Haven't you ever heard the expression "it's so quiet you can hear a pin drop"?

oh

yeah i have

i'm reading suugaku. i'm trying to avoid doing homework. my teacher didn't seem to appreciate my direct proof when he asked for induction.

professors get to be that way, unfortunately

his words were "since the problem asks for induction, you should probably actually use it." aka "if you want points, answer the question"

on the plus side, you won't be in that class forevr

@DavidWheeler that's true. on the negative side it now doesn't affect my major at all so there's no point to it

what class is it?

math 451. "Problem Solving in Number Systems and Discrete Mathematics"

well, you might still learn some useful tricks. could be useful for playing psychic at cocktail parties.

hahahaha :) true. it is reminding me why i love math in the first place, but his grading is killer. i screwed up and switched to k one step later than he wanted and he took off 75% of my score even though everything else was correct

he sounds overly pedantic, but I lack enough context to truly judge.

yeah

my guess: he confuses following directions with paying attention.

perhaps

in any event, discrete math skills are terribly useful in programming

yeah?

well, for example, when counting some collection of things, it's useful to distinguish cases. these correspond directly to subroutine branching (usually some kind of if then else command).

yeah that's true. i've never thought about it. my programming experience has been so goal and language oriented that i've never considered the math

i still cant get the idea about $$e$$

there are direct parallels with the method of attack of a math proof and the method of attack in program design

@SayanChattopadhyay what idea is that?

how does e come into being

Do you mean-why is it special?

yup

@DavidWheeler that make sense. it's very logic based

i have a whole book about the number e

Well, suppose we have a function $f: \Bbb R \to \Bbb R$ with $f(x+y) = f(x)\cdot f(y)$

There are lots of functions like that: $f(x) = 2^x$, for example.

yes that function

i saw i think 8 videos for it but didnt get anything out it

In fact, we might suppose that if $f$ is a SMOOTH function, that any such function is of the form: $f(x) = a^x$, for some number $a > 0$.

$$\lim_{x \to \infty}{{1+1/x}}^{x}=1$$ how is it posiible

What you wrote isn't quite true. Your limit is 2.

its whole to the power of x

what you want is $$\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x$$

yup

$$\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x$$

how is this 1

Now what you get is something "inside" that is getting closer to 1, but being taken to a higher and higher power.

what i thought $$\left(1 + \dfrac{1}{x}\right)$$ let it be a

then the function is $$a^x$$

Have you tried calculating what you get with $x = 1,100,1000,1000000$?

approximate value of e

but what after this

differentiating both sides

@DavidWheeler is what i am doing right

Well, back up a bit-how do you know what the natural logarithm function is?

I mean, if we don't even know "e" exists, I don't see how we can justify using it as a logarithm base.

you get the natural logarithm function while calculating the derivative of $$f(x)=2^x$$

really? how?

How are you going to differentiate it?

let's try that:

$\lim_{h \to 0} \dfrac{k^{x+h} - k^x}{h} = \lim_{h \to 0} \dfrac{k^x(k^h - 1)}{h}$

yes

and then what?

ok, what is $\lim_{h \to 0}\dfrac{k^h - 1}{h}$?

thats what we dont know so thats what we have to find in order to get e

and i am stuck after this but my teacher told me you will get e like this so i started watching videos but didnt understand it

Well, we can rewrite that as: $\lim_{n \to \infty} \dfrac{k^{1/n} - 1}{1/n}$

yup and then

@DavidWheeler "might not be smooth enough to be integrable" :D hilarious

@PaulPlummer Maybe you're right. I did many times lately understand mathematical concepts on a much deeper level and then when I went back to my horrible school text books it suddenly wasn't nonsense.

I'm asking you: how can we do that limit?

@PaulPlummer I'm going to try this.

thats what i dont get @David

@meer2kat You changed your pic again.

@gideon Oh I was confused for a bit, you are talking about trying moving forward with studies rather than staying with algebra 1

@ᴇʏᴇs What are you doing now?

@JasperLoy mhmmmm

@David can i name $$\lim_{n \to \infty} \dfrac{k^{1/n} - 1}{1/n}$$ as a function and then put values and see whats the limit

You could, but the answer might depend on $k$

are you all still discussing e?

and then say $$M(e)=1$$@DavidWheeler

I think you are still missing my basic point: what is $e$?

lets just assume $e$ is a mystery number

we dont know

What facts do we know about it?

