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Na 10

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20 views15 pages

Na 10

Uploaded by

Ezgi Geyik
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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NUMERICAL ANALYSIS

Prof. Dr. Süheyla ÇEHRELİ

10
INTERPOLATİON
 Estimation of intermediate values between precise data
points.
 The most common method used for this purpose is
polynomial interpolation. Recall, that the general
formula for an nth order polinom is;
f(x)=a0+a1x+a2x2+…….+anxn
 For n+1 data points, there is one and only one
polynomial of order n that passes through all the points.
INTERPOLATİON

 Newton’s Forward Difference Interpolation Formula

 Newton’s Backward Difference Interpolation Formula

 Lagrange Interpolation Formula


FORWARD AND BACKWARD DİFFERENCE

r 1 r 1
 yi  
r
yi 1   yi
k 1 k 1
 yi  
k
yi   yi 1 ,
i  n, (n  1),..., k
NEWTON’S FORWARD DİFFERENCE INTERPOLATİON
FORMULA

 Let y = f (x) be a function which takes values f


(x0), f (x0+ h), f (x0+2h), …, corresponding to
various equi-spaced values of x with spacing h,
say x0, x0 + h, x0 + 2h, … .
Suppose, we wish to evaluate the function f (x)
for a value x0 + ph, where p is any real number,
then for any real number p, we have the
operator E such that
NEWTON’S FORWARD DİFFERENCE INTERPOLATİON FORMULA

E f ( x)  f ( x  ph).
p

f ( x0  ph)  E f ( x0 )  (1   ) f ( x0 )
p p

 p( p  1) 2 p( p  1)( p  2) 3 
 1  p       f ( x0 )
 2! 3!
NEWTON’S FORWARD DİFFERENCE INTERPOLATİON FORMULA

f ( x0  ph)  f ( x0 )  pf ( x0 )
p ( p  1) 2 p ( p  1)( p  2) 3
  f ( x0 )   f ( x0 )
2! 3!
p ( p  1) ( p  n  1) n
   f ( x0 )  Error
n!

This is known as Newton’s forward


difference formula for interpolation,
which gives the value of f (x0 + ph) in
terms of f (x0) and its leading differences.
NEWTON’S FORWARD DİFFERENCE INTERPOLATİON FORMULA
 This formula is also known as Newton-Gregory forward
difference interpolation formula. Here p=(x-x0)/h.
An alternate expression is

p( p  1) 2
yx  y0  py0   y0
2!
p( p  1)( p  2) 3
  y0 
3!
p( p  1)( p  n  1) n
  y0  Error
n!
NEWTON’S BACKWARD DİFFERENCE INTERPOLATİON
FORMULA

 Forinterpolating the value of the function y = f (x)


near the end of table of values, and to extrapolate
value of the function a short distance forward from yn,
Newton’s backward interpolation formula is used

p( p  1) 2
yx  yn  pyn   yn
2!
p( p  1)( p  2) 3
  yn 
3!
p( p  1)( p  2) ( p  n  1) n
  yn  Error
n!
 EXP 10-1: Calculate y at x=1,1 by the Newton
forward difference polynomial.

x y
1,0 0,7651977
1,3 0,6200860
1,6 0,4554022
1,9 0,2818186
2,2 0,1103623
 EXP 10-2: Calculate y at x=2 by the Newton
backward difference polynomial.
LAGRANGE INTERPOLATİON FORMULA
 The Lagrange interpolating polynomial is simply a
reformulation of the Newton’s polynomial that avoids
the computation of divided differences:

n
yp ( x)   Li ( x) yi
i0
n x  xj
Li ( x)  
j 0 xi  x j
j i
LAGRANGE INTERPOLATİON FORMULA

x  x1 x  x0
y p ( x)  y0  y1
x0  x1 x1  x0

y p ( x) 
 x  x1  x  x2 
y0 
 x  x0  x  x2 
y1
x0  x1 x0  x 2  x1  x0  x1  x 2 

 x  x0  x  x1 
y2
x2  x0  x2  x1 
 EXP 10-3: Find Lagrange’s interpolation polynomial fitting
the data points. Hence find y (4,5).

x y
1 -1
3 7
6 34

(3 data points 2nd order polynomial!!!)


 EXP 10-4: Find Lagrange’s interpolation polynomial fitting
the points y(1) = -3, y(3) = 0, y(4) = 30, y(6) = 132. Hence
find y(5).

The given data can be arranged as

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