In interpolating with $n + 1$ equally spaced nodes on an interval, we could use $$x_i = a + (2i + 1)\frac h 2,$$ where $0 \leqslant i \leqslant n − 1$ and $h = \dfrac {b-a} n$. What bound can be given for $$\prod_{i=0}^{n}|x − x_i|$$ when $a \leqslant x \leqslant b$?
The solution given is $$\prod_{i=0}^{n-1}|x − x_i| \leqslant \frac {h^n(2n)!} {2^{2n}n!}$$
I know that the property for interpolation error says: $$|f(x)−p(x)| \leqslant \frac{Mh^{n + 1}}{4(n + 1)}$$ where $h = \dfrac{b − a}{n}$ is the spacing between nodes. I believe represents the upper bound of $f^{n+1}$ (derivative here for $n+1$) how would I apply it possibly here?
I think there is a typo and you want the upper bound on $$\prod_{i=0}^{n-1}\left|x-x_i\right|$$ Because there is no $x_n$ defined. I claim it happens when $x=a$ and $$\prod_{i=0}^{n-1}\left|a-x_i\right|=\prod_{i=0}^{n-1}\left[\left(i+\frac12\right)h\right]$$ Because for $0\le i\le n-1$ if $a+ih\le x<a+\left(i+\frac12\right)h$ then $$\begin{align}\prod_{i=0}^{n-1}\left|x-x_i\right|&=\prod_{j=0}^{i-1}\left[x-a-\left(j+\frac12\right)h\right]\prod_{j=i}^{n-1}\left[a+\left(j+\frac12\right)h-x\right]\\ &\le\prod_{j=0}^{i-1}\left[\left(i+\frac12\right)h-\left(j+\frac12\right)h\right]\prod_{j=i}^{n-1}\left[\left(j+\frac12\right)h-ih\right]\\ &\le\prod_{k=n-i}^{n-1}\left[\left(k+i+1-n\right)h\right]\prod_{k=0}^{n-1-i}\left[\left(k+\frac12\right)h\right]\\ &\le\prod_{k=n-i}^{n-1}\left[\left(k+\frac12\right)h\right]\prod_{k=0}^{n-1-i}\left[\left(k+\frac12\right)h\right]\\ &=\prod_{i=0}^{n-1}\left[\left(i+\frac12\right)h\right]\end{align}$$ EDIT: I admit the above was a little impenetrable. As an example, suppose $a=0$, $b=10$, $n=10$, and $x=\frac{13}4$. Then our regrouping of distances may be summarized as: $$\begin{array}{|r|c|c|c|c|c|c|c|c|c|c|}\hline \text{Location}&\color{red}{\frac12}&\color{red}{\frac32}&\color{red}{\frac52}&\color{blue}{\frac72}&\color{blue}{\frac92}&\color{blue}{\frac{11}2}&\color{blue}{\frac{13}2}&\color{blue}{\frac{15}2}&\color{blue}{\frac{17}2}&\color{blue}{\frac{19}2}\\\hline \text{Distance}&\color{blue}{\frac14}&\color{blue}{\frac54}&\color{blue}{\frac94}&\color{blue}{\frac{13}4}&\color{blue}{\frac{17}4}&\color{blue}{\frac{21}4}&\color{blue}{\frac{25}4}&\color{red}{\frac34}&\color{red}{\frac74}&\color{red}{\frac{11}4}\\ \hline\end{array}$$ So we took the points on our left and put them in reversed order at the end of the list so now each new distance is paired with a longer original distance.
For $0\le i\le n-1$ if $a+\left(i+\frac12\right)h<x\le(i+1)h$, $$\begin{align}\prod_{i=0}^{n-1}\left|x-x_i\right|&=\prod_{j=0}^i\left[x-a-\left(j+\frac12\right)h\right]\prod_{j=i+1}^{n-1}\left[a+\left(j+\frac12\right)h-x\right]\\ &\le\prod_{j=0}^i\left[\left(i+1\right)h-\left(j+\frac12\right)h\right]\prod_{j=i+1}^{n-1}\left[\left(j+\frac12\right)h-\left(i+\frac12\right)h\right]\\ &\le\prod_{k=0}^i\left[\left(k+\frac12\right)h\right]\prod_{k=i+1}^{n-1}\left[\left(k-i\right)h\right]\\ &\le\prod_{k=0}^i\left[\left(k+\frac12\right)h\right]\prod_{k=i+1}^{n-1}\left[\left(k+\frac12\right)h\right]\\ &=\prod_{i=0}^{n-1}\left[\left(i+\frac12\right)h\right]\end{align}$$ EDIT: As in our previous example, but now let $a=0$, $b=10$, $n=10$, and $x=\frac{11}4$: $$\begin{array}{|r|c|c|c|c|c|c|c|c|c|c|}\hline \text{Location}&\color{red}{\frac12}&\color{red}{\frac32}&\color{red}{\frac52}&\color{blue}{\frac72}&\color{blue}{\frac92}&\color{blue}{\frac{11}2}&\color{blue}{\frac{13}2}&\color{blue}{\frac{15}2}&\color{blue}{\frac{17}2}&\color{blue}{\frac{19}2}\\\hline \text{Distance}&\color{red}{\frac14}&\color{red}{\frac54}&\color{red}{\frac94}&\color{blue}{\frac34}&\color{blue}{\frac74}&\color{blue}{\frac{11}4}&\color{blue}{\frac{15}4}&\color{blue}{\frac{19}4}&\color{blue}{\frac{23}4}&\color{blue}{\frac{27}4}\\\hline\end{array}$$ This time we had to put the points on our left in reversed order at the beginning of the list to achieve a pairing of shorter new distances with longer old distances.
This maximal product evaluates to $$\begin{align}\prod_{i=0}^{n-1}\left|a-x_i\right|&=\prod_{i=0}^{n-1}\left[\left(i+\frac12\right)h\right]\\ &=\prod_{i=0}^{n-1}\left[\frac{(2i+1)(2i+2)h}{2(2i+2)}\right]\\ &=\frac{(2n)!h^n}{2^{2n}n!}\end{align}$$