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3:00 PM
@Silent define integrable.
 
@Silent that’s not my issue. In what sense is infinity an integral?
 
usaully I see the phrase "uniformly convergent" as the limit of a sequence Leyla
 
the dominated convergence theorem is so cool
 
so perhaps they mean something like a sequence of
$\left \langle { \displaystyle \int_{1/n}^n t^{2k} e^{-xt^2} dt } \right \rangle$
and the claim is that this converges unformly to an integral on $[0..+\infty]$
actually that's a sequence of Riemann integrals, you can go straight to
$\left \langle { \displaystyle \int_{1/n}^{+\infty} t^{2k} e^{-xt^2} d\lambda(t) } \right \rangle$ for Lebesgue integral
then by the DCT you want a sequence of measurable functions $f_n$ converging pointwise to $f(x,t)$ and $|f(x,t)|\le g(x)$ etcetera
 
3:16 PM
oh, i see. if i go by Rudin PMA's definition, then integral is lower and upper partial sums. But rudin defines it for closed interval only.@LeakyNun, @Semiclassical
 
@Silent exactly. what you're asking is an improper integral, which is usually defined as a limit
 
ok,, only that kind of counterexample possible?
 
oh I'm sorry @LeylaAlkan I think it's a sequence in $k$
 
@Silent maybe I can answer your questions more effectively if you make them more precise
(or others can, I don't know everything)
 
Leaky knows everything up to a set of $\lambda$-measure $0$ though
2
 
3:24 PM
@LeakyNun No, thank you very much, I am convinced.
 
{{-1, -8.4, -8.4}, {-1, -4.4, -5.4}, {1.2, 6, 7}}
does this matrix converge?
A^n ?
 
@Jacksoja what is its JNF?
 
Unfortunately I am not familiar with Lebesgue integral yet @GFauxPas
 
okay Leyla then let's use limits of Riemann integrals, and we're saying the following
what book is this from? you probably can do it with Riemann integrals and limits but the notation $\int < \infty$ is usually in context with Lebesgue integrals
but anyway
 
@LeakyNun why do we need that?
 
3:30 PM
@Jacksoja well, converge iff all eigenvalues have magnitude smaller than one
so you don't actually need JNF, you just need its eigenvalues
 
you want a sequence of functions $f_n$ converging to $f$ pointwise
 
Okay let me check that
 
okay they actually are using Lebesgue integration in a sneaky way @Leyla , implicitly
but let's pretend we don't know that and we're using riemann integrals with limits
to use theorem 7.38 you need to find a sequence of functions
$\langle f_n \rangle$
such that for every $n$, $|f_n| < g$ and you can integrate $g$ and get a finite number
 
It's from Folland's Advanced Calculus
 
lol I'm using Folland's Real Analysis to learn this stuff, high five
he's limiting the case where you can do everything with limits of finite integrals; with lebesgue integration you don't need things to have finite limits always and you have more freedom, so he's only dealing with a case where you can write everything as improper riemann integrals. make sense?
that's funny that we're using the same author
so your sequence $\langle f_n \rangle$ to use theorem 7.38 you need to bound EACH $f_n$ with a function integrable in the sense of an improper riemann integral. let me know if you're folllowing
 
3:38 PM
I guess yes
 
well let's do better than "I guess" because I have lots of free time now to make my explanation simpler or more detailed :)
know what a sequence of functions is?
 
hahaha thanks then, yes
 
okay, so
know what uniform convergence is
 
yes
 
@LeakyNun does any entries of that matrix converge?
 
3:41 PM
so we know that if every member of the sequence $f_n$ is integrable on a closed finite interval, $f_n \stackrel{\text{unif}}{\to} f$ implies $f$ is integrable on that interval
right?
 
@Jacksoja no idea
 
besides entries 1,1 and 2,1
Can you use some program?
I need to install these things
 
okay @GFauxPas
 
okay the problem here is that $[0,\infty)$ is neither closed nor bounded so that's why we need extra theorems than just "uniform limit of riemann integrable functions is riemann intergrable", because these are limits of riemann integrals, they're improper integrals, so the old theorems are not enough
okay?
 
Given a set of functions, can you take the limit of the whole set?
 
