Given a sequence $(a_n)_{n\in\mathbb{N}}$, if I prove that for a fixed $k_0 \in \mathbb{N}$ it is $a_{n-k_0} \ge 0$ for each $n \in \mathbb{N}$ can I deduce that $a_n \ge 0$ for each $n\in\mathbb{N}$? I think that this is true, because if $a_{n-k_0}$ for each $n \in \mathbb{N}$ then given an arbitrary $i\in\mathbb{N}$ this holds in particular for $i+k_0 \in \mathbb{N}$; that is, $0 \le a_{i+k_0-k_0}=a_i$ for each $i\in\mathbb{N}$. Is this a correct proof?
@TedShifrin actually, it was just a thought of mine; no exercise. You are right, I should have assumed that $n-k_0$ is nonnegative! So, assuming that, the reasoning works? Thank you.
Bad idea. Do topology later. I advised students for close to 40 years. You need plenty of experience and topology generalizes what you learn in analysis. Take it later.
I recently wrote a Taylor Series Expansion of $e^{a\arcsin x}$ but then a question says to find the Taylor expansion of $\arcsin x$ from the expansion of $e^{a\arcsin x}$ and I have no idea on how to do it?
@SoumikMukherjee and @leslietownes Can you please help me with this? I am stuck with this problem badly.
I want to evaluate
$$\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}\,\mathrm dx$$
I tried to let $$I(a)=\int_0^\infty\frac{\ln(1+a^2x^2)}{e^{\pi x}-1}\,\mathrm dx,$$ and then
$$
\begin{align}
I'(a)&=\int_0^\infty\frac{2ax^2}{(1+a^2x^2)(e^{\pi x}-1)}\,\mathrm dx\\[25pt]
&=2a\int_0^\infty\mat...
all purpose no-spoiler ideas include 'try residue methods over some cleverly chosen something' 'try introducing a parameter and differentiating with respect to it' etc.
it seems to be difficult, or the people who regularly jump on these things would have jumped on the first linked problem above
Leslie, I have one question which is "Is there anything bad in seeing a solution? Or when to give up on a problem? Cuz sometimes when I see solution "I often feel like oh gosh I was almost there.
lucky i dunno, with something like "finding a 'closed form' for [some random definite integral]" i don't see what would be spoiled, exactly, by seeing a fully worked out answer. computing nice forms for definite integrals is a gigantic bag of tricks, and you add to your bag by looking at what others have done.
i see concern about spoilers arising, if at all, in a situation where maybe seeing routine things worked out by other people prevents you from ever bothering to learn basic techniques yourself. but this is deep enough in the weeds that i don't see that popping up
in the interest of complete honesty i also don't know why anyone gives an ess aych eye tee about definite integrals like that
so learn 'em, don't learn 'em, do 'em, don't do 'em, it's all kinda the same to me
@leslietownes yeah! You are right. But I don't know why "my mind doesn't allow me to see the solution." I think I should solve it by myself. When I have full worked out solution I am allowed to see what other people have done. lol,
i guess part of my background apathy is, i regard this subject as one in which clever people can come up with simple-looking problems of essentially arbitrarily high "difficulty" (however you measure it)
so i guess i see it as mostly just a puzzle thing, and you get to make your own rules. i wouldn't look at partial solutions when i work out a crossword puzzle, something like that. not good or bad, just how i puzzle.
i also wouldn't generally expect there to be a "nice" closed form, but that's another thing
the people who do those problems on MSE will be like "oh wow! here is a nice closed form for your integral it is twice the octo-logarithm of the apery constant plus 5 euler's gamma" and i wouldn't regard that as "nice", but they do
@LuckyChouhan i taught for a while but certainly not this kind of stuff
i wonder how often people have taught topics courses where you just do definite integrals all day. not for any contest, just to do them
well if it's anything like the others of these, it's probably on page 123 from some textbook from 1880 and the author does something similar with another problem on page 122.
Manifolds question: is it true that if $M$ is an $n$-manifold and $F\subseteq M$ is a finite set, then there is an open $U\subseteq M$ with $F\subseteq U$ and $U\cong\Bbb R^n$?
