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Math Help - Need help deriving formula for a sum/polynomial stuff

  1. #1
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    Need help deriving formula for a sum/polynomial stuff



    OK, I have no idea what this is asking... I don't know what to do. Can someone help please?? thanks.
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  2. #2
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    Roots of a polynomial

    Hello qtpipi
    Quote Originally Posted by qtpipi View Post


    OK, I have no idea what this is asking... I don't know what to do. Can someone help please?? thanks.
    First of all, let me write out what the question is asking. The first \Sigma expression means:

    f(x) = c_0 + c_1x+c_2x^2 + \dots + c_nx^n

    And to say it has zeros \alpha_1, \dots, \alpha_n means

    f(\alpha_1) = f(\alpha_2) = \dots = f(\alpha_n) = 0

    And then, we are to derive a formula for

    \frac{1}{\alpha_1} + \dots + \frac{1}{\alpha_n}

    So, using the Remainder Theorem, (x-\alpha_1), \dots, (x-\alpha_n) are factors of f(x).

    Since there are n of these factors, and f(x) is a polynomial of degree n, this means that we can write f(x) as:

    f(x) = c(x-\alpha_1)(x-\alpha_2)\dots (x-\alpha_n)

    (This is what the expression f(x) = c\prod (x-\alpha_i) means.)

    Now the coefficient of x^n is c_n. So c = c_n. And the coefficient of x^{n-1} in the expansion of c\prod(x-\alpha_i) is -c\sum \alpha_i which is c_{n-1}. So:

    \sum \alpha_i = -\frac{c_{n-1}}{c_n}

    So how do we find \sum\frac{1}{\alpha_i}?

    Replace x by \tfrac{1}{y}:

    f(\tfrac{1}{y}) = c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n which has zeros at

    y = \tfrac{1}{\alpha_1}, \dots, \tfrac{1}{\alpha_n}

    and c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n = 0 \Rightarrow c_0y^n + c_1y^{n-1} + \dots + c_n = 0

    and, using the same method as before, the sum of the roots of this polynomial is -\frac{\text{coefficient of } y^{n-1}}{\text{coefficient of } y^{n}}

    \Rightarrow \sum_{i=0}^n \frac{1}{\alpha_i} = -\frac{c_1}{c_0}

    Grandad
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