3:49 PM
well if they're indexed by a countable set you essentially have a sequence of functions geocalc
but the concept your looking for is called "limit point"\
a set of functions can have a function as a limit point. not always, and it's not always unique
let me know when to continue Leyla
 
@GFauxPas okay I'm following
 
@geocalc33 It will be similar to treating a sequence of functions, but you will need some notion of topology on the set before limits can be talked about
 
so to show the convergence of the improper riemann integral in the example we use theorem 7.38
 
@Secret what sort of notion of topology would you need?
 
your sequence is $\left \langle { t^{2k} e^{-xt^2} } \right \rangle_{k = 0}^{+\infty} $
 
3:55 PM
At minimum you will need some directed sets, so that some notion of convergence can be talked about. But assuming your set of function is countable, you need converging sequences.
e.g. $x^2, 0.5x^2, 0.25x^2,0.125x^2,...$ converges very slowly to $0$
 
and the interval you're checking is $[\delta,+\infty)$ for $\delta > 0$
with me?
 
yes
 
So, for a set of function to have a limit, it has to be a directed set which eventually converges towards some function
 
we want a function $g$ such that $|f_n(x,t)| < g(x)$ for every $n$ and such that $g$ is integrable as an improper integral over $[\delta,+\infty)$. that's what theorem 7.38 requires, except the theorem uses $c$ instead of $\delta$
 
@Secret basically they are all the same function with different parameter values, and as the parameter tends towards infinity the area under the curve approaches zero
 
4:01 PM
good?
 
right, so your family of functions is arranged under some linear order based on the parameter
 
yeap
 
all the sets are non-negative so we'll ignore the absolute value part
we want an integrable function $g$ such that
 
@Secret yeah
 
$t^{2k} e^{-xt^2} \le g(t)$ for any $k$
the RHS isn't a function of $x$ so you need to find a bound that works for all $x \in [\delta,+\infty)$
any ideas?
does $t^{2k}e^{-xt^2}$ have a maximum on $x \in [\delta,+\infty)$ that might help?
$\dfrac {t^{2k}}{e^{xt^2}}$ has a maximum where the denomenator has a minimum in this case
 
4:07 PM
so that's why we take $\delta$?
Since we can make it as smaller as possible
 
because $e^{xt^2}$ has a maximum at $x = \delta$ so $t^{2k}e^{-xt^2} \le t^{2k}e^{-\delta t^2}$
 
Yeap makes sense
 
the RHS is integrable because $t^{2k}$ increases much much slower than $e^{-\delta t^2}$ decreases
 
So that means, letting $\lambda$ be the parameter, you have $\{\int f_{\lambda}\}$. Thus for $\lim_{\lambda \to \infty} \int f_{\lambda}$ to exists, this set has to have a limit point under the order topology of some subset of $\Bbb{R}$
 
you can prove it formally by inequalities or taylor expansions or whatever but it's integrable
on $[\delta,+\infty)$
 
4:11 PM
I don't know how to articulate that better cause I don't know my topology well enough
 
Then Folland says that because the convergence is uniform on $[\delta,+\infty)$, you can integrate and use limits to let $\delta \to 0^+$
and then you win the prize
the prize here, I'm guessing, is to find the value of $\displaystyle \int_{-\infty}^{+\infty} e^{-t^2}\, \mathrm dt$
or half of it,the integrand is even
 
which is $\sqrt \pi$
 
:)
I'd have to see the previous pages but he's probably justifying a step in a larger proof of that integral
 
So will you study this book of Folland too :)?
 
I'm studying a different once of his
 
4:18 PM
The book that I use
Apart from the real analysis one?
 
Show that a $k$-form $\omega$ is smooth iff it is smooth as a section $\omega : M\rightarrow \Lambda ^k(M)$ of $\pi$.
 
oh, for that I ordered adult rudin, but it didnt come in the mail yet
 
What is your definition for $\omega$ to be smooth?
 
Oh , but I think Folland's way of explaining things is so cool and easy to follow.
 
