chapter 3 interpolation and polynomial approximation11/8/2012 1 chapter 3 interpolation and...

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Chapter 3 Interpolation

and Polynomial Approximation

Numerical Analysis

Problem type

ถ้ามีข้อมูลจากการวัด ซึ่งแทนความสัมพันธ์ของตัวแปรต้นและตัวแปรตาม

แล้ว

◦ ต้องการทราบค่าตัวแปรตาม ณ จุดอ่ืนๆ ในช่วงของการวัด

◦ ต้องการทราบพฤติกรรมของฟังก์ชันที่แทนข้อมูล

Weierstrass 𝑓 𝑎, 𝑏 ε > 0 𝑃 𝑎, 𝑏 𝑓 𝑥 − 𝑃 𝑥 < 𝜀 𝑥 ∈ 𝑎, 𝑏

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Theory of Weierstrass

O x

y

nx 1x 0x

xf x

xPn x

2x

Interpolation: Overview

𝑓 𝑥 = 𝑒𝑥 𝑃𝑖 𝑥 𝑖

𝑥0 = 0

𝑃0 𝑥 = 1

𝑃1 𝑥 = 1 + 𝑥

𝑃2 𝑥 = 1 + 𝑥 +𝑥2

2

𝑃3 𝑥 = 1 + 𝑥 +𝑥2

2+

𝑥3

6

𝑃4 𝑥 = 1 + 𝑥 +𝑥2

2+

𝑥3

6+

𝑥4

24

𝑃5 𝑥 = 1 + 𝑥 +𝑥2

2+

𝑥3

6+

𝑥4

24+

𝑥5

120

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𝑥 𝑓 𝑥 = 𝑒𝑥 𝑃1 𝑥 𝑃2 𝑥 𝑃3 𝑥 𝑃4 𝑥 𝑃5 𝑥

-2.0 0.13533528 -1.00000000 1.00000000 -0.33333333 0.33333333 0.06666667

-1.5 0.22313016 -0.50000000 0.62500000 0.06250000 0.27343750 0.21015625

-1.0 0.36787944 0.00000000 0.50000000 0.33333333 0.37500000 0.36666667

-0.5 0.60653066 0.50000000 0.62500000 0.60416667 0.60677083 0.60651042

0.0 1.00000000 1.00000000 1.00000000 1.00000000 1.00000000 1.00000000

0.5 1.64872127 1.50000000 1.62500000 1.64583333 1.64843750 1.64869792

1.0 2.71828183 2.00000000 2.50000000 2.66666667 2.70833333 2.71666667

1.5 4.48168907 2.50000000 3.62500000 4.18750000 4.39843750 4.46171875

2.0 7.38905610 3.00000000 5.00000000 6.33333333 7.00000000 7.26666667


𝑥0 = 1 𝑓 𝑥 =1

𝑥 𝑛

𝑃𝑛 𝑥 = 𝑓 𝑘 1

𝑘 ! 𝑥 − 1 𝑘𝑛

𝑖=0 = −1 𝑘 𝑥 − 1 𝑘𝑛𝑖=0

𝑓 3 𝑃𝑛 𝑥

𝑃𝑛 3 𝑛 0 1 2 3 4 5 6 7

𝑃𝑛 3 1 -1 3 -5 11 -21 43 -85

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Lagrange Polynomial

O

y

𝑥

Lagrange Polynomial

2 𝑥0 , 𝑓 𝑥0 𝑥1 , 𝑓 𝑥1 ( )

𝑃 𝑥 =𝑥 − 𝑥1

𝑥0 − 𝑥1𝑓 𝑥0 +

𝑥 − 𝑥0

𝑥1 − 𝑥0𝑓 𝑥1

𝑥 = 𝑥0 𝑃1 𝑥0 =𝑥0−𝑥1

𝑥0−𝑥1𝑓 𝑥0 +

𝑥0−𝑥0

𝑥1−𝑥0𝑓 𝑥1 = 1 ∙ 𝑓 𝑥0

+0 ∙ 𝑓 𝑥1 = 𝑓 𝑥0 𝑥 = 𝑥1 𝑃1 𝑥1 =

𝑥1−𝑥1

𝑥0−𝑥1𝑓 𝑥0 +

𝑥1−𝑥0

𝑥1−𝑥0𝑓 𝑥1 = 0 ∙ 𝑓 𝑥0

+1 ∙ 𝑓 𝑥1 = 𝑓 𝑥1 𝑃1 𝑥

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Lagrange Polynomial

𝑥0, 𝑓 𝑥0 𝑥1, 𝑓 𝑥1

𝐿1,0 𝑥 =𝑥−𝑥1

𝑥0−𝑥1 𝐿1,1 𝑥 =

𝑥−𝑥0

𝑥1−𝑥0

𝑥 = 𝑥0 , 𝐿1,0 𝑥0 = 1, 𝐿1,1 𝑥0 = 0 𝑃1 𝑥0 = 𝑓 𝑥0

𝑥 = 𝑥1 , 𝐿1,0 𝑥1 = 0, 𝐿1,1 𝑥1 = 1 𝑃1 𝑥1 = 𝑓 𝑥1

𝐿𝑛 ,𝑘 𝑥

𝐿𝑛 ,𝑘 𝑥𝑖 = 0, 𝑖 ≠ 𝑘1, 𝑖 = 𝑘

𝑘 = 0,1,2, … , 𝑛

Lagrange Polynomial

𝐿𝑛 ,𝑘 𝑥

𝐿𝑛 ,𝑘 𝑥 = 𝑥 − 𝑥0 𝑥 − 𝑥1 … 𝑥 − 𝑥𝑘−1 𝑥 − 𝑥𝑘+1 … 𝑥 − 𝑥𝑛

𝑥𝑘 − 𝑥0 𝑥𝑘 − 𝑥1 … 𝑥𝑘 − 𝑥𝑘−1 𝑥𝑘 − 𝑥𝑘+1 … 𝑥𝑘 − 𝑥𝑛

𝐿𝑘 𝑥 𝐿𝑛 ,𝑘 𝑥 𝑛

𝑛

𝑃𝑛 𝑥 = 𝐿0 𝑥 𝑓 𝑥0 + ⋯ + 𝐿𝑛 𝑥 𝑓 𝑥𝑛 𝑃𝑛 𝑥 = 𝐿𝑘 𝑥 𝑓 𝑥𝑘

𝑛𝑘=0

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Lagrange Polynomial: Example

