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Section 3 Newton Divided-Difference Interpolating
Polynomials
The standard form for representing an n th order interpolating polynomial is straightforward. There are however, other representations of n th order polynomials which on the surface may seem a bit more unwieldy, but require less manipulation to arrive at. One such form is the subject of this section.
Given a set of n +1 data points ( x i , y i ), i = 0, 1, 2, …., n where the xi are all different and the y i are sampled from an underlying function y = f ( x ), the n th order interpolating polynomial can be expressed as
f (^) n ( ) x = b 0 (^) + b 1 (^) ( x − x 0 (^) ) + b 2 (^) ( x − x 0 (^) )( x − x 1 (^) ) + ..... + bn ( x − x 0 (^) )( x − x 1 (^) ) ⋅ ⋅ ⋅ ⋅ ⋅ ( x − xn -1) (3.1)
Before we consider how to determine the coefficients, observe that Equation (3.1) is in fact an nth order polynomial as evidenced by the last term which includes the highest power of x , namely xn. All powers of x are present in Equation (3.1) despite the fact that the coefficients of say x^0 , x^1 , x^2 ,…etc. are not as obvious as when the polynomial is expressed in standard form (see Equation 2.1). Since there are n +1 independent coefficients b (^) i available, i = 0, 1, 2,…, n we can be assured there is at most an n th order polynomial that passes through the given data points.
The rationale for selecting the analytical form in Equation (3.1) will be apparent after we look at a few simple examples.
A) Given ( x 0 , y 0 ), ( x 1 , y 1 ) where y 0 = f ( x 0 ) and y 1 = f ( x 1 )
The first order polynomial for the case when n = 1 through the two data points is
f (^) 1 ( ) x = b (^) 0 + b (^) 1 ( x − x 0 ) (3.2)
Substitution of the two data points ( x 0 , y 0 ), ( x 1 , y 1 ) into Equation (3.2) gives
f (^) 1 ( x (^) 0 ) = b (^) 0 + b 1 (^) ( x (^) 0 − x 0 ) (3.3)
f 1 (^) ( x 1 (^) ) = b 0 (^) + b 1 (^) ( x 1 (^) − x 0 ) (3.3a)
Solving for b 0 and b 1 yields,
b (^) 0 = f (^) 1 ( x 0 ) (3.4)
b
f x f x (^1) x x
1 1 1 0 1 0
(3.4a)
By design, the interpolating function f 1 ( x ) and the actual function f ( x ) from which the data points were obtained are equal at x = x 0 and x = x 1 (see Figure 2.2). As a result, b 0 and b 1 can be expressed in terms of the given data,
b (^) 0 = f ( x 0 ) (3.5)
b
f x f x (^1) x x
1 0 1 0
(3.5a)
Example 3.
The monthly payment on a 30-yr mortgage of $100,000. for two different annual interest rates is given in Table 3.1 below. Use an interpolation formula in the form of Equation (3.2) to estimate the monthly payment corresponding to an interest rate of 8. % per year.
Data Point Number k
Annual Interest Rate i (^) k
Monthly Payment Ak = f ( i (^) k )
0 7 % $ 665. 1 10 % $ 877.
Table 3.1 Monthly Payments for $100,000 30-yr Mortgage with Different Interest Rates: Two Data Points
The first order interpolating polynomial is written
f (^) 1 ( ) i = b (^) 0 + b 1 (^) ( i − i 0 ) (3.6)
where
b f i f
b
f i f i i i
f f
0 0
1
1 0 1 0
The estimated monthly payment is therefore
f 1 (8. 25 ) = 665 3. + 70 76. (8. 25 − 7 ) =753 68.
designed to solve simultaneous equations. Nonetheless, the solutions are obtained in fundamentally different ways.
