14B (Numerical Analysis)

Exactness on the polynomial basis

1, x, x^{2}

requires

a_{0} + a_{1} + a_{2} = 0, - a_{0} + a_{2} = 0, a_{0} + a_{2} = 2.

Solving gives

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

Thus the approximation is the unit-spacing centered second difference,

μ (f) = f (- 1) - 2 f (0) + f (1)

b

The error

e (f) = f^{''} (0) - f (- 1) + 2 f (0) - f (1)

is linear and annihilates all degree-at-most-two polynomials. The Peano kernel theorem with derivative order three gives

e (f) = \int_{- 1}^{1} K (t) f^{'''} (t) d t, K (t) = e_{x} (\frac{( x - t ) _{+}^{2}}{2}) .

For

t \neq = 0

, the second derivative of the truncated quadratic at

x = 0

1_{{t < 0}}

. Its values at

x = - 1, 0, 1

then give

K (t) = 1_{{t < 0}} + (- t)_{+}^{2} - \frac{( 1 - t ) ^{2}}{2} = {\frac{1}{2} (1 + t)^{2}, - \frac{1}{2} (1 - t)^{2}, - 1 < t < 0, 0 < t < 1.

The value at the single point

t = 0

\int_{- 1}^{1} ∣ K (t) ∣ d t = \frac{1}{2} \int_{- 1}^{0} (1 + t)^{2} d t + \frac{1}{2} \int_{0}^{1} (1 - t)^{2} d t = \frac{1}{3} .

Consequently

∣ e (f) ∣ \leq \frac{1}{3} ∥ f^{'''} ∥_{C [- 1, 1]} .

Its changing sign explains why cubic polynomials also have zero error: their third derivative is constant and

\int_{- 1}^{1} K = 0

. No fourth derivative is needed for the claimed bound.

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