𝑓 𝑥 =1

𝑥 𝑥0 = 2, 𝑥1 = 2.5 𝑥2 = 4

𝑥 2 2.5 4

𝑓 𝑥 0.5 0.4 0.25

𝑓 3 𝐿0, 𝐿1, 𝐿2

𝐿0 𝑥 = 𝑥 − 𝑥1 𝑥 − 𝑥2

𝑥0 − 𝑥1 𝑥0 − 𝑥2 =

𝑥 − 2.5 𝑥 − 4

2 − 2.5 2 − 4 = 𝑥2 − 6.5𝑥 + 10

𝐿1 𝑥 = 𝑥 − 𝑥0 𝑥 − 𝑥2

𝑥1 − 𝑥0 𝑥1 − 𝑥2 =

𝑥 − 2 𝑥 − 4

2.5 − 2 2.5 − 4 =

−4𝑥2 + 24𝑥 − 32

3

𝐿2 𝑥 = 𝑥 − 𝑥0 𝑥 − 𝑥1

𝑥2 − 𝑥0 𝑥2 − 𝑥1 =

𝑥 − 2 𝑥 − 2.5

4 − 2 4 − 2.5 =

𝑥2 − 4.5𝑥 + 5

3


𝑃2 𝑥 = 𝐿𝑛 𝑥 𝑓 𝑥𝑘 2𝑘=0

= 0.5 𝑥2 − 6.5𝑥 + 10 + 0.4 −4𝑥2+24𝑥−32

3 + 0.25

𝑥2−4.5𝑥+5

3

= 0.05𝑥2 − 0.425𝑥 + 1.15 𝑓 3 =

1

3= 0.333∙ 𝑃2 3 = 0.325

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𝑓 𝑛+1 𝜉 𝑥−𝑥0 𝑛+1

𝑛+1 ! 𝜉 𝑥 𝑥0

𝑛 𝑥0 , 𝑥1 , … , 𝑥𝑛

𝑓 𝑥 = 𝑃𝑛 𝑥 +𝑓 𝑛+1 𝜉

𝑛 + 1 ! 𝑥 − 𝑥0 𝑥 − 𝑥1 … 𝑥 − 𝑥𝑛

𝜉 𝑥 𝑥0 , 𝑥1 , … , 𝑥𝑛

Divided Difference

𝑓 𝑥𝑖 𝑓 𝑥𝑖 = 𝑓 𝑥𝑖 (recursive) 𝑓 𝑥𝑖 𝑥𝑖+1

𝑓 𝑥𝑖 , 𝑥𝑖+1 =𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖

𝑥𝑖+1 − 𝑥𝑖

𝑘 𝑥𝑖 , 𝑥𝑖+1, … , 𝑥𝑖+𝑘

𝑓 𝑥𝑖 , 𝑥𝑖+1, …𝑥𝑖+𝑘 =𝑓 𝑥𝑖+1, … , 𝑥𝑖+𝑘 − 𝑓 𝑥𝑖 , … , 𝑥𝑖+𝑘−1

𝑥𝑖+𝑘 − 𝑥𝑖

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Divided Difference

( )

𝑃𝑛 𝑥 = 𝑓 𝑥0 + 𝑓 𝑥0, 𝑥1 𝑥 − 𝑥0 + 𝑓 𝑥0, 𝑥1, 𝑥2 𝑥 − 𝑥0 𝑥 − 𝑥1 + ⋯+ 𝑓 𝑥0, 𝑥1, … , 𝑥𝑛 𝑥 − 𝑥0 𝑥 − 𝑥1 … 𝑥 − 𝑥𝑛−1

( )

𝑃𝑛 𝑥 = 𝑓 𝑥𝑛 + 𝑓 𝑥𝑛−1, 𝑥𝑛 𝑥 − 𝑥𝑛 + 𝑓 𝑥𝑛−2, 𝑥𝑛−1, 𝑥𝑛 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 + ⋯+ 𝑓 𝑥0, 𝑥1 , … , 𝑥𝑛 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 … 𝑥 − 𝑥1

Divided Difference

𝒙 𝒇 𝒙 First divided difference Second divided difference 𝑥0 𝑓 𝑥0

𝑓 𝑥0, 𝑥1 =𝑓 𝑥1 − 𝑓 𝑥0

𝑥1 − 𝑥0

𝑥1 𝑓 𝑥1 𝑓 𝑥0, 𝑥1 , 𝑥2 =𝑓 𝑥1, 𝑥2 − 𝑓 𝑥0, 𝑥1

𝑥2 − 𝑥0

𝑓 𝑥1, 𝑥2 =𝑓 𝑥2 − 𝑓 𝑥1

𝑥2 − 𝑥1

𝑥2 𝑓 𝑥2 𝑓 𝑥1, 𝑥2, 𝑥3 =𝑓 𝑥2, 𝑥3 − 𝑓 𝑥1, 𝑥2

𝑥3 − 𝑥1

𝑓 𝑥2, 𝑥3 =𝑓 𝑥3 − 𝑓 𝑥2

𝑥3 − 𝑥2

𝑥3 𝑓 𝑥3

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Newton Divided Difference: Example

𝑥 1.0 1.3 1.6 1.9 2.2

𝑓 𝑥 0.7651977 0.6200860 0.4554022 0.2818186 0.1103623

𝒇 𝟏. 𝟓


xi f(xi)=f[xi] 1st Divided Diff 2nd Divided Diff 3rd Divided Diff 4th Divided Diff

1.0 0.7651977 -0.4837057

1.3 0.6200860 -0.1087339 -0.5489460 0.0658784

1.6 0.4554022 -0.0494433 0.0018251 -0.5786120 0.0680685

1.9 0.2818186 0.0118183 -0.5715210

2.2 0.1103623

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( ) Newton 𝑃4 𝑥 = 0.7651977 − 0.483705 𝑥 − 1 − 0.1087339 𝑥 − 1 𝑥 − 1.3

+ ⋯ + 0.0658784 𝑥 − 1 𝑥 − 1.3 𝑥 − 1.6 + 0.0018251 𝑥 − 1 𝑥 − 1.3 𝑥 − 1.6 𝑥 − 1.9

𝑃4 1.5 = 0.511820

( ) Newton

ค่าประมาณที่ x=1.8 มีค่าเท่าไร

Forward Divided Difference

𝑥0 , 𝑥1 , … , 𝑥𝑛 ℎ = 𝑥𝑖+1 − 𝑥𝑖 𝑖 = 0,1,2, … , 𝑛 𝑥 = 𝑥0 + 𝑠ℎ 𝑥 − 𝑥𝑖 = 𝑠 − 𝑖 ℎ

( ) 𝑃𝑛 𝑥 = 𝑃𝑛 𝑥0 + 𝑠ℎ

= 𝑓 𝑥0 + 𝑠ℎ𝑓 𝑥0 , 𝑥1 + 𝑠 𝑠 − 1 ℎ2𝑓 𝑥0 , 𝑥1 , 𝑥2 + ⋯ +𝑠 𝑠 − 1 𝑠 − 2 … 𝑠 − 𝑛 + 1 ℎ𝑛𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑛

