# Newton’s Backward Difference Interpolation Formla-Numerical Analysis-Lecture Handouts, Lecture notes for Numerical Analysis. Chennai Mathematical Institute

## Numerical Analysis

Description: This course contains solution of non linear equations and linear system of equations, approximation of eigen values, interpolation and polynomial approximation, numerical differentiation, integration, numerical solution of ordinary differential equations. This lecture includes: Newton, Backward, Difference, Interpolation, Formula, Function, Derivation, Equisapced, Shift, Operator, Binomial, Expansion
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Numerical Analysis –MTH603 VU
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( ) ()( ) ()(1 ) ()
pp p
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(1) (1)(2)
()1 2! 3!
(1)( 2)( 1) Error ( )
!
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23
(1) (1)(2)
( ) () () () ()
2! 3!
(1)( 2)( 1) ()Error
!
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(1) (1)(2)
2! 3!
(1)( 2)( 1) Error
!
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"
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(7
7.
.5
5)
).
.
docsity.com
Numerical Analysis –MTH603 VU
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(1)
2!
(1)( 2)
3!
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+
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I
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m,
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7.5 8.0 0.5
1
n
xx
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== =
23
169, 42, 6
nnn
yyy∇= = ∇ =
7.5
(0.5)(0.5)
512 ( 0.5)(169) (42)
2
( 0.5)(0.5)(1.5) (6)
6
512 84.5 5.25 0.375
421.875
y
=+− +
+
=− − −
=
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19
97
79
9
docsity.com
Numerical Analysis –MTH603 VU
S
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In this example,
1979 1982 1.5
2
p
==
and
2
34
5, 1,
2, 5
nn
nn
yy
yy
∇= ∇ =
∇==
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1979
( 1.5)( 0.5) ( 1.5)( 0.5)(0.5)
57 ( 1.5)5 (1) (2)
26
( 1.5)( 0.5)(0.5)(1.5) (5)
24
y
−−
=+− + +
−−
+
57 7.5 0.375 0.125 0.1172=+++
Therefore,
1979 50.1172y
=
Example
Consider the following table of values
x 1 1.1 1.2 1.3 1.4 1.5
F(x) 2 2.1 2.3 2.7 3.5 4.5
Use Newton’s Backward Difference Formula to estimate the value of f(1.45) .
Solution
x y=F(x) y 2y
3y
4y
5y
1 2
1.1 2.1 0.1
1.2 2.3 0.2 0.1
1.3 2.7 0.4 0.2 0.1
1.4 3.5 0.8 0.4 0.2 0.1
1.5 4.5 1 0.2 -0.2 -0.4 -0.5
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Numerical Analysis –MTH603 VU
1.45 1.5 0.5
0.1
n
xx
ph
== =
, n
y= 1 , 2
n
y= .2 , 3
n
y= - .2 , 4
n
y= -.4 ,
5
n
y= -.5
As we know that
23
45
(1) (1)(2)
2! 3!
( 1)( 2)( 3) ( 1)( 2)( 3)( 4)
4! 5!
xn n n n
nn
pp pp p
yypy y y
pp p p pp p p p
yy
+++
=++ ∇ +
++ + ++ ++
+∇+ ∇
()()
(
)()
() ()
() ()
0.5 ( 0.5 1) 0.5 ( 0.5 1)( 0.5 2)
4.5 0.5 (1) (0.2) 0.2
2! 3!
0.5 ( 0.5 1)( 0.5 2)( 0.5 3) 0.5 ( 0.5 1)( 0.5 2)( 0.5 3)( 0.5 4)
0.4 0.5
4! 5!
x
y− −+ − −+−+
=+− + +
− −+−+−+ − −+−+−+−+
+−+ −
= 4.5 0.5 0.025 + 0.0125 + 0.015625+ 0.068359
4.07148
−−
=
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