Mathematics - 2013-10-15 (page 1 of 2)

Bitrex

00:02

@PedroTamaroff I have one of those

Ted Shifrin

That has happened to me, and the converse, too.

Bitrex

0

A: Error accumulation in an approximating numerical algorithm for $y_n =\int_{0}^{1} \frac{x^n}{x+10} dx $

When $y_{n_1} = 0$, $y_{n_1-1} = \dfrac{1}{10n_1} + \alpha_1$, where $\alpha_n$ is the value of the roundoff error at each step (not necessarily the same for each step.) Plugging this back into the formula: $y_{n_1-2} = \dfrac{1}{10} \left(\dfrac{1}{n_1} - y_{n_1 - 1}\right)$ we get: $y_{n_1...

Thanks for the vote :)

Pedro Tamaroff

@KevinDriscoll You're the physicist here.

Do you buy this?

Bitrex

Wow

Alec Teal

@PedroTamaroff don't tease the physicist.

@KevinDriscoll ignore him, Physics is totally worth it really :P

Pedro Tamaroff

00:12

@AlecTeal Tease? How so?

Pedro Tamaroff

00:24

@anon Dude. Not interested in the automorphic forms and reps?

anon

a zero divisor is automatically nonzero

the other day I was talking to you and had to write "nonzero non-zero-divisor"

autos?

I am not at the right level to look into them.

Pedro Tamaroff

@anon Ah, OK.

@anon Maybe you're talking about something else.

Jack M

does anyone here know anything about differential field theory?

anon

a zero divisor is by definition nonzero

it's not a consequence of the definition, it's built-into the definition

Pedro Tamaroff

@anon My fault.

anon

00:28

@JackM like, derivations and stuff? not really.

Pedro Tamaroff

@anon I honestly thought $0$ was also considered a zero divisor. =)

anon

indeed

Pedro Tamaroff

@anon ?

anon

meaning it was clear you were operating under that impression

Fernando Martin

@PedroTamaroff: What's an automorphic form?

Pedro Tamaroff

00:29

@FernandoMartin Dunno.

anon

jgi

it's a generalization of a modular form

Fernando Martin

@anon: The stuff at wikipedia is 2deep4me

Pedro Tamaroff

@anon Ah.

@anon We know that $x^{p-1}=1\mod p$ has exactly $p-1$ solutions. This means that $x^{(p-1)/2}=1\mod p$ must have exactly $(p-1)/2$ solutions, yes?

anon

yes

Fernando Martin

Hey @PedroTamaroff, can I ask you something about characters?

Pedro Tamaroff

00:37

@FernandoMartin You can try.

@anon I should doubt less.

Fernando Martin

Suppose $\chi$ is a character over $\mathbb{F}_p$

Bitrex

@PedroTamaroff I think that golfball vid is real

Pedro Tamaroff

@FernandoMartin Kay.

Fernando Martin

Is it true that $\sum_t \chi(1-t^m)=\sum_t \chi(1-t^{am})$ for any $a$?

Pedro Tamaroff

@FernandoMartin $m,a>0$?

Fernando Martin

00:40

I don't see why it should be, but it should be (otherwise some problem out of I&R is false)

yes

well, maybe for any $a$ is a bit reckless

Pedro Tamaroff

@FernandoMartin What are you looking at?

Fernando Martin

Let me try to explain

anon

clearly true if (a,p-1)=1

Fernando Martin

@anon: since ^a is a group iso then, right?

anon

right

Fernando Martin

00:42

great

well, I'll try to prove that (a,p-1)=1 in this problem I'm working on

anon

for that equality to hold generically, $\sum_{c\in C}\chi(1-c)$ would have to be invariant with respect subgroups $C\le\Bbb F_p^\times$

Pedro Tamaroff

@FernandoMartin Page?

Fernando Martin

Exercise 8 from page 105

I'm trying to reduce the problem to the case $m|(p-1)$, which is easy

Pedro Tamaroff

@FernandoMartin Heh, I'm on page 52, long road.

