#help-49
1 messages · Page 193 of 1
Ohh ok
8 - 10 are all correct.
and factoring
im taking 2 online classes this summer prey for me
i tried taking an online remedial math class last summer and had to withdraw
idk what compelled me to do that
Ohh
omg i just realized this is the help channel 💀
HELP
im so sry for wasting ur time dawg
this guy said u did it all correct tho so ur prolly fine
OH MY GOD
Wait I have question cause I only knew a bit so I also did 12-16 and I’m more lost
Sob sob sob
😿 😭
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Sorry, I thought that a completely different part was your answer.
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YAY
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the twisted cubic curve is isomorphic to P^1, but P^1 has degree 1 and the twisted cubic has degree 3. What's the explanation?
Also, the degree genus formula gives a relationship between genus and degree. Does that mean isomorphic varieties can have different genuses?
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Quick question, what exactly am I supposed to show here
so I consider an arbitrary basis for V and W?
yeah
I was wondering if rank -nullity can be appplied here

I think it can
formalising it may take a while though
i would like to start with a basis of imT then extend that to W
but that wouldn't be all bases of W
think about the rank of the matrix rep, I think 
so this is wrong?
well it might not be wrong, but it’s not what first jumped to my mind
idk, I stupidly avoided matrices like the plague in LA1.
I resorted to linear maps instead
to prove stuff even related to matrices
i mean matrices are kinda ass
you can let j = dim range T and select j linearly independent vectors from the list Tv_1, … , Tv_n
everytime you want to right one down you need to pick a basis
yea, something like that
you’ll get j = dimrange T nonzero columns because of linear independence they obviously are all nonzero
ah that's clever too
that’s pretty much it, yeah
instead of pivots
hence my comment on rank
i don’t think axler even covered rank at this point in the book
hmm, yea, makes sebse
*sense
Axler does all sorts of weird choices imo
I'm not sure if I should use FIS or axler atp imo
axler
for LA 2
i like axler
I love axler too
as long as you stuck to one, it’s fine
but we do dets early on in the course , so 
i’ve opened FIS and it felt too computational
an axiomatic defn
my preference isn’t with Axler, but I think he’s okay
yea, my problem too
my course reccs axler AND FIS
how
det is the unique multilinear alternating blah blah blah
FIS has a section on this, iirc
okie, will get to axler during the summer I guess, FIS during the sem

Thanks
Will hopefully finish axler over the summer
idk if it’s really necessary to go over both
just start learning other math and pick up what you need along the way 
yeah, but when I study dets I suppose I'll use FIS
other than that, axler
I mean I have like 5 LA books with me
💀
we would all still be doing basic shit if we needed to master everything perfectly
$det(v_1, ..., v_n) = \bigwedge dx^i(v_1, ..., v_n)$ this one?
frosst
Axler, FIS, Kumarasan, pooole and planning to get halmos soon
you dont need to cover LA in such detail or you will spend all your waking ours doing LA
but why
this one
that is 4 books too many 💀
I want to do my thesis on LA
😭
why
thesis on linalg is sussy
UG thesis
linalg will be a tool not an end in itself no?
nobody in their right mind reads 5 books on linear algebra
fair, something LA adjacent
this requires you to pick a basis then you need to show this definition of det is well-defined
linear algebra is a chore
idk, axler makes it fun
like you’re supposed to learn it so you can learn fun things
^^that's what i mean by tool
this is the actual theorem given
you prove its existence by showing that it’s given by that formula
but you said you’re using FIS?
I'm THINKING of using FIS for LA 2
Idk yet
axler >
does it really matter that much?
I just started revising LA 1, for which I used axler
nope
what the fuck even is LA1 and LA2
My worry is dets
2 courses in LA
yeah but LA is LA
was the first course not proof based?
it was
🤔
but we only covered until the basics of spectral theorm
this is like saying you read Munkres for topology, but then are worried about the manifolds perspective, so you're gonna pick up Lee too and relearn topology from a different perspective
good on you my uni after 3 semester still haven't covered that
now we'll cover diganolisation, factorisation etc
and won't 🙂
I feel like this is just a waste of your time
spectral theorem was like one of the last things we did
and end with quadratic forms
we didn't do a proof
of spectral
applied math probably
nope
🤔
frosst has said quite a few things about his uni's courses in the past 
is it this
I don't know if I recognize this definition 
my stuff with dets is super rusty though
we've comandeered wai's channel whoops
aren’t you doing abbott too?
