1 & 0 & 0 & | & b^*_{1} \\
0 & 1 & 0 & | & b^*_{2} \\
0 & 0 & 1 & | & b^*_{3} \\
\end{bmatrix} a 11 ′′ 0 0 0 a 22 ′′ 0 0 0 a 33 ′′ ∣ ∣ ∣ b 1 ′′ b 2 ′′ b 3 ′′ ro w o p er a t i o n 1 0 0 0 1 0 0 0 1 ∣ ∣ ∣ b 1 ∗ b 2 ∗ b 3 ∗
Matrix Inversion
สร้าง Matrix [ A ∣ I ] [A|I] [ A ∣ I ] [ I ∣ A − 1 ] [I|A^{-1}] [ I ∣ A − 1 ]
พิสูจน์สไตล์จารย์ 😎
A = [ 2 7 5 8 ] E 1 A = [ 2 7 0 8 ] E 2 E 1 A = [ 2 0 0 8 ] E 3 E 2 E 1 A = [ 1 0 0 8 ] E 4 E 3 E 2 E 1 A = [ 1 0 0 1 ] ∴ E n … E 3 E 2 E 1 A = I A − 1 = E n … E 3 E 2 E 1 \begin{align*}
A = \begin{bmatrix}
2 & 7 \\ 5 & 8
\end{bmatrix} \\
E_1A = \begin{bmatrix}
2 & 7 \\ 0 & 8
\end{bmatrix} \\
E_2E_1A = \begin{bmatrix}
2 & 0 \\ 0 & 8
\end{bmatrix} \\
E_3E_2E_1A = \begin{bmatrix}
1 & 0 \\ 0 & 8
\end{bmatrix} \\
E_4E_3E_2E_1A = \begin{bmatrix}
1 & 0 \\ 0 & 1
\end{bmatrix}
\end{align*} \\
\therefore
{\color{red}E_n \dots E_3E_2E_1}A = I\\
A^{-1} = {\color{red}E_n \dots E_3E_2E_1} A = [ 2 5 7 8 ] E 1 A = [ 2 0 7 8 ] E 2 E 1 A = [ 2 0 0 8 ] E 3 E 2 E 1 A = [ 1 0 0 8 ] E 4 E 3 E 2 E 1 A = [ 1 0 0 1 ] ∴ E n … E 3 E 2 E 1 A = I A − 1 = E n … E 3 E 2 E 1
วิธีทำแบบ Computer ก็คือการทำ Guass Jordan elimination
[ A ∣ I ] ⇒ E n [ I ∣ A − 1 ] \begin{bmatrix}
A | I
\end{bmatrix} \xRightarrow{E_n}
\begin{bmatrix}
I | A^{-1}
\end{bmatrix} [ A ∣ I ] E n [ I ∣ A − 1 ]
จากนั้นคูณ Matrix [ A ] { x } = { B } → [ A − 1 ] { B } = { X } [A]\{x\}=\{B\} \rightarrow [A^{-1}]\{B\}=\{X\} [ A ] { x } = { B } → [ A − 1 ] { B } = { X }
LU Decomposition method
ระเบิด Matrix ออกเป็นสองส่วน บู้มๆๆ 🔥
A = L U [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] = [ l 11 0 0 l 21 l 22 0 l 31 l 32 l 33 ] [ 1 u 12 u 13 0 1 u 23 0 0 1 ] A = LU \\
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{bmatrix} = \begin{bmatrix}
l_{11} & 0 & 0 \\
l_{21} & l_{22} & 0 \\
l_{31} & l_{32} & l_{33} \\
\end{bmatrix} \begin{bmatrix}
1 & u_{12} & u_{13} \\
0 & 1 & u_{23} \\
0 & 0 & 1 \\
\end{bmatrix} A = LU a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 = l 11 l 21 l 31 0 l 22 l 32 0 0 l 33 1 0 0 u 12 1 0 u 13 u 23 1
Step วิธีทำ
สร้าง A x = B {\color{red}A}x=B A x = B
L U x = B {\color{red}LU}x=B LU x = B Y = U x Y=Ux Y = Ux L Y = B LY=B L Y = B หา Y Y Y L Y = B LY=B L Y = B
หา x x x U x = Y Ux=Y Ux = Y
สูตร
l 11 = a 11 u 12 = a 12 l 11 u 13 = a 13 l 11 l 21 = a 21 l 22 = a 22 − l 21 ( u 12 ) u 23 = a 23 − l 21 ( u 13 ) l 22 l 31 = a 31 l 32 = a 32 − l 31 ( u 12 ) l 33 = a 33 − l 31 ( u 13 ) − l 32 ( u 23 ) \begin{align*}
& l_{11} = a_{11} \\
& u_{12} = \dfrac{a_{12}}{l_{11}} \\
& u_{13} = \dfrac{a_{13}}{l_{11}}
\end{align*}\space\space\space\space\space\space
\begin{align*}
& l_{21} = a_{21} \\
& l_{22} = a_{22} - l_{21}(u_{12}) \\
& u_{23} = \dfrac{a_{23} - l_{21}(u_{13})}{l_{22}}
\end{align*}\space\space\space\space\space\space
\begin{align*}
& l_{31} = a_{31} \\
& l_{32} = a_{32} - l_{31}(u_{12}) \\
& l_{33} = a_{33} - l_{31}(u_{13}) - l_{32}(u_{23})
\end{align*} l 11 = a 11 u 12 = l 11 a 12 u 13 = l 11 a 13 l 21 = a 21 l 22 = a 22 − l 21 ( u 12 ) u 23 = l 22 a 23 − l 21 ( u 13 ) l 31 = a 31 l 32 = a 32 − l 31 ( u 12 ) l 33 = a 33 − l 31 ( u 13 ) − l 32 ( u 23 ) Cholesky decomposition method