Fernando Martin

I haven't read the first 100 pages entirely though

Pedro Tamaroff

00:52

@FernandoMartin Got a mail, Number Theory Seminar.

@anon "Hecke/Sturm bounds for classic modular forms."

Fernando Martin

01:05

Let's say that I write $(m, p-1) = am + b(p-1)$ for some $a$ and $b$

can I choose $a$ such that $(a, p-1)=1$?

I can almost sense that I just asked something stupid

Pedro Tamaroff

@FernandoMartin Write $a=k(p-1)+r$, $r<p-1$?

Anthony Carapetis

bloody hell

turn up to give talk at geometry reading seminar

and then robert bryant walks in

yay pressure

Pedro Tamaroff

@AnthonyCarapetis Dunno who that is. Googles.

Fernando Martin

@PedroTamaroff: so?

Pedro Tamaroff

@FernandoMartin If $r\neq 0$, you may take that $r$.

Fernando Martin

01:14

ugh right

that settles it

thanks!

Pedro Tamaroff

Clicks tongue.

@FernandoMartin Well, duh.

Fernando Martin

What?

Pedro Tamaroff

If $r\neq 0$; keep $a$!

I mean I was being dumb.

Fernando Martin

if $r\neq 0$, then $p-1$ does not divide $a$ but it may still be the case that $(a,p-1)\neq 1$

Pedro Tamaroff

Sorry, yes.

Me being dumb again.

Fernando Martin

01:16

Oh

Well, then that's the case for $r$ too

haha, so it's not settled

dang.

Pedro Tamaroff

I was mixing up my primes.

I read $p-1$ as $p$ in my head.

Fernando Martin

@PedroTamaroff: I'm trying to prove that $\sum_t \chi(1-t^m) = \sum_t \chi(1-t^d)$ where $d=(m,p-1)$

I used that $d=am+b(p-1)$, so $t^d=t^{am}$

now it would suffice to show that $(a,p-1)=1$, since in that case ^a would be a group iso and we would be done

Fernando Martin

01:32

@PedroTamaroff: Do you know these guys?

The song at the second part of the video is called Poincaré, haha

Pedro Tamaroff

@FernandoMartin I'm watching $\text{HOM}\exists \text{LAND}$. =)

Fernando Martin

Oh ok

Twink

hi argentinos. excuse me but how do you know how many variables libres are there in a linear equations system?

Fernando Martin

I don't get your question

Twink

for example here

how do we know that we can choose two variables arbitrarily?

Fernando Martin

01:44

In general, a subspace of $\mathbb{R}^n$ defined by k non-redundant linear equations is of dimension $n-k$

Pedro Tamaroff

@Twink $n-\rm rank$.

Twink

Pedro but I still don't know the rank in that example

after I have the basis I will know the rank

I don't even have it in a form of a matrix

Pedro Tamaroff

@Twink The equation simply says that $y=2x+3z$.

You sub that and you're done.

Two free variables.

Twink

sub?

Pedro Tamaroff

There is no difference between that set and $\{(x,2x+3z,z):x,z,\in\Bbb R\}$

@Twink Substitute.

Twink

01:47

so if we had an equation with 4 variables

Pedro Tamaroff

Note that is $x(1,2,0)+z(0,3,1)$ by the way.

Twink

then we could choose 3 arbitrarily?

Pedro Tamaroff

So the span of $\{(1,2,0),(0,3,1)\}$.

@Twink Usually that is the case.

Twink

ok :) thanks

the rank is the number of LI columns right?

Pedro Tamaroff

@Twink Or rows. Yes.

