@N3buchadnezzar I simply used $u=\log(x)$ and it was not too difficult. I just tried to put everything in terms of $\int\frac{\mathrm{d}x}{\log(x)}$ (a term for which I know there is no closed form integral).
I think a website filled with proofs and mathematical problems with solutions would be really cool. As this site seems to be more about you already having a problem and wanting a solution for it. Not looking up new problems to solve, or proofs... But Yeah, I do not know.
I should really start to sort my clothes by their topological propperties!
@KannappanSampath Be careful with using expletives. There seems to be a mixed opinion about their usage on the site, and your message did just receive a flag. I marked it as invalid, since I don't care about their usage so long as they are not directed at other users, but some 10kers / mods will mark them as valid.
I doubt the regulars here would do that, generally we don't care about things like that, but who knows. Good thing for you to give heads up, thanks mate. (I distinctly recall someone calling a formula I used a "random shitty identity." I just lol'd.)
We really don't care, so long as you don't direct them at users (in a non-joking way), but they occasionally still get flagged depending on who happens to be around.
@robjohn I'm having trouble interpreting the phrase: "The lengths in feet of three ropes are consecutive multiples of 3." I think of it as x, 3x, and 6x but this is wrong; it should be x, x + 3, and x + 6. Could you explain to me why please?
the positive multiples of three are, in order, 3, 6, 9, 12, 15, ...
In order for them to be consecutive they need to be spaced by exactly 3, or else there will be multiples of 3 in between them and they won't be consecutive
How would something like {2,6,12} be consecutive multiples of three?
hey hi, @all, I just wanted to say goodbye. I might not come to chat for some time in the future (to the extent I can help it). Hope to meet you all later.
I meant the order of elements. Are there elements of all finite order that dic=vides the cardinality? I do know the cardinality of the group in question!
@Victor: The function is well-defined on $\mathbb{C}$, the complex numbers. The problem is just asking you to restrict your attention to the function defined on the reals $\mathbb{R}$.
Or wait. If $q=p^r$ is a prime power, $r>1$, then that might not be the GL. scratches head
@Victor: It probably would have been better, but most people would automatically understand the question is talking about R in my opinion so it's not vital or anything.
(IOW, the group if infinite matrices which have finitely many non-zeroes outside of the main diagonal, finitely many non-ones on the diagonal, and the dets of sufficiently big principal minors starting on the left top corner is non-zero)
You could join the two first sentences as follows «Since $blah$, there exists an open set $G$ in $R^n$ such that $bleh$ and, as its [or: the] complement $G^c$ is closed and $K$ is compact, we have $\delta\dots$.» (The location of the commas is important! :) )
Or: «Since $blah$, there is an open set $G$ in $R^n$ such that $bleh$ and we have $\delta=blah>0$, because the complement $G^c$ is closed and $K$ is compact.»
You can probably guess how painful the process of writing my thesis was... :P
I disagree with the claim it is necessarily bullshit, I also should point out it was probably a joke anyway, and ultimately there was no point to your being flagged so rest easy.
Question title "Lagrange multipliers" Oh, I can do this! Question begins "Suppose $(M,g)$ is a Riemannian manifold." Okay, nevermind. slumps back in bed
just like....no particular reason..............He once told he was going into financial math side....i guess they do not have time to spare for chatting
@Ilya All work and no play make Ilya a dull boy.All work and no play make Ilya a dull boy.All work and no play make Ilya a dull boy.All work and no play make Ilya a dull boy.All work and no play make Ilya a dull boy.All work and no play make Ilya a dull boy.
I usually work with kernels meaning that this are functions $K:\mathscr X\times \mathscr B(\mathscr X) \to \mathbb R$ on some measurable space $(\mathscr X,\mathscr B(\mathscr X))$
so $K(x,\cdot)$ is a measure for each $x$ and $K(\cdot,B)$ is a measurable function for each $B$
there are lots of this kernels that correspond to the initial stochastic kernel and most of the are useful for the analysis. So since harmonic measure is $$ H:\mathscr X\times \mathscr B(\mathscr X)\times \mathscr B(\mathscr X)\to\mathbb R$$
namely, $H(x,A,\cdot)$ is a measure for each fixed $x,A$. I would like to rather call it the harmonic kernel since I need this dependence and want to lead to a confusion. So my question is - is it appropriate to use such notation?
If Volgograd is one of the best universities in Russia to prepare experts in the field of international relations, how come it looks like crap on the pictures?