The second advantage of the nonstandard representation is more compelling. The expressions for b 0 and b 1 in Equations (3.9) and (3.9a) are the same as in Equation (3.5). Why is this important? Suppose you just completed the process of finding an interpolating polynomial for a given set of data points. There may be some doubt in your mind concerning the accuracy of results based on the use of this polynomial. Later on there will be a discussion of quantitative methods for approximating the errors inherent in interpolation. More data points may be required to reduce the estimated errors. Incorporating additional data points is easier with the nonstandard representation for the interpolating polynomials. This is because each additional data point requires the computation of a single coefficient for the new term in the polynomial. This is illustrated in the following example which extends the results obtained in Example 3.1.
Example 3.
Suppose we obtain two additional data points in the previous example dealing with the estimation of mortgage payments. The new data points correspond to 8 % and 9 % loans. Use one of the two additional points to obtain a second order interpolating polynomial. Estimate the monthly payment for an 8.25 % loan.
We must choose one of the two new data points to find the second order interpolating polynomial f 2 ( i ). Our intuition suggests the data point for an 8 % loan is the wiser choice because its closer to the point where the estimate is required, i.e. 8.25 %. The second order polynomial is obtained from the data points corresponding to k = 0, 1, and 2 in the following table. It is given in Equation (3.10).
Data point Number k
Annual Interest Rate i (^) k
Monthly Payment Ak = f ( i (^) k )
0 7 % $ 665. 1 10 % $ 877. 2 8 % $ 733. 3 9 % $ 804.
Table 3.2 Monthly Payments for $100,000 30-yr Mortgage with Different Interest Rates: 4 Data Points
f (^) 2 ( ) i = b (^) 0 + b 1 (^) ( i − i (^) 0 ) + b (^) 2 ( i − i (^) 0 )( i − i 1 ) (3.10)
Coefficients b 0 and b 1 were determined in Example 3.1 using the first two data points. The remaining coefficient b 2 is obtained from Equation (3.9b) as
b
f i f i i i
f i f i i i (^2) i i
2 1 2 1
1 0 1 0 2 0
733 76 877 57 8 10
L
N
M
O
Q
P −^
L
N
M
O
Q
P
L
NM^
O
QP^
L
NM^
O
QP
The polynomial f 2 ( i ) is therefore,
f (^) 2 ( ) i = 665 3. + 70 76. ( i − i (^) 0 ) + 1148. ( i − i (^) 0 )( i − i 1 ) (3.11)
and the estimate of payments for an 8.25 % mortgage is now
f (^) 2 (8. ).. (8. ). (8. )(8. )
The previous two examples are somewhat academic in nature. Quite obviously, the loan officer at the bank will not be interested in your calculations to estimate the monthly payments. He or she has access to the function A = f ( i ) from which you obtained several data points in the Sunday paper. Let's see how close your two estimates were to the correct answer. Figure 3.1 includes graphs of the true function f ( i ) as well as the interpolating polynomials f 1 ( i ) and f 2 ( i ). The upper plot includes the data points as well.
It is clear from looking at the upper plot that either interpolating function f 1 ( i ) or f 2 ( i ) will provide accurate estimates of the true monthly payment function f ( i ) over the entire range from i = 7 % to i = 10 % and then some. The lower plot is an enlargement of each graph in the region about i = 8.25 %, the interest rate under consideration. Here we see that the quadratic interpolating polynomial f 2 ( i ) and the real function f ( i ) are virtually indistinguishable. You may be interested in knowing that the true function A = f ( i ) is given by
A f i P
i i n i n
1200
where P is the mortgage amount, n is the loan period in months and i is the annual percent interest rate. Perhaps you are surprised at how close the graphs of the low order interpolating polynomials f 1 ( i ) and f 2 ( i ) are to the true function f ( i ) which, based on its analytical form in Equation (3.12), does not appear to resemble a polynomial function.
of various orders are defined in Equations (3.13) - (3.13e). Note how the divided differences are obtained recursively from two divided differences of order one less.
