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12:00 PM
Ok, btw after concluding that it's strictly increasing, you say that it has to have exactly one real root.
 
a challenge question (one that does not count towards marks) from my assignment is to prove that there does not exist a rational number $x$, where $x^3+x+1=0$
and I am supposed to do it by contradiction
 
@micsthepick Start by showing that it would in fact then have to be an integer
 
The very theorem might help. The same method, I mean.
 
I was thinking that it would probably be a proof by infinite descent (right terminology?), but I wondered if there was a way of figuring out it's root
the thing is that that theorem is not part of my course
@TobiasKildetoft do I need that theorem for this step?
 
what theorem?
 
the Rational Zero Theorem ^
 
it'll probably be apart of some future course
 
I am natural, integral, rational, real, and complex, but I ain't transcendental.
 
are we supposed to guess what you are?
 
@JasperLoy you are not irrational
 
12:11 PM
@user1732 No, I am just listing my character traits.
 
there are an infinite number of numbers that share those traits
 
@micsthepick If I am rational, then I am not irrational.
 
that fact that you are rational means that you are not transcendental also
 
@micsthepick I am not a number but a person.
 
12:13 PM
How appropriate!
@user1732 Hello, pal.
 
hello, pal
 
@JasperLoy Some people here don't upload their profile picture on stackexchange server. What should we do with them?
 
@user91500 We should just let them be. I only uploaded a blue square myself.
 
@JasperLoy So you are a regular person!
 
@user91500 Yes.
@user1732 Why don't you use your original username?
 
12:19 PM
i'm protesting 0celo's one year suspension
 
@JasperLoy There is just one person here that doesn't like to upload on SE server, So I'm very sad :(
 
@AnotherJohnDoe There is no pace you need to set. Just make sure you understand every sentence. Take your time. Exercises are important but if you don't have time skip them first.
@user91500 Who is this person?
 
@JasperLoy J S
 
@user91500 Don't know who that is.
 
more importantly, why would it make you very sad?
 
12:24 PM
@user1732 What happened, in one sentence?
 
Anyone here familiar with the Weierstrass approximation theorem?
 
in Physics Meta, May 18 '17 at 23:01, by Shog9
@heather I think he could benefit from some time not in chat. And I think chat can benefit from some time without him. We'll see how that works out. Beyond that... I don't know there's much of value I can say.
 
@user1732 Since his profile picture doesn't load in my browser, So an irregularity has been appeared here!
 
@user91500 You mean has appeared and not has been appeared.
@user1732 Doesn't say anything, but OK, that will do.
Hello @AlexClark.
 
@JasperLoy I mean "has been appeared".
 
12:34 PM
@LeakyNun To reconcile what different people have told you, although there are people working in set theory, of course these are much fewer than people working in algebra, or analysis, or geometry, or topology. That's all.
 
@JasperLoy here
 
@OskarTegby oh we haven't talked since a long time ago
I hope he's still doing well
 
@LeakyNun Did you reply to the wrong line?
 
@JasperLoy no
 
@LeakyNun Are you referring to ocelot?
 
12:48 PM
no
i was making a joke
 
1:08 PM
@JasperLoy, thank you for your advice. I do plan to solve the exercises.
 
@LeakyNun: Really? You and Weierstrass go way back?
 
no, just his theorem
i'm friends with his theorem
 
$💥(Weierstrass)$
Nope I have no idea what happens in a Weierstrass category
 
Hello @Secret I am still thinking what your secret is.
 
lol
I like learning secrets
In mathematical analysis, the Weierstrass approximation theorem states that every continuous function defined on a closed interval [a, b] can be uniformly approximated as closely as desired by a polynomial function. Because polynomials are among the simplest functions, and because computers can directly evaluate polynomials, this theorem has both practical and theoretical relevance, especially in polynomial interpolation. The original version of this result was established by Karl Weierstrass in 1885 using the Weierstrass transform. Marshall H. Stone considerably generalized the theorem (Stone...
$💥(\text{Stone-Weierstrass theorem})$
 
1:17 PM
who's gonna stone weierstrass?
 
