"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 

08132020, 03:50 PM
Post: #22




RE: "Counting in their heads"  1895 oil painting
There is also the Faulhaber polynomials, with sumofpowers formula as function of triangular number.
Let \(\large t = \binom{x}{2},\;{s_{2m} \over (2x1)t}\) and \(\large {s_{2m+1} \over t^2}\) are polynomial of t, degree m1 Example, get s4(x) and s5(x) in terms of t, using divided difference. Note: we start from x=2, instead of 0, to avoid dividebyzero issue. Code: x s4(x)  t s4/(2xtt) divideddiff Redo previous example, using horners rule for the difference. Code: lua> a,b = 50, 151  next line replaced with t's 

08142020, 02:25 PM
(This post was last modified: 08142020 02:29 PM by Albert Chan.)
Post: #23




RE: "Counting in their heads"  1895 oil painting
(08122020 01:32 AM)Albert Chan Wrote: I was wrong. There seems to be a pattern to sum of powers formula after all ... To extend formula for spacings = d, simply replace S_{p} by S_{p}/d^{p}, c by c/d Or, just check the dimensions for each term. All terms must have same units. Example, for sumofsquares, all terms should have unit of c² S2 = n*c² + (n³n)/12 * d² // c/d is dimensionless, thus c, d have same unit Example, for sum of m odd squares 1² + 3² + 5² + ... + (2m1)² // d=2, c = m = n = m*m² + m*(m²1)/12 * 2² = m*(4m²1)/3 = \(\binom{2m+1}{3}\) We can confirm this from sumofnsquares formula \(\begin{align} {n(n+1)(2n+1) \over 6} &= {n(n+1)·[(n1) + (n+2)] \over 6}\\ \binom{2n+2}{3}/4 &= \binom{n+1}{3} + \binom{n+2}{3}\\ \end{align}\) Let n = 2m: LHS = sumofmoddsquares + sumofmevensquares \(\binom{n+2}{3}\) = 4×sumofmsquares = sumofmevensquares ⇒ \(\binom{2m+1}{3}\) = sumofmoddsquares 

08142020, 07:20 PM
Post: #24




RE: "Counting in their heads"  1895 oil painting
C'mon now Albert... Is all this really 'in your head'? Seems like many whiteboards...
Bob Prosperi 

08152020, 02:55 AM
Post: #25




RE: "Counting in their heads"  1895 oil painting
(08132020 03:50 PM)Albert Chan Wrote: There is also the Faulhaber polynomials, with sumofpowers formula as function of triangular number. Indeed there is more than a way to skin a cat (if we still may say that these days). There is also Hurwitz zeta function. For example, ζ(5, 50)  ζ(5, 151) = 1934166665000 But what calculator has that builtin? 

08152020, 05:04 AM
Post: #26




RE: "Counting in their heads"  1895 oil painting
(08152020 02:55 AM)Gerson W. Barbosa Wrote: There is also Hurwitz zeta function. Anybody know which zeta function Prime's CAS uses when Zeta() is given two arguments? Prime evaluates Zeta(2,2) as 1.9892802343, but Wolfram Alpha returns pi^2/61 (approx 0.644934) for HurwitzZeta(2,2). Prime's Help screen for Zeta() does not mention a twoargument syntax, so it's a mystery to me. <0ɸ0> Joe 

08152020, 07:20 AM
Post: #27




RE: "Counting in their heads"  1895 oil painting
(08152020 05:04 AM)Joe Horn Wrote: Anybody know which zeta function Prime's CAS uses when Zeta() is given two arguments? That’s the derivative of Zeta function. The second argument is the derivative order. In WA: (d^2 ζ(x))/(dx^2), x = 2 

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