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Problem:

If we have a polynomial $f$ with a derivative f\,' and quotient $q$ function defined as:

q(x)=\sum_{i=1}^{\infty}a_ix^{-i}=\frac{f\,'(x)}{f(x)},

and the roots of $f$ are $x_1,x_2,\ldots,x_k$, how to prove

$a_i=\sum_{j=1}^{k}x_j^i$

Details:

If $f(x)=x^2-5x+6$, f\,'(x)=2x+5,

$q(x)=2 x^{-1}+5 x^{-2}+13 x^{-3}+35 x^{-4}+97 x^{-5}+\ldots$

  • 0
    Correction: $f'(x) = 2x-5$.2015-11-29

1 Answers 1

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Assume without loss of generality that $f(x)$ is a monic polynomial with $n$ roots $x_1, x_2, \ldots, x_n$ so that we can write $f(x) = \prod_{k=1}^n(x-x_k)$

The product rule for derivatives then gives us $f^{\prime}=\frac{\mathrm d}{\mathrm dx}f(x)=\frac{\mathrm d}{\mathrm dx}\prod_{k=1}^n(x-x_k)= \sum_{k=1}^n\ \,\prod_{i=1,i\neq k}^n(x-x_i)$ where the $k$-th term of the sum on the right is the product of all the $(x-x_i)$ except $(x-x_k)$. Therefore, $\frac{f^{\prime}}{f} = \frac{\sum_{k=1}^n\prod_{i=1,i\neq k}^n(x-x_i)}{\prod_{k=1}^n(x-x_k)} = \sum_{k=1}^n\frac{1}{x-x_k}.$ Now, basic "long division" of $1$ by $x-x_k$ produces a "quotient" $x^{-1} + x_k\cdot x^{-2} + x_k^2\cdot x^{-3} + \cdots $ so that $\sum_{k=1}^n\frac{1}{x-x_k} = \sum_{i=1}^{\infty}\left(\sum_{k=1}^n x_k^{i-1}\right)\cdot x^{-i}$ which is essentially the answer wanted by Dan Garou except that, as noted by Thomas Andrews, it is "off-by-one." The "long division" can be formalized by expanding $(1-x_k\cdot x^{-1})^{-1}$ in a Taylor series in $x^{-1}$or the binomial theorem etc. but I will leave the details to Dan Garou to fill in.

Note: If anyone feels strongly enough about the cavalier treatment of power series in this last part to want to fill in the details, please feel free to edit this answer.

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    Really thx. Very nice proof.2019-03-06