For example if you have $x^x = 2$, can you express $x$ as a numerical expression containing only the addition, multiplication and exponentiation operators?
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2Not really – SBF Apr 04 '13 at 12:19
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See here. – Mhenni Benghorbal Apr 04 '13 at 12:21
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The solution of your equation is $\frac{\ln(a)}{W(\ln(a))},$ where $W$ is the Lambert $W$-function. – Mhenni Benghorbal Apr 04 '13 at 12:42
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I looked at problems like this once and found the Lambert W function. It is not the addition, multiplication and exponential operators you have asked for. I can't see how you would do that since we get the exact answer as follows:
$x\exp(x)=y \iff x=W(y)$
$x^x=2$
$x \log x=\log 2$
$(\log x)\exp(\log x) = \log 2$
$x=\exp(W(\log 2))$
But $\log(2)=W(\log 2)\exp(W(\log 2))=W(\log 2)x$
$x=\frac{\log 2}{W(\log 2)}$
J. M. ain't a mathematician
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I don't really know this function but is there a method to evaluate this at a point? I mean if it has a simple taylor expansion then I guess we can express it using basic operations. – Alraxite Apr 04 '13 at 17:47
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From the Wikipedia page $W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n = x - x^2 + \frac{3}{2}x^3 - \frac{8}{3}x^4 + \frac{125}{24}x^5 - \cdots$ – shilov Apr 05 '13 at 00:41