4:34 PM
the L shape thing is not really a function?
 
the black curves are all functions with different values for the common parameter and the purple line is y=x
 
yup
and it seems they converge to some L shaped thing
 
and as the parameter approaches infinity all the curves go to (0,0)
along y=x that is
 
yup
 
and as the parameter approaches negative infinity all the curves go to (1,1)
 
4:40 PM
I don't know how to set up an order topology of $f(x,\lambda)-x=0$
because you are basically computing $\lim_{\lambda \to \infty} f(x,\lambda) - x$
o wait, the limit will be defined on the usual topology of $\Bbb{R}$ because of how the function is rearranged
 
I don't know how to turn on the math symbols in this chat
 
i clicked on that but it won't let me right click the link to add as a bookmark
 
what browser you are using?
 
chrome
 
4:51 PM
so you don't have a right click box popping up as you click on the link?
 
yeah i can right click
but there's no option to add as a bookmark
 
O, for chrome, you can just drag the link to the bookmark bar at the top
 
okay got it to work thanks
 
5:15 PM
Hi, how do you get the bounds of an iteraded integral from the area? For example let $f: \Bbb R^3 \to \Bbb R$ and $B=\{(x,y,z) \in \Bbb R^3 \mid 0 \le x \le y \le z \le 1\}$. How do I get the bounds for $\int_B f \, \text{dvol} = \int_a^b \int_c^d \int_e^f f \, dxdydz$?
 
x from 0 to 1
y from x to 1
z from y to 1
oops you have them in the reverse order
z from 0 to 1, y from 0 to z, x from 0 to y
 
Guys I'd appreciate help on:
0
Q: Regularization of Exponentially exponential series?

More AnonymousQuestion What are the convergence properties of the last equation: $$ J^{-1}(x) + \delta = \int \frac{dx}{1+ e^{x} + x + \ln{x} + \ln\ln(x) + \dots + e^{J^{-1}(x)}} $$ Is it possible to ignore $e^{J^{-1}(x)}$ under some suitable range? How can one infer the value of $\delta$? Background...

 
5:30 PM
@LeakyNun thanks, but can you explain how you get them?
 
5:41 PM
@Lozansky I get f(x)f(0)=f(x)..
 
6:03 PM
$f(x)$
 
@AlessandroCodenotti hi
 
What does $1+2+\dots+n$ represent here?
Oh, its counted from, top right(1) upto diagonal(n). Got it
@LeakyNun, Let $M_3(\Bbb R)$ be set of $3\times 3$ matrices with real entries, and $V$ be subspace of symmetric matrices with trace zero. So, dimension of quotient space $M_3(\Bbb R)/V$ is $3^2-(\frac{3^2+3}{2}-1)$ , right? Also, if $M_3(\Bbb C)$ is considered, then also, dimension of quotient space remains unchanged, right?
 
6:23 PM
@Silent could you clarify the part with $\Bbb C$?
 
Hello, guys. My ODE textbook has a problem of finding an example of autonomous ODE i.e. $\frac{dy}{dx}=f(y)$ which all of its fixed points isolated.
 
@Silent the first claim is right
 
Now, a trivial solution to this would $\frac{dy}{dx}=0$
But can $\frac{dy}{dx}= \lim{a\to \infty}sin(ay)$ be a non-trivial solution?
I mean as the frequency of the sine function get larger and larger the zeros of the sine function also get dense. So, if the limiting frequency is taken to be infinity then the zeros would be very close.
O. Sorry. I meant the fixed/stationary points would be non-isolated i.e. there can be no open interval which contains only one fixed point.
I think my messages are becoming unclear. Let me re-organize.
My ODE textbook has a problem of finding an example of autonomous ODE i.e. dydx=f(y) which all of its fixed points non-isolated i.e. there can be no open interval which contains only one fixed point..
 
@LeakyNun I am sorry but internet connection was lost. Do you need to rewrite me the question, or make clear what I thought before inferring 2nd claim from first?
 
@Silent rewrite the second claim
 
6:36 PM
ok
 
Now, a trivial solution to this would $\frac{dy}{dx}=0$. But can $\frac{dy}{dx}=\lim_{a \to \infty} sin(ay)$ be a non-trivial solution?
 
Let $M_3(\Bbb C)$ be set of $3\times 3$ matrices with complex entries, and $V$ be subspace of symmetric matrices with trace zero. So, dimension of quotient space $M_3(\Bbb C)/V$ is $3^2-(\frac{3^2+3}{2}-1)$ , right?
 
