@psie the leftover factor, difference between $\lfloor b\rfloor$ and $b$, will converge to zero because $(-1)^n/n$ does. That's why it's irrelevant and the limit is just the series $\sum (-1)^n/n$
You can replace this by any convergent series to see that $\lim_{b\to\infty}\int_0^b f = \sum a_n$
And if the series is absolutely convergent then $\int f$ will exist and equal $\sum a_n$
of course if the series isn't absolutely convergent then the limit is $\int f = \sum a_n$ as well, just in the Henstock-Kurzweil sense
Lebesgue integration is only concerned with absolutely convergent things, so it can't capture conditional convergence well
but of course there's other benefit of it being very general
@Jakobian let $f=\sum a_n$ be the function such that $\int |f|=\sum n^{-1}$. You are first applying $\left|\sum a_n\right|\leq\sum |a_n|$, so by monotonicity $\int|f|=\int \left|\sum a_n\right|\leq\int\sum|a_n|$ and then simply applying additivity of the integral for nonnegative measurable functions, i.e. $\int\sum|a_n|=\sum\int |a_n|=\sum n^{-1}$, but wouldn't that give us $\int |f|\leq \sum n^{-1}$?
Here $a_n=n^{-1}(-1)^n\chi_{(n,n+1]}$.
Hence $\sum\int |a_n|=\sum n^{-1}$.
@psie It's late. My first sentence here should read something like; let $f=\sum a_n$ be the function with $a_n=n^{-1}(-1)^n\chi_{(n,n+1]}$.
Meeh. I think I'll just work with the positive part of $f$ instead. Makes it easier for me to understand why $f$ is not Lebesgue integrable. Anyway, good night!
Not good...I want to be able to partition $\beta \mathbb{N}$ into proper non-empty clopen subsets...
Like, given any proper non-empty clopen subset of $\beta \mathbb{N}$, I want to find other non-empty proper clopen subsets $B$ and $C$ such that $A$, $B$, and $C$ partition $\beta \mathbb{N}$...hmm...maybe I'll have to treat that as a special case...hmm.
So, that's one way $\beta \mathbb{N}$ and the cantor space differ...hmm...maybe there's a way around it...I'll have to let it rattle in my head for a bit.
And since we're at the topic of clopen sets, you can decompose $\beta\mathbb{N}$ into copies homeomorphic to itself by decomposing $\mathbb{N}$ into finite amount of infinite sets $A_1, ..., A_n$ then $\overline{A_1}, ..., \overline{A_n}$ are clopen, homeomorphic to $\beta\mathbb{N}$, disjoint and their union is $\beta\mathbb{N}$.
You can also decompose $\mathbb{N}$ into infinite amount of such sets, but the union won't be necessarily equal to $\beta\mathbb{N}$ then
Yes, definitely helpful! We can do the same thing with the Cantor set, and basically the Thompson groups are groups of homeomorphisms which swap/permute these homeomorphic copies.
Interesting...I wonder if it is possible to construct a free group in $Homeo(\beta \mathbb{N})$ that not in the image of the embedding of $Sym(\mathbb{N})$ into $Homeo(\beta \mathbb{N})$...that sounds hard...
I'm not sure why are you asking this, like I said the two are literally the same as groups in the sense that $f\mapsto f\restriction_\mathbb{N}$ is a map from one to the other which is a group isomorphism
Really? You're saying $Sym(\mathbb{N})$, the group of bijections of $\mathbb{N}$, is isomorphic to $Homeo(\beta \mathbb{N})$? That doesn't sound right to me.
If $f:\beta\mathbb{N}\to\beta\mathbb{N}$ is a homeomorphism then necessarily $f(\mathbb{N}) = \mathbb{N}$ so that $f\restriction_\mathbb{N}$ is a bijection of $\mathbb{N}$
and if $g:\mathbb{N}\to\mathbb{N}$ is a bijection of $\mathbb{N}$, then treated as map $\mathbb{N}\to\beta\mathbb{N}$ and then by Cech-Stone compactification property taking extension $g^\beta:\beta\mathbb{N}\to\beta\mathbb{N}$ we can see that $g^\beta$ is actually a homeomorphism since its inverse is $(g^{-1})^\beta$
@user193319 yes because each $n\in\mathbb{N}$ has countable system of neighbourhoods in $\beta\mathbb{N}$, which is not true for any $p\in \beta\mathbb{N}\setminus\mathbb{N}$
@user193319 to clarify, any closed $G_\delta$ subset of $\beta X$ disjoint from $X$ must contain a copy of $\beta\mathbb{N}$, which is of size $2^\mathfrak{c}$, so that no $p\in\beta X\setminus X$ can be $G_\delta$. Here $X$ is some Tychonoff space.