Though trying to do it without row reduction now
That's for RA
i know but you’re going through multiple at once
I can explain
basically I want to cover everything we will do during the sem by myself first
for multiple reasons
- I can focus more on the details in the sem
- Gives me a head start, allows me to relax more during the sem
school is indeed much easier when you know everything ahead of time
or at least have been exposed to it
yea, which is what I'm doing rn
this is very good
all my courses are very easy now because i spent a lot of time in first year getting the general picture when i audited all the classes
i got a lot more time to spend on the specifics and details of the coursework when i actually took them for credit
Thoough going to be a bit hard doing AA, probability , RA1 and LA 2 over the summer
i just sat in the classes
it gives you the correct path of "general idea" of the content
if you follow a book you'll go down a lot of rabbit holes that the course won't
probability sucks tbh
💀
not enough theory in the intro books
i always got bored trying to read blitzstein
Here, I'm going to start off by showing that if all entires are 1, then the rank is 1.
\
If All entries are 1, we can row reduce it. to eliminate all rows but the first , and similalry column reduce to only get the first entry as 1. This clearly has a rank of 1, we're done
the exercises just felt like high school
after measure theory maybe it will be interesting
next fall 🙏🏻
that's not the fault of probability
How does this sound
idk measure theory still
that probably works... 
I think you can honestly do something simpler though? 
if all the entries in the matrix rep of T are 1, then T(v_1) = T(v_2) = ... = T(v_n) = w_1 + ... + w_m; in that case, they're all the same vector w in W, so range T = span{T(v_1), ..., T(v_n)} = span{w}, which has cardinality 1 
so dim range T = 1
I think that would be enough
though somebody can correct me if I'm messing this up
That's even better
Thanks
Now for the other direction
hhmm
If dim(Range)=1 , the span is 1 D, so a matrix with all entires 1 is one possibility?
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Is this true?
e = 2 + 1/6 + 1/4! + 1/5! + 1/6! .....?
,w e = 2+ sum from n=3 to infty of 1/n!
where are you getting these beginning terms from? you can use the taylor series for e^x to get a true statement that looks similar to what you have.
wait
Yes, is it that:
e = 2 + 1/2! + 1/3! + 1/4! + 1/5! + ......
you should have a 1 in the beginning
what
1 + (2 + 1/2! + 1/3! +....)?
Or 1 + 1/1! + 1/2! + 1/3! + .......................?
thanks
y'all
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what would be the difference between a function from like calculus, f(x)=e^x for example, and a linear transformation?
a linear transformation is a particular kind of function between vector spaces that satisfies f(ax + by) = af(x) + bf(y)
where a,b are scalars
"transformation" and "function" are essentially synonyms here
so we could call them "linear functions" instead and it would mean the same thing
Functions from calculus are usually analytic, or smooth, or k-times differentiable or at least continuous, with ideas of limits and sequences that have more to do with metric and topological spaces, as opposed to vector spaces which can be independent of a metric or topology
linear transformations have in general a vector space as their domain and codomain
rather than R
this is all great answers, i appreciate it
is it true that in calculus all functions are like, endomorphisms? what I mean is like R->R or C->C, or is it possible the domain is restricted but the image not or something? am I making any sense
endomorphism implies "morphism" of some kind (generally homomorphism)
most calculus functions are not homomorphisms
only if you are extremely category theory pilled
and call a function from a set to itself an "endomorphism in Set"
which is stupid
what would be the domain and codomain of modulus function? by modulus function you mean absolute value right? so R -> Z+?
not Z+
Z = integers only
the codomain can be any set containing all the possible values of the function
so any set containing [0,infty)
ok, só in calculus not all functions are like f: K->K
it could also have a restricted codomain/image
i just dont know how else to call it when f : K -> K
there may be exceptions, but in calculus usually the codomain is just taken to be all of R (to the extent that anyone even talks about codomains in calculus classes)
the domain may or may not be all of R, depending on the function
ok i appreciate it a bunch, sorry for the cat theory jargon
no problem!
function from a set to itself
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We can split this into two cases. Case 1: v_1 is in the kernel, in which case the entire first column would be 0( as the vector is mapped to 0). In the Second case, it doesm
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Whats wrong about this to get y? Im somehow missing a c?
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can someone explain the geometric intuition behing determinants calculating the volume of a paralelelepipid
what is @paralelelepipid@?