@LeakyNun Can you help me with this problem?
 
@ParthaSarker $\displaystyle \lim_{a \to \infty} \sin(ay)$ does not exist
 
yo leaky
what is up ? :D
 
6:40 PM
@LeakyNun I know that but what I mean is as the frequency of the sine function get larger and larger the zeros of the sine function also get dense. So, if the frequency is taken to be very large then the zeros would be very close.
 
but they are still discrete
@KasmirKhaan hi
@ParthaSarker you would have to make your question precise in a way that it can be interpreted unambiguously
 
So, we can't take the limiting frequency to be infinity?
@LeakyNun Sorry about that!
 
In finding maximum of functions of 2 var
i take partial with respect to the two var and set them to 0 right?
 
Got disorganised while writing it
 
@LeakyNun I forgot my multi var stuff :D
 
6:44 PM
@LeakyNun Wouldn't Dirac delta function be a non-trivial solution to the problem?
 
@tatan If $f(0) \neq 0$ and $f(x) = f(0) f(x)$, then $f(x)$ must be...?
 
@ParthaSarker it isn't a function
 
it just plays one on TV
 
@GFauxPas Hey
 
@LeakyNun Usually it is said to be a function.
 
6:54 PM
yo
$\delta$ is a functional or a measure
 
@ParthaSarker that doesn't make it a function
 
And it is a generalized function I think @LeakyNun
 
or a distribution/generalized function
 
Generalized function
 
but it's not a function
it's defined by $\langle \delta, \phi \rangle = \phi(0)$ where $\phi$ is a function
that's the easiest way to think of it without using measure theory
 
6:57 PM
Okay. So is there any non-trivial function $f(y)$ for which the autonomous ODE $\frac{dy}{dx}=f(y)$ has all its zeros non-isolated?
*fixed points
Hey, guys?
 
why are closed homotopies more interesting than open ones
 
7:14 PM
Ok, can anyone just check I'm not being stupid. If I have a filter with poles at: 0.5(-1-i) and 0.5(-1+i), and I want to find the length of the vector between each pole at 0Hz, and I get ~ 1.5, but the mark scheme says it should be ~0.707 (sqrt(2)/2) -- can anyone help me?
 
7:29 PM
I am being stupid, fixed it.
 
does anyone want to see a mathematical structure i made
 
8:09 PM
proofwiki.org/wiki/Laplace_Transform_of_Complex_Power finished product! look how pretty it is :3
sure geocalc
 
8:23 PM
@GFauxPas I made it on desmos
click on it to enlarge
I'm trying to determine the manifold that results from these geodesics (the black curves). A very tricky problem
 
8:50 PM
Does a bounded variation path has to be continuous almost everywhere?
Because it is differentiable almost everywhere.
 
Well it's pretty but I don't know manifold theory so good luck
 
9:02 PM
If $R$ is a commutative ring, then it has at least one maximal ideal, cal iit $\frak{m}$. What does $R^n/\mathfrak{m} R^n$ denote and how does it differ from $R^n/\frak{m}^n$?
 
$f(x)$
 
@user193319 it's a $R$-module which is just the direct sum of $n$ copies of $R/\mathfrak{m}$
What exactly do you mean by $R^n/\mathfrak{m}^n$?
When I see $\mathfrak{m}^n$, then I think of the product of ideals, but that's probably not what you mean
Hey @Antonios-AlexandrosRobotis @AlessandroCodenotti
 
hi @MatheinBoulomenos!
 
Not exactly sure. I was thinking it was the same thing as $R^n/\mathfrak{m}R^n$, but you seem to be saying that's wrong.
 
I think you mean the same thing as $R^n/\mathfrak{m}R^n$, but the notation is confusing
 
9:12 PM
Aw man some nerds
How's it going?
 