= 𝑠 𝑠 − 1 𝑠 − 2 … 𝑠 − 𝑘 + 1 ℎ𝑘𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑘

𝑘=𝑛

𝑘=0

𝑃𝑛 𝑥 = 𝑃𝑛 𝑥0 + 𝑠ℎ = 𝑠𝑘 𝑘! ℎ𝑘𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑘

𝑘=𝑛𝑘=0

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𝑃𝑛 𝑛=0∞

∆𝑃𝑛 = 𝑃𝑛+1 − 𝑃𝑛 𝑛 ≥ 0

∆𝑘𝑃𝑛 = ∆ ∆𝑘−1 𝑃𝑛 𝑘 ≥ 2

𝑓 𝑥0, 𝑥1 =𝑓 𝑥1 − 𝑓 𝑥0

𝑥1 − 𝑥0=

1

ℎ∆𝑓 𝑥0

𝑓 𝑥0, 𝑥1, 𝑥2 =1

2ℎ ∆𝑓 𝑥1 − ∆𝑓 𝑥0

ℎ =

1

2ℎ2∆2𝑓 𝑥0

𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑘 =1

𝑘 !ℎ𝑘 ∆𝑘𝑓 𝑥0


𝑃𝑛 𝑥0 + 𝑠ℎ = 𝑠𝑘 𝑘! ℎ𝑘𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑘

𝑘=𝑛𝑘=0

𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑘 =1

𝑘 !ℎ𝑘 ∆𝑘𝑓 𝑥0

Newton

𝑃𝑛 𝑥 = 𝑠𝑘 ∆𝑘𝑓 𝑥0

𝑘=𝑛

𝑘=0

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Backward Divided Difference

( ) 𝑃𝑛 𝑥 = 𝑓 𝑥𝑛 + 𝑓 𝑥𝑛−1 , 𝑥𝑛 𝑥 − 𝑥𝑛

+ 𝑓 𝑥𝑛−2 , 𝑥𝑛−1 , 𝑥𝑛 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 + ⋯+ 𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑛 𝑥 − 𝑥𝑛 𝑥 − 𝑥𝑛−1 … 𝑥 − 𝑥1

𝑥 = 𝑥𝑛 + 𝑠ℎ, 𝑥𝑖 = 𝑥𝑛 − (𝑛 − 𝑖)ℎ 𝑥 − 𝑥𝑖 = 𝑠 + 𝑛 − 𝑖 ℎ 𝑃𝑛 𝑥 = 𝑃𝑛 𝑥𝑛 + 𝑠ℎ

= 𝑓 𝑥𝑛 + 𝑠ℎ𝑓 𝑥𝑛−1 , 𝑥𝑛 + 𝑠 𝑠 + 1 ℎ2𝑓 𝑥𝑛−2 , 𝑥𝑛−1 , 𝑥𝑛 + ⋯ + 𝑠 𝑠 + 1 𝑠 + 2 … 𝑠 + 𝑛 − 1 ℎ𝑛𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑛


𝑃𝑛 𝑥0 + 𝑠ℎ = −𝑠𝑘

𝑘! ℎ𝑘𝑓 𝑥𝑛−𝑘 , … , , 𝑥𝑛−1 , 𝑥𝑛 𝑘=𝑛𝑘=0

𝑃𝑛 𝑛=0∞

∇𝑃𝑛 = 𝑃𝑛 − 𝑃𝑛−1 𝑛 ≥ 1

∇𝑘𝑃𝑛 = ∇ ∇𝑘−1 𝑃𝑛 𝑘 ≥ 2

𝑓 𝑥𝑛−1, 𝑥𝑛 =1

ℎ∇𝑓 𝑥𝑛 , 𝑓 𝑥𝑛−2, 𝑥𝑛−1, 𝑥𝑛 =

1

2ℎ2 ∇2𝑓 𝑥𝑛

𝑓 𝑥𝑛−𝑘 , … , 𝑥𝑛−1, 𝑥𝑛 =1

𝑘!ℎ𝑘 ∇k𝑓 𝑥𝑛

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𝑃𝑛 𝑥0 + 𝑠ℎ = −𝑠𝑘

𝑘! ℎ𝑘𝑓 𝑥𝑛−𝑘 , … , , 𝑥𝑛−1 , 𝑥𝑛 𝑘=𝑛𝑘=0

𝑓 𝑥𝑛−𝑘 , … , 𝑥𝑛−1, 𝑥𝑛 =1

𝑘!ℎ𝑘 ∇k𝑓 𝑥𝑛

– 𝑆𝑘

= −−𝑠 −𝑠−1 … −𝑠−𝑘+1

𝑘 !=

−1 𝑘𝑠 𝑠+1 … 𝑠+𝑘−1

𝑘 !

“ Newton”

𝑃𝑛 𝑥 = −1 𝑘 𝑠𝑘 ∇𝑘𝑓 𝑥𝑛

𝑘=𝑛𝑘=0


𝑥 1.0 1.3 1.6 1.9 2.2

𝑓 𝑥 0.7651977 0.6200860 0.4554022 0.2818186 0.1103623

𝑓 1.1 𝑓 2.0

𝑓 1.1 𝑥 = 1.1 𝑥0 = 1.0

ℎ = 0.3, 𝑠 =1

3 𝑃4 1.1 = 𝑃4 1.0 +

1

3 0.3

𝑓 2.0 𝑥 = 2.0 𝑥4 = 2.2

ℎ = 0.3, 𝑠 = −2

3 𝑃4 2.0 = 𝑃4 2.2 + −

2

3 0.3

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Example (Newton Divided Difference)


