I have a question. You might be familiar with the following expression for the golden ratio:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

X = \sqrt{1 + {\sqrt{1 + {\sqrt{1 + {\sqrt{1 + {\sqrt{1 + ....}}}}}}}}}

[/tex]

For those who haven't seen this before, to solve this we just square both sides (which removes the outermost square root sign from the right hand side), subtract one from both sides of the equals sign, and then substitute X for what remains on the right (since it's identical to the original equation). This gives the simple quadratic equation

[tex]

X^2 - 1 = X or X^2 - X - 1 = 0

[/tex]

which is easilly solved to give [itex]X = \frac{1+\sqrt{5}}{2}[/itex]

Playing around, I generalized this by using "n" instead of "1" in the original equation, so that any (non-negative) value could appear under the root sign, just to see what I'd get. This gave me [itex]X^2 - X - n = 0[/itex] or

[tex]

X = \frac{1 + \sqrt{1 + 4n}}{2}

[/tex]

This gives the expected result for n = 1 and the interesting integer results of X = 2 for n = 2. Additional integer solutions occur for n = 6, 12, 20, etc.

But what I find most curious is when we set n equal to zero! Substitituting into the general solution we find that X = 1, which gives us the remarkable equation that

[tex]

1 = \sqrt{0 + {\sqrt{0 + {\sqrt{0 + {\sqrt{0 + {\sqrt{0 + ....}}}}}}}}}

[/tex]

Whoa!!! Obviously something isn't quite kosher here. I do note that taking the negative sign of the square root from the generalized form does give us the expected result. But I'd like to know if there is some fundamental principle I can use to justify excluding the positive root in this case. Hopefully, something more revealing than "because it just doesn't work" might be offered. But I don't see what it might be.

Can anyone help? Thanks.

Bob

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# Recursive square roots

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