@tidal turret
you should first try to understand why the determinant is the area of a parallelogram in R^2
in particular, what do row operations mean geometrically
afterwards this translates nearly immediately into R^3
how is the determinant the area of a parallelogram for vectors in r2
what?
is like all the possible linear combinations of the vectors or what
i dont really get it tbh
I dont follow, what is ad-bc
where are the vectors
there is something that I am missing here
the formula ad-bc screams determinant
I think thats for r2
,, \begin{pmatrix} a & b \ c & d \end{pmatrix}
renato
the columns are (a,c) and (b,d)
trace?
let me do the drawing
yep but I still dont follow how is the area of the parallelogram ad-bc
maybe I am lacking some parallelogram formulas
base times height
I am not claiming that its obvious that the area of this parallelogram is ad-bc
the key observation is what happens when you do a row operation
geometrically
draw the parallelograms for $\begin{pmatrix} 1 & 3 \ 2 & 8 \end{pmatrix}$ and $\begin{pmatrix} 1 & 3 \ 0 & 2 \end{pmatrix}$
Denascite
what did the row operation do geometrically
hmm ok maybe not the best vectors
when we do a row operation, depending on the operation the determinant changes or maybe it doesnt
for example adding a multiple of one row to another row, doesnt change the determinant
but swapping rows, changes the determinant by a factor of -1
@tidal turret Has your question been resolved?
hi, what is your question? 😄
and what are those row operations geometrically
why do they not change the area
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#1260937723452330155 message Can I have some help formalising this
wai
wai
hi
@twilit field Has your question been resolved?
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为? 
Oh
喂
oops
im tired i just got back to my hotel
nice
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if W1,W2,U are all sub vector spaces of V and i pvoed that W1(+)W2=V, then can i just intersect both LHS and RHS with U and get the statement in the pic?
or is it not enough
i dont get why they put the intersection in each () on RHS instead of just once as a distributive thing
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r u able to prove the claim?
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if I have a line that is entirely contained in a plane. how would you know that the direction of the line is orthogonal to the normal of the plane?
if the dot product of two vectors is zero, the vectors are orthogonal
You can use the direction cosines of the line and the normal vector of the plane
What was said above
definitionally the normal of the plane is orthogonal to all vectors lying in the plane, including the direction vectors of the line
yes, but how?
yeah the definition of the normal is literally it being orthagonal to the plane, therefore lines within the plane aswell
are you asking how the normal is orthogonal to the plane?
Try drawing/imagining one that isn't
like all the vectors present in the plane are orthogonal to the normal of the plane
hmm
thats kinda the point of a normal lmfao
What does it mean for a vector to be normal to a plane
all the vectors inside the plane are well make up the plane, they are all parallels to the plane
good question, idk i don't have a good geometric intuition
wdym?
We conventionally define a normal vector to be one that is orthogonal to every vector in the plane
Now you’re asking why they are orthogonal
why tho?
Well, by definition…
Well because it’s something we want to talk about a lot
A normal vector to a plane is used a lot so we want to name it so we have a shortcut to say “a vector with so and so properties”
How do you define it then?
u can visualize it like this
if u draw a normal to the plane the lines will naturally have the same normal
good point hahaha
wdym "normal vector to a plane"?
u have to be sure of the definition of a normal tho lol..
I mean a vector with the property that it is orthogonal to every vector in the plane
Is a vector we like to use often
the normal of the plane
well i just call it normal vector of the plane
It’s just a name
why
A name just means it is used often enough we want a shortcut to say it
ok
they are literally on the plane
Being “parallel” is not a good idea to think about
is just hard to find an definition about it online
It’s not
A normal vector to a plane is one that is orthogonal to every vector on the plane
In geometry, a normal is an object (e.g. a line, ray, or vector) that is perpendicular to a given object. For example, the normal line to a plane curve at a given point is the infinite straight line perpendicular to the tangent line to the curve at the point.
A normal vector is a vector perpendicular to a given object at a particular point.
A no...
I suppose we also say it has unit length
For convenience
it doesn't mention it's orthogonal to all the vectors lying in the plane
In three-dimensional space, a surface normal, or simply normal, to a surface at point P is a vector perpendicular to the tangent plane of the surface at P.
