Mathematics - 2016-06-20 (page 1 of 3)

Axoren

01:05

What is the type of continuation called when you continue a function defined on the naturals to one on the reals?

Leaky Nun

the-continuation-which-made-the-gamma-function

Axoren

It's not an analytic continuation.

Leaky Nun

> In mathematics, the gamma function (represented by the capital Greek letter Γ) is an extension of the factorial function

They just call it an extension, lol.

Axoren

Yeah, but extensions by themselves are not unique.

I could easily extend the factorial function by splicing a random-function in-between all the naturals

Leaky Nun

Generalization

natural-to-real continuation

coin your own terminology.com

Axoren

01:08

I'm looking for a term that people actually use so that I can find content discussing it.

Leaky Nun

What's your function?

Axoren

I want the same power as an analytic continuation, which boasts the uniqueness of the result of the continuation.

The Harmonic Numbers sequence.

$H_k = \sum_{n = 1}^k \frac 1 n$

Leaky Nun

And what is the continuation?

Axoren

That's just it. I know there exists a continuation that everyone uses all the time, based on the gamma and polygamma functions.

But I can't find out how they found it.

It's something like $\gamma +\psi^{(0)}(1+k)$

Leaky Nun

Then how is it not an analytic continuation?

Axoren

01:11

An analytic continuation requires an analytic function to begin with.

A function only defined on the naturals is not complex differentiable at any point in its domain.

Leaky Nun

ok

Axoren

Not only that, I'm fairly certain that the naturals form a closed subset of the complex plane, since their complement is clearly open.

I don't have much to work with here, at least given my existing knowledge on the matter.

Leaky Nun

Not sure if this helps: $H_n\approx\ln(n)+\gamma+\dfrac1{2n}-\dfrac1{12n^2}$

Axoren

While that's a very useful approximation, I'm not sure if it helps either.

Which means it doesn't help :(

Leaky Nun

you could just use this to the reals, lol

Axoren

01:14

While I could use the approximation, the approximation is not the continuation I'm looking for.

By the sheer nature of being an approximation, it's no longer an extension.

Leaky Nun

I am not sure if the partial sum has an exact representation

Axoren

@LeakyNun I've read through a bit of the page on Wikipedia for the Gamma function, they claim its an analytic continuation, but I can't believe it just yet.

Leaky Nun

(Ignore me) $\int_n^{n+1}f(x)\ \mathrm dx=\frac1n$

Axoren

A function on the naturals is not infinitely differentiable in the space of reals or complex numbers, is it?

Leaky Nun

*$\int_n^{n+1}f(x)\ \mathrm dx=\frac1{n+1},\ n\in\mathbb N$

Axoren

01:24

@LeakyNun What are you using as $f(x)$?

Hey, @TedShifrin

Leaky Nun

I am trying to find such function

Ted Shifrin

Hi @Axoren and @Leaky

Leaky Nun

@TedShifrin Good morning.

Ted Shifrin

It's really (meromorphic or) analytic continuation once you have a functional equation that needs to be satisfied.

Axoren

@TedShifrin Maybe you could answer this swiftly and put an end to my confusion. Are functions only defined on the natural numbers analytic?

Leaky Nun

01:25

@Axoren Of course not.

Ted Shifrin

That makes no sense, really. You need an open domain to talk about analyticity.

Leaky Nun

Analytic continuation makes the function analytic.

Axoren

Then how is $\Gamma(x)$ an analytic continuation of $(n - 1)!$?

Ted Shifrin

But analytic continuation is normally done from one (open) domain to a larger domain.

Leaky Nun

Because $(n-1)!$ isn't analytic and $\Gamma(x)$ is!

Axoren

01:26

That's what I thought.

Now wait, hold on. I thought you could only analytically continue analytic functions.

Ted Shifrin

First of all, the gamma function is only meromorphic. I would not use the word "analytic continuation" therewith.

Axoren

That's just Wikipedia playing a dangerous game, then: en.wikipedia.org/wiki/Gamma_function

Ted Shifrin

That's not the right setting for analytic continuation. You can say it's a meromorphic function extending the factorial function on the positive integers.

