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12:00 AM
when people write something like x = (x_1, ..., x_n) they are implicitly doing this. the n-tuple is also an assignment of i to x_i, i.e., a function on {1,...n}
or i guess x = (x_0, ..., x_{n-1}) in your case :)
at least someone understands the struggle
wait what
12:16 AM
@Thorgott I admire / am in awe of your talents in that area, as you demonstrated by that proof sketch. I'm tryna learn it too!
It's like it's a big gap in my knowledge, and each time I bring a problem to an advanced form, it's like well would homology / cohomology theory apply to this, and the answer for now is always idk
I came up with though an integer-divisor homology - you probably have seen my original post on MSE
Don't know enough NT or homology to take it anywhere, but it does form a standard complex in several ways of cutting it
Let me know and I could teach you it in 2 paragraphs
I also found that if you reverse homology, namely ker $\subset$ image, then exact sequences are the same thing, and you can define "reverse-homology" in that way and it's also functorial in the same senses. But as far as actually doing real homology theory or galois theory, I'm very new.
I say galois because it appears in the cohom section of Weibel
My brain is like a sieve :) And a shitty one. Eratosthenes
@Thorgott knowing about advanced topics as you clearly do. If I say: canonicalize the space of finite directed, acyclic graphs in such a way that is efficient (< 1 second database search time) for up to $N = 256$ nodes say. Then what would you do to start out attacking that problem? I'm curious to know this, but it's not really something I could ask on MSE concretely
Canonicalize to either integer or string output, i.e.
I was thinking a group presentation could canonicalize a group, even an infinite one, but you'd need to canonicalize the list of relations some how, and then you're back in graphs
12:45 AM
@Thorgott @AlessandroCodenotti this is still open and I even gave it a bounty, you can earn 100 rep points if you answer it
@Jakobian that's funny 🤓 a linked answer in the first comment contains an answer I made a while back and I honestly don't know how to reproduce that math unless I did a review 🤦‍♂️
I don't use it enough!
@Jakobian My first guess is try it with modular arithmetic groups and $\Bbb{Z}\times \Bbb{Z}$.
If it's not counter on groups and then let the structures be rings and check then
Reason: the integers / primes conspire and never do what you want them to do
But maybe not because those are finite structs and you usually employ the discrete topology, in which case everything would work out nicely
2 hours later…
2:24 AM
Evaluate $I=\int\int\int_E y^2z^2+z^2x^2+x^2y^2dxdydz$ taken over the domain bounded by the cylinder $x^2+y^2=2ax$ and the cone $z^2=k^2(x^2+y^2). $
Is the region at all bounded? I don't think so...
How to evaluate the integral in this case?
@robjohn can you please help me with this?
@ThomasFinley can you get the first cylinder in a better form?
2:50 AM
@Obliv Yes, that's a circle of radius a centered at a.
Right, so that becomes a cylinder in $\mathbb{R}^3$ as $z$ can be any value. Which way does the cone open?
trick question it's both ways ok so now you wanna find the regions of intersection or something I haven't done multivar calculus in a while :D
the cylinder has radius $a$ and extends infinitely along the z-axis and the cone has curvature $k^2$ intersects the cylinder
@Obliv Yeah, but I think the cone opens on one side only. But then, both the cylinder and cone extends infinitely over the postive z axis and so, the region seems not to be a bounded one!
@Obliv The thing you wrote in your now deleted comment might indeed be a good possibility. Maybe the question wants us to assume those things i.e bounded above by a plane parallel to the xy plane.
I'm going to put it in calcplot3d
a resource my professor used to use and I now see why it's handy :P
@Obliv cool
3:06 AM
this is so confusing LOL
idk how he managed to use this
I added -sqrt(k^2(x^2+y^2)) and sqrt(k^2(x^2+y^2)) as separate functions
Looks like this
I don't know how to add the cylinder yet :P
@Obliv just when wrote about calcplot3d I tried using it, but after a minute I gave up...it's frustrating:P
@Obliv is this the whole graph?
even my professor had to fumble with it to get it to work
No, that's just the cone
@Obliv ok, then it makes sense...
