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Vandermonde determinants (introduced by Alexandre-Th茅ophile Vandermonde). Consider distinct real numbers ${\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{a}}_{\mathbf{n}\mathbf{.}}$. We define$\left(n+1\right)\mathbf{脳}\left(n+1\right)$ the matrix

$\mathit{A}\mathbf{=}\left[\begin{array}{ccc}1& 1& ....1\\ a_0& a_1& ....a_n\\ a_0^2& a_1^2& ....a_1^2\\ a_0^n& a_1^n& ....a_n^n\end{array}\right]$

Vandermonde showed that

$\mathit{d}\mathit{e}\mathit{t}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\underset{\mathbf{i}\mathbf{>}\mathbf{j}}{\mathbf{鈭�}}\mathbf{(}{\mathbf{a}}_{\mathbf{i}}\mathbf{-}{\mathbf{a}}_{\mathbf{j}}\mathbf{)}$

the product of all differences$\left(a_i-a_j\right)$, where exceeds j.
a. Verify this formula in the case of$\mathit{n}\mathbf{=}\mathbf{1}$.
b. Suppose the Vandermonde formula holds for$\mathit{n}\mathbf{=}\mathbf{1}$. You are asked to demonstrate it for n. Consider the function

$\mathbf{f}\left(t\right)\mathbf{=}\mathbf{det}\left[\begin{array}{ccccc}1& 1& ...& 1& 1\\ a_0& a_1& ...& a_{n-1}& t\\ a_0^2& a_1^2& ...& a_{n-1}& t^2\\ 鈰�& 鈰�& ...& 鈰�& 鈰�\\ a_0^n& a_1^n& ...& a_{n-1}^n& t^n\end{array}\right]$

Explain why f(t) is a polynomial of ${\mathit{n}}^{\mathbf{t}\mathbf{h}}$degree. Find the coefficient k of${\mathit{t}}^{\mathbf{n}}$ using Vandermonde's formula for${\mathit{a}}_{\mathbf{0}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$. Explain why

role="math" localid="1659522435181" $\mathit{f}\mathbf{\left(}{\mathbf{a}}_{\mathbf{0}}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}{\mathbf{a}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}\mathit{f}\mathbf{\left(}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Conclude that

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{k}\mathbf{(}\mathit{t}\mathbf{-}{\mathbf{a}}_{\mathbf{0}}\mathbf{)}\mathbf{(}\mathit{t}\mathbf{-}{\mathbf{a}}_{\mathbf{1}}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(}\mathit{t}\mathbf{-}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{)}$

for the scalar k you found above. Substitute$\mathbf{t}\mathbf{=}{\mathbf{a}}_{\mathbf{n}}$ to demonstrate Vandermonde's formula.

Step by step solution

01

(a) By using Vandermonde’s Formula. 

For$n=1$, we have a$2脳2$matrix

$A=\left[\begin{array}{cc}1& 1\\ a_0& a_1\end{array}\right]$

Using Vandermonde's formula, we have

$\underset{i>j}{鈭�}\left(a_i-a_j\right)=a_1-a_0=detA$

02

(b) To  Find the coefficient k of tn using Vandermonde's formula.

By the Laplace expansion along the$\left(n+1\right)$-th column, we see that f is a polynomial of n -th degree, the coefficient ofbeing in fact the Vandermonde's determinant for$n-1$ , which is$\underset{i,j-1i>j}{\overset{n-1}{鈭�}}\left(a_i-a_j\right)$.

For$t=a_m,m=0,1,...,n-1$, the$\left(m+1\right)$-th and the$\left(n+1\right)$-th column will be the same, thus the determinant will be 0 . So,

$f\left(a_m\right)=0,鈭赌m=0,1,...,n$

For k being the Vandermonde's determinant for , we have exactly$f\left(t\right)=k\underset{i=0}{\overset{n-1}{鈭�}}\left(t-a_i\right)$ .

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