#help-49
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which solves for $m = -1$ and $n = 1$
1 divided by 0 equals Infinity
yep, thats the only solution
ty
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How is infinity one of the answers ?
wht is the ques?
Looks odd
Yeah
Took 2 days to reach here 😭
So does anyone have any idea why infinity is one of the solutions
I know that m=infinity implies 90 degrees but how does solving that equation give m=infinity?
By their logic every linear equation has infinity as one of its solutions though!
like put m2 = infinity in the first equation after (2) and it makes somewhat sense
it reduces the degree of the equation
Wdym by that ?
then, If a quadratic equation gets reduced to a linear one then infinity is one of the solutions?
well if the quad is in m the equation of slope
Umm, so if we were supposed to get 2 slopes but only got 1 at the end then infinity is one of the slopes ?
it's abuse of notation
if instead you consider the equation ax + by - 5/2 b = 0 for the line
you end up with the following equation for a and b
,w (a + 3b - 5/2 b)^2/(a^2 + b^2) = 1
in such a case, you arrive at the lines x = 0 and 3x + 4y - 10 = 0
yeah, bcz slope of vertical line is undefined and we can't get to the answer by algebra (someone else might explain better, my internet is horrid tdy)
i mean it is just abuse of notation and when interpreted literally is utter nonsense
one shouldn't set up the equation like that because a vertical line has no representation as y - 5/2 = mx
to set up the line equation that is agnostic to whether it's vertical, you should just use a form like ax + by + c = 0 where there is no problem
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just gave my maths exam
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In my book there's the following problem:\
Suppose that we have $n$ boxes numbered $1,\ldots, n$, and that the $i$th box contains $r_i$ red balls and $n_i$ black boxes, where $r_i,n_i\geq 1$. Imagine that one chooses a box uniformly at random, and then picks a ball (again at random) in the chosen box. Compute the probability that the $i$th box was chosen knowing that a red box was picked.\
Is there a typo in the last sentence? First the author speaks about red balls, then about red boxes. How do I approach this?
psie
yes likely a typo
So should it be red balls in the end instead of red box? Or perhaps black box instead of red box?
red balls
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How should I start in this question?
simplify the Re(trig bs)
maybe writing c and s for cos(θ) and sin(θ) could save you some effort
yes
alright
$Re(\frac{2\cos^2{\theta} + 7\iota \sin{\theta}\cos{\theta} -3\sin^2{\theta}}{9\sin^2{\theta} + \cos^2{\theta}})$
Prathmesh
ohh
yea
thanks
in this question should I first use that property that z + z' = 2Re(z) and then put z = x+iy or is there any better and short way of solving this ?
for the first section of the problem yeah
if would simplify to $Re\left(\frac{z-1}{2z+i}\right) = 1$
and then i guess nothing to do other than z = x+iy
yea
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Can i get a hint about the following integral
I did complete the square but what i do with the x in the numerator
P(D') + Q = x, where D' is the derivative of the denominator
P and Q are real numbers
x/((x+1)² + 1) = (x+1 - 1)/((x+1)² +1) no ? And then you separate and proceed with u = x+1 for the first integral
then you can separate into two integrals. one of them can be solved via a u-sub and the other by trigonometric sub
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how to find residue of $\frac{1}{(x^2+1)^3} $ at $e^{i\frac{\pi}{2}}$
Slowaq
solving this limit seems to he hard
$$\lim _{z\to e^{i\frac{\pi}{2}}}\left((z-e^{i\frac{\pi}{2}})^3\frac{1}{(z^2+1)^3}\right)^{''}$$
Slowaq
nvm i can factor out denominator
oh be careful, you're missing a factor of 1/2! in front
ah of course thanks for pointing it out!
no worries!
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,, (P) \begin{cases} \sum_{i=1}^4 z_i \ra \max & \ 3z_1+z_2+4z_3 &\le 1 \ z_1+2z_2+5z_4 &\le 1 \ z \ge 0 \end{cases}
can somebody spot my mistake
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Help please
Have you set an equation up?
like with modular arithmetic?