0 th order f^ [^ x^ i ]^ =^ f^ (^ xi ) (3.13)
1 st order f [ x x
f x f x x x i j i^ j i^ j i j
, ]
[ ] [ ]
≠ (3.13a)
f x f x x x
i j i j
(3.13b)
nd order [
[ [
2 f x x x
f x x f x x x x i j k i^ j^ j^ k i^ j^ k i k
, , ]
, ] , ]
≠ ≠ (3.13c)
( (^) i ) ( (^) j ) ( (^) j ) ( (^) k ) i j j (^) k i (^) k
f x f x f x f x x x x x x x
(3.13d)
-1 2 1 1 -2 1 0 -1 1 0 0
[ , ,...., , ] [ , ,...., , ]
th order [ (^) n , (^) n ,...., , ] n^ n^ n n n
f x x x x f x x x x n f x x x x x x
= −^ −
(3.13e)
When there are 3 data points ( n = 2), the divided differences f [ x 0 ], f [ x 1 , x 0 ], and f [ x 2 , x 1 , x 0 ] are identical to the coefficients b 0 , b 1 , and b 2. [see Equations (3.14) - (3.14h)]
The coefficient b 0 of the interpolating polynomial f 2 ( x ) is numerically equal to the first of the three zero order finite divided differences, i.e. the one that depends on the data point [ x 0 , f ( x 0 )]. The coefficient b 1 is equal to the first of the two first order finite divided differences, i.e. the one that depends on the data points [ x 0 , f ( x 0 )] and [ x 1 , f ( x 1 )]. Finally, the coefficient b 2 is equal to the first and only second order finite divided difference, i.e. the one that requires all three data points, [ x 0 , f ( x 0 )], [ x 1 , f ( x 1 )] and [ x 2 , f ( x 2 )].
In general, with n + 1 data points there are n +1 zero order divided differences, n first order divided differences, n -1 second order divided differences, etc. up to one n th order divided difference. The first computed divided difference of order " i " is equal to the coefficient b (^) i , i = 0, 1, 2, …, n.
f [ x (^) 0 ] = f ( x 0 ) (3.14)
= (^) b 0 (3.14a)
f x x
f x f x x x
[ , ]
] ]
1 0 1 0 1 0
[ [
(3.14c)
f x f x x x
1 0
(3.14d)
= b 1 (3.14e)
f x x x
f x x f x x x x
[ , , ]
[ , ] , ]
2 1 0
2 1 1 0 2 0
[
(3.14f)
L
N
M
O
Q
P −^
L
N
M
O
Q
P
f x f x x x
f x f x x x x x
2 1
1 0 1 0 2 0
(3.14g)
= (^) b (^) 2 (3.14h)
Table 3.4 is a helpful aid in remembering how to compute the finite divided differences. The coefficients bi , i = 0, 1, 2,…, n of the interpolating polynomial in Equation (3.1) are the first finite divided-difference entries in their respective columns. The polynomial is referred to as the Newton divided-difference interpolating polynomial. The arrows indicate which finite divided differences are used to calculate the higher order ones. Always remember, the denominator of any divided difference is the difference between its first and last argument.
f 1 [ ] f 2 [ ] f (^) n -1 [ ] f (^) n [ ] x 0 f [ x 0 ] f [ x 1 , x 0 ] x 1 f [ x 1 ] f [ x 2 , x 1 , x 0 ] f [ x 2 , x 1 ] x 2 f [ x 2 ] f [ x 3 , x 2 , x 1 ] f [ x 3 , x 2 ] x 3 f [ x 3 ] f [ x 4 , x 3 , x 2 ] ⋅ ⋅ f [ x 4 , x 3 ] ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ f [ xn -1 ,^ xn -2 ,…,^ x 2 ,^ x 0 ] ⋅ ⋅ ⋅ ⋅ ⋅ f [ xn ,^ xn -1 ,^ xn -2 …, x 1 ,^ x 0 ] ⋅ ⋅ ⋅ ⋅ ⋅ f [ xn , xn -1 ,…, x 2 , x 1 ] x n-2 f [ xn -2 ] (^) ⋅ ⋅ f [ xn -1 , xn -2 ] (^) ⋅ x n-1 f [ xn -1 ] f [ xn , x n-1 , xn -2 ] f [ xn , xn -1 ] x n f [ xn ]
Table 3.4 Table of Finite Divided Differences for a Function f ( x ) Known at Discrete Data Points
The true function T = f (ω) is not available to assess the accuracy of this estimate. Conceivably, one could derive a structure or analytical form of the function based on established engineering principles and scientific laws. A number of physical parameters, i.e. constants would likewise have to be known before the function could be used for evaluation of motor performance. Rarely is this ever attempted in situations where all that is required is a reliable estimate of how a particular device, component or system will perform. Conversely, a model of the torque motor based on scientific principles is far more useful to an engineer designing a motor to satisfy specific design criteria.