เงื่อนไขคือ [ A ] [A] [ A ] Symmetric Matrix \color{red}\textrm{Symmetric Matrix} Symmetric Matrix
[ A ] = [ L ] [ L ] T [ a 11 a 12 a 13 a 12 a 22 a 23 a 13 a 23 a 33 ] = [ l 11 0 0 l 21 l 21 0 l 31 l 32 l 33 ] ⋅ [ l 11 0 0 l 21 l 21 0 l 31 l 32 l 33 ] [A]=[L] [L]^T \\
\begin{bmatrix}
a_{11} & {\color{red}a_{12}} & {\color{blue}a_{13}} \\
{\color{red}a_{12}} & a_{22} & {\color{pink}a_{23}} \\
{\color{blue}a_{13}} & {\color{pink}a_{23}} & a_{33}
\end{bmatrix}=\begin{bmatrix}
l_{11} & 0 & 0 \\
l_{21} & l_{21} & 0 \\
l_{31} & l_{32} & l_{33} \\
\end{bmatrix}\cdot\begin{bmatrix}
l_{11} & 0 & 0 \\
l_{21} & l_{21} & 0 \\
l_{31} & l_{32} & l_{33} \\
\end{bmatrix} [ A ] = [ L ] [ L ] T a 11 a 12 a 13 a 12 a 22 a 23 a 13 a 23 a 33 = l 11 l 21 l 31 0 l 21 l 32 0 0 l 33 ⋅ l 11 l 21 l 31 0 l 21 l 32 0 0 l 33 a 11 = l 11 l 11 + 0 l 21 + 0 l 31 l 11 = a 11 . . . \begin{align*}
& a_{11} = l_{11}l_{11} + 0l_{21} + 0l_{31} \\
& l_{11} = \sqrt{a_{11}} \\
& ...
\end{align*} a 11 = l 11 l 11 + 0 l 21 + 0 l 31 l 11 = a 11 ... รวมสูตร
ชื่อ สูตร วิธีทำ Graphical method - - Bisection search x M = x L + x R 2 x_M = \dfrac{x_L + x_R}{2} x M = 2 x L + x R - False-position method x 1 = x L f ( x R ) − x R f ( x L ) f ( x R ) − f ( x L ) x_1=\dfrac{x_Lf(x_R)-x_Rf(x_L)}{f(x_R)-f(x_L)} x 1 = f ( x R ) − f ( x L ) x L f ( x R ) − x R f ( x L ) - One-Point Iteration method x n + 1 = . . . x n x_{n+1} = ... x_{n} x n + 1 = ... x n 1. ย้ายข้างสมการให้ได้ x = . . . x=... x = ... 2. สร้าง x i + 1 = . . . x i x_{i+1}=...x_i x i + 1 = ... x i 3. กำหนด x 0 x_0 x 0 Taylor Series f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + ( x − x 0 2 ) 2 ! f ′ ′ ( x 0 ) + . . . + ( x − x 0 ) n n ! f ( n ) ( x 0 ) f(x)=f(x_0)+(x-x_0)f'(x_0)+\dfrac{(x-x_0^2)}{2!}f''(x_0)+...+\dfrac{(x-x_0)^n}{n!}f^{(n)}(x_0) f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + 2 ! ( x − x 0 2 ) f ′′ ( x 0 ) + ... + n ! ( x − x 0 ) n f ( n ) ( x 0 ) - Newton-Raphson method x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n ) - Secant method x n = x n − 2 − f ( x n − 2 ) ( x n − 1 − x n − 2 ) f ( x n − 1 ) − f ( x n − 2 ) x_{n} = x_{n-2} - \dfrac{f(x_{n-2})(x_{n-1} - x_{n-2})}{f(x_{n-1}) - f(x_{n-2})} x n = x n − 2 − f ( x n − 1 ) − f ( x n − 2 ) f ( x n − 2 ) ( x n − 1 − x n − 2 ) - Cramer's rule x i = d e t ( A i ) d e t ( A ) ; A 1 = [ b 1 a 12 b 2 a 22 ] , A 2 = [ a 11 b 1 a 21 b 2 ] x_i = \dfrac{det(A_i)}{det(A)}\space;\space A_{\color{red}1}=\begin{bmatrix}{\color{red}b_1}&a_{12}\\ {\color{red}b_2}&a_{22}\end{bmatrix},A_{\color{red}2}=\begin{bmatrix}a_{11}&{\color{red}b_1}\\ a_{21}&{\color{red}b_2}\end{bmatrix} x i = d e t ( A ) d e t ( A i ) ; A 1 = [ b 1 b 2 a 12 a 22 ] , A 2 = [ a 11 a 21 b 1 b 2 ] Guass elimination - 1. Forward elimination 2. Back substiution Guass Jordan elimination - 1. Forward elimination 2. Back elimination Matrix Inversion - Guass Jordan elimination LU Decomposition method - 1. A x = B → L U x = B Ax=B \rightarrow LUx=B A x = B → LUx = B 2. หา Y จาก L Y = B LY=B L Y = B 3. หา x จาก U X = Y UX=Y U X = Y Cholesky decomposition method
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