1
$\begingroup$

I'm currently working on this problem:

recursive sequence problem

At first, this looked like a pretty straightforward induction problem. But, once I started working on (b), I ran into an issue.

I can show that my base case is greater than or equal to 2, and then I assume for some n in N that x sub n is greater than or equal to 2. Next, I want to show that x sub n+1 is greater than or equal to 2, and I planned to do so with a simple series of inequalities. However, the 1/(x sub n) in the recursive definition is causing difficulty for me because it breaks my inequality.

Can someone give me some pointers here? Thanks!

$\endgroup$
1
$\begingroup$

maybe you don't need induction. Maybe if you recall the AM-GM inequality

$$ \frac{a+b}{2} \geq \sqrt{ab} $$

For positive $a,b$. Now, as for your problem, notice that

$$ x_{n+1}^2 = \frac{x_n^2}{4} + 1 + \frac{1}{x_n^2} $$

Now, with $a= \frac{x_n^2}{4}$ and $b = \frac{1}{x_n^2}$, one has

$$ x_{n+1}^2 \geq 2 \sqrt{ \frac{x_n^2}{4}\frac{1}{x_n^2} } + 1 = 1+1=2$$

$\endgroup$
  • $\begingroup$ Wow, that is beautiful. Thanks so much. Does the way the recursive sequence was defined preclude me from using induction? $\endgroup$ – Justin Oct 8 '18 at 4:37
  • $\begingroup$ yes, because it will work for any $n$. The proof works for any $n$ so induction is not necesarry. I hope this helps, my friend $\endgroup$ – James Oct 8 '18 at 5:30
  • $\begingroup$ Oops, sorry it looks like I wasn't clear. I was asking if I could also use induction in this proof. I would be curious to see how that works, since I couldn't get the pieces of my inductive proof to work properly. Thanks again. $\endgroup$ – Justin Oct 9 '18 at 0:28
  • $\begingroup$ @Justin AM-GM is commonly used for such, e.g. see this old question on the same problem $\endgroup$ – Bill Dubuque Oct 9 '18 at 1:32
  • $\begingroup$ Thanks All. I used Jimmy's solution for (b), with expanded explanations. I got through (d) by assuming (c), showing boundedness, applying the M.C.T, and solving for the limit. For (c), simple induction is again failing me. Is induction the right way to go? Should I be applying (b) in some way? $\endgroup$ – Justin Oct 9 '18 at 3:46
0
$\begingroup$

$(\frac 12 x_n + \frac 1{x_n})^2 = \frac 14 x_n^2 + \frac 1{x_n}^2 + 1$

If $x_n^2 = 2$ then $\frac 14 x_n^2 + \frac 1{x_n}^2 + 1=2$. But if $x_n^2 > 2$ then

Let $x_n^2 - 2 = \epsilon > 0$

Then $\frac 12 - \frac 1{x_n}^2 = \frac 12 - \frac 1{2+ \epsilon} = \frac {2+\epsilon - 2}{2(2+\epsilon)}=\frac {\epsilon}{4+ 2\epsilon}$

So $\frac 14 x_n^2 + \frac 1{x_n}^2 + 1 = \frac 14(2 + \epsilon) + \frac 12 - \frac {\epsilon}{4+ 2\epsilon}+1=2 + \frac \epsilon 4 - \frac \epsilon {4+ 2\epsilon} > 2 +\frac \epsilon 4 - \frac \epsilon 4=2$

....

In general if $1 \le a < b$ then $b-a > \frac 1b - \frac 1a$. [because $\frac 1b - \frac 1a = \frac {a-b}{ab}< \frac {a-b}1$]. A handy fact to know.

$\endgroup$
  • $\begingroup$ Thanks fleablood! Much appreciated. $\endgroup$ – Justin Oct 9 '18 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.