I should shrink the cover so that I have global sections? But how does that yield a finite sum?
So I start with a finite trivialising cover $U_\alpha$. On $U_\alpha$ we can find a frame $\sigma_i$ such that $f^*\sigma_i$ is a frame on $f^*E\vert_{f^{-1}(U_\alpha)}$. Around each point $U_\alpha$ we can find a global $\overline\sigma_i$ which locally equals $\sigma_i$. However, now I have an infinite cover.
you're missing the following result: if you have an open cover $U_i$, there's an open cover $V_i$ over the same index set s.t. $\overline{V_i}\subseteq U_i$
this holds in any, uh, $T4$-space, if I'm not mistaken
For $z \in \mathbb{C}$, define $T_{z}: \mathbb{C} \to \mathbb{C}$ by $T_{z}(u) = zu$. Characterize those $z$ for which $T_z$ is normal, self adjoint, or unitary.
For $T_z$ to be unitary means $\|T_z(u)\| = \|u\|$. Therefore $\|zu\| = \|u\|$. From which I work out:
Which would mean that $|z|^2\langle u,u \rangle = \langle u,u \rangle$ can only occur iff $|z| = 1$.
What I'm having trouble with is showing normal and self-adjointedness. I know for a linear operator to be normal means $TT^* = T^{\star}T$ . I have linear o…
ok. what is the adjoint of T_z, then? you need a definition there.
you're writing <u,u> up above. what is <a,b> when a and b are not the same?
that's my question. there are two conventional answers, it doesn't really matter which one you pick, except for how you might write things out in ways that don't use < >
i'm not trying to being deliberately obtuse or nitpicky here, but this is exactly the question raised by "seeing somebody somewhere do [blah]." they're implicitly using whatever the definition of the inner product actually is.
okay. so for any z, a, and b, <T_z a, b> = za conj(b) = a conj(conj(z) b) = <a, conj(z) b> = <a, T_{conj(z)} b>. using that definition of the inner product. i'm writing conj( ) because i don't want to overline or use tex or worry about * ruining my text.
so the adjoint of T_z is T_{conj(z)}.
if you were using stars, you could just pop the * from a superscript on the T and move it down to a superscript on the z.
$T_zT_{\overline{z}}^*(u) = z\overline{z}u$, likewise $T_{\overline{z}}^*T_z = \overline{z}zu$. So then this would mean that this operator is normal iff $z \in \mathbb{R}$.
because $z\overline{z} = |z|$ which is a real number
why? you didn't prove that up above. you remarked that whatever z is, T_z T_z* is multiplication by a real number. so what?
T_i T_i^* = T_i^* T_i is the identity. it has to be, because you already told us that T_z was unitary if |z| = 1. so there's a normal T_z where z isn't real.
So if I'm understanding you correct, you're saying because I showed that $T_z$ was unitary then as a consequence I have already that $T_z$ is normal (which is a part of a theorem in the text showing the equivalent statements for something being unitary), now for this all to be true relies on $|z| = 1$ so $z$ can be complex or real just as long as $|z| = 1$ then the condition is satisfied.
I happened to show unitary first because I didn't know how to show normality first. WHat would I have done in the case where I hadn't shown it being unitary first but tried to show normality?
Hi, I am trying to answer a problem. I define $f:=(\sin x)^{\sin x}$. I know that as function $f$ is well defined over $]0,\pi[$. I know that $f\to 1$ as $x\to 0^{+}$, but can I say that $f\to 1$ as $x\to 0^{-}$?
and they should elaborate things in class. For example: for last one hour or more-I have been stuck at proving $\pi(x)\ge \log(\log x)$ (I don't understand one step in the proof given in the book.) but I remember that in class it was not even talked about.
Well....I'm sure that experience would be different, but I guess it depends on where and with whom. The PhDs in the chat would be able to expand on this...
But attending classes here feels like waste of time. The one who has studied the subjects before don't really care. But the ones who are studying it for the first time feel like their time.
Here $\pi(x)=$ number of primes which are less or equal to $x$.
Let $p_n$ denote the $n$th prime. Then since any prime dividing $p_1...p_n+1$ is distinct from $p_1,...,p_n$; it follows that $p_{n+ I}\le p_1...p_n + 1$.
By induction, $p_n\le 2^{2^n}$ for all $n\in \mathbb N$. It follows that $\pi(...
@A.P. Wolfram makes mistakes, so you need to be critical. Some people think the symbol $\lim_{x\to a}$ applies only to those $x$ (approaching $a$) in the domain of the function. But you should ask yourself how $a^x$ is defined in general and for what negative $x$ the quantity $x^x$ is defined.
