meant to say x = (b^2)^-1
#help-39
fluid axle
yea
thick musk
ok all good
thanks for the help
cool cool
š
whats wrong with this derivation?
first line, multiplied both right hand sides by c^-1
second line, multiplied both rhs by 'a'
oh and forgot to add, x^2 = b, and x^5 = e
since x^2 = b, on the fourth line, x^2 becomes b, which results in fifth line
fluid axle
are you even solving the same question ?
how do you involve the a and c?
@thick musk
pearl pondBOT
@thick musk Has your question been resolved?
thick musk
Let me check
fluid axle
Let me check
like I don't understand how you get your very first line
thick musk
That is all the textbook says
Is it missing something, by the way, this pdf version seems to have a lot of typos etc
humble root
I think what aPlatypus is trying to say is that you have made too many unnecessary steps.
fluid axle
right so it was indeed a different question
humble root
In particular, x^2b = xa^-1c
Where did a, c come from? The equation is only in terms of x, b.
o
fluid axle
looks like it at least
humble root
Anyways the main issue here is on the second to last step; what did you do there?
thick musk
according to the solved problem on Quizlet, 1. is x = a^-1 c b^-1 which is what I got
we are just solving for 'x'
we don't care about a or c
humble root
since x^2 = b, on the fourth line, x^2 becomes b, which results in fifth line
Wait, are you sure that x^2 = b?
thick musk
ah ok
humble root
Isn't that an example
thick musk
I see
thick musk
lol
humble root
It's an equation
thick musk
so in the example, x^2 = b, that doesn't apply for 1) and 2)?
humble root
Yeah.
thick musk
ok I thought it did
so x = a^-1 c^-1 b^-1
seems pretty obvious
I think
yet Quizlet says otherwise
humble root
wait how'd you get c^-1
spare lark
<@&268886789983436800> i think its bot, they posted message everywhere at the same time
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fallow cosmos
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normal magnet
Prove that out of any n + 1 numbers chosen from the numbers 1, 2, 3, ...., 2n, there exist two numbers one of which divides the other.
normal magnet
Is this not equivalent to proving that:
the sum of floor(log_2 ((2n/(2k-1)))) from k = 1 to k = n >= n?
toxic lichen
why log_2, and why the sum of these logs
toxic lichen
again why are we summing log_2(floor(...))
toxic lichen
my question stands
how is the original existence proof equivalent to some summation being ā„ n?
pearl pondBOT
@normal magnet Has your question been resolved?
hard cipher
Is this not equivalent to proving that:
the sum of floor(log_2 ((2n/(2k-1)))) fr...
I think you're overthinking it. Here's a hint:
||use Dirichlet's principle to deduce that there is at least one even number chosen||
sharp smelt
pearl pondBOT
sharp smelt
The best I've been able to come up with it $\langle T(v),T(v) \rangle = 0 = \langle T(v), T^*(T(v)) \rangle\iff v =0$
jolly parrotBOT
wai
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buoyant fulcrum
49
Idk how to find C coordinate just with given information
pearl pondBOT
49
Idk how to find C coordinate just with given information
buoyant fulcrum
Here,
AB lies on x-axis
pearl pondBOT
@buoyant fulcrum Has your question been resolved?
iron cobalt
build a right triangle beside the given triangle
stable flame
Here,
AB lies on x-axis
do you know ab the cossine rule?
buoyant fulcrum
do you know ab the cossine rule?
No
stable flame
hmmm
buoyant fulcrum
No
stable flame
theres that aswell
glossy blade
if u dont know that just drop perpendicular bisector and see if u can figure out something with the right angle triangles
stable flame
No
u solve for hip and u got all ur triangle lenghts
then u can find the other coordinates
stable flame
yw
buoyant fulcrum
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buoyant fulcrum
I got answer option A but I was told that the answer is option C
buoyant fulcrum
,rccw
jolly parrotBOT
fallen sundial
I got answer option A but I was told that the answer is option C
how did you get A
rough meteor
JEE Main š„
me too
me as well
buoyant fulcrum
2a+2b=ab
fallen sundial
a hint i would give you would be to use the intercept form of the straight line
fallen sundial
2a+2b=ab
yes
now try to check what point would ts equality you got, would satisfy for the real eqn
use ts
fallen sundial
wlcm
buoyant fulcrum
Secret: I just put coords of stone into 2a+2b=ab idk why
š
@fallen sundial Thank You so much again
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main oriole
<@&268886789983436800> twice in a row decapitate him
vague stump
hes in my help 49 and i was getting help š
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main oriole
thank you
autumn trellis
crypto casino is crazy work
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bitter pelican
find wherer x(sqrt(64-x^2) is increasing and decreasing
pearl pondBOT
find wherer x(sqrt(64-x^2) is increasing and decreasing
fallen sundial
find wherer x(sqrt(64-x^2) is increasing and decreasing
differentiate it and draw a wavy curve of what you get
btw define it's domain first
so you dont commit any war crimes later on 
fallen sundial
(-infinity, -sqrt32) and sqrt32,infinity)
i told you
broken fossil
are you in C
fallen sundial
define the domain first
bitter pelican
and -sqrt32,sqrt32 for the increasing
bitter pelican
so the domain is
x cant be 8 or -8
wait no
x cant be less than -8 or greater than 8
fallen sundial
x cant be less than -8 or greater than 8
yes
fallen sundial
(-infinity, -sqrt32) and sqrt32,infinity)
so ts will change now
bitter pelican
always define the domain first
ahh ok i was wondering why i got this test question wrong :0
ok ty
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loud orbit
why can u1, one of the vectors that form the basis B of a vector space V, be written as a linear combination of the two vectors that are in B'?'