A plane is its own tangent plane
Well, not really but that’s not important here
That’s technically not true by some wacky ass weird thing you don’t have to worry about at all
But just pretend its true
Hey anyone here does MUN’s
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
what i don't really understand is why you guys say "definationally the normal of the plane is orthogonal to all the vectors lying in the plane"
like WHAT definition wtf or is it geometric intuition
Imagine a plane and all the vectors on it
Do you know the vector equation that represents a plane
jewels!
Look at this normal to the plane and image the vectors on this plane
Like the ones already drawn
i only know the cartesian equation of the plane, vector eq for the plane hasn't been covered in my class
i really don't know how you could define "this vector/direction is perpendicular to a plane" other than "it's perpendicular to every line in the plane"
since angles are by default defined between two lines or vectors
yeah i see what you guys mean now
if the normal is orthogonal to the plane is orthogonal to every line or vector inside it, otherwise it wouldn't be a normal vector of the plane because is not orthogonal to the plane
yeah i appreciate the geometric intuition i think i get it now, well idk i think it's more of an geometric intuition or rather a consequence of the definition of normal of the plane (vector that is orthogonal to the plane)
whatever, i didnt attend class when planes where covered i think i was ill
but i will recheck the definitions in the reader, i appreciate everyone who contributed
thanks, i appreciate it
well wasn't covered in my class but this looks like an interesting explanation to it
i appreciate it
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Can you translate it
Let
L₁: X = λ(0, −1, 1) + (4, 2, 1)
L₂: X = λ(0, 1, 2) + (4, −2, −4)
and A = (4, 2, 1).
Find a point B on L₁ and a point C on L₂ such that triangle ABC is right-angled at A.
Gave up? 
@tidal turret Has your question been resolved?
why would you?
wdym?
I appreciate the translation, ty
I mean change the coordinate system so that instead of looking for a right triangle at A, you look for a right triangle at O=(0,0,0)
you need a coordinate system change that doesn't change angles, of course
He was LaTeXing and eventually settled on plain text, that's what I meant
oh lol
think of it like
if you start with a circle:
$$(x-3)^2+(y+1)^2=11$$
then you can define new coordinates
$$x'=x-3$$
$$y'=y+1$$
$$x'^2+y'^2=11$$
gfauxpas
now you have a circle in a cooridnate system such that it's centered at 0, it makes the math easier
and once you find the solution in the changed coordinate system, shift iti back to the original one using
$$x=x'+3$$
$$y=y'-1$$
gfauxpas
does that make sense
but we havent covered circles, there is a simpler path that we are missing
I'm just illustrating the principle, doesnt matter that it's a circle
the point is, by a change of variables, you can make the problem into an easier one
you want a coordinate system such that A becomes the origin, so
$$T(A)=O'$$
gfauxpas
first we need to find the intersection of this two lines L1 and L2 that is exactly B
easier to find the intersection of the two lines L1' and L2' that is exactly B' in the new coordinate system
how about you try it my way, then try doing it the other way, and that will help you see why my way is easier
or is your question, how do you know that the intersection in the new cooridnate system will correspond to the intersection in the original coordinate system?
affine 🔥
If you don't want to, you could just bash through in the current coordinate system
sure
But the idea of moving the origin will help you
it's just ... harder
I like it
why will it help? changing the coordinates system?
sure, I just dont follow
Ease
I could explain it, but, I think moreh elpful than an explanation will be doing it my way, and trying to do it the other way, and seeing what complications there will be
feeling more than reading will give you an intuition of WHY I'm giving this advice
I mean it's pretty immediate isn't it? Since intersection of W1,W2 is only the zero vector, so you can intersect the stuff on RHS like shown with U. If you have a proof/contradiction then show me
WLOG is why you CAN choose the origin. The reason to choose the origin is because things being 0 simplifies math
ok so A = (0,0,0) then?
T(A)=(0,0,0)
T?
$$A = \left( \begin{array}{c}
4 \
2 \
1
\end{array} \right)
\quad \Rightarrow \quad
T(A) = O' = \left( \begin{array}{c}
0 \
0 \
0
\end{array} \right)$$
gfauxpas
a change of coordinates
a linear map
$$\begin{aligned}
x' &= x - 4 \
y' &= y - 2 \
z' &= z - 1
\end{aligned}$$
gfauxpas
this is like the circle example I gave, i'm just using the symbolism of linear algebra
the change of a vector into a new coordinate system is a linear map (why?)
wait no
it's AFFINE not linear
durp
so maybe misleading to use the symbol "T"
since T is usually used for linear mappings, my apologies
but

what you are saying is that for B i will subtract 4 in x1 subtract 2 in x2 and subtract 1 in x3
for B = (x1,x2,x3)
It's just a substitution that makes the job easier for us
and you have to do it to every point that you're analyzing
did you take basic physics yet? did you learn about choosing a frame of reference?
