I'm stuck on this problem, can someone tell me what the simplification of the following is?
$$\ {e^2}^n $$
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$e^{2^n}$...... – Sep 19 '17 at 03:20
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I was thinking that it could just be simplified down to $$\ 2^n $$ , but I'm not sure... – random_x_y_z Sep 19 '17 at 03:22
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No, that is not true – Sep 19 '17 at 03:22
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yeah, k........ – random_x_y_z Sep 19 '17 at 03:23
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@ZacharySelk Could be $e^{2^n}=k^n$ for a $k$.? – Nosrati Sep 19 '17 at 03:25
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1@MyGlasses No, $e^{2^n}$ diverges much more quickly than $k^n$ for any $k>0$. – shalop Sep 19 '17 at 03:26
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$\sum_{i=0}^{\infty} \frac{2^{ni}}{i!}$ – Michael Sep 19 '17 at 04:04
1 Answers
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The default is that ${e^2}^n$ is read as $e^{(2^n)}$, for which there is no simplification. In fact, the reason given for the default is the lack of a simplification. It becomes larger than $k^n$ for any $k$ once $n$ is large enough.
Ross Millikan
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nope, not missing anything, $$\ {e^2}^n $$ is exactly what i meant. – random_x_y_z Sep 19 '17 at 03:41
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