If $a$ is not an integer, show that
$$f(z) =\prod_{n \in\Bbb Z\setminus\{0\}}\left(1+{z\over a+n}\right)e^{-z/(n+a)}$$
is an entire function.
Using the series expansion of $e^{-z/(n+a)}$, we get
\begin{align*}
\left(1+{z\over a+n}\right)e^{-z/(n+a)}& = \sum_{k=0}^\infty \left(1+{z\over a+n}\rig...
When does the 'term-by-term' thing possible for infinite product? For example, $\prod_n (1+a_n)\prod_{n}(1+b_n) = \prod_{n}(1+a_n)(1+b_n)$. For absolute convergent case?
The best case for when $n \ge 2m-1$ or $(n-(m-1)) | n$ has $n(m-1) + k$ distinct integers, where$k=\max\left\{d : d|n \land d\le \frac{n}{n-(m-1)}\right\}$ . The construction works for all $n>m$ cases, but I don't have the proof for all cases. Please edit/comment them out if you get them.
Here is...
Can anybody help me prove this
Currently the only difference between construction and upper bound is when $(n-(m-1)) \not | n$
I think we will be able to improve the upper bound to fit the construction. But please let me know if you can think of a better construction method.
I kind of forgot my multi var calculus, if i have a function $\phi(x,y): \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^m, (x,y) \mapsto f(x) - y$ How can i find $Df(x,y)$ as a matrix? for some function $f$
I'm trying to understand the proof of lemma 4.2 (Page 15) of link.springer.com/content/pdf/10.1007/BF00289616.pdf Why does (2) implies f preserves sources (i.e., $s_{D'} f' = f' s_D$). I think, the injectivity of $c'_1$ is not enough
Hi, this is not the right place to ask LaTeX related questions but I don't want to ask in TeX stack exchange. I wanted to write a LaTeX document in Odia/Oriya language (my mother tongue). I searched the internet and found "https://ctan.org/pkg/oriya". See the page 5 of the file "https://drive.google.com/file/d/1Awjf5iRKuTM6GNlePnEFXKYN_sOiLbQs/view?usp=sharing" . It says we just have to add package "oriya". But LaTeX gives error that "oriya.sty" not found
Since I started working at my current institution, I have had to deal with Office 365 nearly daily. It is the absolute worst. I hate it. I hate it so much. "Oh, hey! It looks like you are making a list! Let me just randomly indent and number the things on your list! That's what you wanted, right?!"
To expand on products again, it's not that the limit can't be zero, it's that after removing a finite amount of first terms, there has to be a point at which the limit is non-zero
Even if the first thousand terms involve zero, as long as we remove them and obtain a finite non-zero limit, the product is considered to still be convergent
I am studying manifolds, and i am just wondering, what in the flying heck is all this formalisim, i amjust swimming in symbols and have little understanding to what i am doing
given that i study physics, i wanted to study differential geometry in next semester, but now i am asking myself, if i really want to do abstract nonesensical mathematics
@XanderHenderson We concur emphatically emphatically! I had to use it for reports for the university administration when I was associate department head, and it drove me to drink.
@TedShifrin It is only 10:30 am here, and this stupid, stupid "procedures" template is nearly driving me to drink. Fortunately, the campus is dry by law, so there is no whiskey in my desk.
@Mad If you start with a concrete curves and surfaces course (for example, download my free text), then you have a lot more intuition and experience before trying to tangle with abstract manifolds. To be honest, the physicists' treatment (using only tensors with indices that magically move up and down) makes it more symbol-pushing and less conceptual.
Frustration: Someone from MSE emailed me a differential geometry question, trying to do something incorrect. I patiently explained why it was impossible to do. Then I get an emailed pdf of a conference abstract saying precisely the same wrong stuff. throws hands in the air
Thankfully, the person forgot to thank me for "helpful conversations."
Suppose that $F$ is a non-decreasing function on R and takes only non negative values. Let $A(\tilde I)$ be the algebra generated by left open and right closed intervals. Suppose that $\mu_F$ is a measure defined on $A(\tilde I)$ is defined as $\mu_F(a,b]=F(b)-F(a)$
How do I prove that $\mu_F\{x\}=F(x)-\lim_{y\downarrow x}F(y)$?
How to write mathjax for 'vertical downward arrow'?
personally, I think there's little insight to be gained from writing down the geodesic equation in coordinates except that it is always locally solvable and perhaps solving it in the two or three cases where that is explicitly reasonable
Calculations/coordinates don't bother me. I am a fan of moving frames computations, which are far better than coordinates, however, when one can do them.
I still say one should do basic curves and surfaces — even differential topology with submanifolds of $\Bbb R^n$ — before even looking at abstract manifolds and Riemannian geometry. That way one has intuition and experience.
When I taught the graduate geometry course I always did (with moving frames) a few weeks of classic curves/surfaces material both to build experience with the moving frames and to give a foundation for the general stuff to come. Plus plenty of exercises, of course.
It's about degree and transversality, which are beautiful and powerful notions one really needs eventually, but they aren't typically covered in any graduate course emphasizing Riemannian geometry.