Twink

01:51

thank you :D

thank you for sharing your knowledge

Karl Kronenfeld

@FernandoMartin I am too lazy to type up the answer to the question here, but there is something that interests me about it. Call the category defined in that question $\mathcal H(V,W)$ and note that it is a full subcategory of the functor category $\text{Vec}^\mathbf{2}$. I wonder if a functor $\mathcal H(V,W)\to\mathcal H(V',W')$ can be represented by an arrow in that larger category.

I thought this may perhaps interest you too.

Pedro Tamaroff

@FernandoMartin Did you sort your problem out?

Maybe try another approach?

masfenix

02:09

HEY GUYS

sorry caps.

I was wondering about a problem, does there exist an unbounded linear map $F: l^2(\mathbb{N}) \rightarrow \mathbb{C}$

Kevin Driscoll

@Pedro Sorry I was away. Yeah at 150 mph (240 kph) I buy it. Its very similar to high speed footage of water balloons at lowish speeds and golf balls are surprisingly elastic

Ted Shifrin

@masfenix: Why do you think there should be one?

Fernando Martin

@KarlKronenfeld: let's see

@PedroTamaroff: No, just took a break

masfenix

@TedShifrin I think there is one because $l^2(\mathbb{N})$ is an infinite-dimensional Hilbert space.

so I remember my professor saying that there can be unbounded maps

Ted Shifrin

I know of discontinuous linear maps on Banach spaces, but not on Hilbert spaces.

masfenix

02:20

So the answer is no?

well then I would have to prove that.

Ted Shifrin

Inner products are powerful things.

Hmm, I take it back.

masfenix

@TedShifrin en.wikipedia.org/wiki/Discontinuous_linear_map says that it may be hard to find a concrete example for a complete space (so Hilbert spaces?)

Ted Shifrin

I am just very rusty ... It's been about 35 years since I've thought about this :P

No, I untake it back. I can think of examples $H\to H$, but not for a linear functional on $H$.

Kevin Driscoll

@Ted @masfenix I dont think the inner product helps you because an unbounded map is precisely one for which $\left| F( \phi ) \right|$ is unbounded even though $\left| \phi \right|$ is finite

but what do I know

Ted Shifrin

Right @Kevin. Can you give a map to $\Bbb C$ ?

Maybe so. What if we take $H=L^2([0,1])$ and something like evaluation at $0$?

No, that's not defined.

Kevin Driscoll

02:27

@Ted No, I can't give an example. @Masfenix talked about this the other day and my instinct was that no such map exists, but I'm totally unsure

Ted Shifrin

I remember loving learning about densely defined discontinuous linear maps in the context of PDEs when I was in grad school. But I'm old and I never use this stuff at all :P

masfenix

Discontinuous linear map

In mathematics, linear maps form an important class of "simple" functions which preserve the algebraic structure of linear spaces and are often used as approximations to more general functions (see linear approximation). If the spaces involved are also topological spaces (that is, topological vector spaces), then it makes sense to ask whether all linear maps are continuous. It turns out that for maps defined on infinite-dimensional topological vector spaces (e.g., infinite-dimensional normed spaces), the answer is generally no: there exist discontinuous linear maps. If the domain of defin...

look at the section under General existence theorem

Anthony Carapetis

The "general existence theorem" section seems to be exactly what you want

masfenix

I think this may help us

yes @AnthonyCarapetis exactly.

Kevin Driscoll

@Ted I see. So it may be that the function is integrable, but its value at a point may still be infinite

Ted Shifrin

02:29

yeah, but values of functions don't make sense, because elements of $L^2$ are equivalence classes.

Hmm, @masfenix @Anthony, I was thinking about an example like that but convinced myself it wouldn't work in a Hilbert space.

masfenix

Now, I understand the example but I could use a little intuition on it.

Ted Shifrin

@Anthony is our resident fancy analyst ... I defer.