1.0 0.7651977 -0.4837057

1.3 0.6200860 -0.1087339 -0.5489460 0.0658784

1.6 0.4554022 -0.0494433 0.0018251 -0.5786120 0.0680685

1.9 0.2818186 0.0118183 -0.5715210

2.2 0.1103623


𝑃4 1.1 = 𝑃4 1.0 +1

3 0.3

= 0.7651977 +1

3 0.3 −0.483705

+1

3 −

2

3 0.3 2 −0.1087339

+1

3 −

2

3 −

5

3 0.3 3 0.0658784

+1

3 −

2

3 −

5

3 −

8

3 0.3 4 0.0018251 = 0.7196480

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1.0 0.7651977 -0.4837057

1.3 0.6200860 -0.1087339 -0.5489460 0.0658784

1.6 0.4554022 -0.0494433 0.0018251 -0.5786120 0.0680685

1.9 0.2818186 0.0118183 -0.5715210

2.2 0.1103623


𝑃4 2.0 = 𝑃4 2.2 + −2

3 0.3

= 0.1103623 −2

3 0.3 −0.5715210 −

2

3

1

3 0.3 2 0.0118183

−2

3

1

3

4

3 0.3 3 0.0680685

−2

3

1

3

4

3

7

3 0.3 4 0.0018251 = 0.2238754

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Hermite Interpolation

การที่พหุนามมีดีกรีสูงข้ึนจะท าให้ค่าประมาณดีข้ึน การประมาณค่าในช่วง

ของ Hermite นอกจากจะใช้ค่าฟังก์ชัน ณ จุดที่ก าหนดแล้ว ยังใช้ค่า

อนุพันธ์อันดับหนึ่ง ณ จุดที่ก าหนดนั้นด้วย ส าหรบัข้อมูลจ านวน n+1 ตัว พหุนาม Hermite จะมีดีกร ี2n+1


Hermite

𝑓 ∈ 𝐶1 𝑎, 𝑏 𝑥0 , 𝑥1 , … , 𝑥𝑛 ∈ 𝑎, 𝑏 𝑓 𝑓 ′ 𝑥0 , 𝑥1 , … , 𝑥𝑛 2𝑛 + 1

𝐻2𝑛+1 𝑥 = 𝑓 𝑥𝑗 𝐻𝑛 ,𝑗 𝑥 𝑛𝑗=0 + 𝑓 ′ 𝑥𝑗 𝐻 𝑛 ,𝑗 𝑥

𝑛𝑗=0

𝐻𝑛 ,𝑗 𝑥 = 1 − 2 𝑥 − 𝑥𝑗 𝐿𝑛 ,𝑗′ 𝑥𝑗 𝐿𝑛 ,𝑗

2 𝑥

𝐻 𝑛 ,𝑗 𝑥 = 𝑥 − 𝑥𝑗 𝐿𝑛 ,𝑗2 𝑥

𝐿𝑛 ,𝑗 𝑥 𝑗 Langrange 𝑛

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Hermite

𝑓 ∈ 𝐶 2𝑛+2 𝑎, 𝑏

𝑓 𝑥 = 𝐻2𝑛+1 𝑥 +𝑓 2𝑛+2 𝜉 𝑥

2𝑛+2 ! 𝑥 − 𝑥0

2 … 𝑥 − 𝑥𝑛 2

𝜉 𝑥 ∈ 𝑎, 𝑏

𝑓 ∈ 𝐶𝑛 𝑎, 𝑏 𝑥0 , … , 𝑥𝑛 ∈ 𝑎, 𝑏 𝜉 𝑥 ∈ 𝑎, 𝑏

𝑓 𝑥0 , 𝑥1 , … , 𝑥𝑛 =𝑓𝑛 𝜉

𝑛 !


Hermite 𝑧𝑘 𝑘=02𝑛+1

𝑧2𝑖 = 𝑧2𝑖+1 = 𝑥𝑖 𝑖 = 0,1, … , 𝑛 𝑧0, 𝑧1, … , 𝑧2𝑛+1 𝑧2𝑖 = 𝑧2𝑖+1 = 𝑥𝑖 𝑖

𝑓 𝑧2𝑖 , 𝑧2𝑖+1 =𝑓 𝑧2𝑖+1 −𝑓 𝑧2𝑖

𝑧2𝑖+1−𝑧2𝑖

𝑓 𝑥0, 𝑥1, … , 𝑥𝑛 =𝑓𝑛 𝜉

𝑛 ! 𝑓 𝑥0, 𝑥1 = 𝑓 ′ 𝜉

𝜉 ∈ 𝑥0, 𝑥1

lim𝑥1→𝑥0𝑓 𝑥0, 𝑥1 = 𝑓 ′ 𝑥0

𝑓 𝑧2𝑖 , 𝑧2𝑖+1 = 𝑓 ′ 𝑥𝑖

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Hermite

𝑓 ∈ 𝐶1 𝑎, 𝑏 𝑥0 , 𝑥1 , … , 𝑥𝑛 ∈ 𝑎, 𝑏

𝐻2𝑛+1 𝑥 = 𝑓 𝑥0 + 𝑓 𝑧0, 𝑧1, … , 𝑧𝑘 𝑥 − 𝑧0 𝑥 − 𝑧1 … 𝑥 − 𝑧𝑘−1

2𝑛+1

𝑘=1

𝑧2𝑘 = 𝑧2𝑘+1 = 𝑥𝑘 𝑓 𝑧2𝑖 , 𝑧2𝑖+1 = 𝑓 ′ 𝑥𝑘 𝑖 =

0,1, … , 𝑛


𝑧 𝑓 𝑧 First divided difference Second divided difference

𝑧0 = 𝑥0 𝑓 𝑧0 = 𝑓 𝑥0 𝑓 𝑧0, 𝑧1 = 𝑓 ′ 𝑥0

𝑧1 = 𝑥0 𝑓 𝑧1 = 𝑓 𝑥0 𝑓 𝑧0, 𝑧1, 𝑧2 =

𝑓 𝑧1, 𝑧2 − 𝑓 𝑧0, 𝑧1

𝑧2 − 𝑧0

𝑓 𝑧1, 𝑧2 =

𝑓 𝑧2 − 𝑓 𝑧1

𝑧2 − 𝑧1

𝑧2 = 𝑥1 𝑓 𝑧2 = 𝑓 𝑥1 𝑓 𝑧1, 𝑧2, 𝑧3 =

𝑓 𝑧2, 𝑧3 − 𝑓 𝑧1, 𝑧2

𝑧3 − 𝑧1

𝑓 𝑧2, 𝑧3 = 𝑓 ′ 𝑥1 𝑧3 = 𝑥1 𝑓 𝑧3 = 𝑓 𝑥1

𝑓 𝑧2, 𝑧3, 𝑧4 =𝑓 𝑧3, 𝑧4 − 𝑓 𝑧2, 𝑧3

𝑧4 − 𝑧2

𝑓 𝑧3, 𝑧4 =

𝑓 𝑧4 − 𝑓 𝑧3

𝑧4 − 𝑧3

𝑧4 = 𝑥2 𝑓 𝑧4 = 𝑓 𝑥2 𝑓 𝑧3, 𝑧4, 𝑧5 =

𝑓 𝑧4, 𝑧5 − 𝑓 𝑧3, 𝑧4

𝑧5 − 𝑧3

𝑓 𝑧4, 𝑧5 = 𝑓 ′ 𝑥2 𝑧5 = 𝑥2 𝑓 𝑧5 = 𝑓 𝑥2

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Hermite Interpolation: Example

