Richardson Extrapolation for Euler Values

Why Combine Two Euler Approximations?

Forward Euler gives an approximation to a solution of an initial-value problem by taking repeated tangent-line steps. When the step size is reduced, the endpoint approximation generally improves. The earlier material on global error establishes that, under suitable conditions, forward Euler has first-order convergence: its endpoint error is proportional to the step size to leading order.

That leading error has a predictable scaling. If the step size is halved, the leading error is approximately halved as well. Richardson extrapolation uses that relationship to combine the two endpoint values so that the leading error cancels. The result is a more accurate endpoint estimate without simply choosing an even smaller step size.

Euler Endpoint Values and the Extrapolation Formula

Consider an initial-value problem \(y'=f(t,y)\), \(y(t_0)=y_0\), and a target time \(T>t_0\). Let \(h=(T-t_0)/N\) for a positive integer \(N\), so the interval is divided into \(N\) equal steps. Write \(Y_h\) for the Euler approximation at \(T\), starting from \(Y_0=y_0\) and using the updates

Now repeat the computation on the same interval with step size \(h/2\), which uses \(2N\) steps. Denote its endpoint value by \(Y_{h/2}\). Both computations approximate the same solution at the same time \(T\), and they start from the same initial data.

The weight 2 is not an arbitrary adjustment. It is chosen to eliminate the first-order error term. This depends on comparing matching computations: if the initial values, differential equation, or target time differ, the cancellation argument does not apply as stated.

Why the Leading Error Cancels

For a sufficiently smooth solution and right-hand side, and with a fixed initial time and endpoint, the Euler endpoint values have a first-order asymptotic expansion of the form

where \(C\) depends on the differential equation and the interval, but not on the step size in this expansion. Here \(O(h^2)\) means a remainder whose absolute value is bounded by a constant times \(h^2\) for sufficiently small \(h\). The same expansion applied with step size \(h/2\) gives

Multiplying the refined value by 2 and subtracting the coarse value gives the cancellation directly:

The remainders still have order \(h^2\): multiplying an \(O(h^2)\) term by a fixed number does not change its order. The first-order term disappears because its coefficient is the same \(C\) in both expansions, while its size is halved in the refined computation.

Where the Expansion Comes From

The scaling of the leading error is linked to the local Taylor expansion of the exact solution. For a smooth solution, one exact step satisfies

Euler takes the first two terms and omits the term involving \(h^2y''(t)/2\). Thus the one-step discrepancy is of order \(h^2\). Across about \((T-t_0)/h\) steps, these local discrepancies accumulate into a global error of order \(h\), as established in the earlier material on local and global error.

For the extrapolation argument, the important refinement is that this global error has a leading coefficient that is stable as the step is refined. Smoothness allows the accumulated leading contributions to be represented by \(Ch\); the remaining terms are of order \(h^2\). Applying this expansion on the same interval with \(h/2\) gives the same leading coefficient \(C\), scaled by one-half. If the solution or right-hand side lacks the regularity needed for this expansion, the claimed second-order improvement need not follow.

Worked Examples

Common Mistake: Reversing the Combination

A frequent error is to subtract the refined Euler value from twice the coarse value, using \(2Y_h-Y_{h/2}\). The first-order terms then reinforce rather than cancel. Substituting the error expansions shows why:

The remaining term \(\frac32Ch\) is still first order. The correct combination is \(2Y_{h/2}-Y_h\): the finer value receives the factor of 2 because its leading error is half as large. A useful check is to substitute \(Y_h=y(T)+Ch\) and \(Y_{h/2}=y(T)+Ch/2\) into any proposed formula and verify that the coefficient of \(C\) is zero.

What Richardson Extrapolation Does—and Does Not Do

Richardson extrapolation produces a new endpoint estimate by combining two completed Euler computations. It is not an Euler update with a modified step, and it does not provide values along the interval between \(t_0\) and \(T\). The two input computations must approximate the same endpoint from the same initial-value problem, with the refined mesh using half the coarse step.

Its second-order conclusion is conditional on a suitable asymptotic error expansion. First-order convergence alone says that the error is bounded in magnitude by a constant times \(h\); the cancellation argument additionally uses a stable leading coefficient \(C\) and a remainder of order \(h^2\). If those assumptions fail, the formula may not improve the order. Even when they hold, the constant in the remaining error can be large, and rounding or other numerical effects can influence computed digits.

The method matters because it extracts more accuracy from the known error pattern of Euler's method. The essential reasoning is not merely “compute twice and combine,” but rather identify the leading error's dependence on \(h\), then choose weights that remove that term.