Transcript Lecture-14-Interpolation

Computers in Civil Engineering
53:081 Spring 2003
Lecture #14
Interpolation
Interpolation: Overview

Objective: estimate intermediate values
between precise data points using simple
functions
y
Interpolation
y
Curve Fitting
x
Curve goes
through data
points

single
value
Solutions
– Newton Polynomials
– Lagrange Polynomials
– Spline Interpolation
x
Curve need not
go through data
points
multiple
values
Example
y
f ( x)  ?
y ( xi )  ?
High-precision data points
xi
x
Quad
Cities
Dresden
LaSalle
Braidwood
Quad-Cities Nuke Station Diffuser Curve
Jain's TMC-1
9
8
DT
7
Poly. (DT)
Poly. (DT)
Delta T (deg F)
6
5
4
3
y = -5.594406E-18x4 - 1.100233E-13x3 + 3.470629E-08x2 - 1.251748E-03x + 1.691667E+01
R2 = 0.9999
2
1
0
0
5000
10000
15000
20000
Miss. R. Discharge (cfs)
25000
30000
35000
Examples of Simple Polynomials
Fist-order (linear)
Second-order (quadratic)
Third-order (cubic)
Newton’s Divided-Difference
Interpolating Polynomials




General comments
Linear Interpolation
Quadratic Interpolation
General Form
Linear Interpolation Formula
By similar triangles:
f ( x1 )
f1 ( x )  f ( x 0 )
f1 ( x )
Rearrange:
x  x0

f1 ( x )  f ( x 0 ) 
f ( x0 )
x0
The notation:
x
x1
f 1 ( x ) means
f ( x1 )  f ( x 0 )
x1  x 0
f ( x1 )  f ( x 0 )
x1  x 0
( x  x0 )
 f ( x0 )  m ( x  x0 )
the first order interpolating polynomial
Example
Problem: Estimate ln(2) (the true value is 0.69)
Solution: We know that:
at x = 1 ln(x) =0
at x = e ln(x) =1 (e=2.718...)
Thus
,
f1 ( x )  f ( x 0 ) 
 f (1) 
0
f ( x1 )  f ( x 0 )
x1  x 0
f ( e )  f (1)
e 1
1 0
2 . 178  1
( x  x0 )
( 2  1)
( 2  1)  0 . 58
Quadratic Interpolation
General form: f 2 ( x )  a 0  a1 x  a 2 x
2
Equivalent form:
f 2 ( x )  b 0  b1 ( x  x 0 )  b 2 ( x  x 0 )( x  x1 )
(f2(x) means second-order interpolating polynomial)
To solve for
b 0 , b1 , and b 2 ,three
points are needed:
( x 0 , f ( x 0 )), ( x1 , f ( x1 ), ( x 2 , f ( x 2 ))
Quadratic Interpolation
f 2 ( x )  b 0  b1 ( x  x 0 )  b 2 ( x  x 0 )( x  x1 )
 (1)
Set x  x 0 in (1) to find b 0  f ( x 0 )
Substitute b 0 in (1) and evaluate at x  x1 to find:
b1 
f ( x1 )  f ( x 0 )
x1  x 0
Substitute b 0 , b1 in (1) and evaluate at x  x 2 to find:
Note: this looks
like a second
derivative…
f ( x 2 )  f ( x1 )
b2 
x 2  x1

f ( x1 )  f ( x 0 )
x2  x0
x1  x 0
Example
Problem Estimate ln(2) (the true value is 0.69)
Solution We know that:
at x = x0 = 1 ln(x) =0
at x = x1 = e ln(x) =1 (e=2.718...)
at x = x2 = e2 ln(x) = 2
b1 
b 0  f ( x 0 )  ln( 1)  0
f ( x 2 )  f ( x1 )
b2 
x 2  x1

ln( e )  ln( 1)
e 1

1 0
2 . 7183  1
f ( x1 )  f ( x 0 )
x2  x0
x1  x 0
  0 . 05
f 2 ( 2 )  b 0  b1 ( x  x 0 )  b 2 ( x  x 0 )( x  x1 )  0 . 62
 0 . 58
How to Generalize This?
It would get pretty tedious to do this for third,
fourth, fifth, sixth, etc order polynominal
We need a plan:
Newton’s Interpolating Polynomials
General form of Newton’s
Interpolating Polynomials
f n ( x )  b 0  b1 ( x  x 0 )    b n ( x  x 0 )( x  x1 )  ( x  x n 1 )
To solve for b 0 , b1 ,  b n , n+1 points are needed:
( x 0 , f ( x 0 )), ( x1 , f ( x1 ),  , ( x n , f ( x n ))
Solution b  f ( x )
0
0
b1  f [ x1 , x 0 ]
b 2  f [ x 2 , x1 , x 0 ]
What does this [ ]
notation mean?