1:28 PM
2
Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

mickLet $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins , and prime twins grow like $O (n * ln(n)^2 ) $ as i...

Any ideas ?
 
Hiiii again
 
hello
 
I'm taking a break. That problem is still driving me nuts
😥
 
@usukidoll this is regarding the "set of points in exactly k subsets" problem?
What I find weird is that, if k>1, then there's seemingly no guarantee that the sets A_i will have any points in common
as such, it hardly seems guaranteed that the resulting space will be nonempty
(e.g. take the integers and partition them into evens/odds. Then the set of integers which are either even or odd but not both is just the integers themselves. So the space of integers which are both even and odd is empty and it seems like that can't be an event space)
 
1:44 PM
What does $\|\cdot\|_\infty$ mean?
 
'Infinity norm'
 
Why would it appear in the sense of a contraction map?
$$|f(x)|=\left|\int_0^1K(x,y)f(y)dy\right|\leq\int_0^1|K(x,y)||f(y)|dy\leq\frac{‌​1}{2}\int_0^1|f(y)|dy\leq\frac{1}{2}\|f(y)\|_\infty$$
 
it's a norm, and I think one often wants to show that a contraction decreases norm
also, the reason it's called the infinity norm is that you start with the p-norm $(\sum_i x_i^p)^{1/p}$ and take $p\to\infty$
which if you think about it will just give the value of the largest $x_i$
 
@Semiclassical yeah that one. I can't even. . ughhhh what is it asking?!!!!! 😠
 
As such, the infinity norm is just the largest value on the set
 
1:48 PM
I don't really see why it would be the largest value on the set, but I see why it's used.
 
Suppose you pick an element $x_k$. then $(\sum_i x_i^p)^{1/p} = x_k (\sum_i (x_i/x_k)^p)^{1/p}$
 
I did the rest of the problems but only a select few gets turned in and this is unfortunately one of them. Like there's 6 problems and I did 5 sooooo this one thing. I can't ... More likely who is the m= 2 and k=1??
 
Yeah
 
If $x_k$ is the largest such element (and for simplicity I'll take it to be unique)
then $x_i/x_k\leq 1$ with equality only when $i=k$
as such, $(x_i/x_k)^p\to 0$ as $p\to\infty$ if $i\neq k$. if $i=k$, then instead $(x_i/x_k)^p = 1^p \to 1$ as $p\to\infty$
 
I wish exercise 1.8-1.10 can be turned in. I understood those. Also got the Venn diagram and proved the P( A or B or C) . It's that one problem left that's making my brain tired 🙄😫😓
 
1:51 PM
so in the limit $p\to\infty$ one just gets $(\sum_i x_i^p)^{1/p}\to x_k$
so the infinity norm is just another name for the supremum norm
 
Okay. Cool! That's a really good explanation.
 
one does have to be a bit careful: what if multiple $x_i$ were the same?
e.g. $x_1=x_2$ both gave the same max value
it turns out not to be a problem. in that case you'd just end up with $(\sum_i x_i^p)^{1/p}\sim x_1 (1+1+0+\cdots +0)^{1/p}\to x_1=x_2$
so it does work regardless, but it's a bit more annoying
Now, this of course doesn't tell you why they want the infinity norm in your particular problem
But that's not something I'd know the answer to :P
 
Okay. :)
 
@usukidoll For the m sets you just need to split them into 2 groups, one group of the k sets and the other group of m-k sets. Then you can take intersection over (not of) the k sets and intersection over (not of) the m-k sets. Then there are m choose k ways to choose k out of m sets. Then you can take union over (not of) this set of m choose k ways. This is how to write your answer in short.
 
I'm gonna cry. This is worse than the heads and tails problem attempt I did tonight 😭
$( m \choose k) (k \ choose m-k)$ ajdhsjshf
 
1:57 PM
It's OK to cry. I hope you feel better soon. =D
 
Draw picture
Explain like I'm five. I've been at this thing for 2 days x.x
 
huzzah for inclusion-exclusion
 
I think if you read some proofs they have this kind of thing also for general numbers m and k and what not. I guess you just haven't come across them.
 
