Polynomial Interpolation
– Part II –
Prof. Dr. Florian Rupp
German University of Technology in Oman (GUtech)
Introduction to Numerical Methods for ENG & CS
(Mathematics IV)
Spring Term 2016
Exercise Session
Reviewing the highlights from last time
(1/ 2)
Reviewing the highlights from last time
Page 174, exercise 4
Verify that the polynomials
p(x) = 5x3 − 27x2 + 45x − 21
q(x) = x4 − 5x3 + 8x2 − 5x + 4
interpolate the data
x
f (x) = y
1
2
2
1
3
6
4
47
and explain why this does not violate the uniqueness part of the existence and
uniqueness theorem on polynomial interpolation.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 3 / 48
Reviewing the highlights from last time
(1/ 2)
We have
x
f (x) = y
p(x)
q(x)
1
2
2
2
2
1
1
1
3
6
6
6
4
47
47
47
and thus
p(x) = 5x3 − 27x2 + 45x − 21
and
q(x) = x4 − 5x3 + 8x2 − 5x + 4
are both interpolating polynomials. Though, their degree differs and thus the
uniqueness part is not violated.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 4 / 48
Reviewing the highlights from last time
(2/ 2)
Reviewing the highlights from last time
Given the table
x
f (x) = y
■
■
0
7
2
11
3
28
4
63
Page 174, exercise 1
Use the Lagrange interpolation process to obtain a polynomial of least
degree that interpolates the above table.
Page 174, exercise 1 (reformulated)
Use the Newton interpolation process to obtain a polynomial of least degree
that interpolates the above table.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 5 / 48
Today, we will focus on further details of
polynomial interpolation
Today’s topics:
■
Calculating the coefficients for Newton’s interpolation ansatz using
divided differences
■
Interpolation with monomials and the Vandermode matrix
Corresponding textbook chapters: 4.1
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 6 / 48
Calculating the Newton
Coefficients Using Divided
Differences
The key idea in gaining the Newton
coefficients (1/ 4)
We now turn to the problem of determining the coefficients a0 , a1 , . . . , an of
the Newton interpolation polynomial
pn (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )(x − x1 ) + . . .
· · · + an (x − x0 )(x − x1 ) · · · (x − xn−1 ) ,
efficiently. Again, we start with a table of values of a function f :
x
f (x)
x0
f (x0 )
x1
f (x1 )
x2
f (x2 )
...
...
xn
f (xn )
The points x0 , x1 , x2 , . . . , xn are assumed to be distinct, but no further
assumption is made about their positions on the real line. We already know
that for each n = 0, 1, . . . there exists an unique polynomial pn such that
■
■
the degree of pn is at most n, and
pn (xi ) = f (xi ) for all i = 0, 1, . . . , n.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 8 / 48
The key idea in gaining the Newton
coefficients (2/ 4)
A crucial observation about the Newton interpolation polynomial
pn (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )(x − x1 ) + . . .
· · · + an (x − x0 )(x − x1 ) · · · (x − xn−1 ) ,
is that its coefficients a0 , a1 , . . . , an do not depend on n.
In other words, pn is obtained from pn−1 by adding one more term, without
altering the coefficients already present in pn−1 itself:
pn (x) = pn−1 (x) + an (x − x0 )(x − x1 ) · · · (x − xn−1 ) .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 9 / 48
The key idea in gaining the Newton
coefficients (3/ 4)
A systematic way of determining the unknown coefficients a0 , a1 , . . . , an from
the table
x
f (x)
x0
f (x0 )
x1
f (x1 )
x2
f (x2 )
...
...
xn
f (xn )
is to set x equal in turn to x0 , x1 , . . . , xn during the construction of the
Newton interpolation polynomial
pn (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )(x − x1 ) + . . .