@user193319 more generally, if $I$ is an ideal in $R$ and $M$ is an $R$-module, then $IM$ which is generated by elements of the form $im$ with $i \in I,m \in M$ is a submodule of $M$. So one can also look at the quotient $M/IM$. In this case we just have $M=R^n$ and $I=\mathfrak{m}$
Hey @Daminark
 
So $\mathfrak{m}R^n$ denotes the direct sum of $n$ copies of $\mathfrak{m}$?
 
it turns out to be that, yeah, but by definition I'd say it denotes the submodule in $R^n$ which is generated by elements $mx$ with $m \in \mathfrak{m}$ and $x \in R^n$
@Daminark Pretty well, thanks. I will give a talk in a seminar organized by my advisor on tuesday
When I said that $R^n/\mathfrak{m}R^n$ is a direct sum of $n$ copies of $R/\mathfrak{m}$, that wasn't the definiton, but more what it turns out to be, if you know what I mean @user193319
 
Yes, I do. Thanks!
 
0
Q: Analysing a sum of products

Mohammad Areeb SiddiquiI am trying to understand that what could be the possible outcomes of this following expression as $k \to \infty$ and as $|t| \to \infty$. $$\sum_{k=0}^{k=n}\dfrac{(\prod_{1}^{t}k)^{-i}}{\sqrt{k}}$$ Considering the numerator first: $$(\prod_{1}^{t}k)^{-i}=\prod_{1}^{t}\frac{1}{k^i} = (\prod_{1}...

 
9:16 PM
@user193319 np. I'm always happy to talk about commutative algebra :)
How's it going for you? @Daminark
@Antonios-AlexandrosRobotis How are you doing? Do you still have 2 TA jobs?
 
just finished those when the semester ended, @MatheinBoulomenos . Teaching linear algebra the summer, and trying to get towards being able to write a compelling master's thesis.
 
Sounds good. By teaching you mean actually giving the lectures?
 
yeah and the grading and so on
 
wow, I don't think the let grad students do that here
certainly not masters students
 
I impressed the right people somehow LOL
I'll be able to feed myself properly now rofl
 
9:21 PM
for LA (which is a mandatory Freshman course here), they always have full professors or at least some experienced post-docs which permanent positions etc.
 
Hi
How to find the max of this function
10 ln (x^2+2y^2+1) -x^2-y^2
f(x,y) =
I did the partial derivative w.r.t both x and y , it did not give me much to work on
x^2+2y^2=9 from I) and x^2+2y^2=19 from II)
 
Anyone know a little about manifold theory
 
on the circle x^2+y^2 =16
forgot to add that part
 
@Antonios-AlexandrosRobotis here's a fun suggestion for treating determinants (it's acutally what one of the profs here did) Let $k$ be a field
1) Define $\Bbb Z/2\Bbb Z$-graded algebras over $k$.
2) Define the twisted tensor product of $\Bbb Z/2\Bbb Z$-graded algebras and show it's again a $\Bbb Z/2\Bbb Z$-graded algebra
3) Define the Clifford algebra associated to a bilinear form by a universal property
4) Use twisted tensor products to show the existence of Clifford algebras (uniqueness follows from abstract nonsense)
 
Is the set of eigenvalues of a $\Bbb R$-linear operator a compact subset of $\Bbb R$?
 
9:31 PM
@AndersonFelipeViveiros in finite dimension, of course, as it is finite. In infinite dimensions, no
 
That's right
Hmm
 
It could be all of $\Bbb R$ even
 
And my space is not finite-dimensional ^^'
 
Consider the space $V=C^\infty(\Bbb R)$ of all smooth functions and the derivative operator $d:V \to V$. Then every real number $\lambda$ is an eigenvalue with the eigenvector $x \mapsto e^{\lambda x}$
 
Exactly. The space I'm working is indeed $C^\infty(\Sigma)$, with $\Sigma$ a surface embedded in a 3-manifold $M$ ^^'
The operator is the Jacobi operator
Is there any hope? ^^'
 
9:35 PM
I know zero geometric analysis
 
Hahaha I'm not an expert either
Thanks anyway
 
@AndersonFelipeViveiros if you know that your operator has some super nice properties (e.g. compact self-adjoint on a Hilbert space), then there might be stuff like spectral theorems
 
@MatheinBoulomenos do you know any manifold thoery
 
@MatheinBoulomenos Hm, thanks. I'll check it
 
@AndersonFelipeViveiros In that case, you know at least that the set of eigenvalues is countable and as a sequence it converges to $0$
but that's pretty specialized
If you have a bounded operator on a Banach space, then the set of eigenvalues is bounded by the operator norm at least
 
9:40 PM
@Jacksoja Lagrange multiplier maybe?
 