@Obliv LOL
Maybe you can use your imagination to draw the cone. It's centered at x=a and extends along the z-axis with radius a
but even if you can't imagine it, I think there is some process like substituting in variables and solving the region of intersection etc
if the cone only opened up or down, then yeah you're right it wouldn't be bounded but since it can open both ways it does bound a region.
I get it.
@Obliv Since the figure is symmetric so, maybe I can only work with the upper part and then multiply the answer by $2.$
But the region should be bounded by two planes even in this case. So, I take take it the question wants the readers to assume that ofc.
3:14 AM
Not sure what you mean by two planes but yeah you could just work half of the volume and multiply it by two.
It has $\Bbb{Z}/2$ symmetry
$\Bbb{Z}/2 \times \text{ circle group}$
Actually has more than that but might lack continuity in your symmetries then
Are you talking about that cone I posted?
Yes the two kissing cones
Oh we're trying to find the volume bounded by that cone + a cylinder $(x-a)^2+y^2=a^2$
so I think even that volume has $\mathbb{Z}/4$ symmetry even but idk
It would have to have more symmetry than that to apply a calculus formula!
3:19 AM
Idk how that notation works :P but like you can cut it in half 3 times I guess
@Obliv any idea what will be the limits of $z$ ? I take the region $D$ to be the disk of the radius a centered at $(a,0)$ ...
all I know is $S_n$ and dihedral group lol
lie groups have to do with continuous symmetries and that jazz right
56 mins ago, by Thomas Finley
Evaluate $I=\int\int\int_E y^2z^2+z^2x^2+x^2y^2dxdydz$ taken over the domain bounded by the cylinder $x^2+y^2=2ax$ and the cone $z^2=k^2(x^2+y^2). $
@DanielDonnelly Really?
3:21 AM
@ThomasFinley if you want to you can cut it in half so start at z=0 and then go up to the function bounded above
@ThomasFinley he's talking about something else, don't worry lol
Considering the original problem above I thought it was really a calculus problem :D
which is the intersection of the two boundaries
@Obliv K, that makes sense!
I think you want to use a cylindrical $r, \theta$ integral
Forgot how they work lo
Honestly I'm really rusty with this I think you plug $z = \sqrt{k^2(x^2+y^2)}$ into the cylinder equation to get the equation of the intersection?
yeah you probably do want to do that lol
3:24 AM
plug it into one of those solving online calculators
symbolab or something
I don't think you can use symbolab on an exam though :P
Can't believe I learned all that calculus then almost never use it
tsk tsk
Calculus = weapons maybe
rockets, fusion devices, etc.
I mean you're not wrong but without it we'd also be in the 1600s still
3:26 AM
I stick to abstract algebra, where the only thing blowing up is the time complexity of an agoritm lol
@Obliv .Umm...the cone is indeed intersecting the cylinder 2 times (one below and the other above) . Note, that, I am talking about the upper part of the cone only
3:41 AM
I got $z=\pm k\sqrt {2ax}$
4 hours later…
7:57 AM
Can anyone make it clear to me as to why it is enough to prove Proposition 1.4.3 for radial functions only?
This is taken from Function Theory in the Unit Ball in $\mathbb C^n$ by Walter Rudin.
4 hours later…
12:04 PM
I can't take on any more projects, @DanielDonnelly, unfortunately. With a couple of conferences coming up, I'm very busy lately. Keep me posted though. I'll offer help where I can.
12:28 PM
I'm reading in my notes about the distribution of the sum of two random variables. In deriving this function, the author writes \begin{align}F_{S_2}(u)&=P(S_2\leq u) \\ &=P(X_1+X_2\leq u) \\ &=P(X_1+X_2\in(-\infty,u]) \\ &=\int_{x_1+x_2\in (-\infty,u]}f(x_1,x_2)\,d\mu(x_1)d\mu(x_2) \\ &=\int_{x_1=-\infty}^\infty\int_{x_2=-\infty}^{u-x_1}f(x_1,x_2)\,d\mu(x_1)d\mu(x_2),\end{align}and says the last equality follows by a "standard multiple integration formula".
I wonder, what formula does the author have in mind?
Is this by any chance related to Fubini-Tonelli's theorem?