· · · + an (x − x0 )(x − x1 ) · · · (x − xn−1 ) ,
i.e.
p0 (x0 ) = f (x0 ) = a0 ,
p1 (x1 ) = f (x1 ) = a0 + a1 (x1 − x0 ) ,
p2 (x2 ) = f (x2 ) = a0 + a1 (x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 ) , etc. . . .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 10 / 48
The key idea in gaining the Newton
coefficients (4/ 4)
The equations
p0 (x0 ) = f (x0 ) = a0 ,
p1 (x1 ) = f (x1 ) = a0 + a1 (x1 − x0 ) ,
p2 (x2 ) = f (x2 ) = a0 + a1 (x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 ) , etc.
can be solved for the ai ’s in turn, starting with a0 :
a0 = f [x0 ] := f (x0 )
f (x1 ) − f (x0 )
f (x1 ) − a0
=
a1 = f [x0 , x1 ] :=
x1 − x0
x1 − x0
f (x2 ) − a0 − a1 (x2 − x0 )
a2 = f [x0 , x1 , x2 ] :=
(x2 − x0 )(x2 − x1 )
=
Prof. Dr. Florian Rupp
f (x2 )−f (x1 )
x2 −x1
−
f (x1 )−f (x0 )
x1 −x0
x2 − x0
, etc.
GUtech 2016: Numerical Methods – 11 / 48
Some remarks on Newton’s divided
differences
The (bracket) notation
ak = f [x0 , x1 , . . . , xk ]
indicates that ak depends on the values of f at the nodes x0 , x1 , . . . , xn .
As the values ak are uniquely determined through the system of equations
shown on the last slide we can view the identity ak =: f [x0 , x1 , . . . , xk ] as
definition of the quantity f [x0 , x1 , . . . , xk ].
The quantity f [x0 , x1 , . . . , xk ] is called divided difference of order k for f.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 12 / 48
Example: Determining divided differences
Example
Determine the quantities f [x0 ], f [x0 , x1 ] and f [x0 , x1 , x2 ] based on the table
x
f (x)
1
3
−4
13
0
−23
The determining systems of equations reads as
p0 (x0 ) = f (x0 ) = 3 = a0 ,
p1 (x1 ) = f (x1 ) = 13 = a0 + a1 (x1 − x0 ) = a0 + a1 (−5) ,
p2 (x2 ) = f (x2 ) = −23 = a0 + a1 (x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 )
= a0 + a1 (−1) + a2 (−1)(4) .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 13 / 48
Example: Determining divided differences
Example [cont.]
From
3 = a0 ,
13 = a0 − 5a1 ,
−23 = a0 − a1 − 4a2 ,
we can read of the coefficients of the Newton interpolation polynomial as
a0 = 3 = f [1]
a1 = −2 = f [1, −4]
a2 = 7 = f [1, −4, 0] .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 14 / 48
Newton’s form of the interpolating
polynomial
With the notation of the divided differences, the Newton form of the
interpolating polynomial takes the form
pn (x) =
n
X
i=0
=
n
X
ai
i−1
Y
(x − xj )
j=0
f [x0 , x1 , . . . , xi ]
(x − xj ) ,
j=0
i=0
with the convention
i−1
Y
Q−1
j=0 (x
− xj ) := 1.
n in p is f [x , x , . . . , x ] because the term
Notice that the coefficient
of
x
n
0 1
n
Q
xn occurs only in n−1
j=0 (x − xj ).
Thus, if f is a polynomial of degree ≤ n − 1, then f [x0 , x1 , . . . , xn ] = 0
leading to a Newton interpolating polynomial of degree ≤ n − 1, too.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 15 / 48
Computing the divided differences (1/ 2)
The determining system of the coefficients (a.k.a. the divided differences)
f (x0 ) = a0 ,
f (x1 ) = a0 + a1 (x1 − x0 ) ,
f (x2 ) = a0 + a1 (x2 − x0 ) + a2 (x2 − x0 )(x2 − x1 ) , etc.
can be rewritten as
f (xk ) =
k
X
i=0
ai
i−1
Y
(xk − xj ) ,
for 0 ≤ k ≤ n .
