How to interpret this class of numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:24:04Z http://mathoverflow.net/feeds/question/100430 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100430/how-to-interpret-this-class-of-numbers How to interpret this class of numbers? Damodar Kulkarni 2012-06-23T07:52:01Z 2012-07-28T11:27:54Z <p>Let's say $f(p)$ is a number defined as shown below:<br></p> <p>$\hspace {10 mm}f(p) = {\sqrt {2} ^ {{\sqrt {2} ^ \sqrt {2}} ^ {... \hspace {1 mm} p\hspace {1 mm} times } }}$</p> <p>What I understand is: <br> We normally assume, presume or impose a restriction on $p$ that $p$ should be a positive integer. <br> <br> My question is: <br> <b>Can we allow $p$ to be any number in general?</b></p> <pre> e.g. Can $p$ be irrational? </pre> <p>If $yes$, how to interpret such a number? <br> If $no$, why? <br> </p> <p>Thanks.</p> http://mathoverflow.net/questions/100430/how-to-interpret-this-class-of-numbers/100589#100589 Answer by Harun Šiljak for How to interpret this class of numbers? Harun Šiljak 2012-06-25T12:51:43Z 2012-06-25T12:51:43Z <p><em>As suggested in meta, I'll turn my comment into an answer (CW).</em></p> <p>At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex values of n. Various approaches are mentioned in <a href="http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights" rel="nofollow">http://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights</a> .</p> http://mathoverflow.net/questions/100430/how-to-interpret-this-class-of-numbers/103373#103373 Answer by Gottfried Helms for How to interpret this class of numbers? Gottfried Helms 2012-07-28T09:57:50Z 2012-07-28T11:27:54Z <p>In addition to the answer of Harun Siljak one should perhaps mention that for the slightly different expression $$\operatorname{dxp}\small([topexponent],[base],[iterationheight])=\operatorname{dxp}(x,t,h)$$ where $$\operatorname{dxp}(x,t,0)= x <br>\\ \operatorname{dxp}(x,t,1)= t^x - 1 <br>\\ \operatorname{dxp}(x,t,2)= t^{t^x - 1 } - 1 <br>\\ \cdots$$ there is a solution for real $h$ based on the power series for the exponential function minus the constant term. Usually this is discussed for the function $$\operatorname{dxp}(x,e,1)= \exp(x) - 1$$ and fractional or even irrational heights $h$ and a parametrization for the coefficients for $$\operatorname{dxp}(x,t,1)= \sum_{k=1}^\infty u^k{x^k \over k!} <br>\\ \operatorname{dxp}(x,t,h)= \sum_{k=1}^\infty \mathcal{P}(u,h,k){x^k \over k!} <br>\\ \text{where I wrote }u \text{ for } \ln(t)$$ where $\mathcal{P}$ denotes a polynomial in iteration-height, $\ln(t)$ and the series-index $k$. </p> <p>For series like this and its iterations it is accepted, that the indicated family of iteration heights form a semigroup, where the height-parameter $h$ can be non-integer and can even be complex. This can already be found in L.Comtet's "advanced combinatorics" but also elsewhere. </p> <p>Unfortunately, although the iterations of <em>dxp()</em> and <em>exp()</em> can be converted into each other (simply by a change of base) for integer heights, this is not uniquely determined for the fractional heights (the reason is, that for the same base in $b^x$ we have multiple bases $t$ in $t^x-1$ and the various $t$ give different results for the same $x$ and height $h$ if $h$ is fractional). Which then leads to the comment in the other answer, that there is not (yet) a commonly accepted interpretation for the noninteger heights in your original problem.</p>