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

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

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3  
Use the fact that $\frac{f'}{f} = (\log f)'$. –  Qiaochu Yuan Dec 7 '11 at 21:20
    
Do you really want $x^{-i}$ rather than $x^i$? –  Thomas Andrews Dec 7 '11 at 21:21
    
@ThomasAndrews, yes because the degree of $f'$ is less then $f$. But if you use $1/x$ we can use $x^i$ instead. –  GarouDan Dec 7 '11 at 21:24
3  
I would have said that (assuming $f(x)$ to be monic) using the product rule for differentiating $\prod_k (x-x_k)$ shows that $$\frac{f^\prime}{f} = \sum_k \frac{1}{x-x_k}$$ in which each term can be expanded via the binomial theorem into a power series in $x$, etc. –  Dilip Sarwate Dec 7 '11 at 21:32
1  
@Dilip should post his comment as an answer, I think. Garou: you might be interested in the Newton-Girard identities. –  J. M. Dec 7 '11 at 22:37

1 Answer 1

up vote 3 down vote accepted

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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If we think of them as formal power series, it doesn't seem so cavalier. +1, of course. –  J. M. Dec 7 '11 at 23:35
    
Really thx. Very nice proof. –  GarouDan Dec 8 '11 at 0:30

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