j=0
It is evident, that the computation of the coefficients can be performed
recursively. E.g. a1 can be computed once a0 is known, a2 can be computed
once a0 and a1 are known and so on.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 16 / 48
Computing the divided differences (2/ 2)
We rewrite f (xk ) =
Pk
f (xk ) = ak
k−1
Y
i=0 ai
Qi−1
j=0 (xk
(xk − xj ) +
− xj ) as
k−1
X
ai
i=0
j=0
i−1
Y
(xk − xj ) ,
j=0
and get
f (xk ) −
ak =
k−1
X
i=0
ai
i−1
Y
(xk − xj )
j=0
k−1
Y
(xk − xj )
j=0
which allows a recursive computation of the coefficients ai = f [x0 , x1 , . . . , xi ].
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 17 / 48
Algorithm for computing the divided
differences
Algorithm (Divided Differences)
1.
2.
Set f [x0 ] = f (x0 ).
For k = 1, 2, . . . , n compute f [x0 , x1 , . . . , xk ] via
f (xk ) −
f [x0 , x1 , . . . , xk ] =
k−1
X
f [x0 , x1 , . . . , xi ]
i−1
Y
(xk − xj )
j=0
i=0
k−1
Y
(xk − xj )
j=0
The divided differences f [x0 ], f [x0 , x1 ], . . . , f [x0 , x1 , . . . , xn ] can thus be
computed at the cost of 13 n(3n + 1) additions, (n − 1)(n − 2) multiplications
and n divisions.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 18 / 48
The first four divided difference
Example
Use our algorithm to write out the divided difference formulas for f [x0 ],
f [x0 , x1 ], f [x0 , x1 , x2 ] and f [x0 , x1 , x2 , x3 ].
We have:
f [x0 ] = f (x0 )
f (x1 ) − f [x0 ]
f [x0 , x1 ] =
x1 − x0
f (x2 ) − f [x0 ] − f [x0 , x1 ](x2 − x0 )
f [x0 , x1 , x2 ] =
(x2 − x0 )(x2 − x1 )
f [x0 , x1 , x2 , x3 ]
=
f (x3 ) − f [x0 ] − f [x0 , x1 ](x3 − x0 ) − f [x0 , x1 , x2 ](x3 − x0 )(x3 − x1 )
(x3 − x0 )(x3 − x1 )(x3 − x2 )
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 19 / 48
Improving the Computation
of the Divided Differences
Improving the computation of the divided
differences
We have just derived an algorithm that computes the divided differences
f [x0 ], f [x0 , x1 ], . . . , f [x0 , x1 , . . . , xn ] at the cost of 13 n(3n + 1) additions,
(n − 1)(n − 2) multiplications and n divisions.
In this section, we will improve this result such that our new algorithm
requires only
■
■
n(n + 1) additions, 12 n(n + 1) divisions, and
the storage of just three statements (!).
The tools that allow for the derivation of this algorithm are
■
■
the recursive property of the divided differences, and
the invariance theorem for divided differences.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 21 / 48
The recursive property of the divided
differences
Theorem (Recursive Property of the Divided Differences)
The divided differences obey the formula
f [x0 , x1 , . . . , xk ] =
f [x1 , x2 , . . . , xk ] − f [x0 , x1 , . . . , xk−1 ]
xk − x0
For instance, this allows the computation of f [x0 , x1 , x2 ] via the scheme
f [x0 ]
i→
f [x0 , x1 ]
f [x1 ]
i→
i→
f [x0 , x1 , x2 ]
f [x1 , x2 ]
f [x2 ]
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 22 / 48
Proof of the recursive property (1/ 3)
Since f [x0 , x1 , . . . , xk ] was defined to be equal to the coefficient ak in the
Newton form of the interpolating polynomial
pk (x) =
k
X
i=0
ai
i−1
Y
(x − xj ) ,
j=0
we say that f [x0 , x1 , . . . , xk ] is the coefficient of xk in the polynomial pk of
degree ≤ k which interpolates the function f at the points x0 , x1 , . . . , xk .
