This document discusses different interpolation methods: - Interpolation finds values of a function between known x-values where the function values are given. - Newton's forward and backward interpolation formulas are presented along with examples. - Newton's divided difference interpolation uses a formula involving differences to find interpolating polynomials. - Langrange's interpolation formula expresses the interpolating polynomial as a linear combination of basis polynomials defined in terms of the x-values. An example computing an interpolated value is shown.

1. S.N.PATEL INSTITUTE OF TECHNOLOGY
AND RESEARCH CENTRE
Computational Method for Mechanical Engineering
Prepared by :
Padhiyar Brijesh R.
(170490728009)

2. Interpolation
Interpolation means to find values of
function f(x) for an x between
different x-values x0, x1…, xn at
which the values of f(x) are given.

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

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

5. 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

6. 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

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

8. NEWTON’S BACKWARD 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

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

10. 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

11. 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 ) = 𝑦 𝑛 (300)
∴ 𝑝 = 𝑥−𝑥 𝑛
= 300−250
= 1
ℎ 50
∴ 𝑦n(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

12. NEWTON’S DIVIDED DIFFERENCE
INTERPOLATION

13. NEWTON’S DEVIDED
DIFFERENCE FORMULA
        0 0 0 1 0 1 0 1 2.....f x y x x x x x x x x x x x     

14. Example :
Find the polynomial of the lowest possible degree which assumes the values 3, 12,
15,-21 when x has the values 3, 2, 1,-1, respectively. Hence find f(0).

15. Solution
X Y Δ Δ2 Δ3
-1 -21
18
1 15 -7
-3 1
2 12 -3
-9
3 3

16. Using Newton’s divided difference formula, we get
f(x)=−21+(x − (−1)) * 18 +(x +1) (x − 1) * (−7) + (x + 1) (x − 1) (x − 2) * 1
= −21 + 18 (x + 1) + (x2 −1) (−7) + (x3 - 2x2 – x +2)
f(x) =x3 - 9x2 + 17x + 6
Hence f(0) = 6.

17. LANGRANGE’S INTERPOLATION
1 2
0
0 1 0 2 0
0 2
1
1 0 1 2 1
0 1
2
2 0 2 1 2
( )( )........( )
( )
( )( )........( )
( )( )........( )
( )( )........( )
( )( )........( )
+ .....
( )( )........( )
n
n
n
n
n
n
x x x x x x
F x y
x x x x x x
x x x x x x
y
x x x x x x
x x x x x x
y
x x x x x x
  

  
  

  
  
  
...

18. EXAMPLE
• Compute f(0.3) for the data
𝑥 0 1 3 4 7
f 1 3 49 129 813

19. 1 2
0
0 1 0 2 0
0 2
1
1 0 1 2 1
0 1
2
2 0 2 1 2
( )( )........( )
( )
( )( )........( )
( )( )........( )
( )( )........( )
( )( )........( )
+ .....
( )( )........( )
n
n
n
n
n
n
x x x x x x
F x y
x x x x x x
x x x x x x
y
x x x x x x
x x x x x x
y
x x x x x x
  

  
  

  
  
  
...
(0.3−1)(0.3−3)(0.3−4)(0.3−7) (0.3−0)(0.3−3)(0.3−4)(0.3−7) (0.3−0)(0.3−1)(0.3−4)(0.3−7)
(0.3−0)(0.3−1)(0.3−3)(0.3−7) (0.3−0)(0.3−1)(0.3−3)(0.3−4)
(−1)(−3)(−4)(−7) (1)(-2)(−3)(−6) (3)(2)(−1)(−4)
(4)(3)(1)(−3) (4)(3)(6)(7)
*1+= *49+*3+
*813*129+
= 1.831

20. Thank you…