There are situations where a combination of empirical and scientific modeling are used in conjunction with each other. Imagine a situation where a theoretical model is known, however it may be of such complexity that data points are expensive to obtain. In this situation, a minimum number of data points can be obtained from solution of equations (differential and algebraic) comprising the scientific model. The data points are then used as a basis to generate an empirical model for interpolation.
Little has been said up to this point about the ordering of the data points, primarily because an n th order polynomial passing through n +1 data points is unique and the ordering is irrelevant. Changing the order of the data points will affect the representation of the polynomial; however the polynomial itself has not changed. The selection of which n +1 data points to draw from a larger sample to obtain an n th order polynomial is a different matter. In this case, some thought should be given as to which n +1 points produce the best interpolating polynomial. Example 3.4 illustrates these points.
Example 3.
Consider the data in Table 3.2. Find the third order Newton divided-difference interpolating polynomial f 3 ( i ) containing the 4 data points using different orderings of the data points. Estimate the function value f (8.25).
The data from Table 3.2 and the divided differences are given in Table 3.6.
k i k Ak = f ( i k ) f 1 [ ] f 2 [ ] f 3 [ ]
0 7 665.30 ( b 0 ) 68.46 ( b 1 ) 1 8 733.76 1.200 ( b 2 ) 70.86 -0.05167 ( b 3 ) 2 9 804.62 1.
3 10 877.
Table 3.6 Finite Divided Differences and Newton Interpolating Polynomial Coefficients for Data in Example 3.
The coefficients b 0 , b 1 , b 2 and b 3 are read directly from the table. The resulting third order Newton divided difference interpolating polynomial is given in Equation (3.18)and the polynomial is then used for the interpolation of f (8.25).
f (^) 3 ( ) i = b (^) 0 + b 1 (^) ( i − i (^) 0 ) + b 2 (^) ( i − i (^) 0 )( i − i (^) 1 ) + b 3 (^) ( i − i (^) 0 )( i − i (^) 1 )( i − i 2 ) (3.17)
f (^) 3 ( ) i = 65 3. + 68 46. ( i − 7 ) + 12. ( i − 7 )( i − 8 ) − 0 05167. ( i − 7 )( i − 8 )( i − 9 ) (3.18)
f (^) 3 25 65 3 68 46 25 7 12 25 7 25 8 0 05167 25 7 25 8 25 9
751 26
(8. ).. (8. ). (8. )(8. ). (8. )(8. )(8. )
.
= + − + − − − − − −
=
In Table 3.7 the data points are reordered and the process of finding the interpolating polynomial is repeated.
k i (^) k Ak = f ( i k ) f 1 [ ] f 2 [ ] f 3 [ ]
0 7 665.30 ( b 0 ) 70.757 ( b 1 ) 1 10 877.57 1.148 ( b 2 ) 71.905 -0.0515 ( b 3 ) 2 8 733.76 1.
3 9 804.