I still don't get itttttttttttttttt 😲😭
 
Well, here's a common way this kind of problem would show up
 
1:59 PM
Just expand what I wrote above and you get the proof you can turn in.
 
Suppose you've got a collection of university students, each taking a set of courses
 
Inclusion-exclusion is like $P( A \cup B)= P(A)+P(B)-P(A \cap B)$
 
Then you could certainly talk about the set of students which are taking 2 out of m possible courses.
or 3 out of m possible courses, etc.
 
Oh I remember the students taking courses problem
It's like some take English ... Others take Math and some are in both English and Math
 
Right.
So in principle you could partition the students into event spaces based on how many courses each of them is taking.
 
2:02 PM
P(A) is the event that students take English and P(B) is the event that students take Math
 
Right. And P(A Delta B) would be the people who are taking exactly one of the two
What bothers me, going to what I said above, is that there's no guarantee that these spaces are nonempty
 
Exactly one English or exactly one math
 
Right
 
If there's no guarantee that the spaces will be nonempty that means that either A will be empty or B will be empty and that breaks one of the definitions of an event space
 
I mean, suppose you have different versions of the same course (e.g. two different profs teaching first-semester intro physics)
then the event space of students taking both versions of the course is presumably empty
in which case it's...not much of an event space
 
2:04 PM
Let x_1,...,x_k be the indices such that the points belong to Ax_1,...,Ax_k and not to the others of A1,...,Am. This is how you can start off @usukidoll, something like that.
 
perhaps, some quick review of high school is in order?
 
So I feel like there's some assumptions being made in order for the problem to make any sense.
 
We don't need any combinatorics to solve that problem, just need to take complements and unions and intersections @Semiclassical.
 
Regardless, I hope the example is clear enough: You'd have $m$ courses being offered, and each student would take some $k$-subset of these courses.
@JasperLoy true enough. I just like being able to ground problems like this in examples.
 
@user1732 highschool was 12 years ago :p
 
2:07 PM
@Semiclassical Oh yeah, that's an exact parallel
 
So if m= 3 and k=2 then out of three courses students take 2 of them.
 
I think the problem with writing that proof for @usukidoll is that she hasn't seen proofs involving unions and intersections taken over sets we define, only the usual 2 or 3 set unions and intersections.
 
There we go
 
Right. Of course, you could have a school where some students take 1 course and other students take 3
 
That's why although the idea is simple, it is hard for her to write the concrete proof.
 
2:08 PM
I usually see a bunch of A and Bs and ABCs things
 
But for this it seems like a string ... A string from the index set
 
Yeah, right, index set.
 
Of unions... Like in definition 1.4
That's what's driving me nutZo
 
It still bothers me that, taking the definition as they've written it, there's no guarantee that "the set of all elements in exactly k subsets" is nonempty
 
2:10 PM
Whether that set is empty or not doesn't matter right?
 
It does matter for the definition of an event space
 
If A is in F then the complement must also be in F. The empty set and Omega must be in there which is what 1.2 says. Like it needs something. Not blank space
 
That's the first condition: that the collection of subsets of the sample space be nonempty
 
Yup
Meaning our sample space must have elements
 
So if you take the definitions as written, then the problem is wrong on trivial grounds.
 
2:12 PM
1.3 is if A is in F then the complement must also be in there so not A is in F
 
@Semiclassical That is saying there is at least one set in the sample space, not that there is at least one element in the set of all elements in exactly k subsets, right?
 
Wut da
 
that's requiring that the collection of subsets be nonempty
but if there's no such subset, then that's empty
 
I'm using Grimetts's book. probability an introduction 2nd edition. Got both an ebook and a paperback
 
Now there is a difference between the collection of subsets being empty and an empty set that is a subset itself.
 
2:13 PM
oh, good point
 
🙃
 
So F would just contain the empty set and therefore be non-empty itself
 
The empty set is always in F, so F is nonempty.
 