You don't have to worry about the geometric intuition, which is shifting the origin
i did
but i didnt passed

yeah it took me 3 tries to pass physics, I had to drop out twice, I had horrible profs , but anyway
Imagine I tell a physics student
but I know what FoR is
to choose a coordinate system to drop a ball off a building
if I'm going to ask them questions about the process of dropping the ball, it might be natural to put the origin at the top of the building
if im going to ask about what happens when it hits the ground, it might make sense to put the origin at the bottom of the building
Do you know how to find the angle between 2 lines
but it's the same process, just a question of, what should I call the starting point?
and here, the point of interest is A, so I want that to be the starting point
Because I would do that and travel 1 unit along 1 line from the intersection, and cos(θ) along the other direction
And those would be my 2 points
"travelling" would be hard computationally
It’s not
guys I swear maybe u guys are overcomplicating it
I still think my idea is the easiest way to do it 
yes but idk physics
If you start at a point on the line eg the intersection then you can just add some copy of (0,-1,1) or (0,1,2)
i was just giving an illustration of why you can choose a frame of reference based on convenience
if the intersection is at the origin!!!
it doesnt matter in principle,. but in computation it does
otherwise it becomes thorny
It’s the same as doing an affine shift to re-centre but the exact formulation is not particularly insightful
anyway, I'mg oing to play skyrim, gl
gl dragonborn!!
guys using affine knowledge is using advanced machinery for this simple problem
he made a substitution lol
thats all it is
dont get bogged down in what its called
but sure, you can go ahead and work as is
but if we translate A to the origin why is it easier?
calculations are simpler, but why?
i mean only for finding the intersection of the lines L1 and L2, only there
or maybe I am missing something 🤔
Assume B is governed by a parameter t, and C is governed by a parameter s
You're gonna need to solve BA.AC = 0
oh that part too
no
B is an exact point, the L1 n L2
C is parametrized yeah
but B is NOT
yeah but A and B we know
.
we only lack C
Fixed
B is in L1 and in L2
but the translation is wrong
let me translate it
Let L1 : X = λ(0,-1,1)+(4,2,1) , L2 : X = λ(0,1,2)+(4,-2,-4) and A = (4,2,1)
Find a point B that is in L1 and in L2, and a point C ∈ L2 such that ABC is a right triangle at A
"Hallar un punto B que pertenezca a L1 y a L2"
And apply the dot product condition on C insead
wdym?
AB.AC=0 still holds
Yeah my bad
I think maybe this is why lad proposed a change of coordinates, the translation was incorrect
hi, first find the interesection of the lines. that point is your point B. then use the fact that prependicular vectors have 0 dot product to find point C 😄
because the problem is very simple
yeah
ok, let's do the number crunching shall we?
yep!
good luck
(4,1,2)=B
yes, that looks correct 😄
the method is correct, and the answer is also looks good !
ty mate, exercise looked complicated but after you dismantle by parts is not bad mate
where you from mate?
glad to help 😄 make sure you can do the dismantling on your own
<@&268886789983436800> this seems to be a scam (fake top level domain, steamcommunity spelled incorrectly) 
i appreciate the preciseness
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is the lower statement really always true?
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How do I show the realtion between them? I was able to prove that they are dependent but I couldn't show u in terms of v or vice vesa.
If someone could just give me a hint, that would be great.
check out the tangent angle sum formula
thanks, lemme try that.
tan u = tan (x+y/1-xy)
but I'm not sure how to go any further
this does look very much like what I have but
how would I get all those x, and y with tan?