Table 3.7 Finite Divided Differences and Newton Interpolating Polynomial Coefficients for Data in Example 3.2 with Reordering of Data Points
The new polynomial based on reordered data points and the interpolated value are
f (^) 3 ( ) i = 65 3. + 70 757. ( i − 7 ) + 1148. ( i − 7 )( i − 10 ) − 0 0515. ( i − 7 )( i − 10 )( i − 8 ) (3.19)
f (^) 3 8 25 653 70 757 8 25 7 1148 8 25 7 8 25 10
0 0515 8 25 7 8 25 10 8 25 8
75126
function f ( x ) happens to be a quadratic polynomial, then f 2 ( x ) and f ( x ) are identical and the estimate of R 1 ( x ) would be exact for any value of x.
A numerical example will help clarify the process of approximating the errors occurring from the use of Newton divided-difference interpolating polynomials.
Figure 3.2 Estimating the error R 1 ( x ) in Linear Interpolation
Example 3.
The solution to the differential equation
d dt
y t ( ) + y t ( )= 1 when y (0) = 0 is
y ( t ) = 1- e- t , t ≥ 0. Let T be the time it takes for the solution y ( t ) to reach the value A , where 0 ≤ A ≤ 1, i.e. y ( T ) = A. Several points ( T , A ) on the solution y ( t ) are tabulated below. A low order polynomial is needed for interpolation of T for a given value of A.
T A T A
x
f 1 ( x )
f 2 ( x )
f ( x )
x 0 x 2 x 1
x 0 = 0 x 1 = 5 x 2 = 3 x = 1
R 1 ( x )
f 2 ( x )- f 1 ( x )
Table 3.8 Data Points for Finding Interpolating Polynomial to Estimate T Given A
The underlined data points were selected to determine a third order Newton divided-difference polynomial which can be used for interpolation over the interval. The divided difference table is shown below.
i Ai T (^) i = f ( Ai ) f 1 [ ] f 2 [ ] f 3 [ ] f 4 [ ]
0 0.0 0.
1 0.3 0.3567 1. 2.1183 4. 2 0.7 1.2040 5.6246 7. 5.4930 8. 3 0.9 2.3026 7.
4 0.5 0.
Table 3.9 Divided Difference Table for Finding f 3 ( A ) and f 4 ( A )
From the table, the third order interpolating polynomial f 3 ( A ) is
f (^) 3 ( A ) =^1189. A +^13275. A A ( −^ 0 3. ) +^ 4 7745. A A ( −^ 0 3. )( A −^ 0 7. ) (3.24)
Suppose we wish to estimate f (0.45), the time required for the solution to reach 0.45, using f 3 (0.45). The result is
f (^) 3 0 45 1189 0 45 13275 0 45 0 45 0 3 4 7745 0 45 0 45 0 3 0 45 0 7
Choosing the additional data point (0.5, 0.6931) because of its proximity to the interpolant value 0.45 allows us to find the 4 th^ order interpolating polynomial f 4 ( A ). Table 3.9 includes the additional finite divided differences calculated from the new data point. The result is
f (^) 4 ( A ) = 1189. A + 13275. A A ( − 0 3. ) + 4 7745. A A ( − 0 3. )( A −0 7. )
+ 7 6675. A A ( − 0 3. )( A − 0 7. )( A −0 9. ) (3.25)
function to approximate f ( A ). The 4 closest points to the interpolant value 0.45 are (0.3,0.3567), (0.4,0.5108), (0.5,0.6931) and (0.6,0.9163,).
Figure 3.3 Third and Fourth Order Interpolating Polynomials and the True Function for Data in Table 3.
At this point it is necessary to introduce a new function f [ x , x 1 , x 0 ], similar to a second order divided difference, defined as
[
[ [
f x x x
f x x f x x x x
, , ]
, ] , ]
1 0
1 1 0 0
f x f x x x
f x f x x x x x
1
1 0 1 0 0
(3.27a)
However, unlike the second order finite divided differences previously encountered, its first argument x is treated as a variable. Of course, once x assumes a numerical value, f [ x , x 1 , x 0 ] becomes an ordinary second order divided difference. The importance of f [ x , x 1 , x 0 ] is its relationship to R 1 ( x ) which is presented here and left as an exercise for later.
