Chapter 3 Interpolation and Polynomial Approximation

Chapter 3
Interpolation and Polynomial
Approximation
Let x 0 , x 1 ,..., x n be distinct real and y 0 , y 1 ,..., y n be associated function values. We will
find a polynomial p(x) that interpolates the given data:
p(x i ) = y i ,
i = 0, 1,...,n.
3.1 Interpolation and Lagrange Polynomial
Theorem 3.1 Given n + 1 distinct points x 0 , x 1 ,..., x n and n + 1 ordinates y 0 , y 1 ,...,
y n , there is a polynomial p(x) of degree ≤ n that interpolates y i at x i , i = 0, 1,...,n. This
polynomial p(x) is unique among the set of all polynomials of degree at most n.
proof : (1) Existence: Denote the Lagrange interpolation polynomial
( )
n∑
n∏ x − xi
p(x) = y j l j (x), l j (x) =
.
x j − x i
j=0
i=0,i≠j
The polynomial p(x) with degree ≤ n satisfies p(x i ) = y i , i = 0, 1,...,n, since l j (x i ) = 0
if i ≠ j and l j (x i ) = 1 if i = j.
(2) Uniqueness: Suppose q(x) is another polynomial of degree ≤ n satisfying
q(x i ) = y i ,
i = 0, 1,...,n.
Define r(x) = p(x) − q(x), then degree r(x) ≤ n and
r(x i ) = p(x i ) − q(x i ) = 0,
i = 0, 1,...,n.
1

Since r(x) has n + 1 zeros, we must have r(x) ≡ 0. Thus p(x) = q(x). This completes
the proof of theorem.
Theorem 3.2 Let x 0 , x 1 ,..., x n be distinct number in [a,b] and f(x) ∈ C n+1 [a,b]. Then
there exists x ∈ [a,b] and ξ(x) ∈ (a,b) with
f(x) −
n∑
j=0
f(x j )l j (x) = (x − x 0)...(x − x n )
f (n+1) (ξ(x)).
(n + 1)!
proof : Assume x does not equal any node point. Since if x is any node point it is
trivially true. Define
n∑
E(x) = f(x) − p n (x), p n (x) = f(x j )l j (x)
j=0
G(t) = E(t) − Φ(t)
Φ(x) E(x), Φ(x) = (x − x 0)....(x − x n )
The function G(t) ∈ C n+1 [a,b]. Also
G(x i ) = E(x i ) − Φ(x i)
E(x) = 0,
Φ(x)
G(x) = E(x) − E(x) = 0
i = 0, 1,...,n
Thus G has n + 2 distinct zeros in [a,b]. Using the mean value theorem, G (j) (t) has
n + 2 − j zeros, for j = 0, 1,...,n + 1. Let ξ be a zero of G (n+1) (t).
G (n+1) (t) = f (n+1) (t) −
(n + 1)!
Φ(x) E(x)
G (n+1) (ξ) = f (n+1) (ξ) −
(n + 1)!
Φ(x) E(x) = 0
E(x) =
This completes the proof of theorem.
Φ(x)
(n + 1)! f(n+1) (ξ)
2

3.2 Divided Difference
Newton’s divided difference formula for the interpolation polynomial:
p 0 (x) = f(x 0 )
p j (x) = p j−1 (x) + (x − x 0 )...(x − x j−1 )f[x 0 ,x 1 ,...,x j ], j = 1, 2,...,n
The coefficient a n = f[x 0 ,x 1 ,...,x n ] is called the n-th-order Newton divided difference of
f(x). Define
f[x 0 ,x 1 ,...,x n ] = f[x 1,...,x n ] − f[x 0 ,x 1 ,...,x n−1 ]
x n − x 0
Lemma 3.1 The n-th-order Newton divided difference have the following theory.
1. Let Φ n (x) = (x − x 0 )...(x − x n ). Then the coefficient a n can be formed as:
f[x 0 ,x 1 ,...,x n ] =
2. For any permutation (i 0 ,...,i n ) of (0, 1,...,n)
n∑
j=0
f(x j )
Φ ′ n(x j )
f[x 0 ,x 1 ,...,x n ] = f[x i0 ,x i1 ,...,x in ]
3. Let f(x) be enough smooth.
f[x 0 ,x 1 ,...,x m ] = f(m) (ξ)
, ξ ∈ ϑ{x 0 ,...,x m } (3.2.1)
m!
Moreover, if x 0 = x 1 = ... = x n then f[x 0 ,x 1 ,...,x n ] = f (n) (x 0 )/n!.
4. Suppose f(x) be enough smooth.
proof :
d
dx f[x 0,x 1 ,...,x n ,x] = f[x 0 ,x 1 ,...,x n ,x,x]
p n (x) = p n−1 (x) + (x − x 0 )...(x − x n−1 )f[x 0 ,x 1 ,...,x n ]
If x is not a node point, we have Lagrange interpolation polynomial
p n (x) =
n∑
j=0
Φ n (x)
(x − x j )Φ ′ n(x j ) · f(x j) (3.2.2)
3

