For example if you have $x^x = 2$, can you express $x$ as a numerical expression containing only the addition, multiplication and exponentiation operators?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ Not really $\endgroup$– SBFApr 4, 2013 at 12:19
-
$\begingroup$ See here. $\endgroup$– Mhenni BenghorbalApr 4, 2013 at 12:21
-
$\begingroup$ The solution of your equation is $\frac{\ln(a)}{W(\ln(a))},$ where $W$ is the Lambert $W$-function. $\endgroup$– Mhenni BenghorbalApr 4, 2013 at 12:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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 $\log(2)=W(\log 2)\exp(W(\log 2))=W(\log 2)x$
$x=\frac{\log 2}{W(\log 2)}$
-
$\begingroup$ 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. $\endgroup$– AlraxiteApr 4, 2013 at 17:47
-
$\begingroup$ From the Wikipedia page $W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n = x - x^2 + \frac{3}{2}x^3 - \frac{8}{3}x^4 + \frac{125}{24}x^5 - \cdots$ $\endgroup$– shilovApr 5, 2013 at 0:41