I've been playing with Simpson's rule and a thing came up to my mind. The rectangular rule is a 0th order polynomial approximation of integration. The trapezoidal rule is 1st. Simpson's rule is 2nd. Then what about nth? I've been working on this issue more than 2 weeks and couldn't get the answer. How could I solve this?
What I did is below.
approach 1
A nth order polynomial function passing $n+1$ points $(x_i, y_i) (0 \leq i\leq n)$ can be written by Lagrange polynomial as
$$ g(x) = \sum_{j=0}^{n} y_i \displaystyle \prod_{i=0 \atop i \neq j}^{n} \frac{x-x_i}{x_j - x_i}$$
So, if $y=f(x)$, $g(x)$ is a polynomial approximation of $f(x)$. So, the integration of $f(x)$ can be written as follow.
$$ \int_{x_0}^{x_n} f(x) dx \simeq \int_{x_0}^{x_n} g(x) dx = \int_{x_0}^{x_n} \sum_{j=0}^{n} f(x_i) \displaystyle \prod_{i=0 \atop i \neq j}^{n} \frac{x-x_i}{x_j - x_i} dx \\ = \sum_{j=0}^{n} f(x_i) \displaystyle \prod_{i=0 \atop i \neq j}^{n} (x_j - x_i)^{-1} \int_{x_0}^{x_n} \prod_{i=0 \atop i \neq j}^{n} (x-x_i) dx $$
Here, we suppose $x_i = x_0 + hi$ where $h$ is constant, $z = \frac{x - x_0}{h}$ and $z_i = \frac{x_i - x_0}{h} = i$ and the equation can be simplified.
$$\sum_{j=0}^{n} f(x_i) \displaystyle \prod_{i=0 \atop i \neq j}^{n} (x_j - x_i)^{-1} \int_{x_0}^{x_n} \prod_{i=0 \atop i \neq j}^{n} (x-x_i) dx \\ = \sum_{j=0}^{n} f(x_i) \displaystyle \prod_{i=0 \atop i \neq j}^{n} (j - i)^{-1} h \int_{0}^{n} \prod_{i=0 \atop i \neq j}^{n} (z-i) dz$$
$f(x_i)$ was left intact intentionally. Solving equation boils down to solving $\int_{0}^{n} \displaystyle \prod_{i=0 \atop i \neq j}^{n} (z-i) dz$ which is quite difficult.
$S(z, j) := \displaystyle \prod_{i=0 \atop i \neq j}^{n} (z-i)$ can be written as follows.
$$S(z, j) = \displaystyle \prod_{i=0 \atop i \neq j}^{n} (z-i) = \sum_{i=0}^{n} a_i z^i$$
$a_i$ is constant. Notice the upper value of $\sum$ is $n$ for the number of $z$s in $\prod$ is $(n - 0 + 1) - 1 = n$ because $i \neq j$.
It is trivial $S(z, j) = 0$ when $z = 0, 1, ... , n$ but $\neq j$. So the following is true.
$$\sum_{i=0}^{n} a_i z^i = 0 \,(z = 0, 1, ... , n \cap z \neq j)$$
The integration can be calculated as below.
$$\int_{0}^{n} \displaystyle \prod_{i=0 \atop i \neq j}^{n} (z-i) dz = \int_{0}^{n} S(z, j) dz = \int_{0}^{n} \sum_{i=0}^{n} a_i z^i dz \\ = \big[\sum_{i=0}^{n} a_i \frac{z^{i+1}}{i+1}\big]_{0}^{n} \\ = \sum_{i=0}^{n} a_i \frac{n^{i+1}}{i+1}$$
To solve this, evaluating $a_i$ is required. $a_i$ can be evaluated as below.
$$\sum_{i=0}^{n} a_i z^i = 0 \,(z = 0, 1, ... , n \cap z \neq j) \\ \Leftrightarrow \displaystyle {\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2^{2} & \cdots & 2^{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & n & n^{2} & \cdots & n^{n} \end{pmatrix}} {\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}} = {\begin{pmatrix}b_{0}\\b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}} \\ (b_i = \begin{cases} S(j,j)\,\,(i = j) \\ 0 \,\,(i \neq j) \end{cases} ) $$
As $a_n = 1$ and $a_0 = \prod_{i=0 \atop i \neq j}^{n} -i$ the equation can be written as follows.
$$\sum_{i=0}^{n} a_i z^i = 0 \,(z = 0, 1, ... , n \cap z \neq j) \\ \Leftrightarrow \sum_{i=0}^{n-1} a_i z^i = -z^n \,(z = 1, 2, ... , n \cap z \neq j) \\ \Leftrightarrow \displaystyle {\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2^{2} & \cdots & 2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & n & n^{2} & \cdots & n^{n-1} \end{pmatrix}} {\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n-1}\end{pmatrix}} = {\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n-1}\end{pmatrix}} \\ (b_i = \begin{cases} S(j,j) - z^j\,\,(i = j) \\ -z^i \,\,(i \neq j) \end{cases} ) $$
The leftmost matrix is Vandermonde's matrix and multiplying its inverse matrix to the left of both sides gives $a_i$. However, the inverse matrix is quite complicated and I find it pretty hard to calculate (Inverse of Vandermonde's matrix).
approach 2
I guessed $i \neq j$ makes the problem difficult. So I went for deleting it.
$$ \int_{0}^{n} \displaystyle \prod_{i=0 \atop i \neq j}^{n} (z-i) dz = \int_{0}^{n} \frac{\displaystyle \prod_{i=0}^{n} z-i}{z-j} dz$$
The limit of $\frac{\prod_{i=0}^{n} z-i}{z-j}$ when $z$ approaches $j$ is given by L'hospital's rule, which is finite.
$$ \frac{(\prod_{i=0}^{n} z-i)'}{(z-j)'}|_{z=j} = (\prod_{i=0}^{n} z-i)'|_{z=j}$$
But after hours of thinking, I couldn't come up with how to solve $\int_{0}^{n} \frac{\displaystyle \prod_{i=0}^{n} z-i}{z-j} dz$. Two $z$s at both the enumerator and denominator. Integration by parts didn't work.
I've read the following articles and still didn't make it to find the answer.
How do I solve this?