slender viper
why can u1, one of the vectors that form the basis B of a vector space V, be wri...
do you recall what it means to be a basis
what it means for B' to be a basis
[yes, find the definition]
loud orbit
yes, that 1) the vectors contained within the basis B span vector space V without possible contradiction, and 2) that the vectors in B are linearly independent
slender viper
replace B with B' in this definition here, since B' is a basis
loud orbit
so because V has multiple possible bases, I must be able to obtain one from the other?
slender viper
yes, that 1) the vectors contained within the basis B span vector space V withou...
[i'm not very sure what's "without possible contradiction" in this definition, just say it spans vector space V]
loud orbit
as in the solution vector and the coefficient matrix must never have a contradiction
slender viper
as in the solution vector and the coefficient matrix must never have a contradic...
[yeah, but see, you have to specify the idea of a coefficient matrix and solution vector for this "without possible contradiction" to make sense]
loud orbit
well, then the vectors wouldnt span V to begin with if there were contradictions between A and b
slender viper
anyway - it's a feature of the definition of basis that you can express any vector v (regardless of whether v comes from a basis) as a linear combination of vectors in a basis
loud orbit
yeah
so it's redundant to say it
youre right
but
why are we using a and c as the coefficients for the linear combination for [u1]B' and not a and b
slender viper
it's just variable choice
you can choose what variable names you want anyway, if you want to use a and b, you can rewrite the argument using a, b for the first vector and c, d for the second vector
loud orbit
so it's saying that the two vectors that make up B can be expressed as linear combinations of the two vectors that make up B'?
slender viper
yeah, because of the fact B' is a basis
loud orbit
why do we use (u)B for coordinate vector, but then [u]B' for... whatever this is
slender viper
hmm, based on what is being written here, it seems that it's just notation picked for this argument
slender viper
it seems to be exactly the same here though
loud orbit
the coordinate vector relative to a basis was taught in the previous section. It's just reminding what it meant in the example
slender viper
ah, I think one is defined as a row vector and the other is defined as a column vector
the round bracket is row vector and square bracket as column vector
loud orbit
trying to comprehend the rest below it
yeah so some vector v from the old basis B can be written as a linear combination of u'1 and u'2. Sure, fine
but then what is [v]B'
underlined statement
sleek sandal
curious, what class is this?
pearl pondBOT
@loud orbit Has your question been resolved?
loud orbit
curious, what class is this?
linear algebra
I dont understand what the matrix at the bottom means
shut vortex
underlined statement
Itās the vector v written with respect to the basis Bā
pearl pondBOT
@loud orbit Has your question been resolved?
loud orbit
Why are we able to pre-emptively write P on the write side of the augmented matrix (before the rref), when we don't initially know that P = B?
It should be like so
shut vortex
It should be like so
Well no not really
shut vortex
Itās given that P is the transition matrix from some basis to the basis S
loud orbit
yes
shut vortex
So the columns of P will be the basis vectors of P written with respect to the basis S,
But S is the standard basis
loud orbit
So when you rref [S | B] you get [I | P_{B->S}]
loud orbit
So the columns of P will be the basis vectors of P written with respect to the b...
what?
shut vortex
A change of basis matrix P from a basis B to Bā
Has columns which are the basis vectors of B written with respect to Bā
loud orbit
nonono. You're changing from basis B to S, and you're given P and you're given S.
shut vortex
I know
B and Bā are placeholders for any basis
Set Bā=S and the statement is true
loud orbit
yeah. So [B' | B] rref [I | P_{B -> B'}]
loud orbit
the notes are putting in the value for B, but we dont know B
we're trying to find B. So you can't put B into the initial augmented matrix
there's nothing to put in
we know S, we know P_{B->S}
so when you write it out you get that
shut vortex
Given the question we know that B = {u1, u2, u3} and {ui}S is the iāth column of P
So yeah, you can write it on the left hand side
They havenāt claimed that itās the change of basis matrix until after the justification of no row reduction necessary
loud orbit
but how are you able to put in u1 u2 u3 when you dont know what they are
we arent given a u1 u2 u3
we're given S
shut vortex
You do know what they are
loud orbit
how?