No just the equations you need to do this
I don't think you'll need any modular arithmetic
ok
so x+y+z=20 and 108=2x+5y+7z where x, y, and z are positive integers greater than or equal to 1
i heard of it but i forgot what it was
let me search it up rq
ok yeah i remember it now
Do you know how to use it in this case
oh ok
From these two, can you get a simpler equation in 2 variables that we can work with
yeah we can take mod x and mod y for both equations
oh wait or you could just get mod x i think that'll work
yeah i am taking the mod with the smaller coefficient to make it easier
you can get it down to 1 variable lol, you have 2 equation for 3 variables
Plotting it into 3d graph gives a line
I assumed this was what his unit's about
Besides it reduces the tedious casework
Ig so
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Jesus christ
<@&268886789983436800>
<@&268886789983436800> idk
It’s not that deep bro calm down
Same thought 🙈
Ok
<@&268886789983436800>
I need help
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hello
indeed
the third vector is 3*first + 2*second
then this is false
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that is indeed false. it is true only for sets of two vectors
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✅ Original question: #help-49 message
It says "others"?
naw cloud is right
it's not a scalar multiple of either
it's a linear combination of the others, which would be a more generally true condition
that works
thanks
a set of more than 2 vectors in R^n is lin dep if at least one of the vectors can be written as a linear combination of the other vectors right?
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hi
Yes
if we deal with n by n matrcies it's just always n then right
yeah cause m = n
no, the co-domain is always greater than or equal to the range
so the range is a subspace of R^m
ok
it can be all of R^m when?
identifiy matrix right
identity matrix from R^m to R^m
but it doesn't have to be identity
you will learn later
it just has to be "surjective"
idk where you are at in linear algebra
but there is a ton of theory to develop
regarding surjectivity
injectivity
and such
the last thing we learned is single value decomp
as I've now realized, the order of math curriculum is far from homogenous
we hit IMT, eigenvectors and markov chains
I haven't hit any of that
yes imt mb
but I am quite knowledgeable in linear maps
for n x n matrices, the codomain will be the same as the image unless the determinant is zero
and other equivalent conditions
ok good
help me out u two ima cover all of linear algebra in 2 hours
let's see what yall know
is your class proof-based
nope
oh great
a lotta shit though
than u probably can cover it all
well idk if u need to know injectivity and surjectivity then for your test
you tell me
and I can explain them
i don't think we're that theory based
wait
I guess my version of it is far different from a non-proof based course
so you've just covered matrix operations, determinants, vector spaces, inner product spaces, eigenvalues, and linear transformations?
does injectivity have different phrases
one-to-one
sounds right
i mean not yet
we covered all that in class, but not in my self review session
also we covered a lot of stuff on orthogonality
then some symettrical matrices stuff
and markov chains (im bad at)
yeah thats typical
ok good
im reviewing everything while doing problems on them
then i'll do some practice tests
still gonna bomb this final but i'll take a B
so image is the specific output vector pretty much
you should be good, the only "proofs" you will do are just showing something is a vector space, or something is an inner product
everything else is just memorizing how to solve a specific kind of problem basically
yeah exactly this
we have hard true or false questions
like conceptual questions (e.g. definitions, properties)?
yeah pretty much
could you give an example
yeah for stuff like this you will also need to remember theorems from your textbook and/or lectures
yeah they're tricky and niche
some of them are easy theorems, others are kinda a stretch on theorems like the 4 sub-spaces one was kinda hard (question 1)
yeah these problems are pretty hard, like for the first one you kind of need to memorize that its true
you got 3, 4, and 6 wrong
oh i misread it nvm
how tf did u know
some of them you can intuitively think through
i see why 4 is false tho DMANNIT
4 is false because if the vectors are linearly dependent, u can't do gram schmidt
yeah, its tricky
oh P^2
it requires you to remeber the definition
but also, this will probably just be one part of your final. since your time is limited, you should focus more on problems that you're struggling with
for sure ima lock in
linear transformations is something I need to master bbecause i don't have a good grasp on it
what textbook are you using?
not as intuitive as eigenvectors
im using my professors slides who uses the book linear algebra and it's applications 6th edition by Lay
oh lord
i thought that since W is a m-subspace of R^n, it must be less or equal to n. I know the dim(W) + dim(Wperp) = n, so i just did n-m. does that make sense or is my approach ass
we were given that W = m btw
yeah
wow i explained that bad i edited it
is consciousness something from our brains or a metaphysical thing
@wraith dirge @hybrid crow
professor said this is very powerful
but what does this mean in words?
