Fixed point iteration is sometimes used for solving the implicit Euler method

$$y_{n+1}^\mathrm{(I)}= y_n + hf(y_{n+1}^\mathrm{(I)})$$

with the iteration procedure starting with explicit Euler

$$y_{n+1}^\mathrm{(E)}= y_n + hf(y_n)$$

My question is, once we achieve convergence on the iteration and we have both the value computed implicitly and the explicit value that we used to start the iteration, cannot we combine both values to get a more accurate result?

$$y_{n+1}^\mathrm{(A)} := \frac{y_{n+1}^\mathrm{(I)} + y_{n+1}^\mathrm{(E)}}{2} = y_n + h \frac{f(y_{n+1}^\mathrm{(I)}) + f(y_n)}{2}$$

because the above average is similar to the trapezoidal method

$$y_{n+1}^\mathrm{(T)}= y_n + h \frac{f(y_{n+1}^\mathrm{(T)}) + f(y_n)}{2}$$

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    $\begingroup$ Error analysis is a pretty complicated subject. There is no simple answer to your question. Is this going to be used on sampled data obtained as the result of an experiment or on exact data? Is the data periodic or exponential? And so on. $\endgroup$ – steven gregory Aug 3 '18 at 10:56
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    $\begingroup$ The trapezoidal method's solution is not the average of the explicit Euler solution and the implicit Euler solution. This becomes clear if you don't abbreviate the right-hand sides: $y_{n+1}^{\rm(T)} = y_n + \frac h2(f(y_n) + f(y_{n+1}^{\rm(T)})) \ne y_n + \frac h2(f(y_n) + f(y_{n+1}^{\rm(I)}))$. Your method is probably also second-order accurate like the trapezoidal method, but unlike the trapezoidal method it is not A-stable. $\endgroup$ – Rahul Aug 3 '18 at 12:24
  • $\begingroup$ @Rahul Nice catch! Edited and eliminated the mistake $\endgroup$ – juanrga Aug 4 '18 at 11:52

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