define converge

in what sense

a sum? series?

what it means

ugh

to converge is

to have a finite, expressable value as it approaches infinity or negative infinity

for example

the sequence $1, 1/2, 1/3$ converges

but it's sum doesn't

that is, $1 + 1/2 + 1/3 + \dots$ is not finite

however,

the sum

$1 + 1/2 + 1/4 + 1/8 + \dots$ does converge

it converges to 2

I thought you just said it doesnt converge

those are 2 different sums

the first one

$1 + 1/2 + 1/3 + \dots$ is not finite

seems to converge to 2 as well?

uhhh

no

in fact

it would have to be greater than 2

because you have the original $1 + 1/2 + 1/4+ 1/8 + \dots$

0.5 + 0.3- + 0.25

true

but then, you have extra, the fractions that aren't powers of 2, like $1/3$, $1/5$, $1/6$

no, not true

stop being so insistent that you're always correct.

laughing my bum off, I was not insisting I was correct I was agreeing with you. My usage of true was pertaining to your explanation. "true" I am wrong you have a point there it doesnt converge.

:P

i'm in snarky mood

because

i have to write a fucking essay

sorry dud

no worries,

this chat channel produces will in me to use new words I do not use

heh

anyways

If you want a proof that $1 + 1/2 + 1/3$ doesn't converge, here it is

Consider the sum

$1/2 + 1/2 + 1/4 + 1/4 + 1/6 + 1/6 + \dots$

because every other term is greater than the other sum

However, when you group those sums together, you actually get the same sum

So we have this contradictory statement, that the sum is greater than itself

well if I pick abitrary terms and measure that there is no pattern of it reaching something closer and closer (well other than infinity) cant I make a decision

This can not occur for finite values, and thus is divergent

You can use a so-called "ratio test"

Anyways I really need to write this essay... adios

I never said hola

Hi guys

I stumbled upon a Math question today.

please fill your mouth with it and spit it out so we can help <3

if it's clean of course

The question was 40/3 * 30

lol im typing..

i see 2 ways to solve this

that's not a question, nor a thing you solve

divide 40 by 3 and then multiply by 30

you are presumably talking about the problem of simplifying it

or multiply 40 and 30 and then divide by 3

@Haunted well

either

Division is actually multiplication by the reciprocal

but the answers are different each time

no they aren't

So we actually have

40 * 1/3 * 30

so, apply commutative and associative rules however you'd like

yes they are.

no... 40 divided by 3, multiplied by 30 is 400

thanks Zach

no, (divide [40] by 3 and then multiply by 30) and (multiply 40 and 30 and then divide by 3) give you the same thing

40/3 is 13.3333333333

40 multiplied by 30, THEN divided by 4 is also 400

@Haunted so? you're multiplying by 30

@Haunted 40/3 is approximately 13.3333333333

so that factor of 1/3 will be canceled out

@arctic and his nitpicking :P

go write your essay Zach

:[

I wrote both body paragraphs

$13.\overline3$ is not a real number so we have to keep it in rational form

all i gotta do is write a dumbass conclusion

@CausingUnderflowsEverywhere yes it is...

@CausingUnderflowsEverywhere yes $13.\overline{3}$ is a real number

what makes you say it isn't?

my calculus teacher

LOL, is it a fake number?

@ZachHauk No i divided 40 by 3 and then multiplied the answer with 30 which is giving 399.9999999 which is not equal to 400

@Haunted yes they are LOL

oh boy, the classic 9.9999 =/= 10

@Haunted that's because you divided 40 by 3 and rounded

@Haunted Bruh

If you round in the middle of your calculations, the end result will be different.

@Akiva Code red, we got a 0.9999 =/= 1

yes $1.\overline9$ equals to 2 but they wont explain why. so from now on Im writing all my limits equal to $1.\overline9$ for instance instead of 2 !

who is "they"?

Someone post the proof in here

This dependan heavily on whether or not you put the three dots at the end of it @ZachHauk

0.9999 =/= 0.9999…

@Akiva Why you nitpick

because logic and truth

i'm hungry as fuck

all my life i was led to believe 0.99999 is not equal to 1 :/

that would be someone I dont want to single out

did i learn it wrong?

