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Chapter 3. Polynomial Interpolation/Function Approximation
3.1. Lagrange Interpolation Formula
For a set of data points: x i , y i , i  1, 2,  , n and n  1, the elementary Lagrange
interpolation formula is
l in ( x ) 
x xj
, i  1, 2, 3, ..., n
x  xj
j 1, j  i i
n

(3.1.1)
lin (x) is a polynomial with degree no greater than n  1 . Its value at any data
point x k within the data set is either 1 or 0
l in ( x k ) 
l in ( x k ) 
n

xk  x j
x  xj
j 1, j  i i
n

xk  x j
x  xj
j 1, j  i i
1,
for i  k
(3.1.2)
0,
for i  k
(3.1.3)
which is equivalent to
1, i  k 
l in ( x k )   ik  

0, i  k 
(3.1.4)
The Lagrange interpolation polynomial of degree n  1 is
n
p n1 ( x )   y i l in ( x )
i 1
(3.1.5)
and its values at the data points are
n
n
i 1
i 1
p n1 ( x k )   y i l in ( x k )   y i  ik  y k , k  1, 2,  , n
(3.1.6)
2
which means p n1 ( x) passes through all the data points exactly.
For the simplest case where n  2 , there are only two data points and p1 ( x) is a
linear function which passes through the two data points. Thus, p1 ( x) is just a straight
line with its two end points being the two data points. From (3.1.6), we have
p1 ( x )  y1
x  x2
x  x1
y  y1
 y2
 y1  2
( x  x1 )
x1  x 2
x 2  x1
x 2  x1
(3.1.7)
For n  3 , p2 ( x) is the quadratic polynomial that passes through three data
points.
p2 ( x )  y1
( x  x 2 )( x  x 3 )
( x  x1 )( x  x 3 )
( x  x1 )( x  x 2 )
 y2
 y3
( x1  x 2 )( x1  x 3 )
( x 2  x1 )( x 2  x 3 )
( x 3  x1 )( x 3  x 2 )
(3.1.8)
An advantageous property of the Lagrange interpolation polynomial is that the
data points need not be arranged in any particular order, as long as they are mutually
distinct. Thus the order of the data points is not important. For an application of the
Lagrange interpolation polynomial, say we know y1 and y 2 or y1 , y 2 and y 3 , then we
can estimate the function y (x) anywhere in [ x1 , x2 ] linearly and [ x1 , x3 ] quadratically.
This is what we do in finite element analysis.
3.2 Newton Interpolating Polynomial
Suppose there is a known polynomial p n1 ( x) that interpolates the data set:
( xi , yi , i  1, 2, , n ) . When one more data point ( xn1 , y n1 ) , which is distinct from all
the old data points, is added to the data set, we can construct a new polynomial that
interpolates the new data set. Keep also in mind that the new data point need not be at
either end of the old data set. Consider the following polynomial of degree n
n
p n ( x )  p n1 ( x )  c n  ( x  x i )
i 1
(3.2.1)
where cn is an unknown constant. In the case of n  1, we specify p0 ( x) as p0 ( x)  y1 ,
where data point 1 need not be at the beginning of the data set.
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At the points of the old data set, the values of p n (x) are the same as those
of p n1 ( x) . This is because the second term in Equation 3.2.1 is zero there. Since we
assume p n1 ( x) interpolates the old data set, p n (x) does so too.
At the new data point, we want pn ( xn1 )  y n1 . This can be accomplished by
setting the coefficient cn to be
cn 
p n ( x n1 )  p n1 ( x n1 ) y n1  p n1 ( x n1 )