@psie I would guess Fubini-Toneli.
ok, me too
12:58 PM
Suppose I have a recursive formula defined for any natural n, T(0) is a constant and T(n) satisfies for large enough n: $T(n) \leq T(\frac{2}{3}n) + c$ for some constant positive number $c$. Now say I amm interested to bound () $T(\frac{2}{3}n) + c$. I thought it could be more convinient if I could define another recursive formula $S(n)$ that will be equal to the formula at () but in terms of S and not T. If someone understands me and can guide me how this can be done I will be glad
1:32 PM
@XanderHenderson how would you interpret the region $x_1+x_2\in (-\infty,u]$? After all, the Fubini-Tonelli theorem states we can break up an integral over say $\mathbb R\times\mathbb R$ into two iterated integrals over $\mathbb R$. Here, the last integral are clearly those two iterated integrals, but I don't know how to interpret the region $x_1+x_2\in (-\infty,u]$ as a Cartesian product.
1:47 PM
In computer science, the Akra–Bazzi method, or Akra–Bazzi theorem, is used to analyze the asymptotic behavior of the mathematical recurrences that appear in the analysis of divide and conquer algorithms where the sub-problems have substantially different sizes. It is a generalization of the master theorem for divide-and-conquer recurrences, which assumes that the sub-problems have equal size. It is named after mathematicians Mohamad Akra and Louay Bazzi. == Formulation == The Akra–Bazzi method applies to recurrence formulas of the form: T ( x )...
1:58 PM
@Anacardium the claim is, that I see, that it follows from taking $g$ defined by $g(x) = \int_S f(\|x\|\cdot \zeta)d\sigma(\zeta)$
You'd have to show integral of this over C^n is equal that of f
@psie never mind, I think I have figured it out by writing out some stuff. We can write $f(x_1,x_2)=x_1+x_2$ and $x_1+x_2\in (-\infty,u]$ as $f^{-1}((-\infty,u])=\{x\in\mathbb R^2: x_1+x_2\in (-\infty,u]\}=\{x\in\mathbb R^2: x_1\in (-\infty,u-x_2]\land x_2\in \mathbb R\}$. The last set describes a Cartesian product.
2:29 PM
Good morning all!
This is not an algebra book so it's not surprising they'd sweep stuff under the rug, but I'd like to understand the underlying algebra for this problem a bit better:
I have the following vague notion: what we are actually speaking of here is the equivalence class $[\langle p, q\rangle ] = [\langle p, 1\rangle ][\langle q, 1\rangle ]^{-1} \in K[X] \times K[X]^\times / \sim$ (with $\sim$ the usual equivalence relation for the field of fractions). I am now trying to convince myself about how I would define this "$r$ evaluated at $x_n$" notion, since my book has sadly not defined it
What I do already have is a notion of a polynomial being evaluated at a given $x_n$
I do know the following. I can specify the unique $p,q$ such that $\langle p,q \rangle \in r$ with $q$ monic and of lowest degree (this is the analogue to $\Bbb Q$ and "lowest terms"). Then I can define $r(x_n) = p(x_n)(q(x_n)^{-1})$. But I feel like I should be able to have some more general definition which I can conclude equals the above
3:08 PM
@psie the proof of deriving the density of $S_2$, the sum of $X_1$ and $X_2$ being continuous, continues as follows. We have $$F_{S_2}(u)=\int_{x_1=-\infty}^\infty\int_{x_2=-\infty}^{u-x_1}f(x_1,x_2)\,d\mu(x_2)d\mu(x_1).$$ Now differentiate this equation with respect to $u$ and move $\frac{d}{du}$ inside the outer integral. The motivation given for this is "because the integrand is positive". I'm confused. Which result/theorem/proposition is this maneuver using?
Here, $f(x_1,x_2)$ is the density of $(X_1,X_2)$.
