nice
#help-28
slate glacier
oh wait, I corelated it with the power, whoops
white karma
np
slate glacier
so what does the theorem tell us?
that for a^(number of relative primes of n), the remainder of it when divided by n is 1?
white karma
Right, so here a=7 and n=11 as you mentioned. So first we should compute φ(11)
and what's the value of that?
slate glacier
okay, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, so 10
white karma
Right, in general φ(p)=p-1 for any prime number p since all positive integers <p are relatively prime with p
Ok so now we know that 7^10\equiv 1 mod 11
slate glacier
yes
white karma
But we want to know what 7^102 is mod 11
white karma
Ok so now we know that 7^10\equiv 1 mod 11
We will use this fact
slate glacier
can we raise both sides to the power 10 (if that works here)?
white karma
Nice, and this works
you can see that it works like this: 7^100=(7^10)(7^10)...(7^10)\equiv 1x1...x1\equiv 1 mod 11
slate glacier
yes I see that
slate glacier
oh wow, this is good
white karma
Yeah this simplifies things nicely
slate glacier
Okay wait, how about I do 77^77 mod(8) step by step, and you correct me if I mess up
white karma
Sure, that's the spirit. After all if you don't try to do it yourself you won't gain much
slate glacier
okay, so we have 77^77 mod(8)
slate glacier
and we have euler's formula, a^(relative primes of n) equi 1 mod(n)
white karma
Nice but remember the condition
slate glacier
so here, a is 77, and n is 8
white karma
There should be something satisfied between a and n right?
Otherwise you can't use euler's theorem
slate glacier
oh yeah, was it that n can't be a divisor of a?
slate glacier
1
white karma
Nice
slate glacier
okay okay, so a = 76 and n = 8 wouldn't have worked?
white karma
Yeah Euler's theorem probably wouldn't have worked here
but you could've done something else
Note that 76=19x4
and 4²=16\equiv 0 mod 8
slate glacier
ooooooo, yes I see what you mean
white karma
so 76^k\equiv 0 mod 8 for any k>=2
slate glacier
oh wait yeah
okay so coming back to the steps
So a = 77, n = 8
function(8) = 4
white karma
correct
white karma
Nice
white karma
Equivalently, 77=80-3
slate glacier
ahhhh yes yes
slate glacier
sure man, damn
Can I ping you if I have some stuff related to this later, as I have to go somewhere
white karma
Sure, ping me whenever you want and I reply whenever I can
I probably won't reply in the next few hours since it's 2:40 am here so I will go sleep after a few mins
white karma
np, I hope this was helpful and not much boring.
slate glacier
np, I hope this was helpful and not much boring.
It was great man, thank you
white karma
Btw are you doing this for a course in uni like discrete math?
Or are you just self studying
slate glacier
I've got an exam on 23rd
white karma
Ohhh I see
slate glacier
We'll talk later, can I add you?
white karma
Yeah sure
slate glacier
Alright, see ya man, and thanks again🫡
white karma
Cya, gn
slate glacier
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scarlet dome
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2c
silk bridge
what have you tried ?
silk bridge
I am confused
!show
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I am confused
Show your work, and if possible, explain where you are stuck.
scarlet dome
Which one to use as principal
The net int of 1yr - tax + principal of 1st year or
CI of 1st year + principal of 1st year
silk bridge
The net int of 1yr - tax + principal of 1st year or
net
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thick hedge
$ \frac{1}{\sqrt{2π}} \int_{0}^{\infty} \frac{x^r}{x} e^{ - \frac{\ln^2(x)}{2}}dx$
\Let $\ln(x)=t$
\$ \frac{1}{\sqrt{2π}} \int_{-\infty}^{\infty} e^{tr} e^{ - \frac{t^2}{2}}dt$
glossy valveBOT
wai
thick hedge
unless I'm missing something , there's no neat way to integrate this?
Would finding the MGF be a better idea here
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spice pilot
Is the following claim true? If $B(x,r)\subset B(y,s)$ and $r<s$, then $\overline{B(x,r)}\subset B(y,s)$. Here's my attempt: Let $z \in \overline{B(x, r)}$. Then, $\forall \epsilon > 0, \exists p \in B (x,r)$ such that $d(p, z) < \epsilon$. Pick e.g. $\epsilon=(s-r)/2$. Hence, $$d(y,z) \leq d ( p, z ) + d ( y, p ) < s + \epsilon.$$ How do we proceed?
glossy valveBOT
psie
split hatch
The claim is false
spice pilot
The claim is false
Ok. That's good to hear.
split hatch
B(1/2, 1/2) = (0, 1) is a subset of B(1, 1) = (0, 2) but the closed ball \bar B(1/2, 1/2) = [0,1]
spice pilot
B(1/2, 1/2) = (0, 1) is a subset of B(1, 1) = (0, 2) but the closed ball \bar B(...