@CausingUnderflowsEverywhere is he a reliable source?

eat words and spit them out til you finish your essay

@Haunted 0.99999<0.99999...=1

like, a professor?

@Haunted If 0.999… were different from 1, there would be a number between them.

no it's you guys <3 xD

you guys told me $4.\overline9$ is 5

(Note that "0.999…1", with the 1 after infinitely many 9s, isn't a number.)

Because it is

when I was figuring out limits

someone post the proof in here

ohh okay, alright, i get it

@AkivaWeinberger send some math books my way

sorry, if this question was kind of dumb..

@CausingUnderflowsEverywhere it is

@Haunted it's really fine

exactly so my point stands :D

@ZachHauk Uh, Game, Set, Math by Ian Stewart

(and answering them, too :P)

or anything by him really

@AkivaWeinberger I mean physically send them

why $4.\overline9$ is $5$ ?

@CausingUnderflowsEverywhere If it was different, then $5-4.\overline{9}$ would be greater than 0, right?

@ZachHauk Sure, when are you next in NYC? I'm conveniently located near a subway station :P

(Too lazy to mail things)

I'm kidding

(Or to go to NJ)

You don't have to give me jack shit

@CausingUnderflowsEverywhere so tell me, what is $5 - 4.\overline{9}$?

@ZachHauk Actually, there's a series of notes on knot theory that's available for free online

If I recall correctly, they're not too advanced or anything

5-$4.\overline9 = 0.\overline01$

hehe

0000000's forevers

@CausingUnderflowsEverywhere ok, so that's where things get fucked up

Let's consider how to treat that "number"

Knots Knotes

Ill meet you akiva

So, how do we write a number with $n$ zeroes ?

@CausingUnderflowsEverywhere If we're doing this: What's that number divided by two? What's that number multiplied by five?

Being a 140wpm typer sucks in this chat cuz it makes me wait before I can post another message

So, how do we write a number with $n$ 9's after the decimal point without causing a buffer overflow and running out of strings in our string theory to repesent more 9's

…What?

So $10^{-n}$ has $n$ zeroes

So, how many zeroes do we want? infinitely many!

But it doesn't make sense to evaluate $10^{-\infty}$... So let's take a limit

What is $\lim_{n\to-\infty}10^n$?

5 * 4.$\overline9 = 24.5\overline4\overline5$

And please don't say $0.\overline{0}1$

@CausingUnderflowsEverywhere false

true

@CausingUnderflowsEverywhere No, I mean what's $5\times0.\bar01$

Answer that limit.

or $0.\bar01\div2$

(hint: it's 0)

holdon there's 2 of you at the same time

I actually pulled that limit thing out of my ass... Just comes to show how many ways you can prove $4.9... =/= 5$ false

the limit does not exist!

@CausingUnderflowsEverywhere no...

what makes you say that?

oh negative infinity?

$n\to-\infty$ in his thing, yeah

thats right it's what I learned from Cady

@AkivaWeinberger Syntax error LOL

@AkivaWeinberger fuck I forgot that whole limit spiel I was given.

I don't understand why any old value doesn't just work. But that's probably because I forgot it.

why is the limit 0? I have a test on limits maybe in a couple hours

essay finished

@CausingUnderflowsEverywhere do you know $\epsilon-\delta$?

wow you're good at multi tasking

(I sure as hell don't fucking know)

not yet I didnt get around to research that

er

yeah

$\lim_{n\to-\infty}10^n=\lim_{m\to+\infty}10^{-m}=$

$\lim_{m\to\infty}\frac1{10^m}=\frac1\infty=0$

using the extended reals for that last step

5 * $0.\bar01$ is easy but once you want to change the place value of that 1, you got an index out of bounds exception flying your way

Stop bringing technology into this... >.>

Even though I know you're joking

okay you're right and Im wrong you cant compute it.. not really

Sorry, not eating sucks

@CausingUnderflowsEverywhere we're not talking about computing

@Cau Wouldn't we just get $5\times0.\bar01=0.\bar05$?

eat some oatmeal with milk

Nope

I've gotten this far, can't eat now

I only had dinner today

And $0.\bar01\div2=0.\bar005$

@AkivaWeinberger I like your style.

