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Romberg Method says:

$$\int_a^bf(x)\,dx\;{\color{red}=}\;T_h^0+c_2 h^2+c_4 h^4+\dots$$

where $T_h^0$ is the Trapezoidal Rule:

$$T_h^0=(b-a)\left[\dfrac{f(b)+f(a)}{2}\right]$$

My question is: why is true the "$\color{red}=$" in the Romberg Method?

yemino
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  • It isn't. The reality of the situation is much more complicated than it seems at first; the relevant "buzzword" is "asymptotic expansion". – Ian Nov 08 '15 at 22:42
  • Also, that's not $T^0_h$; $T^0_h$ is a sum of terms like that one. – Ian Nov 08 '15 at 23:02
  • thanks @Ian for answer, but I don't understand your first comment. About the second comment, I'm agree, you are right. – yemino Nov 08 '15 at 23:04
  • In reality, you don't actually have equality. Instead, you're writing something called an asympotic expansion. What makes these different from an ordinary expansion is that they usually don't converge to the desired quantity; actually, they usually don't converge at all. Instead, for each value of $h$, there is an optimal number of terms on the right side to use to estimate the left side. This number of terms grows without bound as $h$ decreases to zero (under some conditions, in this case assuming that $f$ is infinitely differentiable). – Ian Nov 08 '15 at 23:11
  • But only in very special situations do you have true equality in that equation. – Ian Nov 08 '15 at 23:12
  • I'm not convinced. Why is $c_2h^2+c_4h^4+\dots$ instead of $c_1h+c_2h^2+c_3h^3+\dots$? or that is my choise? – yemino Nov 08 '15 at 23:22
  • That happens because of a symmetry argument which is specific to the trapezoidal rule. Whatever source you are using to discuss Romberg-type refinements of the trapezoidal rule should discuss this. – Ian Nov 08 '15 at 23:26
  • But in trapezoidal rule:

    $$\displaystyle\int_a^b f(x),dx=T_h^0+R_T(f)$$

    where

    $$R(f)=\dfrac{-f''(\xi)}{12}(b-a)h^2$$

    and

    $$T_h^0=h\left[\dfrac{f(a)+f(b)}{2}+\displaystyle\sum_{i=1}^nf(x_i)\right]$$

    and then,

    $$R(f)=\frac{-f''(x_0)}{12}(b-a)h^2{\color{red}=},c_2h^2+c_4h^4+...$$

    There are something that I can't see. (the "$\color{red}=$")

     – yemino Nov 09 '15 at 00:24
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    As I said, the equality does not actually hold. We pretend it does to get a sequence of methods, which for smooth integrands have increasing orders. But actually there is no reason to believe that this sequence of methods converges for any fixed $h$. – Ian Nov 09 '15 at 00:47
  • What is more or less going on in going from one form to the next is that we notice that $x_0$ depends on $h$. If we approximate it as being independent of $h$, then we get a $c_2 h^2$ term. The remaining terms are corrections to the approximation that it is independent of $h$. You should look into this symmetry argument (based on the Euler-Maclaurin summation formula, as I recall) to see why there is no $h^3$ term in the expansion. – Ian Nov 09 '15 at 00:51
  • We can reformulate this using notation which is more tedious but technically correct. Basically, we start with the trapezoidal rule. This is second order. That means that the error in $T_h$ should be approximately four times the error in $T_{h/2}$ (with the same sign). That is, $T_h-I \approx 4(T_{h/2}-I)$. So $T_h - 4T_{h/2} \approx -3I$. So $\frac{4T_{h/2}-T_h}{3}$ should have a higher order than $2$. To get an even higher order method, we need to calculate the order of this method. It turns out to be $4$, so we repeat the procedure, and can continue ad infinitum. – Ian Nov 09 '15 at 01:00

1 Answers1

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The more rigorous way of writing what is usually written in describing Romberg integration and more generally Richardson extrapolation goes like this. We start with the method $T^0_h$. This has second order. Therefore $T^0_h-I \approx 4(T^0_{h/2}-I)$ (here $I$ is the true integral). Therefore $T^1_h := \frac{4 T^0_{h/2}-T^0_h}{3}$ should be a higher order method than $T^0_h$. This method, which is actually the composite Simpson rule, turns out to be 4th order (which you should really prove separately, in part because it requires a different regularity hypothesis than the previous one). So we can go again: $T^1_h-I \approx 16 (T^1_{h/2}-I)$ so $T^2_h := \frac{16 T^1_{h/2} - T^1_h}{15}=\frac{64 T^0_{h/4} - 20 T^0_{h/2} + T^0_h}{45}$ should be a higher order method than $T^1_h$. And so forth. All that is required at each step is knowing what the order of the method you have now is, and whatever additional assumptions are required to ensure that the order of the new method is actually higher. (In the case of integration, these additional assumptions are usually regularity.)

There is no real equality in the expansion of the trapezoidal error that is written in the OP, except under special circumstances. Instead this expansion is merely an asymptotic expansion (and even then, it only holds in the sense of asymptotic expansions for smooth integrands).

Ian
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