But I don't think $C^\infty(\Sigma)$ is such a nice space
@geocalc33 "any" manifold theory? yes, I took courses in diff geo and they also came up in alg top. I'm far, far from an expert, though
 
@Lozansky thanks but I remeber doing this in diff way, i did this long time ago
 
@MatheinBoulomenos do you know if it's possible to take a set of curved geodesics and build up a representation of the manifold that results from the geodesics?
 
@Lozansky is chaning to polar coordinates something that can be done here?
 
Yes might be a good idea
 
9:43 PM
@geocalc33 I don't understand the question
 
like for example if you have a graph of straight lines going across a 2D surface
 
I think you mean: Given a family of 2D curves, find a 3D manifold whose geodesics project to the plane curves
 
you know that you're dealing with flat euclidean space
@Semiclassical I think that's what I'm getting at
 
@Jacksoja You have $10\ln(r^2(1+\sin^2\theta )+1) - r^2$, set $r^2 = 16$ so maximum for $\theta = \pi/2$ and is $10\ln(33)-16$
 
@geocalc33 Certainly your set of geodesics has to have some properties for this to work. (e.g. resulting from uniqueness) What do you know about your geodesics? Do you know how they intersect etc.? Is the union of all those geodesics a manifold?
 
9:53 PM
Here's a picture for reference
 
In any case, I'm really the wrong person to ask for this. I "know" manifolds, but it's a fleeting acquaintance
 
They do intersect
 
@geocalc33 It reminds me of four-leaved rose ie $r=a|\sin 2\theta|$
 
@Lozansky yeah i can see the similarity
It also resembles a penrose diagram rotated
 
@Mathein sorry I was out but yeah things are going quite well!
What are you giving the talk on?
 
10:02 PM
local class field theory
It's the main result of the seminar
I'll prove some version of local class field theory, building on all the results from the previous talks. But it doesn't quite look like the usual formulation, I'll prove another result the week after that to fix this
 
@Semiclassical can you help with the manifold picture
 
that's pretty cool considering that class field theory is probably the most imporant result in 20th century number theory and a lot of modern research is about generalizing it
 
Hi, can I choose the order in which I compute an iterated integral freely in general, I mean if I integrate first over $x$ and then over $y$ or vice versa?
 
@philmcole Sounds like Fubini's theorem?
 
@philmcole not in general, no. The conditions for Fubini/Tonelli are not superfluous
 
10:09 PM
Ok thanks. Indeed I was assuming that the function is reasonable such that Fubini can be applied.
 
in that case, this is precisely the statement of Fubini
 
Okay. While we're at it. Do you know if this is true?
7 hours ago, by philmcole
58 mins ago, by philmcole
For a bounded real-valued function $f(x,y)$ (not necessarily continuous!) Fubini's theorem says that one can compute the double integral as iterated integral first over $y$ and then over $x$ for almost all $x$. Not necessarily for all $x$ since the inner integral over $y$ must not exist for every $x$. This means the problematic points $x$ are a Lebesgue null-set.

I was wondering if this poses any problem in real life when computing iterated integrals. Is it true, that we can just ignore the problematic points since the integral over a null set is zero anyways?
 
To see what kinds of things can go wrong, consider $f(x,y)=\frac{x-y}{(x+y)^3}$. You can compute that $\int_0^1\int_0^1 f(x,y) \mathrm{d}x\mathrm{d}y=-\frac{1}{2}$, but $\int_0^1\int_0^1 f(x,y) \mathrm{d}y\mathrm{d}x=\frac{1}{2}$
 
@Lozansky thanks !
@Lozansky how to find the max inside the circle ?
 
@MatheinBoulomenos could i ask my geodesic/manifold question on math stack exchange do you think?
 
10:15 PM
@geocalc33 I don't find your question very precise as you stated it, to be honest. It might get closed for "unclear what you're asking"
@philmcole as the example shows, you do have to be careful when you interchange the integrals. The function $f(x,y)= \frac{x-y}{(x+y)^3}$ looks innocent enough, but it actually doesn't satisfy the conditions for Fubini
 
That sounds fantastic
 
@Jacksoja Try solving $\dfrac{\partial f(r,\theta)}{\partial r}=0$ with $0 \leq r^2 \leq 16$. Then plug it back in together with $\sin^2 \theta = 1$
 
@Daminark yeah, I'm stoked
 
@MatheinBoulomenos What happens here that this goes wrong? In our theorem the assumption is that the inner integrand $f_x: X \to \Bbb R$ must be Riemann-integrable. Why can you actually compute both versions of the integrals if the inner integral doesn't exists?
 
