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Suppose I have a set of data $(x_i,y_i)$ for $i=0,1,...n$.

I want to prove that if $p(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_0$ then $\displaystyle a_n = \sum_{i=0}^n y_i \prod_{j=0,j\neq i}^n \frac{1}{x_i-x_j}$.

My attempt is that if $\displaystyle p(x) = \sum_{i=0}^n y_i L_i(x)$ then $p(x_j)=y_jL_j(x_j)=y_j \delta_{j,j}=y_j=a_n x_j^n + a_{n-1}x_j^{n-1}+\cdots+a_0$, so we can form the following matrix: \begin{equation} \begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^n \\ 1 & x_1 & x_1^2 & \cdots & x_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^n \end{pmatrix} \begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_n \end{pmatrix} = \begin{pmatrix} y_0\\ y_1\\ \vdots\\ y_n \end{pmatrix} \end{equation} From here I don't know how to proceed. I know that I can get $a_i$ from Cramer's rule as: \begin{equation} a_i = \frac{\det{\begin{pmatrix} 1 & x_0 & \cdots & y_0 & \cdots & x_0^n \\ 1 & x_0 & \cdots & y_1 & \cdots & x_0^n \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_0 & \cdots & y_n & \cdots & x_0^n \\ \end{pmatrix}}}{\det{\begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^n \\ 1 & x_1 & x_1^2 & \cdots & x_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^n \end{pmatrix}}} \end{equation} The determinant from the denominator resembles the one from the Vandermonde's Matrix, but the one from the numerator I don't know how to calculate it.

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1 Answer 1

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My attempt is that if $\displaystyle p(x) = \sum_{i=0}^n y_i L_i(x)$ then $\;\ldots$

That looks like the Lagrange form of the interpolation, in which case $\,\displaystyle L_i(x)= \prod_{j=0, \; j \ne i}^n \frac{x - x_j}{x_i-x_j}\,$.

The coefficient of $\,x^n\,$ in $\,L_i(x)\,$ is $\,\displaystyle b_i = \prod_{j=0, \; j \ne i}^n \frac{1}{x_i-x_j}\,$ and $\,\displaystyle \,a_n = \sum_{i=0}^n y_i b_i\,$ which gives the result.

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  • $\begingroup$ Yes, I was totally confused with what the question asked. I thought I had to find $a_i$ for any $i\in\{0,1,...,n\}$. That's more interesting. $\endgroup$
    – davidaap
    Commented Jun 14, 2018 at 3:20
  • $\begingroup$ @math4everyone All $a_i$'s don't necessarily have simple closed forms, see here or here for example. $\endgroup$
    – dxiv
    Commented Jun 14, 2018 at 3:40

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