# Math Help - recursive sequences!

1. ## recursive sequences!

hey guys...I'm having a big trouble with Recursive Sequences(Sequences and Series in Majority)! I just failed on my first exam because of not having a clear idea about them(their convergence, sum..etc) I was asking if you could send me any link or suggest me a book from where I can get better understanding!

2. Originally Posted by javax
hey guys...I'm having a big trouble with Recursive Sequences(Sequences and Series in Majority)! I just failed on my first exam because of not having a clear idea about them(their convergence, sum..etc) I was asking if you could send me any link or suggest me a book from where I can get better understanding!

3. Originally Posted by Jhevon

Yeah I tried google but I'm not finding the 'core'.
The problem is that my professor didn't make much excersises on this particular issue! On the other hand there are such a probs. on tests!
for e.x. like this one:

or another example:

too messy

4. Originally Posted by javax
Yeah I tried google but I'm not finding the 'core'.
The problem is that my professor didn't make much excersises on this particular issue! On the other hand there are such a probs. on tests!
for e.x. like this one:
show that the sequence is monotonically decreasing and that it is bounded below by 0. this will show it converges. you can do this using induction

for the second part, call the limit $L$ and note that $\lim x_n = \lim x_{n + 1}$

so, $\lim x_{n + 1} = \lim \frac {\sqrt{x_n + 1} - 1}{x_n} \implies L = \frac {\sqrt{L^2 + 1} - 1}L$

now solve for $L$

5. Originally Posted by javax
or another example:

too messy
again, assuming the limit exists, call it $L$

taking the limit of the system $y_n = y_{n - 1}(2 - xy_{n - 1})$ we get:

$L = L(2 - xL)$

now solve for $L$. disregard one of the solutions based on a reasonable cause

6. Originally Posted by Jhevon
again, assuming the limit exists, call it $L$

taking the limit of the system $y_n = y_{n - 1}(2 - xy_{n - 1})$ we get:

$L = L(2 - xL)$

now solve for $L$. disregard one of the solutions based on a reasonable cause
you gotta help me with the $L$ a bit more to finalize!

7. Originally Posted by javax
you gotta help me with the $L$ a bit more to finalize!
help in what way? explain your set back. you mean to solve for L, or are you just not getting the whole L thing altogether?

8. Originally Posted by Jhevon
help in what way? explain your set back. you mean to solve for L, or are you just not getting the whole L thing altogether?
as I understand you're supposing that there exists the limit of Xn, named(L)
if it is for Xn, it means it is for Xn-1 too!
But then?

9. Originally Posted by javax
as I understand you're supposing that there exists the limit of Xn, named(L)
The first step in this is to prove that limits do exist!
You have not done that.

10. Originally Posted by Plato
The first step in this is to prove that limits do exist!
You have not done that.
indeed. as i said javax, induction is the way i usually see this done. and you are to show the sequence is monotonic and bounded, because then, a theorem tells us it converges (by the way, it is in proving the limit exist that you realize which of the answers in the second part to disregard)

Originally Posted by javax
as I understand you're supposing that there exists the limit of Xn, named(L)
if it is for Xn, it means it is for Xn-1 too!
But then?
yes. so when we take the limit, all the $y_n$'s go to $L$, and we get the expression i have there

now we must solve for $L$. begin by expanding the right hand side, and then try to get all the $L$'s on one side to solve for $L$

11. Originally Posted by Plato
The first step in this is to prove that limits do exist!
You have not done that.
assume & then prove(by induction or smth)! I think it goes like that but I better consult with books now...

12. Originally Posted by Jhevon
indeed. as i said javax, induction is the way i usually see this done. and you are to show the sequence is monotonic and bounded, because then, a theorem tells us it converges (by the way, it is in proving the limit exist that you realize which of the answers in the second part to disregard)

yes. so when we take the limit, all the $y_n$'s go to $L$, and we get the expression i have there

now we must solve for $L$. begin by expanding the right hand side, and then try to get all the $L$'s on one side to solve for $L$
Thanks, I think I got your point...I'll see a bit more about induction
Thanks again!

13. Originally Posted by javax
Thanks, I think I got your point...I'll see a bit more about induction
Thanks again!
sure. if you get stuck, come back

what's smth?

14. ahhh, I meant 'something'