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Question Number 101303 by Dwaipayan Shikari last updated on 01/Jul/20
i^i^(i.∞)  =?
$${i}^{{i}^{{i}.\infty} } =? \\ $$
Commented by Dwaipayan Shikari last updated on 01/Jul/20
If there any possible solutions?
$${If}\:{there}\:{any}\:{possible}\:{solutions}? \\ $$
Answered by mr W last updated on 01/Jul/20
say i^i^i^(...)   =x  i^x =x  e^((πix)/2) =x  xe^(−((πix)/2)) =1  −((πix)/2)e^(−((πix)/2)) =−((πi)/2)  −((πix)/2)=W(−((πi)/2))  ⇒x=−(2/(πi))W(−((πi)/2))  ⇒x=((2i)/π)W(−((πi)/2))
$${say}\:{i}^{{i}^{{i}^{…} } } ={x} \\ $$$${i}^{{x}} ={x} \\ $$$${e}^{\frac{\pi{ix}}{\mathrm{2}}} ={x} \\ $$$${xe}^{−\frac{\pi{ix}}{\mathrm{2}}} =\mathrm{1} \\ $$$$−\frac{\pi{ix}}{\mathrm{2}}{e}^{−\frac{\pi{ix}}{\mathrm{2}}} =−\frac{\pi{i}}{\mathrm{2}} \\ $$$$−\frac{\pi{ix}}{\mathrm{2}}={W}\left(−\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{2}}{\pi{i}}{W}\left(−\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{i}}{\pi}{W}\left(−\frac{\pi{i}}{\mathrm{2}}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 01/Jul/20
Thanking you
$${Thanking}\:{you} \\ $$
Commented by mr W last updated on 01/Jul/20

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