How to calculate this determinate

[jdstokes]

Let $n$ be a positive integer, and define the function

$f_n(x_1,x_2,\ldots, x_n) = \left( <br /> \begin{array}{ccccc}<br /> 1 & 1 & 1 & \ldots & 1 \\<br /> x_1 & x_2 & x_3 & \ldots & x_n \\<br /> x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\<br /> \vdots & \vdots & \vdots & \ddots & \vdots \\<br /> x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_n^{n-1}<br /> \end{array}<br /> \right)$

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

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

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

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

Thanks.

James

 

[AKG]

Why do you have

$$f_n(x_1,x_2,\ldots,\mathbf{x_2}) = g_0 + g_1x_1 + g_2x_1^2 + \cdots + g_{n-1}x_1^{n-1}$$

Do you mean

$$f_n(x_1,x_2,\ldots,\mathbf{x_n}) = g_0 + g_1x_1 + g_2x_1^2 + \cdots + g_{n-1}x_1^{n-1}$$

I'm not exactly sure if this will work, but write $f_n$ as a polynomial in $x_1$ with coefficients $g_0,\, \dots ,\, g_{n-1}$. You should be able to write these coefficients (or, at least one of them) in terms of $f_{n-1}$, which you can in turn write as a polynomial in $x_2$. You'll get something with $x_1$ and $x_2$ in there, and hopefully you'll be able to factor it to get the desired results. You should end up with new coefficients for which you can do something similar, working in all the $x_i$ and then hopefully being able to factor.

Alternatively, treat $f_n(x_1,\, \dots ,\, x_n)$ as the single-variable polynomial $f_{n-x_2,\dots ,x_n}(x_1)$. If $f_n$ is supposed to factor as it suggests in part c), then so should this new polynomial. But this polynomial is of degree n-1, and the problem suggests that you can pull out n-1 factors, so if what you are supposed to prove is correct, then the new polynomial should be a constant multiplied by the product of those factors. This doesn't seem exactly right because your coefficients are matrices, but the roots you're finding are whatever $x_1$ is, but you may want to try this approach. Take the product of those factors, and multiply it by $g_{n-1}$ and see if you end up with $f_n$.

 

[thepatientmental]

To calculate the determinant of the given function, we can use the Laplace expansion method. This involves expanding the determinant along one row or column and using cofactors to solve for the determinant. In this case, we will use the first column expansion.

First, we can see that the determinant is a polynomial in x_1, with degree n-1. This means it can be written as g_0 + g_1x_1 + g_2x_1^2 + \cdots + g_{n-1}x_1^{n-1}. We can also see that the coefficients of this polynomial are determined by the cofactors of the first column.

Using the Laplace expansion, we can write the determinant as:

f_n(x_1,x_2,\ldots,x_n) = x_1 \cdot \left(
\begin{array}{ccccc}
x_2 & x_3 & x_4 & \ldots & x_n \\
x_2^2 & x_3^2 & x_4^2 & \ldots & x_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_2^{n-1} & x_3^{n-1} & x_4^{n-1} & \ldots & x_n^{n-1}
\end{array}
\right)

Now, we can see that this is just the determinant of f_{n-1}(x_2,x_3,\ldots,x_n), with x_1 as a common factor. This means that g_{n-1} is equal to (-1)^{n-1} times the determinant of f_{n-1}(x_2,x_3,\ldots,x_n).

For part (c), we can use the same method of Laplace expansion to show that x_i-x_1 is a factor of f_n(x_1,x_2,\ldots,x_n). Expanding along the i-th column, we get:

f_n(x_1,x_2,\ldots,x_n) = (x_i-x_1) \cdot \left(
\begin{array}{ccccc}
1 & 1 & 1 & \ldots & 1 \\
x_1 & x_2 & x_3 & \ldots &