in this book or here on the page $58$ I was trying to obtain the compound Simpson's rule $(2.2.4)$ from the Simple Simpson's approximation as in $(2.2.2)$. I understand that the compound rule on $-1,...,1$ is obtained by the simple rule applied on each interval $-1,...,0$ and $0,...,1$ separately.
To this end I divided the interval like this: $a=x_0<...<x_{2n}=b$ with $x_0=-1=a$, $x_1=0$ and $x_2=1=b$.
Now I was left with these two partial results and got stuck with how to get rid of $f$-values at fractional points. Perhaps some linear transformation would help:
$$\frac{1}{6}(f(-1)+4f(\frac{-1}{2})+f(0))$$ and $$\frac{1}{6}(f(0)+4f(\frac{1}{2})+f(1))$$
If somehow adding these two expressions they should yield $$\frac{h}{3}(f_0+4f_1+f_2),$$ with $h=1$ but I'm unable to get correctly get rid of fractional values.
Also there is an additional term $2\cdot f(0)=f_2$ while summing up.
The notation change might be somewhat disorienting.
The point sequence where the function is evaluated is $a=x_0<x_1<x_1<...<x_{2n}=b$, $x_{i+1}=x_i+h$.
The segmentation for the composite Simpson rule are the $n$ segments with borders at $a=x_0,x_2,x_4,...,x_{2n}=b$, this is why the point sequence has $2n+1$ points. The length of the segments is $2h$, thus giving a factor of $\frac{2h}{6}=\frac{h}3$
Thus for 2 segments in $[a,b]=[-1,1]$ you get $x_0=-1$, $x_2=0$, $x_4=1$ with the midpoints $x_1=-\frac12$ and $x_3=\frac12$. There is no contradiction, the resulting formula is simply $$ \frac{1/2}3(f_0+4f_1+2f_2+4f_3+f_4),~~~f_k=f(x_k). $$