Not the usual statement, Alessandro, but if $\dim M\ge 2$ the isotopy lemma says you can isotope the manifold and move the finite set anywhere you want.
I confused myself by thinking about manifolds with boundary and then missed that I knew the answer to my own question after reducing it to one about manifolds without boundaries!
Because $M\setminus\partial M$ is an honest manifold, and you can move finitely many points wherever you want with a map that becomes the identity at infinity
@TedShifrin well the points are not on the boundary in the case I'm interested in :P
Actually @Ted the situation I'm working with is the following: I have a closed disk $D'$ from which I've removed the interior of finitely many pairwise disjoint closed disks $C_i$ to obtain a manifold with boundary $D$. I have a homeomorphism $h\colon\bigcup\partial C_i\to\bigcup\partial C_i$ so that, for every $j$, $h\upharpoonright C_j$ is either orientation preserving or orientation reversing. I want to extend it to $D$. Now @Astyx swears it can be done, but I'd like a reference
Scalar multiplication $v \mapsto (1 - t)v$ commutes with arbitrary linear transformations, in particular with transition functions.
Consequently, if $E$ denotes the total space of your vector bundle, $x$ denotes a local coordinate on the base, and $v$ denotes a local coordinate in the fibres, th...
Suppose that is true for all vector spaces of dimension $<n$, now take an $n$ dimensional vector space $V$ and consider a subspace $V_1$ of $V$. Now $V_1$ is contained in some maximal subspace $V_2$. Apply the hypothesis on $V_2$
But I am not sure whether this qualifies as a proof by induction, also I don't think everything can be shown by induction
Suppose there is some element $x$ that is not in $V_1$, adjoin it to $V_1$. If the resulting subspace is the total space then $V_1$ is itself a maximal subspace. If the resulting subspace is maximal then you are done, otherwise keep adjoining elements. This process should terminate after finite steps as the whole space is finite dimensional
@SoumikMukherjee but doesn't showing that the process terminates after finitely many steps assume that a subspace of a finite dimensional space is finite dimensional, making the argument circular?
what does 'dimension' mean? it can certainly be circular if that core notion isn't pinned down in the right way. a common way of confronting the key issue is to prove that if a space is spanned by even a single list of m vectors, then any list of more than m vectors in that space is linearly dependent.
and you build your definition of dimension somehow around that.
if that happens to be what axler does, the question of whether an 'inductive' proof exists is maybe an objection to this idea of proofs that work by inductive constructions that produce increasingly long lists of vectors, instead of inductive constructions that reduce somehting about dimension n to something about dimension n-1.
if for example dim V := max {|S|: S is a linearly independent subset of V} it is trivial that a subspace of a finite dimensional space is finite dimensional and it is not something that would naturally lend itself to a proof by induction. but you still need probably need all of those lemmas about lists of vectors, removing things from linearly dependent spanning sets without changing the span, etc. to operationalize the definition.
The usual proof (which you mentioned) doesn't use the axiom of choice at all.
You can prove this directly using induction. Pick some non-zero $w_1\in F$.
Suppose that we chose $w_1,\ldots,w_n$ and $F_n$ is their span. If $F_n=F$ then $F$ is finite dimensional. Otherwise, $F_n$ is a proper sub...
Something about "You don't need AOC because you are only making finitely many choices". So thought maybe the algorithm can be turned into some kind of inductive or recursive argument.
AP: what does 'bounded in X' mean here, where X is a normed space? it isn't necessarily true, for example, that if a function has a finite L^2 norm, then its L^infty norm also has to be finite. e.g. 1/x on [0,1] has finite L^2 norm but is not in L^infty.
prithu: if V has a linearly independent spanning set {v_1, ..., v_n} and w_1, ..., w_{n+1} are n+1 vectors in V, by writing each w_j as a linear combination of the v's and using fairly concrete scalar math (that a homogeneous system of n+1 equations in n unknowns has a nonzero solution, findable by gaussian elimination and such) you can expressly write one of the w's as a linear combination of the others.
and that's enough to rule out [any subspace of] V containing n+1 linearly independent vectors, let alone arbitrarily long lists of linearly independent vectors.
i don't know that you can remove the difficulty of, in general, constructively doing this with lists of vectors will involve the equivalent of matrix computation, and that if n is large this could be unwieldy to do by hand, or unwieldy numerically, or something. but it should be clear that you don't need some new principle of set theory to do it.
