## Thursday, 18 November 2010

### Plus or Minus Infinity

Came across the problem above at the Math Central web page from Canada. I had written before about the values of iterated roots, but had not considered the case of what happens if they are joined by subtracting instead of adding. It turns out that $\sqrt{n+\sqrt{n+\sqrt{n+...}}}=\frac{1+\sqrt{4n+1}}{2}$. The derivation of this can be found in a short paper I wrote here.

By a similar approach we see that $\sqrt{n-\sqrt{n-\sqrt{n-...}}}=\frac{-1+\sqrt{4n+1}}{2}$ making the problem above pretty simple as an aftermath of all this... but I wondered if there was an intuitive approach to justify this without the actual itereation results....nothing jumps out yet....

It also made me wonder about alternating signs... $\sqrt{n-\sqrt{n+\sqrt{n-...}}}=??$.

Playing around with this I came up with a fourth degree equation to solve, with two positive roots...Yikes, but one of them was a spurious solution created by a second squaring of the equation. When I worked it out for n=2, I got a real surprise.. it led me back to the itereated sequence which started me on this exploration way back when...$\sqrt{1+\sqrt{1+\sqrt{1...}}}$... it turns out that the iteration of roots of two using alternating signs is deeply related to the iteration of roots of one using addition. I'll leave it to you to explore...

And their difference, if I have calculated correctly, is the iterated square roots of 3 using alternating signs.... Too much to explore,...... more later, but I just figured out in class today (I hope) that the series with alternating signs will equal $\frac{\sqrt{4n-3}-1}{2}$ and produces an integer value when the value of n is = $x^2+x+1$ for integer x.