What is the value of √(2+√(2+√(2+...))) etc?

This is just a challenge question for which I am curious about. :)

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- Apr 30th 2008, 11:31 PManne.[SOLVED] Just a challenge question
What is the value of √(2+√(2+√(2+...))) etc?

This is just a challenge question for which I am curious about. :) - Apr 30th 2008, 11:37 PMMoo
Hello,

Is it 2 ? :D

Assuming that

If L is the limit, then

-->

The solutions to this are -1 and 2. But L has to be > 0. - Apr 30th 2008, 11:38 PManne.
It could be...I have no idea how to check it. Can you explain how you worked that out? Thanks. :)

- Apr 30th 2008, 11:40 PMMoo
I edited :)

This is, of course, assuming that the series converges. - May 1st 2008, 12:28 AManne.
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. - May 1st 2008, 12:34 AMMoo
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 :)

Square it :

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 (Wink))

Hence

The discriminant (I don't know if you've already studied that, if not, complete the square ;))

Thus the solutions are ,

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 (Bow) - May 1st 2008, 12:44 AManne.
Oh, I understand it now. (Rofl)

Thanks for your help (Bow)

(PS: How much time do you spend on this forum? Every time I ask a question, it's always you who replies so quickly :D) - May 1st 2008, 12:48 AMMoo