Similarly, f [x1 , x2 , . . . , xk ] is the coefficient of xk−1 in the polynomial q of
degree ≤ k − 1 which interpolates the function f at the points x1 , x2 , . . . , xk .
And finally, we can interpret f [x0 , x1 , . . . , xk−1 ] as the coefficient of xk−1 in
the polynomial pk−1 of degree ≤ k − 1 which interpolates the function f at
the points x0 , x1 , . . . , xk−1 .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 23 / 48
Proof of the recursive property (2/ 3)
We have
■
■
■
f [x0 , x1 , . . . , xk ] is the coefficient of xk in the polynomial pk ,
f [x1 , x2 , . . . , xk ] is the coefficient of xk−1 in the polynomial q, and
f [x0 , x1 , . . . , xk−1 ] is the coefficient of xk−1 in the polynomial pk−1 .
These three polynomials pk , q and pk−1 are related as follows:
x − xk
(q(x) − pk−1 (x)) .
pk (x) = q(x) +
xk − x0
!
To establish this, we utilize the interpolation properties of pk , q and pk−1 .
First, we see that the right hand side is a polynomial of degree at most k.
Next, we evaluate the right hand side at xi , i = 1, . . . , k − 1. This gives
xi − xk
xi − xk
q(xi ) +
(q(xi ) − pk−1 (xi )) = f (xi ) +
(f (xi ) − f (xi ))
xk − x0
xk − x0
= f (xi ) .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 24 / 48
Proof of the recursive property (3/ 3)
Similarly, evaluating
q(x) +
x − xk
(q(x) − pk−1 (x))
xk − x0
at x0 and xk gives f (x0 ) and f (xk ), respectively.
Thus, both sides of the identity are interpolating the same k + 1 points and
are polynomials with a degree at most k. Thus, their difference is a
polynomial of degree at most k that vanishes at k + 1 points. Hence, the two
sides are equal.
Finally, as the coefficient of xk−1 in q is f [x1 , x2 , . . . , xk ] and that of xk−1 in
pk−1 is f [x0 , x1 , . . . , xk−1 ], we have that the coefficient of xk on the right
hand side of the equation is
f [x1 , x2 , . . . , xk ] − f [x0 , x1 , . . . , xk−1 ]
.
xk − x0
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 25 / 48
The invariance property of the divided
differences
Notice that f [x0 , x1 , . . . , xk ] is not changed if the nodes x0 , x1 , . . . , xk are
permuted. Thus, e.g., we have
f [x0 , x1 , x2 ] = f [x1 , x2 , x0 ] .
The reason is that f [x0 , x1 , x2 ] is the coefficient of x2 in the quadratic
polynomial p2 interpolating f at x0 , x1 and x2 , whereas f [x1 , x2 , x0 ] is the
coefficient of x2 in the quadratic polynomial q2 interpolating f at x1 , x2 and
x0 . Due to the uniqueness and existence theorem of polynomial interpolation
these two polynomials are of course the same.
This implies the following theorem:
Theorem (Recursive Property of the Divided Differences)
The divided difference f [x0 , x1 , . . . , xk ] is invariant under all permutations of
the arguments x0 , x1 , . . . , xk .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 26 / 48
Generalization of the divided difference
formula ...
As the variables x0 , x1 , . . . , xk and k are arbitrary, the recursive formula can
also be written as
f [xi , xi+1 , . . . , xj−1 , xj ] =
f [xi+1 , xi+2 , . . . , xj ] − f [xi , xi+1 , . . . , xj−1 ]
xj − xi
The first three divided differences are thus
f [xi ] = f (xi )
f [xi+1 ] − f [xi ]
f [xi , xi+1 ] =
xi+1 − xi
f [xi+1 , xi+2 ] − f [xi , xi+1 ]
f [xi , xi+1 , xi+2 ] =
.