Ted lectures: three forms explicit, implicit, parametrized manifolds are considered equivalent. in my class lectures: Parametrized manifolds is defined (but no equivalence of the three definitions is discussed). somewhere else: Suppose M is in X (a topological space), then M is an n dimensional manifold if given any p in M, there exists an open set U containing p, which is homeomorphic to an open set in $\mathbb R^n$. (In Ted's lecture, U = graph of some $C^1$ function.)
@Koro: Remember that I'm teaching a calculus class, so all manifolds are submanifolds of $\Bbb R^n$. When you do manifolds at the graduate level, they're abstract topological spaces, not sitting anywhere. It's still better to master the case of submanifolds first.
Yes, @Feynman_00, that's part of the self-learning thing. Unless you do something more restricted, like linear algebra or group theory. One would like to believe that we professors perform something of a service if we organize material effectively and teach a good class with good exercises.
I had two fabulous professors for a challenging physics course my first year of university. Two of my favorite courses! I did the third semester in self-study format; it just wasn't nearly as engaging or interesting.
Lie groups is a vast subject, too, Feynman. There are all sorts of aspects, some geometric, some algebraic, some analytic.
I see. I needed differential geometry first and I'll need some topology before starting. I think having fun plays an important role too and Lie Theory sure looks exciting
I agree that when one is ready to work with Grassmannians and other homogeneous spaces, then of course one must go the abstract route. But that's not where one should start.
I don't know what you mean by Lie theory, Feynman. As I said, it's a very broad/diverse topic.
OK, Feynman. That's way more restricted and, indeed, useful. You should understand invariant vector fields and invariant differential forms, the structure equations (Maurer-Cartan forms), etc.
But proving these are Lie groups is, first of all, a little bit of differential topology :)
Let me write the solution: Suppose that $x_n\downarrow x$. $\{x\}=\cap_n \{x_n,x]$ so $\mu_F(\{x\})=\mu_F(\cap_n \{x_n,x])=\lim_n \mu_F(x_n,x]=\lim_n F(x)-F(x_n)$.
But I don't really care about it, I study Math to do Physics but I want to study things as if I were a Math student (i.e. not just doing the calculations), so I'm fine with taking things slower but starting from the fundamentals
i always learn new things, that i missed in my studies, i am surprised and embarassed at same time for not being able to answer your questions sometimes
"That doesn’t work. You have to shift down by 10 and then rescale by .5. Shifting and rescaling do not commute." @TedShifrin It does work, it's basically just switching up the order of addition, like f(x) = x^3 - 4, you can start at -4 and treat it as g(x) = x^3 with origin (0|-4), go one to the right, that will be 1, so you go one up, which will be at (1|-3), ...
I think you thought I meant just shifting before stretching and using that value to stretch, which will also stretch the shift
@ILikeMathematics Well, that is pretty much what I remember you saying. Your addition example is too simplistic; it doesn't work with any coefficient other than $1$ in front of the $x^3$ term.
Did I understood it correctly, that every inner product space is a normed space since $||x||=\sqrt{\langle x,x\rangle}$ defines a norm. But on the other hand there are norms $||x||$ which can not be constructed from an inner product (but we use the same notation $||•||$
Let $f : [a, b]\to \mathbb R$ and for any real number $\alpha > 0$, let $E_\alpha:= \{x\in [a, b]: f '(x) \text{ exists and $|f '(x)| < \alpha$} \}$. Then $\lambda^* (f (E_{\alpha}))\le \alpha \lambda^* (E_{\alpha})$.
@Koro: this is a well known result. I posted a solution that uses that result (with a proof) here I also give references to other methods used elsewhere (in one and more dimensions)
Is there an algorithm or heuristic for determining a real number which some rational approximates but requires the least information to represent? Like if I put 1.414213562, to represent this numerical value as a truncation of $2^{\frac{1}{2}}$, it is possible to represent all of those unique digits using a single, much smaller representation (viz. the one I just gave plus a little info about the point of truncation).
Something similar to how Wolfram Alpha works and then, given such a list of possible closed forms, picks from it the one with the least amount of information in its representation. wolframalpha.com/input?i=closed+form+1.4142
If I know that $v=(x\log x, -y\log y)$ is Killing then might $w=\bigg(x{\rm log}x,-y{\rm log}y,-z{\rm log}z \bigg)$ also be Killing? I know that the pullback of $w$ along the embedding $e(x,y)=(x,y,0)$ is $v$
Sphere, for the double dual. I have an idea, what about, if $F\in S_{X^{**}}\setminus X$ and $x\in S_X$ then $x_t = \frac{tx+(1-t)F}{\|tx+(1-t)F\|}$. Since $F$ is not in $X$, the denominator makes sense, and his should converge in norm to $x$ as $t\to 1$.
I think that proves it?
And $x_t\in S_{X^{**}}\setminus X$ for all $t< 1$.
I think this proves that in fact $\overline{S_{X^{}}\setminus X} = S_{X^{}}$
That norm closure of the double dual sphere without X is the double dual sphere
