Recurrence relation for sum of powers of roots of a cubic equation.

Question

Let $f(x) = x^3 −s_1x^2 +s_2x−s_3 = (x−α)(x−β)(x−γ) ∈ \mathbb{Q}[x]$ where $α, β, γ ∈ \mathbb{C}$. Denoting $σ_i = α^ i +β^ i +γ^ i$ for $i ≥ 0$, show that $σ_0 = 3, σ_1 = s_1$ and $σ_2 = s^2_1 −2s-2.$

Show further that $σ_r = s_1σ_{r−1} − s_2σ_{r−2} + s_3σ_{r−3}$ for $r \geq 3$.

It was easy enough to get the results for $σ_0,σ_1$ and $σ_2$, but I'm struggling to see how I can get the further result. Presumably it's by induction but I can't even get the base case of $r=3$, never mind the next step.

Am I missing an obvious trick? Any help would be appreciated.

Answer 1

Just expand $$\begin{align}s_1σ_{r−1} − s_2σ_{r−2} + s_3σ_{r−3}&=(\alpha+\beta+\gamma)(\alpha^{r-1}+\beta^{r-1}+\gamma^{r-1})\\&\quad{}-(\alpha\beta+\alpha\gamma+\beta\gamma)(\alpha^{r-2}+\beta^{r-2}+\gamma^{r-2})\\ &\quad{}+\alpha\beta\gamma(\alpha^{r-3}+\beta^{r-3}+\gamma^{r-3})\\ &=\alpha^r+\beta^r+\gamma^r+(\alpha^{r-1}\beta+\alpha^{r-1}\gamma+\beta^{r-1}\alpha+\cdots)\\ &\quad{}-(\alpha^{r-1}\beta+\alpha^{r-1}\gamma+\cdots)-(\alpha^{r-2}\beta\gamma+\cdots)\\ &\quad{}+\alpha^{r-2}\beta\gamma+\cdots\\&=\alpha^r+\beta^r+\gamma^r\end{align} $$

Answer 2

Recurrence relationship:

$$s_1σ_{k-1} − s_2σ_{k−2} + s_3σ_{k-3}=σ_k \tag{1}$$

can be obtained in a direct, "heuristic" way. Let us start for that from the following identity valid for any root $x$ of the given polynomial :

$$ s_1x^2 − s_2x +s_3 =x^3$$

Multiplying LHS and RHS by $x^{k-3}$, we get:

$$s_1x^{k-1} − s_2x^{k-2} + s_3x^{k-3}=x^{k}.\tag{2}$$

Writing (2) for the different roots $\alpha, \beta, \gamma$ and summing up these 3 equations, one gets (1).