xi+2 − xi
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 27 / 48
... and the divided difference table
As an illustration for n = 3, this leads to the following divided difference table
for an arbitrary function f :
x
x0
f []
f [x0 ]
f[ , ]
i→
x1
i→
x3
f [x0 , x1 , x2 ]
f [x1 , x2 ]
f [x2 ]
i→
i→
i→
f[ , , , ]
f [x0 , x1 ]
f [x1 ]
i→
x2
f[ , , ]
f [x0 , x1 , x2 , x3 ]
f [x1 , x2 , x3 ]
f [x2 , x3 ]
f [x2 ]
In this table, the coefficients along the top diagonal are the ones needed to
form the Newton form of the interpolating polynomial p3 .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 28 / 48
Example: Set-up of a divided difference
table
Example
Construct a divided difference table for the function f given in the following
data table, and write out the Newton form of the interpolating polynomial:
x
f (x)
1
3
3
2
13
4
0
3
2
5
3
We apply the just derived formulas for the divided differences:
f [xi ] = f (xi )
f [xi+1 ] − f [xi ]
f [xi , xi+1 ] =
xi+1 − xi
f [xi+1 , xi+2 ] − f [xi , xi+1 ]
, and so on
f [xi , xi+1 , xi+2 ] =
xi+2 − xi
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 29 / 48
Example: Set-up of a divided difference
table
Example [cont.]
Thus, for
x
f (x)
1
3
3
2
13
4
0
3
2
5
3
the first entry of the divided difference table for f reads as
f [1,
3
2]
=
f [ 23 ] − f [1]
=
3
2 −1
13
4
3
2
− 3)
1
= ,
2
−1
and, for instance, after completion of the columns for f [] and f [ , ] the first
entry f [ , , ] is given as
f [1, 32 , 0] =
Prof. Dr. Florian Rupp
1
1
−
f [ 23 , 0] − f [1, 32 ]
1
6
2
=
=
.
0−1
0−1
3
GUtech 2016: Numerical Methods – 30 / 48
Example: Set-up of a divided difference
table
Example [cont.]
Finally, the complete divided difference table for f reads as
x
1
f []
3
f[ , ]
i→
3
2
i→
3
i→
2
5
3
f[ , , , ]
1
2
13
4
i→
0
f[ , , ]
1
6
− 23
1
3
i→
i→
−2
− 35
Thus, we obtain
p3 (x) = 3 + 12 (x − 1) + 31 (x − 1)(x − 32 ) − 2(x − 1)(x − 32 )x .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 31 / 48
Classroom problem: Set-up of a divided
difference table
Classroom Problem
Construct a divided difference table for the function f given in the following
data table, and write out the Newton form of the interpolating polynomial:
x
f (x)
1
3
−4
13
0
−23
Use the just derived formulas for the divided differences:
f [xi ] = f (xi )
f [xi+1 ] − f [xi ]
f [xi , xi+1 ] =
xi+1 − xi
f [xi+1 , xi+2 ] − f [xi , xi+1 ]
f [xi , xi+1 , xi+2 ] =
.
xi+2 − xi
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 32 / 48
Classroom problem: Set-up of a divided
difference table
Classroom Problem [Solution]
x
1
f []
3
f[ , ]
i→
−4
−2
13
i→
i→
0
f[ , , ]
7
−9
−23
Thus we obtain
p2 (x) = 3 − 2(x − 1) + 7(x − 1)(x + 4) .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 33 / 48
Interpolation with
Monomials and the
Vandermode Matrix
Approximation with basis functions ...
Another view of interpolation is that for a given set of n + 1 data points
x
f (x) = y
x0
y0
x1
y1
x2
y2
...
...
xn
yn
we want to express an interpolating function f (x) as a linear combination of a
set of basis functions ϕ0 , ϕ1 , . . . , ϕn such that
f (x) ≈ c0 ϕ0 (x) + c1 ϕ1 (x) + · · · + cn ϕn (x) ,
where the coefficients c0 , c1 , . . . , cn are to be determined.
