When is a proof too vague? (If $r_1,r_2,\ldots,r_n$ are roots of polynomial $p$ with leading term $x^n$, then $r_1+r_2+\cdots+r_n=(-1)a_{n-1}$)

Question

Theorem. Let $p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ with roots $r_{1}, r_{2}, \ldots, r_{n}$,

i.e

$$ p(x)=\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right), $$ then $$ r_{1}+r_{2}+\cdots+r_{n}=(-1) a_{n-1} . $$

$Proof$.

If we expand out $\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right)$, we can see that the $x^{n-1}$ terms are precisely $x^{n-1}\left(-r_{1}\right), x^{n-1}\left(-r_{2}\right), \ldots, x^{n-1}\left(-r_{n}\right)$. We group them together and get $x^{n-1}\left(-r_{1}-r_{2}-\cdots-r_{n}\right)$, therefore it must be that $-r_{1}-r_{2}-\cdots-r_{n}=a_{n-1}$, from which it follows that $r_{1}+r_{2}+\cdots+r_{n}=(-1) a_{n-1}$.

Im a bit considered that it is too vague. If im being honest I feel like this more of an explanation why the theorem may be true, but it doesn't really convince you that it proves the statement.In general, when is a proof too vague to be considered invalid? Obviously a reader that doesnt know polynomials is not gonna understand this proof, but that doesn't automatically mean its invalid.

Answer 1

As you've hinted at, it's context-dependent. For newbies, a proof might deserve more explanation; for professional mathematicians, enough of a trail of breadcrumbs might be left that they can fill in the details themselves in their head with only a few seconds of thought. For a learner's textbook on that specific topic, it might not be very good; for a published paper, it might be more suitable.

Personally, I find your proof perfectly reasonable (since you have to choose one summand from each factor, and if you want to extract the coefficients of $x^{n-1}$ you need to choose $n-1$ $x$'s and one of the $-r_i$). Other people might prefer that you clarify a bit further, however.

Answer 2

I think you should prove a general theorem first:

$$ \sum_{k=0}^n a_k x^k = \sum_{k=0}^n b_k x^k\quad (\forall\ x\in\mathbb{R})\implies a_k = b_k\quad (\forall\ k\in \{0,\ldots,n\}).$$

Proof sketch: $ \displaystyle\sum_{k=0}^n a_k x^k = \sum_{k=0}^n b_k x^k\quad (\forall\ x\in\mathbb{R})\implies \sum_{k=0}^n (a_k-b_k) x^k = 0\quad (\forall\ x\in\mathbb{R}).$ Now let $j$ be the largest integer in $\{0,\ldots,n\}$ such that $a_j - b_j \neq 0.\ $ Then, $\ \left\lvert \displaystyle\sum_{k=0}^n (a_k-b_k) x^k \right\rvert = \left\lvert \displaystyle\sum_{k=0}^j (a_k-b_k) x^k \right\rvert \to\infty\ $ as $\ x\to\infty,\ \implies \displaystyle\sum_{k=0}^n (a_k-b_k) x^k \neq 0\ \forall\ x\in\mathbb{R},$ a contradiction. Therefore there is no $j$ in $\{0,\ldots,n\}$ such that $a_j - b_j \neq 0.\ $ and the theorem is proved.

Now you can apply this theorem to your question starting with

$$ x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0} = \left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right)\quad \forall x\in\mathbb{R}.$$