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An absurdly interesting radical...

DreamWeaver

Well-known member
Sep 16, 2013
337
Here's an interesting limit I found the other day...


\(\displaystyle \text{limit}_{\, \epsilon \to 0^{+}}\sqrt{\epsilon + \sqrt{\epsilon + \sqrt{ \epsilon + \sqrt{\epsilon + \cdots} } } } = 1\)


It's both obvious and yet elusive... Any ideas on how to prove it...???
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Incidentally, I'm only part-way toward a proof myself, so I don't have the answer yet. But still, it's both a Challenge and a Puzzle, so I thought of this board... (Heidy)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Here's an interesting limit I found the other day...


\(\displaystyle \text{limit}_{\, \epsilon \to 0^{+}}\sqrt{\epsilon + \sqrt{\epsilon + \sqrt{ \epsilon + \sqrt{\epsilon + \cdots} } } } = 1\)


It's both obvious and yet elusive... Any ideas on how to prove it...???
Setting $\displaystyle x = f(\varepsilon)$ the x must satisfy the equation $\displaystyle \sqrt{\varepsilon + x} = x$ and, if You suppose that thye only positive value of a square root must be considered, the explicit expression of f(*) is...


$\displaystyle f(\varepsilon) = \frac{1 + \sqrt{1 + 4\ \varepsilon}}{2}\ (1)$


... and observing (1) it seems to be $\displaystyle \lim_{\varepsilon \rightarrow 0} f(\varepsilon) = 1$. However we have to consider that we are evaluating the 'limit of a limit' like $\displaystyle \lim _{\varepsilon \rightarrow 0}\ \lim_{n \rightarrow \infty} \text {of something}$ and the result can be different if You invert the order of limits and evaluate $\displaystyle \lim_{n \rightarrow \infty}\ \lim_{\varepsilon \rightarrow 0}\ \text {of something}$. For this reason also the alternative...


$\displaystyle f(\varepsilon) = \frac{1 - \sqrt{1 + 4\ \varepsilon}}{2}\ (2)$

... [supposing to consider the negative value of the square root...] should be considered and in this case is $\displaystyle \lim_{\varepsilon \rightarrow 0} f(\varepsilon) = 0$. All that in any case requires more investigation...



Kind regards


$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573

chisigma

Well-known member
Feb 13, 2012
1,704
Here's an interesting limit I found the other day...


\(\displaystyle \text{limit}_{\, \epsilon \to 0^{+}}\sqrt{\epsilon + \sqrt{\epsilon + \sqrt{ \epsilon + \sqrt{\epsilon + \cdots} } } } = 1\)


It's both obvious and yet elusive... Any ideas on how to prove it...???
In order to get a correct answer to the question You have to clarify what the expression $\displaystyle f(\epsilon)= \sqrt{\epsilon + \sqrt{\epsilon + \sqrt{ \epsilon + \sqrt{\epsilon + \cdots} } } }$ means. In effect is $\displaystyle f(\epsilon) = \lim_{n \rightarrow \infty} a_{n}$ where $a_{n}$ satisfy the difference equation...


$\displaystyle a_{n+1} = \sqrt{a_{n} + \epsilon},\ a_{0}= \epsilon\ (1)$

Analysing (1) You imposing the condition $\Delta_{n} = a_{n+1} - a_{n} =0$ You find that that there is an attractive fixed point in $\displaystyle x_{1} = \frac{1 + \sqrt{1 + 4\ \epsilon}}{2}$ and a repulsive fixed point in $\displaystyle x_{0} = \frac{1 - \sqrt{1 + 4\ \epsilon}}{2}$ and that means that for any $\displaystyle x_{0} < \epsilon < x_{1}$ is $\displaystyle f(\epsilon) = x_{1} = \frac{1 + \sqrt{1 + 4\ \epsilon}}{2}$ and $\displaystyle \lim_{\epsilon \rightarrow 0} f(\epsilon)=1$, even if [and that is an important detail...] is $\displaystyle f(0)=0$...


Kind regards


$\chi$ $\sigma$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
I am not sure if this actually works (there might be something missing) but here goes:

Suppose $\sqrt{a + \sqrt{a + \sqrt{a + \cdots}}} = a^*$, $\sqrt{b + \sqrt{b + \sqrt{b + \cdots}}} = b^*$, and $0 < a < b$.

Then $a^* = \frac{1 + \sqrt{1 + 4a}}{2}$ and $b^* = \frac{1 + \sqrt{1 + 4b}}{2}$ therefore it follows that $a^* < b^*$.

Now suppose that $a^* < 1$. This leads to the contradiction that $\sqrt{1 + 4a} < 1$. Therefore $a^* \geq 1$, so $1 \leq a^* < b^*$.

Therefore $a^*$ goes to $1$ as $a$ goes to $0$.

$\blacksquare$