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8h
comment How can I calculate this limit?
Yes. Maybe the question in your textbook was phrased differently, I don't know. But that limit does not exists :)
9h
answered How can I calculate this limit?
2d
comment Does the zeros of an analytic function $f$ form a discrete set?
the zeros of analytic complex function are indeed a discrete set. Don't know about $\mathbb R^n$ though :)
2d
revised Strange integral test for convergence in my Analysis Script (proof flawed ?)
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2d
comment What we get if we add 1/2 infinite times
If you enjoy "naive" speculations (I do too!) there are plenty of problems less "controversial", so to speak, to have fun with.. ie contest math, number theory and the like.
2d
comment What we get if we add 1/2 infinite times
@Serby I am sorry if I've offended you, as it was not my intention. I also like when people care about math, I don't want to scare you off! Also, I still haven't studied analytic continuation well, so those series to me still are largely just divergent series. On the other hand, if you're aware of what divergent means and what analytic continuation is, it certainly does not appear so from you question, as you would know that some properties do not hold. If you want my advice, you can start studying it more rigorously to get to the bottom of the matter.
2d
answered Strange integral test for convergence in my Analysis Script (proof flawed ?)
2d
comment What we get if we add 1/2 infinite times
Well there are a lot of people who flight planes, too, but I wouldn't trust myself and a manual of instruction. I really feel like the only answer at this level for the OP is "the sum is divergent and it's meaningless to attach any value to it. Then come talk to me in a couple of years about analytic continuation " :-) Of course other may disagree! :)
2d
comment Checking if a matrix is in the span of other matrices
yes, it's correct :)
2d
comment What we get if we add 1/2 infinite times
@robjohn Well if they present it like an interesting obeservation / food for thought, I'd say that is an idea worth of more investigation. And upon doing so I believe one comes close to analytic continuation? It's a pity I never got the chance to study that properly, though :) If they present it as rock hard equality without any other comment, I'd be less happy :)
2d
comment What we get if we add 1/2 infinite times
Yes but I doubt he has a clear understanding of what a "divergent series identity" is.. If you ask me this will just confuse him further. I don't mean to discourage from the study of math but I don't like this "sensational" statements without any sort of understanding of what is going on. It makes math appear obscure, patently absurd, and arbitrary
2d
comment What we get if we add 1/2 infinite times
@robjohn Don't get your point.. Of course you can assign value to those expression (analytic continuation, cesaro sums etc) but that requires a solid understanding of what one is doing, and what "equality" means. But laymen don't have this kind of understanding and they get all confused when it turns out that contradictions start to appear as soon as you play with the series and their values ;)
2d
comment What we get if we add 1/2 infinite times
I down voted because the OP is clearly confused about what "equality" even means when it comes to infinite sum. With this answer you're actually reaffirming it's understanding that $1 - 1 + 1 - ... = 1/2$
2d
comment What we get if we add 1/2 infinite times
Can I sue the people who make videos on youtube saying to laymen that since $1 - 1 + 1 - \dots$ can be either +1 or 0, the value of the infinite sum is $1/2$?
2d
answered Why doesn't L'Hopital's rule work in this case?
2d
revised Prove that $E[U(X)] \ge E[U(Z)]$
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2d
reviewed Edit What can we conclude if $A$ and $B$ are events such that $p(A) + p(B) - p(A \cap B) = p(A)$?
2d
revised What can we conclude if $A$ and $B$ are events such that $p(A) + p(B) - p(A \cap B) = p(A)$?
Some Latex fix
Jun
27
comment How can I find $f'(0)$ of this function?
@imranfat That was actually me :P Anyhow I can use the definition of derivative if I want to, the point is that I can't find the error of why that limit does not yield $-3/4$! :)
Jun
27
comment Prove that $E[U(X)] \ge E[U(Z)]$
@SONTO I'm flattered but really, I am one of the most lowly contributors on the site.. There are some pretty impressive people here. The best way to get a question answered is to ask on math.stackexchange! :D Good luck on your studies ;-)