I'm having a little bit of trouble figuring out how exactly to do this.(adsbygoogle = window.adsbygoogle || []).push({});

Prove that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].

How exactly does one go about doing this? I mean, I understand it goes on infinitely, but doesn't that create an infinitely large number? Why would the number be so precise?

Also, it appears to be the solution of a quadratic equation (with the [itex]\pm[/itex] and all). Is that an approach? If so, how does one derive such a quadratic equation?

I tried working backwards, but when I do, all I get is that [itex]\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \frac{1\pm\sqrt{4n+1}}{2}[/itex].

There's one less [itex]n[/itex] in the square root, but I still end up getting the same equation. Can this process by repeated infinitely until I get [itex]\sqrt{n} = \frac{1\pm\sqrt{4n+1}}{2}[/itex]? But this isn't true! Can someone help me?

Any help would be gladly appreciated.

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Endless Square Root Problem

Loading...

Similar Threads for Endless Square Root |
---|

I Can we construct a Lie algebra from the squares of SU(1,1) |

Least Square basic problem |

B ##AB = I \implies BA = I##, for square matricies ##A,B## |

**Physics Forums - The Fusion of Science and Community**