Also I think you should write:

$M(k) = \lim_{n\to\infty} \dfrac{k^{1/n} - 1}{1/n}$

according to my assumption$e$ $$\dfrac{e^{1/n} - 1}{1/n}=1$$

@meer2kat You are not sleeping yet?

So that $M$ does not depend on $n$

@gideon it is definitely worth a try and it sounds like it maybe working out better. Plus the kinds of things that might be in apostol rather than some algebra 1 text is more interesting so you will have more motivation to keep on going and learning. I think it is important to do math your interested in since it is motivating.

@David then what should i do

Now, IF we knew that $M(k)$ DID in FACT EXIST, AND, we knew that $M$ was one-to-one, THEN we could define $e = M^{-1}(1)$.

But these are non-trivial things to prove.

this can show the existence of e

Try this: what happens if you use the binomial expansion on $(1 + 1/n)^n$?

Try defining $a_n = (1 + 1/n)^n$.

but whats the problem in my method

Use the binomial theorem to show that $$a_n = 2 + \sum_{k = 2}^n \dfrac{1}{k!}\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)\cdots\left(1 - \dfrac{k-1}{n}\right)$$

How do you show your $M$ is a WELL-DEFINED FUNCTION, that EVERY such limit exists (is finite) for ANY $k > 0$, and that $M$ is one-to-one? I don't think that's easy to do.

wait @DavidWheeler can we try my way once

I think what David is saying is beyond Sayan's understanding now. He should finish Hammack first.

no @JasperLoy i know what he is trying to say

@JasperLoy maybe so.

but i think this is better

@SayanChattopadhyay Show the sequence $\{a_n\}$ is increasing, and bounded by $3$.

how does one write d/dx

This will show that the limit EXISTS.

something just clicked in my head

If it exists, we can give it a name.

$$d/dx {k^{x}}_{x=0} =M(k)$$ @DavidWheeler

right

$\dfrac{d}{dx}k^x = k^xM(k)$

@DavidWheeler we know the slope at x=0 so we can try to find the slope for the function everywhere

Presumably, you want to try to establish $M(k)$ is the natural log function-good luck with that.

i will show you wait

@JasperLoy eh sorta

What're some interesting results dealing directly with calculus of transcendental functions?

@DavidWheeler from my way i can prove $$\dfrac{d}{dx}e^x=e^x$$

If you can get a grip on $M$.

I'm not saying you can't do it, but I think you are underestimating the difficulty.

$|f(x)|-|f(x_n)|\leq|f(x)-f(x_n)|\leq \frac{|f(x)|}{2}\implies |f(x_n)|\geq \frac{|f(x)|}{2}.$

How does that implication come from the left?

Oh, nevermind.

how does one send a pic here

@DavidWheeler i have written the way find e^x but how do i send it to you

@DavidWheeler whats your email

@PaulPlummer oh yes I do feel stupid learning Algebra 1. I might not get 5 Algebra 1 answers right in a row but I certainly know it. Calculus I WANT TO learn desperately, have wanted to for exactly 4 years now, because I have come across it's applications in physics, machine learning so far and maybe a few more instances in my job and life.

@PaulPlummer I want to learn more about Combinatorics, Formal Logic etc because again as I run deeper into my career field of programming I get stumped at the bottom when I hit the related underlying math. So yes I have lots and lots of motivation to study these things.

And I WILL apply them, one or or another I will find a way to apply them.

..... In programming.

I had a mathematical dream last night. There was the value 99/100*Pi^2 calculated from some operation on a matrix, and then there was the same value 99/100*Pi^2 as the convergent of a series. It does not make sense and I never saw any more details than this.

@MatsGranvik u have any idea of how to post pics on chat

@SayanChattopadhyay there is this upload button to your right next to "Send"

its not there

@SayanChattopadhyay Hmm. Not sure then, search/post on meta.SE. I looked at privileges and you seem to have enough to post links and what not. What browser/OS are you using ?

@gideon i am using mozilla firefox

@gideon for operating system it is windows xp

@SayanChattopadhyay Try pasting a url to your image in here?

Greetings

@robjohn I face a very strange situation with Mathematica. It shows me that Hurwitz zeta function becomes negative for large input. How is this possible?

The flaws are simply dramatic. Mathematica can't even imagine how to compute one of my creations and becomes suddenly crazy.

hello

i want to prove that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$ (or $\mathbb{R}^n$ in general)

if i suppose that there is a homeomorphisme $f: \mathbb{R}\rightarrow \mathbb{R}^2$

how to find the contradiction ?

