Simply: How do I solve this equation for a given $n \in \mathbb Z$?
$x^x = n$
I mean, of course $2^2=4$ and $3^3=27$ and so on. But I don't understand how to calculate the reverse of this, to get from a given $n$ to $x$.
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Have you tried solving this for small $n$ say $n=2$? – Mar 29 '16 at 06:40
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@PraphullaKoushik I have no idea how to solve this. Of course $2^2=4$ and $3^3=27$, but I have no idea what it should be for $n=2$. – Byte Commander Mar 29 '16 at 06:42
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Why the downvote? – Byte Commander Mar 29 '16 at 06:44
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2Some people automatically downvote posts that show no effort of attempt to solve it themselves. To prevent this in the future, I would include your $2^2$ and $3^3$ in your original post so that it looks like you tried to others. – Decaf-Math Mar 29 '16 at 06:48
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The solutions are all $x\in\mathbb{N}$ I think. – Theodoros Mpalis Mar 29 '16 at 07:04
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That one has infinite solutions. – Theodoros Mpalis Mar 29 '16 at 07:04
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@TheodorosMpalis Why should this have infinite solutions? When $n=27$, my only solution is $x=3$. The $n$ is a given constant from which I want to calculate $x$. – Byte Commander Mar 29 '16 at 07:07
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@ByteCommander You need to find "x" that $x^x=n\in\mathbb{N} and NOT "n". – Theodoros Mpalis Mar 29 '16 at 07:14
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Is $x$ also constrained to be in $\mathbb Z$ ? (If not, knowing that $n$ is integer doesn't make the problem simpler.) – Mar 29 '16 at 07:28
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@YvesDaoust No, $x \in \Bbb R$. There's no such simple restriction unfortunately. – Byte Commander Mar 29 '16 at 07:49
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Then $n\in\mathbb Z$ is of no use. – Mar 29 '16 at 07:56
2 Answers
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See this wikipedia article: Lambert W function
If $x^x=n,$ then $$x=\dfrac{\ln n}{W(\ln n)},$$ Where $W$ is the Lambert W function.
Byte Commander
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Bumblebee
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Hey out of curiosity , could you show your steps how you got to x= ? – bigfocalchord Mar 29 '16 at 07:13
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It may required a little knowledge about Lambert W function. You can find this from the given link. – Bumblebee Mar 29 '16 at 07:17
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That requires more than a little knowledge, and I've never heard of the Lambert W function before... I would need to find a way to code this in Python, but I can't use the
scipypackage where this function would be built in... :-/ – Byte Commander Mar 29 '16 at 07:26 -
@ByteCommander Basically, you're not going to find a solution in terms of elementary functions. You either have to use a numerical approximation algorithm (as noted in marty cohen's answer), or a special function such as this (Lambert W function). – Eff Mar 29 '16 at 07:28
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1@ByteCommander If you wish to calculate the Lambert W function, there are some easy methods you may find in the Lambert W tag. – Simply Beautiful Art Apr 02 '16 at 13:02
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Simpler:
If $x^x = n$, then $x\ln(x) = \ln(n) =y$.
Let $f(x) = x\ln(x)-y $. $f'(x) =\ln(x)+1 $.
Applying Newton's iteration, starting with $x = \frac{y}{\ln y}$, $x_{new} =x-\frac{f(x)}{f'(x)} =x-\frac{x\ln(x)-y}{\ln(x)+1} =\frac{x\ln(x)+x-x\ln(x)+y}{\ln(x)+1} =\frac{x+y}{\ln(x)+1} $.
Iterate until cooked.
marty cohen
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Thanks for your explanation. But maybe the formula would be more clear if you don't substitute $y = ln(n)$... – Byte Commander Mar 29 '16 at 07:34
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And I just noticed that your value for $y$ must be manually set to $1$ or something similar if $n<1$. Else we get a problem because it would evaluate to $ln(0)$. – Byte Commander Mar 29 '16 at 07:48
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@ByteCommander Yes, of course, it is recommended that you choose $y$ so that you can avoid such troubles. Interestingly, if you choose different $y$ values, you may experience different complex or real results... (try Google calculator for complex logarithm) – Simply Beautiful Art Apr 02 '16 at 13:01