0
Q: Analysing a sum of products

Mohammad Areeb SiddiquiI am trying to understand that what could be the possible outcomes of this following expression as $n \to \infty$ and as $|t| \to \infty$. $$\sum_{k=0}^{k=n}\dfrac{(\prod_{1}^{t}k)^{-i}}{\sqrt{k}}$$ Considering the numerator first: $$(\prod_{1}^{t}k)^{-i}=\prod_{1}^{t}\frac{1}{k^i} = (\prod_{1}...

 
10:18 PM
@philmcole the integral is improper here
 
Ah right, the left bound is a limit to zero.
We didn't talk about improper integrals in the context of multivariable calculus yet.
 
well, if you do iterated integrals, then it's just single variable calculus
 
We discussed in the context of Fubini's theorem that Fubini is true "almost for all" points in the domain. The problem points are just those where the inner integral doesn't exists. But those are a Lebesgue-null-set (apparently).
 
In general, your argument that removing finitely many points doesn't change the integral doesn't work, I think. When you remove points from the (closed) interval you're integrating over, then you end up with an improper integral for which the theorems about integrals on closed intervals don't necessarily apply
You could also say in the example that $(0,0)$ is the only problematic point
 
Okay, I was being too optimistic :P
 
10:23 PM
I only learned Fubini in a measure theory context, so I can't really comment on what you might do in multivariable calc
 
I didn't see @hippalectryon for a long period of time ...
 
(never even took multivariable calc)
 
Wait how's that possible? For me it's mandatory (part of second semester)
 
I'm from Europe, we do things quite differently
 
Lol ich doch auch.
 
10:28 PM
Du bist Deutscher?
 
ja
studier halt in Zürich grad
 
meinst du Ana 2 mit multivariable calc? Da haben wir Lebesgue-Daniel Integrale, Hilberträume, Fourierreihen und Differentialformen gemacht
aber unser Prof war auch verrückt
 
Hi guys
 
Ja genau. Wow wir machen nicht so viel aber bis zu Jordan-messbaren Mengen gehts schon.
Skript sind auch 800 Seiten
Ana 1 und 2 zusammen aber
 
Okay
Ich kenne Fubini für Riemannintegrale nicht, also kp
 
10:31 PM
Okay, ja ich habs jetzt schon besser verstanden auf jeden Fall!
 
Can anyone please tell me where the "1" in "P + bP − dP = (1 + b − d)P" comes from in this math example?

Let us imagine that we wish to predict the population of a country in 20 years’ time.

One very simple model we might use represents the entire country as a pair of numbers (t, P(t)).

Here, t represents the time and P(t) stands for the size of the population at time t.

In addition, we have two numbers, b and d, to represent birth and death rates. These are defined to be the number of births and deaths per year, as a proportion of the population.
 
@philmcole es wird wahrscheinlich klarer, wenn du das nochmal nächstes Semester in Maßtheorie siehst
 
Ich glaub der Physik Studiengang den ich mache hat keine Maßtheorie. Aber wir haben noch noch Funktionentheorie glaube ich.
 
@JessicaMendoza Factor out $P$?
 
@philmcole echt? Also die Physiker hier in Heidelberg hören alle Ana 3, wo man Maßtheorie macht. Aber Funktheo ist wahrscheinlich sogar nützlicher
 
10:35 PM
Ja ich denk die haben das wichtigste aus Ana 3 schon in Ana 2 gepackt. Jedenfalls kenn ich noch Unis, da kommt die Hälfte was wir in Ana 2 machen erst in Ana 3 dran. Bei euch scheint aber Ana 2 schon krass gewesen zu sein :D
 
@philmcole Wir hatten halt so einen Prof, der total genial ist aber alles immer auf seine Weise macht. Wir haben absolut konvergente Reihen als Lebesgue-integrierbare Funktionen auf $\Bbb N$ bezüglich dem Zählmaß definiert ^^
 
$P=(1)P$
 
Bei andern Profs ist das hier auch nicht so heftig
 
@Lozansky is it because a multiple cannot be 0?
 