@psie write it as integral over $\mathbb{R}^2$ using indicator functions
@psie no
@psie this is just poor math
I think my professor proved this when $X_1, X_2$ are independent
sum of two continuous random variables is not necessarily continuous
@Jakobian what would the sets be for the indicator functions? Would it simply be $1_{g^{-1}((-\infty,u])}$, where $g(x_1,x_2)=x_1+x_2$?
when $X_1, X_2$ are independent then $f(x_1, x_2) = f_1(x_1)\cdot f_2(x_2)$ and you can do some simple algebraic manipulations to show that density is given by convolution of $f_1$ and $f_2$
@psie yes
@psie you might try to follow this argument just to see where it leads you, even though its not a good one
3:23 PM
yeah, I've seen this differentiation under the integral on a couple of sites, but with little motivation
it does make sense under some conditions but I don't think they hold here
but yeah you can say, assume we can do that and we obtain density
that still doesn't show the random variable is necessarily continuous, you just have part of the random variable that is continuous
for it to be continuous this density would have to integrate to $1$
I haven't done this calculation before, maybe it is continuous except for trivial cases
i.e. $X_1+X_2 = \text{const}.$ lets say
I think this is a dupe but can't find it.
This is some sort of a variation of broom space
@SoumikMukherjee I don't know about that, its asking about a pretty specific space
3:33 PM
ohh, but I feel like I have seen this space on this site earlier
maybe its an exercise in some book, I don't know
> This isn't a homework. This was a question I had on my exam: Analytical Topology and was interested wether my answer was correct or not.
who knows what the source is
3:57 PM
I see what you mean now, Jakobian. \begin{align} F_{S_2}(u)&=\int_{\mathbb R\times\mathbb R} 1_{x_2\leq u-x_1} f(x_1,x_2)\, d\mu(x_2)d\mu(x_1) \\ &=\int_{\mathbb R}\int_{\mathbb R} f(x_1,x_2) 1_{x_2\leq u-x_1}\, d\mu(x_2)d\mu(x_1),\end{align} which is the same thing as $F_{S_2}(u)=\int_{x_1=-\infty}^\infty\int_{x_2=-\infty}^{u-x_1}f(x_1,x_2)\,d\mu(x_2)d\mu(x_1).$
It's very tempting to differentiate this...
just assume you can do that
1 hour later…
5:24 PM
did the color of the tags change?
5:53 PM
Q: Seeking feedback on tag colors update

PiperUpdate February 15th, 2024 Thank you for providing the requested feedback on the original post. From the discussion, it seems like Option 3 and Option 4 were more favorable. Therefore we have taken that into account and settled on Option 3 as the base style due to its cleaner aesthetic. Since the...

seems like this is because of feedback from the community
6:11 PM
tags look bolder now
such an eyesore
they should bring dark mode.
7:12 PM
7:27 PM
@Pizza hi
7:38 PM
@SineoftheTime I was trying to solve the 2nd point of that line exercise. However, in the end point 1 was correct. It had no solution.
can i tell you what ive done?
Given in space an orthogonal monometric reference, consider the line $r$ represented by:$$
r: \begin{cases}
x+y = 0 \\
z = 0
$$the point $A(1,1,1)$ and the plane represented by
$$\alpha : x + y + z + 1 = 0$$Represent:
$(2.)$ The line passing through $A$ that intersects and is perpendicular to $r$.
this one
ok, I remember it
ok wait
i write
From $x + y = 0$ and $z = 0$, the direction vector of $r$ is $\vec{d}_r = (1, -1, 0)$.
The line through $A$ should have a direction vector perpendicular to $(1, -1, 0)$. Let $\vec{d} = (a, b, c)$ be the direction vector of the desired line. Then:
$(a, b, c) \cdot (1, -1, 0) = 0 \implies a - b = 0 \implies a = b$
so a simple perpendicular direction vector
can be
$(1, 1, c)$. To ensure the line intersects $r$, set $c = 0$, making the direction vector $(1, 1, 0)$?
can i do this ?
can you elaborate on why $c$ has to be $0$ ?
7:55 PM
wait, maybe c can't be 0
to ensure that do i need to solve
$L: \begin{cases}
x = 1 + t \\
y = 1 + t \\
z = 1 + ct
this ?
I want this line to intersect
$r$ where $z=0$.
can i do like that?
oh yes
so you wrote the parametric equation of $L$ right?
with $c=0$ I have an error
@SineoftheTime yes
can you write the cartesian equation?
do you want me to convert it?