Well, this isn't exactly elementary
another way that I see is to use that every clopen set of $\beta\mathbb{N}$ is of the form $\overline{A}$ for some $A\subseteq \mathbb{N}$ and $\beta\mathbb{N}$ has basis of clopen sets
Since $\mathcal{U}\in\overline{A}$ iff $A\in\mathcal{U}$, the claim can be reduced to that there is no sequence $(A_n)\subseteq\mathcal{U}$ such that for every other ultrafilter, some $A_m$ doesn't belong to it
And you should be able to prove that there's at least two ultrafilters which contain the sequence $(A_n)$
To do this you can for example assume that the sequence $(A_n)$ is strictly decreasing and consider the ultrafilter containing $\bigcup_n (A_{2n}\setminus A_{2n+1})$ and the one containing its complement
That is, consider filters generated by $(A_n)$ and those sets, and extend them to ultrafilters
for the (usual) definition of a manifold, are charts judged to be continuous with respect to the appropriate subspace topologies?
i.e. a manifold has the data $(\mathbb{R}^m, \tau_d)$ and $(X, \tau)$. a chart is then a map $\varphi: U \to K$ where $U \in \tau$ and $K \in \tau_d$. To say that $\varphi$ is continuous, do I consider the subspace topologies induced on $U$ and $K$?
@psie I don't have Folland in front of me right now---my recollection is that this is basically encapsulated in his definition of the Lebesgue integral.
That is, look at how the integral of a measurable function is actually defined.
@XanderHenderson Hmm, ok. Well, the integral of a measurable real-valued function goes back to the definition of nonnegative measurable functions, which in turn goes back to the definition of the integral of simple function (which are measurable by definition). I'm really not sure what he means, but perhaps he is alluding to the fact that a real-valued or complex-valued function can be approximated from below by simple functions and thus we are only concerned with "lower approximations".
I guess one could develop analogously that every nonpositive measurable function can be approximated from above by nonpositive (measurable) simple functions. Then we'd solemnly get an "upper approximation" I guess.
I don't know if it would actually yield an "upper approximation". Now that I'm used to the theory with nonnegative functions, rewriting the theory in terms of nonpositive functions seems...convoluted.
@Jakobian hmm ok 👍 but if we had a nonpositive $f$, then the conclusion of the lemma for $f$ would be that we'd approximate it from above by nonpositive simple function, right?
Find the exact value of sin(y) given that y in in quadrant IV & sec(y)=$\frac{{6\sqrt3}{5}$ My work led me to the answer that sin(y) should be $\frac{\sqrt249}{18}$ does this seem correct?
@Jakobian I skimmed through both videos, the second one by BBC does not seem to explain the idea of partitioning the range. I still think that, if we have a nonpositive function, and we partition its range, then, since the integral will be negative, our approximation will be an upper bound to the integral. In that sense, we can speak of upper approximations when talking about the Lebesgue integral too.
Find the exact value of $sin(y)$ given that $y$ is in Quadrant IV and $sec(y)$=$\frac{6\sqrt3}{5}$ my work led me to the answer $sin(y)$=$\frac{\sqrt249}{18}$. Does this seem correct.
I guess so. Would that make the answer $sin(y)$=$-\frac{\sqrt249}{18}$? That makes as $sin$ should be negative in the fourth quadrant. This is of course depending on whether or not $\frac{\sqrt249}{18}$ is correct anyway.
Hi everyone, am I the only one having this issue with the website? The cursor moves out of the editable text form as soon as I attempt to post a comment or answer.