No, wiki is moderately careful. They say that the gamma function is an extension of factorial and then that function makes sense on $\Bbb C$ if we allow poles.

Axoren

Oh, I see.

Leaky Nun

(Ignore me) $F(n+1)-F(n)=\frac1{n+1}$

Ted Shifrin

01:30

But they do abuse the language, in my opinion. They talk about analytically continuing the gamma function defined just on the positive real axis. That is wrong.

Leaky Nun

@TedShifrin No, that is not what the wiki says

> by analytic continuation to all complex numbers except the non-positive integers

Well, if I was able to solve that equation, I would be world-famous

so I just give up

Ted Shifrin

Oh, you're right. They are correct. They say the integral formula for the gamma function is analytic on all of the right half-plane in $\Bbb C$, and then they analytically continue. Perfect.

Anderson Felipe Viveiros

Hello, guys. I just have a simple question and Googling it didn't help too much. Are the multiplicities of the eigenvalues of a matrix in the minimal polynomial equals to the dimension of the corresponding eigenspace?? It seems I've already seen this result before, but I can't find it anywhere! Do you have some links, or whatever...

Ted Shifrin

@Leaky: Those are the harmonic numbers. I dunno what you mean about "solve."

Leaky Nun

@TedShifrin Functional equation: $F(n+1)-F(n)=\frac1{n+1}$

Axoren

01:33

That's some muddy language, in my opinion. I misunderstood it the first time reading it.

If I made a mistake, it's the source's fault, not mine.

Ted Shifrin

@Anderson: Not really.

For example, for the identity, the minimal polynomial is $p(x)=x-1$, but the eigenvalue has multiplicity $n$.

Anderson Felipe Viveiros

Even for the generalized eigenspaces?

I really think I've already seen something like that...

Ted Shifrin

If the minimal polynomial is $p(x)=(x-1)^k$, that tells you that the dimension of the largest generalized eigenspace is $k$.

But we have to be careful here.

That excludes regular eigenspaces :P

Leaky Nun

What is the current record for the time-complexity of bitwise squaring, in terms of number of bits $n$?

Bananach

Hi guys. Does this question have a positive answer? math.stackexchange.com/questions/1572743/…

Axoren

01:36

@LeakyNun I think it's $n \log n$ because you can alternate adding and shifting based on which bits are 1 and which are 0.

Leaky Nun

@Axoren Multiplication is $n\log(n)$, I thought squaring could be better

Axoren

Wait, hold on.

Anderson Felipe Viveiros

Thank you guys!

Axoren

I don't know if the information that the number is the same contributes to a more efficient multiplication.

It probably does, now that you mention it

But I don't see how immediately.

Ted Shifrin

@Bananach: I don't understand. In the original statement, what norm is being used in the first place?

Leaky Nun

01:39

@Axoren $(2n+1)^2=4n^2\mbox{ (constant time) }+\mbox{ (O(n) time) }(4n\mbox{ (constant time) }+1\mbox{ (constant time) })$

Axoren

@LeakyNun Are you working with truncated results? where the result only needs $n$-bits?

Bananach

@TedShifrin the norm infinity as well

Leaky Nun

@Axoren The result needs $2n-1$ bits

Axoren

When I said "I don't see how", I meant the means by which it attains the lower time-complexity is what eludes me.

Ted Shifrin

@Bananach: Isn't the question answered in the original thread? It looks like it to me.

Bananach

01:42

@TedShifrin I thought maybe they were referring to some counterexample which can be constructed by way of convolutions

Axoren

@LeakyNun That's a shame on the result's size. Were you working in a computer, you could cut the multiplication short because all the carries fall off into oblivion.

Leaky Nun

@Axoren Never mind, my algorithm is still $O(n^2)$ I believe

@Axoren Python. Or BigInteger, for that matter.

Ted Shifrin

No, go back to the original post.

Oh, good, MSE is down.

Leaky Nun

@TedShifrin Not for me.

Ted Shifrin

Well, dinner time for me. :)

Axoren

01:44

@TedShifrin Peace, thanks for the help earlier.

Ted Shifrin

Sure thing.

Axoren

@LeakyNun There was this method for bitwise field arithmetic that worked rather efficiently. I'm trying to remember what it was.

