Sequence of Riccati's roots

Question

Let $(p,q) \in (\mathbb{C}^{*})^2$ and consider the Riccati's polynomial $R \in \mathbb{C}[X]$ by $$ R(X)=X^3+pX+q $$ Let $z_1$, $z_2$ and $z_3$ be the three complex roots of $R$.

Finally, consider for $n \in \mathbb{Z}$ the sequence $(S_n)_{n \in \mathbb{Z}}$ given by $$ S_n=z_1^n+z_2^n+z_3^n $$ I guess it is possible to calculate $S_0(=3)$, $S_1$ and maybe $S_2$ with polynomial's roots theorem.

But how can I show easily that for $n \in \mathbb{Z}$ $$ S_{n+3}+pS_{n+1}+qS_n=0 $$ Is there way to find a simple expression of $(S_n)$? Wolfram gives me a not sympathetical one ...

There's also a question I've seen that is to calculate (precising for which $(p,q)$ it exists) $$ \frac{1}{1+z_1^2}+\frac{1}{1+z_2^2}+\frac{1}{1+z_3^2} $$ How is that possible?

Answer 1

We have \begin{eqnarray*} z_1+z_2+z_3=0 \\ z_1z_2+z_2z_3+z_3z_1=p \\ z_1z_2z_3=-q. \\ \end{eqnarray*} So \begin{eqnarray*} -p S_{n+1}-qS_n &=&-(p(z_1^{n+1}+z_2^{n+1}+z_3^{n+1})+q(z_1^{n}+z_2^{n}+z_3^{n})) \\ &=& z_1z_2z_3(z_1^{n}+z_2^{n}+z_3^{n})-(z_1z_2+z_2z_3+z_3z_1)(z_1^{n+1}+z_2^{n+1}+z_3^{n+1}) \\ &=& -((z_2+z_3)z_1^{n+2}+(z_3+z_1)z_2^{n+2}+(z_1+z_2)z_3^{n+2})=S_{n+3}. \\ \end{eqnarray*}

For the second part square $X^3+pX=-q$ \begin{eqnarray*} X^6+2pX^4+p^2X^2-q^2=0 \\ (1+X^2)^3+(2p-3)(1+X^2)^2+(p^2-4p+3)(1+X^2)+(-q^2-p^2+2p-1)=0 \\ (-q^2-p^2+2p-1)\left(\frac{1}{1+X^2}\right)^3+(p^2-4p+3)\left(\frac{1}{1+X^2}\right)^2 \\+(-q^2-p^2+2p-1)\left(\frac{1}{1+X^2}\right)+1=0 \\ \end{eqnarray*} So \begin{eqnarray*} \frac{1}{1+x_1^2} +\frac{1}{1+x_2^2} +\frac{1}{1+x_3^2}= \frac{p^2-4p+3}{q^2+(p-1)^2}. \end{eqnarray*}

Answer 2

First a detail: the number $S_n$ is defined for negative $n$ only if $q\ne0$.

Just compute it: $$ S_{n+3}+pS_{n+1}+qS_{n}= z_1^n(z_1^3+pz_1+q)+ z_2^n(z_2^3+pz_2+q)+ z_3^n(z_3^3+pz_3+q) =0 $$ Actually, if $T_n=\alpha z_1^n+\beta z_2^n+\gamma z_3^n$, then $T_{n+3}+pT_{n+1}+qT_n$, for any $\alpha,\beta,\gamma$.

Answer 3

For the second sum, the method of @Donald Splutterwit is clever. Here is a alternative method based on Sylvester's resultant (https://en.wikipedia.org/wiki/Sylvester_matrix) whose advantage is in its systematicity and its easiness when using a CAS.

Let us write again, for further reference, the expression of $R$:

$$\tag{0}R(X)=x^3+px+q=0$$

(I will not use letter $z$). Let us set:

$$\tag{1}y=\dfrac{1}{1+x^2}.$$

Our objective is to build an equation with variable $y$ (and roots $y_i=\frac{1}{1+x_i^2}$) (equation (3)) on wich we will apply one of Vieta's formulas.

Here is the computation in 4 steps:

1) Convert (1) into an equivalent polynomial equation in $z$ with coefficients in $y$: $$\tag{2}S(x):=yx^2+0x+(y-1)=0.$$

2) Compute the resultant, which expresses a necessary and sufficient condition on parameters $p,q,y$ for equations (0) and (2) to have a common root:

$$\tag{3}{\frak R}(R,S)=\begin{vmatrix}1& 0& p& q& 0\\ 0& 1 &0& p &q\\ y &0 &(y-1) &0& 0\\ 0& y &0 &(y-1)& 0\\ 0& 0& y& 0 &(y-1)\end{vmatrix}=0$$

3) Expand by a CAS, which gives:

$$\tag{4}{\frak R}(R,S)=Ay^3+By^2+Cy+D=0 \ \ \text{with} \begin{cases}A=p^2-2p+1+q^2\\B=-p^2+4p-3\end{cases} $$

4) Finally use the first Vieta formula in (4)

$$y_1+y_2+y_3=-\dfrac{B}{A}=\dfrac{p^2-4p+3}{p^2-2p+1+q^2}$$

which is (happily!) the same result.