If we demand that y in B(x,r), the claim becomes true, right? 🙂
spice pilot
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vernal rose
I have a line that goes from (1, x1, 0) to (0, x2, 1). Looking from (0,0,0), How can I solve, which point in the line I have "steepest" angle to?
silk bridge
I have a line that goes from (1, x1, 0) to (0, x2, 1). Looking from (0,0,0), How...
can you rephrase
!xy or
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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.
turbid badge
hmm i guess it'd be whichever point is farther to the origin
silk bridge
hmm i guess it'd be whichever point is farther to the origin
or is perpendicular to the origin?
turbid badge
or is perpendicular to the origin?
that would be 90° which is the most, but ig we'll know when the ambiguity is clear
vernal rose
hmm i guess it'd be whichever point is farther to the origin
If I have line that goes through (1, 1, 0) and (0, 1, 1) then the point for steepest angle is in the middle of those points
silk bridge
If I have line that goes through (1, 1, 0) and (0, 1, 1) then the point for stee...
wdym by steep closer to 90?
turbid badge
i forgot the meaning of steep 😭
turbid badge
wdym by steep closer to 90?
probably what it should mean
turbid badge
If I have line that goes through (1, 1, 0) and (0, 1, 1) then the point for stee...
oh
vernal rose
vernal rose
wdym by steep closer to 90?
Yes!
I am trying to find an point with angle closest to 90 degree from origo.
(For extra context: I am trying to find a single point to rotate towards instead of rotating towards x and then z... and I have hard time finding the single point)
y is the height axis in this picture just for clarification
vernal rose
Huh ... was it that simple : DDD
silk bridge
if you want steepest
silk bridge
vernal rose
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✅ Original question: #help-28 message
vernal rose
I have doubts
For line that goes through (0,0,1) and (1,1,0) the steepest point is (1,1,0), which is not perpendicular to the line from origin
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@vernal rose Has your question been resolved?
silk bridge
which plane do you want to maximize it with
vernal rose
which plane do you want to maximize it with
The "ground" plane. I call it xz (and height is y).
silk bridge
ok
vernal rose
Highest this value in the line (y / sqrt((x-z)^2 + (z-x)^2) 😄
My Bachelor's degree was over third just math courses and I can barely remember anything
silk bridge
i'm not university
silk bridge
You are not an university
yh
silk bridge
Highest this value in the line (y / sqrt((x-z)^2 + (z-x)^2) 😄
but you have to maximize this angle
y here is 1
but idk how to find vecotr equation of a line
vernal rose
In the picture y is 1. In reality y can be anything.
silk bridge
yh but we need to find for given
silk bridge
Highest this value in the line (y / sqrt((x-z)^2 + (z-x)^2) 😄
if you find vecotr equation of the line then you just have to maximize this
by finding maximum
vernal rose
Yeah I just forgot how to find maximum value. But I think that it is easily googleable... so I try that
Hmm it's very doable. I can simplify that even further. Then maybe do derivative on it or something similar.
silk bridge
Hmm it's very doable. I can simplify that even further. Then maybe do derivative...
oh, yh hm.. you find critical points and check for maxima
vernal rose
x range is from 0 to 1. and then the line has points (0,x1, 1) and (1, x2, 0). x1 and x2 are some constants.
Then I can try to generate function that tries to find the maxima of (x * x1 + (1 -x) * x2)/sqrt(x^2 + (1-x)^2), if given the x1 and x2 to it.
I have to go now, so I let the bot close this
Good chance to solve this with this amount of knowledge 
.close (Huh it might be that the x = (x1 / x1+x2) have to double check when I get back)
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digital void
,tex
Hello. I am trying to prove the following statement. Is the proof correct?
