Mathematics - 2015-10-05 (page 1 of 3)

to put this abstractly: suppose G acts transitively on two sets X and Y, and f:X->Y is such that f(gx)=gf(x) for all g in G and x in X. then f must be onto. see if you can prove it.

Karim Mansour

oh

actually I proved something like that in set theory before

anon

3:45 AM

you had group actions in (elementary?) set theory?

Karim Mansour

yeah towards end of it

By the way @anon do you like category theory ?

anon

it's a nice condiment.

Karim Mansour

we had this question in topology last week

it was very nice question

I didn't know you can define product in abstract terms not just for sets

anon

lots of stuff is actually categorical

Karl Kronenfeld

Do you agree with this definition of product?

Karim Mansour

3:51 AM

yeah @KarlKronenfeld

I asked this question to my prof

in topology but he told me its hard to find examples of such things

I asked him the following

Is there category of objects that isn't just sets with extra structure on it ?

because it seems that category is just sets with extra structure

I mean is their things that are more general than sets?

anon

consider posets and groupoids

if you want something nonconcretizable, the homotopy category

Karim Mansour

I see

cp101020304

4:07 AM

Hi, I was wondering if someone could help me understand an answer for a particular question

0

Q: language of bitstrings with no more than 3 consecutive zeros generating function

I am trying to find the generating function of a sequence in the language of bitstrings, $X$, where each bitstring contains no more than 3 consecutive zeros. I have come up with the recurrence relation for a sequence $h_n$ of length $n$ as follows: (I am not 100% sure that it is correct, either...

How is the second term $h_1xh_2x^2$

I would assume it to be a1*x

h1*x sorry

Karim Mansour

4:55 AM

I am gonna buy artin group theory book

DF sometimes sucks

Huy

5:39 AM

sell me your DF

Karim Mansour

nah I want it aswell

Huy

but sometimes DF sucks

Karim Mansour

sometimes not all the times

haha

Huy

:(

Anthony

5:55 AM

heyo

gideon

7:09 AM

Glad I could share it with you. He's a guy with 3 MIT degrees who left his to make this site, eventually funded by google and Bill Gates foundation.

His math practice system is all AI based ;)

Understand

9:27 AM

Does anyone here like to discuss algebra sometimes?

I just want to discuss quotient rings

Balarka Sen

what's the question, @Understand?

Understand

Well I just wanted to discuss adjoining roots

So taking $\Bbb R[x] /\langle x^2 + 1\rangle$

Is that the same thing as taking $(\Bbb R[x])(i)$?

Balarka Sen

it's isomorphic to $\Bbb R(i)$

Understand

Oh sorry, yes

So if we quotient a field by an irreducible polynomial, it means the quotient ring is a field

Balarka Sen

that's not true. $\Bbb Z[x]/(x^2+1)$ is isom to $\Bbb Z[i]$ - not a field.

Understand

9:32 AM

$\Bbb R[x] /\langle x^2 + 1\rangle \cong \Bbb C$ and then $\Bbb C$ is an extension of $\Bbb R$, so then $\Bbb C$ is a real vector space of dimension two(which ignores multiplication)

Balarka Sen

what you mean is that if $k$ is a field, then $k[x]/(p(x))$, where $p(x)$ is an irred, is a field. in fact every field extension of $k$ arise in this way.

Understand

Sorry yes, I am not use to instant messaging

Balarka Sen

well, *finite algebraic field extension

grump, finite algebra galois field extension.

Understand

and if we have an degree $n$ irreducible polynomial, and we quotient a field by it as a generator, i.e. $K[x]/\langle f(x)\rangle$ we get an $n$-dimensional $K$-vectorspace as a result.

But I think I can't get that until I understand what happens with the multiplication on $\Bbb C$ with our 'two dimensional real vector space'

Chris's sis the artist

@robjohn Have you seen this one? My Mathematica is not able to calculate it. $$\int_0^1 \frac{\operatorname{arcsin}(x)}{x \sqrt{x+1}} \, dx$$

@robjohn It has a very cute closed form.

Understand

9:38 AM

does it converge chris's?

Balarka Sen

@Understand multiplication on $\Bbb R[x]/(x^2+1)$ is the same as multiplication on $\Bbb C$, so I am not sure what the question is. do you know what the isomorphism between the two is, explicitly?

