7
$\begingroup$

Given the recurrence: $$x_{n+1} = ax_n+\frac{b}{x_n}$$ How do I determine the general solution? Is it possible to solve by generating functions? If I could have a small hint that would be great. Thanks.

$\endgroup$
  • 1
    $\begingroup$ There is not going to be a general solution. Just the asymptotics, maybe. $\endgroup$ – Ivan Neretin Oct 20 '17 at 10:16
  • 1
    $\begingroup$ Nah, there is going to be a solution. Or I will make one. $\endgroup$ – mtheorylord Oct 20 '17 at 10:39
  • $\begingroup$ I like your attitude. As to the solution, let's wait for some third opinion. $\endgroup$ – Ivan Neretin Oct 20 '17 at 10:55
  • $\begingroup$ For $\,a\neq 1\,$ it's enough to discuss the special case $\,\displaystyle z_{n+1} = az_n+\frac{1-a}{z_n}\,$ because with $\,\displaystyle z_n:=x_n/\sqrt{\frac{b}{1-a}}\,$ for all $\,n\,$ one gets the initial recurrence. $\endgroup$ – user90369 Oct 20 '17 at 12:24
0
$\begingroup$

Hint:

Similar to Does this recurrence have a closed form limit $x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$?,

Let $x_n=ku_n$ ,

Then $ku_{n+1}=aku_n+\dfrac{b}{ku_n}$

$u_{n+1}=au_n+\dfrac{b}{k^2u_n}$

$u_{n+1}=a\left(u_n+\dfrac{b}{ak^2u_n}\right)$

Case $1$: $ab>0$

Take $k=\dfrac{\sqrt b}{\sqrt a}$ , the recurrence becomes

$u_{n+1}=a\left(u_n+\dfrac{1}{u_n}\right)$

$\endgroup$
  • 2
    $\begingroup$ I'm sorry, I don't see how this is any easier than the original $\endgroup$ – mtheorylord Oct 23 '17 at 19:39

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.