$\forall \ \text{sequences} \ x_n , \ \lim_{n\to\infty} x_n = \infty : \ \lim_{n\to\infty} f(x_n) = l \implies \ \lim_{x\to\infty} f(x) = l $
\vspace{0.23cm}
Proof: We assume that the negation holds. So:
$$ \left(x_n \to \infty , f(x_n) \to l \right) \land \left( \lim_{x\to\infty} f(x) \neq l\right) $$
\vspace{0.25cm}
Meaning: $$ \forall M_1 >0 , \exists n_0 \in\mathbb{N} : n\geq n_0 \implies x_n >M \ (1) $$
$$\forall \epsilon >0 , \exists n_0 : n\geq n_0 \implies \ x_n >M \implies \lvert f(x_n) -l\rvert <\epsilon \ (2) $$
$$\exists \epsilon ' >0 , \ \forall M>0 : \exists x: x>M \ \text{and} \ \lvert f(x)-l\rvert >\epsilon \ (3) $$
So, take $\epsilon = \epsilon ' $ on (2) and choose a sequence $x_n >M , n\geq n_0 $. We then have that (2) and (3) hold simultaneously (for $x=x_n $ in (3)). Perchance.
glossy valveBOT
fijokazż
digital void
i forgot the epsilon' at the end of (3) btw
rapid rain
"You can't just say Perchance"
thorny isle
You can't choose your sequence. You know that it exists but nothing more. So you use said sequence from the begining. You have the right idea. There exists $\epsilon >0$ such that for all $m \in \mathbb{N}$, there exists $n\ge m$ such that $|f(x_n) - l | > \epsilon$. Now, take $M>0$, there exists $N \in \mathbb{N}$ such that $\forall n \ge N, x_n \ge M$. You will be able to conclude by taking $M \in \mathbb{N}$ and applying the first statement.
glossy valveBOT
bourbactive
rapid rain
Statement (1) I presume
rapid rain
I'll add that the proof as you wrote it is incomplete. You do need to explicitly create a sequence $(x_n)$ such that $\lim_{n\to\infty}x_n = \infty$ (by create I mean show how you pick each term) and $\lim_{n\to \infty}f(x_n) \neq l$. So yes the idea is to apply statement (3) to increasingly bigger values of $M$.
glossy valveBOT
Rafilouyear2026
digital void
can we not create our x_n via induction?
rapid rain
you could, but why do induction when you don't need it particularly
digital void
im not sure i know any other way
rapid rain
**fijokazż**
by property (3), you can fix your epsilon'
so apply it to M = n for all n
rapid rain
$$\forall n\in \bN : \exists x_n: x_n>n : \text{and} , \lvert f(x_n)-l\rvert >\epsilon'$$
glossy valveBOT
Rafilouyear2026
digital void
why do we need it to hold for all n naturals?
rapid rain
why do we need it to hold for all n naturals?
? you pick x_1 with property (3) applied to M = 1
you pick x_2 with property (3) applied to M = 2...
etc
digital void
oh okay thats easier to see
rapid rain
so you pick x_n with property (3) applied to M = n, for any n
digital void
okayokay
how do we connect it to (2)?
like, make them be true simultaneously
we take ε= ε' and M=n for all n
so we dont need the n_0 i assume
rapid rain
I noticed that statement 2 doesn't make sense
digital void
oh lol
rapid rain
where does the M come from
digital void
(2) is true for any positive real, so we take that number to equal M
rapid rain
The definition of $\lim_{n\to\infty} f(x_n) = l$ is the definition of sequence converging to some finite value, with $u_n = f(x_n)$
glossy valveBOT
Rafilouyear2026
rapid rain
$$ \forall \epsilon>0 , \exists n_0 \in\mathbb{N} : n\geq n_0 \implies |f(x_n)-l|<\epsilon$$
glossy valveBOT
Rafilouyear2026
digital void
yes but we also have that x_n >M in the middle
rapid rain
??? what is M
digital void
okay lets not call it M
rapid rain
we don't need what the limit of x_n is to talk about the limit of f(x_n) btw
digital void
$$ \forall \epsilon>0 , \exists n_0 \in\mathbb{N} : \forall N>0 \ n\geq n_0 \implies x_n >P \implies |f(x_n)-l|<\epsilon$$
glossy valveBOT
fijokazż
rapid rain
it makes less and less sense 😭
digital void
huhhh
rapid rain
why do you need to involve this "x_n > P" when it isn't needed
digital void
but for f(x) diverging we say x>P => d(f(x),l)< ε
digital void
why do you need to involve this "x_n > P" when it isn't needed
i think it is implied
rapid rain
diverging?
rapid rain
but for f(x) diverging we say x>P => d(f(x),l)< ε
but that's for a function converging
here f(x_n) is a sequence
digital void
wha
rapid rain
sure, it's formed using the function f, but it doesn't matter
digital void
f(x_n) is a function that takes the terms of a sequence as inputs
rapid rain
f(x_n) is a function that takes the terms of a sequence as inputs
(f(x_n))_n is the sequence of outputs of x_n by the function f
rapid rain
?
