Jim LambersMAT 460/560

Fall Semester 2009-10Lecture 18 Notes

These notes correspond to Section 3.2 in the text.

Interpolation Using Equally Spaced Points

Suppose that the interpolation points x0, x1, . . . , xn are equally spaced; that is, xi = x0 + iℎ forsome positive number ℎ. In this case, the Newton interpolating polynomial can be simplified, sincethe denominators of all of the divided differences can be expressed in terms of the spacing ℎ. If werecall the forward difference operator Δ, defined by

Δxk = xk+1 − xk,

where {xk} is any sequence, then the divided differences f [x0, x1, . . . , xk] are given by

f [x0, x1, . . . , xk] =1

k!ℎkΔkf(x0).

The interpolating polynomial can then be described by the Newton forward-difference formula

pn(x) = f [x0] +n∑

k=1

(sk

)Δkf(x0),

where the new variable s is related to x by

s =x− x0

ℎ,

and the extended binomial coefficient

(sk

)is defined by

(sk

)=

s(s− 1)(s− 2) ⋅ ⋅ ⋅ (s− k + 1)

k!,

where k is a nonnegative integer.

Example We will use the Newton forward-difference formula

pn(x) = f [x0] +

n∑k=1

(sk

)Δkf(x0)

to compute the interpolating polynomial p3(x) that fits the data

1

i xi f(xi)

0 −1 31 0 −42 1 53 2 −6

In other words, we must have p3(−1) = 3, p3(0) = −4, p3(1) = 5, and p3(2) = −6. Note that theinterpolation points x0 = −1, x1 = 0, x2 = 1 and x3 = 2 are equally spaced, with spacing ℎ = 1.

To apply the forward-difference formula, we define s = (x− x0)/ℎ = x + 1 and compute(s1

)= s

= x + 1,(s2

)=

s(s− 1)

2

=x(x + 1)

2,(

s3

)=

s(s− 1)(s− 2)

6

=(x + 1)x(x− 1)

6,

f [x0] = f(x0)

= 3,

Δf(x0) = f(x1)− f(x0)

= −4− 3

= −7,

Δ2f(x0) = Δ(Δf(x0))

= Δ[f(x1)− f(x0)]

= [f(x2)− f(x1)]− [f(x1)− f(x0)]

= f(x2)− 2f(x1) + f(x0)

= 5− 2(−4) + 3,

= 16

Δ3f(x0) = Δ(Δ2f(x0))

= Δ[f(x2)− 2f(x1) + f(x0)]

= [f(x3)− f(x2)]− 2[f(x2)− f(x1)] + [f(x1)− f(x0)]

= f(x3)− 3f(x2) + 3f(x1)− f(x0)

= −6− 3(5) + 3(−4)− 3

2

= −36.

It follows that

p3(x) = f [x0] +

3∑k=1

(sk

)Δkf(x0)

= 3 +

(s1

)Δf(x0) +

(s1

)Δ2f(x0) +

(s2

)Δ3f(x0)

= 3 + (x + 1)(−7) +x(x + 1)

216 +

(x + 1)x(x− 1)

6(−36)

= 3− 7(x + 1) + 8(x + 1)x− 6(x + 1)x(x− 1).

Note that the forward-difference formula computes the same form of the interpolating polynomialas the Newton divided-difference formula. □

If we define the backward difference operator ∇ by

∇xk = xk − xk−1,

for any sequence {xk}, then we obtain the Newton backward-difference formula

pn(x) = f [xn] +n∑

k=1

(−1)k(−sk

)∇kf(xn),

where s = (x− xn)/ℎ, and the preceding definition of the extended binomial coefficient applies.

Example We will use the Newton backward-difference formula

pn(x) = f [xn] +n∑

k=1

(−1)k(−sk

)∇kf(xn)

to compute the interpolating polynomial p3(x) that fits the data

i xi f(xi)

0 −1 31 0 −42 1 53 2 −6

In other words, we must have p3(−1) = 3, p3(0) = −4, p3(1) = 5, and p3(2) = −6. Note that theinterpolation points x0 = −1, x1 = 0, x2 = 1 and x3 = 2 are equally spaced, with spacing ℎ = 1.

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To apply the backward-difference formula, we define s = (x− x3)/ℎ = x− 2 and compute(−s1

)= −s

= −(x− 2),(−s2

)=−s(−s− 1)

2

=s(s + 1)

2

=(x− 2)(x− 1)

2,(

−s3

)=−s(−s− 1)(−s− 2)

6

= −s(s + 1)(s + 2)

6

= −(x− 2)(x− 1)x

6,

f [x3] = f(x3)

= −6,

∇f(x3) = f(x3)− f(x2)

= −6− 5

= −11,

∇2f(x3) = ∇(∇f(x3))

= Δ[f(x3)− f(x2)]

= [f(x3)− f(x2)]− [f(x− 2)− f(x1)]

= f(x3)− 2f(x2) + f(x1)

= −6− 2(5) +−4,

= −20

∇3f(x3) = ∇(∇2f(x3))

= ∇[f(x3)− 2f(x2) + f(x1)]

= [f(x3)− f(x2)]− 2[f(x2)− f(x1)] + [f(x1)− f(x0)]

= f(x3)− 3f(x2) + 3f(x1)− f(x0)

= −6− 3(5) + 3(−4)− 3

= −36.

4

It follows that

p3(x) = f [x3] +

3∑k=1

(−1)k(−sk

)∇kf(x3)

= −6−(−s1

)∇f(x3) +

(−s1

)∇2f(x3)−

(−s2

)∇3f(x3)

= −6− [−(x− 2)](−11) +(x− 2)(x− 1)

2(−20)− −(x− 2)(x− 1)x

6(−36)

= −6− 11(x− 2)− 10(x− 2)(x− 1)− 6(x− 2)(x− 1)x.

Note that the backward-difference formula does not compute the same form of the interpolatingpolynomial pn(x) as the Newton divided-difference formula. Instead, it computes a different Newtonform of pn(x) given by

pn(x) =

n∑j=0

f [xj , xj+1, . . . , xn]

n∏i=j+1

(x− xi)

= f [xn] + f [xn−1, xn](x− xn) + f [xn−2, xn−1, xn](x− xn−1)(x− xn) +

⋅ ⋅ ⋅+ f [x0, x1, . . . , xn](x− x1)(x− x2) ⋅ ⋅ ⋅ (x− xn−1)(x− xn).

The divided-differences f [xj , . . . , xn], for j = 0, 1, . . . , n, can be obtained by constructing a divided-difference table as in the first example. These divided differences appear in the bottom row of thetable, whereas the divided differences used in the forward-difference formula appear in the top row.□

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