tommy's "linear" summability method tommy1729 Ultimate Fellow     Posts: 1,742 Threads: 382 Joined: Feb 2009 12/30/2022, 11:44 PM tommy's summability method ************************** I might not be the first to consider it so the name is perhaps disputable or temporarily anyway , This method is for divergeant strictly increasing positive integer sums consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x now we see 2 x = 2 + 4 + 8 + ... notice x - 1 = 2 + 4 + 8 + ... so 2 x = x - 1. thus x = -1. lets generalize for real a > 1 : 1 + a + a^2 + ... = x then a x = x - 1 so x = -1/(a-1) = 1/(1-a). This is all old stuff. So whats next ? consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ... such that f(i+1) > f(i) for every integer i > 0. Notice f(0) = 0. Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x Here is how sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x sum_n sum_i  f_i (a^n)^i = x so x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n = sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i) As example consider dexp(x) := exp(x) - 1. dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x Then x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) ) this clearly converges. (I got an estimate of x = -1.19355 btw) I call this a linear summability method because it has some linear properties. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,742 Threads: 382 Joined: Feb 2009 01/20/2023, 12:45 AM I want to point out some problems : 1) SUMMABILITY METHODS WORK GREAT TO GET A VALUE OF AN ANALYTIC CONTINUATION BUT OFTEN FAIL TO DETECT ( TRUE ) DIVERGEANCE !! we still want to know about divergeance , also beyond the analytic continuation ! 2) the entire function is an interpolation  so f(2^n) is defined as an f that grows like say f(x) = exp(x). but almost any entire f that agrees with exp(x) at integer x can be used ... but might give different results !! So a very disturbing thing. 3) the sum 1 + 2 + 4 + ... is associated to the equation 2 x = x - 1 or 4 x = x - 3 8 x = x - 7  etc ALL of them give x = -1. HOWEVER the sum 1 + 3 + 9 + ... has equations 3 x = x - 1 9 x = x - 1 - 3 = x - 4 etc all with DIFFERENT solutions. --- these " problems " are quite universal with summability methods. Which is why " the master forbids it " ( being Weierstrass ) I do not forbid it , but be very careful. regards tommy1729 jacob Junior Fellow  Posts: 2 Threads: 0 Joined: Aug 2018 01/24/2023, 11:26 AM (This post was last modified: 01/24/2023, 11:28 AM by jacob. Edit Reason: 0 ) 1 + 3 + 9.... = -0.5 « Next Oldest | Next Newest »

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