@Vrouvrou Is this a homework assignment?

and do you want a hint

What are your thoughts on the problem so far?

no it is not a homework

i found in a book and i try to prove it

I am not sure what book but normally this sort of problem follows something that happened in the chapter you found it in, what happened in that chapter?

no it is not a problem they just say (a book in arabic) that R and R^n are not homeomorphic

Okay. Would you know how to show that $(0,1)$ is not homeomorphic to $(0,1/2) \cup (1/2,1)$?

yes why not

If $f: \mathbb{R} \to \mathbb{R}^2$ is a homeomorphism, would you now how to show that there is a homeomorphism $\mathbb{R}\setminus \{x\} \to \mathbb{R}^2 \setminus \{f(x)\}$

why this is true please ?

Why what is true

?

If you look at the restriction of $f$ to $\mathbb{R} \setminus \{x\}$ you would have such a homeomorphism

if R is homeomorphic to R^2 then R-{x} is homeomorphic to R^2-{f(x)}

Yes

it is always true for any set for any homeomorphisme ?

Yes, that is a good exercise.

i don't know how to prove this

Do you have any background in topology (or even analysis)?

i have i have

but i am a big banana lol

and so where is the contradiction

if R-\{x\} is homeomorphic to R^2-{f(x)} ?

Maybe you should try drawing a picture it should be clear

How about the other direction, then there would be a homeomorphism $g: \mathbb{R}^2 \setminus \{f(x) \} \to \mathbb{R} \setminus \{x\} $

@DavidWheeler I got the way to prove the derivative

Hi, off topic. I'm studying stereographic projection (Riemann sphere). Let $z=x+iy$. The line in 3-D from $z$ to $N$ is given by the set of points $tN+(1-t)z$. How is this set equivalent to $\lbrace ( (1-t)x, (1-t)y, t) : -\inft < t < \infty \rbrace $? Why isn't $N$ present in the latter set?

How do you paste the URL of an image

@Vrouvrou got it yet?

How do you do paste the URL anyone knows

yes, you have [name] followed directly by (url)

with those brackets

@PaulPlummer so what should I do if I want to post a pic

So that the pic shows up in the chat room or as a url? @Sayan

Yup

As a pic@PaulPlummer

[Xy5HJ](url)

Do you have the upload button right next to send

?

Its not happening

I don't have

Not sure, I think if you post the plain url, a lot of sites get formated and previewed in chat, but I don't know if that works for all sites. Maybe give that a try

Can you show me an example@PaulPlummer

abstrusegoose.com

@Sayan Not sure how to get an image to show in chat

How did you do this

If you get a url to the actuall picture file it works @Sayan

>http://abstrusegoose.com/strips/another_fun_game_is_comic_tac_toe.png

If you the path to the file it will load

without the > sign @Sayan

But I have the picture in a file.....

Oh on your computer?

Yup

Then what should I dk

How did you do it

For me there is an upload button next to the send button. are you on a computer or mobile device

nevermind it was a softball.

In my computer also I didn't have an upload button and also on my mobile device

What's your OS @PaulPlummer

Linux

Me windows

so you don't have this

upload at the far right

No I dont

I think there is a reputation score minimum :-)

What is it

Yah you don't get all the listed privledges till you reach 1000, but none of them say anything about pictures

at 100 you can create your own chat room, that may be it, I am not sure

100 I am on 86

Awww.......

Get busy my friend ;-)

Can you give reputation

Wait if I can take my pic to any website then I can just write the URL right?

Alright you should be good to go, I upvoted 3 of your questions, maybe leave, exit the browser and come back and see if that works (it might take a bit, not sure if there is some lag)

It is not in my phone I have to see in my pc

Hello @robjohn @DavidWheeler @DanielFischer !! Could you explain to me how we find the characteristic curves of a initial value problem??

@PaulPlummer $\mathbb{R}\setminus\{0\}$ is not connected

@Vrouvrou yup, basically that is how the contradiction arrises

to get a contradiction we must have that $\mathbb{R}^2\setminue\{f(0)\}$ is not connected

and i don't know if this is true

To get a contradiction we need that it is connected actually (again draw a picture)

$\Bbb R^2 - \{(x,y)\}$ is connected for any point $(x,y)$. It's even path-connected (just drive around the hole).

