I set x=1 and n=2 Then I apply the recursive relationship: x=(n^(1/x) + x)/2 As it stands, this method converges fairly quickly to the more general problem x^x = n for some range of real values of n. It has been many years since I worked on this problem, and I don't remember much of the details, except that the above method undergoes many changes depending on the range of values of n. The boundaries of these values, if I remember correctly, depend on the constant e. In general, this method and its variants also work for complex values of n.
Cool beans! Never knew about the Lambert W function. This was a very helpful introductory video to the function. Thanks.
Simply assume x = 2
Thus square root both side you get the answer...
Which is x = √2
I.e 1.414 approximately
So, that's what we call magic!
How do you find 16 in Lambert's formula? Not how you find the indices 0 and -1 of W.
how is that difficult tho
I set x=1 and n=2
Then I apply the recursive relationship:
x=(n^(1/x) + x)/2
As it stands, this method converges fairly quickly to the more general problem x^x = n for some range of real values of n. It has been many years since I worked on this problem, and I don't remember much of the details, except that the above method undergoes many changes depending on the range of values of n. The boundaries of these values, if I remember correctly, depend on the constant e. In general, this method and its variants also work for complex values of n.