shut vortex
Itās given that P is the transition matrix of some basis B to S
loud orbit
yeah
shut vortex
The columns of ANY transition matrix from B to another basis Bā
Are the basis vectors B
Written with respect to the basis Bā
But in your case Bā=S
So the columns of P must be the basis vectors of B
loud orbit
Are the basis vectors B
B is a basis not a vector thhough
shut vortex
A basis is a set of vectors
shut vortex
I wrote it wrong
loud orbit
sorry i got confused
shut vortex
No problem
loud orbit
the columns of any transition matrix from B to B' are the basis vectors B? Is that true? One second
shut vortex
The basis vectors B written with respect to the basis Bā
loud orbit
but in other examples that isn't true
loud orbit
this doesnt make sense..
shut vortex
[v]Bā and [v]B arenāt specific vectors
Itās true for all vectors
Again, we do have B
Because of the structure of P
loud orbit
but B is what we're trying to find
shut vortex
Yes
loud orbit
so how can we have B if we're trying to find it
shut vortex
We have B written with respect to the basis weāre sending it to
loud orbit
????
loud orbit
you mean a coordinate vector?
loud orbit
if I'm given a basis and a vector, I find the coefficients of the linear combination that I would need to write the vector in terms of the basis vectors
shut vortex
Correct
So when youāre constructing a transition matrix
You do that with the basis vectors of the basis youre in, with respect to the basis youre trying to go to
But the basis weāre going to is the standard basis
loud orbit
You do that with the basis vectors of the basis youre in, with respect to the ba...
what do you mean? you use rref to convert from B to B'
because then you find the matrix that when multiplied by B gives you B'
loud orbit
listen. If I know the basis I'm trying to go to, and I know the matrix that takes me there, I need to find the original basis. I can't just state that the columns of the transition matrix equal the original basis at the beginning, because the new basis might not necessarily be I
shut vortex
The columns of the transition matrix
Are the basis vectors of B
Written with respect to S
By definition
loud orbit
what does this mean though? And how do you know this?
shut vortex
Itās by definition
If youāre working with transition matrices you have to know this
That is what a transition matrix literally is
loud orbit
I mean in the general case, not when B' = S
loud orbit
shut vortex
Yep
loud orbit
so I can't state that the operations I do to B' will make the vectors of B equal the column vectors of the transition matrix, because I don't know what operations I'm performing on B'
my notes say that the vectors of B in the basis of B' equal the column vectors of the transition matrix, but this isn't the same as the vectors of B itself. That's the vectors of B with respect to B'. As far as I'm concerned, the vectors of basis B are with respect to nothing
shut vortex
If you have a basis ānot respect to anythingā it is implicitly with respect to the standard basis
Which is the basis S
Like the vector in your transition matrix [1 2 1], that 1* [1, 0, 0] +2* [0, 1, 0] + 1*[0, 0, 1]
shut vortex
Which bit
loud orbit
the matrix in pink
shut vortex
Itās probably best to do it with an example
loud orbit
what is k1 and k2
oh it's the old vector
I understand the procedure from converting the basis B to B'. You put them in an augmented matrix and rref it to get the transition matrix. That transition matrix when multiplied with some vector w that is relative to the original basis B will equal w relative to the new basis B'. It get that
I understand the procedure. I don't get why it works but I know that it works
I also know that reversing the operation, P B'->B = the inverse of the transition matrix P B->B'
shut vortex
Good
loud orbit
rihgt, but then why can you say that the column vectors ofthe transition matrix is the vectors of the original basis
because this isnt true in that example
3 != 2
shut vortex
In what example
shut vortex
Itās the vectors of the original basis, written with respect to the new basis
Letās look at the first column u1
u1= (2, 1, 1)
In the transition matrix we have (3, -2, 5)
loud orbit
ok
shut vortex
So what was my statement, the columns of the transition matrix are the basis vectors of the old basis written with respect to the new basis
So how do we write u1 with respect to the other basis?
loud orbit
oh, if you rref it, it makes the vector with respect to B'... it's the same as just doing it with a single vector
[B' | u1] rref is [u1]B'
wait so then why is the transition matrix all three vectors of the original basis with respect to the new basis?
shut vortex
Well thatās a slightly more complicated question
A handwavey answer is that the image of every vector is determined by the image of the basis
loud orbit
like why is it that when you smush all the vectors of the old basis B with respect to the new basis B', you get a matrix P {B->B'} that you can multiply with vectors w that in span{B} to get [w]B'?
oh, it's complicated
shut vortex
Youāre asking why transition matrices work, itās hard to answer using text
No itās not too complicated
But there are hundreds of proofs online Im sure
loud orbit
We havent laerned transformations yet, if that's what oyu mean
there was nothing about multiplying transition matrices in the notes...
shut vortex
Well it sort of just makes sense anyway
loud orbit
putting them in a single matrix or multiplying transition matrices?
shut vortex
Well no the fact that a transition matrix has such columns isnāt very intuitive immediately
loud orbit
oh
shut vortex
But multiplying them is
shut vortex
Like if I have 2 transition matrices P1:A->B and P2:B-C then do P2 * P1(v) where v is some vector written wrt A
Then P1(v) sends it to B
Then P2 sends it from B to C
loud orbit
uhhh?
oh
because P times a column vector is a column vector
so P1 * v = a column vector, times P2 is a column vector
so (P2 * P1) v = a column vector
but why would that product be A->C and not C->A?
oh because it's to the left
no but that shouldnt make sense
because
P_{B' -> B''} = P_{B->B''} P_{B'->B}
shut vortex
but why would that product be A->C and not C->A?