Anthony Carapetis

hey don't be calling me things like that

Ted Shifrin

@masfenix, $\ell^2$ has an obvious orthonormal basis :P

Anthony Carapetis

I think the key thing to notice is that it's an algebraic basis, not a Schauder basis

Kevin Driscoll

02:32

@Ted equivalence classes......... yes, of course............ quite right

Ted Shifrin

well, I'm thinking Hilbert space basis ... not algebraic

masfenix

@TedShifrin, right which is $(1, 0, 0 \ldots), (0, 1, 0 \ldots), (0, 0, 1, \ldots)$

Ted Shifrin

I've been spending all night thinking about Chern classes and degeneracy classes. I should go back to that ... :P

yes @masfenix

Anthony Carapetis

If you try to define a map like that using only a Hilbert space basis

Ted Shifrin

So, mapping the $n$th one to $n$ and extending by linearity is an example.

Anthony Carapetis

02:33

Then you're only defining it on a dense subset

Ted Shifrin

Well, unbounded operators are usually densely defined, anyhow :P

Anthony Carapetis

Usually we extend by continuity to get a bounded linear map, but obviously in this example we don't want to

So you need a full algebraic basis to "complement" it I guess

Ted Shifrin

UGH at algebraic basis.

I don't do that stinking uncountable stuff.

Anthony Carapetis

Hahaha

Karl Kronenfeld

@FernandoMartin I think we may lose the vector space structure present in the hom-set, but if there is a way to represent addition and scalar multiplication "categorically", or to otherwise show that functors are linear we'd be making good progress.

masfenix

02:36

So if I understand this correctly, we take the usual orthonormal set of $l^2$ and define a linear functional by $F(e_n) = n\cdot |e_n|$

so how do I show this map is discontinuous?

Kevin Driscoll

@Ted what happened to your geometrical answer for the 'building a bridge' question? Did you find an error?

masfenix

because the distance between two points is always one?

Fernando Martin

@KarlKronenfeld: sorry, I got distracted and just finished reading

Anthony Carapetis

@masfenix: that's only defined for the subspace spanned by the $e_n$

Fernando Martin

let's see

Ted Shifrin

02:37

Just show it's unbounded, @masfenix. Continuity is equivalent to boundedness.

masfenix

right, but I am not really sure how to show its unbounded. Isn't it easier to show discontinuity?

Ted Shifrin

Well, @Kevin, they convinced me it was wrong. And I had convinced myself it was right based on symmetry, and there was a 7th grade algebra flaw in that. So I think my "obvious" triangle inequality step was wrong.

masfenix

@AnthonyCarapetis, the orthonormal basis $e_n = \delta_n$ where $\delta_n$ is the Kronicker delta function spans the entire space, no?

Fernando Martin

@KarlKronenfeld: I don't know much about this, but wouldn't this have to do with enriched categories?

Ted Shifrin

You have elements of length $1$ whose images have arbitrarily large lengths.

masfenix

02:43

@TedShifrin, thank you. So I've formulated something like the following: Define $F(e_n) = n |e_n| $. Recall that a unbounded map is exactly one for which, $\forall x \in H$ where H is a Hilbert space, $|F(x)|$ is unbounded whereas $|x|$ is bounded, and so since the map defined above takes elements (that are bounded) to images with increasingly larger lengths, it follows that the image is not bounded and hence the map is unbounded.

Fernando Martin

@KarlKronenfeld: How would you represent, for instance, the trivial endofunctor?

Kevin Driscoll

@Ted ah okay. That's unfortunate because I didn't understand your answer, and maybe there was something there I need to understand

Ted Shifrin

I still think the path should consist of parallel beginning and end, @Kevin, but my proof was flawed (since dilating the inner ball distorts distance).

@masfenix: First, $|e_n|=1$, so it's just $F(e_n)=n$. But I guess you should address @Anthony's point of whether this extends to give a well-defined linear map on all of $\ell^2$.

Anthony Carapetis

@masfenix: Hilbert space orthonormal bases only span the whole space only if you allow infinite linear combinations. This normally works fine to extend the definition of a linear functional to the whole space (by continuity); but since this one is unbounded that won't work. For example $\sum_k e_k / k^2 \in \ell_2$ but $F$ would send it out of $\ell_2$.

Ted Shifrin

@Anthony: We're defining a linear functional! So what do you mean?

Anthony Carapetis

02:48

@Ted: what's the issue? We've only defined it on a dense subspace, and we clearly can't extend by continuity (since it's not even bounded on the subspace)

Ted Shifrin

Right. But my experience (years ago) with unbounded operators is that they're essentially always only densely defined (e.g., differentiation on $L^2$). I dunno.

Anthony Carapetis

Oh right, forgot about that

Yeah if that's the convention being used then you don't need to worry about extending it

masfenix

@AnthonyCarapetis could you explain "extending" a little bit more. I want to understand it a bit more

Anthony Carapetis

@masfenix: the example we have is defined only for those $x \in \ell_2$ with finitely many non-zero components, so it is defined only on a subspace of $\ell_2$. It is conventional that an "unbounded linear operator $X \to Y$" actually means a linear map $Z \to Y$ for some subspace $Z \subset Y$; so if you are using this convention then the example suffices.

If not (i.e. if you want the map to have domain $X$) then you need to define it for sequences with infinitely many non-zero components; i.e. "extend it to all of $\ell_2$".

I always found that terminology confusing... "unbounded operator" really means "partially-defined operator" rather than "operator that is not bounded"

Pedro Tamaroff

@KevinDriscoll Physics approved!.

masfenix

03:01

@AnthonyCarapetis wow, thanks.

Kevin Driscoll

03:13

@Pedro Indeed. There's some cool videos of water dropltets in space that NASA did that show the same sort of thing

jsmith95

04:52

Are linear maps the same as linear operators? Wikipedia says the two are the same, while Axler's "Linear Algebra Done Right" (which I've only just started reading) says that an operator is a linear map from a vector space back onto itself.

Anthony Carapetis

@jsmith95: different people use different terminology/conventions. In general I would not assume "operator" implied self-mapping unless specifically defined that way in the article/book.

jsmith95

Ok, thanks Anthony.

anon

"linear operator" has various slightly different conventions. I'd avoid it for general linear transformations because the most important usage is that of operators of functions (e.g. differentiation, fourier/laplace transforms, etc.).

jsmith95

Thanks Anon.

what'sup

05:35

who's downvoting my answers ?

J. W. Perry

06:17

@what'sup Not me, I have only downvoted one answer in my life, and it really needed it. Interesting though, I looked, and someone definitely slammed a few of your old answers. There is, as you may have read an automatic catcher for this. Alternatively you can get the attention of a mod.

I am sorry this happened to you, and I hope it works out. The fact that you had 3 answers downvoted in short order, one from October 6 tells me someone is being a harsh jerk.

Did you make someone mad? hehe

Tobias Kildetoft

@what'sup it certainly looks like targeted downvotes, as they are within one minute.

Ramana Venkata

Can Somebody give me hint How to derive this lemma imgur.com/CbUG8rE ??

J. W. Perry

@Tobias Indeed! Hopefully the algorithm will catch this and restore the deal for what'sup.

Tobias Kildetoft

@J.W.Perry unfortunately I am not sure 3 downvotes is enough for that to kick in (though I am not that knowledgeable about the algorithm)

J. W. Perry

@TobiasKildetoft Yeah I was thinking the same. I do not have personal experience (yet), but that would definitely hurt my feelings. maybe a Mod will catch this and look at it?

Kevin Driscoll

06:25

@what'sup you can always ask @robjohn to take a look at it

J. W. Perry

There ya go.

robjohn

@KevinDriscoll can't tell that.

If there is something fishy, there are automatic scripts to handle it.

Ramana Venkata

@robjohn imgur.com/CbUG8rE Can you give a hint to prove this lemma..

robjohn

@what'sup Since they were in rapid succession, it does look suspicious. However, there is nothing more that I can tell than you can tell by looking at your reputation page

J. W. Perry

@RamanaVenkata Why don't you post that as a question with everything you have tried in the usual manner. It is definitely a defendable question.

robjohn

06:34

@RamanaVenkata I'd look at the Mean Value Theorem...

what'sup

06:45

@robjohn @J.W.Perry @KevinDriscoll @TobiasKildetoft thank you for helping

J. W. Perry

@what'sup My (our) pleasure, and thanks for your contribution.

what'sup

:-)

goodbye now i'm in MSE but not in the chat . :-)

J. W. Perry

Cya around :-)

Ethan

08:43

blah

Flaw

09:45

I'm having trouble trying to think about volume of an ellipsoid

Can I imagine it as a sphere first, as if the axes of the graphs were scaled such that it looks like a sphere?

robjohn

09:58

@Flaw Consider the effect that scaling of the axes would have on $x^2+y^2+z^2=r^2$

Flaw

I can imagine the individual axes being scaled by a b and c

but the volume... r^3 term I'm not sure how

skullpatrol

What is your specific question?

Flaw

10:24

I'm trying to get some kind of spherical polar form of a volume element of a ellipsoid.

frx08

10:52

hello, I need support in a math problem about euclidean space and rototrasformations.. I'm really poor at it, does anyone know where I can get support? Thanks

Jack M

rototransformations?

Karl Kronenfeld

11:58

@FernandoMartin I don't know about enriched categories, but it seems to be enriching the category of all $\mathcal H(V,W)$ over $\text{Vec}^\mathbf 2$. I do not see why the $0$ arrow $(V\to V,W\to W):f\to f$, with $f$ anything would not represent the trivial endofunctor. However, I do see a problem with my general idea, which I will try to turn into a counterexample later. The fact that one has to choose representing objects as well as arrows seems to be too restrictive.

Michael Albanese

12:24

I'm new to chat so I apologise if this is inappropriate. Could someone take a look at my post here (math.stackexchange.com/questions/518855/how-to-find-liminf-f-n/…) and tell me if there is something wrong with it? I've had two downvotes but no indication of why.

Pedro Tamaroff

@MichaelAlbanese Weird.

Dunno.

Michael Albanese

Maybe people don't like answers which post such short hints. In this case it is really all you need though.

Pedro Tamaroff

@MichaelAlbanese Yep, the question is pretty straightforward. People should try harder =/

Michael Albanese

@PedroTamaroff Thanks for taking a look.

frx08

12:48

@JackM 3d rotation and translation.. in italian we have this word but maybe you don't

Mats Granvik

13:40

Because the Lambert W function can be defined from the rows in this triangle:
$$\displaystyle \begin{bmatrix} 0=0 \\ +1&-1=0 \\ +1&+1&-2=0 \\ +1&+1&+1&-3=0 \\ +1&+1&+1&+1&-4=0 \\ +1&+1&+1&+1&+1&-5=0 \\ +1&+1&+1&+1&+1&+1&-6=0 \end{bmatrix}$$

I would like to define a prime version of the Lambert W function from the rows in this triangle:
$$\displaystyle \begin{bmatrix} 0=0 \\ +1&-1&=0 \\ +1&+1&-2&=0 \\ +1&-1&+1&-1&=0 \\ +1&+1&+1&+1&-4&=0 \\ +1&-1&-2&-1&+1&+2&=0 \\ +1&+1&+1&+1&+1&+1&-6&=0 \end{bmatrix}$$

I know how to define it, but I can't extend the range of convergence.

If the latter prime version really converges, that is.

Mats Granvik

13:55

from the first triangle:
$$\text{x/LambertW(x)} = \int \left(\sum _{n=1}^{\infty } (-x)^n \exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}\right)+1\right) \, dx+1$$

and from the second triangle:
$$\text{x/PrimeLambertW(x)} = \int \left(\text{Expand}\left[\sum _{n=1}^{\infty } (-x)^n \exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{\text{If}[n=1,0,\text{Table}[\text{DivisorSum}[m,\text{$\#$1} \mu (\text{$\#$1})\&],\{m,\text{nn}\}][[\gcd (n,k)]]]}{k^{s-1}}\right)\right]+1\right) \, dx+1$$

Mats Granvik

14:28

The prime version is close to x/LambertW(x):
x = (1 - 11/8)/Exp[1]
2*Pi*Exp(1)*x/PrimeLambertW(x) = 14.5228821632539947115667526619...
2*Pi*Exp(1)*x/LambertW(x) =14.5213469530656281679750582094...

and:
x = (2 - 11/8)/Exp[1]
2*Pi*Exp(1)*x/PrimeLambertW(x) = 20.6316183974506150286988447032...
2*Pi*Exp(1)*x/LambertW(x) = 20.6557403556995572026029091483...

skullpatrol

But we need a guidance

device

N3buchadnezzar

yo

skullpatrol

yo

Mats Granvik

14:47

@skullpatrol At least thinking feels harder, so it must be better. No pain, no gain. Burn the fat, feed the muscle. Your brain is your muscle. Use it or loose it. But asking can give you the answer easily. Knowledge is power. But on the other hand, nothing worth knowing can be taught.

I don't know why I channeled a fitness guru, but it seemed fitting...

skullpatrol

True, you do need to practice... Use it or loose it.

but

Education is what remains after one has forgotten everything he learned in school

Mats Granvik

@skullpatrol How would you go about finding a formula for the zeta zeros? If there is one.

@skullpatrol Yes, one should not let your education interfere with your learning, someone said.

skullpatrol

@MatsGranvik I would go to the library and look it up.

Mats Granvik

@skullpatrol But that book has not been written yet.

skullpatrol

@MatsGranvik Someone must have done something about it.

Mats Granvik

15:00

Yes there is a an approximate formula that is very good on average. But it is far from exact for a individual zero.

http://mathoverflow.net/questions/82635/explicit-formula-for-riemann-zeros-counting-function
http://www.archive.org/stream/proceedingsofthe032881mbp#page/n116/mode/1up

Those links above are about how to count them. But not what they are.

the zeta zeros, that is.

@skullpatrol

skullpatrol

@MatsGranvik Thanks for sharing.

Mats Granvik

@skullpatrol Others say the path of least resistance is the best way to creativity.

skullpatrol

15:16

@MatsGranvik "creativity" is generally frowned upon on during the lower grades of math

anon

Chat Guidelines | $\LaTeX$ in chat

6

Josué Molina

@skullpatrol On the basis that "you're new, so your ideas must suck."

skullpatrol

@JosuéMolina yes and inexperienced...

Josué Molina

@skullpatrol I felt that in college creativity was seen as a sign of plagiarism.

I often hesitated solving problems in ways disparate from course content.

skullpatrol

@JosuéMolina Indeed, the Professor gets to make the rules.

Josué Molina

15:27

The rules were often tacit.

anon

sounds like a pity party to me

Josué Molina

I like to naïvely surmise that professors stop underestimating you upon reaching graduate school.

Though the opposite could be just as true, as xkcd often displays it.

skullpatrol

16:14

What's an experiment?
:D

Kevin Driscoll

16:33

Experiment? never heard of it.....

Mats Granvik

17:29

Probably way outside the radius of convergence but the logarithms of the power series are close to each other, according to the plot above.

This when comparing x/PrimeLambertW(x) and x/LambertW(x)

nsanger

17:51

In problem 9 of chapter 6 in Spivak's calculus (4th edition), it asks to prove that if some function $f$ is defined at $a$ but not continuous at $a$, then prove that for some number $\varepsilon > 0$ there are numbers $x$ arbitrarily close to $a$ with $|f(x) - f(a) > \varepsilon$.

