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Just a relatively simple question; I'm just wondering what would be the proper notation to use to express an infinite power tower that has each repeated exponent increasing by a value of $1$, like such;

$$a^{{{{{{(a+1)}^{(a+2)}}^{(a+3)}}^{.}}^{.}}^{.}}$$

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  • $\begingroup$ Just write $\infty$. $\endgroup$ – azimut Dec 2 '19 at 15:43
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    $\begingroup$ @azimut It's unfair for $a\leq 1$. $\endgroup$ – WhatsUp Dec 2 '19 at 15:45
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    $\begingroup$ It would just be not very interesting function. $\endgroup$ – mathreadler Dec 2 '19 at 16:13
  • $\begingroup$ The order of evaluation is not clear. Usually, $a^{b^c} = a^{(b^c)}$. But I don't see how this evaluation order makes sense with the dots. If you want it the other way round, you have to use parentheses. So maybe you want $(((a^{a+1})^{a+2})^{a+3})^\cdots$? $\endgroup$ – azimut Dec 3 '19 at 8:02
  • $\begingroup$ @azimut Sorry about the confusion; Latex seems to be having some trouble parsing the notation I'm using. The dots are meant to be displayed as if they were continuing up the tower linearly, just like you typically see in the notation used to describe $a \uparrow \uparrow \infty$ $\endgroup$ – Aussie Mathematician Dec 3 '19 at 8:41
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for $0\leq a<1$ the expression become $0$. for $a=1$ it's $1$. For the rest it is $+\infty$ or $-\infty$
But, I read somewhere $n¡=n^{{n-1}^{{{n-2}^.}^.}}.$

note: $n¡$ is the factorial notation $!$ turned upside down. $¡$ keeps exponentiating while $!$ keeps multiplying. Here is an example of it.
It seems you are interested in infinite Power Tower $($Tetration with infinite height$)$. For further information visit here and wikipedia also.

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  • $\begingroup$ Thanks for the reply. I am very much interested in finding out more about this inverted exclamation point notation but I haven't been able to find anything online about it (probably due to not being sure how to properly word my queries). If possible could you please point me to where I can find out more about it? $\endgroup$ – Aussie Mathematician Dec 3 '19 at 2:42
  • $\begingroup$ If my answer satisfied your question then you can accept it by click the check mark @AussieMathematician $\endgroup$ – emonHR Dec 20 '19 at 15:00
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When $a>1,$ $(a+1)^{(a+2)^{(a+3)^{\cdots}}}\ge 2^{3^{4^{\cdots}}}$ will obviosuly go to infinity, and thus $a^{(a+1)^{(a+2)^{(a+3)^{\cdots}}}}$ will also go to infinity.

When $a=1,$ the power tower will clearly go to $1.$

When $0<a<1,$ $(a+1)^{(a+2)^{(a+3)^{\cdots}}}$ will go to infinity, as seen above. Thus, $a^{(a+1)^{(a+2)^{(a+3)^{\cdots}}}}$ will go to $0.$

When $a=0,$ the power tower will clearly go to $0.$

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