my daughter was working on a magazine that had a list of shapes, where you were supposed to write in their name, and their numbers of sides and corners, respectively. the very first shape is a filled-in circle.
it got easier from there (square, triangle, 'diamond' aka tilted square)
in her view, a filled in circle has one side and no corners, although this didn't match the other shapes, whose numbers of sides matched their numbers of corners.
Hm, I suppose I can go about rigorously defining "closed form" actually despite using it as a poorly-defined adjective/noun. A numerical expression is in closed form provided it can be expressed as a uniformly convergent sum or product whose terms are also expressible as uniformly convergent sums or products provided that this nesting of sums/products eventually terminates in expressions of a finite number of operations and/or symbols.
I don't think I would consider a series to be in closed form, but some closed forms are nothing but a wrapper for a function that is not much more than a series or an integral.
Well I mean I'm trying to cover transcendental relations that require either a limit or infinite sum by necessity, but I suppose it's worth pointing out that there should be a way to represent these in terms of more fundamental relations like the exponential and logarithm, and at times, even by integer domain expressions.
If I just wanted to define closed form for what it is, I would just say, "that which is representable with a finite number of non-unique or 'serial' symbols".
So you could define exp as a closed form here by taking its limit definition, but I mention uniformly convergent series because they're convenient for numerical approximations.
Something like erf representing a subset of the incomplete gamma function comes to mind as an example.
erf, Hypergeometric functions, etc. are common, but not as common as some other functions. Certainly, when an answer can be expressed using them, it is easier to use them, but sometimes calling them a closed form feels like cheating.
Yeah speaking of, I'm still looking for a closed form for Gamma of all things expressible through some form of the exponential. Digamma of all things has a closed form which doesn't feel like cheating because it consists of constants and a finite sum of a form of the exponential, and integrating would give a closed form for Gamma, but my problem is that it isn't computable in $O(1)$...
...unless there's some trickery with $\frac p q$ I don't know about to guarantee the sum is constant while permitting $\psi_0(z)$.
I mean I'm getting ahead of myself--it needs to be generalized to the reals and then the complexes for that, but I don't imagine it's too hard to construct the algebraic numbers from the rationals.
I have computed an asymptotic formula for $H_n$ in one of my answers. That is just a constant different from digamma. One can integrate that asymptotic formula and exponentiate to get Gamma.
How do you get digamma from the harmonic numbers for all complexes?
See I was trying to do something crazy like use the functional equation relations for Gamma and try to get Gamma out of that somehow by various different closed form identities.
Sounds promising though. Even the formula given to me by Plouffe (via a paper he gave me) only gets me as far as 48 bits of accuracy for Gamma, and then you still have to compute in linear time based on which pair of integers your value sits between.
Yeah. Something which gives a set of all possible inputs to $\Gamma(z)$ such that at some point $\{z, \Gamma(z)\}$, it might be called "locally inverse" as a sort of $\Gamma^{-1}(z)$.
I don't really know the notation for complex inputs and outputs smh
The fact that it's a self-similar fractal should help
At least for real-valued Gamma, I can easily say $\Gamma(x) \propto \Gamma(x + n)\forall n\in \Bbb Z$ is what I mean by self-similar fractal.
So if we compute an inverse for just one of those cases, we can compute an inverse for all of them.
Something like how arctan works, only if it gave us some answer like $2\pi f(n)$ as opposed to $2\pi n$.
@robjohn Thanks. I didn't try to calculate any areas, I was just playing with the plotting. I don't have much experience with Bessel functions, although I did write some code to approximate them using Simpson's rule, several years ago. stackoverflow.com/a/33715116/4014959 They're used in a Fourier expansion for Kepler's equation, but I usually solve Kepler's equation using Newton's method.
@PM2Ring I set up the equations, but had to let Mathematica do the grunge work. Since the answer came out pretty simple, I might try to evaluate it by hand.
