"Counting in their heads"  1895 oil painting

08122020, 01:50 PM
(This post was last modified: 08152020 07:26 PM by Albert Chan.)
Post: #21




RE: "Counting in their heads"  1895 oil painting
There is also a pattern for s_{p} (see Benoulli Number thread)
\(s_p(n) = \sum_{x=0}^{n1}x^p = \large {n^{p+1}\over p+1}  {n^p \over 2} + {p\over12}(n^{p1}) + k_{p3}(n^{p3}) + k_{p5}(n^{p5}) + \cdots \) s_{p}(1) = s_{p}(0) + 0^p = s_{p}(0) = 0 ⇒ above formula does not have a constant term ⇒ when p is odd, p>1, s_{p}(n) has factor n² Redo previous example: s5(n) = n^6/6  n^5/2 + 5/12*n^4 + k2*n^2 s5(1) = 1/6  1/2 + 5/12 + k2 = 0 → k2 = 1/12 → s5(n) = (2*n^6  6*n^5 + 5*n^4  n^2) / 12 T = 50^5 + 51^5 + 52^5 + ... + 150^5 = s5(151)  s5(50) = 1936617185625  2450520625 = 1934166665000 Update: we may use this to help mental calculation: For n ≥ 0, s_{p}(n) = (1)^{p+1} * s_{p}(1n) Redo above example, using s5(151) = s5(150), and horner's rule Code: b = 150 Update: we are better off using EulerMaclaurin formula, which work for any f(x) Note: coefficients were B(1)/1! = 1/2, B(2)/2! = 1/12, B(4)/4! = 1/720, ... Σf = Δ^{1} f = (e^{D}1)^{1} f = (D^{1}  1/2 + D/12  D³/720 + ...) f Example, with f = x^5 s5 = ∫f dx  f/2 + f'/12  f'''/720 + ... = x^6/6  x^5/2 + (5x^4)/12  (60x^2)/720 

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