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This question already has an answer here:

If Euler's formula states that $e^{it} = \cos(t) + i\sin(t)$, then how would I use it to express $e^{e^i}$?

Edit: I didn't even see that $e^{e^i}$ could be rendered as $e^{cos(1)+isin(1)}$. I mishandled the problem and thought that $e^{e^i}$ could be rendered as $e^{ie}$, but can now clearly see that isn't the case. Thank you all for your input, this makes a lot of sense now.

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marked as duplicate by Trevor Gunn, Leucippus, Claude Leibovici, Henrik, J. M. is a poor mathematician Aug 6 '17 at 7:39

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    $\begingroup$ I'm not sure this is so low effort as to warrant so many downvotes. At one point this wasn't obvious to me, either. $\endgroup$ – GPhys Aug 5 '17 at 23:12
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    $\begingroup$ Hm, I see a differential equations tag but no sign of any differential equations. @GPhys also note that many users frown upon problem statement questions. I'd probably classify this as one, noting there is only one sentence in the entirety of the post, which indicates lack of attempt to even write any context. Simply put, I don't see this as a good question and neither do the downvoters. If the OP had no idea where to start, they should see the guide for not having a clue. $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:22
  • $\begingroup$ For some basic information about writing math at this site see here, here, here and here. $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:24
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    $\begingroup$ @GPhys To cut it short, you say that this is not so low effort as to warrant so many downvotes, but from my perspective, there is no effort to begin with, so what are you to expect? $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:26
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    $\begingroup$ My apologies for not conforming to posting standards. I was at a loss and didn't really even know how to proceed. Edited the original post to adhere to community standards. $\endgroup$ – Alex Dymovsky Aug 6 '17 at 1:31
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You would start down from the top because that is how we associate powers by default. $e^i=e^{i \cdot 1}=\cos(1)+i \sin(1)$. Now you want $e^{\cos(1)+i \sin(1)}$. You can plug the $i \sin(1)$ into Euler's formula. The $e^{\cos(1)}$ is a real exponential.

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  • $\begingroup$ This makes a lot of sense, thank you. $\endgroup$ – Alex Dymovsky Aug 6 '17 at 1:33
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Well $e^i = \cos 1 + i \sin 1$ soo $e^{e^i} = e^{\cos 1 + i \sin 1} = e^{\cos 1}[\cos (\sin 1) + i \sin(\sin 1)] \approx e^{0.54030230586813971740093660744298}(\cos 0.8414709848078965066525023216303 + i \sin 0.8414709848078965066525023216303) = 1.7165256995489035179801866917897(0.66636674539288052633780454726237 +i0.55595536063449993982642571881368)$.

There's nothing particularly significant to this number.


I suppose one way thinking about it is $1 \approx \frac {\pi}3$ so $e^i \approx \frac 12 + i\frac {\sqrt{3}}2$ (give or take a rather large degree of error).

So $e^{e^i} \approx e^{\frac 12}(\cos \frac {\sqrt {3}}2 + i\sin \frac {\sqrt{3}}2)$. $\frac{\sqrt {3}}2 \approx \frac 78 \approx \frac {\pi} 4$ (give or take a margin large enough to get you fired from any government job).

So $e^{e^i} \approx \sqrt{e}(\frac {\sqrt{2}}2+ i \frac {\sqrt{2}}2)$ (with a margin of error that if you were aiming for downtown tokyo you would still remain in the solar system).

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    $\begingroup$ Thanks, but could you provide a few more pages of digits? The first twenty clearly aren't satisfying enough for me ;) $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:23
  • $\begingroup$ Heh!. What if I round to one quarter of a significant digit instead. $\endgroup$ – fleablood Aug 5 '17 at 23:28
  • $\begingroup$ Lol, nicely comical :) $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:31
  • $\begingroup$ I wanted to impress that 1) this isn't "oooh, imponderable" question; it has a specific answer and 2) the answer isn't particularly interesting or significant and 3) these numbers aren't actually mumbo-jumbo either. 1 is a number a little under a 6th of the way around a circle so ... the weird and silly and non-significant number will have some actual meaning. $\endgroup$ – fleablood Aug 5 '17 at 23:32
  • $\begingroup$ Quite fun capturing what my screen looks like before the MathJax fully aligns itself $\endgroup$ – Simply Beautiful Art Aug 5 '17 at 23:36

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