Let $I_n$ denote the approximation of $$I = \int_a^b f(x)dx$$ obtained by applying Simpson's rule with $2n$ intervals of uniform length. Define a new approximation $$J_n=\frac{16I_{2n}-I_n}{15}.$$ Find the formula for the error $\int_a^b f(x)dx-J_n$?
We have studied Simpson's rule and proved it's error term $$-\frac{(b-a)^5}{2880n^4}f^{(4)}(c)$$ by applying Cauchy's mean value theorem repeatedly to the quotent $\frac{E(x)}{x^5}$ where $$E(x)=\int_{-x}^x f(t) dt - \frac{x}{3} \left[f(-x)+4f(0)+f(x)\right]$$ I tried apply same method but it was very confused.
Let define $$E_*(x)=\int_{-x}^{x} f(x)dx-J_n$$
Choose an interval [-c,c] for two intervals we have Simpson's approximation $$I_n=\frac{c}{3}\left(f(-c)+4f(0)+f(c)\right)$$ The we can check that $E_*(x)$ is quite smooth function, it's first, second and third derivative is zero at $x=0$
By applying Cuachy's mean value theorem repeteadly we get $$\frac{E_*(x)}{x^5}=\frac{E_*^{(1)}(u)}{5u^4}=\frac{E_*^{(2)}(v)}{20v^3}=\frac{E_*^{(3)}(z)}{60z^2}=\frac{E_*^{(4)}(t)}{120t}$$ which $0<t<z<v<u<x$ $${\frac{E_*(x)}{x^5}={\frac{\frac{1}{45}[-7tf^{(4)}(-t)-2tf^{(4)}(-\frac{t}{2})-2tf^{(4)}(\frac{t}{2})-7tf^{(4)}(t)-17f^{(3)}(-t)+16tf^{(3)}(-\frac{t}{2})-16tf^{(3)}(\frac{t}{2})+17f^{(3)}(t)]}{120t}}}$$
Another approach would be $$\int_a^b f(x)dx-I_n =-\frac{(b-a)^5}{2880n^4}f^{(4)}(c)$$ For some $c$ in $[a,b]$ $$\int_a^b f(x)dx-I_{2n} =-\frac{(b-a)^5}{2880n^4*16}f^{(4)}(k)$$ For some $k$ in $[a,b]$ $$J_n=-\frac{(b-a)^5}{2880n^4}(f^{(4)}(c)-f^{(4)}(k))$$ I couldn't proceed whatever these are right approachs.
I have looked up some texts which describes Richardson extrapolation, Taylor series and $O(h)$ notation. And nowhere I have seen eroor terror in explicit form. I am not familiar with them. So could anybody help me please.
