@leslietownes You have lead a very happy life, then.
Although, now that you have pointed it out, it really, really annoys me. Particularly since they had to go through the extra step of using \limits to get it to typeset the indices correctly. :(
we've all been guilty of defining unorthodox macros, although i usually defined them as a placeholder. "i don't know how to do this right, and i don't want to do it wrong a million times in some un-fixable way, so"
good idea. if you've seen raiders of the lost ark, what happened at the end of that to that guy is what happened to me when i looked at \Pi in that paper.
Let $G$ be a finite simple group with $p\mid |G|$. Suppose the number of Sylow $p$ subgroup of $G$ is 7. The conjugation action $G$ on the set of Sylow $p$ subgroup induces an injective homomorphism $G\to S_7$. My TA deducted a point because the induced homomorphism could be trivial.
@Koro woops. The statement of the lemma should have been: $f \nearrow \implies \int_{-\infty}^x f \nearrow$. I had it backwards.. I use this lemma correctly in the argument that follows..
Second fundamental theorem of Calculus: Suppose that $F$ is differentiable on [a,b] and there exists a $g$ such that $F'=g$ on [a,b]. If $g$ is Riemann integrable on [a,b], then $\int_a^b g =F(b)-F(a)$.
Is differentiability of F on the whole [a,b] necessary?
I think that differentiability of F on (a,b) is sufficient. g is integrable so given any $\epsilon>0$ there is a P ={a=x_0<x_1<x_2<...<x_n=b} such that $U(P,g)-L(P,g)<\epsilon$.
$F(x_i)-F(x_{i-1})= g(z_i)(x_i-x_{i-1})$ by MVT. Adding both sides to get: $F(b)-F(a)=\sum g(z_i)(x_i-x_{i-1})$. It follows that $|\sum g(z_i)(x_i-x_{i-1})-\int g|<\epsilon$
That is, $|F(b)-F(a)-\int g|<\epsilon$. It may be noted here that nowhere did I use the fact that F is differentiable at the end points.
hey , im stuck in this problem mind sharing some hints
$y^{1/m}+y^{-1/m}=2x$ then prove that $(x^2-1)y_3+3xy_2+(1-m^2)y_1=0$ here m is arbitrary constant and $y_n$ is the nth order derivative of y with respect to x
Does anyone here know how to pick a smooth timelike vector field, given the following metric $g=-\cos(2\varphi)(dt)^{2}+2\sin(2\varphi)dtdx+\cos(2\varphi)(dx)^{2}$ ?
@AkivaWeinberger Fun (but difficult depending how much you know about the Hilbert cube) question: does $[0,1]^\Bbb N$ support a topological group structure?
It's usual knot diagrams but no vertical tangencies, only singular points are cuspidal, and overcrossings have less slope than undercrossing, at each crossing
@AkivaWeinberger the point is that every continous $f\colon [0,1]^\Bbb N\to[0,1]^\Bbb N$ has a fixed point, while topological groups are full of fixed point-free self homeomorphisms (the translation maps)
@AkivaWeinberger Even determining whether two Legendrian unknots are Legendrian isotopic are hard. That's some classical result
There's an invariant which would relate to counting cusps, as you can imagine
I think that's called the Thurston-Bennequin number?
Ok, the Thurston-Bennequin number for a knot diagram $D$ is $w(D) - 1/2 c(D)$ where $w(D)$ is the writhe of the knot, and $c(D)$ is the number of cusps
bruh
yeah its obvious that this is an invariant
an increase of one to the writhe corresponds to an increase of 2 to the cusps, because of the first legendrian reidemeister move
How do you define differential operators on say $L^2(\mathbb{R}^d)$?
trying to understand the setting that Villani starts his book with, but he doesnt seem to mention back to the Schwartz space and just carries out operations on $\mathcal{H}$ (which is a weighted $L^2$ space).