8:10 PM
ok but wait
I got $c=1$
looks correct
now to convert
From the equation $x = 1 + t$, i can solve for $t$ as $t = x - 1$.
$y = 1 + (x - 1) = x$
$z = 1 + (x - 1) = x$
The Cartesian form of the line $L$ is $x=y=z$
it seems correct, no?
This describes a line in 3D space where all coordinates are equal, thus passing through points like $(1, 1, 1)$, $(2, 2, 2)$, $(3, 3, 3 )$, etc.
8:19 PM
- The line intersects with $r$ which is defined as $x + y = 0$ and $z = 0$, specifically at the point where $t = -1$ leading to the intersection point $(0, 0, 0)$.
- The line is perpendicular to $r$. The dot product of the direction vector of the line $(1, 1, 1)$ and the direction vector of $r$ $(1, -1, 0)$ is zero, confirming perpendicularity:

$(1, 1, 1) \cdot (1, -1, 0) = 1 - 1 + 0 = 0$
so there is a solution
Can I ask you for advice
i have a math book
but sometimes I don't understand what I read, and I get stuck
do you have some advice
because I don't want to learn things without knowing them
are you talking about theory or exercises?
what book are you following?
8:26 PM
Lezioni di Analisi matematica due, sbordone, marcellini,fusco
lo conosco
è normale fare fatica, perché molti risultati sono dimostrati nel volume 1
e soprattutto non so che base hai in analis 1
@Koro the Italian?
io personalmente quando non capisco qualcosa, ci sto un po' su anche per molto tempo e dopo chiedo a chi ne sa più di me
@Jakobian yes
8:29 PM
una cosa da non fare è arrendersi subito, cioè se non capisci subito devi cercare di arrivarci da solo perché lo studio consiste in questo
si esatto, ma poi tu vai avanti o ti fermi proprio a quel punto?
cioè se non capisci subito, rimanici su e non chiedere subito
@Pizza delle volte bisogna andare avanti perché magari quel passaggio lo capisci dopo
però non ha senso saltare tutto quello che non capisci, quindi magari vai avanti di qualche riga e poi torna indietro
se proprio non riesci, metti da parte e poi chiedi a qualche tuo compagno di corso, o al prof oppure cerca in rete
legend says that if you say pizza five times in the mirror at midnight, you'll become Italian
ma tu personalmente come studi un libro, ti appunti qualcosa su un quaderno , tab ecc ?
no, leggo direttamente e vicino a me ho sempre un foglio in cui dimostro le cose lasciate per esercizio oppure scrivo i passaggi che non capisco a modo mio oppure facendo esempi
8:33 PM
@Jakobian you've tried?
hopefully the Italians in chat will forgive me for making a stupid joke about Italians
@SineoftheTime e invece per le dimostrazioni? tu consigli di farle subito o di lasciarle alla fine?
dipende, stai studiando per preparare un esame oppure per conto tuo?
le dimostrazioni sono l'essenza della matematica, quindi non ha tanto senso saltarle. poi molte di queste sono costruttive
@Pizza I tried and penso che mi stia succedendo qualcosa
aspetta cosa sta succedendo
8:50 PM
@SineoftheTime preparare l'esame di analisi ii
scusa se non ti ho risposto stavo facendo una cosa
se sono dimostrazioni tecniche, in cui ti devi ricordare tanti passaggi, se le studi un mese prima probabilmente te le dimentichi
quindi puoi studiare un po' prima dell'esame
poi dipende quanto tempo prima inizi a prepararlo
has anyone ever eaten croquembouche? How is it?
@SineoftheTime io vorrei dare algebra lineare e geometria ed analisi ii
geometria una data è a giugno
analisi ii a luglio
@DanielDonnelly I mean, this sounds like a CS question, so it probably depends on what application you have in mind and what context you are working in. A standard CS textbook probably has better advice than I do on this matter.
@EE18 no reason for this amount of complication, they're just dividing two functions
@SineoftheTime tu mi consiglieresti di fare 1 materia al giorno oppure entrambe ogni giorno? oppure magari mi concentro di più su algebra lineare che è più vicino rispetto ad analisi?
9:02 PM
quanti crediti è analisi 2?