Chris's sis the artist

@Understand Yeah, note that $\displaystyle \lim_{x\to 0} \frac{\arcsin(x)}{x}=1$. Near $0$ the integrand behaves like $\displaystyle \frac{1}{\sqrt{x+1}}$

Understand

@BalarkaSen I know of a construction that proves it must be the case, but I haven't got an explicit map

Balarka Sen

the map takes $p(x) \pmod{x^2+1}$ to $p(i)$.

Understand

I like it

Thank you Balarka

Balarka Sen

9:44 AM

no problem

robjohn

@Chris'ssistheartist Mathematica 8 gives an answer with lots of Hypergeometric functions.

I will have to go to my other computer to check Mathematica 10

Chris's sis the artist

@robjohn Well, yeah, my Mathematica too, but I didn't want to call that one an answer. :-)

robjohn

@Chris'ssistheartist I don't like people using Hypergeometrics as closed forms. Most things that can be written with a decent series can be written as Hypergeometrics

Chris's sis the artist

@robjohn Agree.

robjohn

10:05 AM

@Chris'ssistheartist I did a bit of manipulation, but Mma 10 gets $4G-\pi \operatorname{coth}^{-1}(\sqrt2)$

Chris's sis the artist

@robjohn Yeah :D

robjohn

I should go back to 8 and see if that manipulation helps there

@Chris'ssistheartist since it involved $\frac{\arcsin(x)}{x}$, I am not surprised with the appearance of $G$

Chris's sis the artist

@robjohn I came across this one while I was working on a problem by professor Bradley that I wanted to calculate it in a different way. I'll show you my work later when I put all in latex. I also wanna include it in my book.

@robjohn But that $\sqrt{x+1}$ makes things look bad.

robjohn

@Chris'ssistheartist I was looking at handling that with a series...

Chris's sis the artist

$$\sum _{n=0}^{\infty } \binom{-\frac{1}{2}}{n} x^{n-1}=\frac{1}{x\sqrt{x+1}}$$

robjohn

10:11 AM

@Chris'ssistheartist If I ask Mma 10 Integrate[x/Sin[x]/Sqrt[1 + Sin[x]], {x, 0, Pi/2}] it gives the answer I quoted above. Mma 8 gives something with some complex polygammas

Chris's sis the artist

@robjohn Mma 10 seems to come up with a lot of improvements in terms of integration.

robjohn

@Chris'ssistheartist well, Mma 10 gave the Hypergeometric garbage with the original statement of the question.

Chris's sis the artist

@robjohn Ah, OK

@robjohn Using that series (as you suggested) is enough to calculate the integral. There are some boring computations, and then you're done.

Understand

11:01 AM

@PantelisSopasakis Your hair is out of control

Pantelis Sopasakis

@Understand Haha! Cheers :)

@Understand that photo of mine is a bit old... now I got short hair

Understand

@PantelisSopasakis Hahaha my hair isn't that long, precisely so I don't have to worry about it

Oh I once had crazy hair too, but people worried I looked 'unprofessional'

Pantelis Sopasakis

but still it's dishevelled

no, quite the contrary... in the mathematics community a fuzzy hair style is considered perfectly professional

Understand

Tell me please @PantelisSopasakis, how do you approach problems, if you were to write it like an algorithm

@PantelisSopasakis Oh well in that case, I am excited :)

Pantelis Sopasakis

if you look like an Einstein, people tend to think you will be as clever as one

what do you mean by that?

Understand

11:05 AM

I mean the naive problem solving method is:

1) Read definitions
2) Attack problem

if solved, end, else go to 1)

Have you thought about precisely what your study method is? I only asked because I looked in your profile and you have masters + PHD :P

Pantelis Sopasakis

Do I?

Ah, yes...

Well, I have a focus or two (currently for example I'm mainly working on model predictive control and optimization methods based on proximal operators), so I read articles and books to, first of all, state an interesting problem

and then I'm looking for a solution... But I don't have a very algorithmic approach. I study, discuss, post questions on stackexchange,

try things on MATLAB to get a better understanding...

I would say

Understand

Discuss is what is missing for me I think, I do think that's really important. When did you gain a network to discuss with?

Pantelis Sopasakis

that when trying to develop new things -- is it theory or algorithms -- intuition is more important that rigor

Well, that's what stackexchange is for...

but I also have my colleagues at work

Understand

But stackexchange doesn't let me 'discuss' unfortunately. With your colleagues, do you ever worry what they will think of you? Do you verify things before you will say them? I have trouble with the other students because I worry they will think I am dumb

Thank you for your advice @PantelisSopasakis !

