Post on 05-Mar-2021
11/8/2012
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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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Interpolation: Overview
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
Interpolation: Overview
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
Lagrange Polynomial: Example
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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Lagrange Polynomial: Example
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
Newton Divided Difference: Example
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 Divided Difference: Example
( ) 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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Forward Divided Difference
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
Forward Divided Difference
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
Backward Divided Difference
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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Backward Divided Difference
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
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 1.1 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)
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
Newton Divided Difference: Example
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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Newton Divided Difference: Example
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
Newton Divided Difference: Example
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 Interpolation
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 Interpolation
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 Interpolation
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 Interpolation
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
Hermite Interpolation
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
Hermite Interpolation: Example
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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Hermite Interpolation: Example
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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Cubic Spline Interpolation
การประมาณโดยพหุนามเป็นส่วนๆ ที่พบบ่อยท่ีสุดคือ การใช้พหุนามก าลัง
สามระหวา่งคู่ของจุด ที่เรียกว่า Cubic Spline
พหุนามก าลังสาม มีค่าคงตัว 4 ค่า โดยท่ัวไปแล้วอนุพันธ์ของ Cubic
Spline ไม่จ าเป็นต้องเท่ากับอนุพันธ์ของฟังก์ชันจริง แม้ที่จุดนิยาม
Cubic Spline Interpolation
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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Cubic Spline Interpolation
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
Cubic Spline Interpolation
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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Cubic Spline Interpolation
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
Cubic Spline Interpolation
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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Cubic Spline Interpolation
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
Cubic Spline Interpolation
𝑏𝑗 𝑏𝑗−1 5
7 ℎ𝑗−1𝑐𝑗−1 + 2 ℎ𝑗−1 + ℎ𝑗 𝑐𝑗 + ℎ𝑗 𝑐𝑗+1 =3
ℎ𝑗 𝑎𝑗+1 − 𝑎𝑗 −
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
𝑆𝑗 𝑥 = 𝑎𝑗 + 𝑏𝑗 𝑥 − 𝑥𝑗 + 𝑐𝑗 𝑥 − 𝑥𝑗 2
+ 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
Cubic Spline: Example
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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Cubic Spline: Example
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
Cubic Spline: Example
6 𝑏𝑗 =1
ℎ𝑗 𝑎𝑗+1 − 𝑎𝑗 −
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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Cubic Spline: Example
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
Cubic Spline: Example
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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Cubic Spline: Example
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
𝑆𝑗 𝑥 = 𝑎𝑗 + 𝑏𝑗 𝑥 − 𝑥𝑗 + 𝑐𝑗 𝑥 − 𝑥𝑗 2
+ 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
Cubic Spline: Example
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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Cubic Spline: Example
6 𝑏𝑗 =1
ℎ𝑗 𝑎𝑗+1 − 𝑎𝑗 −
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
Cubic Spline: Example
(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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Cubic Spline: Example
(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
Cubic Spline: Example
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
𝑖=1 − 𝑥𝑖𝑚𝑖=1
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
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
𝑖=1 − 𝑥𝑖𝑚𝑖=1
2
𝑥𝑖𝑚𝑖=1 , 𝑦𝑖
𝑚𝑖=1 , 𝑥𝑖
2𝑚𝑖=1 , 𝑥𝑖𝑦𝑖
𝑚𝑖=1
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Least Square Method: Example
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
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
𝑖=1 − 𝑥𝑖𝑚𝑖=1
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
Least Square Method: Example
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
Least Square Method: Example
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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Least Square Method: Example
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