By looking at each n-th-degree term in (3.2.2), we obtain
This completes the proof of (1).
f[x 0 ,x 1 ,...,x n ] =
n∑
j=0
f(x j )
Φ ′ n(x j )
n∑
j=0
f(x j )
n
Φ ′ n(x j ) = ∑ f(x ij )
Φ ′ n(x ij )
Then f[x 0 ,x 1 ,...,x n ] = f[x i0 ,x i1 ,...,x in ]. This completes the proof of (2).
j=0
p n+1 (x) = p n (x) + (x − x 0 )...(x − x n )f[x 0 ,...,x n ,t], p n+1 (t) = f(t)
By Theorem 3.2 we have
Let t = x n+1 , n = m − 1 then
This completes the proof of (3).
f(t) − p n (t) = (t − x 0 )...(t − x n )f[x 0 ,...,x n ,t]
f[x 0 ,...,x n ,t] = f(n+1) (ξ)
(n + 1)!
f[x 0 ,...,x m ] = f(m) (ξ)
m!
d
dx f[x f[x 0 ,x 1 ,...,x n ,x + h] − f[x 0 ,x 1 ,...,x n ,x]
0,x 1 ,...,x n ,x] = lim
h→0 h
f[x 0 ,x 1 ,...,x n ,x + h] − f[x,x 0 ,x 1 ,...,x n ]
= lim
h→0 (x + h) − x
= f[x,x 0 ,x 1 ,...,x n ,x] = f[x 0 ,x 1 ,...,x n ,x,x]
This completes the proof of (4).
= lim
h→0
f[x,x 0 ,x 1 ,...,x n ,x + h]
Theorem 3.3 (Hermite-Gennochi) Let x 0 , x 1 ,..., x n be distinct and f(x) ∈ C n on
ϑ{x 0 ,...,x n }. Then
∫ ∫
f[x 0 ,...,x n ] = ... f (n) (t 0 x 0 + ... + t n x n )dt 1 ...dt n
τ n
4

where
τ n =
{
(t 1 ,...,t n )| ∀ t i ≥ 0,
}
n∑
n∑
t i ≤ 1 , t 0 = 1 − t i .
i=1
i=1
proof : We show that it is true for n = 1 and n = 2.
(1) n = 1. Then τ 1 = [0, 1] and t 0 = 1 − t 1
=
∫ 1
0
f ′ (t 0 x 0 + t 1 x 1 )dt 1 =
∫ 1
1
f(x 0 + t 1 (x 1 − x 0 ))
∣
x 1 − x 0
0
f ′ (x 0 + t 1 (x 1 − x 0 ))dt 1
t 1 =1
t 1 =0
= f(x 1) − f(x 0 )
x 1 − x 0
= f[x 0 ,x 1 ]
(2) n=2. Then τ 2 is the triangle with vertices (0, 0), (0, 1) and (1, 0).
=
=
=
=
∫ ∫
τ 2
f ′′ (t 0 x 0 + t 1 x 1 + t 2 x 2 )dt 1 dt 2
∫ 1 ∫ 1−t1
0
∫ 1
0
f ′′ (x 0 + t 1 (x 1 − x 0 ) + t 2 (x 2 − x 0 ))dt 2 dt 1
0
1
[f ′ t 2 =1−t 1
(x 0 + t 1 (x 1 − x 0 ) + t 2 (x 2 − x 0 ))]
∣ dt 1
x 2 − x 0 t 2 =0
∫
1
1
]
f ′ (x 2 + t 1 (x 1 − x 2 ))dt 1 − f ′ (x 0 + t 1 (x 1 − x 0 ))dt 1
[∫ 1
x 2 − x 0 0
1
x 2 − x 0
{f[x 1 ,x 2 ] − f[x 1 ,x 0 ]} = f[x 0 ,x 1 ,x 2 ]
This completes the proof of theorem.
• Forward difference and backward difference
Let the note points x i = x 0 +ih, i=0,1,2,3,.... Then define the forward difference operator
△:
△f(x i ) = f(x i+1 ) − f(x i )
0
△ r+1 f(z) = △ r f(z + h) − △ r f(z),
△ 0 f(z) = f(z).
The term △ r f(z) is the rth-order forward difference of f at z.
Lemma 3.2
f[x 0 ,x 1 ,...,x k ] = 1
k!h k △k f 0 , k ≥ 0.
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proof : For k = 0, it is trivially true. For k = 1,
f[x 0 ,x 1 ] = f(x 1) − f(x 0 )
x 1 − x 0
= 1 h △ f 0.
Assume the result is true for all k ≤ r. Then for k = r + 1
f[x 0 ,x 1 ,...,x r+1 ] = f[x 1,...,x r+1 ] − f[x 0 ,x 1 ,...,x r ]
x r+1 − x 0
[
1 1
=
(r + 1)h r!h r △r f 1 − 1 ]
r!h r △r f 0
1
=
(r + 1)!h r+1 △r+1 f 0
This completes the proof of lemma.
Lemma 3.3
△ r f(x i ) = h r f (r) (ξ i ), x i ≤ ξ i ≤ x i+r .
proof : By (3.2.1) and Lemma 3.2 we obtain
△ r f(x i ) = r!h r f[x i ,x i+1 ,...,x i+r ] = r!h rf(r) (ξ i )
r!
This completes the proof of lemma.
= h r f (r) (ξ i ).
From Lemma 3.2 we have the Newton forward difference form of the interpolating polynomial
n∑
p n (x) = f 0 + (x − x 0 )...(x − x k−1 )f[x 0 ,...,x k ]
k=1
n∑
= f 0 + (x − x 0 )...(x − x k−1 ) 1
k=1
k!h k △k f 0
( )
n∑ µ
=
△ k f
k 0 , µ = x − x 0
.
k=0
h
Define the binomial coefficients,
( )
µ µ(µ − 1)...(µ − k + 1)
= , k > 0,
k
k!
Define the backward difference by
▽f(z) = f(z) − f(z − h)
(
µ
0
)
= 1.
▽ r+1 f(z) = ▽ r f(z) − ▽ r f(z − h), r ≥ 1.
6

