An absurdly interesting radical....

[DreamWeaver]

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]

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]

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...?

Spoiler

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]

A so called crank used this to prove that $1 = 0$ quite a few days ago in MMF : http://www.mymathforum.com/viewtopic.php?f=40&t=44831

 

[chisigma]

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 $\chi$ $\sigma$

 

[Nono713]

I am not sure if this actually works (there might be something missing) but here goes:

[sp]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 $0$.

$\blacksquare$[/sp]