The proof to 1.4 was super condensed. I was trying to replicate the proof in the book but only understood 1.2 and 1.3 so when I saw that string in exercise 1.11 it was like related to 1.4 and I was like ... 🙃
 
hmm
That sounds like a reasonable resolution, but it's not obvious to me that "the set of points in the sample space which belong to exactly k of the subsets" would contain the empty set
 
2:16 PM
@Semiclassical No, the set of points being discussed here is just considered as a subset of omega.
 
what?
 
And you just wanna show that this set of points is in F.
 
ugh, you're right
you're not trying to show that "the set of points which are in exactly k subsets" are an event space
you're trying to show that "the set of points which are in exactly k subsets" are a subset of the event space
 
Isn't that the ... I have the book . One of the examples is $\Omega$ is any nonempty set and $F= \{ \emptyset, A, \Omega \backslash A, \Omega \}$ where A is a given subset of $\Omega$
 
So we just need to take the complements and unions and intersections and we are done.
 
2:18 PM
Oh crap
 
So only the writing is hard because it involves index sets.
 
Flash floods warning on my phone
I'm indoors
 
yikes
 
But yeah we need the union, intersection, and complement
Just that when I did it to 1.8-1.10 I was able to see it... But I'm dealing with a string so I'm like ughhhhhhh
 
So you just write things like x1,...,xk and things like take the union over this index set and bla bla bla.
 
2:20 PM
Which is painful
X1 U X2 U,..., U xK and then take complement?
 
If you understand the idea completely, then it's only tedious, not hard, now that we know how to write using index sets.
 
"tedious but straightforward" as one textbook would put it
 
Therefore left as an exercise for the reader.
 
😥
 
2:22 PM
If I didn't have to type on a keyboard I could just show you the proof on the blackboard now.
But typing in this chat is hard enough, lol.
 
This is the kind of problem I have a hard time sustaining interest in nowadays
 
Cool a blackboard. Pictures help me visualize things
 
But there is a caution.
Sometimes we think that something is obvious and we don't bother writing it out.
 
I know what it is.
 
And then when we actually write it out we realise we actually don't know the proof.
That means we thought we solved the problem when we actually did not!
 
2:24 PM
Point.
 
Or I can throw this problem in a furnace... Nah. It's starting to sink in a little
😓😥
 
So although many answers on Math SE are hints, I wonder if the answerer actually knows the proof or not.
 
Someone suggested induction. And I tried to do it in my head
 
induction is not always the most intuitive approach, but it does keep you honest.
 
Induction would be the more formal way to do it.
 
2:26 PM
Base case
K case
 
But I think in this case, using index sets and ... will suffice.
 
K+1 case
D:
The burnout is real
 
Honestly I can't stand using induction here.
It is highly non-intuitive and non-natural to think this way.
 
for you :-)
 
Sleepy
 
2:28 PM
sleep
 
i'm reminded of the old anecdote
"The mathematician is lecturing about a topic. He writes a preliminary lemma on the board and then says “This result is trivial to prove.” At that point, he stops, pausing pensively. After a few minutes, he says “Excuse me” and leaves the room. Ten minutes later, he returns and says “Yes, it is trivial, so let’s move on,” continuing the lecture from where he had left off."
6
 
If $M$ is a surface of genus $g$ with $k$ punctures (i.e. $k$ points removed), and $N \to M$ is a $d$-sheeted covering, can we say something about the genus and punctures of $N$?
 
K uses user1732 as a pillow
 
:D
 
(full disclosure: I blatantly c/p'd that statement of the anecdote from the internet)
 
2:30 PM
we know
 
XD
 
Let the k sets be Ax_1,...,Ax_k and the remaining m-k sets be Ay_1,...,Ay_(m-k). Now consider Ax_1 intersect...intersect Ax_k intersect (Ay_1 complement) intersect ... intersect (Ay_(m-k) complement) @usukidoll for example.
Call this set B_i. Now there are altogether (m choose k) ways to choose k sets out of m sets, and we now let C be the union of the B_i's where i goes from 1 to (m choose k). @usukidoll
Since F is closed under intersections and complements, each B_i is in F. Since F is closed under unions, C is in F. QED. @usukidoll
Just tidy up what I wrote above a little bit or two and you will be fine.
 
in other news, the complementary error function has a surprisingly-nice continued fraction representation:
$$\sqrt{\pi}e^{x^2}\text{erfc }x=\frac{2x}{2x^2+1-}\frac{1\cdot 2}{2x^2+5-}\frac{3\cdot 4}{2x^2+9-}\cdots$$
 
@usukidoll Are you still here?
Maybe fell asleep already.
Oh she left the chat room. Hope there are no flash floods there.
I am leaving this room too, bye bye.
 