AP: yes, if |f_n(t)| <= 1 for all n and t then each f_n is in L^oo, and ||f_n||_oo <= 1
[i meant n equations in n+1 unknowns above, if that wasn't clear.]
prithu, you might compare your feelings about the abstract vector space situation with your feelings about whether you really believe that something like row echelon form of a matrix actually exists in general, or has just been shown to exist in small examples
i don't mean that as a joke, i think it's fundamentally the same kind of indeterminacy where as a logical matter, of course it exists, but actually doing it (even in fairly concrete cases) is enough of a pain in the ass that you might wonder
i don't think so. there's certainly a separate adjective for such spaces ("completely metrizable"), and there probably wouldn't be if it were coextensive with metrizable
If a function on average is greater than another over the same interval, this always implies that the integral of the bigger function is greater, The converse not true right? if the function had a jump discontinuity?
But what is both the functions are continuous without jump discontinuities , will the converse hold?
Find the Taylor Series expansion of $e^{a\arcsin x}$ and hence deduce, the Taylor series expansion of $\arcsin x$.
I could find the Taylor Series expansion of $e^{a\arcsin x}$ as $$1+ax+\frac{a^2x^2}{2}+\frac{a^3+a}{6}x^3+\cdots $$
However, I have no idea how to deduce the series for $\arcsin x$ ...
I am knowing Abstract Algebra things; I am searching aims of Abstract Algebra and origins of parts of Abstract Algebra. I thought original initial textbooks have explicit links to aims and origins of the parts of Abstract Algebra.
Can we list major textbooks on Abstract Algebra in the order of ti...
If $G=GL_2(\Bbb{R})$, $H=\{g\in G: g_{21}=0\}$ and $X=\Bbb{R}\cup \{\infty\}$. How can I find a map $f:G/H\rightarrow X$ s.t. $f(gy)=gf(y)$ for $y\in G/H$? I'm a bit confused since $y=g'H$ for some $g'\in G$ but then the map should end into the real line. The beginning of the exercise we showed that $g\cdot x=\frac{ax+b}{cx+d}$ where $g$ is the matrix with entries $a,b,c,d$ is a group action so I think I need to do somthing with it
Rudin's Functional analysis is more readable than I expected. That's good.
maybe I should try RCA later.
The thing is that I don't know what would be a good time distribution between self-studying topics I'm interested in and coursework. I used to spend most of my time on the former but maybe I need to care about coursework too.
Find the values of $p$ and $q$ such that $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13.$ Assume, that L' Hospital's rule is applicable.
I tried solving the problem as follows:
Given, $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13$ and according to the given information, that L'...
@Thorgott I mean that $\tau_{G/H}\subset \tau_{d}$ where $\tau_{d}$ is the discrete topology is clear. But I mean If I take $U\in \tau_{d}$ , i.e. $U$ is any subset of $G$. Then to show that it is open in $G/H$ we need to show that $\pi^{-1}(U)$ is open in $G$ right?
A naive question: we know that the composition of two bijections is a bijection. Why should I not be surprised that one can form a bijection out of two functions, neither of which is a bijection?
This comes up in the following context: **Lemma:** If a set $X$ injects into $\mathbb{N}$, then it's countable.
*Proof.* If $X$ finite we are done, so let $X$ be infinite. Let $f : X \to \mathbb{N}$ be the relevant injection injection.
Being a subset of $\mathbb{N}$, the set $f(X) = \{ f(x)\,:\, x \in X \}$ is well-ordered, and in particular it has a least element, a 2nd-least element, a 3rd-least element, and so on. Define $g : f(X) \to \mathbb{N}$ by sending the $n^{\text{th}}$-least element of $f(X)$ to $n \in \mathbb{N}$. Then the composite $g \circ f : X \to f(X) \to \mathbb{N}$ is i…
The somewhat confusing bit is that $h:= g \circ f$ is a bijection, but $f$ is not. I think $g$ may be a bijection to by its definition. So my confusion is that (1) we get a bijection out of an injection and a bijection (i guess this is OK?) and (2) if $f$ is not a bijection, we can still write $f = g^{-1} \circ g \circ f = g^{-1} \circ h$, which shows $f$ is a bijection? Where am I going wrong?