consciousness is physically located in our brains
and is just like electrical impulses and neurons fired collectively
oh yeah this is op
i disagree but im not gonna argue that
because i sent this in the wrong chat mb
this basically means every linear map is uniquely determined by its action on basis vectors
because T is a linear map and has the properties of both additivity and homogeneity
knowing where the basis vectors goes
allows us to derive
where any other vector goes
since every vector is just a linear combination of those basis vectors
naw you lost me hold on
i'ma give u an example rn
the map of any vector is a combination of the maps of the basis vectors
geometrically, if you know where the basis vectors are mapped by T, you can figure out where any vector will be mapped
Suppose $v_1,...,v_n$ is a basis of $V$. Define $T \in \mathcal{L}(V,W)$ by $T(v_k) = w_k$ for each $k = 1,...,n$. Now, let $v$ be an arbitrary vector such that $v \in V$. Since $v_1,...,v_n$ is a basis, we know that $v = a_1v_1+\dots + a_nv_n$ for $a_j \in \textbf{F}$. From this, we know that $T(v) = T(a_1v_1+\dots + a_nv_n) = a_1Tv_1+\dots + a_nTv_n = a_1w_1+\dots + a_nw_n$.
holo_morph
Essentially, this shows that you can define a linear map T just based on its action on basis vectors
then you can "linearly extend" it across the domain
naw you lost me
to see what it does to any other vector
is V a subspace?
V is an arbitrary vector space
like is it R^n
bc you can represent any vector as a linear combination of the basis vectors right
as is W
yeah this
let's say u have a mystery transformation matrix right
since the transformation map is linear, you can also represent the map of any vector as a linear combination of the maps of the basis vectors
you know what the outputs are for e1 and e2
and u wanna find the output for an arbuitrary vector in R^2
you just do the first entry times e1 and second times e2 right
ok
v = a_1e_1 + a_2e_2
where the a's are scalars
then apply the linear map/matrix to both sides of the equation
Tv = T(a_1e_1+a_2e_2)
now the right side can be simplified
that's essentially the matrix A
using the properties of a linear map
Tv = a_1Te_1 + a_2Te_2
so we have expressed Tv
as a linear combination
damn ur proofs class got u messed up if u know this shit like the back of ur hand
of the images of e_1 and e_2 under the map T
I know it better than that
are u a TA or something
ok need to grind thanks tho
can't lie i heard gpa doesn't matter that much
so idc if I get a b here
I've never heard that
in college not high school
homogeneity?
btw Mittens
it depends what you do after college
that second property
you've probably seen it used in integrals
integral of 3x = 3 * integral of x
typa deal
or when differentiating
integral of 3x is 3x^2/2 right
and the first property
integral of (f + g) = integral of f + integral of g
anyways what I am trying to highlight here
is that differentiation and integration are both linear maps
because they follow the properties
and you can even
represent them
as matrices
is this a multivariable thing
probably
I see green's theorem typing now
Property of a linear transformation can be summed up as follows:
$T(a\mathbf v + b\mathbf u) = aT(\mathbf v) + bT(\mathbf u)$
Just to leave visual reference
thanks
You can see that line combines both core concepts.
And V and U are vectors
Always remember that vectors are in reality mostly anything
looked up green's thorem it looks like multivariable is gonna be hard
Functions can be vectors, etc...
yeah i see it
you dont need to prove greens theorem or anything in multi
applying it isnt that complicated
for calc3, greens and stokes are just a tool
the problems will be simple unless you have a bs professor
if youre doing real analysis tho 
do electrical engineers need to do real analysis
no
well kind of but not really
Depends on your college and country, but most engineering carreers settle with Calculus instead of Analysis
Luckily.
Thats calculus too.
the theoretical side of electrical engineering definitely needs real analysis, but if you are working as a regular electrical engineer you dont. i would assume you will fall into the latter category
couldn't tell you
They have a lot in common even if they have different scopes, and Analysis is certainly a lot harder since its centered around proving what Calculus has been doing since Newton.