I understand that, wolfram makes mistakes, for example when it calculates limits, but I'm a little more curious, since the problem seems to be very similar to wondering whether or not $\lim_{x\to 0}x^{x}$ there exists or not.
I know that the answer is yes, always that $x>0$.
And using an argument that I don't know if it is correct, one can say that $\lim_{z\to 0}z^{z}=1$.
They are typically multivalued. However, if use complex numbers to define $x^x$ for negative real $x$, it will be multivalued as well, but the limit will in fact be 1 regardless.
I finally understood it. For $2\le x\le e^{e^3}$, suppose on the contrary that there is some x such that $\pi(x)<\log\log x\implies \pi(x)<3\implies \pi(x)=1$ or $\pi(x)=2$
If $\pi(x)=1$ then $x\in [2,3)$. So $\log \log x\lt \log(\log 3)\le 1$ hence $\pi(x)\ge \log(\log x)$ which is not possible. So $\pi(x)=2$. It follows that $x\in [3,5)$, whence $\log (\log x)\le \log\log 5\le 2=\pi(x)$ which is also not possible. This is a contradiction.
So the inequality holds for 2<=x<=e^{e^3} as well.
@Koro: I was reading your messages related to the futility of attending classes. Do they allow you to ask questions in a lecture? I think it is difficult to ignore genuine questions (asked politely) irrespective of the nature of the lecturer.
This is especially true if number of students in class is small (and not like typical undergrad class).
@ParamanandSingh It is allowed but the problem is this. This is an example of the situation in class: Product topology was covered in like 15 mins, no example no nothing whatsoever. There are some people who have studied Topology before but then there are some who were meeting the subject the first time (like me). What question will you ask there?
Attending class should have won my heart over, should have put the product topology in my head in such a way that I would think about it even when I am outside class doing something else. But the reality is I did not learn product topology in the class. I studied it myself and asked questions here as and when I got stuck.
but it seems that it was done in the class just to 'complete the syllabus' with the attitude that 'atleast the term product topology was mentioned in the class so the job is done'.
but anybody could speedrun things on board. What's the point? Project book on a screen and skip at 1.5x speed instead. Why bother writing on board?
@TedShifrin he was doing Tychonoff and then the class told 'product topology is not done yet'. He was shocked and then covered product topology to prove Tychonoff's.
Good morning everyone. Currently, I'm reading the Handbook of Matrices by H. Lutkepohl. The author refers to square matrices as quadratic. This is the first time I've seen this term in a linear algebra book I've read. In terms of quadratic forms x^TAx, the matrix A is implicitly assumed to be symmetric but not every square matrix is a symmetric one.
yes, common to put definitions in boldface, nothing goofy about that
my point was that a translator who knew english math terminology would not have translated it that way, not that it isn't a fine phrase in some other language
one time i was helping someone edit a mathematical 'appendix' in an economics article, and the journal they sent it to for some reason had a professional but non math educated copy editor go through it and they ruined all of the prose
so we had to send it back and say "do not make any of those changes to the appendix, otherwise OK thanks"
I am trying to prove the following:
$\pi(x)\ge \frac{\ln x}{2\log 2}, x\ge 1$.
I know that $\pi(x)\ge \ln\ln x$ for all $x\ge 2$.
So take $x\ge 2$.
Suppose on the contrary that there is some $x\in \{x:x\ge 2\}$ such that $\pi(x)\lt \frac{\ln x}{2\ln 2}$.
That is, $x\gt e^{2(\ln 2)\pi(x)}=e^{(\ln ...
I just need another pair of eyes to see this. What Am I overlooking in looking at this paper? They have equation (2), then below they says put C=0, and they show the result. But this does not match at all equation (2) when C=0. They did this in other places in the paper. I will put screen shot also link if someone wants to check more. No need to know anything about what the paper is taking about. This is just algebra
Should not the result of (2) when C=0 be 1/(A x' + B) ? At first I thought it was a typo, but in other places they do something similar.
Here is a direct link to the paper in PDF the above page is page 3.
I will share another very confusing thing in this paper. On page 3 above they put the form as you see in (2). But on page 6, talking about the same thing, they, for some reason, change the letters used. Why would authors do this? Here is screen shot
This is so confusing. using different letters for same thing in different parts of the paper.
It looks to me like one author wrote one section , and another author wrote another sections, but they never agreed first what letters to use for the formula.
Hello! There is something I'm wondering. In the context of matrix Lie groups, is there a specific reason the exponential map $\exp:\mathfrak{g}\rightarrow G$ is used? Something like $A(t)=I+tX$ would be a curve such that $A'(0)=X$ as it happens for $\exp(tX)$. I understand that the curve I proposed is $G$-valued but I mean: is there some kind of uniqueness theorem that states the exponential map is the only possible map?
