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CHAPTER 8
Interpolation
Let a physical experiment be conducted and the outcome is recorded only at some finite
number of times. If we want to know the outcome at some intermediate time where
the data is not available, then we may have to repeat the whole experiment once again
to get this data. In the mathematical language, suppose that the finite set of values
{f (x i ) : i = 0, 1, · · · , n}
of a function f at a given set of points
{x i : i = 0, 1, · · · , n}
is known and we want to find the value of f (x), for some x ∈ (xm, xM ), where xm =
min
j=0,1,··· ,n
x j and x M = max j=0,1,··· ,n x j
. One way of obtaining the value of f (x) is to
compute this value directly from the expression of the function f. Often, we may not
know the expression of the function explicitly and only the data
{(x i , y i ) : i = 0, 1, · · · , n}
is known, where y i = f (x i ). In terms of the physical experiments, repeating an ex-
periment will quite often be very expensive. Therefore, one would like to get at least
an approximate value of f (x) (in terms of experiment, an approximate value of the
outcome at the desired time). This is achieved by first constructing a function whose
value at xi coincides exactly with the value f (xi) for i = 0, 1, · · · , n and then finding
the value of this constructed function at the desired points. Such a process is called
interpolation and the constructed function is called the interpolating function for the
given data.
Section 8.1 Polynomial Interpolation
In certain circumstances, the function f may be known explicitly, but still too dif-
ficult to perform certain operations like differentiation and integration. Thus, it is
often preferred to restrict the class of interpolating functions to polynomials, where the
differentiation and the integration can be done more easily.
In Section 8.1, we introduce the basic problem of polynomial interpolation and prove
the existence and uniqueness of polynomial interpolating the given data. There are at
least two ways to obtain the unique polynomial interpolating a given data, one is the
Lagrange and another one is the Newton. In Section 8.1.2, we introduce Lagrange form
of interpolating polynomial, whereas Section 8.1.3 introduces the notion of divided
differences and Newton form of interpolating polynomial. The error analysis of the
polynomial interpolation is studied in Section 8.3. In certain cases, the interpolating
polynomial can differ significantly from the exact function. This is illustrated by Carl
Runge and is called the Runge Phenomenon. In Section 8.3.4, we present the example
due to Runge and state a few results on convergence of the interpolating polynomials.
The concept of piecewise polynomial interpolation and Spline interpolation are discussed
in Section 8.5.
8.1 Polynomial Interpolation
Polynomial interpolation is a concept of fitting a polynomial to a given data. Thus, to
construct an interpolating polynomial, we first need a set of points at which the data
values are known.
Definition 8.1.1.
Any collection of distinct real numbers x 0 , x 1 , · · · , x n (not necessarily in increasing
order) is called nodes.
Definition 8.1.2 [Interpolating Polynomial].
Let x 0 , x 1 , · · · , x n be the given nodes and y 0 , y 1 , · · · , y n be real numbers. A polyno-
mial pn(x) of degree less than or equal to n is said to be a polynomial interpolating
the given data or an interpolating polynomial for the given data if
p n (x i ) = y i , i = 0, 1, · · · n. (8.1)
The condition (8.1) is called the interpolation condition.
Section 8.1 Polynomial Interpolation
This leads to the system of linear equations for a i s given by
a 0
2
0
a 2 +... + x
n
0
a n = y 0
a 0
2
1
a 2 +... + x
n
1
a n = y 1
a 0
2
n
a 2 +... + x
n
n
a n = y n
The above system may be written as
V a = y,
where V is the Vandermonde matrix
V =
1 x 0 x
2
0 · · · x
n
0
1 x 1 x
2
1
· · · x
n
1
1 x n x
2
n
· · · x
n
n
a = (a 0 , a 1 ,... , an)
T , and y = (y 0 , y 1 ,... , yn)
T .
Let the j
th column of V be denoted as vj. Then we claim that the vectors v 0 , v 2 ,
... , vn are linearly independent.
Let the constants c 0 , c 1 ,.. ., c n be such that
c 0 v 0 + c 1 v 1 +... + cnvn = 0.
The k
th equation of the above system is
c 0
2
k
+... + c n x
n
k
= 0, k = 0, 1,... , n.