Leaky Nun

@Axoren Fast Fourier transform?

Axoren

Nope, simpler.

Leaky Nun

nice, because I don't even understand FFT

Axoren

01:46

Essentially, you would iterate over the bits of a number. If it was a 1, you add the original number to the result buffer. If it was a 0, you double the result buffer.

FFT is cool. Luckily, it's not needed here. Cool things are scary to use.

The benefit of the mechanisms you have there is that multiplication by 2 is essentially a bit-shift left.

Leaky Nun

@Axoren If it is a 1, you add the original number to the result buffer after doubling the result buffer

I just wrote a recursive online algorithm based on $(2n+1)^2=4n^2+4n+1$

Axoren

Right, I missed that part.

Leaky Nun

but it was $O(n^2)$ time I believe

Axoren

That may have been because of improper tail-recursion.

Leaky Nun

What do you mean?

Axoren

01:48

Python's interpreter will optimize your recursive functions if they're in tail-recursive form.

If the recursive call is the last statement in the function, you don't accrue additional variable assignments of function stack space.

So calculating your time and space complexity empirically will show higher time spent and higher memory spent.

Leaky Nun

So what should be the time complexity?

Oh, and by "I just wrote", I really mean "I just thought of"

Axoren

Depends on how you made your recursive function.

Leaky Nun

@Axoren What's your native (programming) language?

Axoren

Java for 10 years, Python for no where near as main.

C somewhere in there, back in 2009.

C++ is a mess. Touched it once, sent it to the pits never to return.

Leaky Nun

Then I'll write it in Java then

Axoren

01:52

Why?

Leaky Nun

because Java is your main language

Axoren

That's a terribly idea.

Leaky Nun

??

Axoren

Why would you write a function in someone else's native programming language?

Leaky Nun

Because I speak both Java and Python

and Java is my main language

Axoren

01:53

That's a better reason.

Java's compiler does have tail-recursive optimizations as well.

However, Java requires a lot more boilerplate before getting a single function to operate in it.

I like it because I like boilerplate, but I recognize that it's terrible for just getting down-and-dirty with a single function.

Leaky Nun

public long square(int i){
	if(i==0 || i==1) return i;
	if((i&1) == 0) return square(i>>>1)<<2;
	long temp = square(i>>>1);
	return (temp<<2) + ((i<<2) | 1);
}

definitely not tail-recursive, right

Axoren

Of course.

The recursive call is in the middle of the function, rather than the tail-end of it.

Leaky Nun

I don't quite care, I just count the number of iterations

in this case it is $n$

Axoren

Personally, I'd write recursive code in Haskell if I get to choose.

Leaky Nun

Let's get back to the main topic

This code is $O(n^2)$ right

Axoren

01:57

Let's ignore the degenerate cases of 0, 1, etc.

Your main work-horse if-statement has two cases.

Leaky Nun

I could write it in closed-form

Axoren

The first case performs $square(x/2)*4$

Leaky Nun

but it would be a pain in the butt

Axoren

The second case performs $square(x/2)*4 + c$

Leaky Nun

$+4x+1$

Axoren

01:59

Yeah, that thing.

That's $c$

Leaky Nun

and then?

Axoren

What's the cost of multiplying by 4?

O(n), right?

But every time you do it, you do it on linearly fewer $n$.

So it's as if at each iteration, you did it on $n/2$

Leaky Nun

@Axoren O(0.5 n^2) is still O(n^2)

Axoren

This becomes $n^2$

Leaky Nun

@Axoren Why not O(1)?

Axoren

02:02

What's the sum of $n$ from $n_0$ to $1$?

Leaky Nun

@Axoren 0.5(n0)(n0+1), which is still n^2

Axoren

Right.

Maybe I misinterpreted you, given that you were so quick to identify that relationship.

What did you mean by $O(1)$?

Which part?

Leaky Nun

I could use little-endian to make multiplying by 4 constant time

@Axoren multiplying by 4

Axoren

Hmm.

If that's the case, then you're left with $O(n)$ just considering the $square(x/2)*4$

But that's too amazing.

Leaky Nun

no, I analyzed it completely before

Axoren

02:05

You now have to consider case two in your workhorse if-statement.