You started in A, then went to B, then P2 took you from B to C
loud orbit
so the right matrix, the first basis is the input, and then it goes to the second basis. then in the left matrix, the first basis is the second basis, and then it goes to the third basis which is the output
I appreciate you @shut vortex
sorry for being difficult..
shut vortex
Itās tough to wrap your head around itās no problem
Yeah because with functions (matrices) you do it right to left remember
loud orbit
yeah..
i have an exam in a week and a half and I still haven't learned this section and 4 other sections
19th
so I have 10 days to study
well, 9 now that today is almost over
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hazy lily
Having a hard time understanding why this last part is incorrect. If anybody could help me out with some guidance
sly garden
are a, b, c, d all correct?
hazy lily
yes
sly garden
also you win bonus points because your handwriting is legible and nice :)
hazy lily
thank you I tried my best lol š
sly garden
looking at pt e again
sly garden
I also tried inputting it as S286.83^oE
i think you might need to rethink this
because given the way the initial question formats it iām not sure you want a number larger than 90
like 290Ā° got formulated as S20Ā°E
hazy lily
okay I see what you're saying
Okay! I did -73.134...+90=S16.87E and it was correct
Thank you so much for the help!
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sharp smelt
pearl pondBOT
sharp smelt
Here I need $\iint_{S} \frac{xy}{6Ļ}dxdy - \int_{-2}^{2} \frac{x}{6Ļ} \int_{-3}^{3} \frac{y}{6Ļ}$
jolly parrotBOT
wai
sharp smelt
unsure of how to neatly integrate over the ellipse
polar is an option, but won't that be messy
I suppose I try elliptical coordinates?
steep saddle
how would polar be messy
autumn trellis
well, it would seem so that you use elliptical coordinates, considering you are integrating over an ellipse
sharp smelt
how would polar be messy
We're integrating over an ellipse
Let $x=2r cos(t); y=3r sin(t)$
the jacobian would be 6r
jolly parrotBOT
wai
sharp smelt
although im not sure how you got the two integrals in the second term
E[X] ; E[Y]
autumn trellis
i know that
steep saddle
**wai**
i dont see how this is messy? its doable and like if you dont want to, you dont have to
do you even have to do the integral? it's over an ellipse and the integrand is odd so then ||it's 0||
sharp smelt
i dont see how this is messy? its doable and like if you dont want to, you dont ...
oops
didn't notice that
right
so it's just 0
tq
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buoyant fulcrum
,rccw
61st
pearl pondBOT
,rccw
61st
jolly parrotBOT
toxic lichen
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
ok
buoyant fulcrum
What is tried is making an equation of line of slope 60Ā°
toxic lichen
bruh.
buoyant fulcrum
Passing through (2,3)
toxic lichen
that sounds like status 2 mate why do you lie
buoyant fulcrum
I mean i don't think if that's a true or actual approach
toxic lichen
the dumb way is roughly:
- find the eq of the line parallel to sqrt(3)x - y + 1 = 0 passing thru (2,3)
- intersect that with the line 2x-3y+28=0
- find the distance from (2,3) to that intersection point
buoyant fulcrum
Yea
toxic lichen
maybe there is some prebaked formula they expect you to mug up and mug up and mug up
buoyant fulcrum
I just got that thought and got ur message
toxic lichen
i think my way must be considered unsuitable as it takes more than 2 minutes
compact ridge
if you're studying for JEE, you should know the formula for the shortest (perpendicular) distance)
toxic lichen
if you're studying for JEE, you should know the formula for the shortest (perpen...
but we're not dealing with perp distance.
we're dealing with distance measured along a specific line.
compact ridge
then use trig
ah but then the slope isn't neat for 2x - 3y + 28 = 0
I think it would be easiest to find the intersection point then distance formula, can't really see any other way
buoyant fulcrum
if you're studying for JEE, you should know the formula for the shortest (perpen...
Idk formulae for these i just derive on the spot and get fucked up in mains
But perform no bad in Advanced
buoyant fulcrum
the dumb way is roughly:
1. find the eq of the line parallel to sqrt(3)x - y + 1...
