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[2016-11-19. : 6:19 pm]  O)FaRTy1billion[MM] -- which is |x| < 1, then just tested the endpoints which is sum (-1)^n/n^2 and sum 1/n^2, so [-1,1] ;o[2016-11-19. : 6:15 pm]  Vrael -- I think it still converges moose, but your argument for why is wrong, consider n = 10: 1/(2^10) < 1/10^2[2016-11-19. : 6:13 pm] O)FaRTy1billion[MM] -- I already turned the assignment in though,but I think I did it right[2016-11-19. : 6:13 pm] O)FaRTy1billion[MM] -- my question on it was mostly for the derivatives though xD[2016-11-19. : 6:09 pm] O)FaRTy1billion[MM] -- VraelVrael shouted: because n is discrete? pretty sure that still conveges doesnt it n is integers, if that's what you mean[2016-11-19. : 6:07 pm] Vrael -- well you guys can keep your answers then because I solved a completely different problem[2016-11-19. : 6:05 pm] Vrael -- or is it sum from n = 1 to n = +infinity because x^0/0^2 is undefined[2016-11-19. : 6:04 pm] O)FaRTy1billion[MM] -- FaRTy1billionFaRTy1billion shouted: f(x) = sum of x^n/n^2 for n=1 to infinity [2016-11-19. : 6:03 pm] Moose -- It's like Dem0n, Pr0nogo, and dumbducky all had a child with a learning disability.[2016-11-19. : 6:02 pm] O)FaRTy1billion[MM] -- VraelVrael shouted: why didn't you just write f(n) = sum(x^n/n^2) because x is the argument of the function, and n is the sum value[2016-11-19. : 6:02 pm] IskatuMesk -- fuck(nuts) = sum(porn(circumflexaccent)nerds/negros(circumflexaccent)poop)[2016-11-19. : 5:54 pm] Moose -- IskatuMeskIskatuMesk shouted: math is for dorks Dem0n, get off Mesk's account[2016-11-19. : 5:51 pm] O)FaRTy1billion[MM] -- first derivative was like sum x^(n-1)/n, second was (n-1)x^(n-2)/n or something[2016-11-19. : 4:57 pm] Vrael -- if you had said for example, f(x,n) = sum(x^n/n^2) I would have made different assumptions, the problem would be a lot harder, and all my answers would be wrong[2016-11-19. : 4:55 pm] Vrael -- also I'm assuming all these sums are with respect to x and not n since you said f(x) = sum(x^n/n^2)[2016-11-19. : 4:54 pm] Vrael -- ok so if f(x) = sum(x^n / n^2), we don't care about the n^2 part for convergence, so f'(x) ~ sum(n*x^(n-1)) ~ n*sum(x^(n-1)) which converges for (n-1) <= -2[2016-11-19. : 4:48 pm] Vrael -- which is convergent, but I doubt we're considering negative values of X simply because they always use x > 0 in these classes[2016-11-19. : 4:47 pm] Vrael -- if x is just all values of the reals then sum(x^1) is just 0 by symmetry[2016-11-19. : 4:43 pm] Vrael -- the convergence of sum(x^n / n^2) is equivalent to the convergence of C*sum(X^n) since n^2 is a constant for any value of n[2016-11-19. : 4:42 pm] Vrael -- Mini Moose 2707Mini Moose 2707 shouted: Er, [-1, 1] since when does the sum( x^1 / 1^2) converge?[2016-11-19. : 4:19 pm]  Oh_Man -- Dem0nDem0n shouted: My friend's dog keeps licking my pants/socks Yeah that's not gonna hold up in court |
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