Dominic Michaelis

Hello everyone

skill patrol

11:17 AM

Hi pal

Dominic Michaelis

someone said that $L^\infty$ is connected, but normally $L^\infty$ is kind of weird, what do I need for a measure and a space such that it is connected?

Pantelis Sopasakis

Posing a question, admittedly, requires some effort from our side as well. And it has happened to me more than once to write down a question carefully and clearly, look at it and find the solution...

Agawa001

11:32 AM

@gideon keep working on your maths, all my biggest scores in bitwise/arithmetics/datastructs challenges are reached because of simplified mathematical equations, especially combinatorics

i like simplifying boolean expressions too

Mats Granvik

What is this number sequence?
1, 1, 2, 4, 10, 24, 66, 178, 508, 1464, 4320, 12886, 38992, 119030, 366740, 1138036, 3554962, 11167292, 35259290, 111825840, 356100044,...
It must have something to do with the Riemann zeta function. Possibly some kind of reverse function.

Agawa001

@MatsGranvik oeis.org/A239605

Mats Granvik

@Agawa001 Yes I know. Thanks for the link though.

Understand

If I work over an algebraically closed field $K$, and $f(x)\in K[x]$, then $f(x)$ always factorises into a product of linear polynomials, being the roots, is that correct?

I.e. $f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)$?

Understand

12:27 PM

We say the polynomial splits and it is up to a product of linear terms and a constant. The linear polynomials are $(x-a_i)$ where $a_i$ is the root, and it is important to careful distinguish this(obvious after thinking about it for a sec) fact

$f(x)=b(x-a_1)(x-a_2)\cdots (x-a_n)$

Kartik Sharma

12:44 PM

Hey everybody! I posted a problem about half a month ago and it is still unanswered. here Can anybody see into it and help me?

GBeau

1:06 PM

I am trying to show $\{\frac{n^2-1}{n^2}\}$ is cauchy using definition...is this supposed to be composing in an easy way!

I feel I am doing wrong

GBeau

1:20 PM

I was indeed being silly - it is just inverse squares when written under cauchy...

Balarka Sen

1:44 PM

@Understand Yes, because $K$ contains all the algebraics over itself, by definition of being "algebraically closed".

@GBeau Yes, the sequence is just $\{1-1/n^2\}$, so it is clear that you can get arbitrarily close pair of points in your sequence at some point.

GBeau

well, right, the question was asking to show it formally using the definition (i.e., explicitly construct the cauchy criterion)

it was easy once I realized it was 1/n^2-1/k^2

yeah

:)

Balarka Sen

correct. it's not hard to derive that it's Cauchy once you visualize what the sequence is, though.

I really shouldn't have said "clear" up there, though. Just meant you can see what is going on.

Moses

Hi @DanielFischer

Daniel Fischer

@Moses Hi.

Balarka Sen

woo. I just got a mail with an excellent article on reciprocity laws and galois representations

GBeau

1:53 PM

$\sum _{n\in\mathbb{N}} x_n$ converges $\implies $ $\sum _{n\in\mathbb{N}} (x_{2n}+x_{2n+1})$ converges (this is easy since those are subseries.) My book says the converse fails - what kind of examples should I be looking at?

Balarka Sen

$x_n = 1$ when $n$ even and $-1$ when $n$ odd?

Daniel Fischer

@GBeau Something where $x_n$ doesn't converge to $0$.

(Exercise: If $\sum (x_{2n} + x_{2n+1})$ converges and $x_n \to 0$, then $\sum x_n$ converges.)

GBeau

I will work to prove this ^

Moses

@DanielFischer What do you think of my proof. Remember you said that the spectrum must be restricted to a subalgebra of $\mathcal{A}$ when writing $\sigma(a_{k}) = \sigma_{k}$ in the Riesz Decomposition Theorem. But I'm proving that $\sigma_{\mathcal{A}}(a_{k}) = \sigma_{k}$ holds. Let me know if you can see a problem with the proof.

GBeau

thanks

Daniel Fischer

1:59 PM

@Moses Small problem at the start, you need to require $\sigma_1$ and $\sigma_2$ to be closed, otherwise disjoint $\Omega_k \supset \sigma_k$ don't exist.

Moses

@DanielFischer Oh yes I see. So we can use that $\mathbb{C}$ is normal and hence there exists two disjoint open neighbourhoods $\Omega_{1}$ and $\Omega_{2}$.