digital void
u wrote it like that just now
rapid rain
what do you think $\lim_{n\to\infty} f(x_n) = l$ means
glossy valveBOT
Rafilouyear2026
rapid rain
what is "f(x_n)" here
digital void
**Rafilouyear2026**
for all epsilon >0 and for all P>0 : x_n >P => d(f(x_n),l)< epsilon
rapid rain
for all epsilon >0 and for all P>0 : x_n >P => d(f(x_n),l)< epsilon
So you think lim(f(x_n)) = l interprets "f(x_n)" as a function
and not a sequence
digital void
what is "f(x_n)" here
im not sure what wed call it anymore
digital void
So you think lim(f(x_n)) = l interprets "f(x_n)" as a function
yeah
rapid rain
I can't do much if you can't see that "$\lim_{n\to\infty} f(x_n)$" is the limit of the sequence $u_n = f(x_n)$, as $n$ goes to $\infty$
glossy valveBOT
Rafilouyear2026
digital void
i can see your point
but i always thought that was a function
im kinda flabberghasted
rapid rain
f is a function, but $\lim_{n\to\infty} f(x_n)$ isn't the limit of the function as the input goes to $\infty$
glossy valveBOT
Rafilouyear2026
digital void
yeah that makes sense tbh
glossy valveBOT
Rafilouyear2026
rapid rain
and $x_n = (-1)^n$
glossy valveBOT
Rafilouyear2026
rapid rain
what is $\lim_{n\to \infty}f(x_n)$
glossy valveBOT
Rafilouyear2026
digital void
1 prolly
rapid rain
well not "probably", exactly
each term f(x_n) = 1
so it's a constant sequence, so it converges to that value
did we require at any point that x_n converges to some limit
digital void
we didnt
rapid rain
and if so, did we use the limit of $f(x)$ as $x$ converges to that point
glossy valveBOT
Rafilouyear2026
digital void
x_n didnt converge anywhere so thatd be pointless
rapid rain
we didnt
ok, so the definition of $\lim_{n\to\infty} f(x_n)$ shouldn't require any knowledge about the limit of $x_n$ and/or how $f$ behaves generally around the limit of $x_n$
glossy valveBOT
Rafilouyear2026
digital void
yeah that makes sense
rapid rain
**Rafilouyear2026**
So we just define $\lim_{n\to \infty}f(x_n) = l$ as we would do with any sequence $(u_n)$, applying it to $u_n = f(x_n)$
glossy valveBOT
Rafilouyear2026
rapid rain
now, of course, if you also know that $x_n$ also goes to $\infty$. Then for any $M$ and $\epsilon$, you can find a rank $n_0$ such that you have both $x_n > M$ and $|f(x_n)-l| < \epsilon$
glossy valveBOT
Rafilouyear2026
rapid rain
but the definition of "f(x_n) goes to l" doesn't depend on that fact
digital void
hmm okay
that doesnt hurt the proof then cuz we dont need one to imply the other
ill try to do it and ill come back
thank you for all of that i had no idea i was doing it wrong
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twin wolf
hello where does the part underlined in red come from
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hello where does the part underlined in red come from
rapid rain
hello where does the part underlined in red come from
You want to find an affine transformation to describe how R is mapped to D
so, you're looking for $F(u,v) = A\begin{pmatrix}u\v\end{pmatrix} + B$
glossy valveBOT
Rafilouyear2026
rapid rain
where A is a 2x2 matrix
and B is 2 x 1, representing F(0,0)
Say I want the bottom-left edge of R to be mapped to the leftmost edge of D
then F(0,0) = (1,1)
that solves for B
then, pick enough points to completely fix F
twin wolf
**Rafilouyear2026**
hi rafilou!! im a little confused where this comes from
rapid rain
turns out you need two more points
rapid rain
hi rafilou!! im a little confused where this comes from
An affine transformation is a linear transformation up to some additive constant
so suppose $G(u,v) = F(u,v) - B$ is linear for some $B$
glossy valveBOT
Rafilouyear2026
rapid rain
Then, since we're in finite dimension, $G(u,v) = A\begin{pmatrix}u\v\end{pmatrix}$ for some matrix $A$
glossy valveBOT
Rafilouyear2026
twin wolf
so are we writing the parallelogram D as some linear combination of vectors?