But $0.\bar005=0.\bar05$, since they're both infinitely many zeroes.

@AkivaWeinberger isn't this a concept in like the "surreal numbers"

Contradiction; we can't have five times the number (which is greater than $0.\bar01$) be equal to half the number (which is less than $0.\bar01$)

he's asking me to go over my work and make literal sense of it but I already discovered that even doing 5x doesnt really work because the number is undefined

…unless that number equals zero.

Conclusion: The only way to make sense out of $0.\bar01$ is to make it equal zero.

@CausingUnderflowsEverywhere In math, never say "fine whatever I get it." Make sure you understand why it occurs, instead of just accepting it

And so, we get $4.\bar9=5$, since their difference is zero.

so why is 4.$\bar9$ a number then if we cannot define it?

It is a number

It's 5

@Cau The real thing we need to define first are real numbers in general.

OW

There are many ways to do this, two famous ways being equivalence classes of Cauchy sequences and Dedekind cuts.

my headphones compress my ears against my glasses

and it hurts like shit

(Which sound complicated, but that's just 'cause there are people's names in them)

The simpler one is Dedekind cuts, so I recommend you look those up

What's a real number? something ducks walk on?

if i have xn -> (-infinity) for n -> infinity can this still be written using the Lim syntax ? my book seems to not do so in this example - presuming there is a reason for that

@ZachHauk Zach.... this is my exact way of learning. It disgusts me to just accept things. Im honestly so behind in my calc class because I couldnt wrap my head around the limit as x approaches 5 is 5 when it reeally is 4.$\bar9$ so please help me why is that 5 D: and also when I didnt understand what limit meant at all and asked that question about how to tell the different between when the limit is a defined and undefind point in the fxn

@ZachHauk No those are Quacky sequences

@CausingUnderflowsEverywhere i'm glad

@CausingUnderflowsEverywhere But $4.\overline{9}$ is $5$!!!!!!

@Cau Hm. Maybe we need to look at this question: when is a limit of a sequence equal to zero?

For example, take the following sequence:

$(1,1/2,1/3,1/4,\dots)$

Is its limit equal to zero? What does that even mean?

@AkivaWeinberger bringing out the harmonic series

can you teach Zach how to "nickflash" me without typing my entire name? Makes me feel bad that you have to type that much Zach

@ZachHauk I'm not adding them up or anything

@CausingUnderflowsEverywhere no

math.stackexchange.com/questions/2173711/… anyone knows how to answer this question?

All I do is press @C then press tab

you just press tab

oh good

the wonders of javascript

@Cau Actually, here's a better sequence to illustrate what I'm gonna say on:

I'll leave this to @Akiva

$(1,0.1,0.01,0.001,\dots)$

he's smarter

@ZachHauk No, I just know epsilon-delta

AUGH

feels like the limit is equal to .... haha $0.\bar01$

@CausingUnderflowsEverywhere cuz it is

@Cau Note that, from the third number in that sequence onward, everything is less than a tenth.

I was reading up on topological definition of limit

cuz epsilon delta is ugly

AND THE PAPER I FOUND IS TYPESET IN MS WORD

maa.org/sites/default/files/pdf/pubs/cmj_supplements/…

@Z now that you've done your essay you can eat some brain food then you can study epsilon-delta

No I can't eat food

@Cau Also, from the seventh number in the sequence onwards ($0.0000001$, $0,00000001$, etc.), everything in that sequence is less than a millionth.

drink food then (milk)

Trying to lose more weight

less pi then

skim milk

In fact, if you give me any positive rational number, I can give you a place in that sequence where everything after it is less than that rational. @Cau

I've already lost a few pounds

Does that seem plausible?

93 -> 90 in the past week or so

The trouble with that approach is that it's one thing to lose weight, and another to keep it off.

hmmm

@Cau Like, eventually, every number in that sequence will be less than a billionth.

(From the thirteenth number onwards, I think)

And I can do this for any positive rational you throw at me.

if it's not finite then you can't challenge that idea with anything

Let's look at another sequence, though:

$(3,3.1,3.14,3.141,\dots)$

can I get some froot loops first

This sequence does not go to zero.