2:44 PM
later
 
Maybe it wasn't that trivial if it took him ten minutes to figure out why.

https://chat.stackexchange.com/transcript/message/46461396#46461396
Okay. That doesn't work.
 
When doing Gaussian e
oops
when doing Gaussian elimination, let’s say on a real 3x3 matrix, what stops you from doing the following row operations and concluding the rows are linearly dependent. R1 +R2 and R2+R1
Then followed by R1 x R2 or vice verse to get a 0 row
@TedShifrin this seems up your alley
@Semiclassical also just because you’re my fave physicist
Is it because ultimately what I am doing is the original r1 minus itself,
just hiding it behind a veil?
 
R1+R2 isn't a row operation. Do you mean R2->R1+R2?
the point will be that, once you add row 1 to row 2, you've changed what row 2 is
i.e. you go from rows (R1,R2,R3) to (R1,R1+R2,R3)
and if you now add row 2 to row 1, you'll get (2R1+R2,R1+R2,R3)
 
3:01 PM
yes sorry you’re right
 
so the new first row is 2R1+R2, not R1+R2
 
what about adding the origins, row 2 to row 1
original*
 
You do successive row operations on the rows you have, not on the rows you started with
 
why can’t you do it?
 
Because that won't be an elementary row operation anymore.
 
3:02 PM
(I see that it leads to the problem I have but is there a more concrete reason)
 
because it isn't reversible, if you're looking for a moral reason
 
Do you know how elementary row operations are implemented in terms of matrices?
 
Yes I think so
 
other than that, I don't know what kind of reason you're looking for
 
Then what you should note is that, if you have a set of elementary matrices A1,A2,A3 acting on a matrix M
you would implement row ops in the order 1,2,3 as A3.A2.A1.M
and the point is that, when you do the third row op, you're really doing A3.(A2.A1.M)
which is to say, A3 is formulated as an elementary row operation on the rows of A2.A1.M
i.e. on the rows you have access to after having done the first two row ops
 
3:06 PM
@Semiclassical gonna change my previous answer to no
 
Sorry :’)
 
Suppose you want to implement the row operation R2-> R1+R2 on a matrix $M=\begin{pmatrix} R1 \\ R2 \\ R3\end{pmatrix}$
 
5 mins ago, by Semiclassical
You do successive row operations on the rows you have, not on the rows you started with
this is the reason, and Semi is just rephrasing this, which I don't see the need to
 
the way you'd do that is to multiply on the left by the matrix $A_1 = \begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$
 
3:08 PM
@LeakyNun trying to figure out why you must do it this way though.
 
because it's how it's defined
 
since then $$A_1 M = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}R1 \\ R2 \\ R3\end{pmatrix} = \begin{pmatrix} R1\\ R1+R2\\ R3\end{pmatrix}$$
siiigh
 
@JakeRose you do the row operations one by one
 
@Semiclassical ah I see what you mean now
Mhmm I get it
 
whew
yeah
once you write it like that, it's obvious that you lose access to the old rows once you do an elementary row op
 
3:11 PM
Yeah I get it
 
right
 
And if you do bring the old rows back in you’re essentially doing a row - itself which proves nothing
@Semiclassical thanks for the help @LeakyNun
 
right
it's not that the old rows are no longer valid---they still represent sensible equations--- it's that they'll be redundant
and once you have redundant equations, you definitely risk doing the equivalent of equation - equation = 0
 
 
1 hour later…
4:19 PM
Is a totally geodesic embedding of $S^1$ in a Riemannian manifold $M$ the same as what is called a "closed geodesic" in $M$ or rather what is called a "geodesic loop" in $M$ in the literature?
Or rather "simple closed geodesic"? I'm confused about these concepts.
 