Thus, we obtained a polynomial q(x) = c 0
2 +... + c n x
n of degree less than
or equal to n, and having n + 1 distinct roots, namely, x 0 , x 1 ,.. ., x n
. This shows
that the polynomial q(x) is the zero polynomial. That is, c 0 = c 1 =... = c n
This shows that the vectors v 0 , v 2 ,... , v n are linearly independent. This implies
that the Vandermonde matrix is invertible and hence the above system has a unique
solution a = (a 0 , a 1 ,... , an)
T
. Thus, we obtained all the coefficients of pn(x) and are
unique. This proves the existence and uniqueness of the interpolating polynomial of
a given data set.
Section 8.1 Polynomial Interpolation
Remark 8.1.5.
A special case is when the data values yi, i = 0, 1, · · · , n, are the values of a function
f at given nodes xi, i = 0, 1, · · · , n. In such a case, a polynomial interpolating the
given data
x x 0 x 1 x 2 x 3 · · · x n
y f (x 0 ) f (x 1 ) f (x 2 ) f (x 3 ) · · · f (xn)
is said to be the polynomial interpolating the given function or the interpolating
polynomial for the given function and has a special significance in applications of
Numerical Analysis for computing approximate solutions of differential equations
and numerically computing complicated integrals.
Example 8.1.6.
Let the following data represent the values of f :
x 0 0.5 1
f (x) 1.0000 0.5242 −0.
The questions are the following:
- What is the exact expression for the function f?
- What is the value of f (0.75)?
We cannot get the exact expression for the function f just from the given data,
because there are infinitely many functions having same value at the given set of
points. Due to this, we cannot expect an exact value for f (0.75), in fact, it can be
any real number. On the other hand, if we look for f in the class of polynomials of
degree less than or equal to 2, then Theorem 8.1.4 tells us that there is exactly one
such polynomial and hence we can obtain a unique value for f (0.75).
The interpolating polynomial happens to be
p 2 (x) = −1.9042x
2
and we have
p 2
The function used to generate the above table of data is
f (x) = sin
π
e
x
Section 8.1 Polynomial Interpolation
8.1.2 Lagrange’s Form of Interpolating Polynomial
Definition 8.1.7 [Lagrange’s Polynomial].
Let x 0 , x 1 , · · · , x n be the given nodes. For each k = 0, 1, · · · , n, the polynomial l k (x)
defined by
l k (x) =
n ∏
i= i! =k
(x − xi)
(x k − x i
is called the k
th Lagrange polynomial or the k
th Lagrange cardinal function.
Remark 8.1.8.
Note that the k
th Lagrange polynomial depends on all the n+1 nodes x 0 , x 1 , · · · , x n .
The Lagrange polynomials l 0 , l 1 , · · · , l n form a basis for the space of polynomials
of degree ≤ n.
Theorem 8.1.9 [Lagrange’s form of Interpolating Polynomial].
Hypothesis:
- Let x 0 , x 1 , · · · , x n be given nodes.
- Let the values of a function f be given at these nodes.
- For each k = 0, 1, · · · , n, let l k (x) be the k
th Lagrange polynomial.
- Let p n (x) (of degree ≤ n) be the polynomial interpolating the function f at
the nodes x 0 , x 1 , · · · , x n
Conclusion: Then, p n (x) can be written as
p n (x) =
n ∑
i=
f (x i )l i (x). (8.3)
This form of the interpolating polynomial is called the Lagrange’s form of In-
terpolating Polynomial.
Proof.
Firstly, we will prove that q(x) :=
n
i=
f (x i )l i (x) is an interpolating polynomial for
Section 8.1 Polynomial Interpolation
the function f at the nodes x 0 , x 1 , · · · , x n
. Since
li(xj ) =
1 if i = j
0 if i %= j
we get q(xj ) = f (xj ) for each j = 0, 1, · · · , n. Thus, q(x) is an interpolating poly-
nomial. Since interpolating polynomial is unique by Theorem 8.1.4, the polynomial
q(x) must be the same as p n (x). This completes the proof of the theorem.
Remark 8.1.10.
The set of Lagrange polynomials {l 0 (x), l 1 (x),... , l n (x)} forms a basis for the
space of all polynomials of degree less than or equal to n.
Example 8.1.11.