But if $f: \Bbb R \to \Bbb R^2$ was a homeomorphism, then the partition $\{(-\infty,0),(0,\infty)\}$ would map to a partition $\{U,V\}$ of $\Bbb R^2 - \{(x,y)\}$ for $(x,y) = f(0)$.

@Sayan Are you seeing the upload button on your computer?

@Sayan on my android tablet I don't have the upload button, so it may have something to do with the mobile version.

@DavidWheeler, @PaulPlummer the contradiction is that $\mathbb{R}\setminus\{0\}$ is not connected where $\marhbb{R}^2\setminus{f(0)}$ is connected right ?

@Vrouvrou Yes

but why $\mathbb{R}^2\setminus\{(x,y)\}$ is connected ?

@Vrouvrou Like I said it should be clear if you draw a pic and David gives a pretty good description, you can always draw a path around the hole

Draw a straight line between any two point, and draw a circle of radius 1 around (x,y).

Either the straight line does not intersect the circle, and can be our path, or it does, and we just use a circular arc as a detour.

If the point lies within the circle, draw a smaller circle.

That's the funny thing about open sets-there's always "wiggle room"

You can't hit them "right at the edge" they have no edges

Have you ever heard of Zeno's Paradox?

I feel I need something better than Mathematica. Is there such a thing on market?

@DavidWheeler I thought of a series.....

What was your thought?

$${1/3!+1/4!+1/5!..........1/n!+.....} <{1/2^2+1/2^3+1/2^4+......1/2^{n-1}+...}$$

Adding $${1+1/1!+1/2!}$$ two both the sides of the inequality @DavidWheeler

So $e < 3$ right?

Yup but I was stuck at a place

And clearly, $2 < e$

In fact, we can do a bit better: $2.5 < e < 3$

$${1+{1/1+1/2^2+1/2^3+........1/2^n-1$$ what should I do after thus @DavidWheeler

So it makes sense to define $e = \sum_{k = 0}^{\infty} \dfrac{1}{k!}$, since we can see it converges.

@DavidWheeler I know the left hand sides inequality converges but what should I do the right hand side

What you added to both sides is just 2.5, can you see what the RHS side is?

You can re-write it as: $\dfrac{1}{4}(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +\cdots)$

Do you know what the stuff in the parentheses sums to?

2@DavidWheeler

and what is (1/4)*2?

$\dfrac{1}{4}\cdot 2 = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = $?

I didn't get what you wrote

Earlier

You just proved that $e < 2.5 + 0.5 = 3$

A

B

C

D

E

The great

aitcheyejaykay elemenopi

What textbook to learn conformal mappings and mobius

@DavidWheeler I also got the value for $e^x$ using the slope of the $a^x$

At different points

Is anyone familiar with characteristic curves of PDE's ??

A

OCD

All nilpotent 2x2 matrices are similar to $J = \begin{bmatrix} 0&1\\0&0\end{bmatrix}$

But how do I find all of these matrices?

$A=P^{-1}JP$

I do think that the only such cases are $J$ and $J^T$

I am not 100% sure what you want but I am pretty sure all you have to do is calculate $P^{-1}JP$ for generic values in $P$ $$P=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ @Committing

Although its almost 6am here and I havn't gone to bed (this is not normal behavior for me) so I would take any advice I give with a mountain of salt

I got the same advice when I posted my question

That is good to see, I don't remember if 0 matrixes are considered nilpotent (I think it depends on the source) but there is also the zero matrix

@Committingtoachallenge

They are

The trivial case xD

I was reading some math in a book, it was looking at the symmetry group of the octahedron proving that it was generated by these two subgroup the author defined. Anyways part of the reason I am up so late/early Is that I was trying to figure how this group the author defined was isomorphic to $S_3$, I thought it was a typo and maybe meant $S_4$ because it seemed to be much larger than $S_3$.

So I have been sitting around doing a bunch of calculations to see if the author actually meant $S_4$, only to come to the conclusion that the author did mean $S_3$ but the author actually meant to say it was generated by rotations rather than reflections!

I am mildly irritated in part because of the mistake the author made and in how long it took me to figure out the mistake. I should have caught it sooner.

As I have said before, the times I learn best are when I am trying to prove something that is false(without knowing it)

Do you @paul consider that mistake as a reason to change books?

Haha yes that is always a concern isn't it? Losing faith in your author

there are some mistakes that I consider "unforgivable" by an author

others not so much ...

