14
$\begingroup$

I would appreciate any help with this problem:

If

$$y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$$

Then how do I find $y^2 - y$?

I'm not sure whether this is an arithmetic or geometric series.

$\endgroup$
2

3 Answers 3

28
$\begingroup$

Note that $(y^2-5)^2=5-y$. It is also clear that $ y^2 \geq 5$ and $0<y$.

We have $0=y^4-10y^2+y+20=(y^2-y-4)(y^2+y-5)$.

We have $y^2+y-5>0$, so $y^2-y=4$.

P.S. Incidentally, if you want to find $y$, it is the positive root of $y^2-y-4=0$, which is $\frac{1+\sqrt{17}}{2}$.

$\endgroup$
3
  • 10
    $\begingroup$ One really needs to show that the sequence converges. If it doesn't converge, the solution to $y=\sqrt{5+\sqrt{5-y}}$ is meaningless. $\endgroup$
    – robjohn
    Commented Feb 28, 2013 at 21:29
  • $\begingroup$ Honestly, I didn't think I would get so many upvotes... Number of upvotes seems to have a negative correlation with amount of time spent solving the problem, but that's another matter. Of course convergence needs to be shown for $y$ to even be defined, which I seemed to have skipped over while busy with other stuff. I would edit it into my answer, but it seems that @robjohn has provided a better one (+1 for him), so I'll leave it at that. $\endgroup$
    – Ivan Loh
    Commented Mar 1, 2013 at 1:36
  • 1
    $\begingroup$ @IvanLoh: as I tell people all the time, voting is capricious. Don't worry about votes that you may not think you've earned, because there will be plenty of cases where you won't get votes when you do deserve them. $\endgroup$
    – robjohn
    Commented Mar 1, 2013 at 1:53
26
$\begingroup$

Define $$ a_0=0\quad\text{and}\quad a_{k+1}=\sqrt{5+\sqrt{5-a_k}} $$ Show that for $k\ge1$, $\sqrt5\le a_k\le\sqrt{5+\sqrt5}$.

Initially, $0\le a_0=0\le5$.

Suppose that $0\le a_k\le 5$. Then $\sqrt5\le a_{k+1}\le\sqrt{5+\sqrt5}$. $$ \begin{align} \left|\,a_{k+1}-a_k\,\right| &=\frac{\left|\,\left(5+\sqrt{5-a_k}\right)-\left(5+\sqrt{5-a_{k-1}}\right)\,\right|}{a_{k+1}+a_k}\\ &=\frac{\left|\,\sqrt{5-a_k}-\sqrt{5-a_{k-1}}\,\right|}{a_{k+1}+a_k}\\ &=\frac{\left|\,a_k-a_{k-1}\,\right|}{(a_{k+1}+a_k)\left(\sqrt{5-a_k}+\sqrt{5-a_{k-1}}\right)}\\ &\le\frac{\left|\,a_k-a_{k-1}\,\right|}{\left(\sqrt5+\sqrt5\right)\left(\sqrt{5-\sqrt{5+\sqrt5}}+\sqrt{5-\sqrt{5+\sqrt5}}\right)}\\ &\le\frac{\left|\,a_k-a_{k-1}\,\right|}{13} \end{align} $$ Thus, $a_k$ converges since $$ \lim_{n\to\infty}\sum_{k=1}^n(a_k-a_{k-1})=\lim_{n\to\infty}a_n-a_0 $$ converges absolutely; that is, $$ \sum_{k=1}^\infty\left|\,a_k-a_{k-1}\right|\le\frac{13}{12}\sqrt{5+\sqrt5} $$

Set $a=\lim\limits_{k\to\infty}a_k$. Since $\sqrt{5+\sqrt{5-x}}$ is continuous for $x\le5$, we have that $a=\sqrt{5+\sqrt{5-a}}$, which means $$ \begin{align} 0 &=a^4-10a^2+a+20\\ &=(a^2-a-4)(a^2+a-5) \end{align} $$ The roots of $a^2-a-4$ are $\frac{1\pm\sqrt{17}}{2}$ and the roots of $a^2+a-5$ are $\frac{-1\pm\sqrt{21}}{2}$. The only one that is between $\sqrt5$ and $\sqrt{5+\sqrt5}$ is $\frac{1+\sqrt{17}}{2}$. Therefore, $$ \lim_{k\to\infty}a_k=\frac{1+\sqrt{17}}{2} $$

$\endgroup$
2
  • 3
    $\begingroup$ This is an answer. A real answer. $\endgroup$ Commented Feb 28, 2013 at 21:58
  • $\begingroup$ Thank you! You provided explanations as well. :) $\endgroup$ Commented Feb 28, 2013 at 23:40
0
$\begingroup$

It's not a series at all, just a recursively defined sequence. Notice the successive terms alternating in sign means by constructing the terms recursively you have for any infinitieth term a different expression for the next one.

Compare $\{1, -1, 1, -1, ...\}$ which does not converge.

So, the sequence can only have a limit if the values are the same regardless of the expression. To test that, though, you need a way to make sure the first time you test it the infinitieth value has one sign, and the second time it has the other sign.

You need to test two sequences for convergence:

$a_{k+1}=\sqrt{5+\sqrt{5-a_k}}$ and $a_{k+1}=\sqrt{5-\sqrt{5+a_k}}$.

Suppose they converge. Since you defined the outermost sign to always be positive, if the second of the two converges to $a$, take $\sqrt{5+a}$. If the limit of the 1st one is the same as $\sqrt{5+a}$, then it doesn't matter whether the sign at the limit is positive or negative, so the original sequence you defined converges.

$\endgroup$
2
  • $\begingroup$ At least the way I interpret it, the sequence defined by $b_{k+1}=\sqrt{5-\sqrt{5+b_k}}$ is encountered as an intermediate value when computing the given sequence, but the terms of the sequence in the question are defined by $a_{k+1}=\sqrt{5+\sqrt{5-a_k}}$ since the only sequence mentioned starts $\sqrt{5+\sqrt{5-\sqrt{5+\dots}}}$. $\endgroup$
    – robjohn
    Commented Feb 28, 2013 at 22:33
  • $\begingroup$ The sequence as defined doesn't discern which is the intermediate. The sequence starts with the same sign either way, it just isn't well-defined which sign it ENDS with, just like alternating 1 & -1 starting from 1. To put it another way, is infinity odd or even? I'd leave the other sequence as an exercize. $\endgroup$
    – Loki Clock
    Commented Feb 28, 2013 at 22:42

You must log in to answer this question.

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