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

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

  • 4
    Use the fact that $\frac{f'}{f} = (\log f)'$.2011-12-07
  • 1
    Do you really want $x^{-i}$ rather than $x^i$?2011-12-07
  • 0
    @ThomasAndrews, yes because the degree of $f'$ is less then $f$. But if you use $1/x$ we can use $x^i$ instead.2011-12-07
  • 0
    @QiaochuYuan, very interesting. I will think about this sentence.2011-12-07
  • 0
    Just because $f'(x)$ is lower degree than $f(x)$ doesn't mean we can't write $f'(x)/f(x)$ as a power series of $x$. For example $\frac{1}{1-x} = \sum x^i$. However, I see why you need it in this problem.2011-12-07
  • 4
    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.2011-12-07
  • 0
    Also, it seems like $a_i = \sum_j x_j^{i-1}$2011-12-07
  • 0
    Huhuh. I just changed this becuase the series expansion of that $\frac{f'(x)}{f(x)}$, using Mathematica, is $-\frac{5}{6}-\frac{13 x}{36}-\frac{35 x^2}{216}-\frac{97 x^3}{1296}-\frac{275 x^4}{7776}-\frac{793 x^5}{46656}+\ldots$ instead of $2 x+5 x^2+13 x^3+35 x^4+97 x^5+\ldots$. But I think you got the question spirit ^^. Thx about the comments.2011-12-07
  • 0
    Incidentally, you'll see that the coefficient of $x^i$ you got from Mathematica are of the form $-(2^{-i-1} + 3^{-i-1})$2011-12-07
  • 0
    @DilipSarwate, interesting point. Looks a bit hard prove your sentence but find it I think is done, because, as you explain $\frac{1}{x-x_k}=\displaystyle\sum_{i=1}^{\infty}x^{-i}{x_k}^{i-1}$. Very nice.2011-12-07
  • 2
    @Dilip should post his comment as an answer, I think. Garou: you might be interested in the [Newton-Girard identities](https://en.wikipedia.org/wiki/Newton%27s_identities).2011-12-07
  • 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.

  • 1
    If we think of them as formal power series, it doesn't seem so cavalier. +1, of course.2011-12-07
  • 1
    Really thx. Very nice proof.2011-12-08