Leaky Nun

26 mins ago, by Leaky Nun

@Axoren $(2n+1)^2=4n^2\mbox{ (constant time) }+\mbox{ (O(n) time) }(4n\mbox{ (constant time) }+1\mbox{ (constant time) })$

The $O(n)$ time is $+$

Axoren

Right, from case 2.

The only place where you do it.

Leaky Nun

yep.

So the best case is $O(n)$ and the worst case is $O(n^2)$

Axoren

Yup.

You have to calculate that out.

What does it mean for the average time complexity to be $O(f(n))$?

In a nice problem like yours, we can consider good-enough estimates of the average-time complexity.

Leaky Nun

In particular, if $m$ is the number of $1$s in the number, the complexity is $O(nm)$

Axoren

02:09

You're already ahead of me on that.

So, now consider $E_m(O(nm))$

Leaky Nun

So we fix the number of digits?

Axoren

$\sum_{m = 0}^n m O(nm)$

Leaky Nun

$\displaystyle2^{-n}\sum_{m=0}^n\binom nmO(nm)$

Axoren

I'm fairly certain we can just do it this way.

By the average number of flipped bits.

Rather than by the average of all possible bit-strings

Yours is more precise.

Leaky Nun

$\sum\binom nr=2^n$

Oh!

$r\dbinom nr=r\dfrac{n!}{r!(n-r)!}=n\dfrac{(n-1)!}{(r-1)!(n-r)!}=n\dbinom{n-1}{r-1}$

$\displaystyle2^{-n}\sum_{m=0}^n\binom nmO(nm)=2^{-n}\sum_{m=0}^{n-1}\binom {n-1}mO(n^2)=0.5O(n^2)=O(n^2)$

@Axoren right?

Axoren

02:24

I'm fairly certain you're on the money.

Leaky Nun

@Axoren What does "on the money" mean?

Axoren

Correct.

Leaky Nun

$\displaystyle\sum_{r\mbox{ odd}}^{2n}\binom{2n}r = \sum_{r\mbox{ even}}^{2n}\binom{2n}r$

Axoren

on the money

Leaky Nun

Thanks

Prove it by induction, lol

(just horsing around)

Axoren

02:26

I prefer proof by definition, myself.

It is because I say it is.

Leaky Nun

nice.

Axoren

So, I found out more about the harmonic numbers.

Apparently, there's a good way to define them with a continuous function.

Leaky Nun

care to share?

Axoren

$\frac{1 - x^n}{1 - x} = 1 + x + x^2 \dots x^{n-1}$

Leaky Nun

$\displaystyle\sum\frac1k=\sum\int_0^1(x^{k-1}\ \mathrm dx)=\int_0^1\frac{x^n-1}{x-1}\ \mathrm dx$

Axoren

02:29

Yup.

Leaky Nun

I just thought of that, lol

Axoren

It's disgusting yet elegant.

An abhorrent mess of beauty.

That function extends the Harmonic Number to the reals.

Leaky Nun

nice.

Leaky Nun

02:45

Is anyone interested in this?

Find $\displaystyle\sum_{n\mathop=0}^\infty\frac1{\tbinom{3n}{n}}$

Semiclassical

@LeakyNun Mathematica returns that as a hypergeometric-3F2 function, and doesn't simplify further

Leaky Nun

@Semiclassical Could you test Mathematica on $3n$ replaced with $2n$?

Semiclassical

heh, you read my mind

it comes back with $\frac{2}{27}(18+\pi \sqrt{3})=1.7364$

Leaky Nun

Good.

Could you test $4n$?

Semiclassical

in that case it comes back with a hypergeometric-4F3

so it seems to follow that pattern of $kn$ giving a hypergeometric-kF(k-1)

Leaky Nun

02:55

and I have no idea what a hypergeometric is.

Semiclassical

there might be a simplification, but mathematica doesn't know it.

ehh. it's just a fairly generic power series sum

Leaky Nun

and what is the power series?

Semiclassical

see here: en.wikipedia.org/wiki/Generalized_hypergeometric_function

Leaky Nun

I mean, what is the series that it returns?

Semiclassical

it's in there. that's pretty much the definition.

so when mathematica returns that, it's basically just a rewriting. doesn't really give anything deep.