@MatheinBoulomenos Gut, da versteh ich noch nichts von :D
Aber bin schon gut beschäftigt mit unserer Analysis VL. Die legen so ein Niveau vor... Vor allem fehlt irgendwie jegliche praktische und anschauliche Herangehensweise. In den Übungen macht man 90% Beweise und vielleicht mal ein Rechenbeispiel. Gut für die Mathematiker aber nicht für die Physiker...
 
10:40 PM
@philmcole grob gesprochen, wenn du den Begriff von "Integral" allgemein genug fasst, kannst du für eine Funktion $f:\Bbb N \to \Bbb R$, die Summe $\sum_{n \in \Bbb N} f(n)$ als Integral auffassen. Das macht auch irgendwie Sinn, du könntest dir ja vorstellen z.B. du machst aus der Funktion $f:\Bbb N \to \Bbb R$ einfach eine funktion auf $\Bbb R_{\geq 0}$ wenn du immer auf Intervallen der Länge 1 konstant bist
(das ist nicht wirklich der Grund, warum das als Integral zählt, nur als Motivation)
 
Ok cool
 
Also wenn es dir um Anschaulichkeit geht, würde Maßtheorie auch nicht helfen. Eher im Gegenteil, das ist dann noch mehr abstrakte Theorie
 
Ich glaube ehrlich gesagt auch nicht mehr dran dass es irgendwann besser wird :P
 
@JessicaMendoza What do the terms in $P+bP-dP$ have in common?
 
Hier ist auch so, dass viele Physikstudenten die paar Mathevorlesungen, die die so hören müssen, als ziemlich hart empfinden
(LA1 + Ana1-3 wäre das)
 
10:44 PM
So geht's mir auch
 
Ich hab sogar von Physikstudenten gehört, die LA1 bis kurz vor die Bachelorarbeit aufgeschoben haben und dann daran gescheitert sind
 
@Lozansky P
 
Ja ich bin so ein Kandidat. Bei uns ist LA1 halt auch wieder super anspruchsvoll weil jedes Thm gleich in seiner allgemeinsten Form gegeben wird, also z.B. zwischen endl. und unendl. dim. VR unterschieden werden muss. Dann kommt z.B. eine MC-Frage in der Prüfung ob irgendwas allgemein gilt was in endl-dim. VR immer gilt und dann kreuzt du fröhlich "ja" an und liegst falsch...
 
@JessicaMendoza Now apply the distributive law
 
@philmcole so sind wir Mathematiker halt :P
 
10:48 PM
Die ETH macht es den Physikern echt schwer :P
 
@Lozansky P + bP − dP = b * P + b * P
 
What?
 
@philmcole aber Funktionentheorie wird höchstwahrscheinlich wieder besser: man schränkt sich da von vorneherein auf eine (wie sich herausstellt) sehr enge Klasse von Funktionen ein, die überraschend viele schöne Eigenschaften hat. Also es wird mehr "Wow, gilt das wirklich? Das ist doch zu schön um wahr zu sein"-Momente als "Wtf, ohne 50 Annahmen gibt es immer ein seltsames Gegenbeispiel"-Momente geben.
Und man kann mit der Theorie tatsächlich wirklich schön einige knifflige Integrale ausrechnen, das finden oft die Physiker viel cooler als die Mathematiker
 
@JessicaMendoza I mean $a(b+c) = ab+ac$, apply this (but from right to left)
 
@Lozansky P(b + d) = bP + dP.
 
10:56 PM
Das ist cool. Man kann wirklich paar schöne Dinge berechnen. Ich fand z.B. Teilmannigfaltigkeiten in R^3 ganz cool z.B. den Schnitt zweier Zylinder. Hab die dann immer in Mathematica geplottet und mir ein bisschen Intuition verschafft.
 
@JessicaMendoza so what if you add P to both sides?
 
Ja, Mannigfaltigkeiten ist für Physiker ganz cool
 
@Semiclassical what do you mean?
 
What happens to the expression you just had if you add P to both sides of it?
 
P(b + d) + P = (bP + dP) + P
 

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