Most techniques sat "unproven" for a while > more like not a formal proof of them actually working
it depends on the professor you have and how easily you understand the concepts. multi will probably be around the same difficulty though
multi and differential equations involve a lot of calculus, so it will depend a lot on how well you did in calc 1-2
linear algebra is (mostly) unrelated
ask me a question in calc 1 and 2 real quick
uhh integrate 1/(x^3sqrt(x^2 - 1)) from sqrt(2) to 2
wdym by sqrt 2 to 2
the limits of the integral
it is
i failed bc but took a placement exam in my uni and passed
but in bc we didn't learn trig sub
you didnt?
yeah i don't recall it
im pretty sure thats on the exam
is it secant or sin
secant
naw bro this is kinda hard i'd have to re-learn it
but re-learning is very easy if u learned it once already
that was just an example problem, i was just saying if you did okay in calc 1-2 you will probably do okay in multi and diffeq
that's good
differential equations is lowkey easier than multi, a lot of it is just memorizing specific techniques
if you're an ee major, you probably are only required to take those 5 classes (calc 1-3, linalg, diffeq)
depends on the program though
yeah we do need to do all that
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this matrix is size 3 by 3
it is

does this mean its a dumb question
missing some details here
also are you sure co-domain is the word you're looking for?
should i say range
that seems better yes
i think co-domain is better actually
and by 2 and 3 you mean dimension 2 and dimension 3 right
i think it would be better to say the codomain of a 3x3 matrix is R^3 regardless of its range
2-dimensional is not the same as R^2
oh so it's range is R^2
no
its range is like, a subspace of R^3 that looks 2 dimensional (but still made of elements of R^3)
see how in this case, it actually outputs vectors which have 2 components, they are literally in R^2
Ax is linear combination of columns of A with components of x being coefficients. So the range is simply the column space
wait this kinda makes sense add more stuff so it clicks
Its dimension thus is the dimension of the column space, namely the rank
i gotchu thanks bro
Okay, Ax is this
Okay
so it's dim(col A) which is basically rank
Yes
range
range = column space, they are exactly the same
what? really
the co-domain isn't something you compute
well for matrices yes
codomain is just what the target of the map is, can be as large as you want.
the codomain is just set of all vectors of the type the matrix spits out. so because the matrix outputs vectors in R^3, its codomain is R^3, even though it will not output every possible vector in R^3
When you give a map you have to give a domain and a codomain. The codomain can be anything, image is another thing. Like f: R->C; f(x)=x. Codomain is C but image(f)=R
wait we talking about complex stuff now?
domain and codomain are things that apply to all functions
I just gave an example saying that codomain doesn’t mean anything
so i can ignore co-domain
Yes
all i need to know is every output vector b is in R^m
It’s just the target of a mapping. You have to give it for the sake of definition
Doesn’t imply anything
if every output vector b is in R^m, then the most consistent thing to do is to say the codomain is R^m
it's standard for the codomain to be all of R^m because then you don't have to do any calculations, it is evident purely from the matrix's size
Put it this way, change your linear map to this: R^3->R^100
x|->Bx, B=(your A, 3 times 97 size 0 matrix). Range is still of dimension 2, codomain is R^100, It doesn’t matter.
Domain and codomain is for the sake of definition. We compute range/image, codomain we leave it behind knowing there is such a thing we don’t need to think about it.
ok i guess i get it
(Mapping (x_1, x_2, x_3) to (x_1, x_2, 98 many 0 in the example above)
range and image are the only important things because it's the actual answers we get or outputs we get
Yeah
range is the output space i mean
where all the images live
or the span of all images?
Same thing, image of a linear map is still a linear space
It is correct
but range is bigger than image right
no?
Synonym
i thought image is the specific output whereas range is where all the outputs live
I treat range as image, it’s like in some fields, people give special names. But image is for all cases
guess im confusing myself. main take away is forget about codomain for this final
well you can say "the image of vector x under T is T(x)" but you also say "the image of T" which is the set of all images of elements of the domain, i.e. the range
well you should still be able to tell what the codomain of a matrix is given the matrix
just don't overthink its importance
Exactly. It’s just something you have to give when you define a map.
ight
this is also important to know right
v= c1e1 + c2e2
Tv = T(c1e1 + c2e2)
Tv = c1T(e1) + c2T(e2)
i'm proud of you mittens
naw thank you @hybrid crow LOL
as long as the domain of T is 2-dimensional, sure
range of T = image of T, both are subsets of codomain
the same subspace?
yes
although the math doesn't change much for higher dimensions other than making the linear combination longer
image can also be used for an individual vector
namely, the image of v under T is Tv
ye it is
Exactly the same thing. Synonym, literally no difference in my mind, range and image.