In Italian, 'you' has forms?
dipende da quanto sei preparato, però se vuoi dare subito geometria dedica più tempo a quello
like in francais, there are 'tu, vous' for 'you'.
e magari ogni giorno fai 3 o 4 esercizi di analisi così rimani in forma
@Koro also in italian
9:04 PM
Ok. So how do you know when to say which?
tu or vous?
@SineoftheTime e ma comunque per fare gli esercizi devo leggermi un po la teoria
'tu' and 'voi' for 'you'. 'tu' is singular, 'voi' is plural
l'orale di algebra e geometria però si conserva fino a marzo
@SoumikMukherjee @Jakobian it works!
@Pizza dipende da tante cose, anche da quante ore al giorno riesci a studiare. In questi casi, devi gestirti tu perché conosci te stesso
I see, thanks. The usage that I learnt was: voi for showing respect and also for plural, but tu for informal usage.
9:06 PM
voi for showing respect was used a century ago I think
now we use "lei"
so for example when talking with professor we don't use "tu"
@SineoftheTime si vedrò io come gestirmi con le ore
personalmente tu come hai trovato analisi ii?
One of the reasons I love english is because it doesn't have these forms of 'you'.
@Pizza beh io avevo 12 crediti quindi era abbastanza tosta
ma negli esami gli es erano sempre gli stessi più o meno
ah tu frequenti mate immagino?
integrali doppi, tripli, coordinate polari, sferiche, integrali di superficie e di linea più successioni di funzioni e edo
@Pizza sì, tu? Ingegneria o fisica?
@Koro yes, english is so pragmatic as a language
9:11 PM
@SineoftheTime ing informatica
in my native language as well, there are forms of 'you'. And that's why I tend to avoid it unless I'm talking to close people.
analisi ii ho fatto:
in english, I say 'you' to anyone ranging from a child to an elderly and no one gets offended.
successioni di funzioni e serie di funzioni, eq differenziali del primo e secondo ordine, integrali doppi, tripli, funzioni in due variabili, e ora stiamo facendo le forme differenziali
oh. I totally forgot that EE18 had a question
9:12 PM
sì è abbastanza standard
penso faremo anche gli integrali di superficie
But in the languages where 'you' has forms say tu et vous, usage of tu could offend some.
similarly, vous could offend some.
@Pizza eh sì, per gli ingegneri sono fondamentali. Poi farai anche tutti i teoremi degli integrali di superficie
best imo is to avoid them altogether to avoid any trouble.
@SineoftheTime da te hanno fatto prove intracorso?
9:15 PM
sometimes it happened to me that when showing respect and not saying "tu", people got offended cuz it's used when talking with old people
@Pizza sì, però il nostro corso è annuale
@SineoftheTime riguardo ai libri , conosci il libro Algebra lineare e geometria di Enrico Schlesinger della Zanichelli
What are all the languages that don't have formal and informal versions of you?
English is one.
apart from it?
@Pizza no, ho studiato dal libro che ha scritto la mia prof
Enrico Schlesinger è il tuo prof?
we use equivalent of "Sir" to be respectful, but its also used to distance yourself from the person you're speaking to i.e. you're speaking to a stranger lets say
@SineoftheTime no è l'autore del libro
a me hanno detto che va bene qualsiasi libro
9:19 PM
well, you don't use it to respect your elders or anything its more when you're distant to the speaker
or Ma'am I suppose
I hear polish is one such language.
no forms of "you".
there are, "ty" and "wy"
based on if you're describing multiple people or not
sì, algebra lineare più o meno sono sempre le stesse cose. matrici, spazi vettoriali, applicazioni lineari, diagonalizzazione,...
@Koro sign language
oh like formal vs informal. Yeah I guess we don't have them
you'd just use equivalent versions of "Sir" or "Ma'am"
io al secondo anno avrò come materia metodi matematici per l'ingegneria, sarebbe una specie di analisi iii?
so che serve sapere abbastanza bene analisi ii e algebra lineare
9:23 PM
or versions for multiple people of those words
@Pizza sì è vero
Pan, Pani, Panowie, Panie, Państwo
@Jakobian I see. So Sir = Mr. = Pan there.
i'm going , bye!
thanks for the advice @SineoftheTime
Polish is interesting :) No forms of 'you'.
