GFR Wrote: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
Since 0^0 = 1 then \( ssqrt(1) = 0 \) The reason for 0^0=1 is explained here ( it is not 100% based on real number limits, actually, not at all, its based on symbolic calculations ( binomial theorem):
D.Knuth 2 notes on Notations
Quote:Evidently Libri’s main purpose was to show that unlikely functions can be expressed in algebraic terms, somewhat as we might wish to show that some complex functions can be computed by a Turing Machine. “Give me the function 0^0^x , and I’ll give you an expression for [x divides m].”
Most mathematicians agreed that 0^0 = 1, but Cauchy [5, page 70] had listed 0^0 together with other expressions like 0/0 and ∞−∞ in a table of undefined forms. Libri’s justification for the equation
0^0 = 1 was far from convincing, and a commentator who signed his name simply “S” rose to the attack [45]. August Mobius [36] defended Libri, by presenting his former professor’s reason for believing that 0^0 = 1 (basically a proof that limx→0+ x^x = 1). Mobius also went further and presented a supposed proof that limx→0+ f(x)^g(x) = 1 whenever limx→0+ f(x) = limx→0+ g(x) = 0.
Of course “S” then asked [3] whether Mobius knew about functions such as f(x) = e−1/x and g(x) = x. (And paper [36] was quietly omitted from the historical record when the collected works of Mobius were ultimately published.) The debate stopped there, apparently with the conclusion that 0^0 should be undefined.
But no, no, ten thousand times no! Anybody who wants the binomial theorem to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0
and y = 1 to get 1 on the left and 0^0 on the right.
The number of mappings from the empty set to the empty set is 0^0. It has to be 1.
On the other hand, Cauchy had good reason to consider 0^0 as an undefined limiting form, in the sense that the limiting value of f(x)^g(x) is not known a priori when f(x) and g(x) approach 0 independently. In this much stronger sense, the value of 0^0 is less defined than, say, the value of 0+0. Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth
was on their side.
If we use the formula:
\( ssqrt(y) = \ln(y) / W(\ln(y)) \)
for y=1
\( ssqrt(1) = \ln(1) / W(\ln(1)) \)
\( \ln(1) = 0 \)
\( W(0) = 0 \)
Then
\( ssqrt(1) = 0/0 \)
Which not what x^x=y as x=0 says. It says\( ssqrt(1)=0 \)
Differentiating at 1 according to l'Hopitals rule gives ( I hope I did not make a mistake):
\( ssqrt(1) = ln(1)/W(0) + ln(1) =0/0 + 0 \)
No big help here. May be if we divide the powerseries for ln(1+x) and W(x) near 0 we get clear 0, since in these new powerseries the coefficient at \( x^0 \) would be 0, only powers of 0 remain, so the resulting series will equal 0 at x=0.
\( x^{(x^x)} = 1 \) then as \( 0^0=1 \), \( x^1=1 \) in this case 3rd superroot of y=1 is NOT=0 , but 1.
4th superroot is again ln(1) =0 , 5th is 1 or ( perhaps) ln(e) etc.
So here even/odd superroots of 1 clearly divide, and it is consistent with formula for ssqrt(1).
Ivars