(It does go to $\pi$, but that's irrelevant for the moment)

@AkivaWeinberger oh, it goes to $3.14111111$ right :D

So, clearly, all of the numbers in the sequence are less than $4$.

But let's say you throw the rational "1" at me; I won't be able to choose anywhere in the sequence where from there on everything is less than 1.

So, this does not have the property of the last sequence;

brought you some frootloops

Ok so is this correct? The limit of $f(x)$ as $x$ approaches $a$ is $L$ iff for every $(p,q)$ containing $L$, there exists some open set containing $a$ whose image is a subset of $(p,q)$

It is not true for this sequence that "for every positive rational you give me I can find a place in the sequence where from then on it's less than that rational."

Why do we even choose them to be open?

@ZachHauk That's not the limit of a sequence, it's the limit of a function, slightly different. But I'm sure you can find counterexamples if you don't require them to be open

is (p,q) a coordinate? and why would a coordinate have a subset unless subset means subset

I believe in you

@AkivaWeinberger sorry that's what i meant

@Cau No, it's an open interval (confusing notation, sorry)

A third sequence:

We should really use ]p,q[ instead of (p,q)

$(500,-1,-1,-1,\dots)$

I use this just to make a slight change to the property from before

oh

@AkivaWeinberger and so taking a function as it approaches infinity is just taking the subsets $(x,\infty)$ right?

[p,q] as in the interval includes the point p and the point q?

Clearly, if you throw any positive rational at me, I can say, "from the second number onwards, that sequence is always less than that rational"

And yet we don't want to say that this sequence approaches zero

So, I'll make the following definition:

We'll say a sequence approaches zero if, "no matter what positive rational you throw at me, I can find a place in the sequence where from then on, the absolute values of the things in the sequence are less than your rational."

So, for example, for that last sequence,

limits in math arent like your grocery store limits like toilet paper on special limit 5, you can buy up to 5. Nope limits are the number right after the actual limit because the number that is a limit is in many cases complex.

if you throw 0.5 at me, I won't be able to find a place in the sequence where from then on the absolute values are less than 0.5.

BTW Im sure when calculating limits algebraicly this will all come to light for me..

@Cau Am I explaining this well?

Do you follow what I'm saying?

"Approaches zero" is the same as "has a limit equal to zero," they're both defined by the above thing

yeah but is 0/0 a positive rational

No, you don't define division by zero

In symbols: $\lim_{n\to\infty}a_n=0$ means that, for any $\epsilon\in\Bbb Q$ such that $\epsilon>0$, there exists an $N$ such that for all $m>N$, we have $|a_m|<\epsilon$

So now, what does it mean to say that two sequences have the same limit?

For example, $(3,3.1,3.14,\dots)$ has the same limit as $(4,3.2,3.15,\dots)$

Definition: $a_n$ and $b_n$ have the same limit if $a_n-b_n$ has limit equal to zero.

So, in the above example, $a_n-b_n$ would be $(-1,-0.1,-0.01,\dots)$, which approaches zero by the above definition.

And so $(3,3.1,3.14,\dots)$ has the same limit as $(4,3.2,3.15,\dots)$.

thats odd

One approaches $\pi$ from below; the other approaches $\pi$ from above.

But they both approach $\pi$.

Now, another sequence to consider:

What does $(5,5,5,5,\dots)$ approach?

…Well, $5$, clearly.

yeah right

it already is 5

it approaches divide by zero error

No, it approaches $5$. ("Approaches $5$" is the same as "has limit equal to $5$).

@Cau This is essentially gonna be part of our definition of limits:

A constant sequence $(a,a,a,\dots)$ has limit $a$.

Or, in symbols, $\lim_{n\to\infty}a=a$.

Is this definition reasonable?

@CausingUnderflowsEverywhere (We never divided by zero or anything)

pretty much because my teacher gave us a sheet that states that the limit of a constant is equal to the constant

I probably should have put that first in this discussion

Im so sleepy

So, let's ask a new question:

Is the limit of $(4,4.9,4.99,4.999,\dots)$ equal to $5$? Let's see if we can use our definitions.