9:28 PM
@Jakobian male singular, female singular, male plural, female plural, plural (but both)
one interesting thing with Polish is that there's no pronouns and there is a strong distinction between "he" and "she"
@Pizza g'night
pronouns are pretty much something you find in English language because of the nature of the language itself... not necessarily in different languages
and pronouns would hardly work in our language since gender is embedded into words
its not like in English where you just have pronoun at the start, but whole sentence depends on the gender
among others
One problem that I faced speaking Hindi, (which is not my mother tongue) is that verbs are different gender wise.
@Jakobian same in Hindi and in French too.
@SoumikMukherjee yeah I think that's precisely what's happening in our language
9:34 PM
So in Polish, verbs are different too
Oh I see
also in Arabic
each little detail changes prefixes and suffixes of words in a sentence
@SoumikMukherjee many people from east have this issue but it's temporary really.
'train' is the most confusing one I hear.
'train' is feminine in Hindi :)
people think it's complicated but in Polish there is a lot of patterns to it, albeit you need to get used to it I guess
@Koro Yes, I have fixed it now:)
9:37 PM
But I avoid speaking hindi to people that I don't know well. I told the reasons above.
@Jakobian so there are articles too I suppose which vary based on gender.
Le fille
La garcon
Languages are interesting :).
no, there is no articles in Polish
actually something you need to get used to when learning English as a Pole, I've been trying to explain the concept to my dad not that long ago (he barely speaks English)
One great think about Bengali is that there is no order. e.g In English people say I am eating salad but nobody says salad is being eaten by me. But in Bengali it is perfectly fine to say either.
passive vs active form
9:42 PM
same in Polish, we have a lot of freedom with order of words
@Koro haha
Assamese sounds awesome. I think it would be easier for me to learn it.
@SoumikMukherjee Many languages are like that. But such languages typically have other features which mark the grammatical role of a word.
I think if one knows Assamese, then they can understand Bengali too.
9:44 PM
yeah, prefixes and suffixes thing is what makes it so that our language have freedom of order in which you say the words
@XanderHenderson Ohh
Cat is eating chicken.
Chicken is eating cat.
english is a sensitive language
@Koro what do you mean?
I mean one has to be very careful with the ordering, punctuations etc. while writing english.
@Koro This, or something similar, is true of all languages.
9:47 PM
one thing in Polish is that we skip pronouns, and as far as I know this is also present in Russian for example, but way more common in Polish, you almost always skip pronouns
@Koro And vice versa. I think Bengali is an easy language to understand. We had some friends from Odisha who could understand what we were talking.
@Jakobian Common in Russian, too. The verb congugations make the meaning clear.
I just said that its common in Russian
Jakobian: I see. Polish as well as Russian is a slavic language.
@SoumikMukherjee Odisha has similarity to Bengali, I think. I have also observed what you said.
Even I can understand some written Bengali but that's because sometimes they look similar to Hindi.
when I hear Russians speak Polish, they often don't skip the pronouns and this makes them sound a bit odd sometimes
9:50 PM
like Baranagar.
or Dunlop
(subway names)
One thing I wonder that whether a native English speaker and a non native speaker would decipher an online tone differently
@Jakobian (1) You said it occurs. (2) I was agreeing with you, and expanding on your point.
lol, you two are always fighting :)
@Koro ooh
@Koro Yes, the alphabets are similar looking
there's also more words for "to know" in Polish than in Russian, which is some of the differences that were striking to me
9:54 PM
@SoumikMukherjee Some of bengali comes from sanskrit but not all of it.
And I think because of this similarity, I can understand some words-written or spoken.
'Sati pratha' -same word in hindi and bengali
@Jakobian that feels very Worf-Sapir...
'padchaari' (pedestrians) -same in both the languages. Even though, it's used in formal settings in hindi.
suprabhat -same in both the languages
कोरो কোরো
there are no 'maatra' in Bengali?
@Koro There are
@Koro pathchaari
9:57 PM
this is some of my experience with Russians that I had experience with that were trying to learn Polish, so it is a biased view
@SoumikMukherjee yes :). Someone was saying it the other day.
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