And we obtain the Newton backward difference form of the interpolating polynomial
with the notes x 0 , x −1 , ..., x −n , and x −j = x 0 − jh
p n (x) =
( )
n∑ µ
j
j=0
▽ j f 0 , µ = x − x 0
.
h
3.3 Hermite Interpolation
The Hermite interpolation polynomial p(x) of f(x) with the distinct notes x 1 , x 2 ,..., x n
satisfies
p(x i ) = f(x i ) = y i , p ′ (x i ) = f ′ (x i ) = y ′ i, i = 1, 2,...,n.
Then the polynomial p(x) is existence and uniqueness with degree ≤ 2n − 1.
• The Newton divided difference form
p 2n−1 (x) = f(x 1 ) + (x − x 1 )f[x 1 ,x 1 ] + (x − x 1 ) 2 f[x 1 ,x 1 ,x 2 ] + ...
+ (x − x 1 ) 2 (x − x 2 ) 2 ...(x − x n )f[x 1 ,x 1 ,...,x n ,x n ]
• Example
f(x) − p 2n−1 (x) = (x − x 1 ) 2 (x − x 2 ) 2 ...(x − x n ) 2 f[x 1 ,x 1 ,...,x n ,x n ,x]
1. Find p ∈ P 3 such that p(x 0 ) = f 0 , p ′ (x 0 ) = f ′ 0, p ′′ (x 0 ) = f ′′
0 , p ′′′ (x 0 ) = f ′′′
0 .
x f
x 0 f 0
x 0 f 0 f 0
′
x 0 f 0 f 0
′
x 0 f 0 f 0
′
1
f ′′
2 0
1
f ′′
2 0
1
f ′′′
6 0
Then the diagonal entries are the coefficients of the power series:
p(x) = f 0 + f ′ 0(x − x 0 ) + 1 2 f ′′
0 (x − x 0 ) 2 + 1 6 f ′′′
0 (x − x 0 ) 3
which is the Taylor polynomial of degree 3.
2. Find p ∈ P 3 such that p(x 0 ) = f 0 , p(x 1 ) = f 1 , p ′ (x 1 ) = f ′ 1, p(x 2 ) = f 2 .
7