$\exp(tX)$ is a homomorphism in $t$, your $A$ is not
in fact, $t\mapsto\exp(tX)$ is the unique "one-parameter subgroup", meaning a Lie group homomorphism $\mathbb{R}\rightarrow G$, whose derivative at $0$ is $X$
Oh, it's that theorem that states that any one-parameter subgroup in $\mathrm{GL}(n,\mathbb{C})$ can be expressed as $e^{tX}$. So that's why the exponential mapping, because that's the only way a curve of matrices can also be a one-parameter subgroup
I guess I got confused about something dummy, thanks for helping
oh lol, I didn't realise it was a cut-off when I saw it yesterday
btw, @Thorgott, I think I got the argument now
about the pullback sections
Let $\sigma\in\Gamma(E)$ and let $U_\alpha$ be a finite trivialising cover for $E$. As already remarked, we have a finite sum $\sigma\vert_{U_\alpha}=\sum_i f_i f^*\sigma_i$ with $f_i\in C^\infty(f^{-1}(U_\alpha))$, $\sigma_i\in\Gamma(E\vert_{U_\alpha})$.
We know that there exists an open cover $V_\alpha$ of $M$ such that $\overline{V_\alpha}\subset U_\alpha$. By the extension lemma for vector bundles we can extend $\sigma_i\vert_{\overline{V_\alpha}}$ to a global section $\tilde{\sigma_i}\in\Gamma(E)$. We now choose a partition of unity $\psi_\alpha$ subordinate to the cover $f^{-1}(V_\alpha)$ of $N$. Then we have a finite sum $\sigma=\sum_{\alpha}\psi_\alpha\sigma$.
Note that $\psi_\alpha\sigma=\sum_i\psi_\alpha f_i f^*\sigma_i$. We can extend $f_i\vert_{\text{supp}(\psi_\alpha)}$ to $\tilde{f_i}\in C^\infty(N)$. Then $\psi_\alpha\sigma=\sum_i \psi_\alpha\cdot\tilde{f_i}f^*\tilde\sigma_i$.
o i see a typo, I should say $\sigma\vert_{f^{-1}(U_\alpha)}=\sum f_i f^*\sigma_i$
I think you can make it work slightly shorter by making the extensions implicit: you can first choose a pou $(\varphi_{\alpha})$ subordinate to $(U_{\alpha})$, then $(\varphi_{\alpha}\circ f)\sigma=\sum_i(\varphi_{\alpha}\circ f)f_if^{\ast}(\varphi_{\alpha}\sigma_i)$, $(\varphi_{\alpha}\circ f)f_i$ is a smooth function on all of $N$ and $\varphi_{\alpha}\sigma_i$ is a section of $E$ on all of $M$
nvm, that doesn't work. ignore the second message.
I will note that there are legitimate reasons that one might want to flag one's own post. For example, if someone repeatedly vandalizes a post you have written, or if you would like to suggest that the post be marked community wiki, or if you would like the post migrated, etc.
I cannot think of any real reason why anyone would ever want to kick themselves from a room.
That being said, you can quite easily kick yourself from a room:
@XanderHenderson the problem is, in my textbook, the author uses the ratio test to determine the convergence of the series and then claims it is convergent for all positive $x$, which I find confusing
@schn I still don't understand the confusion... your text author has shown that the series converges for all positive $x$. They have not said a single thing about negative $x$.
They have not said that the series diverges for negative $x$, nor have they said that it converges. They have remained silent on the issue.
@Gokuカカロット Sure. And then they vandalize it again.
At some point, you probably want a moderator to lock the post, or to suspend the user who is vandalizing the post. How do you call it to moderator attention if you can't flag your own post?
In any event, I need to go teach a class in a hot second, so I'm outie.
@schn whether it converges for all real $x$ or not (which it does), it is not apparent from that image exactly how they stated the ratio test, and how they possibly extend it to non-negative terms. Stating that it converges for positive $x$ does not mean that it doesn’t converge elsewhere.
@Koro since $\log(n)$ is asymptotically much smaller than $\frac{n}{\log(n)}$, what you state is true. However, you seem to be asked this question with some early set of results (getting a weaker result with fewer tools) to which we are not privy.
Let $\pi$ be the prime counting function. Then
$\pi(x)\geqslant\log x/(2\log2)$ for all $x\geqslant2.$
Maybe I am missing something pretty evident, but, so far, I have proved that $\pi(x)\geqslant\log{\lfloor x\rfloor}/(2\log2)$ using a method Paul Erdős used to prove that there are infinit...
Is this a valid proof? What would you need to add?
