Question Number 220642 by SdC355 last updated on 17/May/25
$$\mathrm{Calculate} \\ $$$$\boldsymbol{{i}}\uparrow\uparrow^{\infty} =?? \\ $$$$\boldsymbol{{i}}\uparrow\uparrow^{\infty} =\boldsymbol{{i}}^{\boldsymbol{{i}}^{\boldsymbol{{i}}^{\iddots} } } \\ $$
Answered by Ghisom last updated on 17/May/25
![x=i^x ln x =xln i [i=e^(i(π/2)) ∧x=re^(iθ) ] ln r +iθ=((πr)/2)(−sin θ +i cos θ) ⇒ (1) ((πr)/2)sin θ =−ln r (2) ((πr)/2)cos θ =θ ⇒ tan θ =−((ln r)/θ) r=e^(−θtan θ) ⇒ (1) = (2) 2θ−πcos θ e^(−θtan θ) =0 approximating we get infinite solutions the first one is θ≈.688453227 ⇒ r≈.567555163 x≈.438282937+.360592472i this is the value you get when iterating u_1 =i u_(n+1) =i^u_n](https://www.tinkutara.com/question/Q220658.png)
$${x}=\mathrm{i}^{{x}} \\ $$$$\mathrm{ln}\:{x}\:={x}\mathrm{ln}\:\mathrm{i} \\ $$$$\:\:\:\:\:\left[\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \wedge{x}={r}\mathrm{e}^{\mathrm{i}\theta} \right] \\ $$$$\mathrm{ln}\:{r}\:+\mathrm{i}\theta=\frac{\pi{r}}{\mathrm{2}}\left(−\mathrm{sin}\:\theta\:+\mathrm{i}\:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\frac{\pi{r}}{\mathrm{2}}\mathrm{sin}\:\theta\:=−\mathrm{ln}\:{r} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\frac{\pi{r}}{\mathrm{2}}\mathrm{cos}\:\theta\:=\theta \\ $$$$\Rightarrow \\ $$$$\mathrm{tan}\:\theta\:=−\frac{\mathrm{ln}\:{r}}{\theta} \\ $$$${r}=\mathrm{e}^{−\theta\mathrm{tan}\:\theta} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:=\:\left(\mathrm{2}\right)\:\:\:\:\:\mathrm{2}\theta−\pi\mathrm{cos}\:\theta\:\mathrm{e}^{−\theta\mathrm{tan}\:\theta} \:=\mathrm{0} \\ $$$$\mathrm{approximating}\:\mathrm{we}\:\mathrm{get}\:\mathrm{infinite}\:\mathrm{solutions} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{one}\:\mathrm{is} \\ $$$$\theta\approx.\mathrm{688453227}\:\Rightarrow\:{r}\approx.\mathrm{567555163} \\ $$$${x}\approx.\mathrm{438282937}+.\mathrm{360592472i} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{you}\:\mathrm{get}\:\mathrm{when}\:\mathrm{iterating} \\ $$$${u}_{\mathrm{1}} =\mathrm{i} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{i}^{{u}_{{n}} } \\ $$