Consider the case n = 1 in which we have two distinct points x 0 and x 1. Thus, we
have
l 0 (x) =
x − x 1
x 0 − x 1
, l 1 (x) =
x − x 0
x 1 − x 0
and therefore,
p 1 (x) = f (x 0 )l 0 (x) + f (x 1 )l 1 (x)
= f (x 0
x − x 1
x 0 − x 1
x − x 0
x 1 − x 0
f (x 0 )(x − x 1 ) − f (x 1 )(x − x 0
x 0 − x 1
After a rearrangement of terms, we arrive at
p 1 (x) = f (x 0
f (x 1 ) − f (x 0
x 1 − x 0
(x − x 0
This is the linear interpolating polynomial of the function f. Similarly, if we are
given three nodes with corresponding values, then we can generate the quadratic
interpolating polynomial and so on..
Example 8.1.12.
Section 8.1 Polynomial Interpolation
Remark 8.1.14.
Let x 0 , x 1 , · · · , xn be nodes, and f be a function. Recall that computing an in-
terpolating polynomial in Lagrange’s form requires us to compute for each k =
0, 1, · · · , n, the k
th Lagrange’s polynomial lk(x) which depends on the given nodes
x 0 , x 1 , · · · , x n
. Suppose that we have found the corresponding interpolating poly-
nomial p n (x) of f in the Lagrange’s form for the given data. Now if we add one
more node x n+ , the computation of the interpolating polynomial p n+ (x) in the
Lagrange’s form requires us to compute a new set of Lagrange’s polynomials cor-
responding to the set of (n + 1) nodes, and no advantage can be taken of the fact
that p n is already available.
An alternative form of the interpolating polynomial, namely, Newton’s form of
interpolating polynomial , avoids this problem, and will be discussed in the next
section.
8.1.3 Newton’s Form of Interpolating Polynomial
We saw in the last section that it is easy to write the Lagrange form of the interpolating
polynomial once the Lagrange polynomials associated to a given set of nodes have been
written. However we observed in Remark 8.1.14 that the knowledge of p n (in Lagrange
form) cannot be utilized to construct p n+ in the Lagrange form. In this section we
describe Newton’s form of interpolating polynomial , which uses the knowledge of p n in
constructing p n+
Theorem 8.1.15 [Newton’s form of Interpolating Polynomial].
Hypothesis:
- Let x 0 , x 1 , · · · , xn be given nodes.
- Let the values of a function f be given at these nodes.
- Let p n (x) (of degree ≤ n) be the polynomial interpolating the function f at
the nodes x 0 , x 1 , · · · , x n
Conclusion: Then, p n (x) can be written as
pn(x) = A 0 +A 1 (x−x 0 )+A 2 (x−x 0 )(x−x 1 )+A 3
2 ∏
i=
(x−xi)+· · ·+An
n− 1 ∏
i=
(x−xi) (8.5)
where A 0
, A
1
, · · · , A
n are constants.
This form of the interpolating polynomial is called the Newton’s form of inter-
polating polynomial.
Section 8.1 Polynomial Interpolation
Proof.
We show that the interpolating polynomial can be written in the form (8.5) using
mathematical induction.
If n = 0, then the constant polynomial
p 0 (x) = y 0
is the required polynomial and its degree is less than or equal to 0. Thus, by taking
A
0 = y 0 , we see that p 0 (x) is in the form (8.5) as required.
Assume that the result is true for n = k. We will now prove that the result is true
for n = k + 1.
Let the data be represented by
x x 0 x 1 x 2 x 3 · · · x k x k+
y y 0 y 1 y 2 y 3 · · · y k y k+
By the assumption, there exists a polynomial p k (x) of degree less than or equal to k
such that the first k interpolating conditions
p k (x i ) = y i , i = 0, 1, · · · , k
hold. Define a polynomial p k+ (x) of degree less than or equal to k + 1 by
p k+ (x) = p k (x) + c(x − x 0 )(x − x 1 ) · · · (x − x k
where the constant c is such that the (k + 1)
th interpolation condition pk+1(xk+1) =
y k+ holds. This is achieved by choosing
c =
y k+ − p k (x k+
(x k+ − x 0 )(x k+ − x 1 ) · · · (x k+ − x k
Note that p k+ (x i ) = y i for i = 0, 1, · · · , k and therefore p k+ (x) is an interpolating
polynomial for the given data. This proves the result for n = k + 1 with A k+ = c.
By the principle of mathematical induction, the result is true for any natural number
n.
Remark 8.1.16.