Hermite 𝑓 1.5

𝑥 𝑓 𝑥 𝑓 ′ 𝑥 1.3 0.6200860 -0.5220232 1.6 0.4554022 -0.5698959 1.9 0.2818186 -0.5811571


𝑧 𝑓 𝑧 1st D Diff. 2nd D Diff. 3rd D Diff. 4th D Diff. 5th D Diff. 1.3 0.6200860

-0.5220232 1.3 0.6200860 -0.08977427

-0.5489460 0.0663657 1.6 0.4554022 -0.0698330 0.0026663

-0.5698959 0.0679655 -0.0027738 1.6 0.4554022 -0.0290537 0.0010020

-0.5786120 0.0685667 1.9 0.2818186 -0.0084837

-0.5811571 1.9 0.2818186

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𝐻2𝑛+1 𝑥 = 𝑓 𝑥0 + 𝑓 𝑧0, 𝑧1, … , 𝑧𝑘 𝑥 − 𝑧0 𝑥 − 𝑧1 … 𝑥 − 𝑧𝑘−1 2𝑛+1𝑘=1

𝐻5 1.5 = 0.6200860 + 1.5 − 1.3 −0.5220232

+ 1.5 − 1.3 2 −0.0897427 + 1.5 − 1.3 2 1.5 − 1.6 0.0663657 + 1.5 − 1.3 2 1.5 − 1.6 2 0.0026663 + 1.5 − 1.3 2 1.5 − 1.6 2 1.5 − 1.9 −0.0027738 = 0.5118277

Cubic Spline Interpolation

ความแม่นย าในการประมาณอาจสูงขึ้น เม่ือใช้พหุนามที่มีดีกรีสูง แต่เมื่อ

ดีกรีที่สูงข้ึนมากอาจจะมีการกวัดแกว่งของเส้นโค้งสูงข้ึนด้วย ซึ่งจะส่งผลให้

ค่าประมาณมีความคลาดเคลือ่นมากข้ึนก็ได้ วิธีหนึ่งที่ใช้แก้ปญัหาคือ แบ่ง

ช่วงทั้งหมดออกเป็นช่วงย่อยๆ แล้วสร้างพหุนามประจ าแตล่ะช่วงย่อย

เรียกว่า “การประมาณโดยพหุนามเป็นช่วงๆ””

ถ้าให้ทุกสองคู่ของจุดแทนช่วงหนึ่งช่วง การเช่ือมจุดของข้อมูลด้วยเส้นตรงก็

คือวิธีที่ง่ายที่สุด แต่ก็จะท าใหเ้ส้นโค้งไม่เรียบ

แนวทางอ่ืนคือ การใช้พหุนาม Hermite แต่ก็ต้องมีข้อมูลของอนุพันธ์อันดับหนึ่งของทุกจุด

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การประมาณโดยพหุนามเป็นส่วนๆ ที่พบบ่อยท่ีสุดคือ การใช้พหุนามก าลัง

สามระหวา่งคู่ของจุด ที่เรียกว่า Cubic Spline

พหุนามก าลังสาม มีค่าคงตัว 4 ค่า โดยท่ัวไปแล้วอนุพันธ์ของ Cubic

Spline ไม่จ าเป็นต้องเท่ากับอนุพันธ์ของฟังก์ชันจริง แม้ที่จุดนิยาม


𝑓 𝑎, 𝑏 𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑛 = 𝑏 𝑆 𝑓

1. 𝑆 𝑆𝑗 𝑥𝑗 , 𝑥𝑗+1 , 𝑗 =

0,1, … , 𝑛 − 1 2. 𝑆 𝑥𝑗 = 𝑓 𝑥𝑗 ( 𝑗 = 0,1, … , 𝑛 )

3. 𝑆𝑗+1 𝑥𝑗 +1 = 𝑆𝑗 𝑥𝑗+1 ( 𝑗 = 0,1, … , 𝑛 − 2 )

4. 𝑆𝑗+1′ 𝑥𝑗 +1 = 𝑆𝑗

′ 𝑥𝑗+1 ( 𝑗 = 0,1, … , 𝑛 − 2 )

5. 𝑆𝑗+1′′ 𝑥𝑗 +1 = 𝑆𝑗

′′ 𝑥𝑗 +1 ( 𝑗 = 0,1, … , 𝑛 − 2 )

6.

a. 𝑆′′ 𝑥0 = 𝑆′′ 𝑥𝑛 = 0 ( )

b. 𝑆′ 𝑥0 = 𝑓 ′ 𝑥0 𝑆′ 𝑥𝑛 = 𝑓 ′ 𝑥𝑛 ( )

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𝑆 𝑥𝑗 = 𝑓 𝑥𝑗 𝑗 = 0,1, … , 𝑛

𝑆𝑗+1 𝑥𝑗+1 = 𝑆𝑗 𝑥𝑗+1 𝑗 = 0,1, … , 𝑛 − 2

𝑆𝑗+1′ 𝑥𝑗+1 = 𝑆𝑗

′ 𝑥𝑗+1 𝑗 = 0,1, … , 𝑛 − 2 𝑆𝑗+1

′′ 𝑥𝑗+1 = 𝑆𝑗′′ 𝑥𝑗+1 𝑗 = 0,1, … , 𝑛 − 2

𝑥0 𝑥1 𝑥𝑗 𝑥𝑗+2 𝑥𝑗+1 𝑥𝑛

𝑆𝑗 𝑆𝑗 +1


𝑆𝑗 𝑥 = 𝑎𝑗 + 𝑏𝑗 𝑥 − 𝑥𝑗 + 𝑐𝑗 𝑥 − 𝑥𝑗 2

+ 𝑑𝑗 𝑥 − 𝑥𝑗 3 𝑗 = 0,1, … , 𝑛 − 1

2 𝑆𝑗 𝑥𝑗 = 𝑎𝑗 = 𝑓 𝑥𝑗

3 𝑎𝑗+1 = 𝑆𝑗+1 𝑥𝑗+1 = 𝑆𝑗 𝑥𝑗+1

= 𝑎𝑗 + 𝑏𝑗 𝑥𝑗+1 − 𝑥𝑗 + 𝑐𝑗 𝑥𝑗+1 − 𝑥𝑗 2

+ 𝑑𝑗 𝑥𝑗+1 − 𝑥𝑗 3 𝑗 = 0,1, … , 𝑛 − 2

ℎ𝑗 = 𝑥𝑗+1 − 𝑥𝑗 𝑗 = 0,1, … , 𝑛 − 1

𝑎𝑛 = 𝑓 𝑥𝑛

1 𝑎𝑗+1 = 𝑎𝑗 + 𝑏𝑗ℎ𝑗 + 𝑐𝑗ℎ𝑗2 + 𝑑𝑗ℎ𝑗

3

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𝑆𝑗′ 𝑥 = 𝑏𝑗 + 2𝑐𝑗 𝑥 − 𝑥𝑗 + 3𝑑𝑗 𝑥 − 𝑥𝑗

