Fun fact: e^W(lnx) is the formula to find the *Super Square Root!* Notice how x^x is a tetration equation, it's x tetrated to 2 or x^^2. The solution to x^^2 = a is the super square root (ssqrt) of a, or e^W(lna)
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.
@@buckyjennings8854 : I was concerned about this initially as well, but rather than thinking of it as “needing a computer”, think of it “as needing a table of the values of the W function”. In other words this is no different from, for example, if the solution were of the form “X = ln(5)” or of the form“X = cos(5dg)” (or, for that matter, “x = sqrt(5)”). In all these these cases you’ve just expressed your answer in terms of a constant value of a previously well-defined function.
Cool beans! Never knew about the Lambert W function. This was a very helpful introductory video to the function. Thanks.
Fun fact: e^W(lnx) is the formula to find the *Super Square Root!*
Notice how x^x is a tetration equation, it's x tetrated to 2 or x^^2.
The solution to x^^2 = a is the super square root (ssqrt) of a, or e^W(lna)
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.
Simply assume x = 2
Thus square root both side you get the answer...
Which is x = √2
I.e 1.414 approximately
x^x=sqrt(2)^sqrt(2) can't be 2 as by using the Gelfond-Schneider theorem, we can prove that sqrt(2)^sqrt(2) is transcendental
What is the point if you still need a computer program to give you the value?
@@buckyjennings8854 : I was concerned about this initially as well, but rather than thinking of it as “needing a computer”, think of it “as needing a table of the values of the W function”.
In other words this is no different from, for example, if the solution were of the form “X = ln(5)” or of the form“X = cos(5dg)” (or, for that matter, “x = sqrt(5)”).
In all these these cases you’ve just expressed your answer in terms of a constant value of a previously well-defined function.
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