If the bound is to hold with finite

c

, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of

t = x + 1

gives

a_{- 1} + a_{0} + a_{1} + a_{2} a_{0} + 2 a_{1} + 3 a_{2} a_{0} + 4 a_{1} + 9 a_{2} a_{0} + 8 a_{1} + 27 a_{2} = 0, = 0, = 2, = 0.

(1)

Solving,

a_{- 1} = 2, a_{0} = - 5, a_{1} = 4, a_{2} = - 1 .

(2)

This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional

L

annihilates all polynomials of degree below

r

, then for

f \in C^{r} [a, b]

L (f) = \int_{a}^{b} K (t) f^{(r)} (t) d t, K (t) = L (\frac{( x - t ) _{+}^{r - 1}}{( r - 1 )!}) .

(3)

Here

r = 4

and

L = e

. Under the stated nonnegativity assumption,

∣ e (f) ∣ \leq (\int_{- 1}^{2} K (t) d t) max_{[- 1, 2]} ∣ f^{(4)} ∣.

(4)

The integral equals

e (q)

for

q (x) = (x + 1)^{4} /24

, since

q^{(4)} = 1

. Now

q^{''} (- 1) = 0

and

η (q) = \frac{- 5 + 4 \cdot 16 - 81}{24} = - \frac{11}{12} .

(5)

Therefore

\int_{- 1}^{2} K (t) d t = e (q) = \frac{11}{12} .

(6)

Equality is attained by

q

, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

c = \frac{11}{12} .

(7)

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