@JackRod No. But a month or two ago I was reading about the work that Richard Brent and the Borwein brothers did on pi using the arithmetic-geometric mean (AGM). There's some info here: arxiv.org/abs/1802.07558 Jonathan Borwein's book Pi and the AGM is available free online. It's a bit above my level, but I still learned a few things from it. It has a lot of great info.
@Obliv if someone cared about this, they'd want the basis for the row space expressed as a list of row vectors, and the basis for the column space expressed as a list of column vectors, but if you got the right n-tuples, i imagine a lot of instructors wouldn't care. note that if you have already written it out, you can change from one to the other by placing a superscript T next to it.
e.g. oops i wrote the row vector [1,2,3]. un-oops, i wrote the column vector [1,2,3]^T, i just happened to write it horizontally.
why is "The set of all first-degree polynomial functions ax, a ≠ 0, whose graphs pass through the origin" not a vector space
nvm i get it
$x - x = 0$ isn't $ax = 0$ where $a \ne 0$
wait so the set of all $m \times n$ matrices are vector spaces? not just for square matrices?
for nonstandard operations on $\mathbb{R}^3$ given by $(x_1,x_2,x_3) + (y_1,y_2,y_3) = (0,0,0)$ and normal scalar multiplication, is it a vector space? I think so
it's just the zero vector, which is a vector space i'm guessing
i'm using . for the scalar multiplication, trying to make it look very axiomatic. but if you like, 2x generally isn't x + x in that thing, so it's not a vector space.
obliv: they are implicitly defining what 'even' means there. presumably the set of all functions from the reals to the reals, satisfying f(-x) = f(x) for all x.
a polynomial is 'even' in that function sense if it involves only even powers of x (where i'm including 1 as an even power of x there). same for 'odd' in place of 'even.'
i'm guessing that's where it comes from, but i dunno for sure.
it's the only thing expressly required by the axioms to be in there. it might be the only thing in there. if there's a nonzero vector v, then there's also all the multiples of v (which are required by the axioms to be different for different scalars, although maybe this isn't immediately clear from the axioms).
seeing as how you can have $(a,b)$ span a line in $\mathbb{R}^2$ can you "over span" $\mathbb{R}^2$ with the standard basis of $\mathbb{R}^3$ for example
obliv: a common way of doing something like that would be to write something like "for 1 \leq i \leq m and 1 \leq j \leq n let E_ij denote the mxn matrix having a 1 in its row i, column j entry and zeros in every other entry. then {E_ij: 1 \leq i \leq m, 1 \leq j \leq n} is a basis for M_{mn}."
Let there be some metric space and let $S$ be a subset of said MS. I thought by the definitions of isolated point and interior point of a set $S$ that every isolated point is an interior point. Because an isolated point is a point such that there exists a neighborhood which contains only said point. This neighborhood is obviously a subset of $S$. Hence this point is also an interior point.
e.g. if S is a subset of the real numbers R, and T is a subset of S, you can ask about the interior of T in S, or the interior of T in R, and they might not be the same set of real numbers.
which is probably what you are saying, just more boring sounding
choosing real numbers just to be concrete, no particular importance to that choice there.
Oh I see. I guess given a metric space $(X, d)$ and subset $S$, I assume that "$p$ is an interior point of $S$" means that there exists a neighborhood $N = N_r(p)$ such that $N \subset S$; at least if I am following Rudin's definition properly. Is this an unusual interpretation of the quoted phrase?
just fill in both blanks of "interior of __ in __" every time, if it bothers you. even if rudin doesn't do that.
i do remember someone going haywire in here a few months ago because they kept switching which space they thought rudin was working in. rudin thought it would be enough to say "in what follows, everything is a subset of a fixed metric space X," and at least for that person it wasn't enough. they wanted "and when i ask about the concepts just described, i mean as they exist in the metric space X"
@DLeftAdjointtoU Nice diagram. But I don't know Quiver. And I've never touched Django. OTOH, I occasionally do vector graphics in Python, by generating SVG code.
silly: you also see this with stuff like 'open' and 'closed,' if S is a subset of a metric space, you can ask, open in X, vs. open in S, and maybe get different notions. the default presumption (some books are better at signaling it than others) is that if you say something like, "we consider __ as a subset of the metric space X," then whenever the choice of ambient space matters for some concept within __, you mean to choose X, and not __.
and so that if you intend to depart from the presumption, you do so by saying so explicitly. instead of having silence mean 'switch to what it means in __' or 'uh oh, who knows what to do.'
of course one way to avoid all of this is just to say what you mean every time. but books get really hard to read if they do that.