and B is some corner we pick
rapid rain
D is the image of R through an affine transformation
so each vector/point in the square R is mapped to some vector/point in D
twin wolf
so can we construct the parallelogram using F(u.v) = <x,y> = <x0,y0> + u <a0,a1> + v <b0,b1>
and i can pick x0 and y0 as a corner on the parallelogram
say 1,1
rapid rain
yep
twin wolf
and u and v vectors are direction vectors im assuming
twin wolf
oh
rapid rain
but if you make "u" bigger, meaning moving horizontally in the positive direction,
that corresponds to moving in direction (a0,a1) by that much on the parallelogram
and changing v, meaning moving vertically when you're on the square
corresponds to moving in direction (b0,b1) on the parallelogram
twin wolf
so i have the vertices
(1,1)
(2,0)
(3,2)
(2,3)
i can pick a0,a1 to be say from (2,0) to (1,1) and subtract these to get <-1,1>
and then b0,b1 to be from (2,3) to (1,1), subtracting we get <1,2>
these are the edges of the parallelogram?
so F(u,v) = (1,1) + u(1,-1) + v(1,2) maybe
rapid rain
so i have the vertices
(1,1)
(2,0)
(3,2)
(2,3)
i can pick a0,a1 to be say from...
you should rather say "from (1,1) to (2,0)" since you're starting from (1,1) which is your base point
twin wolf
ohh yeah
rapid rain
and you get <1,-1>
rapid rain
so i have the vertices
(1,1)
(2,0)
(3,2)
(2,3)
i can pick a0,a1 to be say from...
same thing for b0,b1, even though you got the correct vector
twin wolf
so F(u,v) = (1,1) + u(1,-1) + v(1,2) maybe
and i expand this and i get the stuff underlined in red
rapid rain
exactly the F they got
twin wolf
ohhh okayyy thank you 🙂
rapid rain
Notice we could have picked a different edge to map (0,0) to, and we could have decided to switch (a0,a1) and (b0,b1) together
that would result in different transformations that still map R to D
but it shouldn't really impact your computation of the integral in the end
twin wolf
ohh so like we could have picked (2,0) to be x0,y0
rapid rain
so i have the vertices
(1,1)
(2,0)
(3,2)
(2,3)
i can pick a0,a1 to be say from...
and if you had kept x0,x1 to be (1,1), and flipped the directions to (a0,a1) = (1,2), (b0,b1) = (1,-1)
you'll make a minus appear in the determinant, because you inverted the orientation of the space
twin wolf
you'll make a minus appear in the determinant, because you inverted the orientat...
and we can just take the absolute value of this right
if F doesnt preserve the mapping orientation
twin wolf
yea like |JF(u,v)|
twin wolf
i have written in my notes | JF(u,v) | = | (x_u)(y_v) - (x_v)(y_u) |
| (x_u)(y_v) - (x_v)(y_u) | and the stuff inside this is the stuff you get from calculating the determinant
ad - bc stuff
rapid rain
very weird notation, I wouldn't recomment
either write |det(JF(u,v))| or smthg like ||JF(u,v)||, idk
no not that either, will be mixed up with the norm
rapid rain
my body is reacting viscerally to that, but ok
twin wolf
lmao is this not normal notation i js started this topic so idk
rapid rain
The determinant of a matrix A (no absolute values) can also be written as |A| in linear alg
twin wolf
ya
rapid rain
but here |A| = |ad-bc|
twin wolf
then how would you indicate absolute value if u cannot use | | signs
rapid rain
write det(...) for the determinant
|det(...)| becomes clear then
If your book tells you to write the absolute value of the determinant of the jacobian as |JF(u,v)| then I won't put it on you for following the book's notation
but I will definitely put it on the book
rapid rain
well the jacobian usually refers to the jacobian matrix
and the jacobian determinant is not referred to as the jacobian
Also the notation 💀
$JF(u,v)$ or $J_F(u,v)$ also usually refers to the jacobian matrix
glossy valveBOT
Rafilouyear2026
twin wolf
oh idk then Lol
rapid rain
It's not your fault
your book just decided to go wild on notations and go against all notation conventions generally accepted
twin wolf
lmao
rapid rain
so on your own exam, you'll be fine using the book's wild notation system
but when talking with other people, expect the same concerns to be risen
rapid rain
ok so $\frac{\partial(x,y)}{\partial(u,v)}$ is also the jacobian determinant and they also call it "Jacobian"
glossy valveBOT
Rafilouyear2026
twin wolf
Am I the crazy one?
lmao idk 
rapid rain
lmao idk <a:1412pixelcatsobsad:1172204185056452719>
Ok so there are indeed some sources which prefer using "Jacobian" for "Jacobian determinant" instead of "Jacobian matrix", but they're in the minority
rapid rain
But the notation JF(...) for the determinant is still wild to me
twin wolf
hello higher!
rapid rain
maybe for undergrad level they just interchange
interchanging notations between a matrix and its determinant is the scariest thing ever
twin wolf
But the notation JF(...) for the determinant is still wild to me
maybe i will continue using it then to be different 
anyways thanks for the help!