2

𝑆𝑗′ 𝑥𝑗 = 𝑏𝑗 𝑗 = 0,1, … , 𝑛 − 1

𝑏𝑛 = 𝑆′ 𝑥𝑛 4. 𝑆𝑗+1′ 𝑥𝑗+1 = 𝑆𝑗

′ 𝑥𝑗+1 𝑥 = 𝑥𝑗+1

𝑆𝑗+1′ 𝑥𝑗+1 = 𝑆𝑗

′ 𝑥𝑗+1 = 𝑏𝑗 + 2𝑐𝑗 𝑥𝑗+1 − 𝑥𝑗 + 3𝑑𝑗 𝑥𝑗+1 − 𝑥𝑗 2

2 𝑏𝑗+1 = 𝑏𝑗 + 2𝑐𝑗ℎ𝑗 + 3𝑑𝑗ℎ𝑗2


𝑆𝑗′ ′ 𝑥 = 2𝑐𝑗 + 6𝑑𝑗 𝑥 − 𝑥𝑗

𝑥 = 𝑥𝑗 𝑆𝑗′′ 𝑥𝑗 = 2𝑐𝑗

𝑐𝑛 =1

2𝑆′′ 𝑥𝑛 5.

3 𝑐𝑗+1 = 𝑐𝑗 + 3𝑑𝑗ℎ𝑗 𝑗 = 0,1, … , 𝑛 − 1

3 𝑑𝑗

𝑑𝑗 = 𝑐𝑗+1−𝑐𝑗

3ℎ𝑗

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𝑑𝑗 1 2

4 𝑎𝑗+1 = 𝑎𝑗 + 𝑏𝑗ℎ𝑗 +1

3ℎ𝑗

2 2𝑐𝑗 + 𝑐𝑗+1

5 𝑏𝑗+1 = 𝑏𝑗 + ℎ𝑗 𝑐𝑗 + 𝑐𝑗+1 𝑗 = 0,1, … , 𝑛 − 1

4 𝑏𝑗

6 𝑏𝑗 =1

ℎ𝑗 𝑎𝑗+1 − 𝑎𝑗 −

ℎ𝑗

3 2𝑐𝑗 + 𝑐𝑗+1

𝑏𝑗−1 =1

ℎ𝑗−1 𝑎𝑗 − 𝑎𝑗−1 −

ℎ𝑗−1

3 2𝑐𝑗−1 + 𝑐𝑗


𝑏𝑗 𝑏𝑗−1 5

7 ℎ𝑗−1𝑐𝑗−1 + 2 ℎ𝑗−1 + ℎ𝑗 𝑐𝑗 + ℎ𝑗 𝑐𝑗+1 =3


3

ℎ𝑗−1 𝑎𝑗 − 𝑎𝑗−1 𝑗 = 1, … , 𝑛 − 1

𝑐𝑗 𝑗=0

𝑛 ℎ𝑗 𝑗=0

𝑛−1 𝑎𝑗 𝑗=0

𝑛

𝑐𝑗 𝑗=0

𝑛 𝑏𝑗 𝑗=0

𝑛−1 6 𝑑𝑗 𝑗=0

𝑛−1

3 𝑆𝑗 𝑗=0

𝑛−1

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Cubic Spline: Example

𝑥 1 2 3 4 5

𝑓 𝑥 0 1 0 1 0

Spline 1,2 , 2,3 , 3,4 4,5


+ 𝑑𝑗 𝑥 − 𝑥𝑗 3 ( 𝑗 = 0,1,2,3)

𝑆 = 𝑆0 ∪ 𝑆1 ∪ 𝑆2 ∪ 𝑆3

𝑆0 𝑥 = 𝑎0 + 𝑏0 𝑥 − 1 + 𝑐0 𝑥 − 1 2 + 𝑑0 𝑥 − 1 3 1,2

𝑆1 𝑥 = 𝑎1 + 𝑏1 𝑥 − 2 + 𝑐1 𝑥 − 2 2 + 𝑑1 𝑥 − 2 3 2,3

𝑆2 𝑥 = 𝑎2 + 𝑏2 𝑥 − 3 + 𝑐2 𝑥 − 3 2 + 𝑑2 𝑥 − 3 3 3,4

𝑆3 𝑥 = 𝑎3 + 𝑏3 𝑥 − 4 + 𝑐3 𝑥 − 4 2 + 𝑑3 𝑥 − 4 3 4,5


2. 𝑆𝑗 𝑥𝑗 = 𝑎𝑗 = 𝑓 𝑥𝑗 𝑗 = 0,1,2,3,4

𝑎0 = 0, 𝑎1 = 1, 𝑎2 = 0, 𝑎3 = 1, 𝑎4 = 0

ℎ𝑗 = 𝑥𝑗+1 − 𝑥𝑗 𝑗 = 0,1,2,3 ℎ0 = ℎ1 = ℎ2 = ℎ3 = 1

, 𝑐0 =1

2𝑆′′ 𝑥0 = 0, 𝑐𝑛 =

1

2𝑆′′ 𝑥𝑛 = 0

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7 𝑖 = 1,2,3

ℎ0𝑐0 + 2 ℎ0 + ℎ1 𝑐1 + ℎ1𝑐2 =3

ℎ1 𝑎2 − 𝑎1 −

3

ℎ0 𝑎1 − 𝑎0

ℎ1𝑐1 + 2 ℎ1 + ℎ2 𝑐2 + ℎ2𝑐3 =3

ℎ2 𝑎3 − 𝑎2 −

3

ℎ1 𝑎2 − 𝑎1

ℎ2𝑐2 + 2 ℎ2 + ℎ3 𝑐3 + ℎ3𝑐4 =3

ℎ3 𝑎4 − 𝑎3 −

3

ℎ2 𝑎3 − 𝑎2

𝑎𝑗 , ℎ𝑗

4𝑐1 + 𝑐2 = 3 0 − 1 − 3 1 − 0 = −6 𝑐1 + 4𝑐2 + 𝑐3 = 3 1 − 0 − 3 0 − 1 = 6 𝑐2 + 4𝑐3 = 3 0 − 1 − 3 1 − 0 = −6