Leaky Nun

02:57

It's impossible that Mathematica just doesn't know it.

Mathematica knows everything.

Semiclassical

hahahahahaha

yeah no

Leaky Nun

$\displaystyle\binom{3n+3}{n+1}=\frac{(3n+3)!}{(n+1)!(2n+2)!}=\frac{(3n)!(3n+1)(‌3n+2)(3n+3)}{n!(2n)!(n+1)(2n+1)(2n+2)}=\frac{3(3n)!(3n+1)(3n+2)}{2n!(2n)!(n+1)(2n‌+1)}$

@Semiclassical Could you help me ask mathematica this?

1.41432204432182039186500394383124895084527274214395277647

Semiclassical

03:18

that does seem to match what mathematica gives for your earlier series, yes

Akiva Weinberger

Is there any deeper significance to the fact that $|z|$ and $\operatorname{Re}(z)$ — at least, modifications of them — are the elementary symetric polynomials of $z$ and $\bar z$?

$|z|^2$ and $2\operatorname{Re}(z)$

Leaky Nun

@Semiclassical It still returns the geometric whatever?

Semiclassical

yep.

Leaky Nun

Fascinating.

Akiva Weinberger

I guess it means that anything symetric with $z$ and $\bar z$ can be rewritten in terms of $|z|$ and ${\rm Re}(z)$. Like, $z^2+\bar z^2=4\operatorname{Re}(z)^2-2|z|^2$.

arctic tern

03:55

@MikeMiller I don't think that's correct. The definition I gave said $g(\gamma)$ and $\gamma$ do the same thing to the particular element $\overline{m}$, but didn't say anything about doing the same thing to the whole fiber. Indeed if I understand correctly, principal $G$-bundles over $\Bbb S^1$ give counterexamples to [$g(\gamma)$ not depending on choice of lift $\overline{m}$] and to [$g(\gamma)$ acting the same as $\gamma$].

I collected all my thoughts into a question.

Mike Miller

04:32

Obviously you're right, changing the lift of basepoint conjugates the homomorphism by whatever g you changed the lift by.

Fix your lifted basepoint x, pick loops $\gamma$ and $\eta$ downstairs, abuse notation and call their lifts by the same letter. Say $\gamma(1)=gx$, $\eta(1)=hx$. Concatenate $\gamma$ and $g\eta$ upstairs to get the lift of $\gamma * \eta$ downstairs.

Then upstairs you end at $g\eta(1) = gh$ as desired.

My nonsense claim from before would imply that by changing the basepoint upstairs to $gx$, $g\eta$ would end at $hg$, which is silly as you note, arctictern.

@arctictern Meant to ping. If there's stuff in the question unanswered, let me know and I'll take a more careful look as my penance.

Tobias Kildetoft

06:49

Paper done. Now to read through it a final time, missing all sorts of obvious errors and typos because my brain knows what it was supposed to say :)

user243301

07:05

I've deleted a recent post, since a second deduced statement was obvious. The first one is

**Lemma.** If $M=2^p-1$ and $M=2^{p'}-1$ are two distinc **Mersenne primes,** thus we can take $p<p'$ then $$\sigma(M) \left( 1+\sigma \left( \frac{M'+1}{M+1} \right) \right) =2\sigma(M').$$
I don't know if this is interesting or was in the literature. Now if you take in your hands to think and take new taks about it, you are welcome.

Tobias Kildetoft

@user243301 what is $\sigma$?

sum of divisors?

Mats Granvik

The zeta function associated with a symmetric matrix.

Tobias Kildetoft

@MatsGranvik Pretty colours (I guess...?)

Mats Granvik

yes

user243301

I've deleted a recent post. The statement that still could be interesting, is the following:

**Lemma.** Tt is possible to show for a **Markov triple** $(a,b,c)$, that $$rad \left( \frac{a^2+b^2+c^2}{3} \right)=rad \left( \frac{a^3+ab^2+ac^2}{3bc} \right)rad \left( \frac{b^3+ba^2+bc^2}{3ac} \right)rad \left( \frac{c^3+ca^2+ab^2}{3ab} \right),$$
where $rad(1)=1$ and for $n>1$, $rad(n)$ is defined as the product of distinct primes dividing $n$.