X is just so weird ewe
toxic lichen
I think it would be easiest to find the intersection point then distance formula...
impossible
jolly parrotBOT
buoyant fulcrum
Ohk
I'm doing everything with approxing stuff and ticking up the closest option
I got A
Let's check
And yea it's exactly wrong
Bruh
Answer is D
Hell
Oh god I did a freaking sign mistakes hate
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pearl pondBOT
buoyant fulcrum
Thank You @toxic lichen
I'll do it further
Yea this time even my approx answer matched
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brave mesa
hello
steep saddle
first i would get rid of negative exponents
toxic lichen
progress?
brave mesa
i got on paper but i dont know how to write it here
toxic lichen
send a photo
toxic lichen
and the 16 should also be a 16^(1/4) (a.k.a. 2)
but also uhhh the notation is a bit troublesome i think
wanting to simplify one fraction at a time is a good idea but dont mix them too early yet
also separate the "4" that designates the question number from the q itself
so then: $\left( \frac{2c^2b^{-1}}{a^3b^2} \right)^2 = \frac{4c^4}{a^6b^6}$ and then \textbf{on the next line}\ write $\frac{(b^{1/2}c^{3/2}a)^2}{(16c^4)^{1/4}} = \frac{bc^3a^2}{2c}$
jolly parrotBOT
Ann
toxic lichen
i dont get how b^1/2 becomes 3
what do you mean "becomes 3"
in $bc^3$ the 3 applies to the $c$ only...
jolly parrotBOT
Ann
toxic lichen
do you understand the index law (b^x)^y = b^(x*y)
brave mesa
ahhh i get it
toxic lichen
this law applies even when x and y are scary like fractions or negatives or NEGATIVE FRACTIONS or whatever
brave mesa
yes ok ive written down ill try the third fraction
toxic lichen
im gonna have to go
jolly parrotBOT
coehen1231
pearl pondBOT
@brave mesa Has your question been resolved?
pearl pondBOT
@brave mesa Has your question been resolved?
midnight haven
hello
midnight haven
i need help with some arithmetics
unkempt yacht
ask your question
feral olive
!da2a
pearl pondBOT
No need to ask āCan I askā¦?ā or āDoes anyone know aboutā¦?āāitās faster for everyone if you just ask your question!
midnight haven
i have barely delt with arithmetics i need some guidance to solve this
translation : let a and p be 2 natural numbers bigger than 2
show that if a^p - 1 is prime
then a=2 and p is rime
lean oak
let a and p be naturals larger than 2, then show.... then a = 2... (?)
is that a strict inequality in the initial question statement? because otherwise the question contradicts itself.
midnight haven
is that a strict inequality in the initial question statement? because otherwise...
i guess is isn't
crimson valve
1=1^p, so a^p-1 =a^p-1^p=(a-1)[a^(p-1) ....]
midnight haven
cause 3 is prime so not strictly i think
lean oak
the problem with the question statement is about a, not p. but either way I think you've been given a hint.
pearl pondBOT
@midnight haven Has your question been resolved?
midnight haven
the problem with the question statement is about a, not p. but either way I thin...
got it
thanks everyone
midnight haven
the problem with the question statement is about a, not p. but either way I thin...
yeah i agree they didn't specify enough
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lucid moth
pearl pondBOT
toxic lichen
you're approaching it from the left in your pic
lucid moth
Ahh that was gonna be my question
So when the graph is like this its not possible to approach the limit from the RHS?
toxic lichen
when the graph has no part to the right of the red line? yeah
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meager hornet
does anyone know what ''Reference to the 1st quadrant'' is? i cant find a single good explanation of it, it only shows that sin(-Ļ)=-sin(Ļ) and others similar to that one
meager hornet
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Can some one help.me
tardy anchor
When we find the determinant of a matrix so we just cross multiply the elements given in the bracket but why we use minus (-) in b/w two elements?
tardy anchor
So minus sign is in the formula ?
tropic saddle
I could tell you this follows from the leibniz formula but I assume that is not satisfactory either
the formula is ad-bc
thats just what it is
tardy anchor
But in our book their is no formula for determent
wary vault
well the determinant of a 2 by 2 matrix is the same as finding the area under the parallelogram the two vectors create right? i think you could evaluate it graphically and itd be the same. also sorry for intruding
tardy anchor
well the determinant of a 2 by 2 matrix is the same as finding the area under th...
Yee but their is a question that just find the determent tbh
wary vault
Yee but their is a question that just find the determent tbh
im just telling ya what it evaluates when you have two vectors. in general tho it can mean a lot of things so idk
tropic saddle
But in our book their is no formula for determent
well shitty book I guess
tardy anchor
the formula is ad-bc
Ok so the formula is ad -bc ok thanks
tardy anchor
im just telling ya what it evaluates when you have two vectors. in general tho i...
Hmm
pearl pondBOT
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torn willow
are these formulas correct? smth feels off because of the f(x-n)^2 term and the f(x+n)^2 term
torn willow
i found them in this link - https://germanna.edu/sites/default/files/2024-04/Calculus Disk-Washer and Shell Methods.pdf
and i don't understand why they subtract x inside the function
pearl pondBOT
@torn willow Has your question been resolved?
torn willow
<@&286206848099549185>
pearl pondBOT
@torn willow Has your question been resolved?
torn willow
<@&286206848099549185>
pearl pondBOT
@torn willow Has your question been resolved?
humble root
are these formulas correct? smth feels off because of the f(x-n)^2 term and the ...