Daniel Fischer

@Moses But the real error is when you say $(g_k\cdot f)(\sigma(a)) = \sigma_k$. For $j \neq k$ and $z \in \sigma_j$, you have $(g_k\cdot f)(z) = g_k(z)\cdot f(z) = 0\cdot z = 0$, so $(g_k\cdot f)(\sigma(a)) = \sigma_k \cup \{0\}$.

Moses

2:18 PM

@DanielFischer Damn. Yeah I see, so close...

@DanielFischer How could we alter this proof to coincide with the discussion we previously had. Can we restrict the spectrum to some subalgebra of $\mathcal{A}$ so that my proof still gives a result?

Daniel Fischer

@Moses Well, let $\mathscr{A}_k = \overline{\operatorname{Alg} (E_k,a_k)}$. Then $\sigma_{\mathcal{A}_k}(a_k) = \sigma_k$. I don't think you can use the spectral mapping theorem to see the latter, though.

Moses

@DanielFischer Okay so I would have to change the proof almost entirely?

Daniel Fischer

I think so.

Moses

2:33 PM

@DanielFischer Could I just ask what the exact reasons are for choosing $\mathscr{A}_k = \overline{\operatorname{Alg} (E_k,a_k)}$ and if other subalgebras would also work?

Daniel Fischer

@Moses We need a Banach algebra in which $E_k$ is the identity, and that contains $a_k$. That is the smallest such. Above that, it has the advantage of being commutative. But any closed linear subspace $\mathcal{B}$ of $\mathcal{A}$ with $E_k,a_k \in \mathcal{B}$ such that the restriction of the multiplication to $\mathcal{B}\times \mathcal{B}$ makes $\mathcal{B}$ an algebra with identity $E_k$ would work.

Moses

@DanielFischer Okay thanks. Just to confirm, the $\mathscr{A}_k = \overline{\operatorname{Alg} (E_k,a_k)}$ is the norm closure of the algebra formed from taking the closure of $E_{k}$ with $a$ in terms of multiplication, addition and scalar multiplication?

Daniel Fischer

@Moses It should be $a_k$ rather than $a$ there. Apart from that, yes.

Chris's sis the artist

@robjohn do you wanna see a marvellous jewel? :-)

Moses

@DanielFischer Yeah typo. Thanks.

@DanielFischer Will try to redo proof.

GBeau

2:47 PM

gah my textbook always throws in putnam problems with the easy problems

Chris's sis the artist

the jewel ^^^

Agawa001

i saw this somewhere

Chris's sis the artist

@Agawa001 I just created it. :-)

Agawa001

nearby different

Chris's sis the artist

@Agawa001 hehe, that one is a different stuff :D

I presently have enough stuff to write tons of books. However I need to do my best for my first book, and publish it in an amazing form.

Hjorthenify

2:57 PM

Hey guys

Agawa001

yes , good start is a good sign of success

GBeau

if $0<C<1$, why should $\{x_n \}$ converge when $\lvert x_{n+1}-x_n\rvert\leq C \lvert x_n-x_{n-1}\rvert$? It's in the Cauchy section of my book and I don't see an obvious way to attack the problem with the other methods

Hjorthenify

I've been trying to figure out how (1)/(2e^(6i)) can become equal to one of these:

GBeau

(of the section)

Hjorthenify

Had the option 0.5/(e^(6i)) been there it would've been trivial, but no

Can someone help me or point me in the right direction?

Agawa001

3:05 PM

@Chris'ssistheartist if you are you concerned of anything i do now, i m trying to generalize a rule of congruence

and i see that it s not less beautiful than integrals

Chris's sis the artist

@Agawa001 :D

Daniel Fischer

@GBeau $x_n = x_0 + (x_1 - x_0) + \dotsc + (x_n - x_{n-1})$.

GBeau

@DanielFischer This is good suggestion, I will look this now :)

UserX

How do I prove that if a bouned set A has a maximum M then M=sup(A)?

Daniel Fischer

@Hjorthenify Unless you misread the value, the poser(s) of the exercise made a mistake. None of the listed values equals $\dfrac{1}{2e^{6i}}$.

Hjorthenify

3:09 PM

@Daniel Just realized I made a mistake. It's (1)/(4e^6i)

Daniel Fischer

@UserX Check the definitions. Is $M$ an upper bound? Can anything smaller be an upper bound?

@Hjorthenify I expect that you then see the right value?

Hjorthenify

No :(

Is it option nr. 2?

Daniel Fischer

$\dfrac{1}{4e^{6i}}$, what looks like it may be it?

UserX

No, so it's the LUP thus the sup. Thanks.