4:39 PM
A geodesic loop need not have the same tangent at the start and end @abenthy
Consider a cone (with a sufficiently small angle)
 
@IceInkberry
21:57

I found magic. Magic to use these: ∫ x² dx = αβγδ⁹⁸ ± ∮x⁴ dx please elaborate
 
1 message moved from JEE/High School Chemistry Problems
Sorry not meant to be in this room. moved here by mistake.
Please ignore it.
 
5:18 PM
Hey @TedShifrin :)
 
hi Perturb
 
I have a quick question on the definition of a Riemannian metric. So Lee defines a Riemannian metric as the following: "Let $M$ be a smooth manifold with or without boundary. A Riemannian metric on $M$ is a smooth symmetric covariant $2$-tensor field, $g$ on $M$ that is positive definite at each point. " So I just wanted to spell out the details of that and make sure they're correct.
So here's my attempt at that: Let $M$ be a smooth manifold and let $g$ be a Riemannian metric on $M$. The bundle of covariant $2$-tensors on $M$ is given by $$T^2T^*M = \bigsqcup_{p \in M}T^{(0, 2)}\left(T_p^*(M)\right) = \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M) \right).$$
A smooth covariant $2$-tensor field on $M$ is a section of the smooth vector bundle $$(T^2T^*M, \pi, M)$$ so $g$ is a smooth map $$g : M \to \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M)\right)$$ such that $\pi \circ g = \operatorname{Id}_M$.
Now note that $g$ is also symmetric, by which we mean that for each $p \in M$, $g(p) \in \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M)\right)$ is a symmetric tensor and by positive definite we mean that $g(p)(X_p, X_p) > 0$ for all $X_p \in T_p(M)$ (again we make the usual identifications of tensors products with multilinear maps and elements identified with their counterparts in their biduals).
 
Using $T^2$ for tensors is terrible. Is that Lee's notation? $T$ is for tangent bundle.
Some people use a script $T$.
 
Yep that's Lee's notation
 
Ugh.
 
5:28 PM
 
I think that's horrible. He should have used a different font.
$T$ is the tangent functor.
But, yeah, it looks like you regurgitated fine.
 
@micsthepick I'll be trying to solve the problem you asked. When it ends, I'll ping you.
 
Thanks @Ted. Also sorry about this question, our intro diff geom course (if you can call it that) at my university is taught by a physicist and done without any sort of formality and we're doing stuff on metric tensors and Riemannian metrics so I wanted to get a rigorous understanding of them
 
Hey everyone!
 
Hey! @Daminark
 
5:32 PM
@Perturb: Lee's book should be fine. You can also look at Spivak's first volume (maybe eventually second and third).
yo Demonark
 
@Perturbative Since you mention Lee and Riemannian metric, I would like to inform you that Lee's Introduction to Riemannian Manifolds, second edition, will be published on 29 Sep 2018 by Springer.
 
@Ted Lee's book will be good for now, I'll take a look at Spivak's one too when I have some time
@JasperLoy Yeah I remember seeing you say something about that on the chat a day or two back
 
@Perturbative Yes, the date has changed many times on springer.com.
 
I've gotten annoyed a bit with some of the grad students posting geometry questions on MSE. They have things wrong and then argue with me when I give them a correct answer :P Who do they think they are?!
 
5:41 PM
@TedShifrin some questions are very bad lately
but perhaps I have forgotten what it was like to be young
 
LOL
 
If you have forgotten what it was like to be young, then what about Ted?
 
Right.
 
Ted has forgotten what it's like to be middle-aged.
 
There was a good one about Chern's proof of Gauss Bonnet. Too easy, though ... :P
Remember that I'm teaching 3 10th graders and 1 11th. :P
 
5:43 PM
Spivak's fifth volume proves the Chern-Gauss-Bonnet theorem.
 
So do I in my courses.
Hi Oskar.
 
Hi, Ted!
I'm nervous. My exam in real analysis is tomorrow morning.
 
No point worrying.
 
A little stress is actually good for performance.
 
No. It's just that there's a thing with the Weierstrass approximation theorem that I still have to figure out.
 
5:45 PM
You'd better not mess up uniform convergence or I'll smack you :)
 
@TedShifrin I'm teaching 100 something-graders across 3 classes.
Just not in person.
 
Mike, my calc was 1, so 4 is great. But I'd rather have 15.
 
I used to have 30 naughty kids in my classes.
It was more of managing their behaviour than proving the quadratic formula.
 
May I bother you with one last question on real analysis?
 
The only naughty students I had at university were the future elementary school teachers. They were horrible.
8
 
5:48 PM
"Show that if a continuous function $f:[0,1]\to\mathbb{R}$ satisfies
\begin{align}\label{1}
\int_0^1f(x)x^{1/(2n+1)}=0
\end{align}
for $n=1,2,\dots$, then $f(x)=0$ for all $x\in[0,1]$. Does the statement hold true if the interval $[0,1]$ is replaced by $[-1,1]$?"
 
Bother everyone, @Oskar.
 
@TedShifrin OMG, they must be as naughty as the elementary school students.
 
Worse, Jasper. Spoiled anti-intellectual brats. Trained by the education school faculty to believe that a teacher is a good teacher iff all the students "get" A's.
10
What's your proof, @Oskar?
 
lol, if not because of youtube recommendations, never heard of this shape before
 
I don't know what the corresponding $g(x)$ will be in this problem. I asked a question here on StackExchange and got grasp of the most of the general method.

https://math.stackexchange.com/questions/2896001/showing-that-a-bounded-continuous-function-is-zero-if-an-integral-with-it-is-zer/2896239?noredirect=1#comment5984063_2896239
 
5:50 PM
@TedShifrin blinded leading the blind
 
I'm not looking. We're talking. @Oskar
 
Not looking on the link?
Okay.
I'll copy paste my proof from the other problem that I solved.
 
I have a lot of cynicism with math education tho
 
So, by Weierstrass, you approximate $f$ by a polynomial. What do you learn?
 
at least at the K12 level
 
5:51 PM
Semiclassic: You may be more cynical than I.
Oh, a lot of college level sucks rocks.
 
This is what I did for the problem when $\int_1^\infty f(x)x^{-n}dx=0$.

"If $\int_1^\infty f(x)x^{-n}dx=0$ for all $x\in[1,\infty)$, then we define $g(x)$ such that $g(x)=x^{-8}f(x)$. Thus, $g(x)$ is bounded and continuous on $[0,\infty)$, and consequently integrable on $[0,\infty)$ as well. Now
\begin{align*}
\int_1^\infty g(x)x^{-n}dx=\int_1^\infty x^{-8}f(x)x^{-n}dx=\int_1^\infty x^{-n-8}f(x)dx=0
\end{align*}
for all $n\geq0$. The Weierstrass approximation theorem now gives us that as this holds for all $n\geq0$ we have that
 
@Oskar: I don't want to talk about that one.
 
I'm trying to describe what I don't get, sir.
As you wish.
 
You're thinking about the wrong thing.
 
Okay.
 
5:53 PM
So what happens if you take any polynomial for $f$ on $[0,1]$?
 
@TedShifrin yeah, but K12 is where the bad habits/mindset gets laid
 
Oh, I see the issue.
 
I'm not sure what you mean, professor.
 
So I tend to take more issue with K12 than college
 
Do you know a more general statement of Stone-Weierstrass?
 
5:55 PM
(though that probably also reflects my general disdain for high school vs. my general enthusiasm for college)
 
Yes. It's more or less that you can approximate any function with polynomials under certain circumstances.
 
Well, that's the usual Weierstrass theorem. Did you discuss something about approximating by other families of functions?
 
I don't recall.
 
Stone generalized to a collection of continuous functions that separates points.
If your class didn't do this, I don't think you can do this problem. However, you can answer the second question ($[-1,1]$) by inspection.
 

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