x f
x 0 f 0
x 1 f 1 f[x 0 ,x 1 ]
x 1 f 1 f 1 ′ f[x 0 ,x 1 ,x 1 ]
x 2 f 2 f[x 1 ,x 2 ] f[x 1 ,x 1 ,x 2 ] f[x 0 ,x 1 ,x 1 ,x 2 ]
If denote the diagonal entries as before by a 0 , a 1 , a 2 , a 3 , Newton’s formula takes
the form:
p(x) = a 0 + a 1 (x − x 0 ) + a 2 (x − x 0 )(x − x 1 ) + a 3 (x − x 0 )(x − x 1 ) 2
Then the error formula becomes
f(x) − p(x) = (x − x 0 )(x − x 1 ) 2 (x − x 2 )f[x 0 ,x 1 ,x 1 ,x 2 ,x]
= (x − x 0 )(x − x 1 ) 2 (x − x 2 ) f(4) (ξ)
, ξ ∈ [x 0 ,x 2 ]
4!
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3.4 Cubic spline Interpolation
Consider a function f defined on [a,b] and a = x 0 < x 1 < ... < x n = b, we say S(x) is a
cubic spline interpolant if it satisfies the following conditions:
(1) S(x) is a cubic polynomial denoted S j (x) on [x j ,x j+1 ], j = 0, 1,...,n − 1.
(2) S(x j ) = f(x j ), j = 0, 1,...,n.
(3) S j+1 (x j+1 ) = S j (x j+1 ), j = 0, 1,...,n − 2.
(4) S ′ j+1(x j+1 ) = S ′ j(x j+1 ), j = 0, 1,...,n − 2.
(5) S ′′
j+1(x j+1 ) = S ′′
j (x j+1 ), j = 0, 1,...,n − 2.
(6) One of the following boundary conditions is satisfied
(i) Natural boundary conditions: S ′′ (x 0 ) = S ′′ (x n ) = 0.
(ii) Clamped boundary conditions: S ′ (x 0 ) = f ′ (x 0 ) and S ′ (x n ) = f ′ (x n ).
Note: The cubic spline interpolation S(x) has 4n undetermined coefficients and satisfies
(n+1)+3(n−1) = 4n−2 condition equations then it must add two boundary conditions,
natural boundary conditions or clamped boundary conditions.
To construct the cubic spline interpolant for a given function f(x) satisfying the
conditions (1)−(6). First, let x ∈ [x j ,x j+1 ], x j+1 − x j = h j and
S j (x) = a j + b j (x − x j ) + c j (x − x j ) 2 + d j (x − x j ) 3 , j = 0, 1,...,n − 1.
From the condition (2), we obtain
a j = f(x j ), j = 0, 1,...,n − 1.
Defined a n = f(x n ). From the condition (3), we obtain the equations
a j+1 = a j + b j h j + c j h 2 j + d j h 3 j, j = 0, 1,...,n − 1. (3.4.1)
Defined b n = S ′ (x n ) and observed that S ′ j(x) = b j + 2c j (x − x j ) + 3d j (x − x j ) 2 . From the
condition (4), we obtain the equations
b j+1 = b j + 2c j h j + 3d j h 2 j, j = 0, 1,...,n − 1. (3.4.2)
Defined c n = S ′′ (x n )/2 and observed that S ′′
j (x) = 2c j + 6d j (x − x j ). From the condition
(5), we obtain the equations
c j+1 = c j + 3d j h j , j = 0, 1,...,n − 1.
9

Then
d j = 1
3h j
(c j+1 − c j ), j = 0, 1,...,n − 1. (3.4.3)
Substituting the value, d j , into (3.4.1) and (3.4.2) gives the new equations
a j+1 = a j + b j h j + h2 j
3 (2c j + c j+1 ), j = 0, 1,...,n − 1, (3.4.4)
From the equation (3.4.4), we obtain
b j+1 = b j + h j (c j + c j+1 ), j = 0, 1,...,n − 1. (3.4.5)
b j = 1 h j
(a j+1 − a j ) − h j
3 (2c j + c j+1 ), j = 0, 1,...,n − 1. (3.4.6)
Substituting these values into the equation (3.4.5), when the index is reduced by 1, gives
the linear system of equations
⇒
b j = b j−1 + h j−1 (c j−1 + c j )
1 (a j+1 − a j ) − h j
h j 3 (2c j + c j+1 )
= 1
h j−1
(a j − a j−1 ) − h j−1
3 (2c j−1 + c j ) + h j−1 (c j−1 + c j )
⇒ h j−1 c j−1 + 2(h j−1 + h j )c j + h j c j+1 = 3 h j
(a j+1 − a j ) − 3
h j−1
(a j − a j−1 ) (3.4.7)
for each j = 1,...,n − 1. The system (3.4.7) involves, as unknowns, only {c j } n j=0. Note
that once the values of {c j } n j=0 are known, it is a simple matter to find the remainder
values {b j } n−1
j=0 from Eq. (3.4.6) and {d j } n−1
j=0 from Eq. (3.4.3) and to construct the cubic
polynomials {S j } n−1
j=0.
• Natural spline interpolant: S ′′ (x 0 ) = S ′′ (x n ) = 0.
The boundary conditions imply that c n = S ′′ (x n )/2 = 0 and
0 = S ′′ (x 0 ) = 2c 0 + 6d 0 (x 0 − x 0 ) ⇒ c 0 = 0.
10

