Romberg Method (quadrature) 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?
 A: 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).
