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

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The discussion focuses on demonstrating that the ratio of determinants involving coefficients from the series expansion of the function $(x^2 + ax + b)^{-1}$ is independent of the index $k$. By comparing coefficients of $x^{k+2}$ and $x^{k+3}$ in the equation derived from the series, two simultaneous equations are established. Solving these equations leads to the expression for $b$ as the ratio of the determinants. The independence of $k$ is confirmed since both sides of the derived equation do not depend on $k$. This establishes the desired result regarding the ratio of determinants.
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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 $k$.
 
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lfdahl said:
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 $$(x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k$$ then $$1 = (x^2+ax+b)\sum_{k=0}^{\infty}c_kx^k$$. 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]
 
Opalg said:
[sp]If $$(x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k$$ then $$1 = (x^2+ax+b)\sum_{k=0}^{\infty}c_kx^k$$. 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)
 
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