Newton’s Forward & backward interpolation

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This document discusses Newton's forward and backward interpolation formulas. Newton's forward interpolation formula is used to find the value of tan(0.12) given values of tan(x) at other x values between 0.1 and 0.3. Newton's backward interpolation formula is then introduced and an example shows how to use it to determine the value of y(300) given y-values at x-values between 50 and 250.

NEWTON’S FORWARD
&
NEWTON’S BACKWARD
INTERPOLATION

NEWTON’S FORWARD INTERPOLATION
• Formula of Newton’s Forward Interpolation
𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝚫𝑦0 +
𝑝(𝑝 − 1)
2!
𝚫2 𝑦0 +
𝑝 𝑝 − 1 𝑝 − 2
3!
𝚫3 𝑦0 + ⋯
+
𝑝 𝑝 − 1 𝑝 − 2 … (𝑝 − 𝑛 + 1)
𝑛!
𝚫 𝑛 𝑦0
Here :- 𝑝 =
𝑥−𝑥0
ℎ

EXAMPLE
• Find The value Of 𝒕𝒂𝒏 0.12
𝑥 0.10 0.15 0.20 0.25 0.30
𝑦 = 𝑡𝑎𝑛 𝑥 0.1003 0.1511 0.2027 0.2553 0.3093

SOLUTION
X Y Δ Δ2 Δ3 Δ4
0.10 0.1003
0.0508
0.15 0.1511 0.0008
0.0516 0.0002
0.20 0.2027 0.0010 0.0002
0.0526 0.0004
0.25 0.2553 0.0014
0.0540
0.30 0.3093

Applying Newton’s Forward Difference Interpolation Formula.
𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝚫𝑦0 +
𝑝(𝑝 − 1)
2!
𝚫2 𝑦0 +
𝑝 𝑝 − 1 𝑝 − 2
3!
𝚫3 𝑦0 +
𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)
4!
𝚫4 𝑦0
Here 𝑦 𝑛 𝑥 = tan(0.12)
∴ 𝑝 =
𝑥−𝑥0
ℎ
=
0.12−0.10
0.05
=
0.02
0.05
= 0.4
∴ 𝑦 𝑛 𝑥 = 0.1003 + 0.4 0.0508 +
0.4 0.4−1
2
0.0008 +
0.4 0.4−1 0.4−2
6
0.0002 +
0.4 0.4−1 0.4−2 (0.4−3)
24
0.0002
𝑦 𝑛 𝑥 = 0.1205

NEWTON’S BACKEARD INTERPOLATION
• Formula of Newton’s Backward Interpolation
𝑦 𝑛 𝑥 = 𝑦 𝑛 +
𝑝𝛁𝑦 𝑛 +
𝑝(𝑝 + 1)
2!
𝛁2 𝑦 𝑛 + ⋯ +
𝑝 𝑝 − 1 … (𝑝 + 𝑛 − 1)
𝑛!
𝛁 𝑛 𝑦 𝑛
Here :- 𝑝 =
𝑥−𝑥𝑛
ℎ

EXAMPLE
Consider Following Tabular Values Determine y (300)
𝑥 50 100 150 200 250
𝑦 618 724 805 906 1032

SOLUTION
X Y 𝛁 𝛁2 𝛁3 𝛁4
50 618
106
100 724 -25
81 45
150 805 20 -40
101 5
200 906 5
126
250 1032

Applying Newton’s Backward Difference interpolation Formula.
𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝛁𝑦 𝑛 +
𝑝(𝑝 − 1)
2!
𝛁2 𝑦 𝑛 +
𝑝 𝑝 − 1 𝑝 − 2
3!
𝛁3 𝑦 𝑛 +
𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3)
4!
𝛁4 𝑦 𝑛
Here:- 𝑦 𝑛 𝑥 = 𝑦𝑛 300
∴ 𝑝 =
𝑥−𝑥𝑛
ℎ
=
300−250
50
= 1
∴ 𝑦 𝑛 𝑥 = 1032 + 126 +
1(1+1)
2!
25 +
1 1+1 1+2
3!
5 +
1 1+1 1+2 1+3
4!
(−40)
= 1032 + 126 + 25 + 5 − 4
𝑦 𝑛 300 = 1148