b n  f [ x n , x n  1 ,  , x1 , x 0 ]
Finite Divided Differences
First finite divided difference:
f [ xi , x j ] 
f ( xi )  f ( x j )
xi  x j
Second finite divided difference:
f [ xi , x j , x k ] 
f [ xi , x j ]  f [ x j , x k ]
xi  xk
nth finite divided difference:
f [ x n , x n  1 ,  , x1 , x 0 ] 
f [ x n , x n  1 ,  , x1 ]  f [ x n  1 , x n  2 ,  , x 0 ]
xn  x0
f n ( x )  b 0  ( x  x 0 ) f [ x1 , x 0 ]  ( x  x 0 )( x  x1 ) f [ x 2 , x1 , x 0 ]
   ( x  x 0 )( x  x1 )  ( x  x n 1 ) f [ x n , x n 1 ,  , x 0 ]
Finite Divided Differences
f n ( x )  b 0  ( x  x 0 ) f [ x1 , x 0 ]  ( x  x 0 )( x  x1 ) f [ x 2 , x1 , x 0 ]
   ( x  x 0 )( x  x1 )  ( x  x n 1 ) f [ x n , x n 1 ,  , x 0 ]
Finite divided difference table, case n = 3:
i
xi
f ( xi )
first
second
0
x0
f ( x0 )
f [ x1 , x 0 ]
f [ x 2 , x1 , x 0 ]
1
x1
f ( x1 )
f [ x 2 , x1 ]
f [ x 3 , x 2 , x1 ]
2
x2
f ( x2 )
f [ x3 , x2 ]
3
x3
f ( x3 )
third
f [ x 3 , x 2 , x1 , x 0 ]
b 0 , b1 , b 2 , b 3
Divided Differences Pseudo Code
i
xi
f ( xi )
first
second
0
x0
f ( x0 )
f [ x1 , x 0 ]
f [ x 2 , x1 , x 0 ]
1
x1
f ( x1 )
f [ x 2 , x1 ]
f [ x 3 , x 2 , x1 ]
2
x2
f ( x2 )
f [ x3 , x2 ]
3
x3
f ( x3 )
third
f [ x 3 , x 2 , x1 , x 0 ]
b 0 , b1 , b 2 , b 3
do i=0,n-1
fdd(i,1)=f(i)
enddo
do j=2,n
do i=1,n-j+1
fdd(i,j)=(fdd(i+1,j-1)-fdd(i,j-1))/
&
(x(i+j-1)-x(i))
enddo
enddo
Example – ln(2) again
i
xi
f ( xi )
first
second
third
0
x0
f ( x0 )
f [ x1 , x 0 ]
f [ x 2 , x1 , x 0 ]
1
x1
f ( x1 )
f [ x 2 , x1 ]
f [ x 3 , x 2 , x1 ]
2
x2
f ( x2 )
f [ x3 , x2 ]
3
x3
f ( x3 )
f [ x 3 , x 2 , x1 , x 0 ]
b 0 , b1 , b 2 , b 3
f [ x1 , x 0 ] 
ln( 4 )  ln( 1)
4 1
 0 . 4621
i
xi
f ( xi )
first
second
third
0
1
0
0 . 4621
 0 . 0518
0 . 0079
1
4
1 . 3863
0 . 2027
 0 . 0204
2
6
1 . 7918
0 . 1823
3
5
1 . 6094
f [ x 3 , x 2 , x1 , x 0 ] 
 0 . 204  (  0 . 0518 )
5 1
 0 . 0079
f n ( x )  b0  ( x  x 0 ) f [ x 1 , x 0 ]
 ( x  x 0 )( x  x 1 ) f [ x 2 , x 1 , x 0 ]
 ( x  x 0 )( x  x 1 )( x  x 2 ) f [ x 3 , x 2 , x1 , x 0 ]
i
xi
f ( xi )
first
second
third
0
1
0
0 . 4621
 0 . 0518
0 . 0079
1
4
1 . 3863
0 . 2027
 0 . 0204
2
6
1 . 7918
0 . 1823
3
5
1 . 6094
f n ( x )  0 . 4621 ( x  1)  0 . 0518 ( x  1)( x  4 )  0 . 0079 ( x  1)( x  4 )( x  6 )
f n ( 2 )  0 . 4621 ( 2  1)  0 . 0518 ( 2  1)( 2  4 )  0 . 0079 ( 2  1)( 2  4 )( 2  6 )
 0 . 6289
Newton Interpolation Pseudo Code
See the textbook!
Features of Newton Divided-Differences to
get Interpolating Polynomial




Data need not be equally spaced
Arrangement of data does not have to be
ascending or descending, but it does influence
error of interpolation
Best case is when the base points are close to
the unknown value
Estimate of relative error:
R n  f n 1 ( x )  f n ( x )
Error estimate for nth-order polynomial is the difference
between the (n+1)th and nth-order prediction.
Relative Error As a Function of Order
Example 18.5 in text
Error
0.5
Determine ln(2) using the
following table
True Error
Estimated error
0
5
Order
Estimated error (reversed)
-0.5
x
f(x )= ln (x )
1
4
6
5
3
1 .5
2 .5
3 .5
0
1 .3 8 6 3
1 .7 9 1 8
1 .6 0 9 4
1 .0 9 8 6
0 .4 0 5 5
0 .9 1 6 3
1 .2 5 2 8
MATLAB function interp1 is very
useful for this
Midterm 2
Tuesday 15 April