You shouldn't really memorise these formulas. You only need to memorise the first one, and that apply transformations to get rotation around a certain axis.
That being said, I think the horizontal and vertical axis rotations are a little off. The rotation of f(x) about y = n should be f(x) - n rather than f(x - n). The reason being is because you want to shift everything down so that it's with respect to the x axis. Same reasoning with respect to y.
torn willow
You shouldn't really memorise these formulas. You only need to memorise the firs...
Oh no no noā¦I never āmemoriseā anything without actually understanding and deriving the formula. I just wanted some confirmation about whether those formulas were correct bc I thought they were a little off
Anyways ty for confirming
.close
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wary vault
hello. if we prove via induction that the sum from i=1 to n of "n choose i" is 2^n, can we explain that |P(A)| = 2^n, where n is |A|? Or would it be considered kinda informal?
wary vault
(and P(A) is the powerset)
tulip ore
hello. if we prove via induction that the sum from i=1 to n of "n choose i" is 2...
you need to mention n choose i
and specifically how adding it would be 2^n
just saying or implying it would be so is not enough
also, you can press Shift+Enter on PC to insert a line break into your message
I dont know what youre trying with the spaces
tulip ore
what do you think
what would "it" refer to in these pair of messages
knowing that this is your problem
wary vault
we proved it earlier via induction supposedly
my question refers to whether we can explain with words how it connects to the powerset of A
tulip ore
please screenshot the problem you are trying to solve
wary vault
well i had an exercise referring to n choose i, but the question here is based on interest not on necessity
tulip ore
your question depends on the content and context of the problem you got the idea from
plush bramble
!original
pearl pondBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
wary vault
hmm?
tulip ore
for example, if the question was only ask to prove the statement by induction,
wary vault
okay lemme put it in a tex i guess
tulip ore
thats not the original content
tulip ore
for example, if the question was only ask to prove the statement by induction,
mentioning the power set would be to ask "is there a more direct way to prove this using a combinatorial argument involving the power set?"
in that case, you need to mention more than just the power set
are you just adding a note that 2^n is equal to the power set of A?
wary vault
,tex
Suppose we already have proven that $\sum_{i=0}^{n} \binom{n}{i} =2^n $. Can we then explain that the amount of members of P(A), where A is a set of n members, is $2^n $? Or would it be informal as a proof?
tulip ore
please listen to my instructions closely
wary vault
sorry i was writing lemme read
tulip ore
,tex
Suppose we already have proven that $\sum_{i=0}^{n} \binom{n}{i} =2^n $. Ca...
this is also not how you latex this
use \binom{top}{bottom} to write the binomial coefficient
wary vault
well i didnt know damn
tulip ore
you can drop the braces if only one "thing" is being processed, for example: \binom ni
this will do: $\binom ni$
jolly parrotBOT
mtt
jolly parrotBOT
mtt
tulip ore
you can see here the other letters arent considered part of it since theres no braces to include them
jolly parrotBOT
fijokazż
jolly parrotBOT
mtt
tulip ore
commonly the powerset is denoted with this special P, which is latexed as \mathcal{P} or (dropping the braces) \mathcal P
you are also forgetting to typeset the A
and the n
wary vault
i just stated my question clearly. i dont mind if its not perfectly notated as long as it stays within the confines of logic
wary vault
can i not just be interested in it?
tulip ore
you posted it as an answer to two questions which youve now just ignored
now I have to mention them again
wary vault
wait im confused sorry
tulip ore
please screenshot the problem you are trying to solve
as I was saying,
you need to screenshot the original problem that you got this idea from
wary vault
i texed it because riemann kinda said that i didnt provide a clear question
tulip ore
right, well sorry to break to you but you misread what riemann said
please stop typing until Im done explaining what is happening
the problem with answering your question is that we don't know what kind of answer youre adding this note to
if its just a note that 2^n happens to be 2^|A| = P(A), to add to the end of your induction proof as a bit of a side note, then thats alr
this is called a "remark"
youre adding an extra fact but not elaborating on it, its noteworthy that theyre the same
but you wont put in any additional words to it besides mentioning it
again, please stop typing
there are two different problems at play here
there is first the question you asked
then there is the original homework problem that, after solving, you got this question from
can you understand that these are two separate problems?
yes or no
wary vault
yes
tulip ore
we will denote the original homework problem as P1
and the question youre currently asking as P2
again, please stop typing
it seems pretty clear to me that the note youre going to add is harmless, you should just be able to add this
but now youve added a whole lot of confusion here that makes me real suspicious
wary vault
can i pls say smth
tulip ore
Please show the original problem, **exactly as it was stated to you**, with the ...
that is what riemann is referring to:
youll distinctly notice a flaw in the bot comment
it cannot distinguish between P1 or P2
but commonly it refers to P1
and almost never to P2
that is also what I am referring to
I want a screenshot of P1
not a texed reimagining of P2
what purpose would that even serve?
