Question

How do you find a limit of \[\lim{\sqrt{2+\sqrt{2 + \sqrt{2 + \sqrt{...}}}}} =\,\, ?\] ?

Answer 1

is there a variable or it just limit of a const

Answer 2

Yes - the variable is the number of nested operations
e.g.\[ a_2= \sqrt{2+\sqrt{2}}\]

Answer 3

so \[ \lim_{n\to \infty}a_n \]

Answer 4

aha, so its basically testing the convergence of divergence of this sequence using limit

Answer 5

use the root test to do so

Answer 6

more than that : THE QUESTION IS THE LIMIT VALUE,
Also I am quite scepical as to the easiness of root test application here ...

Answer 7

It is surely more than 1.414 !!!!!

Answer 8

It is involved, but yes.

Answer 9

HEllo Messrs. @experimentX and @CliffSedge

Answer 10

Uggh ... stop calling me Messrs ... I'm not a woman!!

Answer 11

It looks like it converges to 2 (and rather quickly).

Answer 12

LOL @experimentX He's not calling us women, he's calling us French!

Answer 13

French ...
Woops!! @Mikael how many language do you know??

Answer 14

\[a_{n+1}=\sqrt{2+a_n}\]
Recursion?

Answer 15

How do you prove the convergence itself ???

Answer 16

Hmm, trying to find a way to explain it . . .

Answer 17

Well here is an idea
1) Show that the sequence is increasing
2) Show it is bounded by \[\sqrt{2+2}\]

Answer 18

Then fundam. theorem says it converges

Answer 19

2) is "un peu" trickeeee...

Answer 20

Good night guys!!

Answer 21

Sleep well and dream of mathless bliss !

Answer 22

I guess I'm thinking of it this way:
\[a_{n+1}=\sqrt{2+a_n}\]
\[\rightarrow (a_{n+1})^2 = 2+a_n\]
And a_n is always the same, and it began as 2.
Not very rigorous, but that's how my mind sees what's happening.

Answer 23

OK and the "coup de grace" is : what is the Limmit?

Answer 24

You wrote it !

Answer 25

IT IS CALLED FIXED POINT EQUATION\[a_n = a_{n+1}\]

Answer 26

\[ LIMIT=\sqrt{2+LIMIT}\]

Answer 27

Yes, it converges to 2 like I said before.

Answer 28

Aha, and some courtesy to the guide .....

Answer 29

My guide was MS-Excel, actually, but thanks for the fun question.

Answer 30

Tahnks

Answer 31

Thanks @hartnn - I am glad that we share a taste for the esthetic

Answer 32

me too :D
good question!

Answer 33

I plan to create and post a generalized idea , which in mathematics is ubiquitious and very effective - the idea of \[\Large\bf\text{Fixed Point Solution}\] can you guess and suggest some directions of generalization of that ?

Answer 34

sorry, heard that for first time.....

Answer 35

Well e.g.
Newton-Rapson algorithm is in fact a fixed point solution to any f(x)=0 equation using derivatives

Answer 36

Here is the MAJOR IDEA:
Theorem: Fixed point of Contracting Mappings in Banach space
Assume that \[dist(f(x_1),f(x_2))<r*dist(x_1,x_2)\] where r=const<1

Answer 37

Then the mapping f possesses a fixed point solution, and that solution is unique
\[\Large f(x)= x\]

Answer 38

One has to assume that the mapping f acts on a compact set K in some complete space B, and it is into that set K
\[f:K\,\rightarrow K\]

Answer 39

That's it

Answer 40

i m not qualified enough to understand that....never studied such things.......sorry.

Answer 41

Hey you and I are not judges - we are mutual knowledge "growers" just pointed you the idea that leads to many more places, and can be easily read upon.

Answer 42

http://en.wikipedia.org/wiki/Banach_fixed-point_theorem
http://www.math.ucdavis.edu/~hunter/book/ch3.pdf
http://en.wikipedia.org/wiki/Contraction_mapping

Answer 43

oh..i will surely go through them.....thanks for the references...

Answer 44

That is just some very popular down-to-applications-earth application . Taken from the second reference above

Answer 45

http://www.wolframalpha.com/input/?i=plot+y%3Dsqrt%282+%2Bx%29%2C+y+%3Dx+from+x%3D1+to+2

Answer 46

Thanx @Mikael I really enjoy when you post some thing......

Answer 47

@sauravshakya it includes this topic
http://www.math.ucdavis.edu/~hunter/book/ch3.pdf
But MUCH BETTER TO READ FROM THE BEGINNING
http://www.math.ucdavis.edu/~hunter/book/

Answer 48

Thanks , friend !

Answer 49

Can I ask a question?

Answer 50

Ask away !

Answer 51

WELL what is the solution of this:
I am tired of doing this..........