makes sense ye
Your case domain is R^3, codomain is R^3, image is isomorphic to R^2, kernel is isomorphic to R
My case domain is R^3, codomain is R^100, image is still isomorphic to R^2, kernel is still isomorphic to R
ok
i would describe the image as being a 2-dimensional subspace of R^3
same thing but i'm not sure isomorphism has been introduced yet
yeah I don't think mittens knows that
never heard of that
image is a 2d subspace of R^3
Anyway, we think now you get the idea that codomain is not important.
basic idea of an isomorphism is a link between two things with the same underlying structure
that's only for rank 2 right
so image's subspace is the R^Rank A
rank = dimension of image
the image is always a (rank)-dimensional subspace of (codomain)
so image is related to column space?
image = range = column space
You construct any linear map from R^3 to R^a million, rank, which is the dimension of image, will still be <=3. Codomain can be as large as you like.
Same thing
all three are exactly the same subspace
what about null space
null space = kernel
don't worry about it then
but kernel is always a (nullity) - dimensional subspace of (domain)
never heard of kernel in my life. when do i need to
yeah frick kernel we only fw null space around here
like what math is that
for m x n matrix domain is R^n and codomain is always R^m
sometimes kernel is used instead of null space but they mean the same thing
ok good that's why we don;t know it
Linear algebra itself is just a special case of algebra, like linear spaces, just modules over PID. I think reading algebra first can completely cover linear algebra. The latter seems abundant.
indeed
explain single value decomposition in terms of linear transformations
didn't learn SVD yet but i need to later
That involves a decomposition, A=UDU^H
For any complex square matrix A, there exists a hermitian matrix U and a upper triangular matrix D such that A=UDU^H
why is x in R^m not in R^n
You can prove this by induction. This will be a lemma in SVD. You need to know this decomposition first
typo, it should be R^n
isn't it just a rotation, an axis dilation, and a reflection
ok thanks
You will need to use this decomposition to prove existence of SVD, I believe it’s called Schur decomposition or something: A^H A=VDV^H, for hermitian matrix V and diagonal matrix with non-positive real entries D. Thus D=Λ^2 , taking square root of entries. Let U=AVΛ^-1, clearly U is hermitian, thus A=UΛV^H, the SVD you wanted
I view it as coincidence
no idea about hermitian dude
You wanted to know SVD then you need to know those definitions and the Schur decomposition first
It involves those
When A is real, U and V can be orthogonal matrix instead of hermitian matrix. Parallel proof in real case. That’s why you said rotation and stretch. U and V will represent rotation, Λ will be the stretch
we only know about the orthogonal matrix part
U and V will represent rotation or rotation and reflection I mean, depending on the determinant of U and V being 1 or -1
so we find eigen vectors of A^TA
ooh intresting
Yes, which let you obtain D
ngl i need to lock tf in rn this is a lot of stuff
tryna do a whole semester in a day is a bad idea
A^T A=VDV^T, you need to keep this V btw, since you will define U=AVΛ^-1. So don’t throw away V when you got D
It’s a good idea I think. Your thoughts will be continuous right, no interruption by taking breaks
shoot taking one right now is that bad
I think you already got what you needed. A symmetric real (Hermitian complex) matrix B equals UDU^T (UDU^H) for orthogonal real (hermitian complex) U and diagonal non-positive real matrix D is a special case of this Schur decomposition. So you already know it. It just when B is regular (BB^H=B^H B, when B is real it becomes BB^T=B^T B), D, the upper triangular matrix, by calculation has to be diagonal.
You don’t need the general Schur decomposition if you want to skip it for now. It seems you already know the special cases so good to go. I didn’t use the general case in my proof.
naw we didn't dive into the depths
u sure did
we learned one way to calculate it
it's with orthogonal matrices
I prefer to read the general cases first, then take a glimpse at the special cases and say, oh I knew it already, skip.
By a row of skipping, linear algebra is done. The reading part appears in algebra, my case.
Many cases like this, like Jordan form is a special case of Jacobson form. Depending on what you need. If your interest is not math but something else, chemistry, engineering… it’s perfectly fine to know the usual linear algebra version. General forms won’t be used at all.
yeah im an engineer
Then your way is more efficient. You are doing well.
