5
$\begingroup$

Two real sequences $(x_n)$ and $(y_n)$ are defined by $$x_{n+1}=x_n-(x_ny_n+x_{n+1}y_{n+1}-2)(y_n+y_{n+1})$$ $$y_{n+1}=y_n-(x_ny_n+x_{n+1}y_{n+1}-2)(x_n+x_{n+1})$$ with $x_0=1$ and $y_0=2007.$ I need to show that $|x_n|\lt \sqrt{2007}$ for all $n\in\mathbb{N}.$
I proved that $$x_{n+1}^2-x_n^2=y_{n+1}^2-y_n^2\,\,\,\,\,\,\,\,\,\,\forall n\in\mathbb{N},$$ which implies $|x_n|\lt|y_n|$ and $$x_n^2=y_n^2-2007^2+1.$$

Also I would like to know that, Is $x_n$ convergent? Any Idea?

$\endgroup$
5
  • $\begingroup$ $$x_1=-0.99900323990439297512...\approx-1$$ and $$y_1=2006.9999995036057148...\approx 2007.$$ $\endgroup$
    – Bumblebee
    Jun 11, 2015 at 5:10
  • 1
    $\begingroup$ where from with problem? and why this result is hold? $\endgroup$
    – math110
    Jun 11, 2015 at 5:27
  • $\begingroup$ One of my friend gave me this problem. Most probably it can be some past contest problem. $\endgroup$
    – Bumblebee
    Jun 11, 2015 at 5:30
  • 1
    $\begingroup$ you can also have $$(x_{n}y_{n+1}-x_{n+1}y_{n})=(x_{n}y_{n}+x_{n+1}y_{n+1}-2)(y_{n}y_{n+1}-x_{n} x_{n+1})$$,then maybe use AM-GM inequality $\endgroup$
    – math110
    Jun 11, 2015 at 6:50
  • $\begingroup$ It should be $$(x_{n}y_{n+1}-x_{n+1}y_{n})=(x_{n}y_{n}+x_{n+1}y_{n+1}-2)(y_n^2+y_{n}y_{n+1}-x_n^2-x_{n} x_{n+1}).$$ Is n't it? $\endgroup$
    – Bumblebee
    Jun 11, 2015 at 6:59

1 Answer 1

Reset to default
0
$\begingroup$

Some idea:

let $A=x_{n}y_{n}+x_{n+1}y_{n+1}-2$,since $(2)^2-(1)^2$ then we have $$y^2_{n+1}-x^2_{n+1}=y^2_{n}-2Ay_{n}(x_{n+1}+x_{n})+A^2(x_{n}+x_{n+1})^2-x^2_{n}+2(y_{n}+y_{n+1})x_{n}A-A^2(y_{n}+y_{n+1})^2$$ since $$y^2_{n+1}-x^2_{n+1}=y^2_{n}-x^2_{n}$$then we have $$A(y_{n}y_{n+1}-x_{n}x_{n+1})=y_{n+1}x_{n}-x_{n+1}y_{n}$$ or $$y_{n+1}x_{n}-x_{n+1}y_{n}=(x_{n}y_{n}+x_{n+1}y_{n+1}-2)(y_{n}y_{n+1}-x_{n}x_{n+1})\tag{2}$$

as your comment $$(2)x_{n}:y_{n+1}x_{n}=x_{n}y_{n}-A(x_{n}+x_{n+1})x_{n}\tag{3}$$ $$(1)y_{n}:x_{n+1}y_{n}=x_{n}y_{n}-A(y_{n}+y_{n+1})y_{n}\tag{4}$$ so $(3)-(4)$ $$x_{n}y_{n+1}-x_{n+1}y_{n}=A[y^2_{n}-x^2_{n}+y_{n}y_{n+1}-x_{n}x_{n+1})\tag{6}$$ for $(2),(6)$ you can have ?

$\endgroup$
6
  • $\begingroup$ Consider $(2)x_n-(1)y_n,$ it give a different result. $\endgroup$
    – Bumblebee
    Jun 11, 2015 at 7:59
  • $\begingroup$ if you not wrong, note $y_{n}\neq x_{n}$, then $y_{n}y_{n+1}-x_{n}x_{n+1}=0$ and $y_{n+1}x_{n}=x_{n+1}y_{n}?$ $\endgroup$
    – math110
    Jun 11, 2015 at 8:33
  • $\begingroup$ I beg your pardon. As I can see, In your second line, there is a wrong with $(y_n+y_{n+1})^2$ and $(x_n+x_{n+1})^2.$ $\endgroup$
    – Bumblebee
    Jun 11, 2015 at 8:39
  • $\begingroup$ Oh,Now I have edit it $\endgroup$
    – math110
    Jun 11, 2015 at 8:49
  • $\begingroup$ If both statements were true, then we have that $y_n^2-x_n^2=0.$ $\endgroup$
    – Bumblebee
    Jun 12, 2015 at 3:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .