Mathematics

Hippalectryon

17:00

Way 1 : $(x,y)+(x',y')=(x+x',y+y')=(x'+x,y'+y)=(x'+y')+(x+y)$

Way 2 : $2(A+B)=(2x+2x',2y+2y')$, and we see that $2(B+A)=(2x'+2x,2y'+2y)$

There are two ways, but they're both equivalent. No problem here. @KellyBlunie

Kelly Blunie

Yes i see that but in my 2 i sub back in to the equation, and that is wrong proving?

Hippalectryon

No

I told you, both ways are ok

quidstone

https://math.stackexchange.com/questions/136831/factorial-number-of-digits/688304#688304?newreg=3a91eb63e4fc4291be2e0cc5b686cba3

can anyone explain the 2nd comment...please...

Hippalectryon

@quidstone The 2nd comment is just telling you that 20*19*...*2 can be done by hand

quidstone

no its not telling me that...

Kelly Blunie

17:05

IThe addition could be defined as $(x_1,y_1)+(x_2,y_2)=(0,5x_1*x_2)$

Hippalectryon

@quidstone gyazo.com/11d7c0485fa9d1d15d6a7bb683905a9c

quidstone

its telling me the number of digit in a multiplication but i have reasons not to believe it...

Hippalectryon

@KellyBlunie But it isn't. Addition over $R^2$ is defined as $(x,y)+(n,w)=(x+n,y+w)$

Kelly Blunie

Yes

quidstone

@Hippalectryon my bad!! i mean the 2nd answer...

Hippalectryon

17:06

Oh ok

This one right ? @quidstone

3

A: Factorial number of digits

I come from a background in computers, so here's my two cents. Taking the logarithm to the base 10 of n!. If the log comes out to be x, it is not hard to see that the number of digits must be the lowest integer greater than or equal to x, i.e, $floor(x)+1$. Now the question comes down to approxim...

quidstone

yeah!! exactly...

Kelly Blunie

@Hippalectryon thank you for your help i appreciated

Hippalectryon

@quidstone What part of the answer don't you understand ? Do you know base 10 log ?

@KellyBlunie No problem :)

quidstone

@Hippalectryon It is possible to prove by induction that n! lies between (n2)n and (n3)n. this part

Hippalectryon

@quidstone

2

Q: Factorial lower bound

A professor in class gave the following lower bound for the factorial $$ n! \ge {\left(\frac n2\right)}^{\frac n2} $$ but I don't know how he came up with this formula. The upper bound of $n^n$ was quite easy to understand. It makes sense. Can anyone explain why the formula above is the lower bou...

factorial

quidstone

17:16

@Hippalectryon thanks...get it...

Hippalectryon

:-)

quidstone

@Hippalectryon appreciate it...:)

Khallil

Have you done any Analysis, @Hippa?

Hippalectryon

@Khallil Some

Khallil

What kinds of things have you done?

Hippalectryon

17:18

@Khallil Define Analysis :)

I'm in France

Khallil

Mmmmm ...

Hmmmmm ...

quidstone

Hey @Hippalectryon i want to know about the first answer now...sorry for not mentioning...

Hippalectryon

@quidstone No problem :) what is your question ?

Khallil

Defining it is pretty hard, but I can tell you what I've done so far, @Hippa.

Hippalectryon

@Khallil Ok

quidstone

17:20

math.stackexchange.com/questions/136831/…

Khallil

Is that similar to what you've done so far, @Hippa?

Hippalectryon

@quidstone I know that :) I meant, where do you need help ?

@Khallil I've done 1

@Khallil I've done 2

@Khallil I've done 3

And i've done 4

quidstone

"As a rough approximation, multiplying an n-digit number by an m-digit number yields a result with about n+m digits"......this doesn't work for lower numbers....

what i understand is 1*2 should be a two digit number...well...it not....

Huy

@Khallil: Are you writing a summary of your Calc1 lecture?

Hippalectryon

@quidstone It's an approximation :) it doesn't work with $1$

Nor with $0$

Khallil

17:25

Awesome! If I ever have any questions, would you mind giving me a few hints for them, @Hippa?

quidstone

@Hippalectryon again....i am messing up for 20! too i am getting answers of having 34 digit...

Hippalectryon

@Khallil Sure. $5/question

:D

Khallil

Nope. I found them online, @Huy!

Hippalectryon

@quidstone Give me an example

Huy

@Khallil: I see.

quidstone

17:26

@Hippalectryon yeah i totally get it...but from what number it starts working! 10* 12 should be 4 digit i guess..!!!

Khallil

Hahaha! It took me a while to figure that out, @Hippa.
I thought you were writing some $\LaTeX$ code at the beginning and a \ had gone awry!

Chris's sis

Why would one like to become a moderator for free? What's the gain?

Hippalectryon

@quidstone It works for any number, with a 1 number possible error.

@Chris'ssis Ask @robjohn

Huy

@Chris'ssis: I assume some people want to give back to the community.

quidstone

@Hippalectryon yeah!! right!! its a bad guess then....okey going on with it....again how he makes a guess about (20!) so close....!!! i dont know...i am again lost here...

Chris's sis

17:29

@Huy If you wanna give some back to the community you can post answers and questions (that I did already).

@Hippalectryon Yeah, good point.

Huy

@Chris'ssis: I don't, it was an assumption.

Chris's sis

@Huy I see, OK.

Hippalectryon

@quidstone for instance 5!=5*4*3*2 ~= (1+1+1+1) digits

Huy

@Khallil: Let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence and $\sum_{n=1}^\infty a_n$ conditionally convergent. Prove that for all $a \in \mathbb{R}$ there exists a bijection $\phi: \mathbb{N} \to \mathbb{N}$ such that $$\sum_{n=1}^\infty a_{\phi(n)} = a.$$

Khallil

Conditional convergence means that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges, but $\displaystyle \sum_{n=1}^{\infty} \left| a_n \right|$ doesn't, right @Huy?

Huy

17:31

@Khallil: Yes.

Khallil

Ah, nice. Then absolute convergence is the opposite, right?

quidstone

@Hippalectryon what i am doing is...5!= 5*4(=2digit)*3(=3digit)*2(=4digit)...

yeah it same...

Hippalectryon

:)

quidstone

but
20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2(2+2+2+2+2+2+2+2+2+2+2+nine 1s = 31)

is it okey!! @Hippalectryon

why i am getting 31 digit....as a guess @Hippalectryon

Hippalectryon

Did you see the answer's second comment ?

20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2 => 11*2 digits + 8*1 digits = 29 digits

user2179021

17:36

hi

anyone got any ideas about math.stackexchange.com/questions/1059379/… ?

Huy

@Khallil: Absolute just means that the sum of the absolute values converges. Nothing is said about the normal sum.

@Khallil: It's not a very beautiful proof but I think the result is really cool.

Khallil

I think I've seen a proof of it before and it was a monstrosity.

I don't know where to begin.

Huy

@Khallil: It's not a monstrosity, I think we did it on a single page in our first semester.

Khallil

How does one go about proving the existence of something, @Huy?

(I mean, is there a general way of doing so?)

Huy

@Khallil: Sometimes you can only show that there exists something, sometimes you can explicity construct it (which is more helpful then).

quidstone

17:44

ohh!! sorry i didn't see that... and i should have!!! by the way can you vote it down it preciously deserve!!

@Hippalectryon

Hippalectryon

@quidstone ok :)

quidstone

thanks...:)

Venus

17:55

Have a look what have been discussed on

$Mathematics$ Mathematics 2014 Moderator Election

Open Discussion for math.stackexchange.com/election/5

It's all about Pedro & you guys

teadawg1337

Aaaand I'm back. I'm part of a study conducted by the NIH and FDA, sorry I took so long

Khallil

Who are the NIH, @teadawg1337?

teadawg1337

@Khallil National Institutes of Health

Basically, I'm part of a Federal study

Enough about that, let's get back to maths. Has anyone seen any interesting questions lately?

Mike Miller

No.

teadawg1337

Me neither...

Khallil

18:04

Huy's one just a small scroll up is pretty cool.

Mike Miller

Riemann's rearrangement theorem; it's a cute little trick.

Khallil

Is it Hilbert's, @Mike?

Huy

No, it's mine, @Khallil.

Mike Miller

No, it's not. Apologies.

Huy

@Khallil: A little hint in case you don't know how to start: I can do it in one line, without pen and paper.

Khallil

18:08

Its Riemann, right?

Mike Miller

Yes, I edited it.

Alizter

Hi @MikeMiller

Mike Miller

hi

Khallil

Ah, just saw that.

I can make a bijection too.

teadawg1337

Hello yourself, @Alizter

Alizter

18:09

Hi @teadawg1337

@MikeMiller I feel much better today. I have clearer thoughts.

Khallil

$1 \mapsto 2$, $2 \mapsto 1$ and all other elements of $\mathbb{N}$ map to themselves, right @Huy?

Mike Miller

That's good.

Alizter

I am currently tutting at my notes riddled with yesterdays jibberish.

Mike Miller

If you only rearrange finitely many terms, the sum is the same, @Khallil.

Khallil

Oh. So we're talking about a bijection that swaps infinitely many terms and ensures that the sum retains it's value, @Mike?

Mike Miller

18:12

@Khallil I don't quite understand. I thought the question was to start with a conditionally convergent series and change it so that it converges to some arbitrarily chosen real $a$.

I'm saying that if you want to change the value it converges to, you need to rearrange infinitely many terms.

Khallil

Ohhh.

I misread the original question!

Thank you!

Huy

@Khallil: That's a lot more powerful, no? (sorry, I'm cooking dinner)

Khallil

What's a lot more powerful, @Huy? (Also, what are you making? ^_^)

Hippalectryon

@Chris'ssis I was wondering, if I ever need to use some of your results in one of my works, under which pseudo do I refer to you ?

Chris's sis

@Hippalectryon I'll let you know my name one day. Anyway, I'm not a famous person as many of you here, I'm totally unknown.

Hippalectryon

18:23

@Chris'ssis ok

@Chris'ssis I'm not famous at all

Khallil

There are famous people here?

Hippalectryon

@Chris'ssis No one knows me irl here

Huy

@Khallil: The theorem is more powerful the way you read it correctly.

Khallil

Ah, yep. Indeed, @Huy!

Huy

@Khallil: I made some bread and schnitzel. :D

Chris's sis

18:24

@Hippalectryon I think r9m knows me (if so, I wonder how it was that possible). I have reasons to believe that.

Hippalectryon

@r9m :O

Huy

@Khallil: You ever had schnitzel?

Khallil

Nope. Never. What is it, @Huy?

Huy

It's sort of breaded meat.

@Khallil: google.ch/…

Chris's sis

@Hippalectryon The MSE policy is to remain anonymous if you wanna be anonymous. When I asked him for more informations, he said it was just a misunderstanding. Let it be so ... (but I have doubts).

Khallil

18:26

Oh, I've had that before! I never knew it was called Schnitzel, @Huy. It's really nice. ^_^

Hippalectryon

@Chris'ssis I know :)

@Chris'ssis I do it that way too

Studentmath

Indeed @Balarka, thanks! :) And Thanks @Mike. I needed to practice a bit with direct products, and this also practiced up homomorphisms and so on, it was real good.

Chris's sis

@Hippalectryon I wanna learn one thing: every time when I have reasons to get annoyed to try to make fun of that situation. :-)

Hippalectryon

@Chris'ssis Or, do as I do, try not to get annoyed :D

Chris's sis

@Hippalectryon I'm annoyed again looking at the list of those who want to be moderators. :-))))

Huy

18:33

@Chris'ssis: Look at me instead, then.

I'm Huy.

Hippalectryon

^

Haha

Chris's sis

@Huy Great! :-)

Huy

See, already better. :D

Hippalectryon

@Chris'ssis I'm not even looking at it

Chris's sis

@Huy hahaha, yeah! :-))))

Hippalectryon

18:33

Problems solved

Chris's sis

@Hippalectryon That's better :-)

Venus

@Chris'ssis Is that Pedro, the trivial guy? :D

Huy

@Khallil: Yeah, it used to be my favourite meal as a kid (schnitzel with chips). I think I didn't fry it for the required amount of time though, it's a bit chewy. One day I'll be a good cook. :3

Balarka Sen

@Venus Trivial guy? :P

There is a stronger version of Riemann rearrangement theorem, although proved similarly, @Khallil

It says that every divergent series can be rearranged to give a convergent series. In fact, given a value, the series can be rearranged to give exactly that value

teadawg1337

@Balarka I wrote a proof for that a few years ago, wasn't too hard. Don't ask me to find it, I've recycled most of my papers from high school

Huy

18:45

@BalarkaSen: So it's stronger because it applies to every divergent series and not only every conditionally convergent one?

Venus

This statement made my day so far: "Coming from a strictly neutral country - Switzerland - I think I have the ideal predisposition to be a moderator :)"

Huy

@Venus: I too live in Switzerland.

Venus

@Huy I think you have to consider yourself to nominate as a candidate

Huy

@Venus: I'm a frayed knot.

Balarka Sen

@Huy Right.

@Huy What's your fundamental group?

Huy

18:48

@BalarkaSen: It's trivial. But I'm not a sphere.

Balarka Sen

@Huy Interesting.

I don't think that's even possible.

Hippalectryon

@Huy A ball maybe ?

Huy

@BalarkaSen: But can you prove it?

Balarka Sen

@Huy Not sure. I can try, but I have barely studied fundamentals of algebraic topology.

Khallil

Balarka Sen

18:51

My intuition comes from the fact that the complement of a nontrivial link can never be simply connected.

Huy

@BalarkaSen: Neither have I. I barely remembered what a fundamental group was and thus could only come up with a sphere.

Balarka Sen

I think similar works for knots.

FreeMind

Give me a independent branch of mathematics

which focuses on new things than set theory

and is calculus independent

I'm tired of integrals :(

Balarka Sen

@FreeMind Abstract algebra, maybe?

It's calculus independent.

FreeMind

I'm tired of bad notations, confusions

Khallil

18:53

Abstract Algebra is beautiful. It ... freed my mind.

Balarka Sen

You're a free group, @Khallil? :P

Huy

@FreeMind: Complex analysis?

FreeMind

@BalarkaSen Anything else? Is it independent of any prerequisite?

Alizter

@BalarkaSen No.

Khallil

Ahh, what's a free group?

FreeMind

18:53

@Huy No No nO!

Khallil

I know what a group is, but not a free one.

Balarka Sen

@FreeMind It requires a bit of set theory.

Huy

@FreeMind: Why not? It's not as ugly as real analysis.

In fact, it is rather beautiful.

Balarka Sen

No more prerequisites needed.

Khallil

$\mathbb{R}$ analysis is eugh.

Balarka Sen

18:54

Totally, @Huy. Complex analysis is cool stuff.

Huy

@FreeMind: Have you studied linear algebra yet?

FreeMind

@BalarkaSen I want something out of mechanical stuff ( not completely ) , I mean I'm tired of watching inequalities :|

+ absolute values

then epsilon !!!!

Khallil

Abstract Algebra is the way to go, @FreeMind.

It's so nice.

Balarka Sen

@FreeMind Abstract algebra is free of inequalities, absolute value and epsilons.

Huy

@FreeMind: Are you at all interested in theoretical physics?

FreeMind

18:55

@Huy I have studied linear algebra, just on eigenvalues and diagnolization.

I suck at transformation , specially notation used for stuff.

I am tired :D

Balarka Sen

I think abstract algebra would suit your tastes, @FreeMind

FreeMind

@huy yeah physics

Balarka Sen

Try picking up Dummit-Foote.

FreeMind

@BalarkaSen What is that? I am tired dude :D

Huy

@FreeMind: I'd suggest studying quantum mechanics, it has some rather beautiful proofs and constructions, which are purely mathematical.

Balarka Sen

18:57

It's a book. If you're so tired that you can't open up a book, you should forget about studying math @FreeMind.

FreeMind

@BalarkaSen That's the truth.

@BalarkaSen I think, calculus kills innovation.

Balarka Sen

I don't like calculus. So, instead of whining about it, I started studying algebra.

FreeMind

calculus does not sharpen your mind, it just solidifies your creativity then restricts it badly.

Huy

@FreeMind: Complex analysis is the opposite, imo. And quantum mechanics is just mind-blowing.

Mathematics 2014 Moderator Election

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$Mathematics$ Mathematics 2014 Moderator Election

$Mathematics$ Mathematics