Find x:

x = sq.root[1 + sq.root( 1 + sq.root(1 + sq.root(1 +....

"The dots mean that the series go till infinity.

By attempt the answer should be (1 +-sq.root(5))/2 but that is not the answer.(Headbang)

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- Dec 21st 2012, 10:52 PMAditya3003Root problem
Find x:

x = sq.root[1 + sq.root( 1 + sq.root(1 + sq.root(1 +....

"The dots mean that the series go till infinity.

By attempt the answer should be (1 +-sq.root(5))/2 but that is not the answer.(Headbang) - Dec 21st 2012, 11:03 PMjakncokeRe: Root problem
/the answer is $\displaystyle \frac{1 \frac{+}{-} \sqrt(5)}{2} $

The infinite nested radicals, $\displaystyle \sqrt(1 + \sqrt(1 + \sqrt(1 + .... $ = golden ratio.

Since golden ratio says $\displaystyle \frac{a+b}{a} = \frac{a}{b} = x $ then $\displaystyle 1 + \frac{b}{a} = 1 + \frac{1}{x} = x $ so solving $\displaystyle x^2 - x - 1 = 0 $ - Dec 21st 2012, 11:08 PMibduttRe: Root problem
Could you please write as to what is the answer given. The answer you have got appears to be correct.

Attachment 26322 - Dec 21st 2012, 11:10 PMjakncokeRe: Root problem
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- Dec 22nd 2012, 05:06 AMSorobanRe: Root problem
Hello, Aditya3003!

Quote:

$\displaystyle \text{Evaluate: }\:x \:=\: \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}} $

By attempt the answer should be: $\displaystyle \frac{1\pm\sqrt{5}}{2}$, but that is not the answer.

Since $\displaystyle x$ is obviously positive,

. . you must reject the negative root: $\displaystyle \tfrac{1-\sqrt{5}}{2}$

While solving, youthe equation, right?*squared*

. . This can introduce an extraneous root.