4
$\begingroup$

$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$

$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$

Why they are different?

  • 0
    The second thing must be $\lim_{n\to \infty} x_n$ where $x_0=0$ and $x_{n+1}=\sqrt{x_n+2}$. But expressions like the first one always confuse me a bit. It's obviously supposed to be some positive solution to $\sqrt{2+x}=x$, but it feels a bit ambiguous2011-03-09
  • 1
    A proof of convergence can be found here: http://math.stackexchange.com/questions/11945/limit-of-sqrt7-sqrt7-sqrt7-cdots/11969#11969. With 7 replaced by 2, the proof basically carries over.2011-03-09
  • 0
    @Rasmus Do you know how to rigorously 'define' that first thing? To me it seems both limiting points would satisfy the same equation $x=\sqrt{x+2}$.2011-03-09
  • 0
    @user7992 Answer: They are not, since both numbers equal $2$.2011-03-09
  • 0
    @Myself: I deleted my previous comment because I misread the question.2011-03-09
  • 0
    @user7992: How do you define the first expression?2011-03-09

2 Answers 2

4

Define a sequence $(x_n)_{n\geq1}$ so that $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2+x_n}$. Then the second formula you give can be said to colorfully describe the limit $\lim_{n\to\infty}x_n$. If we suppose the limit $L$ does exist (and it is not difficult to show it does exist!), then since for all $n$ we have $x_{n+1}^2=2+x_n$, passing to the limit we see that $L^2=2+L$. The polynomial $x^2-x-2$ has two roots, $-1$ and $2$: since all the $x_n$ are positive, the only possible value for $L$ is $2$.

Can you do the other one?

  • 0
    Thank you very much. But I still can't solve the first one, because I can't write the recursive formula for the first one as you gave for the second one. My teacher told me that they were different. But from the others' replies, it seems they are the same. I'm confused.2011-03-09
  • 0
    You *could* ask your teacher :)2011-03-09
  • 0
    I wonder what motivated the downvote...2011-03-09
4

Suppose that the first converges to some value $x$. Because the whole expression is identical to the first inner radical, $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=x=\sqrt{2+x}$ and solving for $x$ gives $x=2$. Of course, I haven't justified that it converges to some value.

The second can be thought of as starting with $\sqrt{2}$ and repeatedly applying the function $f(x)=\sqrt{2+x}$. Trying this numerically suggests that the values converge to 2. Solveing $f(x)=x$ shows that $2$ is a fixed point of that function.

Looking at the second expression is actually how I'd justify (though it is perhaps not a rigorous proof) that the first expression converges.