.solved
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upbeat venture
how would i type this into a ti 30 calculator
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how would i type this into a ti 30 calculator
upbeat venture
i keep getting the wrong exponent
its supposed to be like 2.60x10^-9
but i keep getting like 2.60x10^-49
frank gale
did you type 10^17
upbeat venture
no i typed 10^-17
frank gale
hmm
lime trellis
,w calculate ((6.626*10^(-34))(3(10^8)))/(7.65(10^-17))
glossy valveBOT
lime trellis
just spam brackets
torn jolt
Hi
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muted token
Where am I going wrong here?
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Where am I going wrong here?
robust slate
Where am I going wrong here?
muted token
I should be getting $-\frac{x}{12y^{14}}$
glossy valveBOT
Vortac
grizzled basin
$-\frac{3}{36}=-\frac{1}{12}$
glossy valveBOT
JamR_71111
grizzled basin
just needed to reduce the fraction a bit, everything else is good
muted token
but $x^{4-4} = $x^0$?
glossy valveBOT
Vortac
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
robust slate
I'm ngl I thought that was an 8 cause it was so zoomed out lol
muted token
just needed to reduce the fraction a bit, everything else is good
Ohh okay
I need to remember to always reduce
grizzled basin
yeah, easy to forget sometimes
muted token
I was thinking somehow 36 needed to be in numerator and 3 in divisor
thanks everyone!
grizzled basin
Np 😁
muted token
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timid dune
Can i get some help on this?
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Can i get some help on this?
timid dune
stiff pier
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@timid dune Has your question been resolved?
opal wolf
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opal wolf
In the proof , why do we require |xn|< 1+ |x|
Can't we just write , M to be Supremum of {|x1|,....,|xn| }
Since if, M is the Supremum of absolute value
All |xn| ≤ M
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split hatch
split hatch
Can't we just write , M to be Supremum of {|x1|,....,|xn| }
What's |xn| here?
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random sky
If every closed and bounded subset of a metric space $M$ is compact, does it follow that $M$ is complete?
glossy valveBOT
Kasli, the Scourge
random sky
Allow me to write up my solution to this problem. I want somebody to review it to make sure there are not flaws in my reasoning.
This may take a few minutes.
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@random sky Has your question been resolved?
glossy valveBOT
Kasli, the Scourge
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
Kasli, the Scourge
random sky
That took a lot longer than I anticipated.
Perhaps now is also a good time to mention that the definition of compactness I am using is that of sequential compactness.
atomic valve
i think it's good modulo typos
atomic valve
your TeX is good
foggy vapor
**Kasli, the Scourge**
Compile Error! Click the <:errors:504266556054962196> rea...
It should be k > N
But anyways its fine
random sky
Guh. Always the little mistakes.
Anyhow, thank you Lex and _gmz for reviewing this.
I'll close the channel now, if there are no other concerns.
.close
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foggy vapor
Anyhow, thank you Lex and _gmz for reviewing this.
Just one last thing: the closure of an open ball isnot always identical to the closed ball d(x,y) <= r. I mean ir is fine here but I feel ike relying on the definition of the closed ball is more appropriate than using thr closure of the open ball
atomic valve
Just one last thing: the closure of an open ball isnot always identical to the c...
did they assume this? i mean they showed cl(B_r(x)) ⊆ B_(r+1)(x)
well they didn't but it's true
foggy vapor
did they assume this? i mean they showed cl(B\_r(x)) ⊆ B\_(r+1)(x)
'...\overline{B_r(x)} is therefore a closed set containing S'
foggy vapor
well they didn't but it's true
Yeah its fine
random sky
Just one last thing: the closure of an open ball isnot always identical to the c...
You raise a good point. I will amend that part of the proof.
Thank you.
atomic valve
i suppose if you were to spell out why cl(B_r(x)) ⊆ B_(r+1)(x) it would be the same amount of work to show cl(S) ⊆ B_(r+1)(x) already
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shut copper
Find the number of integer solutions, where each 0 <= x_i <= 10
x_1 + x_2 + x_3 +x_4 = 21
split hatch
Let A_k, 1≤k≤4, be the number of solutions where x_k > 11, ...
shut copper
so we have to find S - (A_1 and A_2 and A_3 and A_4)
shut copper
so the final expression would be
|S| - (|A1| + |A2| + |A3| + |A4|) + ( |A1 U A2| + ...) - (|A1 U A2 U A3| + ... ) + |A1 U A2 U A3 U A4|?
split hatch
The unions should be changed to intersections
Here's one thing to note
Is it possible that two variables exceed 11 at the same time?
