Infinite tetration and superroot of infinitesimal
#86
Ivars Wrote:
bo198214 Wrote:
Ivars Wrote:Barrows said:
"... if x^x=1/2 has no solutions in real and complex numbers ... "
But ... :
\( \frac{1}{2}=x^x \)
\( \ln\left(\frac{1}{2}\right)=x\ln(x)=e^yy \)
\( W\left(\ln\left(\frac{1}{2}\right)\right)=y=\ln(x) \)
\( x=e^{W\left(\ln\left(\frac{1}{2}\right)\right)} \)
He actually wrote he does not know if it has. "There is no square superroot of 1/2, " . And it was Bromer in 1987 article. So it has solution in complex numbers.

x= e^W(W(-ln2/2)).
Actually, the beautiful formula I propose is:
ssqrt(x) = ln(x) / W(ln(x)), which, for x = 1/2, gives:

ssqrt(1/2) = ln(1/2) / W(ln(1/2)) = 0.26289282802173525.. + 0.4996694356833174.. i

Perfectly coherent with Henryk's formula.

GFR
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