Let us recall the equation (8.6) from the proof of Theorem 8.1.4 now.
Section 8.2 Newton’s Divided Differences
form. Thus the quantity f [x 0 , x 1 , · · · , x n ] is well-defined.
- More generally, the divided difference f [x i , x i+ , · · · , x i+k ] is the coefficient
of x
k in the polynomial interpolating f at the nodes x i , x i+ , · · · , x i+k .
- The Newton’s form of interpolating polynomial may be written, using divided
differences, as
p n (x) = f [x 0 ] +
n ∑
k=
f [x 0 , x 1 , · · · , x k ]
k− 1 ∏
i=
(x − x i ) (8.9)
Example 8.2.3.
As a continuation of Example 8.1.11, let us construct the linear interpolating polyno-
mial of a function f in the Newton’s form. In this case, the interpolating polynomial
is given by
p 1 (x) = f [x 0 ] + f [x 0 , x 1 ](x − x 0
where
f [x 0 ] = f (x 0 ), f [x 0 , x 1
] =
f (x 0 ) − f (x 1
x 0 − x 1
are zeroth and first order divided differences , respectively. Observe that this polyno-
mial is exactly the same as the interpolating polynomial obtained using Lagrange’s
form in Example 8.1.11.
The following result is concerning the symmetry properties of divided differences.
Theorem 8.2.4 [Symmetry].
The divided difference is a symmetric function of its arguments. That is, if z 0 , z 1 , · · · , zn
is a permutation of x 0 , x 1 , · · · , x n , then
f [x 0 , x 1 , · · · , x n ] = f [z 0 , z 1 , · · · , z n
] (8.11)
Proof.
Since z 0 , z 1 , · · · , z n is a permutation of x 0 , x 1 , · · · , x n , which means that the nodes
x 0 , x 1 , · · · , x n have only been re-labelled as z 0 , z 1 , · · · , z n , and hence the polynomial
interpolating the function f at both these sets of nodes is the same. By definition
f [x 0 , x 1 , · · · , x n ] is the coefficient of x
n in the polynomial interpolating the func-
tion f at the nodes x 0 , x 1 , · · · , x n , and f [z 0 , z 1 , · · · , z n ] is the coefficient of x
n in the
Section 8.2 Newton’s Divided Differences
polynomial interpolating the function f at the nodes z 0 , z 1 , · · · , z n
. Since both the
interpolating polynomials are equal, so are the coefficients of x
n in them. Thus, we
get
f [x 0 , x 1 , · · · , x n ] = f [z 0 , z 1 , · · · , z n
].
This completes the proof.
The following result helps us in computing recursively the divided differences of higher
order.
Theorem 8.2.5 [Higher-order divided differences].
Divided differences satisfy the equation
f [x 0 , x 1 , · · · , xn] =
f [x 1 , x 2 , · · · , x n ] − f [x 0 , x 1 , · · · , x n− 1
]
x n − x 0
Proof.
Let us start the proof by setting up the following notations.
- Let p n (x) be the polynomial interpolating f at the nodes x 0 , x 1 , · · · , x n
- Let p n− 1 (x) be the polynomial interpolating f at the nodes x 0 , x 1 , · · · , x n− 1
- Let q(x) be the polynomial interpolating f at the nodes x 1 , x 2 , · · · , x n
Claim: We will prove the following relation between p n− 1 , p n , and q:
p n (x) = p n− 1 (x) +
x − x 0
x n − x 0
q(x) − p n− 1 (x)
Since both sides of the equality in (8.13) are polynomials of degree less than or equal
to n, and pn(x) is the polynomial interpolating f at the nodes x 0 , x 1 , · · · , xn, the
equality in (8.13) holds for all x if and only if it holds for x ∈ { x 0 , x 1 , · · · , x n } and
both sides of the equality reduce to f (x) for x ∈ { x 0 , x 1 , · · · , x n }. Let us now verify
the equation (8.13) for x ∈ { x 0 , x 1 , · · · , x n
- When x = x 0
pn− 1 (x 0 ) +
x 0 − x 0
x n − x 0
q(x 0 ) − pn− 1 (x 0 )
= pn− 1 (x 0 ) = f (x 0 ) = pn(x 0 ).