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charred carbon
There are 14 identical metal cubes in terms of shape and size. Out of these, 7 cubes are made of aluminum and the other 7 are made of steel. It's known that all the aluminum cubes weigh the same, all the steel cubes weigh the same, and that aluminum cubes are lighter than the steel ones. You have a balance scale. The least number of weighings needed to guarantee to correctly identify which cubes are aluminum and which are steel is
charred carbon
My mind is kinda fried rn, I can't think of anything
silk bridge
There are 14 identical metal cubes in terms of shape and size. Out of these, 7 c...
we weigh 2 at a time?
grave elm
There is surely a lower bound of 8 (here is a hint on how to obtain it: consider the n.o. possible distributions of the aluminium cubes and the number of possible "states" you could obtain by using the scales n times)
charred carbon
we weigh 2 at a time?
wdym 2 at a time?
silk bridge
two cubes at a time on balance scale
charred carbon
two cubes at a time on balance scale
any amount is fine
Don't have to be one each side
silk bridge
ok'
charred carbon
There is surely a lower bound of 8 (here is a hint on how to obtain it: consider...
hmm
grave elm
if we do 1-1 comparisons, either:
one is heavier one is lighter -> we are done
they are the same weight -> we put them aside
We can do this 7 times, at worst we will have 6 pairs put aside. Then we can form pairs of those pairs in 3 comparisons and at most 2 pairs of pairs will be equal. Last comparison solves that.
So that's an upper bound of 11.
charred carbon
I get the upper bound
charred carbon
There is surely a lower bound of 8 (here is a hint on how to obtain it: consider...
I'm still trying to understand this one
grave elm
<:pandathink:524885344995508235> I'm still trying to understand this one
There are 14C7 possible distriubtions of the aluminium and steel balls. There are 3 possible results of each weighing. That means that after n weighings, there'll be 3^n possible states. You need to map injectively from those states to distributions to decide, that means 3^n >= 14C7 meaning n >= 8
it actually has a wikipedia
charred carbon
The distributions thing is messing up my mind
charred carbon
https://en.wikipedia.org/wiki/Balance_puzzle#Generalization_to_multiple_unknown_...
Imma give it a look
charred carbon
https://en.wikipedia.org/wiki/Balance_puzzle#Generalization_to_multiple_unknown_...
Okay those are kinda high level stuffs
to me
grave elm
it seems like a pretty difficult problem
charred carbon
It's supposed to be a uni entrance exam problem, shouldn't be that hard
charred carbon
There are 14C7 possible distriubtions of the aluminium and steel balls. There ar...
I get the idea of mapping but why are there 14C7 distributions
grave elm
I get the idea of mapping but why are there 14C7 distributions
You have to pick 7 aluminium balls out of the 14, the rest will be steel
grave elm
<:breadpensive:556701306169982977> It's supposed to be a uni entrance exam prob...
Can u use computer to solve it? And do they require a proof of optimality or is it like a logic thing where you just come up with some good solution that works?
do they perhaps have a file with intended solution somewhere
charred carbon
Can u use computer to solve it? And do they require a proof of optimality or is ...
no, I'm allowed to use calculator
But computer is not
charred carbon
do they perhaps have a file with intended solution somewhere
Yeah they will give me the solution today or the next sunday, I'm not sure
Hmm, I have a different problem to ask
Each cell of a 100×100 square grid is colored with one of 20 colors. A cell is considered 'lonely' if its color is different from the color of every other cell in the same row and column. The maximum number of 'lonely' cells that can exist on the grid is
charred carbon
Each cell of a 100×100 square grid is colored with one of 20 colors. A cell is c...
as my friend suggested, we color 2 diagonal lines first , it's possible to make all of the them "lonely" so the lower bound is 200 for now
Now to color each column or row is tricky, I don't have a good strategy for that
-# if these kind of questions keep appearing on the test I might be cooked 🥀
grave elm
,calc 100C19
glossy valveBOT
The following error occured while calculating:
Error: Undefined symbol C19
grave elm
We pick 1 color which will just color whatever remains, so we wont expect this color to ever be lonely. We are left with 19 colors to use on lonely cells, my idea is that we can color 19 cells in each row s.t. they dont repeat in columns
that'd give 19*100 lonely cells
the question is, whether thats possible
charred carbon
We pick 1 color which will just color whatever remains, so we wont expect this c...