Rob, do you know where I can find a bunch of advanced exercises at infinite series for real analysis 2? The university I study at doesn't provide much and most of what I see in books are ones that are very common
Hi, loosely speaking, if I had a ball with a dot on it, then deciding which direction that dot should point could be thought of as choosing an element of SO(3), right?
Is it true that a vector-valued function is Lipschitz continuous if and only if all of its components are? If so, is there any formula for the Lipschitz constant of the function over a given set in terms of the Lipschitz constants of the components?
If the above is true, it seems like it should be a proposition/theorem in every other lecture notes on Lipschitz continuity, but I haven't found this in any sources I've been looking at.
Actually, my question is a bit strange, since the Lipschitz constant is not unique, but I was wondering if there is a formula, say $L=L_1+L_2$, such that $L$ is a Lipschitz constant of the function when $L_1$ and $L_2$ are for the respective components.
Consider $F(x)=(x,\sqrt{1+x^2})$. Is this function globally Lipschitz on $\mathbb R$? Using the above result, its components are Lipschitz continuous for any $x,y\in\mathbb R$ and hence the function is, however, I can not determine whether this is global or local Lipschitz continuity.
@TedShifrin Hey Ted. Couple of days ago, you said to me that if a sequence na_n approaches zero at infinity then it roughly means the terms decay 1\n faster. I think I did not understand it, do you know what can help me to intuitively grasp it?
There is something called reinforcement learning wherein an "agent" must take an "action" at every discrete time step. Decisions for how to take an action are made via function that takes in an element of the state space and gives a probability distribution over action space. If an agent is controlling, say, a camera and has to decide where to look, then I thought this might connect with rotation groups (and ultimately, probability on groups).
But as you might be able to tell, I am a little over my skis
(for extra clarity: the "action space" would be a group, e.g. $\textit{SO}(3)$, although I'm apparently wrong about that)
Well, the group is the group of isometries of the unit sphere, but there’s a whole circle worth of choices of group element that take the north pole to any given point of the unit sphere. That’s a lot of ambiguity. But maybe you’re ok with that.
I don’t see why you don’t just use the unit sphere itself.
I just thought that there might be a connection with group theory, but I guess I have to think about it more, and learn more group theory too, when I find the time. Thanks for your help
musing aloud: I guess if I forgot about the groups and just thought of the action space as the unit sphere, then I would have a probability distribution over the sphere (a manifold)
I want to evaluate
$$\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}\,\mathrm dx$$
I tried to let $$I(a)=\int_0^\infty\frac{\ln(1+a^2x^2)}{e^{\pi x}-1}\,\mathrm dx,$$ and then
$$
\begin{align}
I'(a)&=\int_0^\infty\frac{2ax^2}{(1+a^2x^2)(e^{\pi x}-1)}\,\mathrm dx\\[25pt]
&=2a\int_0^\infty\mat...
The usual proof (which you mentioned) doesn't use the axiom of choice at all.
You can prove this directly using induction. Pick some non-zero $w_1\in F$.
Suppose that we chose $w_1,\ldots,w_n$ and $F_n$ is their span. If $F_n=F$ then $F$ is finite dimensional. Otherwise, $F_n$ is a proper sub...
Define $m = \sum_{d \mid p_n\#} \frac{(-1)^{\omega(d)}2^{\omega(d) -(2\mid d)}}{d}$. Where $(2\mid d) = 1 \iff 2 \mid d, \text{ else } 0$ is shorthand.
Conjecture.
$$
m(p_{n + 2}^2 - p_{n + 1}^2) \geq \log_{p_{n + 1}}(p_{n + 2}^2 - 2)
$$
for all $n \geq 1$.
Is it true?
Is there a way to prove t...