@robjohn FWIW, ChatGPT can draw diagrams in Tikz. And SVG. But since it's trained to generate streams of tokens it's got a slightly odd sense of 2D structure. It can also do ASCII art, but the results are often bizarre, since the ASCII art in its training data got mangled by whitespace compression. ;)
ooh, i was wondering why it was so bad at ascii art.
beyond it just being bad at 2d structure, i was getting stuff that was just gibberish, even from simple requests. whitespace compression would do that...
(In Riemann surface context) we define an integral divisor $D$ is special if there is an integral divisor $D^*$ with $DD^* = Z$ where $Z$ is a canonical divisor. Riemann-Roch implies if $\deg D\leq g-1$ then $D$ is special. Since $D^*$ is also special and $\deg D+\deg D^* = \deg Z = 2g-2$, I can conclude for any Riemann surface $S$, there is a special divisor of degree $n$ for $0\leq n\leq 2g-2$. Am I correct?
Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.
I only know that $[K:F] \le n!$, but how can I show that $[K:F]$ divides $n!$?
I want to show the following inequality for absolutely convergent series $$\left|\sum_{k=1}^\infty u_k\right|\leq\sum_{k=1}^\infty |u_k|.$$ The triangle inequality proves the statement when the sums are finite, but what about when they are infinite?
I've tried working from the definition of a convergent series, i.e. if $\sum_{k=1}^\infty |u_k|$ converges to $L$, then $\forall \epsilon>0$, $\exists N>n$ such that $\left|\sum_{k=1}^\infty |u_k|-L\right|<\epsilon$, but I am unsure where to go from here.
@shintuku thanks for the reply. If it doesn't contain negative components, then obviously we have equality, however, I am little unsure how to go about it when there are negative components.
I guess you have to be careful with taking limits on strict inequalities of sequences, since, if the sequences converge, they are turned into non-strict inequalities, see here.
X and Y are 2 random variables that are not necessarily independent. Both have standard normal distribution. Can you evaluate Cov(X,Y) without knowing anything else?
I am trying to compute $$\sum_{k=1}^\infty t_{2k},$$ where $t_k=\sum_{j=2}^\infty j^{-k}$. I am struggling a bit with the indices. Is the sum I am trying to compute the following $$\sum_{k=1}^\infty\sum_{j=2}^\infty j^{-2k}?$$
Anyone please have a look at my problem. I have determined to look in the direction that $F(x, y) = \{ f(x) + g(y): f, g \in L^2 \}$ closed in $L^2$. What approach would you suggest?
L= Q( sqrt 2, sqrt 3) is normal extension of Q or not?
I noted that L= $Q(\sqrt 2+\sqrt 3)$ using tower law.
$[L:Q]=4\ne 2$ so can't conclude from 'quadratic extensions are normal.'
Suppose that the extension is normal. Then Irr($\sqrt 2+\sqrt 3)=p(x)$ must split in L.
$p(x)=(x^2-2)^2-24$. So L must contain splitting field of $p$ o/ Q, which is just $Q(\sqrt 2+\sqrt 3)$. But this also does not give me a contradiction.
So it seems that the extension is normal.
Take $a\in L$. Consider $q(x)= Irr (a, Q)$. It must be shown that q splits in L. But I am not sure how to show this.
Can we say that $(-1,1)$ is not closed in $\mathbb{R}$ with the Euclidean topology because the sequence $x_n=1-1/n$ is such that $x_n \in (-1,1)$ for each $n\in\mathbb{N}$ but $\lim_{n\to+\infty} \left(1-\frac{1}{n}\right)=1 \notin (-1,1)$?
