6. Interpolation
6.3. Newton’s Divided Difference Formula
1. Compute the following divided differences:
(a) f[x0, x1, x2, x3], where xi=i,f(x) =x2, (b) f[x0, x1, x2], where xi = 0.2i,f(x) = sinx,
(c) f[x0, x0], where x0 = 0,f(x) = sinx.
2. Letf ∈C1[a, b], andx0, x1∈(a, b),x0 ̸=x1. Show that there existsξ∈ ⟨x0, x1⟩such that f[x0, x1] =f′(ξ).
3. Let x0< x1< x2 < x3 and
P(x) =a0+a1(x−x0) +a2(x−x0)(x−x1) +a3(x−x0)(x−x1)(x−x2).
Show that
a0=P[x0], a1 =P[x0, x1], a2=P[x0, x1, x2], and a3 =P[x0, x1, x2, x3].
6.3. Newton’s Divided Difference Formula
The disadvantage of formula (6.3) is that if we add an additional mesh point, then the whole formula (6.3) must be recomputed. In this section we define a new formula for the Lagrange polynomial, and in this form it will be easy to add a new mesh point to the formula.
Suppose function values yi = f(xi) are given for i = 0,1, . . . , n. First consider the relation
Ln(x) = L0(x) + (L1(x)−L0(x)) + (L2(x)−L1(x)) +· · ·+ (Ln(x)−Ln−1(x)).
By definition,L0(x) = f(x0). Consider the differenceLi(x)−Li−1(x). It is a polynomial of degree at most i, and sinceLi and Li−1 both satisfy the interpolating equations atx0, . . ., xi−1, we have Li(xj)−Li−1(xj) = f(xj)−f(xj) = 0 (j = 0,1, . . . , i−1). But then the Fundamental Theorem of Algebra yields
Li(x)−Li−1(x) = ai(x−x0)(x−x1)· · ·(x−xi−1),
where ai ∈R. If we substitute x=xi into this relation and use for Li−1(xi) the formula (6.3), we get
f(xi)−
i−1
∑︂
k=0
f(xk) (xi−x0)· · ·(xi−xk−1)(xi−xk+1)· · ·(xi−xi−1) (xk−x0)· · ·(xk−xk−1)(xk−xk+1)· · ·(xk−xi−1)
=ai(xi−x0)· · ·(xi−xi−1).
So from this we get for ai that ai = f(xi)
(xi−x0)· · ·(xi −xi−1)− 1
(xi−x0)· · ·(xi −xi−1)
·
i−1
∑︂
k=0
f(xk) (xi−x0)· · ·(xi−xk−1)(xi−xk+1)· · ·(xi−xi−1) (xk−x0)· · ·(xk−xk−1)(xk−xk+1)· · ·(xk−xi−1)
=
i
∑︂
k=0
f(xk)
(xk−x0)· · ·(xk−xk−1)(xk−xk+1)· · ·(xk−xi)
=f[x0, x1, . . . , xi].
Therefore, the Lagrange interpolating polynomial can be written as
Ln(x) =f[x0] +f[x0, x1](x−x0) +f[x0, x1, x2](x−x0)(x−x1) +· · ·
+ f[x0, x1, . . . , xn](x−x0)(x−x1)· · ·(x−xn−1). (6.6) We have to emphasize that this is the same polynomial as (6.3), only it is given by a different formula. The polynomial given by (6.6) is called Newton’s divided difference formula or shortlyNewton polynomial.
The advantage of formula (6.6) compared to (6.3) can be seen immediately. It is easy to add a new mesh point to the formula, we have the simple correction term:
Ln+1(x) =Ln(x) +f[x0, x1, . . . , xn+1](x−x0)· · ·(x−xn).
Another advantage is that a polynomial of the form (6.6) can be easily evaluated using the Horner’s method. Furthermore, the degree of the polynomial can be determined in this form easily. If, for example, f[x0, x1, . . . , xn] ̸= 0, then the polynomial is of degree n. In Algorithm 6.13 we present the computation of the coefficients of the Newton polynomial, i.e., the values ai =f[x0, . . . , xi]. In Algorithm 6.14 we formulate a method to evaluate the Newton polynomial using Horner’s method.
Algorithm 6.13. Computation of the coefficients of the Newton polynomial INPUT: n - number of mesh points − 1
xi, (i= 0,1, . . . , n) - mesh points yi, (i= 0,1, . . . , n) - function values
OUTPUT:ai, (i= 0,1, . . . , n) - coefficients of the Newton polynomial, where ai is the coefficient of the ith-order term
for i= 0,1, . . . , ndo ai ←yi
end do
for j = 1,2, . . . , n do
for i=n, n−1, . . . , j do
ai ←(ai−ai−1)/(xi−xi−j) end do
end do
output(a0, a1, . . . , an)
6.3. Newton’s Divided Difference Formula 123 Note that Algorithm 6.13 was organized so that only those divided differences are stored by the end of the algorithm which are needed for the Newton polynomial.