Theorem: Let $x_0$ be a zero of $f'$ of even degree. Then, $f$ has a saddle point at $x_0$
Proof: If $x_0$ is a zero of even degree of $f'$, then $f'(x_0 + \epsilon)$ and $f'(x_0 - \epsilon)$ for some small $\epsilon > 0$ are of the same sign. Since $f'(x_0) = 0$, this is the definition of a saddle point. Thus, $f$ has a saddle point at $x_0$.
Suppose on the contrary that $2n \choose n$ is not divisible by $p\in (n,2n)$. There exis $k$ and $0\ne r\lt p$ such that
$(2n)\cdots (n+1)=kp \,n!+r \, n!$. The second term on RHS is not divisible by $p$ as $p>n$. I'm not sure how to conclude from here that $r=0$. Any hints on this are much appr...
I'm surmising it is a point $x_0$ so that every small neighborhood contains points $x$ with $f(x)>f(x_0)$ and other points $x$ so that $f(x)<f(x_0)$. Do we mention the tangent line? I don't.
@TedShifrin I changed the proof, I first prove that at $x_0$, there is an inflection point, and since $f'(x_0) = 0$, that inflection point is a saddle point
That's a valid way to define saddle points, right?
@robjohn A former student of mine is newly teaching AP calculus and sent me a handout he was given to use in his class. There are questions that contain graphs of derivatives, second derivatives, etc., and ask various questions. They say "Assume that there is no hidden behavior." Any idea what this means?
@ILike Don't ask me. What does your textbook give as a definition?
@TedShifrin "With the derivative, we can find possible extremas. In figure 2, you can see that when we have an extrema, the derivative is zero. In figure 3, you can see that $f'(x) = 0$ is not a sufficient condition though. We call these points saddle points."
@TedShifrin I'd guess that that means the graphs embody the whole picture, so to speak. That is, you don't need to worry if the graph is $x^4$ when it looks like $x^2$.
Ah @robjohn. My guess was that if it looked continuous/differentiable, then it is. And that behavior we see as we go to the ends of the interval pictures persist beyond.
@ILike First demerit for the book. "extrema" is plural. You can't have "an extrema." So their definition of a saddle point is a critical point (derivative vanishes) which is not a local extreme point.
Me too, @robjohn. I think yours is good, too.
@ILike Make sure you also understand that an inflection points is NOT defined by $f''(x_0)=0$.
At any rate, @ILike, it looks to me that you have to argue pretty much what I said. Why is the point not a local maximum or a local minimum? Why does being an inflection point tell you that?
@robjohn I disagree. You can have an inflection point where the function isn't even differentiable, let alone twice differentiable. But if $f''$ exists, it must be $0$.
The function $f(x)=\begin{cases} x^2, & x\ge 0\\ -x^2, & x<0\end{cases}$ has an inflection point at $0$, @ILike. I can even give you one that isn't differentiable. Why is $0$ an inflection point?
@TedShifrin Still it changes sign, even though its not continuous at the point, but I'm not sure I would call that an inflection point. I'm not sure how to classify some of the non-differential points. I don't know if I've talked about inflection points of non-differentiable functions.
Well, not that one does this is all in a standad Calc I class, but I did it teaching out of Spivak. One defines convexity in terms of the function relative to the chord. One then proves that for a differentiable function, this is equivalent to increasing derivative. Although Spivak disagrees with me, I define an inflection point to be a point of continuity where the concavity changes.
It seems to be true that if a point $P(x, f(x))$ is a point of inflection and $f'(x) = 0$, then that point is a saddle point (see this answer), but not in the other direction
This week I told my students that the critical points of $f(x,y) = x^3$ were not saddle points because a saddle point is a point that looks like a maximum in some direction and a minimum in another
@Astyx How do you define a saddle point? I've always understood a saddle point to be a point where the derivative vanishes in all directions, but the point is not an extremum. So $(0,0)$, for example, is a saddle point of $f(x,y) = x^3$ (at least, per the definition I know).
But I define it precisely the way I stated above. $x$ is a saddle point iff there is $v_1, v_2 \in \mathbb R^n$ (or whatever vector space your function is defined on) such that $t\mapsto f(x+tv_1)$ has a maximum at $t=0$ and $t\mapsto f(x+tv_2)$ has a minimum at $t=0$
Well, $x$ is assumed to be a critical point as well of course
Let $\pi$ be the prime counting function. Then
$\pi(x)\geqslant\log x/(2\log2)$ for all $x\geqslant2.$
Maybe I am missing something pretty evident, but, so far, I have proved that $\pi(x)\geqslant\log{\lfloor x\rfloor}/(2\log2)$ using a method Paul Erdős used to prove that there are infinit...