I don't know if this is interesting or was in the literature. Now if you take in your hands to think and take new taks about it, you are welcome.

Tobias Kildetoft

07:13

@user243301 I am not sure I understand the point of posting these here. Also, what do you mean by taks?

user243301

@TobiasKildetoft yes the sum of divisor function, for example $\sigma(8)=1+2+4+8$

Tobias Kildetoft

@user243301 so both sides are just suitable powers of $2$

user243301

@TobiasKildetoft since I am deleting posts (marked as favorites of people), but were poor in their contents, perhaps it is interesting to some people. In the other hands I say that perhpas some people want try do more experiements. I believe that my two previous staments, as I've said are poor, but perhaps in hands of other user, that can modify those or it is a start point to deduce new things. Thanks

Bye Tobias and Mats good morning.

Taks was a typo, I want to say tasks

Cornelis

07:52

I have a question about deleting posts :D I have asked a question recently and figured it was full of miscalculations. So I think it´s of no use to anybody anymore. Do you think I should delete it? Is it considerd a right thing to do on this forum?

Leaky Nun

@Cornelis You can't delete a post if it already has answer

Cornelis

@lea

Leaky Nun

@cor

Cornelis

@LeakyNun ok, that solves the issue :D Thanks anyhow

Leaky Nun

welcome

user174558

09:24

@Cornelis You can always click delete and see what happens. There is no right or wrong thing, just do it. This is not a crime, lol.

Mats Granvik

"You have your way. I have my way. As for the right way, the correct way, and the only way, it does not exist." Friedrich Nietzsche

Mats Granvik

10:36

Is it possible to analytically continue an eigenvalue?

Tobias Kildetoft

@MatsGranvik an eigenvalue of what? Seen as a function from where to where?

Mats Granvik

@TobiasKildetoft An eigenvalue of a matrix.

Tobias Kildetoft

@MatsGranvik That is just a number. It does not really make sense to continue it analytically

Szabolcs

11:14

This is so strange: math.stackexchange.com/q/1832867/12384 Don't teachers use $\neq$ in school anymore? What is happening? Do they just use PowerPoint these days and type != ??

Juan Fran

lol @Szabolcs nope

it seems like that guy is just someone that got so used to programming

that he forgot the neq altogether

Tobias Kildetoft

12:14

Finally. Paper completed and submitted to arXiv

Andrew Thompson

12:32

@TobiasKildetoft Congrats!

Tobias Kildetoft

@AndrewThompson thanks

Tobias Kildetoft

12:51

Now to figure out a suitable journal to submit it to.

SteamyRoot

13:49

If you want to be 100% sure about the outcome of submitting your paper, there's always the JofUR

MickLH

@SteamyRoot Your name. It's not in my list. What root finding algorithm does SteamyRoot refer to?

SteamyRoot

None that I know of

It's just an anagram of my real name

MickLH

Invent one that exploits some insight about the way steam condenses into droplets in a closed space or something...

SteamyRoot

But if I ever come up with such algorithm, I now have a name to give it

Hahahaha

Agawa001

14:42

hey mathers

Obliv

14:53

@agawa I just realized you were talking to the chatroom as a whole. thought someone in here was named mathers :D

Leaky Nun

@Agawa001 bonjour.

Obliv

hi

Agawa001

@Obliv lol

@LeakyNun salut

Leaky Nun

@Agawa001 ça va?

Agawa001

@LeakyNun pourquoi pas ?

et toi tu te portes bien ?

Leaky Nun

14:57

oui

tu viens là avec une question?

Agawa001

je suis regulier dans ce chat et je frequente cette place par coutume

Leaky Nun

d'accord

Agawa001

parfois je viens pour poser des questions, d'autres fois je reponds aux questions, mais maintenent j'ai rien a contribuer

Leaky Nun

d'accord

Balarka Sen

@anon nah, i am a dumb person. or are you saying you're temporarily being me as in being dumb?

;)

Agawa001

15:03

@LeakyNun tu etudies le français dans un programme scolaire ou volontairement ?

