New Sum of Powers Log Function

03312021, 04:06 PM
Post: #15




RE: New Sum of Powers Log Function
(03312021 01:27 PM)Namir Wrote: I do beleive we must use the solution path of 1 with the Lambert function. The W0(x) does not give the same answers.Although not a proof, I had shown the reason for this here Here is another way. If x = 0, we have s = Σ(1, k=1..n) = n This is when both branches of LambertW gives the same value. p = W(ln(N)/s) / ln(N), where N = n+1/2 With s=n ⇒ p=x+1=1, we do not need the +1/2 correction: s = ∫(1, t=1/2 .. n+1/2) = ∫(1, t=0 .. n) p = W(ln(n)/n) / ln(n) = ln(n) / ln(n) = 1 // W both branches, see identities For n≠s, we have p≠1, but 2 solutions for p. Dropped term 0.5**p/p is a decreasing function, so we want the maximum p. In other words, we pick most negative value of W(ln(N)/s), i.e. 1 branch. 

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