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algorithmshark's question from Mathematics Stack Exchange

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
This question was posed by algorithmshark and not yet solved on math.stackexchange.com:

Evaluate the following infinite radical:

$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \cdots}}}}$$
He has posted some thoughts on his question: real analysis - Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ - Mathematics

Various members (including myself) have shown this expression converges, and a few have expressed doubts about the existence of a closed form solution, but I was curious to see what the MHB community could come up with!

EDIT: huh! An analysis subforum popped into existence a few minutes ago. Can we move this there, please?

EDIT [Ackbach]: Your wish is my command.

EDIT: Thank you!
 
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chisigma

Well-known member
Feb 13, 2012
1,704
This question was posed by algorithmshark and not yet solved on math.stackexchange.com:



He has posted some thoughts on his question: real analysis - Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ - Mathematics

Various members (including myself) have shown this expression converges, and a few have expressed doubts about the existence of a closed form solution, but I was curious to see what the MHB community could come up with!

EDIT: huh! An analysis subforum popped into existence a few minutes ago. Can we move this there, please?

EDIT [Ackbach]: Your wish is my command.
The procedure described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/#post2492

... allows a comfortable solution of the problem. The expression $\displaystyle \sqrt{1+\sqrt{2 + \sqrt{4 +\sqrt{8+...}}}}$ seems to be the limit of the sequence solution of the difference equation...

$\displaystyle a_{n+1}= \sqrt{1+\sqrt{2\ a_{n}}},\ a_{0}=1$ (1)

The (1) can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = \sqrt{1+\sqrt{2\ a_{n}}} - a_{n} = f(a_{n})$ (2)


... and the function f(x) is illustrated here...


MSP8291a56e9b16i2d7fh9000035h1acf6bd64114c.JPG

There is only one 'attractive fixed point' in $\displaystyle x_{0} \sim 1.68377$ and pratically any initial value $a_{0}>0$ produce a sequence converging to $x_{0}$...


Kind regards


$\chi$ $\sigma$

As pointed out by IlikeSerena there is an error in (1) and the right solution is published in a successive post... sorry!...
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
The procedure described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/#post2492

... allows a comfortable solution of the problem. The expression $\displaystyle \sqrt{1+\sqrt{2 + \sqrt{4 +\sqrt{8+...}}}}$ is the limit of the sequence solution of the difference equation...

$\displaystyle a_{n+1}= \sqrt{1+\sqrt{2}\ a_{n}},\ a_{0}=1$ (1)

The (1) can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = \sqrt{1+\sqrt{2}\ a_{n}} - a_{n} = f(a_{n})$ (2)


... and the function f(x) is illustrated here...

MSP6151a56g2a1eg40157i00000gf75e4fab5b6a8b.JPG


There is only one 'attractive fixed point' in $\displaystyle x_{0}= \frac{1+\sqrt{3}}{\sqrt{2}} \sim 1.93185$ and pratically any initial value $a_{0}>0$ produce a sequence converging to $x_{0}$...


Kind regards


$\chi$ $\sigma$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Good approach chisigma, though it doesn't work because your recurrence isn't correct. It seems to produce this:

$$\sqrt{1 + \sqrt{2 + \sqrt{8 + \sqrt{64 + \cdots}}}}$$

I think a similar attempt was made in the original thread, but people noted it didn't work because you couldn't make the recurrence compatible with both the doubling at each successive level and the square roots, since they don't cancel at the same rate. I could have calculated your recurrence incorrectly, though.

The actual result is around 1.783 (by computation)
 
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