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Math.SE seems to be going the stackoverflow way. Pity.


Jan
9
comment Prove that $2^{2^{\sqrt3}}>10$
Why is it obvious that any adequate answer would be computation-heavy? Do you have a proof lying around somewhere? :-P
Jan
9
comment monotonicity by examining the sign of derivative
$f(x) = (1 + k^x)^{1/x}$ seems reasonable. Did you try computing the derivative? What did you get?
Jan
9
reviewed Reject Question about integral of the product of two continuous functions.
Jan
9
answered Question about integral of the product of two continuous functions.
Jan
9
comment Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
@EricStucky: Once the closure is marked properly (dupe, after mod flagging), we can always edit that stuff out. I really don't see the issue. Anyway...
Jan
9
comment Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
@EricStucky: Huh? Without the edit, the "reopen queue" would have no clue that it is actually a dupe, and would actually reopen. I don't understand what you are trying to get at. Anyway, your latest edit is fine. Let's not waste anymore time on this.
Jan
9
comment Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
@EricStucky: FYI. This was already closed as off topic. The edit was made after the closure, for the reopen folks, not for the OP. In fact, I was trying to be nice to OP, saying that it is not really off-topic but a dupe (and such questions are indeed on-topic). I will do it again if something similar happens. Feel free to take it up on meta.
Jan
9
comment Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
@EricStucky: Huh? This is a dupe and ought to be closed as such. Reopen votes make no sense at all. I have flagged for mod attention to close as a dupe. There is no tone there. Just drawing attention that it is a dupe. You are reading something that isn't there.
Jan
9
revised Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
rolled back to a previous revision
Jan
9
revised Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
added 7 characters in body
Jan
9
revised Prove that $2^{2^{\sqrt3}}>10$
edited tags
Jan
9
revised Show that $\displaystyle \displaystyle 2^{2^{\sqrt3}}>10 $ without a calculator
added 136 characters in body
Jan
8
comment Find $\sum_{k_1+\cdots+k_n=m}\frac{1}{k_1!\cdots k_n!}$.
Please don't accept answers so quickly. This will prevent people from adding other good answers, which would be useful to not just you, but future folks who have this question...
Jan
8
comment Find $\sum_{k_1+\cdots+k_n=m}\frac{1}{k_1!\cdots k_n!}$.
This approach can be used to prove binomial/multinomial theorem, if you so choose to...
Jan
8
answered Find $\sum_{k_1+\cdots+k_n=m}\frac{1}{k_1!\cdots k_n!}$.
Jan
8
comment 2N digit number with exactly N ones and N zeros. Prove there exists a number divisible by N
@Andres: I believe number-theory tag is appropriate for this (at least based on one answer)...
Jan
8
awarded  Good Answer
Jan
8
comment 2N digit number with exactly N ones and N zeros. Prove there exists a number divisible by N
The set could be a multiset.
Jan
8
revised Why do we have $ab=ba$?
edited tags
Jan
8
revised 2N digit number with exactly N ones and N zeros. Prove there exists a number divisible by N
edited tags