𝑐1 = −15

7, 𝑐2 =

18

7, 𝑐3 = −

15

7


6 𝑏𝑗 =1


ℎ𝑗

3 2𝑐𝑗 + 𝑐𝑗+1

𝑏0 =1

ℎ0 𝑎1 − 𝑎0 −

ℎ0

3 2𝑐0 + 𝑐1 = 1 − 0 −

1

3 0 −

15

7 =

12

7

𝑏1 =1

ℎ1 𝑎2 − 𝑎1 −

ℎ1

3 2𝑐1 + 𝑐2 = 0 − 1 −

1

3 −

30

7+

18

7 = −

3

7

𝑏2 =1

ℎ2 𝑎3 − 𝑎2 −

ℎ2

3 2𝑐2 + 𝑐3 = 1 − 0 −

1

3

36

7−

15

7 = 0

𝑏3 =1

ℎ3 𝑎4 − 𝑎3 −

ℎ3

3 2𝑐3 + 𝑐4 = 0 − 1 −

1

3 −

30

7+ 0 =

3

7

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3 𝑑𝑗 = 𝑐𝑗+1−𝑐𝑗

3ℎ𝑗

𝑑0 =1

3ℎ0 𝑐1 − 𝑐0 =

1

3 −

15

7− 0 = −

5

7

𝑑1 =1

3ℎ1 𝑐2 − 𝑐1 =

1

3

18

7+

15

7 =

11

7

𝑑2 =1

3ℎ2 𝑐3 − 𝑐2 =

1

3 −

15

7−

18

7 = −

11

7

𝑑3 =1

3ℎ3 𝑐4 − 𝑐3 =

1

3 0 +

15

7 =

5

7


Spline

𝑆0 𝑥 = 0 +12

7 𝑥 − 1 + 0 −

5

7 𝑥 − 1 3

𝑆1 𝑥 = 1 −3

7 𝑥 − 2 −

15

7 𝑥 − 2 2 +

11

7 𝑥 − 2 3

𝑆2 𝑥 = 0 + 0 +18

7 𝑥 − 3 2 −

11

7 𝑥 − 3 3

𝑆3 𝑥 = 1 +3

7 𝑥 − 4 −

15

7 𝑥 − 4 2 +

5

7 𝑥 − 4 3

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Spline 𝑓 𝑥 = 3𝑥𝑒𝑥 − 𝑒2𝑥 . 𝑥 = 1.03

𝑥 1.0 1.02 1.04 1.06

𝑓 𝑥 0.765789386 0.795366779 0.822688170 0.847522258

𝑓 ′ 𝑥 1.5315787 1.1754977

𝑛 = 3


+ 𝑑𝑗 𝑥 − 𝑥𝑗 3

𝑥0 = 1.0, 𝑥1 = 1.02, 𝑥2 = 1.04, 𝑥3 = 1.06

ℎ = 𝑥𝑗+1 − 𝑥𝑗 = 0.02 ( 𝑗 = 0,1,2) 𝑆 = 𝑆0 ∪ 𝑆1 ∪ 𝑆2


2. 𝑆𝑗 𝑥𝑗 = 𝑎𝑗 = 𝑓 𝑥𝑗 𝑗 = 0,1,2,3 𝑎0 = 𝑓 𝑥0 = 0.7657894 𝑎1 = 𝑓 𝑥1 = 0.7953668 𝑎2 = 𝑓 𝑥2 = 0.8226882 𝑎3 = 𝑓 𝑥3 = 0.8475223 𝑆′ 𝑥0 = 𝑓 ′ 𝑥0 𝑆 ′ 𝑥𝑛 = 𝑓 ′ 𝑥𝑛 𝑏0 = 1.5315787 𝑏3 = 1.1754977

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6 𝑏𝑗 =1


ℎ𝑗

3 2𝑐𝑗 + 𝑐𝑗+1

𝑗 = 0,1,2 𝑎0 , 𝑎1 , 𝑎2 , 𝑎3 𝑏0 , 𝑏3 (1) 1.5315787 = 𝑏0 = 1.47887 − 0.01333𝑐0 − 0.006667𝑐1 𝑏1 = 1.36607 − 0.01333𝑐1 − 0.006667𝑐2 𝑏2 = 1.241722 − 0.01333𝑐2 − 0.006667𝑐3 5 𝑏𝑗+1 = 𝑏𝑗 + ℎ𝑗 𝑐𝑗 + 𝑐𝑗+1 𝑗 = 0,1,2 𝑏0 , 𝑏3 (2) 𝑏1 = 1.5315787 + 0.02𝑐0 + 0.02𝑐1 𝑏2 = 𝑏1 + 0.02𝑐1 + 0.02𝑐2 1.1754977 = 𝑏3 = 𝑏2 + 0.02𝑐2 + 0.02𝑐3


(1) (2) 3 𝑐0 + 0.5𝑐1 = −4.2687857 𝑐0 + 0.33333𝑐1 − 0.33333𝑐2 = 8.275435 𝑐2 − 𝑐3 = 9.9305985 𝑐0 − 2𝑐2 − 𝑐3 = 17.80405 (3) 𝑐0 = 46.109654, 𝑐1 = 100.75688

𝑐2 = 12.745401, 𝑐3 = 2.814803

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(2)

𝑏0 = 1.5315787, 𝑏1 = 2.6245232

𝑏2 = 0.8642936, 𝑏3 = 1.1754977

3 𝑗 = 0,1,2

𝑑0 = −2447.7756 , 𝑑1 = 1891.7047, 𝑑2 = −165.50997


𝑆 = 𝑆0 ∪ 𝑆1 ∪ 𝑆2

𝑆0 𝑥 = 0.7657894 + 1.5315787 𝑥 − 1 + 46.109654 𝑥 − 1 2

− 2447.7756 𝑥 − 1 3

𝑆1 𝑥 = 0.7953668 + 2.6245232 𝑥 − 1.02 + 100.75688 𝑥 − 1.02 2 + 1891.7047 𝑥 − 1.02 3

𝑆2 𝑥 = 0.8226882 + 0.8642936 𝑥 − 1.04 + 12.745401 𝑥 − 1.04 2 − 165.50997 𝑥 − 1.04 3

