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I had problems finding the proof of an equation in G. Polya "Mathematics and Plausible Reasoning" p. 18 that upon a little bit of research turns out to be Viète's Theorem:

Given a polynomial,

$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$ with roots $\alpha_1, \alpha_2,...\alpha_n,$

$$a_{n-1}= -a_n\,(\alpha_1+\alpha_2\,+...+\,\alpha_n).$$

The solution is tied to the critical result $a_0=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n$ in this post, but I wanted to spell out the more complete formula in Q&A style.

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Antoni Parellada
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1 Answers1

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Given the equality:

$$a_0\,+\,a_1\,x\, +\,a_2\,x^2 \cdots +\,a_{n-1}\,x^{n-1}\,+\,a_n\,x^n =a_n\,(x-\alpha_1)\,(x-\alpha _2)\,\cdots (x-\alpha_n)$$

resulting from the fundamental theorem of algebra, the expansion of the RHS of the equation will yield,

$$a_n\,x^n\, - a_n\, x^{n-1}\,(\alpha_1\,+\,\alpha_2\,+\cdots\, +\,\alpha_n)+$$

$$+\,a_n\,x^{n-2}\,(\alpha_1\,\alpha_2+\,\alpha_1\,\alpha_3\,+\cdots\, +\,\alpha_1\, \alpha_n\,+\,\alpha_2\,\alpha_3\,+\,\cdots\,+\alpha_{(n-1)}\,\alpha_n)\, +$$

$$-\, a_n\,x^{n-3}\,(\alpha_1\,\alpha_2\,\alpha_3+\,\alpha_1\,\alpha_2\,\alpha_4\,+\cdots\, +\,\alpha_1\,\alpha_2\, \alpha_n\,+\,\alpha_2\,\alpha_3\,\alpha_4+\,\cdots\,+\,\alpha_{(n-2)}\alpha_{(n-1)}\,\alpha_n)\, +$$

$$+\,\cdots\,+$$

$$+\,a_n\,(-1)^n\,\alpha_1\,\alpha_2\,\alpha_3\cdots\alpha_n.$$

And comparing coefficients between the LHS of the initial equation and this expansion, we see that for $x_0$ (constant term):

$a_n\,(-1)^n\,\alpha_1\,\alpha_2\,\alpha_3\cdots\alpha_n =a_0,$

replicating the equality in the penultimate line of the proof in here. And explaining why $a_0=0$ is incompatible with all the roots being $\neq\,0$.

Similarly, for $x^n$,

$a_n = a_n$

And from here we can see that when $x^{n-1}$,

$-\,a_n\,(\alpha_1\,+\,\alpha_2\,+\cdots\, +\,\alpha_n)\,=\,a_{n-1}.$

And when $x^{n-2}$,

$a_n\,(\alpha_1\,\alpha_2+\,\alpha_1\,\alpha_3\,+\cdots\, +\,\alpha_1\, \alpha_n\,+\,\alpha_2\,\alpha_3\,+\,\cdots\,+\alpha_{(n-1)}\,\alpha_n) = a_{n-2}...$

Perhaps more interesting is the link between this derivation, and the parallel application to the alternative expression of the fundamental theorem of algebra in here:

$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$

$$\,= a_0\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right)...\left(1-\frac{x}{\alpha_n}\right)$$

Analogously, the expansion of the RHS of the equation would lead to $a_0 \cdot1^n$ in the constant component of the expansion, while for the coefficient of $x^1$ the equivalent on the expanded form would include any of the $n$ possible $-\,\frac{x}{\alpha_i}$ times $1^{n-1}$, such that

$a_1 = \,-\,a_0 \left(\frac{1}{\alpha_1}\,+\frac{1}{\alpha_2}\,\cdots+\,\frac{1}{\alpha_n}\right), $ which is the last equation on page 18 of G. Polya's "Mathematics and Plausible Reasoning", setting up the stage to prove Basel's problem.

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