n
 ( x n1  x i )
i 1
(3.2.2)
n
 ( x n1  x i )
i 1
which definitely exists since x n 1 is distinct from x i ( i  1, 2, , n) . Now p n (x) is a
polynomial that interpolates the new data set.
For any given data set: ( xi , yi , i  1, 2, , n ) , we can obtain the interpolating
polynomial by a recursive process that starts from p0 ( x) and uses the above construction
to get p1 ( x) , p2 ( x) , , p n1 ( x) . We will demonstrate this process through the
following example.
Example 3.2.1. Construct an interpolating polynomial for the following data set using the
formula in Equations 3.2.1 and 3.2.2
i
x
y
1
0
0
2
5
2
3
7
-1
4
8
-2
5
10
20
Step1: for i  1
p0 ( x )  c 0  y1  0
Step 2: adding point #2
p1 ( x )  p0 ( x )  c1 ( x  x1 )  c1 x
Applying p1 ( x2 )  y 2 , we get c1  0.4 . So
p1 ( x )  p0 ( x )  c1 ( x  x1 )  0.4 x
Step 3: adding point #3
p2 ( x)  p1 ( x)  c 2 ( x  x1 )( x  x 2 )  0.4 x  c 2 x( x  5)
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Applying p2 ( x3 )  y3 , we get c2  0.2714 . So
p2 ( x)  0.4x  0.2714x( x  5)
Step 4: adding point #4
p3 ( x )  p2 ( x )  c 3 ( x  x1 )( x  x 2 )( x  x 3 )  p2 ( x )  c 3 x( x  5)( x  7)
Applying p3 ( x4 )  y 4 , we get c3  0.0548 . So
p3 ( x)  p2 ( x)  0.0548 x( x  5)( x  7)
Step 5: adding point #5
p4 ( x )  p3 ( x )  c 4 ( x  x1 )( x  x 2 )( x  x 3 )( x  x 4 )  p3 ( x )  c 4 x( x  5)( x  7)( x  8)
Applying p4 ( x5 )  y5 , we get c4  0.0712 . So
p4 ( x)  p3 ( x)  0.0712 x( x  5)( x  7)( x  8)
which is the final answer.
If we expand the recursive form, the r.h.s of Equation (3.2.1), we obtain the more
familiar form of a polynomial
pn1 ( x)  c0  c1 ( x  x1 )  c2 ( x  x1 )( x  x2 )    cn ( x  x1 )( x  x2 ) ( x  xn )
(3.2.3)
which is called the Newton’s interpolation polynomial. Its constants can be determined
from the data set: ( xi , yi , i  1, 2, , n )
p 0 ( x 1 )  y1  c 0
p1 ( x 2 )  y 2  c 0  c1 ( x 2  x1 )
(3.2.4)
p2 ( x 3 )  y 3  c 0  c1 ( x 3  x1 )  c 2 ( x 3  x1 )( x 3  x 2 )

which gives
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c 0  y1
y  c 0 y 2  y1
c1  2

x 2  x1 x 2  x1
y  c 0  c1 ( x 3  x1 )
c2  3
( x 3  x1 )( x 3  x 2 )
(3.2.5)

We should note that forcing the polynomial through data with no regard for rates
of change in the data (i.e., derivatives) results in a C 0 continuous interpolating
polynomial. Alternatively, each data condition p( xi )  yi is called a C 0 constraint.
Let’s use the following notation for these constants
c0  [ x1 ] y  y1
c1  [ x1 , x2 ] y 
[ x 2 ] y  [ x1 ] y
x 2  x1
c i 1  [ x1 , x 2  x i ] y, i  1, 2, 
(3.2.6)
which has the following property
[ x1 x 2  x i ] y 
[ x 2 , x 3  x i ] y  [ x1 , x 2  x i 1 ] y
x i  x1
(3.2.7)
so that it is called the divided difference. For the proof of this formula, please refer to
Gautschi (1997), but for example:
x1 , x 2 y  y 2  y1 , x1 , x 2 , x 3 y  x 2 , x 3 y  x1 , x 2 y , etc. Using this
x 2  x1
x 3  x1
recursion property, a table of divided differences can be generated as follows
x1
y1
x2
y2
x3
y3
x4
y4

[ x1 x 2 ] y
[ x 2 x 3 ] y [ x1 x 2 x 3 ] y
(3.2.8)
[ x 3 x 4 ] y [ x 2 x 3 x 4 ] y [ x1 x 2 x 3 x 4 ] y




