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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© Copyright Virtual University of Pakistan 1
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( ) ()( ) ()(1 ) ()
pp p
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−− −
+= = =
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23
(1) (1)(2)
()1 2! 3!
(1)( 2)( 1) Error ( )
!
n
n
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pp pp p
fx ph p
pp p p n
f
x
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+=++ ∇+ +
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+∇+
"
"
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23
(1) (1)(2)
( ) () () () ()
2! 3!
(1)( 2)( 1) ()Error
!
nnn n n
n
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pp pp p
fx ph fx pfx fx fx
pp p p n fx
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+
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+= +∇ + + +
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+∇+
"
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(1) (1)(2)
2! 3!
(1)( 2)( 1) Error
!
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yypy y y
pp p p n y
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=++ ∇ + +
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"
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(
(7
7.
.5
5)
).
.
docsity.com
Numerical Analysis –MTH603 VU
© Copyright Virtual University of Pakistan 2
S
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2
3
(1)
2!
(1)( 2)
3!
xn n n
n
pp
yypy y
pp p y
+
=++ ∇
++
+∇
I
In
n
t
th
hi
is
s
p
pr
ro
ob
bl
le
em
m,
,
7.5 8.0 0.5
1
n
xx
ph
== =
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
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Numerical Analysis –MTH603 VU
© Copyright Virtual University of Pakistan 3
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
∇= ∇ =
∇==
N
Ne
ew
wt
to
on
n
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s
i
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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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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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