I've only been able to prove that there are numbers x such that $|f(X) - f(a) \geq \varepsilon$, but not the strict inequality, and I don't think it's possible to prove the strict one (i.e. there's an error in the book). Can anyone confirm this?

robjohn

18:21

@nsanger This is just the inverse of the definition of continuity

nsanger

@robjohn, Yeah I realize that, but this just means that there is at least one $\varepsilon > 0$ such that no $\delta > 0$ exists satisfying $|x -a| < \delta \implies |f(x) - f(a)| < \varepsilon$. So for that $\varepsilon > 0$ this tells you that there there are $\delta > 0$ such that there exists numbers $x$ where $|x-a| < \delta \implies |f(x) - f(a)| \geq \varepsilon$. How do you make this inequality strict?

robjohn

$f$ is continuous at $a$ iff $\forall\epsilon\gt0,\exists\delta\gt0:|x-a|\le\delta\implies|f(x)-f(a)|\le\epsilon$

@nsanger start with $\epsilon/2$

nsanger

Do you mean $2\varepsilon$?

robjohn

or that

Bitrex

Triangle inequality?

nsanger

18:28

Alright, that works then, thanks.

$|f(x) - f(a)| \geq 2\varepsilon > \varepsilon$, thanks.

Bitrex

18:38

choose a number $|A| < \epsilon$, choose x so that $B = |A + f(x) - f(a)| <2\epsilon$, then $|A + B| - |A| < \epsilon < |f(x) - f(a)|$ ?

Bitrex

18:49

No, I don't think that's right.

Delete, kthnx :)

Enjoys Math

19:33

Can anyone help with this convex analysis question? Should be easy: texpaste.com/n/v73fs0gw

Here it is actually: texpaste.com/n/he0sp4p8

Vaughan Hilts

Hey folks, a basic math question here: http://i.imgur.com/0grknFb.png

Is it sufficent for A to say: A = { x: R, x < 8}

and for B to say A = {1, 2}

Charlie

@VaughanHilts isn't A ={x in IR : x<3}?

Vaughan Hilts

Well, it's a union.

So shouldn't all the numbers from below 8 be there, too?

-2 to 8; and since that falls within below 3, it ends up just being < 8

That was my reasoning, anyway.

Charlie

@VaughanHilts you used letter A to everything :P anyway, yes (a) A U B= {xin IR:x<8}

Vaughan Hilts

Haha, yeah. that was a mistake :)

Charlie

19:47

By def, union is all the elements that are in A or in B

Vaughan Hilts

Yep, I'm enrolled in a proofs class - and I recall that. I just wanted to make sure I'm not doing any slip ups. :) I'm doing a mock test with no answers provided.

So having it verified is a bit tricky ;)

Charlie

@VaughanHilts :)

Vaughan Hilts

Is something is a disjoint non-empty subset, then how can they also be subsets?

The rule of a subset is you must contain at least one element of the other, yet disjoin t means you mustn't contain any.

Is this just a vacuous proof?

Charlie

@VaughanHilts disjoint means A intersection B = Ø

Vaughan Hilts

Oh, so they share no components.

Or elements rather.

Charlie

19:56

@VaughanHilts suppose you have a superset, remove two subsets, they are sybsets, but their intersection is empty

Vaughan Hilts

But even then, don't you require at least one element in common to have a subset?

A intersection B = empty set clearly indicates that they share no elements, no?

Charlie

@VaughanHilts nit necessarily A={1,2,3,4,5,6} B={1,2} C={4,5,6}

Vaughan Hilts

What is C?

Charlie

@VaughanHilts B,C subset A

Vaughan Hilts

I'm not getting it quite... so I'm going to post a question. :)

Kevin Driscoll

19:58

@VaughanHilts A is a subset of B if all of the elements in A are also in B

Vaughan Hilts

Yes, that is correct. However, A and B are disjoint sets of each other, which means they musn't share any elements.

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