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I am given the following results:

  1. If $f(t) = \prod_{i=1}^n(t-\alpha_i)$, then $f'(t) = \prod_{j\neq i}(\alpha_j - \alpha_i)$
  2. The discriminant of such an $f$ is then $disc(f) = (-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^nf'(\alpha_i)$

To prove:

If $f(t) = t^n + bt + c,\; n\geq 2$ then $disc(f) = (-1)^{\frac{n(n-1)}{2}}((1-n)^{n-1}b^n + n^nc^{n-1})$

I'm not sure how to prove this though as I keep getting lost in the algebra and in considering various terms. What I have so far is:

$f(t) = t^n + bt + c \Rightarrow f'(t) = nt^{n-1} + b$

$disc(f) = (-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^n(n\alpha_i^{n-1}+b)$

So we are left to show that: $\prod_{i=1}^n(n\alpha_i^{n-1}+b) = ((1-n)^{n-1}b^n + n^nc^{n-1})$

My first thought was to use proof by induction as the base case $n=2$ is clear. However, since with each successive case we change the roots of the polynomial and thus how they interact, I'm not sure how the previous cases can help you deduce the next case.

Because of this I thought that maybe we are just supposed to prove it directly, though I don't know how to just group a lot of the terms in $(n\alpha_1^{n-1}+b)\dots (n\alpha_n^{n-1}+b)$ to get our result. For example:

  • By choosing the $b$ term in every linear factor we get: $b^n$

  • By choosing the $n\alpha_i$ term in every linear factor we get: $n^n(\alpha_1 \dots \alpha_n)^{n-1} = n^nc^{n-1}$

Now we just need to show the terms other than $b^n$ in $(1-n)^{n-1}b^n$ occur in the expansion of the above formula.

It is tempting then to think that the coefficient of $b^m$ in $\prod_{i=1}^n(n\alpha_i^{n-1} +b )$ is equal to $({n \choose n-m}(-n)^{n-m})b^{n-m}$

But I can't see why this would be true as we keep choosing a different set of roots for each other term in the expansion. What is it that we know about the roots that makes this true? Is there any easier way for me to think about this problem?

user366818
  • 2,653

1 Answers1

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You can simplify $f'(a_i)=na_i^{n-1}+b$ by taking $$a_if'(a_i)=na_i^n+ba_i=-nba_i-nc+ba_i=(1-n)ba_i-nc$$(this uses $a_i^n=-ba_i-c$)

Then you know that the product of the roots is $(-1)^nc$ and most of the basic symmetric polynomials in the roots are zero.

Can you take it from there.

To take this further, now write $$\prod a_if'(a_i)=\prod (1-n)b\left(a_i-\frac{nc}{(1-n)b}\right)=(-1)^nc\prod f'(a_i)=$$$$=(1-n)^nb^n(-1)^n\prod\left(\frac{nc}{(1-n)b}-a_i\right)$$

Now the product is in the form $\prod (t-a_i)=t^n+bt+c$, with $t=\cfrac{nc}{(1-n)b}$ using the original polynomial, and substituting back for $t$ we get $$(1-n)^nb^n(-1)^n \left(\frac {n^nc^n}{(1-n)^nb^n}+b\frac{nc}{(1-n)b}+c\right)=(-1)^n(n^nc^n+nb^nc(1-n)^{n-1}+b^nc(1-n)^n)$$

Cancelling the factor $(-1)^nc$ then gives $$\prod f'(a_i)=n^nc^{n-1}+b^n(1-n)^{n-1}$$ which is what you need.

Mark Bennet
  • 100,194