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Jun
29
answered Is there something between summation and integration?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
@AsafKaragila I specify this was done with zeta function regularization. Would it be more precise to ask for the analytic continuation $\zeta(0)$ ?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
Could you prove Ramanujan summation as Euler-Maclaurin to 1st order?
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
hmm? that seems plausible that $\sum (0+2) = \sum (1+1)$ and yet $\sum (0+2) = 2 \sum 1$.
Jun
28
comment How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
@AsafKaragila this false statement has its own wikipedia article
Jun
28
revised How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
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Jun
28
asked How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization
Jun
28
comment What we get if we add 1/2 infinite times
@Ant There is nothing wrong with this kind of experimentation. Euler did it, Physicists do it routinely (via their lack of rigour). He just needs to understand these series come with at the expense of associativity. We can't invoke $(a+b)+c = a + (b+c)$ infinitely many times and always get away with it. Watch his parentheses carefully.
Jun
28
comment What we get if we add 1/2 infinite times
@Ant his question seems to be whether these divergent series identities can be found via elementary rearrangements.
Jun
28
revised Proof $\{(x,y,z)|4x^2+9y^2+16z^2<1\}$ is an open set
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Jun
28
answered What we get if we add 1/2 infinite times
Jun
28
answered Proof $\{(x,y,z)|4x^2+9y^2+16z^2<1\}$ is an open set
Jun
27
awarded  Populist
Jun
25
awarded  Good Answer
Jun
25
comment Integral relations in Fricke and Klein
@glebovg I think they want to give you the option of not awarding any bounty at all. In the case nobody answers the way you wanted.
Jun
24
comment Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse?
I really like your numerical approach to this problem.
Jun
24
revised Arnold Trivium Problem 39
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Jun
24
revised Arnold Trivium Problem 39
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Jun
24
comment Arnold Trivium Problem 39
@Potato since the Gauss integral does not change under deformations, we can deform $A$ into a straight line or (even easier) flatten $B$ into a circle, where we going around twice. Then the integral might be doable by hand. This is line when we integrate complex functions using Cauchy Residue Formula.
Jun
24
comment Arnold Trivium Problem 39
@Potato I wouldn't dream of doing the integrals by hand... unless you truly enjoy long and tedious integrals :-)