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I am attempting to prove that the discriminant of some polynomial $f$ of degree $n$ with roots $\alpha_1, \dots, \alpha_n$ is given by $$ \Delta_f = (-1)^{\frac{n(n-1)}{2}} \prod\limits_{k = 1}^n f'(\alpha_k) $$

I found a solution in the first part of problem 1 here, but I am struggling to understand some of the logic. In particular, they state that $$ \frac{f'(x)}{f(x)} = \sum\limits_{i=1}^n \frac{1}{x - \alpha_i} \implies f'(\alpha_j) = \prod_{\substack{i = 1 \\ i \neq j}}^n (a_j - a_i). $$

I understand how they derive the initial summation, but I fail to see how the product follows. Any help would be appreciated!

sToxic5
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  • Multiply through that bottom displayed equation by $\frac{f(x)}{C(x-\alpha_j)}$, where $f(x) = C \cdot \Pi_i (x - \alpha_i)$, and then plug in $x = \alpha_j$. – John Hughes Oct 07 '17 at 00:34
  • I should have explained more, but that was what I originally had thought about. What I was struggling with was that seems to be division by zero, because when you substitute in $a_j$ you're dividing by $a_j-a_j=0$. Why is that not an issue? – sToxic5 Oct 07 '17 at 01:34
  • @JohnHughes : Could you write $\prod_i(x-\alpha_i)$ instead of $\Pi_i(x-\alpha_i$? The difference is even more conspicuous in a "displayed" setting, thus: $$ \prod_i (x-\alpha_i) \quad \text{ versus } \quad \Pi_i (x-\alpha_i) $$ – Michael Hardy Oct 07 '17 at 01:42
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    Thanks, @MichaelHardy. I knew there was a better way to write it, but could not recall it on short notice. I'll try to remember that in the future. – John Hughes Oct 07 '17 at 02:29

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What appears below assumes the leading coefficient is $1.$ \begin{align} \frac{f'(x)}{f(x)} & = \sum_{i=1}^n \frac 1 {x-\alpha_i}. \\[10pt] \text{Therefore } f'(x) & = f(x) \sum_{i=1}^n \frac 1 {x-\alpha_i} \\[10pt] & = (x-\alpha_1)\cdots(x-\alpha_n)\sum_{i=1}^n \frac 1 {x-\alpha_i} \\[10pt] & = \sum_{i=1}^n \frac{(x-\alpha_1)\cdots(x-\alpha_n)}{x-\alpha_i} \\[10pt] & = \overbrace{(x-\alpha_2)(x-\alpha_3)\cdots(x-\alpha_n)}^{\Large\text{omitting }\alpha_1} \, + \, \overbrace{(x-\alpha_1)(x-\alpha_3) \cdots(x-\alpha_j)}^{\Large\text{omitting }\alpha_2} \,+\, \cdots\cdots \\[10pt] \text{So } f'(\alpha_1) & = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)\cdots(\alpha_1-\alpha_n) + 0 + 0 + \cdots\cdots + 0. \end{align}

Michael Hardy
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  • When you cancel $x-a_i$ aren't you assuming that it is nonzero? – sToxic5 Oct 07 '17 at 04:06
  • @sToxic5 : Certainly it is nonzero in the case in which $x=\alpha_1$ when $i\ne 1,$ except when there is a multiple root. – Michael Hardy Oct 07 '17 at 04:26
  • When you say a factor is "omitted" from a term in the summation, it is because you divided out that factor. If we take $x = \alpha_1$, the only term which is nonzero in the summation is the term where $x - \alpha_1$ is omitted. In order for that factor to be divided out though, weren't you dividing by $x - \alpha_1= \alpha_1 - \alpha_1 = 0$? – sToxic5 Oct 07 '17 at 19:40
  • @sToxic5 : Suppose $f(x) = (x-2)(x-3)(x-4).$ Then $$\frac{f'(x)}{f(x)} = \frac 1 {x-2} + \frac 1 {x-3} + \frac 1 {x-4}$$ with problematic points at $2,$ $3,$ and $4.$ Multiplying both sides by $f(x)$ we get $$ f'(x) = (x-2)(x-3)(x-4) \cdot \frac 1 {x-2} + \text{two more terms}. $$ That term has $\dfrac{x-2}{x-2}$ as a factor, and that is equal to $1$ if $x\ne 2.$ What to do with it when $x=2$ is the question. Clearly it approaches $1$ as $x\to2,$ and recall that $f$ is a continuous function, so the value of $f$ at $2$ should be just what you get if you replace that fraction with $1. \qquad$ – Michael Hardy Oct 07 '17 at 22:21