02/13/2011, 10:27 AM
Oh, now with the picture I see what you mean.
To get a formula we can use the Lambert W function.
The Lambert W function is the inverse of the function M(x)=x*e^x
and we can express the self root x^(1/x) with help of M:
y = x^(1/x) = exp(-M(-ln(x)))
You can verify this with a bit of calculation.
Then we can obtain the inverse:
exp(-W(-ln(y))) = x
More exactly x can be two values, left and right from e, which correspond to the two branches of W:
\( x_L = \exp(-W_0(-\ln(y))) \)
\( x_R = \exp(-W_{-1}(-\ln(y))) \)
So when you want to get the left value - as in your case - you choose x_L and get:
\( x_L = \exp(-W_0(-\ln(3^{1/3})))\approx 2.47805268028830 \)
\( x_R \) would be simply 3 again.
To get a formula we can use the Lambert W function.
The Lambert W function is the inverse of the function M(x)=x*e^x
and we can express the self root x^(1/x) with help of M:
y = x^(1/x) = exp(-M(-ln(x)))
You can verify this with a bit of calculation.
Then we can obtain the inverse:
exp(-W(-ln(y))) = x
More exactly x can be two values, left and right from e, which correspond to the two branches of W:
\( x_L = \exp(-W_0(-\ln(y))) \)
\( x_R = \exp(-W_{-1}(-\ln(y))) \)
So when you want to get the left value - as in your case - you choose x_L and get:
\( x_L = \exp(-W_0(-\ln(3^{1/3})))\approx 2.47805268028830 \)
\( x_R \) would be simply 3 again.