$L= Q( \sqrt 2, \sqrt 3)$ is normal extension of $Q$ or not?
I noted that $L= Q(\sqrt 2+\sqrt 3)$ using tower law.
$[L:Q]=4\ne 2$ so can't conclude from 'quadratic extensions are normal.
Suppose that the extension is normal. Then Irr($\sqrt 2+\sqrt 3)=p(x)$ must split in $L$.
$p(x)=(x^2-2)^2-24$....
it's been a long time since i've studied this, but i think i agree with the commenter who said - i'd solve this by proving that the given definition of 'normality' is equivalent to one that is easier to check, and then verifying that other condition. and not by approaching the exercise with 'bare hands.'
@robjohn now, suppose I have $t_k=\sum_{j=2}^\infty \frac{(-1)^j}{j^{k}}$ and we are interested in calculating $\sum_{k=1}^\infty t_{2k}$. Can we still switch the order of summation then?
so today, some people complained to our functional analysis teacher that they don't understand what he 'teaches' in class and that they need some time to absorb things. He then said you have plenty of time.
Lethargy is so widespread here that some teachers verify exam answerscripts on the same day when they have to submit it as per deadline.
So you can imagine how they verify answer scripts and give marks.
In a question for example, I was given 0 out of 5. I went to see my answer script and told him that my answer is correct. He said oh, I see and then gave me 5 marks there.
Have you ever seen nonsense like this anywhere?
But suppose that I were not in campus on the day answer scripts were shown, say I were in my hometown, then I would have lost marks!
@Jakobian: In some cases, some mark their attendance and as the teacher turns to the board, they slip outside from the back gate. It's funny to see them do that.
There is one teacher who never looks back and always faces board.
But the real question is how to determine for any real-valued ratio $\frac a b$ such that $a + b = 2^c$ what $a_1 + b_1 = 2^{c_1}$ such that $2^{c_1}$ divides $2^c$.
Let $(X, \mathcal{M}, \mu)$ be a finite measure space and $\left\{f_n\right\}$ a sequence of measurable functions on $X$ that converges pointwise a.e. on $X$ to a function $f$ that is finite a.e. on $X$. Then for each $\epsilon>0$, there is a measurable subset $X_\epsilon$ of $X$ for which $$ \left\{f_n\right\} \rightarrow f \text { uniformly on } X_\epsilon \text { and } \mu\left(X \sim X_\epsilon\right)<\epsilon $$
Let $f$ be a real-valued measurable function on $E$. Then for each $\epsilon>0$, there is a continuous function $g$ on $\mathbf{R}$ and a closed set $F$ contained in $E$ for which $$ f=g \text { on } F \text { and } m(E \setminus F)<\epsilon $$
So F(x, y) can be approximated by continuous function on unit square?
@robjohn thank you. Is it correct then that we are trying to compute $\sum_{k=1}^\infty (\eta(2k)-1)$? Reminder; we had $\sum_{k=1}^\infty t_{2k}$ where $t_k=\sum_{j=2}^\infty \frac{(-1)^j}{j^{k}}=\left(1-2^{1-k}\right)\zeta(k)-1$.
hmm, $1-\frac{3}{4}=\frac{1}{4}$, but how is $\sum_{k=1}^\infty 1-\left(1-2^{1-2k}\right)\zeta(2k)$ the negative of $\sum_{k=1}^\infty\left(\zeta(2k)-1\right)$?
@HashNuke I guess that this follows from the following: let $H$ be a Hilbert space (maybe also assumed to be separable), and $E,F$ two closed subspaces of $H$. Then the sum $E+F\subseteq H$ is also closed.
late to this, but that's not true in general. it's true if you have extra info about the relative position of the subspaces, which maybe we do here. or if we don't, that's maybe a direction to go in to prove the sum isn't closed. i haven't scrolled back. (kinda in and out this afternoon)
Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.
I only know that $[K:F] \le n!$, but how can I show that $[K:F]$ divides $n!$?