Leaky Nun

@Agawa001 les deux

mais j'ai arrete le programme ilya longtemps

Obliv

guys is $(1~2~3) \circ (1~3~2) = (1~3)(2~3)$?

and is $(1~3~2)\circ(1~2~3) = (1~2)(1~3)$ if not, why not?

Leaky Nun

@Obliv What is $\circ$?

Obliv

function composition for cycle notation

these are elements of $S_n$ the symmetry group

Leaky Nun

never mind

Balarka Sen

15:09

@MikeMiller "But nobody likes me ..." I object. I think you're quite popular among the topology crowd here, a great teacher as you are.

Samuel Yusim

j'ai etudie le francais pour un moment il y a longtemps, bien que je ne sais pas comment on utilise ce langue coloquialmente parce que mes professeurs etaient enneyeuse. ...aussi, mes verbes sont probablement horrible apres 3 ou 4 ans

Agawa001

cette langue

ennuyés ou ennuyants*

Samuel Yusim

):

I'm at least satisfied I can still get a point across modulo a couple tiny problems

Leaky Nun

Tu sais, les agrements, ils sont horribles

Agawa001

quant tu te sers d'un pluriel qui regroupe m / f , l'adjectif qui le qualifie ne vient jamais en féminin

@SamuelYusim vos professeurs sont de quel sexe ? masculin ou feminin ou metissés ?

Leaky Nun

15:27

@Agawa001 On etudie francais?

en France

Krijn

@BalarkaSen Hey Balarka! I thought you died or something

Agawa001

@LeakyNun oui en effet

Samuel Yusim

actuelment, elles etaient tous femmes. je ne realisais, etrangement

Leaky Nun

@Agawa001 On etudie quoi?

Agawa001

@LeakyNun ce qui te plait

Leaky Nun

15:29

@Agawa001 J'ai voulu dire, on etudie quoi au sujet de francais en France?

Agawa001

@SamuelYusim mmm oui bizzare ! d'abord a cet occurence tu devrais dire "ennuyeuses" ou "ennuyantes" (more strange thing is how french beauty could ever annoy you lol)

Samuel Yusim

I was also alternatingly taught quebequois french and french french which is kind of weird

Agawa001

le quebecois est un accent non plus une langue !

Leaky Nun

@Agawa001 "quebequois french" is a valid term

Agawa001

hmm lemme check it out

Leaky Nun

15:36

just like British English and American English

Krijn

@Agawa001 What does "plus" do here?

Leaky Nun

@Krijn not "at all"?

Krijn

Ah, okay

My French is not parfait

Samuel Yusim

oui, mais il y a les differences. par example, en francais quebequois, on n'utilise un espace avant la ponctuation

Leaky Nun

@SamuelYusim par exemple?

Samuel Yusim

15:38

quebequois: "par exemple?", francais: "par exemple ?"

Leaky Nun

hein... interessant

je ne le sais jamais

Samuel Yusim

je pense que l'espace est vraiment bizarre

Agawa001

@SamuelYusim trop futile!

not as much as the two english accents differ from

Samuel Yusim

yeah, verbally they're definitely closer to each other

Agawa001

like "z" vs "s" prounonciation

Leaky Nun

15:40

@Agawa001 pronunciation*

for example?

Agawa001

in french it is spelled "prononciation"

Leaky Nun

in English it is spelled "pronunciation"

Agawa001

these two langs are overlapping in my head

Leaky Nun

in English, you also capitalize the language's name

Tandis qu'en francais, on ne capitaliser pas des noms des langues

Agawa001

@LeakyNun like booze and boose

Leaky Nun

15:47

@Agawa001 what is a booze and what is a boose?

Agawa001

its to be hungover

both has same meaning, but in different accents

Leaky Nun

I see, thanks

Agawa001

well there is much like this i hope someone more versed in english language can point out more differences here

Leaky Nun

@Agawa001 Well, color vs colour, capitalize vs capitalise

(AmE vs BrE)

Agawa001

oh yes

Leaky Nun

15:50

mais ils sont tous differences d'orthographe

Agawa001

but they are different

flavor and flavour counts ?

Leaky Nun

@Agawa001 same difference as color vs colour

Agawa001

ok thanks

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$Mathematics$ Mathematics