we can already read P2
wary vault
can i please talk now?
tulip ore
almost
wary vault
hahah
tulip ore
almost does not mean yes
wary vault
oke lemme explain briefly
tulip ore
its a bold statement
wary vault
mtt do you hate me? what did i do to you?
wary vault
we couldve resolved this problem in 2 mins
ancient elk
amirite
tulip ore
mtt do you hate me? what did i do to you?
you interrupted me like what, three times?
wary vault
im sorry
tulip ore
and each time we had to resolve whatever you said
whatever mistake you made
as before, its just a remark, you can add it no problem
but now I dont know if you even mean that remark
or youre going to pull out some other thing that you never elaborated on that I dont agree to
wary vault
can i explain my motivation?
tulip ore
buddy
you already explained your motivation
your motivation is just that you noticed these two things are equal, so youll try to add it at the end of a proof as like a side note
wary vault
no
thats why monologues dont help with two ppl
its just for fun. i dont intend to use it or show it to my professor. i just wondered abt it.
wary vault
quote me
tulip ore
hello. if we prove via induction that the sum from i=1 to n of "n choose i" is 2...
look at P2 here
there are 4 words here
point out the 4 words that indicates youre not just wondering
but writing
wary vault
i honestly do not know
tulip ore
thats a shame
"can we explain that" are the words
commonly this indicates that you are going to write an explanation to your proof
rough forge
ellinas re
wary vault
yeee
tulip ore
no?
wary vault
lemme finish please
when i started wondering abt the powerset, i was thinking of the hypothetical: what if someone were to ask me to prove that |P(A)| = 2^{|A|} ?
thats why i asked abt explaining it
am i getting it twisted?
like sorry if i was misinterpreted
tulip ore
then let me repeat what I said before
you need to mention n choose i
and specifically how adding it would be 2^n
however,
if youre just treating it as a question on its own, there is an easier approach to proving this
you still havent exactly said the chain of events that linked one idea to another,
wary vault
you still havent exactly said the chain of events that linked one idea to anothe...
i can explain that
tulip ore
well before you do that, a quick question to maybe save some time
Im guessing you already know 2 separate approaches (binary, and (n choose i)) to proving that |P(A)| = 2^|A|?
wary vault
nope
tulip ore
thats good, keep going then
wary vault
So when we learnt induction, our teacher proved |P(A)| in a weird way that i didnt quite grasp
and now that i saw this problem, it spawned in my head that if i were asked that question i could use this result with n choose i to prove it
tulip ore
So when we learnt induction, our teacher proved |P(A)| in a weird way that i did...
now youre gonna have to elaborate on this weird way
wary vault
if i say i didnt like it will it be enough?
tulip ore
you need to say what the way actually was
wary vault
i just wanna find a way that i can remember better
tulip ore
theres a particular portion here that you havent elaborated on, Ill need to know what the way actually is to see if your approach holds up or not
seems unrelated at first but trust me on this
wary vault
do you need to know when i went to the toilet today?
like damn
im sorry but
you got me stressin
versed mica
do you need to know when i went to the toilet today?
yea, what time?
tulip ore
do you not know that the "weird way" could 100% be the simplest way but delivered incorrectly?
wary vault
yea, what time?
trick question, didnt go
versed mica
man is constipated
wary vault
do you not know that the "weird way" could 100% be the simplest way but delivere...
for me it is not. and it interests me not
tulip ore
well thats no good
because theres a simple way to prove that 2^|A| = P(A) that I think is easier to remember than this (n choose i) way
wary vault
i disagree with you that so much context would be needed
tulip ore
I cant fit every problem I can mention about your approach in short, fast messages
wary vault
because theres a simple way to prove that 2^|A| = P(A) that I think is easier to...
if i saw that way and didnt like it it means my opinion is i prefer n choose i lol
tulip ore
thats an
extremely strange and bold thing to say
thats like monologue levels of bold
you havent even seen the way yet and youve thrown a wild if statement at me
do I also need to stop typing too now
wary vault
huh? you said i mightve already seen it and i told you if i saw it and didnt like it it means i prefer my way
tulip ore
you havent even seen the way yet, and already youre trying to compare it to the (n choose i) way
some ways arent very comparable because their usefulness are in entirely different areas
wary vault
is this argument 101?
tulip ore
you tell me, you started the statement
wary vault
i rlly did not
i had a simple question regarding validity and you made me go thru my archive and shit like damn
tulip ore
well yea
you inserted 4 words that werent supposed to be there
and now it looks like your approach is a bit shaky
wary vault
in your opinion
tulip ore
Im going to repeat what I said before:
wary vault
its just my way of expressing a hypothetical
tulip ore
so far your proof seems to be:
inductive proof that these terms add up to 2^n
=> 2^n = |P(A)|
and you want to ask about this => step, whether itll work or not or be too informal
wary vault
and now it looks like your approach is a bit shaky
thats cool. you couldve said that 100 million lines earlier
tulip ore
the problem is
I sort of
did
multiple times
in many different ways
but I will say it again for you inca se you missed it
your approach is missing a crucial element which is to involve the (n choose i) stuff in P(A)
you need to explicitly mention that youre choosing i out of n elements then adding them together
wary vault
you said it and then started correcting the way i wrote every single thing
tulip ore
...dry?
