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I came up with this and I am wondering if it is true, because it seems illogical that $i$ can be made from an infinite power tower of reals.

The way I found this is the following: $$i=e^{\frac{\pi}{2}i}$$ From this rather useless definition of $i$ you can see that $i$ is in its own definition, this definition is circular. Therefore on the right handside I can substitute $i$ for $e^{\frac{\pi}{2}i}$, thus giving me $$i=e^{\frac{\pi}{2}e^{\frac{\pi}{2}i}}$$ And I can keep doing this indefinitely, resulting in and infinite power tower, or better defined:$$a_0=e^{\frac{\pi}{2}}$$ $$a_n=a_{0}^{a_{n-1}},n>1$$ $$i=\lim_{n \to \infty}a_n$$ My question: Is the above limit really equal to i?

[EDIT]

I also came up with a proof that for an infinite power tower $x^{x^{x^{.^{.^{.}}}}}$ to equal $i$, $x$ must equal $e^{\frac{\pi}{2}}$:$$x^{x^{x^{.^{.^{.}}}}}=i$$ $$\ln(x)x^{x^{x^{.^{.^{.}}}}}=\ln(i)$$ $$\ln(x)i=\frac{\pi}{2}i$$ $$\ln(x)=\frac{\pi}{2}$$ $$x=e^{\frac{\pi}{2}}$$ Therefore: $$i = e^{\frac{\pi}{2}e^{\frac{\pi}{2}^{.^{.^.}}}}$$

(Just a reminder I am 13 years old)

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closed as off-topic by Yves Daoust, Mark, ArsenBerk, Leucippus, Lord Shark the Unknown Oct 20 '18 at 4:53

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  • $\begingroup$ The problem is that in your identity there are a finite number of levels in the tower, followed by the exponent $i$. You are assuming that the limit of the finite towers equals the infinite tower - effectively ignoring the effect of last exponent $i$. Why does it disappear? $\endgroup$ – Umberto P. Oct 19 '18 at 19:38
  • $\begingroup$ It disappears because it is an infinite tower therefore it never reaches the end which is i. $\endgroup$ – stavros panagiotidis Oct 19 '18 at 20:18
  • $\begingroup$ Well stated. That's why you can't take the limit of the finite towers and expect to get the infinite tower as a result. $\endgroup$ – Umberto P. Oct 19 '18 at 20:49
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All the terms of your sequence are real, so no. (The sequence diverges to infinity.)

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  • $\begingroup$ At the end there is supposed to be and i making it complex, but since the tower is infinite it never reaches that i. Thats why all the terms are real. But look at this: $$x^x^x^x^...=i$$ $$ln(x)*x^x^x^x^x^x^...=ln(i)$$ $$iln(x)=(pi/2)*i$$ $$x=e^{pi/2}$$ $\endgroup$ – stavros panagiotidis Oct 19 '18 at 19:05
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    $\begingroup$ @stavrospanagiotidis: you may say it in any language, but all terms are real, full stop. $\endgroup$ – Yves Daoust Oct 19 '18 at 19:09
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    $\begingroup$ @stavrospanagiotidis: I'd be glad to read a proof if I see one. Did you care to read mine ? $\endgroup$ – Yves Daoust Oct 19 '18 at 19:19
  • $\begingroup$ Ok, but then could you please explain to me why my 2 different methods lead to my result? $\endgroup$ – stavros panagiotidis Oct 19 '18 at 19:20
  • $\begingroup$ @stavrospanagiotidis: which two "methods", please be specific. All I have seen is a real series that diverges plus unreadable fragments. $\endgroup$ – Yves Daoust Oct 19 '18 at 19:21
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I notice you have taken to harassing other people who have responded to you, but I'll try to take your question seriously. If you haven't yet been banned, please read the code of conduct: https://math.stackexchange.com/conduct

An infinitary notation like $e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi\cdots}{2}}}$ requires an interpretation. The usual and implied way is as a limit of a sequence. In this case, the standard sequence interpretation would be as the limit of the sequence:

$\left\{e^\frac{\pi}{2},{e^{\frac{\pi}{2}e^\frac{\pi}{2}}},{e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi}{2}}}},\ldots \right\} \hspace{3cm} (\mbox{I})$

This is a divergent sequence that approaches $+\infty$ (the terms are getting larger and larger, without bound).

The interpretation you proposed is:

$\left\{e^\frac{\pi i}{2},{e^{\frac{\pi}{2}e^\frac{\pi i}{2}}},{e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi i}{2}}}},\ldots \right\} \hspace{3cm} (\mbox{II})$

which is a constant series

$\left\{ i, i, i, \ldots \right\} \hspace{3cm} (\mbox{II})$

and converges to $i$.

Sometimes it can be helpful to algebraically manipulate a sequence to divine that the terms get close to the terms of a constant series, but the terms in $(\mbox{I})$ and $(\mbox{II})$ are not getting close to each other, so the algebraic manipulation you proposed doesn't help in this case.

There are various non-standard ways to "assign" values to divergent sequences, but I'm not personally aware of any one that would assign the value $i$ to the sequence $(\mbox{I})$.

Here's an excellent video I'd recommend: https://www.youtube.com/watch?v=leFep9yt3JY.

The famous mathematician Ramanujan was known to establish convergence of some sequences, like nested roots, by successively manipulating a constant sequence (like his famous constant sequence for $3$) to show that its terms get close to the terms of some other sequence. Often times he omitted this last critical part, making his arguments frequently lacking in mathematical rigor. He was also famous for studying divergent sequences, especially divergent series.

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By definition $e^{it}=\cos t+i \sin t$. Thus $\large{e^{\frac{\pi}{2}i}=\cos \frac{\pi}{2}+ i \sin \frac{\pi}{2}=i}$. That's it, there is no really infinite power or circular definition, $e^{it}$ is just a useful notation.

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