The two equations c 0 = 0 and c n = 0 together with the equations in (3.4.7) produce a
linear system described by AX = b, where A is the (n + 1) × (n + 1) matrix
⎡
A =
⎢
⎣
1 0 0 · · · ... · · · · · · 0
h 0 2(h 0 + h 1 ) h 1 0 · · · · · · · · · 0
0 h 1 2(h 1 + h 2 ) h 2 0 · · · · · · 0
.
...
...
0 · · · · · · 0 h n−3 2(h n−3 + h n−2 ) h n−2 0
0 · · · · · · · · · 0 h n−2 2(h n−2 + h n−1 ) h n−1
0 · · · · · · · · · · · · 0 0 1
...
.
⎤
⎥
⎦
⎡
b =
⎢
⎣
0
3
h 1
(a 2 − a 1 ) − 3 h 0
(a 1 − a 0 )
.
3
h n−1
(a n − a n−1 ) − 3
h n−2
(a n−1 − a n−2 )
0
⎤
⎥
⎦
⎡
X =
⎢
⎣
c 0
c 1
.
c n−1
c n
⎤
⎥
⎦
• Clamped spline interpolant: S ′ (x 0 ) = f ′ (x 0 ), S ′ (x n ) = f ′ (x n ).
From the two boundary conditions, we obtain the following two equations:
f ′ (x 0 ) = S ′ (x 0 ) = b 0 = 1 h 0
(a 1 − a 0 ) − h 0
3 (2c 0 + c 1 )
⇒ 2h 0 c 0 + h 0 c 1 = 3 h 0
(a 1 − a 0 ) − 3f ′ (a) (3.4.8)
f ′ (x n ) = S ′ (x n ) = b n = b n−1 + h n−1 (c n−1 + c n )
= 1 (a n − a n−1 ) − h n−1
h n−1 3 (2c n−1 + c n ) + h n−1 (c n−1 + c n )
= a n − a n−1
h n−1
+ h n−1
3 (c n−1 + 2c n )
⇒ h n−1 c n−1 + 2h n−1 c n = 3f ′ (b) − 3
h n−1
(a n − a n−1 ) (3.4.9)
11

Equations (3.4.7), together with the equations (3.4.8) and (3.4.9) determine the linear
system described by AX = b, where A is the (n + 1) × (n + 1) matrix
⎡
A =
⎢
⎣
⎤
2h 0 h 0 0 · · · · · · · · · · · · 0
h 0 2(h 0 + h 1 ) h 1 0 · · · · · · · · · 0
0 h 1 2(h 1 + h 2 ) h 2 0 · · · · · · 0
.
...
...
...
.
0 · · · · · · 0 h n−3 2(h n−3 + h n−2 ) h n−2 0
⎥
0 · · · · · · · · · 0 h n−2 2(h n−2 + h n−1 ) h n−1 ⎦
0 · · · · · · · · · · · · 0 h n−1 2h n−1
⎡
b =
⎢
⎣
3
h 0
(a 1 − a 0 ) − 3f ′ (a)
3
h 1
(a 2 − a 1 ) − 3 h 0
(a 1 − a 0 )
.
3
h n−1
(a n − a n−1 ) − 3
h n−2
(a n−1 − a n−2 )
3f ′ (b) − 3
h n−1
(a n − a n−1 )
⎤
⎥
⎦
⎡
X =
⎢
⎣
c 0
c 1
.
c n−1
c n
⎤
⎥
⎦
Exercise 1. A natural cubic spline S on [0, 2] is defined by
S(x) =
{
S0 (x) = 1 + 2x − x 3 , if 0 ≤ x < 1
S 1 (x) = a + b(x − 1) + c(x − 1) 2 + d(x − 1) 3 , if 1 ≤ x ≤ 2
Find a, b, c, d.
Exercise 2. A clamped cubic spline S on [1, 3] is defined by
S(x) =
{
S0 (x) = 3(x − 1) + 2(x − 1) 2 − (x − 1) 3 , if 1 ≤ x < 2
S 1 (x) = a + b(x − 2) + c(x − 2) 2 + d(x − 2) 3 , if 2 ≤ x ≤ 3
Given f ′ (1) = f ′ (3), find a, b, c, d.
12