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For a pair of real numbers $\alpha$ and $\beta$, I need to prove that for the sequences

$a_n = (-\alpha)a_{n-1} +b_{n-1}$

$b_n = (-\beta)a_{n-1}$

an upper bound exists with a form similar to

$|a_n| \leq cA^n$

$|b_n| \leq dA^n$

for all natural numbers $n$ where $c$, $d$, and $A$ are positive real numbers.

I have previously guessed

$|a_n|\leq \alpha^n$

$|b_n|\leq \beta\alpha^{n-1}$

but this seems to fail if $\beta \gt \alpha^2$. Help?

*This is part of a longer problem I am working on (Taylor expansion convergence). I need to prove $a_n F'(x) + b_n F(x) \leq A^n(cF'(x)+dF(x))$

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1 Answer 1

The recursion can be rewritten as $$ \begin{pmatrix}a_n\\b_n\end{pmatrix}=\underbrace{\begin{pmatrix}-\alpha&1\\-\beta&0\end{pmatrix}}_{=:M}\begin{pmatrix}a_{n-1}\\b_{n-1}\end{pmatrix}.$$ The eigenvalues of matrix $M$ are the solutions of $$\lambda^2-\operatorname{tr}(M)\lambda+\det(M)=\lambda^2+\alpha\lambda+\beta=0,$$ i.e. $\lambda_{1,2}=\frac{-\alpha\pm\sqrt{\alpha^2-4\beta}}{2}$. For the corresponding eigenvectors, $M$ acts simply as multiplication by $\lambda$. Therefore, if $\lambda_1\ne\lambda_2$, we have $a_n=u\lambda_1^n+v\lambda_2^n$ for suitable $u,v$ and similarly for $b_n$. If $\lambda_1=\lambda_2$ (i.e. if $\alpha^2=4\beta$), we have to use the ansatz $a_n=(u+nv)\lambda^n$ instead.

At any rate, if $A>\max\{|\lambda_1|,|\lambda_2|\}$, then there indeed exist $c,d$ with $a_n\le cA^n$, $b_n\le dA^n$.

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