That's the same as asking of $(5,5,5,5,\dots)$ and $(4,4.9,4.99,4.999,\dots)$ have the same limit, by our last thing

And, that's the same as asking if their difference, $(1,0.1,0.01,0.001,\dots)$, has limit zero (by our definition of when two sequences of numbers have the same limit).

And that's the same as asking the following: If you give me any positive rational, call it $\epsilon$, will I be able to give you a place in that sequence (call it the $N$th place) such that, from the $N$th place onwards, everything in that sequence has absolute value less than $\epsilon$?

(Note that the "absolute value" part here is irrelevant since they're all positive. I just need the things to be less than $\epsilon$.)

For example, as previously noted, if $\epsilon=\frac1{10}$, then $N=2$.

, since $0.01$, the second number in the sequence, is less than $\epsilon$, and so is everything after it.

there's an infinite sum formula?

I have to go to sleep :(

I haven't touched infinite sums yet...

...but if you want me to, the definition is this:

thanks akiva Im sorry that I have to go to sleep

The infinite sum $a_0+a_1+a_2+\dots$ is taken to be the limit of the partial sums $(a_0,~a_0+a_1,~a_0+a_1+a_2,~\dots)$

(that is, the infinite sum is the limit as you add more and more things.)

@CausingUnderflowsEverywhere

Question about proofs. Why did we add $+1$ to the number $a$?

can you tag me so when I login tomorrow I can jump to this location

So, for example, $4+0.9+0.09+0.009+\dotsb$ is the limit of $(4,4.9,4.99,4.999,\dots)$, which we previously established to be equal to $5$.

@CausingUnderflowsEverywhere

@Dragneel The idea is that $(2)(3)(5)(7)(11)+1$ is not a multiple of $2$, nor is it a multiple of $3$, nor is it a multiple of $5$, nor $7$, nor $11$

That's why the $+1$ is important. Without it, it's a multiple of all of those.

If $2,3,5,7,11$ were all the primes that existed, $(2)(3)(5)(7)(11)+1$ would have to be a multiple of one of them

(since every number either is a prime or is a multiple of a prime)

and that gives us the contradiction.

So, by adding the $1$, we will be unable to find any prime divisor?

Right,

since we're assuming, for contradiction, that $p_1,\dots,p_n$ are all the primes that exist.

Wow. That's really clever actually. I wasn't even aware you could do such thing.

Yea.

I understand now. Well, thanks as always @AkivaWeinberger

@Dragneel By the way,

Hmm

this doesn't mean that if you multiply the first few prime numbers and add one, you'll get a new prime number.

For example

$2\cdot3\cdot5\cdot7\cdot11\cdot13+1=30031$ is not prime.

So, the addition of $1$ doesn't represent an additional prime number for $a$

It's equal to $59\cdot509$.

Ahhh

I see. That's why they made use of the parentheses.

The parentheses don't seem to be relevant

The point is,

The divisors of $30031$ aren't in the set $\{2,3,5,7,11,13\}$.

All we need is for this new number to have a divisor not in your list of primes.

If the $1$ DID represent an additional prime number to the set, then we wouldn't result in a fraction. Correct?

(Remember that every number is a divisor of itself)

@Dragneel The fractions came from dividing each side of the equation by $p_k$

Yeah I understand that

I mean, sometimes (or even usually) you get a new prime number

$2\times3\times5+1=31$ is prime, for example

Hmm

The contradiction comes from saying that, if $2$, $3$, and $5$ were the only primes that existed, $31$ would have to be a multiple of one of them

(since every number is a multiple of some prime).

The rest of the proof (where we divide and stuff) is just them proving that $2\times3\times5+1$ is not a multiple of $2$, $3$, or $5$, essentially.

I see

The rest of the proof is very important too though

On the exam (for instance), I can not just omit it

I guess

@AkivaWeinberger I've seen you around in this room for a while. You seem to be familiar with every field of Mathematics haha. Are you a professor, researcher, or..?

hello

which part do you need clarification?

Well it's basically how I would use the remainder theorem really.

I am not sure how I would apply it here... I can see the result directly but I am not sure how to reach that point.

I got it. Never mind.