Calculus & Analysis

Richardson Extrapolation

Combine two central differences to cancel their leading error and see when a higher-order error remains.

By Riano · Published · Updated · Report a correction

Richardson Extrapolation

Richardson extrapolation combines two central-difference derivative approximations evaluated at nested step sizes h and h/2 to eliminate the leading O(h2) truncation error term:

D(h) = [f(x + h) - f(x - h)] / (2h) = f'(x) + c2h2 + c4h4 + ...

D(h/2) = [f(x + h/2) - f(x - h/2)] / h = f'(x) + c2(h2/4) + c4(h4/16) + ...

Taking the weighted combination R = (4D(h/2) - D(h)) / 3 cancels the c2h2 term identically, leaving fourth-order remainder R = f'(x) - (1/4)c4h4 + ....

Worked Example: Cubic Cancellation vs. Quintic Fourth-Order Remainder

Compare the two-level Richardson formula R = (4D(h/2) - D(h)) / 3 on a cubic polynomial (where fourth-order terms vanish) versus a quintic polynomial:

  1. Cubic cancellation (f(x) = x3 at x = 1, h = 1):
    D(1) = (23 - 03) / 2 = 4, and D(1/2) = (1.53 - 0.53) / 1 = 3.375 - 0.125 = 3.25 = 13/4.
    Extrapolating gives R = (4 · (13/4) - 4) / 3 = 9 / 3 = 3, which matches f'(1) = 3(1)2 = 3 with Absolute error = 0 because c4 = f(5)(x)/120 = 0.
  2. Quintic fourth-order residual (f(x) = x5 at x = 1, h = 1):
    For f(x) = x5, D(h) = 5x4 + 10x2h2 + h4, so R = 5x4 - h4/4. At x = 1, h = 1: D(1) = 16, D(1/2) = 121/16 = 7.5625, and R = 5 - 1/4 = 19/4 = 4.75 vs. exact derivative 5, leaving error 1/4 = 0.25. Even at x = 0 where f'(0) = 0, setting h = 1/4 gives R = -(1/4)4/4 = -1/1024 = -0.0009765625.

Interactive Experiment

Test stencil cancellation and fourth-order error scaling in the Richardson Extrapolation interactive workspace:

  1. Select Function = Cubic, Probe x = 1, and Step h = 1. Verify that Central h = 4 and Central h over2 = 3.25 combine into Extrapolated derivative = 3 with Absolute error = 0.
  2. Switch Function to Quintic with Probe x = 1 and Step h = 1. Observe that Extrapolated derivative = 4.75 underestimates Exact derivative = 5 by 1/4 = 0.25.
  3. Set Function = Quintic, Probe x = 0, and Step h = 1/4. Note that Extrapolated derivative is negative (-1/1024 = -0.0009765625) even though Exact derivative = 0.
  4. Select Function = Sine at Probe x = 0 (h = 1/2) and Probe x = 1 (h = 1/4) to see how Richardson extrapolation reduces trigonometric derivative error by orders of magnitude.

Limitations and Assumptions

Richardson extrapolation requires f to be at least five times continuously differentiable so the Taylor expansion with even powers of h holds. It is a fixed-step numerical difference formula, not symbolic differentiation: higher-order terms Beginning at -(1/4)c4h4 survive whenever f(5)(x) ≠ 0. In floating-point arithmetic, shrinking h indefinitely eventually amplifies subtractive cancellation error at rate O(εmach / h), so smaller steps do not improve accuracy without bound.

References

The asymptotic Taylor error expansion of symmetric central-difference stencils and the derivation of Richardson extrapolation coefficients are detailed in the following reference:

Experience it interactively

Adjust parameters, observe in real time, and build deep intuition with Riano’s interactive Richardson Extrapolation module.

Try Richardson Extrapolation on Riano →

More in Calculus & Analysis