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Just to clarify
U',V' aare the dual spaces
tilde thing and U^0 definitions please
What do you mean annihilator of U?
c) comes from a) and b), as i’ tilde: V’/ker(i’) \cong im(i’). By a) ker(i’)=U^0, by b) im(i’)=U’. Thus we only need to show a) and b):
a) directly comes from definition.
b) you simply need to show that any linear f: U->F can be extended to V->F, i.e. there exists g:V->F such that g|U=f. And you can use Zorn’s lemma to show this
Why do I need zorn's lemma for a fin dim case
Consider the set of (W, g), where U<=W<=V, and g:W->F such that g|U=f. This set is not empty since (U,f) belongs to it. Any chain has upper bound: a chain of some (W_j, g_j), you let W=union of all W_j, and g(x)=g_j(x) if x is in W_j. Done
I mean it's not necessary is it
Okay I guess we can. No zorn, induction on dim(V)-dim(U)
Lemme send my proofs?
Sure
I'll start with (a)
f: U->F choose a v in V not in U, and expand f to U’=U+Fv ->F mapping v to zero, dim(V)-dim(U’)=dim(V)-dim(U)-1, thus b) holds by induction on dim(V)-dim(U). Done, no Zorn
for (c).Can't I just use the 1st iso theorm
(Actually on second thought, you can just take W to be the direct summand complement of U in V, i.e. V=U \oplus W, then f can extend to V simply by mapping W to 0)
This is what I said
Anyway, direct sum or induction on dim(V)-dim(U), both work for b). Though latter requires this finite dimension
(b) can probably be done by double inclusion too
Whatever works
Sure
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G is a finite abelian group and a positive integer n is prime to o(G) . Prove that every element $ x \in G $ has an nth root in G.
How do I solve this
Good
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how would we graph this on paper if we wanted to
$$ f(x,y) = (x^2 - 1)(y^2-1) $$
Ved
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I want to ask for the green part, which one is correct.. or both are accepted
Karnaugh, logic gates
Both are correct, also should be accepted i think
Wait in the draw you need to add a not
And change D by B
3rd times the charm
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yo
I'm not sure I get teh question
Like if G is of order 2
all H will be a subgroup
If |H|>2, consider {e,x} where x is an element of some order less than the order of the group abnd x isn't its own inverse
but what if G isn’t order 2
then consider this
Z mod something?
well, We don't know if G is finite
1 or even
ha
😭 is this ur ritual now lmaoo
his ritual is 
do we just have to get a subgroup H that is not abelian?
You want elements of even order that multiply to an element of odd order hmm
all subgroups will be abelian
OHHH
i think those aren't abelian

i think you should Z/6
Huh idt that works
what about S_1
shhhh
Nvm it does
snow is so smart
Who is snow
snow is @silent dock
I mean, looking at their roles, they're definitely a prodigy 
I don’t get the question. H seems to always be a subgroup
Z_6 is a counter example I believe
bluds never heard of snow 
But H={0,3}, isn’t it a subgroup?
mfw 6 is even but has an odd factor
abelian't
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it was all pure
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Can someone pls help with these I'm so stuck
have you started with anything?
and are you allowed to use graphing tools or do you have to sketch this by hand
I have these facts also
U have to actually draw it
perfect
then what would be helpful is that you actually set a value for 'a' and A and work with those values
'a' here is just a distance so doesn't matter what we chose, the graphs will take a similar form . use something simple like a=5 and A=1
I don't know how precise it wants it but that looks like what I have on my graphing calc 😞
maybe the ends matter
the sketching thing is really bad so I can't really get a levelling line
hm this does look okay.
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Can anyone give me a brief explanation of the chain rule? And I have an example question right here: find the derivative of $2^x$
1 divided by 0 equals Infinity
do you wanna know why chain rule works or how to apply it?
so far im just trying to learning derivatives by myself
and i proved some stuffs through the definition
Aren't you 9th grade?
it's called i want to learn ahead
Have you learnt limit?
