Calculating Ratio of Determinants for $a,b$ Real Coefficients

[lfdahl]

Let $a$ and $b$ be real coefficients ($b \ne 0$), and let $(x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k$
for sufficiently small $|x|$.

Show, that the ratio of determinants:

$\begin{vmatrix} c_k & c_{k+1} \\ c_{k+1} & c_{k+2} \end{vmatrix} / \begin{vmatrix} c_{k+1} & c_{k+2} \\ c_{k+2} & c_{k+3} \end{vmatrix}$

- is independent of $k$.

 

[Opalg]

Let $a$ and $b$ be real coefficients ($b \ne 0$), and let $(x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k$
for sufficiently small $|x|$.

Show, that the ratio of determinants:

$\begin{vmatrix} c_k & c_{k+1} \\ c_{k+1} & c_{k+2} \end{vmatrix} / \begin{vmatrix} c_{k+1} & c_{k+2} \\ c_{k+2} & c_{k+3} \end{vmatrix}$

- is independent of $k$.

[sp]If \(\displaystyle (x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k\) then \(\displaystyle 1 = (x^2+ax+b)\sum_{k=0}^{\infty}c_kx^k\). Compare the coefficients of $x^{k+2}$ and $x^{k+3}$ on both sides: $$\begin{aligned}c_k + ac_{k+1} + bc_{k+2} &= 0, \\ c_{k+1} + ac_{k+2} + bc_{k+3} &= 0.\end{aligned}$$ Solve those simultaneous equations for $a$ and $b$ to get $$b = \begin{vmatrix} c_k & c_{k+1} \\ c_{k+1} & c_{k+2} \end{vmatrix} \bigg/ \begin{vmatrix} c_{k+1} & c_{k+2} \\ c_{k+2} & c_{k+3} \end{vmatrix}.$$ Since the left side of that equation is independent of $k$ so is the right side.[/sp]

 

[lfdahl]

[sp]If \(\displaystyle (x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k\) then \(\displaystyle 1 = (x^2+ax+b)\sum_{k=0}^{\infty}c_kx^k\). Compare the coefficients of $x^{k+2}$ and $x^{k+3}$ on both sides: $$\begin{aligned}c_k + ac_{k+1} + bc_{k+2} &= 0, \\ c_{k+1} + ac_{k+2} + bc_{k+3} &= 0.\end{aligned}$$ Solve those simultaneous equations for $a$ and $b$ to get $$b = \begin{vmatrix} c_k & c_{k+1} \\ c_{k+1} & c_{k+2} \end{vmatrix} \bigg/ \begin{vmatrix} c_{k+1} & c_{k+2} \\ c_{k+2} & c_{k+3} \end{vmatrix}.$$ Since the left side of that equation is independent of $k$ so is the right side.[/sp]

Thankyou for the correct answer, Opalg! - and for your participation!(Handshake)