Hjorthenify

@DanielFischer I dont have a clue, sorry.. This really bothers me

@DanielFischer (1/4)*(4e^(-6i)) That?

GBeau

3:16 PM

@DanielFischer Oh I think the geometric sum formula for $C$ gets used, I think $|x_n-x_k|$ ends up something like ~<= $(1-C)/(1-C^{n+1})$ I just need to work out specific now

or uh flip that

sorry I was just doing it in my head

Daniel Fischer

@Hjorthenify That one isn't among the options. (Yes, it's option 2.)

Hjorthenify

Then I dont know.

robjohn

@Chris'ssistheartist sure. Sorry, I am afk frequently today.

Chris's sis the artist

@robjohn $$\int_0^1 \int_0^1 \int_0^1\frac{dx \ dy \ dz}{(1+x) (1+y) (1+z) (1+x+y+z+x y+x z+yz+9 x y z)}$$

@robjohn Can your Mathematica calculate it?

ADG

hello everyone

i need help

Daniel Fischer

3:21 PM

@GBeau something like. Yes, we use the geometric sum formula, you get $\lvert x_n - x_k\rvert \leqslant \frac{C^{n}}{1-C}\cdot \lvert x_1 - x_0\rvert$ for $k \geqslant n$.

robjohn

@Chris'ssistheartist I'll have to go to that computer and check it out.

Daniel Fischer

@Hjorthenify Option 2, $\frac{1}{4} e^{-6i}$.

Chris's sis the artist

@robjohn OK, when you have time. No hurry.

Hjorthenify

@DanielFischer Oh right. That was the one I meant with chat.stackexchange.com/transcript/message/24520901#24520901

Forgot to leave out the 4 :P

Ted Shifrin

hi @DanielF, @robjohn

ADG

3:22 PM

if $\psi_0(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{ikx}{\rm d}k$

Daniel Fischer

Hi @Ted.

Hjorthenify

What is the mathematical rule behind it though?

I mean I could see that was it, but I completely understand why

robjohn

@TedShifrin Hey, Ted! Did you get any rain this weekend? We were supposed to get a good deal, but all we got was some drizzle on Sunday morning.

Daniel Fischer

@Hjorthenify $\frac{1}{a^b} = a^{-b}$

ADG

what is the relation between $\phi(k)$ and $\psi_0(x)$ ??

Ted Shifrin

3:23 PM

It's raining right now, @robjohn, just in time for me to take my car to the body shop and walk 2 miles home :P

Daniel Fischer

@ADG Fourier transform

Ted Shifrin

Ah, @DanielF typed faster than I did.

Daniel Fischer

@TedShifrin Just 'cause you wasted time chatting with robjohn.

ADG

what is it? and how people concluded that this implies $\phi(k)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi_0(x)e^{-ikx}{\rm d}x$??

Ted Shifrin

One of us resembles that remark.

It's the Fourier inversion formula, @ADG.

robjohn

3:25 PM

@DanielFischer Yeah, and it kept me from answering, too :-p

@ADG That is the inverse Fourier transform.

ADG

i don't know all this but need to know just this for studying quantum physics, what could be a good source

Daniel Fischer

@robjohn Well, doesn't matter so much who answers, as long as it gets a correct answer.

robjohn

@ADG Any book on Fourier Analysis will go over this stuff

Ted Shifrin

You probably should look at a math for physicists type book, @ADG, like Arfken or something.

If your math is stronger, Körner's book on Fourier Analysis is beautiful and full of fascinating applications.

OK, I'm done now :D

ADG

does my reputation indicate that?

Ted Shifrin

3:27 PM

Indicate what?

ADG

im asking wheteher my reputation indicate my math strength?

Ted Shifrin

Reputation doesn't indicate much of anything.

ADG

hmm

i think that FT is a process of converting a number into sum of periodic sine and cosine functions?

Ted Shifrin

You need to decide if you know analysis (the theory behind calculus) or not. But my original suggestion is what most undergraduate physics majors learn their math from.

That's Fourier series, not transform.

ADG

*number is incorrect but it should be function

Ted Shifrin

3:29 PM

periodic function, yes.

ADG

im studying real analysis too now

and linear algebra

Sampark Sharma

Hello guys. Interrupting for a second in this conversation. i know little about Maths but had a couple of questions. All I ask is for one of you to point me to the right direction. Any fine fellows willing to help?

Ted Shifrin

oh, you definitely should know linear algebra before you do Fourier analysis :P

Try both and see what you can read and understand.

