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Forward and central differences have first- and second-order error bounds

Statement

Let I⊆R be open, x∈I, and h>0. For the forward bound assume [x,x+h]⊂I, f∈C2 on a neighborhood of that interval, and ∣f′′(y)∣≤M2 there. For the central bound assume [x−h,x+h]⊂I, f∈C3 on a neighborhood of that interval, and ∣f′′′(y)∣≤M3 there. Then

∣Dh+f(x)−f′(x)∣≤M2h2,∣Dh0f(x)−f′(x)∣≤M3h26.

If each function evaluation has absolute error at most δ≥0 as in Forward and central difference stencils with bounded evaluation errors, then

∣D^h+f(x)−f′(x)∣≤M2h2+2δh,∣D^h0f(x)−f′(x)∣≤M3h26+2δh.

For the central stencil the evaluation contribution is in fact at most δ/h; the displayed 2δ/h is a uniform conservative bound for both stencils.

Facts & Assumptions

Given: The interval, step, smoothness, derivative bounds and evaluation-error hypotheses in the statement.

[F1]

Forward and central difference stencils with bounded evaluation errors: Dh+f(x)=(f(x+h)−f(x))/h and Dh0f(x)=(f(x+h)−f(x−h))/(2h); each inexact value is f(y)+ey with ∣ey∣≤δ.

[F2]

The Lagrange and Cauchy forms of Taylor's remainder and Taylor polynomials and their remainders: for a Cn+1 function on the interval, f(y) is its degree-n Taylor polynomial at a plus f(n+1)(ξ)(y−a)n+1/(n+1)! for some intermediate ξ.

[F3]

Absolute value in an ordered field and Basic properties of the absolute value: absolute value is nonnegative, and ∣−r∣=∣r∣ for real r.

[F4]

The triangle inequality: ∣r+s∣≤∣r∣+∣s∣ for real r,s.

Proof

Given: The hypotheses listed above.

1.1givenF1F2algebra

Apply [F2] with n=1, a=x, and y=x+h. Then f(x+h)=f(x)+hf′(x)+h2f′′(ξ)/2 for some ξ∈(x,x+h). Substituting into [F1] gives Dh+f(x)−f′(x)=hf′′(ξ)/2, so the forward truncation bound follows.

1.2givenF1F2F3F4algebra

Apply [F2] with n=2 to y=x+h and y=x−h. Subtracting the two expansions cancels the equal even-order terms and gives Dh0f(x)−f′(x)=h2(f′′′(ξ+)+f′′′(ξ−))/12 for intermediate points ξ+∈(x,x+h) and ξ−∈(x−h,x). The derivative bound and [F3]–[F4] give magnitude at most M3h2/6.

2.1givenF1F3F4step 1.1algebra

Write the forward evaluation errors as e+=ex+h and e0=ex. By [F1], D^h+−Dh+=(e+−e0)/h; [F3]–[F4] bound its magnitude by 2δ/h. The triangle inequality with step 1.1 proves the forward inexact bound.

3.1givenF1F3F4step 1.2algebra∎

Write the central errors as e+=ex+h and e−=ex−h. Then D^h0−Dh0=(e+−e−)/(2h), whose magnitude is at most δ/h by [F1], [F3] and [F4], hence at most 2δ/h. Combining with step 1.2 proves the stated central inexact bound.

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