This table can be viewed as part of an n  (n  1) matrix for a data set that has n
points. The first column is the x values of the data set and the second column is the y or
function values of the data set. For the rest of the (n  1)  (n  1) lower triangle, the rule
for the construction of its elements, say, element(i, j), is as follows
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(1) It takes the form of a division ( j  3, i  j  1 )
element ( i , j ) 
element ( i , j  1)  element ( i  1, j  1)
element ( i , 1)  element ( i  j  2, 1)
(3.2.9)
where
Element(i, j-1) is the element in the matrix immediately left of element(i, j);
Element(i-1, j-1) is the element above and immediately left;
Element(i, 1) is the x on the same row;
Element(i-j+2, 1). To find it, going from element(i, j) diagonally upward and leftward,
when reaching the second column, it is the element to the left.
(2) The denominator is easier to see in this form:
[ x k x l  xq ] y 
element on left  element above left
xq  xk
Example 3.2.2. Using the table of divided differences, construct the Newton interpolating
polynomial for the data set in Example 3.2.1.
0
0
5
2
7
1
8
2
10
20
20
 0.4
50
1 2
 1.5
75
 2 1
 1
87
20  2
 11
10  8
 1.5  0.4
 0.271
70
 1  1.5
 0.167
85
11  1
4
10  7
0.167  0.271
 0.055
80
4  0.167
 0.767
10  5
0.767  0.055
 0.071
10  0
where the number of decimal digits is reduced so that the table fits the page here. The
bracketed terms <…> are the coefficients of the Newton polynomial. Then
p 4 ( x)  0  0.4 x  0.2714 x( x  5)  0.0548 x( x  5)( x  7)  0.0712 x( x  5)( x  7)( x  8)
 0  x(0.4  ( x  5)( 0.2714  ( x  7)(0.0548  ( x  8)0.0712)))
The second line above is called a nested form that can save computation operations and
produces a more accurate solution numerically. Note in passing how similar this table is
7
to Romberg quadrature (or Richardson’s extrapolation), only in here the formula for
calculating the rightmost terms is a simple ratio of differences.
A MATLAB code “newton_example.m” is written to incorporate the divided
difference scheme and calculate the Newton interpolation polynomial using the nested
form. The algorithm is as follows
(1) Input and plot data set points;
(2) Declare c(n, n + 1) as the matrix for the table; Note: c = element in Equation
(3.2.9);
(3) c(i, 1)  x; c(i, 2)  y;
(4) Calculate the rest of the table due to Equation (3.2.9);
(5) Calculate the polynomial using c(i, i +1) as the coefficients of the polynomial.
Figure 3.2.1 shows the resulting polynomial for the above example.
Newton Interpolating Polynomial
20
data points
Newton Polynomial
15
y
10
5
0
-5
-10
-2
0
2
4
6
8
10
12
x
Figure 3.2.1 Newton interpolation polynomial.
Through the above example, we can see the advantages of the divided difference
table over the algebraic approach we have in Example 3.2.1. First, it has less
computational operations. We do not need to write the polynomial and then use the C0
8
condition to calculate the constants. Second, it is much easier to incorporate it in a
computer code.
It is important to realize that both the Lagrange and Newton polynomials are C0
continuous and each would generate the same result. One effect of C0 continuity can be
seen in the large dip that occurs between the first two data points in the figure. Should it
really be there?
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3.3 Hermite Interpolation Polynomial
The Hermite interpolation accounts for the derivatives of a given function. For
example, in the case of a beam finite element, suppose we need to obtain cubic
polynomials that satisfy the following cases:
(1) Consider: y = ax3 + bx2 + cx + d in [0, 1].
(2) Apply conditions
@x=0