Algorithm 6.14. Evaluation of the Newton polynomial INPUT: n - number of mesh points − 1
xi, (i= 0,1, . . . , n) - mesh points
ai, (i= 0,1, . . . , n) - coefficients of the Newton polynomial x - the value where we evaluate the Newton polynomial OUTPUT:y - function value of the Newton polynomial at x
y←an
for i=n−1, n−2, . . . ,0 do y←y(x−xi) +ai end do
output(y)
When we do the computation of the divided differences by hand, it is recommended to list the values of the divided differences in a triangular table as it can be seen in Table 6.1.
The numbers in the first two columns are the input data, the rest of the numbers must be computed: a number is obtained so that we take the difference of the number to the left and above, and it is divided by the difference of the appropriate mesh points xk. The numbers in frames in the diagonal of the table give the coefficients of the Newton polynomial in (6.6).
Table 6.1: Computation of the divided differences by hand x0 f(x0)
x1 f(x1) f[x0, x1]
x2 f(x2) f[x1, x2] f[x0, x1, x2]
x3 f(x3) f[x2, x3] f[x1, x2, x3] . ..
... ... ... ...
xn f(xn) f[xn−1, xn] f[xn−2, xn−1, xn] · · · f[x0, x1, . . . , xn]
Example 6.15. Consider again Example 6.2. We computeL3(x) in Newton’s divided difference form, and we evaluateL3(0). First we compute the table of divided differences:
−1 −3
1 1 2
2 3 2 0
3 29 26 12 3 This yields that
L3(x) =−3 + 2(x+ 1) + 3(x+ 1)(x−1)(x−2),
and so L3(0) =−3 + 2·1 + 3·1(−1)(−2) = 5. We can simplify this formula ofL3 and we get the same form of the polynomial as in Example 6.2: L3(x) = 3x3−6x2−x+ 5.
□
Next we study again the truncation error of the interpolation. In Section 6.1 we obtained that it has the form f(n+1)(n+1)!(ξ)(x−x0)(x−x1)· · ·(x−xn). This is certainly the same for the Newton’s divided difference form of the interpolating polynomial, but here we give a different form of the same truncation error.
Theorem 6.16. Let xi ∈ (a, b) (i = 0, . . . , n) be pairwise different mesh points and yi =f(xi)(i= 0, . . . , n). LetLn(x)be the correspondingnth degree Lagrange interpolating polynomial. Then
f(x) =Ln(x) +f[x0, x1, . . . , xn, x](x−x0)(x−x1)· · ·(x−xn).
Proof. Fixx∈(a, b) which is different from each mesh points. (Ifx=xi for somei, then the statement is clearly true.) Addx to the mesh points together with the function value f(x). Let Ln+1 be the Lagrange interpolating polynomial corresponding to the extended data set. Then we have
Ln+1(t) =Ln(t) +f[x0, x1, . . . , xn, x](t−x0)· · ·(t−xn).
Now substitution t =x proves the statement, since f(x) = Ln+1(x).
□
This form of the truncation error has no practical importance, since in order to com-pute f[x0, . . . , xn, x] the exact value of f(x) is needed. But its consequence is important.
Comparing it to Theorem 6.5 we get the following result.
Corollary 6.17. If f ∈Cn[a, b]and xi (i= 0, . . . , n) are pairwise different mesh points, then there exists ξ ∈ ⟨x0, x1, . . . , xn⟩ such that
f[x0, x1, . . . , xn] = 1
n!f(n)(ξ).
Exercises
1. Repeat Exercise 1 of Section 6.1 using the Newton’s divided difference form of the Lagrange interpolating polynomial.
2. Show that ifP is a polynomial of degreen, then P(x) =
n
∑︂
i=0
P[x0, . . . , xi]
i−1
∏︂
k=0
(x−xk).
3. Let x0, . . . , xnbe pairwise different numbers. Show that if P is a polynomial of degreen, then P[x0, . . . , xm] = 0 for all m > n.
6.4. Hermite Interpolation 125
7. Let f ∈C2. Define the following divided differences:
f[x0, x0, x1] := lim Show that the limits above exist, and the second divided differences satisfy:
(a) f[x0, x0, x1] = f[x0, x1]−f[x0, x0]
8. Check that Algorithm 6.13 gives back the coefficients of the Newton polynomial.
6.4. Hermite Interpolation
In this section we generalize the basic problem of interpolation. Let f be a differentiable function, and given mesh points xi (i = 0, . . . , n). The so-called Hermite interpolation asks to find a polynomial g(x) = c0 +c1x+· · ·+cmxm which interpolates not only the function values yi = f(xi), but also the derivative values y′i :=f′(xi). Therefore, we are looking for a polynomial g of degreem which satisfies the interpolation conditions
g(xi) = yi, g′(xi) = yi′, i= 0,1, . . . , n.