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How to calculate this determinate

  1. Jun 22, 2005 #1
    Let [itex]n[/itex] be a positive integer, and define the function

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

    (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).


  2. jcsd
  3. Jun 23, 2005 #2


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    Homework Helper

    Why do you have

    [tex]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}[/tex]

    Do you mean

    [tex]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}[/tex]

    I'm not exactly sure if this will work, but write [itex]f_n[/itex] as a polynomial in [itex]x_1[/itex] with coefficients [itex]g_0,\, \dots ,\, g_{n-1}[/itex]. You should be able to write these coefficients (or, at least one of them) in terms of [itex]f_{n-1}[/itex], which you can in turn write as a polynomial in [itex]x_2[/itex]. You'll get something with [itex]x_1[/itex] and [itex]x_2[/itex] 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 [itex]x_i[/itex] and then hopefully being able to factor.

    Alternatively, treat [itex]f_n(x_1,\, \dots ,\, x_n)[/itex] as the single-variable polynomial [itex]f_{n-x_2,\dots ,x_n}(x_1)[/itex]. If [itex]f_n[/itex] 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 [itex]x_1[/itex] is, but you may want to try this approach. Take the product of those factors, and multiply it by [itex]g_{n-1}[/itex] and see if you end up with [itex]f_n[/itex].
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