a little bit
Have u checked khanacademy & 3b1b? khanacademy shows how to apply it and the proof through definition, 3b1b offers great intuition
If you're learning by yourself instead of as a requirement for a course, you should take the time to learn the concepts right and how and why they work
You haven't answer his question tho
i want to know why chain rule works and how to apply it
If I were you, well, I'd first make sure I know everything before it, that is, the basics, like the palm of my hand
then try this, calculus with khanacademy + 3b1b is a pretty good combo
Okay lol, just watchi 3b1b videos
And then look for a somewhat gentle introductory text with proofs
lel
$f'(x) = \frac{dy}{dx} f(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
1 divided by 0 equals Infinity
not really scary for me but at least it can scare some secondaries
there are simpler ways to scare ur classmates than learning calculus lol
lol
Pull out some IMO geometry problem, it can even scare a 12th grader
Hell, that'd scare me as well
hell, IMO already scares me
these kiddos in my class only finished through quad formula and intersections between $y = mx + b$ and $y = ax^2$
1 divided by 0 equals Infinity
sounds normal
As should be
mhm
One thing you should keep in mind, not by learning faster are you learning better
i try to make sure i can prove smth first
like, learning by rote is just a worse way to kill yourself
If you want my opinion, you'd get more benefit out of making sure you've got what you should know covered, and maybe proving some results you've seen in class or playing a bit around with them
am i allowed to choose courses in khan academy?
Yea
No, they force u to take one
im cooked
Jumping straight to learning calculus will, most likely, force you to learn by rote as you're saying, do keep that in mind
geometry
You're Vietnamese, right?
get a triangle ABC, and get 3 median lines of each side in ABC, get the interesection point of it. and the distance between A, B, C with the intersection is always = 2/3 vertex line
how do you know
I do know what vietnamse ( the language )looks like
hm
the only thing is that i don't think if you see i use the language before
d/dx, not dy/dx
I have a friend, he always complaint that 9th grade geometry is crazy diffucult there
So you probably should get better at those first
my bad
Idk at what level you're rn, but maybe to test ur understanding of algebra and trig, try taking the unit test here. If you pass it without any issue, you should be more or less ready to learn calculus. That said, there is probably still a ton of simpler topics to be learnt in more depth.
https://www.khanacademy.org/math/precalculus
i agree with that
Calc isnt hard
the diagram gets messy in my eyes
If a 14 year old can learn it by himself then why cant you
aight
1/0 is NOT infinity
Oh but it's definitely annoying
A type of annoyance I believe people should deal with in due time
Tbh, you should only learn calcs only if you confident with your work at school
And you have time
what 14yo
You're studying for exam mostly
Someone green does real anal at 14, so i guess there'll be quite a few 14yo's doing calc in here
Nah thats the best part of math
Me 2 yrs ago
You and I have different tastes in mathematics then, I guess
oh yeah ok because expecting yourself to be the model for all ppl your age worldwide is such a reasonable thing to think
I'm starting to get the feeling the guy came here with some strange intention to flex
Hey, if you love hard problems, try probability and combinatorics instead
@lusty python any questions left?
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,calc 1/0
Result:
Infinity
i let you say it again
1/0 is NOT infinity
Checkmate
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,w 1/0
Look
i forgor the factord
da2a i think
!da2a
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!
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I accidentally opened bro
then do .close
I already did
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HI
hello and welcome to the server
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lim x to a of ((ln(x)-ln(a))/(x-a))
The answer is 1/a but I don't understand my profs justification. This is the justification:
ln(x)=y
y'=1/x
y'(a)=1/a
I'm confused what this means since we haven't solved problems like this before
They've just recognized the limit definition of the derivative
And consequently used the fact that the derivative of lnx is 1/x
A bit circular I suppose
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if you had f(x)=1 if x has more 1's than 0's in its digits and f(x)=0 if x has more 0's than 1's in its digits, or 0.5 if the amount is equal, would f(x) be defined for irrationals?
,calc 1/99
Result:
0.01010101010101
true
oh my god stop changing the message lol
sorry
You'd need to define what f's domain is
thats a rational number tho
yeah i forgot rational numbers can also have infinite digits
Sure, but you can't explore irrationals here without having this checked first
Because, like I said, ^
are there meaningful ways to compare the size of infinite sets other than can/cant have an injective function?
im curious if you could have numbers with infinite digits be in f's domain
See 10-adic numbers
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to find U in an LU decomposition, you perform gaussian elimination but only the adding rows to other rows part, not any other row operations
i gotchu
i learned the theory i think it makes sense
you put A in echelon form without row swapping, and you construct u as the inverse of every row operation
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