𝑓 1.03 𝑆1 1.03

= 0.7953668 + 2.6245232 1.03 − 1.02 + 100.75688 1.03 − 1.02 2 + 1891.7047 1.03 − 1.02 3

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Least Square Method

ฟั ภ ฟั ็

n+1 ็ n ภ ฐ ( )

Least Square Method

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32

Least Square Method

1

𝑎𝑥𝑖 + 𝑏 𝑖 𝑦𝑖 𝐸 𝑎, 𝑏 𝑎, 𝑏

𝐸 𝑎, 𝑏

(Minimax Error Function) 𝐸∞ 𝑎, 𝑏 = max𝑖=1,2,…,10 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏 (Absolute derivation Error Function) 𝐸𝑙 𝑎, 𝑏 = 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏 10

𝑖=1

Least Square Method

ฟั (Total Square Error)

𝐸2 𝑎, 𝑏 = 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏 210

𝑖=1

0 =𝜕

𝜕𝑎 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏

210𝑖=1 = 2 𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏 −𝑥𝑖

10𝑖=1

0 =𝜕

𝜕𝑏 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏

210𝑖=1 = 2 𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏 −1 10

𝑖=1

𝑎 𝑥𝑖

2𝑚𝑖=1 + 𝑏 𝑥𝑖

𝑚𝑖=1 = 𝑥𝑖𝑦𝑖

𝑚𝑖=1

𝑎 𝑥𝑖𝑚𝑖=1 + 𝑏𝑚 = 𝑦𝑖

𝑚𝑖=1

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Least Square Method

𝑎 𝑏

𝑎 =𝑚 𝑥𝑖𝑦𝑖

𝑚𝑖=1 − 𝑥𝑖

𝑚𝑖=1 𝑦𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚

𝑖=1 − 𝑥𝑖𝑚𝑖=1

2

𝑏 = 𝑥𝑖

2𝑚𝑖=1 𝑦𝑖

𝑚𝑖=1 − 𝑥𝑖𝑦𝑖

𝑚𝑖=1 𝑥𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚


2

ฟั 𝑃1 𝑥 = 𝑦 = 𝑎𝑥 + 𝑏

Least Square Method: Example

𝑥𝑖 1 2 3 4 5 6 7 8 9 10

𝑦𝑖 1.3 3.5 4.2 5.0 7.0 8.8 10.1 12.5 13.0 15.6



𝑚𝑖=1 𝑦𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚


2

𝑏 = 𝑥𝑖



𝑚𝑖=1 𝑥𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚


2

𝑥𝑖𝑚𝑖=1 , 𝑦𝑖

𝑚𝑖=1 , 𝑥𝑖

2𝑚𝑖=1 , 𝑥𝑖𝑦𝑖

𝑚𝑖=1

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𝑥𝑖 𝑦𝑖 𝑥𝑖2 𝑥𝑖𝑦𝑖

1 1.3 1 1.3 2 3.5 4 7.0 3 4.2 9 12.6 4 5.0 16 20.0 5 7.0 25 35.5 6 8.8 36 52.8 7 10.1 49 70.7 8 12.5 64 100.0 9 13.0 81 117.0 10 15.6 100 156.0 55 81.0 385 572.4



𝑚𝑖=1 𝑦𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚


2

𝑏 = 𝑥𝑖



𝑚𝑖=1 𝑥𝑖

𝑚𝑖=1

𝑚 𝑥𝑖2𝑚


2

𝑎 =10 572.4 −55 81

10 385 − 55 2 = 1.538

𝑏 =385 81 −55 572.4

10 385 − 55 2 = −0.360

𝑃 𝑥𝑖 = 1.538𝑥𝑖 − 0.360

1.18 2.72 4.25 5.79 7.33 8.87 10.41 11.94 13.48 15.02

𝐸 = 𝑦𝑖 − 𝑃 𝑥𝑖 2

10

𝑖=1

≈ 2.34


0 1 2 3 4 5 6 7 8 9 10-2

0

2

4

6

8

10

12

14

16

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Least Square Method

𝑃𝑛 𝑥 = 𝑎𝑘𝑥𝑘𝑛

𝑘=0 𝑛 < 𝑚 − 1 𝑎0, … , 𝑎𝑛

𝐸 = 𝑦𝑖 − 𝑃𝑛 𝑥𝑖 2𝑛

𝑖=1

𝜕𝐸

𝜕𝑎𝑗= 0 𝑖 = 0,1,2, … , 𝑛 𝑛 + 1

𝑎𝑗

𝑎0 𝑥𝑖0

𝑚

𝑖=1

+ 𝑎1 𝑥𝑖1

𝑚

𝑖=1

+ 𝑎2 𝑥𝑖2

𝑚

𝑖=1

+ ⋯ + 𝑎𝑛 𝑥𝑖𝑛

𝑚

𝑖=1

= 𝑦𝑖𝑥𝑖0

𝑚

𝑖=1

𝑎0 𝑥𝑖1

𝑚

𝑖=1

+ 𝑎1 𝑥𝑖2

𝑚

𝑖=1

+ 𝑎2 𝑥𝑖3

𝑚

𝑖=1

+ ⋯ + 𝑎𝑛 𝑥𝑖𝑛+1

𝑚

𝑖=1

= 𝑦𝑖𝑥𝑖1

𝑚

𝑖=1

⋮

𝑎0 𝑥𝑖𝑛

𝑚

𝑖=1

+ 𝑎1 𝑥𝑖𝑛+1

𝑚

𝑖=1

+ 𝑎2 𝑥𝑖𝑛+2

𝑚

𝑖=1

+ ⋯ + 𝑎𝑛 𝑥𝑖2𝑛

𝑚

𝑖=1

= 𝑦𝑖𝑥𝑖𝑛

𝑚

𝑖=1


ฟั ็

𝑥𝑖 0 0.25 0.50 0.75 1.00

𝑦𝑖 1.0000 1.2840 1.6487 2.1170 2.7183

𝑛 = 2, 𝑚 = 5

5𝑎0 + 2.5𝑎1 + 1.875𝑎2 = 8.7680 2.5𝑎0 + 1.875𝑎1 + 1.5625𝑎2 = 5.4514 1.875𝑎0 + 1.5625𝑎1 + 1.3828𝑎2 = 4.4015

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𝑎0 = 1.0052, 𝑎1 = 0.8641, 𝑎2 = 0.8437

𝑃2 𝑥 = 1.0052 + 0.8641𝑥 + 0.84370.8437𝑥2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

chapter 3 interpolation and polynomial approximation11/8/2012 1 chapter 3 interpolation and...

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