two limits: 0<b<a: x_0=1, x_1=a+b, x_{n+2}=(a+b)*x_{n+1}-ab*x_n

Vali

Junior Member
Joined
Feb 27, 2018
Messages
57
Let \(\displaystyle 0<b<a\) and \(\displaystyle (x_{n})_{n\in \mathbb{N}}\) with\(\displaystyle x_{0}=1, \ x_{1}=a+b\)
\(\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}\)
a) If \(\displaystyle 0<b<a\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}\) then \(\displaystyle L= ?\)
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If \(\displaystyle 0<b<a<1\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}\) then \(\displaystyle L= ?\)
the right answer is \(\displaystyle L=\frac{1}{(1-a)(1-b)}\)
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,977
Let \(\displaystyle 0<b<a\) and \(\displaystyle (x_{n})_{n\in \mathbb{N}}\) with\(\displaystyle x_{0}=1, \ x_{1}=a+b\)
\(\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}\)
a) If \(\displaystyle 0<b<a\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}\) then \(\displaystyle L= ?\)
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If \(\displaystyle 0<b<a<1\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}\) then \(\displaystyle L= ?\)
the right answer is \(\displaystyle L=\frac{1}{(1-a)(1-b)}\)
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
I cheated due to the limitation of time. Look at this.

Once the recursion has been solved, you do the rest. Post your results.
 
Last edited:

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,351
Let \(\displaystyle 0<b<a\) and \(\displaystyle (x_{n})_{n\in \mathbb{N}}\) with\(\displaystyle x_{0}=1, \ x_{1}=a+b\)
\(\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}\)
a) If \(\displaystyle 0<b<a\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}\) then \(\displaystyle L= ?\)
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If \(\displaystyle 0<b<a<1\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}\) then \(\displaystyle L= ?\)
the right answer is \(\displaystyle L=\frac{1}{(1-a)(1-b)}\)
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
x0 = 1, x1=a+b. x2= x0+2 (so n=0)= (a+b)x1-abx0 = (a+b)2-ab
x3=x1+2 (so n=1) = (a+b)2-ab -ab(a+b) continue (if you like)

What is \(\displaystyle \frac{x_{n+1}}{x_{n}}\)?
 
Last edited:

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,977
x0 = 1, x1=a+b. x2= x0+2 (so n=0)= (a+b)x1-abx0 = (a+b)2-ab
x3=x1+2 (so n=1) = (a+b)2-ab -ab(a+b) continue (if you like)
What is \(\displaystyle \frac{x_{n+1}}{x_{n}}\)?
I am just curious. Are you saying that this can be solved without first solving the recursion?
 

Vali

Junior Member
Joined
Feb 27, 2018
Messages
57
I solved both limits!I used the characteristic equation to find \(\displaystyle x_{n}\)
Thanks!
 
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