@infinitesimal Not necessarily, If the book was filled with errors like that maybe. I have read more of this book but I guess the last time I looked at this chapter I didn't spend much time thinking about that example.

Maybe I will email the author though, it doesn't look like there is an errata. (and that part was worded in a strange way)

sounds reasonable :)

@infinitesimalsimplicio have you come across unforgivalbe mistakes and what were they?

Maybe he was up reallly late typing it, like I was up really late trying to decipher it

hmm...not recently

finding typos can be fun

@Committingtoachallenge Definitely, persistently failing is a great way to learn

maybe I should ask on main "what are your favorite typos?"

That sounds like it could be an interesting soft question, not sure how it will be taken though

yeah, it'll probably get closed

profs typos in notes could be included

or even stories of catching mistakes during a lecture?

after all we all are only human :-)

Yah. Not a typo but all this talk about spelling a such reminds me of an MO post, I think it was just asking for funny or clever things in papers and books and I guess one of them the author was looking at the categories of Banach analytic manifolds, and following category theory traditions of using the first three letters of the words the category was called BanAnaMan

BanAnaMan

i got it @PaulPlummer thanks

We have that $w_t(x, t)=0$ and $w(x, 0)=0$. To find $w(x, t)$ we have to integrate the relation $w_t(x, t)=0$, right?? Is it $$\int_0^t w_t(x, t)dt=0$$ or do have to use an other variable, for example, $$\int_0^t w_{\tau}(x, \tau) d\tau =0$$ ??@robjohn

Hi. I need to prove using the definition that the limit of (sqrt(x)-2)/(x-4) when x approaches 4 is 1/4. I have done it, but I am not sure it is correct. Is it acceptable to publish my prove as a question so that more knowledgeable people can verify it for me?

yes it is @Romildo

hi @Ramanewbie

hi @sayan

Hello, is there a difference between "Cluster Point" and "Limit Point" in general?

Well that was a stupid formulation for sure, I know that the concepts are different, but I don't know good translations for those :D

@teadawg1337 I have many problems to do today. :<

Hi @Owatch

Hello Eyes

@Owatch How many problems

fifty

@Owatch Wow lol

Professor is irrational.

& inconsistent.

With assignments

( I feel)

TODAY?

Fifty problems today???

Hi @teadawg1337

Hi @iwriteonbananas

Hello @ᴇʏᴇs

@teadawg1337 Yes

I don't know why he thinks is can be done.

I will maybe do 10 if I can.

Then call it quits till tomorrow.

@Owatch I cannot help you, I have a major test tomorrow that I need to study for

Hello @JonasMeyer !!! Could you take a look at my question? math.stackexchange.com/questions/1200782/…

That's okay, study for it.

More important

I have calc test Wednesday

Waht test tee dog?

Music Appreciation, it's going to have a listening portion where we are expected to identify the title and composer of each composition played

Hello @DavidWheeler ! Could you take a look at my question? math.stackexchange.com/questions/1200965/…

@DavidWheeler Any thoughts?

morning, @Ted

good night, @Mike

hi @teadawg, @Owatch

Morning @Ted

Hi @Ted

hi mr eyeglasses :)

Hello @TedShifrin :) How are you?

hi, @evinda ... Trying to figure out someone's projective geometry question ...

Aha! @TedShifrin

@Committingtoachallenge yes but that doesn't mean you should learn PHP

Hello @VincenzoOliva

what's up eyes?

Hi @meer2kat

How are you doing this morning

Could someone here please take a look at my question? math.stackexchange.com/questions/1200965/…

@ᴇʏᴇs I'm well, thanks. And you?

@meer2kat Same old studying as every day

@evinda Don't know. What is a "thin heap"?

@ᴇʏᴇs I feel. Watcha workin on?

@meer2kat Real analysis

@ᴇʏᴇs fun

@meer2kat What are you working on

@ᴇʏᴇs I'm reading up on the Peace Corps route for my masters and scheduling a time to go in and talk to recruiters

One of my teachers from high school told me to never join Peace Corps because it's more dangerous than working in the military

@ᴇʏᴇs I think it really depends on where you go. For example, I'm not about to sign up for where there are drug cartels everywhere or where there are currently wars

Oh, you can guess where you want to go I guess? I don't know anything about it

Yeah you pick where you want to go before you apply, as well as what you want to do