What is the value of √(2+√(2+√(2+...))) etc?
This is just a challenge question for which I am curious about.
Hello,
Is it 2 ?
Assuming that $\displaystyle f(n+1)=\sqrt{2+f(n)}$
If L is the limit, then $\displaystyle \lim_{n \to \infty} f(n+1)=\lim_{n \to \infty} f(n)=L$
$\displaystyle L=\sqrt{2+L}$
--> $\displaystyle L^2=2+L \Longleftrightarrow L^2-L-2=0$
The solutions to this are -1 and 2. But L has to be > 0.
Thanks for that...
I, however, being an average tenth grader, do not understand that thing with the n+1 and the limit/infinity stuff. (basically, the entire explanation).
I'm thinking that the n+1 is something got to do with a series of numbers of something...but otherwise, I don't understand it.
If it helps, the challenge question provides a hint on how to start it:
Let x be √(2+√(2+√(2+...))) and square both sides.
If you could possibly explain it starting with a simpler method (for dummies), then perhaps I will be able to understand.
Thanks.
Oh, sorry for that
The n, n+1 thing means "the nth term of the sequence, etc..."
Well, with the hint, I can explain better (I hope), and it's quite alike with my method but it will be more clear
$\displaystyle x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\dots}}}}$
Square it :
$\displaystyle x^2=2+{\color{red}\sqrt{2+\sqrt{2+\sqrt{\dots}}}}$
Since the dots mean "infinity", the red part is also equal to x (this is, imho, the most difficult part of the process, so if you understand it, it's great )
Hence $\displaystyle x^2=2+x$
$\displaystyle x^2-x-2=0$
The discriminant $\displaystyle \Delta=1+8=9$ (I don't know if you've already studied that, if not, complete the square )
Thus the solutions are $\displaystyle x_1=\dots$, $\displaystyle x_2=\dots$
But, since x is the square root of something, it has to be a positive number. This means that you can take only one of the solutions