I also found this formula. I'm not sure how this was derived but by any chance did you mean for me to use this one instead of the one I mentioned earlier?
well theyre consequences of each other
I see. I haven't come this far in my trig lessons so I didn't know that.
in [ \tan(\alpha + \beta) = \f {\tan\alpha + \tan\beta} {1 - \tan\alpha\tan\beta} ] simply $\tan^{-1}$ both sides and use $\alpha = \tan^{-1} x$, $\beta = \tan^{-1}y$
ohh. yes that makes sensee.
wait dam how did you get the latex to render white
is that in the settings?
yea
I should be able to solve it now. let me get another crack at it
@gray zenith Has your question been resolved?
alomost there
need another min. my math slow :-0
yep got it.
thanks @hard umbra
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if Null(S)=Null(T), then if S=ET, E has to be injective, right
Here's my proof for that
S(v)=ET(v). Let v not in Null(S), but T(v)\in Null E, we then have S(v)=0, which is a contradiction
nah, this "proof" is wrong, is it not
oh, existance
E is the identity matrix
Does that work
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Prove $\sum_{i=1}^{n} (-1)^{i} \binom{n}{i}= 0$
\
Proof :We can split this sum into $\sum_{i=1}^{n} \binom{n}{2i} - \sum_{i=1}^{n} \binom{n}{2i+1}$
\
Now I wonder if shifting the index of the second sum will quickly allow me to conclude it's zero
wai
I mean
Damn 😔
its the binomial expansion of (1 - 1)^n = 0
Can we not use binomial?
Now you're damned 😔
fr
thanks
what happens when n->inf
sheldon?
if you like proofs
LA is gonna be easier for you
but
I LOVE proofs
if you dont
No 0^n is always going to be 0
0^{\infty} is indeterminate
Wut
said who
sums of partial sums is 0
1^{\infty} is as well
0^x as you take x large will be 0
Ugh what is the compiled list of indeterminate forms
,w indeterminate forms list
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L
Oh well
calculus for ants, volume 1
okay, i'll eat and then get back to combi

Can't wait to do RA / LA tomorrow
Thanks a lot again everyone!
what do you abreviate as RA
Real analysis
Easier than this fr
ok then
search up combinatorial proofs
youll love
actually
just do graph theory for real
3rd year elective
also where did you even get such a question from
This? A probability book
Sheldon ross
well you can do olympiad graph theory problems until then
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i thought the final result would be
$$(arcsin(x^2))^2 * 1/4$$
Snek
yes, thats what it should be. (and don't forget the +C)
oh so its the writers mistake
yes
it looks like a plus in the denominator lol
Yes exactly
wait shit youre right lmao
the first line of work implies its intended to be -
im sending a complaint to the proff telling him to compencate for the time i wasted solving this
hopefully he raises my grades
or expells me
either way im happy
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How do I get k<=9 from this
?
nvm, rational root theorem
Then why did alexheinsis say that at the finale
is k only natural numbers
okay makes sense then
You can lowkey guess and check
with the IVT
since its only naturals
Nah I want correct proof
IVT is valid
I mean
Ehehe
Basically if you have a continuous function f : R --> R, such that f(x) < 0 and f(y) > 0, then there must exist a root of f between x and y
So you show how it's positive for k = 10
And negative for k = 9
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@solemn silo Has your question been resolved?
<@&286206848099549185>
if you're gonna ask about someone's answer, you should just provide the link
by induction?
||for x>= 10 (8x^3 + 16x^2 +16x)/x^4 = 8/x + 16/x^2 + 16/x^3 <= 0.8 + 0.16+0.016 = 0.976 < 1
hence (8x^3+16x^2+16x) < x^4 for x > 10, hence the polynomial is positive||
idk how to give hints for this
dont open spoiler
@solemn silo Has your question been resolved?
rewrite this as $$k^4 + 1\leq 8k^3 + 16k^2 + 16k$$ The right hand side is a cubic, while the left is quartic. both are polynomials with positive integer coefficients. so just by looking at the degrees, there exists an $N$ such that for all $n \geq N$, we have $n^4 + 1 > 8n^3 + 16n^2 + 16n$. you can check that for $n = 9$, the inequality does not hold true but for $n=10$, it does. and by the degree, you can conclude that this is true for all $n\geq 10$, so $k\leq 9$
fastrack_and_backtrack
the general idea is that a polynomial of higher degree with positive integer leading coefficient will outgrow a polynomial of lower degree as we approach +infinity 😄
Guys i did it
I rewrited it so lhs would have only k^4, and then I multiplied all by 1/k3 , and it gives smth true only if k>=9

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hi, probably stupid question
but what does it mean for a system to be "solvable"
is it the same distinction as consistent and inconsistent?
like a linear system of equations?
yea it means the same as consistent, i.e. a solution exists
yeah
Well i've seen this
It's kinda informal but 😭
that makes no sense, right?
