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If $k,x_1$ are positive, and $x_{n+1} = \frac{k}{1+x_n}$ . Show that $x_1,x_3,x_5,$··· is monotone increasing, and $x_2, x_4, x_6,$ · · · is monotone decreasing, and that they both have the same limit.

My attempt:

Since $x_n>0$ for all $n$ then $x_{n+1} = \frac{k}{1+x_n} \leq k$

So $x_n-x_{n-1} <k$

Any hint as to what I can do from here?

Thanks!

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  • $\begingroup$ Hint 1: use $x_{n+1}\over x_n$ or subsequence form to show increasing/decreasing. $\endgroup$ – abiessu Sep 30 '13 at 8:43
  • $\begingroup$ Use induction for monotonic claims. $\endgroup$ – Vishal Gupta Sep 30 '13 at 8:59
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Sorry for writing an answer, but I can't write comments yet... Are you sure this is true?

If $x_1 = k = 1$ then $x_2 = 1/2$, $x_3 = 2/3$ and $x_4 = 3/5$, so $x_3 < x_1$ and $x_4 > x_2$. And if $x_1=1/2$ ($k=1$), then $x_2 = 2/3$, $x_3 = 3/5$ and $x_4 = 5/8$, so $x_3 > x_1$ and $x_4 < x_2$. Of course in any situation where $x_1,x_3,...$ is increasing and $x_2,x_4,...$ is decreasing then you can say $\hat{x}_1=x_2$ and the signs would be exchanged.

So apparently more conditions on $x_1$ and $k$ are required... Or maybe what you want to say is that $\frac{x_{2k+1}-x_{2k-1}}{x_{2k+2}-x_{2k}} < 0$ ?

Anyway, I would go for the subsequence $x_{n+2} = \frac{k(1+x_n)}{k+1+x_n}$ and see what you get from there.

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  • $\begingroup$ Thanks for the hints everyone!! I'll come back when I have something $\endgroup$ – wwbb90 Sep 30 '13 at 9:01
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Define the function $$ f(x)=\frac{k}{1+x} $$ Then the equation $f(x)=x$ is equivalent to $x^2+x-k=0$. This equation only has solutions in case $d=1+4k\geq 0$ so it follows that $k\geq-\frac{1}{4}$. So in particular for positive $k$'s we have the solutions $x=-\frac{1}{2}\pm \frac{\sqrt{1+4k}}{2}$. The largest of these two solutions $x=-\frac{1}{2}+\frac{\sqrt{1+4k}}{2}$ is positive for $k>0$. This shows that $f(x)$ has $x_\infty=-\frac{1}{2}+\frac{\sqrt{1+4k}}{2}$ as a positive fixed point for $k>0$.

Clearly $f$ is a strictly decreasing function. So if $x>x_\infty$ then $f(x)<x_\infty$ and vice versa. This shows that the sequence $$ x_n=f^{n-1}(x_1) $$ jumps forth and back from each side of $x_\infty$ and has $x_n<x_\infty$ for odd $n$ if $x_1<x_\infty$ and similarly $x_n>x_\infty$ for even $n$ in that case.

Now consider the function $g(x)=f(f(x))=\frac{k(1+x)}{1+x+k}$. Clearly $x_\infty$ must be a fixed point of this function as well. So $g(x)=x$ has $x_\infty$ as solution. Now $g(x)>x$ leads to $$ -x^2-x+k>0 $$ which has positive solution interval $(0,x_\infty)$ showing that the sequence $$ x_1,g(x_1),g^2(x_1),...=x_1,x_3,x_5... $$ is monotone increasing if $x_1\in(0,x_\infty)$. Then by similar arguments the sequence $$ x_2,g(x_2),g^2(x_2),...=x_2,x_4,x_6... $$ will be monotone decreasing. The limit of both will be $x_\infty=-\frac{1}{2}+\frac{\sqrt{1+4k}}{2}$.

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