> Consider $z\in\mathbb C$ and $\mathrm{Re}(z)>0$. Define $f_z:(0,\infty)\to\mathbb C$ by $f_z(t)=t^{z-1}e^{-t}$ (here $t^{z-1}=\exp[(z-1)\log t]$). Since $|t^{z-1}|=t^{\mathrm{Re}(z)-1}$, we have $|f_z(t)|\leq t^{\mathrm{Re}(z)-1}$, and also $|f_z(t)|\leq C_ze^{-t/2}$ for $t\geq1$ (the precise value of $C_z$ can easily be found by maximizing $t^{\mathrm{Re}(z)-1}e^{-t/2}$).
Why does $|f_z(t)|\leq C_ze^{-t/2}$ hold for $t\geq1$ but not otherwise? To be honest, I don't see why it holds at all.
psie: as a threshold thing, which is actually really important, he never says "but not otherwise," does he? i would basically never expect an analysis book to expect you to infer significance from what cases they weren't saying things about in an estimate like this.
psie: if your first instinct when reading a statement like that is to wonder what's happening with what the author isn't talking about, you're going to be distracted a lot. (in this particular example, there's nothing too important about 1, any positive number would do independently of z, which isn't even beginning to talk about what you might be able to use for some z but not others to get an estimate like that)
psie: for what you actually ought to have been asking about, see jakobian. f_z(t) := t^(z-1) e^(-t) = [t^(z-1) e^(-t/2)] e^(-t/2), the first equality by definition and the second by basic properties of the exponential, and he's saying that the thing in brackets is bounded
In probability and statistics such bump would be called a mode, so that in general this function should be unimodal that is, with a single bump, is what I expect
psie: the general fact folland may be appealing to here is that a continuous function on an interval like [a, +oo) with a finite limit at +oo will be bounded. it is possible to just 'see' the continuity and the limiting behavior, this is t^p e^(-t) for some fixed p. the fact that the interval [1, +oo) excludes 0 simplifies the analysis because it won't matter whether p is negative
You can extend your function $g(t) = t^{\text{Re}(z)-1}e^{-t/2}$ to be $g(\infty) = \lim_{t\to\infty} g(t) = 0$ then $g$ is continuous on a compact set $[1, \infty]$ and so attains maximum
@Jakobian I was thinking we simply find $N\in[1,\infty)$ so large such that $g(t)<g(1)$ for all $t>N$. Then the maximum of $g(t)$ on $[1,\infty)$ will be attained on $[1,N]$.
You might appreciate this: If a function $f:X\to\mathbb{R}$ is vanishing at infinity, then $f$ is uniformly continuous for any uniformity on $X$, in particular one might take compactification of $X$ to show $f$ must be bounded
Well... alternatively, if $f$ is vanishing at infinity then of course there is a compact $K$ such that $|f(x)| < 1$ for all $x\notin K$, so this also works to show its bounded :P
Obviously I'm doing analysis 2 first, otherwise it wouldn't make sense also because there is the propaedeucy
@SineoftheTime Yes, there are mathematical methods, signal theory which has a part of probability, and electronics (which would be fundamentals of circuits)
For the new subjects, for example electronics, it requires that one knows how to solve differential equations for example, so it's a bit useless to follow if I don't take analysis 2 first or not?
Just to give an example, I'm not just referring to differential equations, but the professor mentioned them.
but since you will probably not start studying electronics from tomorrow, you may go to the lessons since you'll need the knowledge of DE when preparing the exam
it depends on how many hours of lesson you have during the week, how much time do you need to go to the uni, how many hours you can study, how difficult the course is, ...
I wasn't in this problem, because I had passed the test, but I still had some difficulties having studied business economics. connected to computer science
yes, what should not happen is for example making a mistake due to distraction that makes you fail the whole exercise, I say this because it happened to me with a sign
Why is it so difficult in this day and age to connect a frontend web gui text input to a backend database item, using ajax.... It's ridiculous! So flimsy too
@SineoftheTime yes, apart from solving the exercises, I have to remember the various cases, so I will try to do the exercises without checking before finishing
well, since I followed the course I used to do the exercises on a topic while the professor was explaining but to prepare the exam I did mixed exercises
@Pizza If you already have a good understaning and knowledge of a topic it's ok