tulip ore
I havent even begun on the explanation yet
versed mica
š
pearl pondBOT
tulip ore
what the hell am I supposed to do with this? you havent even explained your question properly and the parts you did explain have so many holes I cant fix them all
wary vault
.reopen
pearl pondBOT
ā Original question: #help-39 message
wary vault
i dont wanna be disrespectful
tulip ore
then please understand that your actions have consequences
wary vault
okay? and youve kinda annoyed me even tho i know im at fault
tulip ore
.close
tired narwhal
I'm stuck on how to start a proof for this question. I've got
$P(n):n=2^k\cdot l, \quad k\ge 0, l$ odd.
But then, I'm not sure. I'm very new to induction proofs also and this type of question is unfamiliar
jolly parrotBOT
charnixe
pearl pondBOT
@tired narwhal Has your question been resolved?
stable flame
lemme think about this
it sounds interesting
im assuming u gotta separate the problem for where n is odd or even
when its odd the obvious way to write it would be k = 0 l = n
stable flame
no idea what that means
lemme search it up
ohhhhhhh
wait so that brings something to mind
tired narwhal
ye what
tired narwhal
I think I separate for positive and negative N, and do the base cases for each
stable flame
i dont think positive or negative changes
tired narwhal
yeah idk
stable flame
because a change in sign would just indicate that l is negative/positive
as 2^k cant be negative
if n is even
then its divisible by 2
stable flame
why
tired narwhal
because thatās not referring to the formula or proving that it works
stable flame
take 76
stable flame
because thatās not referring to the formula or proving that it works
yes it is
tired narwhal
I wanna know the base cases
tired narwhal
ye but this isnāt an induction proof
stable flame
lemme formulate an awnser
tired narwhal
it could be a step but like how do I write this
stable flame
idk ive never studied types of proofs before
didnt know it had to be a specific type
lemme search about it
tired narwhal
Actually no yours does work but my teacher said to use strong induction
stable flame
i think i got how induction works
lemme try making up some logic with that
i might be wrong tho
first time even doing an induction proof
stable flame
whats ih
tired narwhal
Inductive hypothesis
stable flame
is that not it
tired narwhal
You canāt do an induction proof in one line
stable flame
i thought proving that something works for N and N+1 was induction
tired narwhal
u havenāt proved base cases
stable flame
i mean u can do the rest
tired narwhal
I mean not really
stable flame
if n is odd is above
tired narwhal
usually in induction you try to GET to 2x FROM 2^(k+1)l too, you donāt just start with the final cuz itās a proof
tired narwhal
i mean the orders of the eggs doesnt change the omelet
Lmfao this guy
now Iām hungry
stable flame
lmao
tired narwhal
i started with 2^k.l
Yeah yeah brotato you got the logic but the issue is how to get there like idk what base cases to use cuz itās positive and neg integers 
stable flame
that doesnt change the proof tho
if n is negative
l is negative
if n is positive
l is positive
which is described by my odd logic
if n is odd k = 0 n = l
tired narwhal
Yes ofc but you need to quickly show this with n = -m, m = |n|
And also cuz itās strong induction my base case is 1 but then what š I have to assume for something but I forgot how
My burger has everything except the buns
somber gull
please help
tired narwhal
Open a new channel this is occupied
oh hang on I get it now
m is in interval of 1, n
.close
pearl pondBOT
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thorny agate
hi, can someone do this problem? i did most of it in my head i fear, but generally i used l'hopital's rule to get 16/4 which i simplidied to 4
thorny agate
ive redone it a few times im so sure its 4
jolly parrotBOT
pearl pondBOT
pearl pondBOT
Available help channel!
brazen night
Hello I need help getting the work for this answer key
pearl pondBOT
Hello I need help getting the work for this answer key
brazen night
This is the work Iāve done so far
Bruh this WiFi
The work I done
The answer key
What steps do I take next cuz idk how to
brazen night
I think so
sacred nimbus
okay
brazen night
Yes
sacred nimbus
let me see ur steps
brazen night
The work I done
The wise are the steps I took so far
sacred nimbus
The work I done
i see an error
brazen night
Yes please tell me
brazen night
Heās
sacred nimbus
first line u wrote in the 2nd bracket
while simplyfiyng the exponent of m
m / m^-3
brazen night
54:
sacred nimbus
u wrote as m^-2
brazen night
Yes
sacred nimbus
incorrect
brazen night
Oh shoot
sacred nimbus
it should be??
brazen night
M^4
sacred nimbus
no
u again did an error
i would highly advise speaking out what u write
in the same speed as u write
brazen night
Yes