You should ask your question, @Sampark.

Sampark Sharma

This here is a paper of Philosophy:- biolbull.org/content/215/3/216.full. Scrolling a little down will bring you to a series of diagrams and equations. I've only studied Maths till 10th grade and thus my knowledge is limited. It does seem an awful lot like French to me.

What I wanted to know was what these equations are.

Don't explain them to me.

Just give me a brief overview and point me somewhere if you can.

Ted Shifrin

I don't know any information theory (which is what this probably is). You need basics on logs, exponentials, and probability, it looks like. And you need to take the time to understand their notation.

Maybe @robjohn can shed more light ...

Sampark Sharma

3:34 PM

It's some heavy stuff, eh?

And I suppose the diagrams mean little to you too?

Ted Shifrin

Right.

But there are lots of people (not necessarily here) who understand this stuff.

Sampark Sharma

And would be able to explain it to me?

Ted Shifrin

I have no idea.

Sampark Sharma

Perfectly all right! Thanks mate. :)

Ted Shifrin

Good luck~

robjohn

3:46 PM

Terrible network trouble to mathjax.org causing terrible lag to SE

Chris's sis the artist

Research is so great these days, pure heaven (excepting my allergy sessions).

Agawa001

@Chris'ssistheartist its autumn

everything is great in autumn

Chris's sis the artist

@Agawa001 Yeah, I know. :-)

Hjorthenify

@DanielFischer Of cause! Why didn't i think of that..

Agawa001

@Chris'ssistheartist except the fig tree in the backyard which gets progressivley unleaved, i see the autumn in a cheerful sens, not a sad portrait like everyone.

Chris's sis the artist

3:58 PM

@Agawa001 When I'm depressed the season doesn't matter much. Autumn is fine. :-)

UserX

What's up with tex rendering lately?

And hey @Chris'ssistheartist long time no see

Agawa001

@Chris'ssistheartist there s a meaninful sens of relation between automn and moral situation, automn is the outset of many things, work/studies/spousal life etc

wait, people dont usually marry in summer in your country or do they ?

Chris's sis the artist

@Agawa001 Yeap

@UserX Hey :-)

Agawa001

so it is a common sens

Chris's sis the artist

@Agawa001 It depends, I didn't check these details, I've never been really interested in it. :-)

UserX

4:08 PM

Depression goes way deeper than seasons unfortunately. @Chris'ssistheartist how's your research? Still dealing with integrals or did you switch to something else?

Chris's sis the artist

@UserX Yeap. I'm trying to finish my first book and publish it: a collection of integrals, series and limits (around 500 problems). :-)

@UserX You?

UserX

Just got back from my first day in a math uni :)

Chris's sis the artist

@UserX Oh, first day is always nice ... :-)

UserX

Yup we did calc 1 and got into sup/inf pretty fast. I like the pace.

Chris's sis the artist

nice

Addem

4:18 PM

In my first Calc class we didn't ever mention sup/inf--is this an honors class or something like that? Or is that just typical at your school?

UserX

You have to study R on the first day right? You want to do it rigorously so you start stating axioms. How do you rigorously state the axiom of completeness? @Addem

Addem

We didn't study R on the first day, per se--we tend to do that in an Analysis class, where of course we discuss completeness, sup, inf, etc. But typically in Calculus things are pitched at a higher level of study, assuming these intuitive features of R.

UserX

Assuming for simplicity(by that I mean it has to be easy to be in calc 1)you construct reals by including it as an axiom. If you want it to be a theorem then you're in the wrong classroom.

Agawa001

oh man, these numbers go thru cycles, its like a circus of numbers

UserX

To be honest I couldn't really tell the difference between calc and analysis. This might sound weird but they do overlap a lot. The textbook I use for the calc 1 class is named "Differential and integral calculus; an introduction to analysis" and it was in the bibliography of the books that cover what I'll be taught in calc 1

robjohn

4:31 PM

going to this link for me takes me to my post, then after the MathJax refresh, I am looking at someone else's post.

Does anyone else see this?

Addem

Well, my understanding is that Calc is where you develop your intuitions, and then in Analysis you go back over the same material but prove everything.

UserX

Nope, just your post. @robjohn

robjohn

@UserX Hmm.. I've tried using both HTML-CSS and Common HTML and I keep ending up in the middle of another post.

UserX

@robjohn what post?

robjohn

The middle of ADG's post

Agawa001

4:34 PM

@robjohn yes,like comments of the upper post ?