@x=1
Case 1:
y = 1, y = 0 y = y= 0
Case 2:
y = 0, y= 1 y = y = 0
Case 3:
y = 0, y= 0 y = 1, y= 0
Case 4:
y = 0, y = 0 y = 0, y = 1
(3) Solve each case for a, b, c, d.
We recall the standard Newton form for a cubic polynomial
y( x)  c0  c1 ( x  x1 )  c2 ( x  x1 )( x  x2 )  c3 ( x  x1 )( x  x2 )( x  x3 )
(3.3.1)
This clearly is not what the Newton interpolation is meant for, but we can employ it as
follows: approximate y by using the divided difference between two points, then letting
one approach the other in the limit. For example, if y(xi) is used, consider adding another
point xm. Letting these two points converge to one point, we obtain
[ xi x m ] y 
y m  yi
 y ( xi ), as xm  xi
x m  xi
From this we discover that the divided difference is an approximation of a derivative.
Then in the divided difference table, we will put two entries for xi in the data set and do
not calculate xi , xi y in its original form (which will overflow numerically), but rather
simply put the y (xi) value there.
For the case mentioned above, the table would be
x1
x1
y1
y1
x2
y2
x2
y2
y1
[ x1 x 2 ] y [ x1 x1 x 2 ] y
y 2
[ x1 x 2 x 2 ] y [ x1 x1 x 2 x 2 ] y
The tables for the four cases are best determined by hand. Then substitution of the
diagonal values for the c i ’s and for the xi ’s into Equation 3.3.1 yield the polynomials.
The results are:
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Case 1 :
0 1
0 1
0
1 0 1 1
1 0 0 1 2
y ( x)  1  0 x  1x 2  2 x 2 ( x  1)  2 x 3  3 x 2  1
Case 2 :
0 0
0 0 1
1 0 0 1
1 0 0 0 1
y ( x)  0  1x  1x 2  1x 2 ( x  1)  x 3  2 x 2  x
Case 3 :
0 0
0 0 0
1 1 1 1
1 1 0 1  2
y ( x)  0  0 x  1x 2  2 x 2 ( x  1)  2 x 3  3 x 2
Case 4 :
0 0
0 0 0
1 0 0 0
1 0 1 1 1
y ( x)  0  0 x  0 x 2  1x 2 ( x  1)  x 3  x 2
It is not hard to modify the code “newton_example.m” and incorporate these
changes. The resulting code is called “hermite_example.m”. It can compute the above
tables. The polynomials are plotted in Figure 3.3.1.
For cases involved with higher order derivatives, the principle is the same (see
Gautschi, 1997). One thing worth noting here is that when y(n)(xi) is used, all lower
derivatives and y(xi) itself must be included in the constraints. For example, you can not
have y(xi) as a constraint but not y(xi), nor y(2)(xi) but not y(xi) and y(xi).
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0.8
1
0.8
0.4
y
y
0.6
0.4
0.2
0
0.2
0
case 1
0
-0.2
0.5
x
-0.4
1
case 2
0
0.5
x
1
0.5
x
1
0.8
1
0.6
0.8
0.4
y
y
0.6
0.4
0.2
data points
Hermite Polynomial
0.6
0
case 3
-0.2
0
0
0.2
0.5
x
1
-0.4
case 4
0
Figure 3.3.1 Hermite interpolation.
Example 3.3.1. Constructing displacements in a beam element from Hermite polynomials
Consider the beam of length L shown in Figure 3.3.2. The Hermite polynomials are:
x
x
N1 ( x)  2( ) 3  3( ) 2  1
(3.3.2)
L
L
x3
x2
N 2 ( x)  2  2  x
(3.3.3)
L
L
x
x
N 3 ( x)  2( ) 3  3( ) 2
(3.3.4)
L
L
x3 x2
N 3 ( x)  2 
(3.3.5)
L
L
These polynomials interpolation functions may be thought of as the fundamental modes
of deflection, two of which are shown in Figure 3.3.2. The deflection w(x ) of any
statically loaded beam can be written in terms of these modes as
w( x)  N1W1  N 2 1  N 3W2  N 4 2
(3.3.6)
where the subscripts associate quantities with positions (or nodes) 1 and 2 on the beam
and Wi , i , i  1, 2, are the deflection and slope, respectively, at each node.
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1
N1
N3
N
x/L
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 3.3.2 Static bending modes N1 and N3 in a beam.
Can you physically explain why N1 + N3 = 1 for all x in Figure 3.3.2? …And why N2 +
N4  1?
References
Gautschi, Walter, “Numerical Analysis,” Birkhauser, Boston, 1997.