R 3 (0.45)
f 3 ( A ) f ( A )
f 4 ( A ) f ( A )
R 1 (^) ( ) x = ( x − x (^) 0 )( x − x (^) 1 ) f [ , x x (^) 1 , x 0 ] (3.28)
If f ( x ) is a slow varying function over the interval ( x 0 , x 1 ), then f [ x , x 1 , x 0 ] will likewise vary slowly over the same interval. Under these conditions, for any value of x 2 in the interval ( x 0 , x 1 ),
f [ , x x (^) 1 , x (^) 0 ] ≈^ f [ x (^) 2 , x (^) 1 , x 0 ] (3.29)
Replacing f [ x , x 1 , x 0 ] with its approximation f [ x 2 , x 1 , x 0 ] in Equation (3.28) gives
R (^) 1 ( ) x ≈ ( x − x (^) 0 )( x − x (^) 1 ) f [ x (^) 2 , x (^) 1 , x 0 ] (3.30)
which is identical to Equation (3.23). It was previously stated that if f ( x ) happens to be a quadratic polynomial, then the estimate of R 1 ( x ) in Equation (3.22) would be exact. What does this imply about the function f [ x , x 1 , x 0 ] under the same conditions?
We can now generalize the preceding analysis about the error in Newton divided- difference interpolating polynomials. With n +1 data points [ xi , f ( xi )], i = 0, 1,2, …., n from a function f ( x ), the Newton divided-difference interpolating polynomial f (^) n ( x ) that passes through each point is related to the function f ( x ) in the following way.
f ( ) = x f (^) n ( ) x + R (^) n ( ) x (3.31)
The error term Rn ( x ) accounts for the difference between the true function f ( x ) and the interpolated value f (^) n ( x ). See Equation (3.20) for the case when n =1. Equation (3.28) is a special case ( n =1) of a general expression for Rn ( x ) which follows.
R (^) n ( ) = ( x x - x (^) 0 )( x - x (^) 1 )( x - x (^) 2 ) ⋅⋅⋅⋅( x − x (^) n - 1 )( x − x (^) n ) f [ x x , (^) n , x (^) n - 1 ,...., x (^) 1 , x 0 ] (3.32)
The ( n +1)st order finite divided-difference term f [ x x , (^) n , x (^) n - 1 , ...., x (^) 1 , x 0 ] in
Equation (3.32) cannot be evaluated at the interpolant value of x unless the function f ( x ) is known. Instead, an estimate of f [ x x , (^) n , x (^) n - 1 , ...., x (^) 1 , x 0 ]is possible if an additional data
point [ xn +1 , f ( xn +1 )] is available. The estimate is
f [ x x , (^) n , x (^) n - 1 , ...., x (^) 1 , x (^) 0 ] ≈ f [ x (^) n + 1 , x (^) n , x (^) n - 1 ,...., x (^) 1 , x 0 ] (3.33)
The right hand side of Equation (3.33) is of course bn +1 , the coefficient of the ( n +1)st order term of f (^) n +1 ( x ). Using this approximation in Equation (3.32) provides a means of estimating the error term Rn ( x ). The result is
R (^) n ( ) x ≈ ( x - x (^) 0 )( x - x (^) 1 )( x - x (^) 2 ) ⋅⋅⋅⋅ ( x − x (^) n - 1 )( x − x (^) n ) f [ x (^) n + 1 , x (^) n , x (^) n - 1 ,...., x (^) 1 , x 0 ] (3.34)
R (^) n ( ) x ≈ bn (^) + 1 ( x - x (^) 0 )( x - x (^) 1 )( x - x (^) 2 ) ⋅⋅⋅⋅ ( x − x (^) n - 1 )( x − xn ) (3.35)
R (^) 2 ( ) x ≈ ( x - x (^) 0 )( x - x (^) 1 )( x - x (^) 2 ) f [ x (^) 3 , x (^) 2 , x (^) 1 , x 0 ] (3.37)
R (^) 2 ( ) x ≈ ( x - )( 1 x - 3 )( x - 6 ) ⋅ 1 (3.38)
i xi f ( x (^) i ) f 1 [ ] f 2 [ ] f 3 [ ] f 4 [ ] f 5 [ ]
0 1 0 13 1 3 26 10 63 1 2 6 215 14 0 91 1 0 3 5 124 13 0 39 1 4 2 7 11 28 5 4 63
Table 3.11 Divided Difference Table - 6 Data Points From A Cubic Function f ( x ) = x^3 -
Using f 2 ( x ) for interpolation at x = 3.5,
f (^) 2 35 13 35 1 10 35 1 35 3
and estimating the error R 2 (3.5) from Equation (3.38),
