@BrownNinja yeah, except that your components of $\vec{x}$ is aribtrarily ordered, thus what stopping me to have an $\vec{x}$ that look like (1,2,6,5,4,8,9,0,0,1,12,etc.) and then you can have component sums of a "subvector" of this vector that are larger than the component sum of a subvector with more components
whlie for majorisation, you will expect all k-1 component sums of $\vec{x}$ starting from the smallest element to be $\leq$ that of the k component sums, for example
Yes. Additionally, a necessary and sufficient condition for a vector $\vec{x}$ to majorize other vector $\vec{y}$ is if $\vec{y} = \vec{x} P$ where $P$ is a doubly stochastic matrix
@Semiclassical @Astyx Okay. With truth table we can show whether a $\implies$ statement is true or false. It doesn't matter what hypothesis and conclusion is involved.
@BrownNinja ok, still I don't quite see how this all fits together though, $T$, being a permutation matrix, must be doubly stochastic, not sure if that is relevant
So, $\vec{x}_{n,k} = x^{\downarrow} T S$, and $x^{\downarrow}_{n,k} = x^{\downarrow} S$. So the condition for majorization is $x^{\downarrow} T S = x^{\downarrow} S P$
Or we can simply write $\vec{x} S = x^{\downarrow} S P$. Since $S$ is not orthogonal, so $S^{T} \neq S^{-1}$? Also since $S$ is not square, inverse does not exist, we need to look at the pseudo-inverse?
regardless, I think your majorisation condition is still preserved since $SPS^T$ is doubly stochastic as proven earlier
but $SPS^T$ is doubly stochastic only if $S$ is a block permutation matrices of blocks $\vec{1}$ or $\tilde{1}$ (well, that's the simplest case where it will hold)
I know that there is 400 messages limit for a conversation. But in rooms with low usage it is possible that an interesting conversation spans more than one day.
Is it somehow possible to bookmark a conversation which starts one days and ends the second day? If not, could something like this be ...
uh, I will expect the opposite happens, since $S_N1$ needs two steps while $S_N2$ needs only 1. It is inperceptibly slow for bulky nucleophiles, though
I forgot which step in $S_N1$ is the rate determining step...
@BalarkaSen Linear algebra question: Consider $\Bbb R^{4n}=\Bbb H^n$ ($\Bbb H$ acts from the RIGHT) and the inclusion $Sp(n)\times Sp(1)\hookrightarrow SO(4n)$ (which identifies $(-1,-1)$ with $(1,1)$). How can I see that the image (denoted by $Sp(n)\cdot Sp(1)$) can be characterized as the subgroup of $SO(4n)$ which preserves the subspace of $\operatorname{End}(\Bbb R^{4n})$ genered by right-multiplication by $i,j,k$?
An element $L$ of $Sp(n)\cdot Sp(1)$ is given by an $n\times n$ quaternionic matrix $A\in Sp(n)$ and $q\in Sp(1)=S^3$ and sends $v\mapsto Avq$ (order not important because $A$ commutes with $q$ by definition). Then it is not hard to show that if $p$ is an imaginary quaternion (also acts from the RIGHT!), then $L\circ p\circ L^{-1}=q^{-1}pq$, which is again an imaginary quaternion
Quaternionic matrices that preserve the std inner product on $\Bbb H^n$ @BalarkaSen
In that case, I think it depends. If you have a bad leaving group like OH (or the resulting carbocation is not very substituted hence very unstable), and a small nucleophile, $S_N2$ can be faster, though my undergrad year 2 lecture notes said it is quite sensitive to the reaction conditions
guys, I kinda forgot how to show whether it's holds that $\sin x=O(x)$. Should I write $\sin x=x+O(x^3)$? And then check if $x=O(x)-O(x^3)$? If so, any hints on that?
@Leaky @Secret I see now, maybe. The context I was thinking about was preparing haloalkanes from alcohols; so even though the leaving group -OH is bad, there is an H^+ ion flying around which attacks the leaving group so that it becomes -OH_2^+; that's a good leaving group so would want to just leave. SN1 is best in that context.
I think one needs something weird like a modified version of $\Bbb{C}$ such that the units are zero divisor to work, but that is impossibel since units cannot be zero divisors
for example, suppose $a=j,b=k$ and $j^2=1,k^2=1$ but $jk=0$ then it might work, but I guess we will end up in octonion territory when we do this
Well, that answers that weird question back when I was 1st year, on whether you can have rectangular matrices $S$ such that $S^T$ is really a legitimate inverse such that $SS^T=I_n$ and $S^TS=I_m$. We have shown the first case is impossible unless we are in some wacky rings
as for the second case, Brown ninja's block diagonal matrix is an example (and I am too lazy to work out the general case)
Hey if I fourrier transform a function a function, afaik $\widehat{f}(\omega)$ tells me the amplitude of of $e^{i\omega t}$ "within" f. My script further states that it tells us the phase of $\omega$ as well I'm not sure formulation exactly means? Does $\widehat{f}(\omega)$ tell us the phase of the $e^{i\omega t}$ part of f?
@Felix.C Well, you could look at it as the mean of $e^{i\omega t}$ weighted by $\frac{f(t)}{\int f(x)\,\mathrm{d}x}$ times $\frac1{\sqrt{2\pi}}\int f(x)\,\mathrm{d}x$.
@Felix.C its the mean of $e^{i\omega t}$ weighted by $f(t)$. If $f$ is bigger when $\omega t$ is some particular value mod $2\pi$, then $\hat{f}(\omega)$ will be larger.
Well, it's really a 4/4 piece that could be turned into a 5/4 piece if you wanted. Or a 3/4 piece. I dunno, just change the length of the first note of each bar :P
[Chemistry] After another 10 hours (36000 seconds) finally finish reading the 3rd journal article. This is much better than the previous two cause at least it is finite (jking)
I know that there is 400 messages limit for a conversation. But in rooms with low usage it is possible that an interesting conversation spans more than one day.
Is it somehow possible to bookmark a conversation which starts one days and ends the second day? If not, could something like this be ...
