J. M. is back.
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 Jul 14 comment Find the roots of this 6th degree polynomial Well, you know what the three cube roots of $1$ look like, no? You can multiply those with the cube root you already have. Jul 14 comment Divergent series whose terms converge to zero In any event: have you seen this? Jul 14 comment Estimating the value of an improper integral numerically @TanMath, if you must, just cite Takahasi and Mori. Jul 14 comment Analytic continuation of primality function (Sorry, it got too long for a comment…) Jul 14 comment Find the roots of this 6th degree polynomial So, you take all three cube roots of $-8$, twice. Jul 14 comment Analytic continuation of primality function Without any other restrictions, @Michael, an analytic continuation is not in general unique. Consider the gamma function: you need the additional machinery of Bohr-Mollerup just to have uniqueness. Jul 14 answered Analytic continuation of primality function Jul 14 comment Analytic continuation of primality function It's possible, @Michael, but why exactly do you need a continuation? What do you actually want to do? Jul 13 awarded Popular Question Jul 12 awarded Popular Question Jul 12 revised Cute Determinant Question deleted 4 characters in body Jul 12 comment Are all mathematicians human calculators? I sometimes like doing math on paper, but I am thankful for the computer, as it lets me make mistakes much faster. ;) Jul 11 revised Why do siamese magic squares have real eigenvalues, symmetric around zero? added 578 characters in body Jul 11 comment Alternating sum of product of Fibonacci numbers What happens if you use Binet? Jul 11 comment Change of radix without using radix 10 @Potato, I'm seeing how people can get confused; they think that the arithmetic with all the carrying and borrowing has to be done in decimal base when the same concepts exactly carry over to other bases. Jul 11 comment Cosh and Sinh analogs In effect, we are taking a discrete Fourier transform here. Jul 11 revised Optimize multi-step calculation into one step? edited tags Jul 11 comment Estimating the value of an improper integral numerically The variable substitution in this answer has in fact been recommended by Takahasi and Mori as an excellent way of dealing with infinite integrals, after which one can then use the trapezoidal rule. But, since the trapezoidal rule is being used anyway, the double exponential substitution might possibly give better results. Jul 11 comment is true that $\sum_{n=0}^\infty B_n(-1)^n=\frac{\pi^2}{6}$? @Herick, you really should have included that information in your original question. What you did is to convert a convergent improper integral into a series that is formally divergent, but regularizable. Jul 11 comment is true that $\sum_{n=0}^\infty B_n(-1)^n=\frac{\pi^2}{6}$? Well, you can re-express the sum as $$\frac12-2\sum_{k=0}^\infty \frac{(-1)^k (2k)! \zeta(2k)}{(2\pi)^{2k}}$$ where we used the relationship between Riemann's function and the Bernoulli numbers, and you should now see why it is divergent as it stands. But, again, maybe it can be regularized...