Section 8.3 Error in Polynomial Interpolation
8.2.2 Divided Differences Table
Given a collection of (n + 1) nodes x 0 , x 1 , · · · , x n and the values of the function f at
these nodes, we can construct the Newton’s form of interpolating polynomial pn(x) using
divided differences. As observed earlier, the Newton’s form of interpolation polynomial
has the formula
p n (x) = f [x 0
] +
n ∑
k=
f [x 0 , x 1 , · · · , x k
]
k− 1 ∏
i=
(x − x i
One can explicitly write the formula (8.15) for n = 1, 2, 3, 4, 5, · · ·. For instance, when
n = 5, the formula (8.15) reads
p 5 (x) = f [x 0
]
- f [x 0 , x 1 , x 2 ](x − x 0 )(x − x 1
f [x 0 , x 1 , x 2 , x 3 ](x − x 0 )(x − x 1 )(x − x 2 )
f [x 0 , x 1 , x 2 , x 3 , x 4 ](x − x 0 )(x − x 1 )(x − x 2 )(x − x 3 )
f [x 0 , x 1 , x 2 , x 3 , x 4 , x 5 ](x − x 0 )(x − x 1 )(x − x 2 )(x − x 3 )(x − x 4
For easy computation of the divided differences in view of the formula (8.12), it is
convenient to write the divided differences in a table form. For n = 5, the divided
difference table is given by
x 0 f (x 0 )
f [x 0 , x 1 ]
x 1 f (x 1 ) f [x 0 , x 1 , x 2 ]
f [x 1 , x 2 ] f [x 0 , x 1 , x 2 , x 3 ]
x 2 f (x 2 ) f [x 1 , x 2 , x 3 ] f [x 0 , x 1 , x 2 , x 3 , x 4 ]
f [x 2 , x 3 ] f [x 1 , x 2 , x 3 , x 4 ] f [x 0 , x 1 , x 2 , x 3 , x 4 , x 5 ]
x 3 f (x 3 ) f [x 2 , x 3 , x 4 ] f [x 1 , x 2 , x 3 , x 4 , x 5 ]
f [x 3 , x 4 ] f [x 2 , x 3 , x 4 , x 5 ]
x 4 f (x 4 ) f [x 3 , x 4 , x 5 ]
f [x 4 , x 5 ]
x 5 f (x 5 )
Comparing the above divided differences table and the interpolating polynomial p 5
given by (8.16), we see that the leading members of each column (denoted in bold font)
are the required divided differences used in p 5 (x).
8.3 Error in Polynomial Interpolation
Let f be a function defined on an interval I = [a, b]. Let p n (x) be a polynomial of
degree less than or equal to n that interpolates the function f at n + 1 nodes x 0 , x 1
Section 8.6 Exercises
8.6 Exercises
Polynomial Interpolation
- Let x 0 , x 1 , · · · , x n be distinct nodes. If p(x) is a polynomial of degree less than or
equal to n, then show that
p(x) =
n ∑
i=
p(x i )l i (x),
where l i (x) is the i
th Lagrange polynomial.
- Show that the polynomial 1 + x + 2x
2 is an interpolating polynomial for the data
x 0 1 2
y 1 4 11
Find an interpolating polynomial for the new data
x 0 1 2 3
y 1 4 11 -
Does there exist a quadratic polynomial that satisfies the new data? Justify your
answer.
- The quadratic polynomial p 2 (x) =
3
4
x
2
1
4
x +
1
2
interpolates the data
x − 1 0 1
y 1
1
2
3
2
Find a node x 3 (x 3 ∈/ {−1, 0, 1}), and a real number y 3 such that the polynomial
p 3 (x) interpolating the data
x − 1 0 1 x 3
y 1 1 / 2 3 / 2 y 3
is a polynomial of degree less than or equal to 2.
- Let p(x), q(x), and r(x) be interpolating polynomials for the three sets of data
x 0 1
y y 0 y 1
x 0 2
y y 0 y 2
, and
x 1 2 3
y y 1 y 2 y 3
respectively. Let s(x) be the the interpolating polynomial for the data
x 0 1 2 3
y y 0 y 1 y 2 y 3
If
p(x) = 1 + 2x, q(x) = 1 + x, and r(2.5) = 3,
then find the value of s(2.5).
- Obtain Lagrange form of interpolating polynomial for equally spaced nodes.
- Find the Largrange form of interpolating polynomial for the data:
x -2 -1 1 3
y -1 3 -1 19