R 2 35 35 1 35 3 35 6 1
How good is this estimate of R 2 (3.5)? We can answer this question because the function f ( x ) is known. The true error is
R (^) 2 f f 2
3
[(. ) ]
The estimate of R 2 (3.5) is exact and f (3.5) will be computed exactly from f 3 (3.5). An expected result, since f 3 ( x ) and f ( x ) are different representations of the same third order polynomial function.
In this example, the estimate of R 2 ( x ) is exact, regardless of where the interpolation occurs. The reason is simple. We learned that Rn ( x ) in Equation (3.34) is exact whenever the ( n +1)st divided difference (^) f [ x x , (^) n , x (^) n - 1 ,...., x (^) 1 , x 0 ] is constant, i.e.
independent of x. This is precisely what happened in this example, namely f [ x , x 2 , x 1 , x 0 ] is constant See if you can show why this is true whenever f ( x ) is a cubic polynomial.
Table 3.11 includes three additional data points despite the fact that only one extra point is needed to approximate R 2 ( x ). Observe that all three third order divided differences f [ x 3 , x 2 , x 1 , x 0 ], f [ x 4 , x 3 , x 2 , x 1 ], and f [ x 5 x 4 , x 3 , x 2 ] are equal.
Let's reconsider this example from a different perspective.. Suppose the function f ( x ) is in unknown and we have obtained the six data points [ xi , f ( xi )], i = 0, 1, 2, 3, 4, 5 from it given in Table 3.11. Unaware of the true function, we set out to compute all the finite divided differences, intending to employ a Newton divided-difference interpolating polynomial of order as high as 5, if necessary. In the process of completing the table, however, we notice that the column of third order divided differences has the same numerical value. The unknown function f ( x ) is exhibiting a property characteristic of third order polynomials, namely f [ x , x 2 , x 1 , x 0 ] is constant.
However, we need to be careful before concluding that the function f ( x ) is a cubic polynomial. It is correct to say there is a third order polynomial that contains our six data points. It is given by
f (^) 3 ( ) x = 13 ( x − 1 ) + 10 ( x − 1 )( x − 3 ) + 1 ( x − 1 )( x − 3 )( x − 6 ) (3.39)
Assuming that f 3 ( x ) and the true function f ( x ) are one in the same is not justified simply because f 3 ( x ) passes through the six data points sampled from f ( x ). The higher order interpolating polynomials f 4 ( x ) and f 5 ( x ) reduce to f 3 ( x ) because b 4 and b 5 are both zero.
If f 3 ( x ) and f ( x ) are the same function then subsequent points obtained from f ( x ) would produce the same result as we saw in Table 3.11, namely all third order finite divided differences would equal one. If there exists a function f ( x ), not a cubic polynomial, that happens to pass through four, five or even all six data points of the function x^3 - 1, then an additional data point from f ( x ) would result in a different value of the third order divided difference. Can you think of such a function?
Figure 3.4 is a graph of two possible functions f ( x ) along with f 3 ( x ). Note that one of the possible f ( x ) functions passes through four points of f 3 ( x ) and the other passes through five points of f 3 ( x ). The first function was generated in MATLAB by specifying
