4

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

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

Why they are different?

Timothy
  • 176
  • 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 ambiguous – Myself Mar 09 '11 at 05:45
  • 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. – Aryabhata Mar 09 '11 at 06:05
  • @Rasmus Do you know how to rigorously 'define' that first thing? To me it seems both limiting points would satisfy the same equation $x=\sqrt{x+2}$. – Myself Mar 09 '11 at 07:10
  • @user7992 Answer: They are not, since both numbers equal $2$. – Did Mar 09 '11 at 07:35
  • @Myself: I deleted my previous comment because I misread the question. – Rasmus Mar 09 '11 at 10:50
  • @user7992: How do you define the first expression? – Rasmus Mar 09 '11 at 10:53

2 Answers2

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?

Mariano Suárez-Álvarez
  • 135,076
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.

Isaac
  • 36,557