Jules
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 Mar8 comment Understanding the integral of $x^a$ In Taylor series and Laurent series we think of $x^0 = 1$. I wonder if there is an alternative series expansion which uses the logarithm for the 0th power. If I remember correctly there is a problem with the logarithm in conventional complex analysis, because the singularity at 0 is a strange kind of singularity. Or maybe I'm talking nonsense... Mar8 comment Understanding the integral of $x^a$ I'll ask a follow-up question if you don't mind. I hope it is not too vague. Does this mean that in some sense the logarithm ought to belong to some extension of the set of polynomials with integer exponents, where the logarithm is related to $x^0$? Mar8 comment Understanding the integral of $x^a$ Thanks, that does help, but it does not completely resolve my confusion. I feel like there is something going on with the order of the limits, because if you have the limit of $3^a/a$ as $a$ goes to 0 then it's not $\log 3$. What am I missing? edit: oh, I already see my mistake, thanks! I was confused because I only looked at the indefinite integral. May7 comment Numerical optimization with nonlinear equality constraints Thanks, that's exactly what I was looking for! Oct25 comment Which one is bigger $2^{n!}$ or $(2^{n})!$? What can we say about f(g(n)) relative to g(f(n)) in general? Feb22 comment Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition) A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.