In retrospect, there probably was a way to do it using the mod function
(You can turn on/off the grid and axes in the settings menu in the top-right)
@JoeShmo Neat thing about that second image: it consists of a straight line (blue) and a curvy line (red). There's a path from the top of the image to the bottom that only crosses the blue line once, and there's a path from the top of the image to the bottom that only crosses the red line once, but there's no path from the top to the bottom that does both simultaneously
I wonder if the way I drew it is the simplest way to achieve that (with the restriction that one of the two lines must be straight)
> A shipowner was about to send to sea an emigrant-ship. He knew that she was old, and not overwell built at the first; that she had seen many seas and climes, and often had needed repairs. Doubts had been suggested to him that possibly she was not seaworthy.
> These doubts preyed upon his mind, and made him unhappy; he thought that perhaps he ought to have her thoroughly overhauled and refitted, even though this should put him at great expense. Before the ship sailed, however,
> he succeeded in overcoming these melancholy reflections. He said to himself that she had gone safely through so many voyages and weathered so many storms that it was idle to suppose she would not come safely home from this trip also. He would put his trust in Providence,
> which could hardly fail to protect all these unhappy families that were leaving their fatherland to seek for better times elsewhere. He would dismiss from his mind all ungenerous suspicions about the honesty of builders and contractors.
> In such ways he acquired a sincere and comfortable conviction that his vessel was thoroughly safe and seaworthy; he watched her departure with a light heart,
> and benevolent wishes for the success of the exiles in their strange new home that was to be; and he got his insurance-money when she went down in mid-ocean and told no tales.
> What shall we say of him? Surely this, that he was verily guilty of the death of those men. It is admitted that he did sincerely believe in the soundness of his ship; but the sincerity of his conviction can in no wise help him,
> because he had no right to believe on such evidence as was before him. He had acquired his belief not by honestly earning it in patient investigation, but by stifling his doubts. And although in the end he may have felt so sure about it that he could not think otherwise,
> yet inasmuch as he had knowingly and willingly worked himself into that frame of mind, he must be held responsible for it.
I just thought the philosophically-minded here would enjoy that story (@Eth)
> Let $$ f(x)=\left\{\begin{array}{cl} \frac{a(1-x \sin x)+b \cos x+5}{x^{2}}, & x<0 \\ 3, & x=0 \\ \left\{1+\left(\frac{P(x)}{x^{2}}\right)\right\}^{1 / x}, & x>0 \end{array}\right. $$ where $P(x)$ is a cubic function and $f$ is continuous at $x=0$. The value of $P''(0)$ is
This is a standard thing to do for unbounded operators. For example, taking derivative is not a bounded operator $L^2(\Bbb R) \to L^2(\Bbb R)$. Or multiplication by $x$. On the dense Banach space of compactly supported smooth functions $C^\infty_c(\Bbb R) \subset L^2(\Bbb R)$, it's all good.
so I guess he is implicitly using some density argument with the schwartz space which he mentions right at the start of the memoir but only has a sentence on it. So Im not really following the whole argument
I agree it is unbounded operator. May I ask, why is the unboundedness 'the big deal'
for me it is just the fact that not all elements of $L^2$ are differentiable...
@AkivaWeinberger Yes, but if you complete it with respect to $L^2$ norm you get $L^2(\Bbb R)$.
@Monty Right - the reason is Fourier analysis works better in the Schwartz space compared to $C^\infty_c$. You don't want functions to completely die off at infinity, but have a decay.
@BalarkaSen so just to confirm, Villani assumes that the operator $B$ 'sends $V$ continuously into $V$' i.e. $B:V\to V$ is continuous, not $B:V\to V^*$
that's interesting, sounds right. i have no idea what this space is.
Ok the other way $C^\infty_c(\Bbb R)$ fails to be a Banach space doesn't exist. it exists for $C^\infty(\Bbb R)$ -- which is that $C^0(\Bbb R)$ is already not Banach
it's again defined by a family of seminorms (actually strictly semi this time), $\|f\|_n = \|f\|_{C^0[-n, n]}$.
