<s>[INST] Smallest number of points required to estimate value of integral Find the smallest number of points required to estimate the value of $$\int_{0}^{1} sin(x^2\pi)dx$$ with error less than $10^{-8}$ using the Composite Simpson method. In the estimation n, use the absolute maximum of the corresponding derivative of $(1+x^2)^{-1}$

In this case I would find the maximum of the fourth derivative of $sin(x^2\pi)$ and then plug it into the following formula for Composite Simpson error: $ 10^{-8} ≥ $ $\dfrac {|f^4|(b-a)h^4}{180} $ and solve for h. Then, I would use $h= \dfrac{(b-a)}{n}$ where n is the number of points, and solve for n to get the number of points. 
HOWEVER, the last sentence of the question confuses me because I do not know what it means to use the absolute maximum of the corresponding derivative for n.
 [/INST] It might be easier than a direct computation of the sequence of derivative expressions to find the coefficient of $s^4$ in $f(x+s)=\sin(\pi(x+s)^2)$.
\begin{align}
\sin(\pi x^2 + \pi(2x+s)s)
&=\sin(\pi x^2)\cos(\pi(2x+s)s)+\cos(\pi x^2)\sin(\pi(2x+s)s)
\\
&=\sin(\pi x^2)\left[1-\frac12\pi^2(2x+s)^2s^2+\frac1{24}\pi^4(2x)^2s^4+O(s^5)\right]
\\&\qquad
+\cos(\pi x^2)\left[\pi(2x+s)s-\frac16\pi^3(2x+s)^3s^3+O(s^5)\right]
\end{align}
Thus
\begin{align}
\frac1{24}f^{(4)}(x)
&=\sin(\pi x^2)\left[-\frac12\pi^2+\frac1{24}\pi^4(2x)^2\right]
  +\cos(\pi x^2)\left[-\frac12\pi^3(2x)^2\right]
\\
|f^{(4)}(x)|&\le4\pi^2\sqrt{(3-\pi^2x^2)^2+(12\pi x^2)^2}
\\
&\le4\pi^2\sqrt{9+138\pi^2+\pi^4}\le160\pi^2\le1600
\end{align}
giving $n\sim \sqrt[4]{10^8\cdot 1600/180}\le100\cdot\sqrt{3}\le 175$. An adaptive strategy finds a sequence of $147$ nodes with a minimal step of $0.00390625$ ($n=256$ if filled out) sufficient to get this accuracy.

</s>