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soandos
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Is it possible to solve x = y^y for y?
Since y is unknown, the yth root of x is unknown.SW VandeCarr said:I'm sure I'm missing something here, but isn't the solution simply y equals the yth root of x?.
soandos said:also, how is the following true (or is it false?):
Sqrt[x]^Sqrt[x]==1/x^Sqrt[x]
soandos said:is there a way to get an exact value, or just a numeric approximation?
soandos said:I am not sure that i understand. if the root is negative, then it will never work, as the derivatives of the function are x^x (1 + \text {Log}[x]) for the first derivative, and x^{-1 + x} + x^x (1 + \text {Log}[x])^2 for the second derivative. i believe that both of these are non-continuous at any negative. is this right, or am is making a mistake?
soandos said:i am not sure that i understand. i thought from what i read that the secant method has the same kind of problem that Newton's method does, namely that if there is no continuous second derivative, then there is no guarantee that it will converge.
also, from what i understand of Halley's function, is that it is dependent on Newtons method, so for -x it will also fail.
soandos said:in fact, i believe that all of these methods are for continuous functions only.
soandos said:i thought that it is not continious at all negative x.
soandos said:also, how could someone compute W using halleys mehtod?
soandos said:thanks. but what about the previous question?
soandos said:I think (based on a a mathematica Table[If[Re[x^x]==x^x,x,0],{x,-200,-.01,.01}]
but am not sure that the only time that it will be continuous is if the number is an integer.
Can someone find a counter-example or prove this?