We want the function f to interpolate the data (xi , yi ), i.e., we want to have
linear equations in c0 , c1 , . . . , cn of the form
f (xi ) = c0 ϕ0 (xi ) + c1 ϕ1 (xi ) + · · · + cn ϕn (xi ) = yi ,
for each i = 0, 1, . . . , n.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 35 / 48
... leads naturally to linear systems
This is a system of linear equations
Ac = y ,
where the entries of the coefficient matrix A are given by ai,j = ϕj (xi ), which
is the value of the jth basis function evaluated at the ith data point. The
right-hand side vector y contains the known data values yi , and the
components of the vector c are the unknown coefficients ci .
Thus, in principle interpolation is nothing else than solving a linear system . . .
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 36 / 48
Monomials as basis functions for
polynomial interpolation (1/ 2) ...
The natural basis for the vector space Pn of all polynomials up to degree n
consists of monomials
ϕ0 (x) = 1 ,
ϕ1 (x) = x ,
ϕ2 (x) = x2 ,
ϕn (x) = xn .
...
Consequently, a interpolation polynomial pn has the form
pn (x) = c0 + c1 x + c2 x2 + · · · + cn xn ,
and the associates linear

1 x0 x20 . . .
 1 x1 x2 . . .
1

 1 x2 x2 . . .
2

 .. ..
.. . .
 . .
.
.
1 xn x2n
Prof. Dr. Florian Rupp
system Ac = y reads


n
c0
x0
 c1 
xn1 


n


x2   c2 
 =
..   .. 
.  . 
cn
. . . xnn
as







y0
y1
y2
..
.
yn




.


GUtech 2016: Numerical Methods – 37 / 48
Monomials as basis functions for
polynomial interpolation (2/ 2) ...
Sketch of the first few monomials on [−1, 1]:
φ0 = 1
1
2
0.8
φ2 = x
0.6
0.4
0.2
φ4 = x4
0
−0.2
3
−0.4
φ3 = x
−0.6
−0.8
φ1 = x
−1
−1
Prof. Dr. Florian Rupp
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
GUtech 2016: Numerical Methods – 38 / 48
... may not be the smartest choice
The coefficient matrix is the well-known Vandermonde matrix. We already
know that it is regular provided the points x0 , x1 , . . . , xn are distinct. So, in
principle, we can solve the polynomial interpolation problem.
Although, the Vandermonde matrix is invertible, we know that it is
ill-conditioned as n increases: For large n, the monomials are less
distinguishable from one another (see the figure on the previous slide). Thus,
the columns of the Vandermonde matrix become nearly dependent in this
case. Moreover, high-degree polynomials often oscillate widely and are highly
sensitive to small changes in the data.
Up to now, we have discussed three choices for the basis functions of Pn : the
Lagrange polynomials li (x), the Newton polynomials πi , and the monomials.
It turns out that there are better choices for the basis functions; namely, the
Chebysev polynomials have more desirable features.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 39 / 48
The Chebysev polynomials ...
The Chebysev polynomials play an important role in mathematics because
they have several special properties such as the recursive relation
T0 (x) = 1 ,
T1 (x) = x
Ti (x) = 2xTi−1 (x) − Ti−2 (x)
for i = 2, 3, 4, . . . . Thus, the first six Chebysev polynomials are
T0 (x) = 1 ,
T1 (x) = x ,
T4 (x) = 8x4 − 8x2 + 1 ,
T2 (x) = 2x2 − 1 ,
T3 (x) = 4x3 − 3x ,
T5 (x) = 16x5 − 20x3 + 5x .
The Chebysev polynomials are usually employed on the interval [−1, 1]. With
change of variable, they can be used on any interval. E.g., substituting x by
cos(t) allows for their use on the interval [0, 2π].
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 40 / 48
... their graphs ...
