- 523

- 0

[itex]f_n(x_1,x_2,\ldots, x_n) = \left(

\begin{array}{ccccc}

1 & 1 & 1 & \ldots & 1 \\

x_1 & x_2 & x_3 & \ldots & x_n \\

x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\

\vdots & \vdots & \vdots & \ddots & \vdots \\

x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_n^{n-1}

\end{array}

\right)[/itex]

(b) By considering the first column expansion of the determinant, show that

[itex]f_n(x_1,x_2,\ldots,x_2) = g_0 + g_1x_1 + g_2x_1^2 + \cdots + g_{n-1}x_1^{n-1}[/itex], and, in particular, [itex]g_{n-1}=(-1)^{n-1}f_{n-1}(x_2,x_3,\ldots,x_n)[/itex]

(c) Show that [itex]f_n(x_1,x_2,\ldots,x_2)[/itex] has [itex]x_i-x_1[/itex] as a factor, for all values of [itex]i[/itex] from 2 to [itex]n[/itex].

I think I've sorted part (b) out. I don't have a clue about how to proceed in part (c).

Thanks.

James