INTERPOLATION

SARDAR PATEL COLLEGE
OF
ENGINEERING,BAKROL
HUMANITY & SCIENCE DEPARTMENT

TOPIC:-INTERPOLATION
SUBJECT:-N.S.M(2130606)
GUIDED by:- A.K.D SIR
SR. NO. NAME ENROLLMENT
NO.
1 PRAJAPATI TIRATH A. 151240106066
BRANCH : CIVIL-B

TYPES OF INTERPOLATIONS
• NEWTON’S FORWARD INTERPOLATION
• NEWTON’S BACKWARD INTERPOLATION
• GAUSS FORWARD INTERPOLATION
• GAUSS BACKWARD INTERPOLATION
• STIRLING’S INTERPOLATION
• LAGRANGE’S INTERPOLATION
• NEWTON DIVIDED DIFERENTIAL INTERPOLATION

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−3)
4!
𝚫4 𝑦0

NEWTON’S BACKWARD INTERPOLATION
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

LAGRANGE’S INTERPOLATION FORMULA
• LAGRANGE’S FORMULA IS APPLICABLE TO PROBLEMS WHERE THE
INDEPENDENCE VARIABLE OCCUR WITH EQUAL AND UNEQUAL
INTERVALS, BUT PREFERABLY THIS FORMULA IS APPLIED IN A SITUATION
WHERE THERE ARE UNEQUAL INTERVALS IN THE GIVEN INDEPENDENCE
SERIES. LET THE VALUES OF THE INDEPENDENCE VARIABLES (X) ARE
GIVEN AS A,B,C,D,.... ETC., AND THE CORRESPONDING VALUES OF THE
FUNCTION (DEPENDENT VARIABLE) AS F(A),F(B),F(C),F(D),...

GAUSS FORWARD INTER POLATION FORMULA:-
8
...)(
3
)1)(1(
)1(
!2
)1(
)0()0()(
2,1,0
32
1,01




 df
xxx
f
xx
fxfxf
GAUSS BACKWARD INTERPOLATION FORMULA:-
...
2
)2()1(
!3
)1(
)1(
22
)1()0(
)0()(
332
2
2






ffxx
f
xff
xfxf

• IS THE SIMPLEST FORM OF INTERPOLATION,
CONNECTING TWO DATA POINTS WITH A STRAIGHT
LINE.
)(
)()(
)()(
)()()()(
0
0
01
01
0
01
0
01
xx
xx
xfxf
xfxf
xx
xfxf
xx
xfxf









NEWTON’S DIVIDED DIFFERENTIAL INTERPOLATION

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
EXAMPLES

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

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

Find the unique Polynomial of degree 2 such that
P(1)=1, P(3)=27, P(4)=64
Use Lagrange’s method of Interpolation [2002-2003]
Sol: Here x = 1 3 4
P(x) = 1 27 64
Now by Lagrange’s formula

18
)34)(14(
)3)(1)(4(
)43)(13(
)4)(1)(3(
)41)(31(
)4)(3)(1(
)(









xxPxxPxxP
xP
)1)(3(
)3)(1(64
)1)(2(
)4)(1(27
)3)(2(
)4)(3(1
)(








xxxxxx
xP
3
)34(64
2
)45(27
6
127
)(
222







xxxxxx
xP

19
48198
)28811448(
6
1
)38451212810840581127(
6
1
2
2
222



xx
xx
xxxxxx

INTERPOLATION

More Related Content

What's hot (20)

Similar to INTERPOLATION (20)

More from tirath prajapati (20)

Recently uploaded (20)

INTERPOLATION