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Newton’s Forward & backward interpolation

1. NEWTON’S FORWARD & NEWTON’S BACKWARD INTERPOLATION

2. NEWTON’S FORWARD INTERPOLATION • Formula of Newton’s Forward Interpolation 𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝚫𝑦0 + 𝑝(𝑝 − 1) 2! 𝚫2 𝑦0 + 𝑝 𝑝 − 1 𝑝 − 2 3! 𝚫3 𝑦0 + ⋯ + 𝑝 𝑝 − 1 𝑝 − 2 … (𝑝 − 𝑛 + 1) 𝑛! 𝚫 𝑛 𝑦0 Here :- 𝑝 = 𝑥−𝑥0 ℎ

3. EXAMPLE • Find The value Of 𝒕𝒂𝒏 0.12 𝑥 0.10 0.15 0.20 0.25 0.30 𝑦 = 𝑡𝑎𝑛 𝑥 0.1003 0.1511 0.2027 0.2553 0.3093

4. SOLUTION X Y Δ Δ2 Δ3 Δ4 0.10 0.1003 0.0508 0.15 0.1511 0.0008 0.0516 0.0002 0.20 0.2027 0.0010 0.0002 0.0526 0.0004 0.25 0.2553 0.0014 0.0540 0.30 0.3093

5. Applying Newton’s Forward Difference Interpolation Formula. 𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝚫𝑦0 + 𝑝(𝑝 − 1) 2! 𝚫2 𝑦0 + 𝑝 𝑝 − 1 𝑝 − 2 3! 𝚫3 𝑦0 + 𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3) 4! 𝚫4 𝑦0 Here 𝑦 𝑛 𝑥 = tan(0.12) ∴ 𝑝 = 𝑥−𝑥0 ℎ = 0.12−0.10 0.05 = 0.02 0.05 = 0.4 ∴ 𝑦 𝑛 𝑥 = 0.1003 + 0.4 0.0508 + 0.4 0.4−1 2 0.0008 + 0.4 0.4−1 0.4−2 6 0.0002 + 0.4 0.4−1 0.4−2 (0.4−3) 24 0.0002 𝑦 𝑛 𝑥 = 0.1205

6. NEWTON’S BACKEARD INTERPOLATION • Formula of Newton’s Backward Interpolation 𝑦 𝑛 𝑥 = 𝑦 𝑛 + 𝑝𝛁𝑦 𝑛 + 𝑝(𝑝 + 1) 2! 𝛁2 𝑦 𝑛 + ⋯ + 𝑝 𝑝 − 1 … (𝑝 + 𝑛 − 1) 𝑛! 𝛁 𝑛 𝑦 𝑛 Here :- 𝑝 = 𝑥−𝑥𝑛 ℎ

7. EXAMPLE Consider Following Tabular Values Determine y (300) 𝑥 50 100 150 200 250 𝑦 618 724 805 906 1032

8. SOLUTION X Y 𝛁 𝛁2 𝛁3 𝛁4 50 618 106 100 724 -25 81 45 150 805 20 -40 101 5 200 906 5 126 250 1032

9. Applying Newton’s Backward Difference interpolation Formula. 𝑦 𝑛 𝑥 = 𝑦0 + 𝑝𝛁𝑦 𝑛 + 𝑝(𝑝 − 1) 2! 𝛁2 𝑦 𝑛 + 𝑝 𝑝 − 1 𝑝 − 2 3! 𝛁3 𝑦 𝑛 + 𝑝 𝑝 − 1 𝑝 − 2 (𝑝 − 3) 4! 𝛁4 𝑦 𝑛 Here:- 𝑦 𝑛 𝑥 = 𝑦𝑛 300 ∴ 𝑝 = 𝑥−𝑥𝑛 ℎ = 300−250 50 = 1 ∴ 𝑦 𝑛 𝑥 = 1032 + 126 + 1(1+1) 2! 25 + 1 1+1 1+2 3! 5 + 1 1+1 1+2 1+3 4! (−40) = 1032 + 126 + 25 + 5 − 4 𝑦 𝑛 300 = 1148

Newton’s Forward & backward interpolation

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