i think better language would be "a unique solution exists"
i wouldnt deep it too much tbh
if theyre saying "its all very sad" in their answers lmao
lolll
😭
yeah the class isn't really a formal math class
but things become very hazy
when the definitions aren't precise
so now i'm left wondering what "solvable" entails
i'm guessing i just go with conventions here?
like uh solvable -> unique or infinite solutions
who wrote this? it's totally wrong
and unsolvable -> no solutions
yeah that makes the most sense
yeah 😭 but also the other students haven't learnt lin alg
so ig maybe it's a pedagogy thing
idk yeah
that system is solvable, it's just that the second equation doesn't introduce any new solutions nor does it remove any
indeed
i mean ig it's fine to define solvable your own way
but unless i'm missing something
the two images contradict themselves, right?
yeah
yeah so ig
but again, not too big of a deal
that's the crux of the issue perhaps
yes they do, they're basically the same scenario with slightly different numbers
right yeah
like i doubt youre ever going to be asked to define solvability
correct
it should always be clear from context what theyre asking you for
so just go with the flow
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guys so d is suppose to be tan theta but im having trouble understanding how to get there
,rccw
oh yea i forgot to label that one of the triangles the one with the angle theta has a hypotenuse of 1
in terms of soh cah toa, think about what side "d" represents in the triangle
hypotenuse
is it?
im assuming
the triangles are similar and d is tangent to the unit circle and perpendicular to the other hypotenuse which is 1
but i would just use the big blue triangle here
we are interested in this triangle
yes
because it is a triangle that has both side d and angle theta
im assuming yes
which side is the hypotenuse? remember that the hypotenuse is always opposite the right angle
the long ahh line at the bottom
yes, so in that case what does side d represent relative to angle theta
opp
1 im assuming
yes, and what side is relative to theta is that (for this triangle specifically)
uh i dont know maybe adjacent or like cos
but tge thing is that those 2 are similar triangles
we don't need to worry about any triangle similarity here. just pure soh-cah-toa on the blue triangle indicated here
i mean you can also do it by similarity but i think that would be more work
i guess but i am watching along for a youtube video and it says the triangles are similar so their ratios gotta be the same and the guy wrote d/1 = sin theta/cos theta
im just having trouble understanding the similarity part and the equation
so then d = tan
let's start by doing it the way i am suggesting and then we can get to the youtube method
like i kind of get it but it feels like i have a concussion
aight bet so whats your suggestion im listening
we have a blue triangle marked here, and we know two sides: d and 1
we know that d is the opposite side, and the bottom is the hypotenuse
so side is 1 for this triangle?
wait what rephrase “so side is 1 for this triangle”
like out of opposite, adjacent, or hypotenuse, which one is the side with length 1
adjacent
ok. so we have two known side lengths, d which is the opposite side and 1 which is the adjacent side
tangent is o/a which is d/1
yes
sure
so that long ahh side is suppose to be 1/cos theta which is sec theta im assuming if i got my trig right but how do i get there
well you know that that side is the hypotenuse, and the adjacent side is 1
i dont know the hypotenuse
right now i got (cos theta)+x
cuz from the 2 triangles we dont know the under side from the second one only cos
ok let's say that the bottom side is "x"
so the bottom side x we said is the hypotenuse, and the adjacent side we said is 1
so we can use a similar strategy as before to solve for x
but how is it just x if we also got a cos
imma send another pic to show u what im seeing
why is my shit sideways
,rccw
ok you have a different x than me
but remember in the first one since they are 2 triangles we combined into on the unit circle before we got the d and the other traingle we just had the plain triangle w the adjacent side was cos
let's instead call the entire long bottom side "y"
the entire long bottom side "y" is the hypotenuse of the big triangle
solve for y?
from trig using only the big triangle, yes
how do i use trig if its a side do i do inverse
you have a big triangle:
- we have said previously that 1 is the adjacent side
- and d is the opposite side
- and y is the hypotenuse
that is all the information we need for this problem
you have a known side: 1 (adjacent) and an unknown side: y (hypotenuse)
what trig function relates the adjacent side and hypotenuse?
yes
fuh do i do with that info
oh
1/cos theta = y
which is sec
damn thats cool af
imma try to find the other function using this knoweledge
ion know how to do ts
i think itll be easier when i actually reach trig
but thanks cuz u made me get some things
thanks a lot
like fr not sarcastix
how do i close ts
use .close
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😭