Sketch of the first few Chebysev polynomials on [−1, 1]:
T0(x)
1
T1(x)
0.8
0.6
T2(x)
0.4
0.2
0
−0.2
T (x)
3
−0.4
−0.6
T4(x)
−0.8
T5(x)
−1
−1
Prof. Dr. Florian Rupp
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
GUtech 2016: Numerical Methods – 41 / 48
... and the equal oscillation property
One of the important properties of the Chebysev polynomials is the equal
oscillation property. Notice in the previous figure that successive extreme
points of the Chebysev polynomials are equal in magnitude and alternate in
sign.
This property tends to distribute the error uniformly when Chebysev
polynomials are used as basis functions.
In polynomial interpolation for continuous functions, it is particularly
advantageous to select as the interpolation points the roots or the extreme
points of a Chebysev polynomial. This causes the maximum error over the
interval of the interpolation to be minimized. (We will illustrate this effect
next time.)
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 42 / 48
Summary & Outlook
Major concepts covered today (1/ 4):
Newton’s form
■
The Newton form of the interpolation polynomial is
pn (x) =
n
X
i=0
ai
i−1
Y
(x − xj ) .
j=0
with divided differences
ai
■
f [x1 , x2 , . . . , xi ] − f [x0 , x1 , . . . , xi−1 ]
= f [x0 , x1 , . . . , xi ] =
xi − x0
The Lagrange and the Newton form are just two different
representations of the unique polynomial p of degree n that
interpolates a table of n + 1 pairs of points (xi , f (xi )) for
i = 0, 1, . . . , n.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 44 / 48
Major concepts covered today (2/ 4):
polynomial interpolation example
■
We illustrate the Lagrange and Newton form with the aid of a small
table for n = 2:
x
f (x)
x0
f (x0 )
x1
f (x1 )
x2
f (x2 )
The Lagrange interpolation polynomial is
(x − x1 )(x − x2 )
(x − x0 )(x − x2 )
p2 (x) =
f (x0 ) +
f (x1 )
(x0 − x1 )(x0 − x2 )
(x1 − x0 )(x1 − x2 )
(x − x0 )(x − x1 )
+
f (x2 )
(x2 − x0 )(x2 − x1 )
Clearly, p2 (x0 ) = f (x0 ), p2 (x1 ) = f (x1 ) and p2 (x2 ) = f (x2 ).
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 45 / 48
Major concepts covered today (3/ 4):
polynomial interpolation example
■
Next, we form the divided difference table
x
x0
f []
f [x0 ]
f[ , ]
i→
x1
f [x0 , x1 ]
f [x1 ]
i→
i→
x2
f[ , , ]
f [x0 , x1 , x2 ]
f [x1 , x2 ]
f [x2 ]
Using the divided difference entries from the top diagonal, we have
p2 (x) = f [x0 ] + f [x0 , x1 ](x − x0 ) + f [x0 , x1 , x2 ](x − x0 )(x − x1 ) .
Again, it can be easily shown that p2 (x0 ) = f (x0 ), p2 (x1 ) = f (x1 )
and p2 (x2 ) = f (x2 ).
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 46 / 48
Preparation for the next lecture (1/ 2)
Please, prepare these short exercises for the next lecture:
1. Page 175, exercise 10 a
Use a divided differences table to construct Newton’s interpolation polynomial for the following data:
x
f (x)
0
7
2
11
3
28
4
63
2. Page 175, exercise 10 b
Without simplifying the Newton interpolation polynomial obtained above,
write this polynomial in nested form for easy evaluation.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 47 / 48
Preparation for the next lecture (2/ 2)
Please, prepare these short exercises for the next lecture:
3. Page 176, exercise 22
Find a polynomial of least degree that takes these values:
x
f (x) = y
1.73
0
1.82
0
2.61
7.8
5.22
0
8.26
0
Hint: Rearrange the table so that the non-zero value of f (x) is the last
entry, or think of some better way.
Prof. Dr. Florian Rupp
GUtech 2016: Numerical Methods – 48 / 48