Limits of Recurrence Relations with $0<b<a$

In summary, the two relations are equivalent and the function is $f(y) = \frac{1}{(ay - 1)(by-1)} = \sum_{j \ge 0}a^j y^j$.
  • #1
Vali
48
0
Let $0<b<a$ and $(x_{n})_{n\in \mathbb{N}}$ with $x_{0}=1, \ x_{1}=a+b$
$$x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $0<b<a$ and $L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$ then $L= ?$
b) If $0<b<a<1$ and $L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$ then $L= ?$
I don't know how to start.I tried to write the first terms $x_{2},x_{3}...$ but I didn't get too far.
 
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  • #2
Define $\displaystyle f(y) = \sum_{n \ge 0} x_n y^n$. Multiply both sides of your relation by $y^n$ and sum over $n \ge 0$:

$\displaystyle \mathcal{LHS} = \sum_{n \ge 0}x_{n+2}y^n = \frac{1}{y^2}\left(\sum_{ n \ge 0} x_n y^n - x_0y^0-x_1y^1\right) = \frac{1}{y^2}\left(f(y) - (a+b)y-1\right)$

$\displaystyle \mathcal{RHS} = (a+b)\sum_{n \ge 0}x_{n+1}y^n-(ab)\sum_{n \ge 0}x_{n}y^n = \frac{a+b}{y} \left(f(y)-1\right)-ab f(y), $ and:

$\displaystyle \mathcal{LHS} = \mathcal{RHS} \implies f(y) = \frac{1}{(ay - 1)(by-1)} = \sum_{j \ge 0}a^j y^j \cdot \sum_{\ell \ge 0}b^{\ell}y^{\ell} = \sum_{n \ge 0}\sum_{0 \le j \le n}a^j y^j b^{n-j}y^{n-j}$

Since the inner sum equals $(a-b)^{-1}(a^{n+1}-b^{n+1})y^n$ going back to the definition of $y(n)$ we've:

$\displaystyle \sum_{n \ge 0}{x_n y^n} = \sum_{n \ge 0} \frac{a^{n+1}-b^{n+1}}{a-b} y^n $ and we can readily read off from the coefficient: $\displaystyle x_n = \frac{a^{n+1}-b^{n+1}}{a-b}.$

Can you do (a) and (b) now? You can also find $x_n$ by solving the characteristic equation.
 
Last edited:
  • #3
I solved both limits!I used the characteristic equation to find [tex]x_{n}[/tex]
Thanks!
 

FAQ: Limits of Recurrence Relations with $0<b<a$

1. What is a recurrence relation?

A recurrence relation is a mathematical equation that defines a sequence by relating each term to one or more of the previous terms.

2. How do you determine the limit of a recurrence relation?

The limit of a recurrence relation can be determined by finding the pattern in the sequence and using algebraic techniques to simplify the equation. In some cases, the limit may be found by taking the limit as n approaches infinity.

3. What is the significance of having $0

Having $0

4. Can a recurrence relation with $0

Yes, it is possible for a recurrence relation with $0

5. Are there any practical applications for recurrence relations with $0

Yes, recurrence relations with $0

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