I thought of that, but is that possible
Yes
grave elm
put 1,2,3,4,5,6,...,19 in the first row
then shift by one in the row under it
shift by one again, ...
brazen lark
ohh
charred carbon
Yes I just thought of tha
brazen lark
thats a famous way to color
charred carbon
It seems possible
brazen lark
i think so
grave elm
the question is whether its optimal
charred carbon
hm
grave elm
it feels like it both should and shouldnt be
charred carbon
try add one more color to that, and see if we can get one more "lonely" cell
charred carbon
it feels like it both should and shouldnt be
idk doing these problems fried my head 
grave elm
the issue is that if we try to do the same thing with all 20 colors in a row, suddenly we have no way to color the rest of the row
without eliminating one color by using it
putting us back to 19
the usage of rows in this strategy is already perfect, if anything could be optimized, its the usage of columns
actually, per each row there can be at most 19 lonely cells
which proves optimality (100 rows * 19 cells per row)
so 1900 is indeed optimal
brazen lark
i think this is kind of sudoku
charred carbon
actually, per each row there can be at most 19 lonely cells
Oh wao you're right
How come I didn't think of that
grave elm
im asking the same question, + "earlier"
its very simple, but there are just so many possible ways to approach the problem, which look very promising that the trivial one isnt so easy to find
charred carbon
its very simple, but there are just so many possible ways to approach the proble...
True I meant
19*100
I instantly thought this's wrong
grave elm
what uni is this for btw? Something with math? Is it like a very prestigious uni or sth? Those question seem quite non-standard for an entrance exam
charred carbon
what uni is this for btw? Something with math? Is it like a very prestigious uni...
It's a national one
Ha Noi University Of Science and Technology
brazen lark
WOWZ
charred carbon
I meant it's the top 1 engineering uni
grave elm
Oh, interesting
charred carbon
In my country
brazen lark
that kinda question is high school entrance test here
grave elm
There are 14 identical metal cubes in terms of shape and size. Out of these, 7 c...
I wonder if this can be solved simply as well
brazen lark
letsgo
grave elm
that kinda question is high school entrance test here
good to know i wouldnt manage to get to HS in ur country
charred carbon
There are 100 passengers waiting in line to board a plane with a maximum capacity of 100 people. The seats on the plane are all numbered from 1 to 100. Alice, the first passenger in line, forgot her seat number, so when she boards, she randomly chooses a seat. Each passenger boarding after Alice will sit in their assigned seat if it is available, or choose a random seat if their assigned seat has already been taken. What is the probability that the 100th passenger will sit in their assigned seat (seat 100)?
charred carbon
There are 100 passengers waiting in line to board a plane with a maximum capacit...
My answer is 1/2
Just want a clarify
brazen lark
wait im doing it
grave elm
There are 100 passengers waiting in line to board a plane with a maximum capacit...
There'll always be exactly one person sitting at the wrong seat, unless someone sits back to alice's seat and closes the cycle
so we just need to find the probability that someone sits back to alice's seat
grave elm
There'll always be exactly one person sitting at the wrong seat, unless someone ...
ig i phrased this quite imprecisely, its not what i meant
i cant even find a way to phrase it correctly
charred carbon
so we just need to find the probability that someone sits back to alice's seat
I think the last person the choose a wrong seat have to choose either Alice's seat or the 100th seat
grave elm
I think the last person the choose a wrong seat have to choose either Alice's se...
If the wrong seat is still there though, which is where you went wrong i believe
If someone went and picked alice's original seat to sit to, then they'd close the cycle and everyone else would just get to seat themselves normally
i might be wrong tho
silk bridge
that kinda question is high school entrance test here
Lmao where do you live
brazen lark
it depends wether the 99th person chooses the right seat or the wrong seat
brazen lark
Lmao where do you live
korea
charred carbon
it depends wether the 99th person chooses the right seat or the wrong seat
it's not
brazen lark
wait hmmmm
silk bridge
Wait north/south
brazen lark
nein
grave elm
oh i kinda get the 1/2
charred carbon
I don't know how to phrase my argument correctly 🥀
grave elm
say alice's og seat was 1 and the 100th guy's seat is 2. Then at the 100th guy's turn, exactly one of 1 and 2 are free. But they are both "not-my-seat" to all the passengers 2-99. So by symmetry, both have the same probability of being taken
brazen lark
Which korea is left
south lol
grave elm
and that probability must therefore be 50%
brazen lark
oh yeah makes sense
grave elm
its basically because from the perspective of passengers, alice's and last guy's seat are indistinguishable
charred carbon
say alice's og seat was 1 and the 100th guy's seat is 2. Then at the 100th guy's...
By symmetry is not so satisfying to me tbh
But my answer is correct so
yay
brazen lark
yay
grave elm
By symmetry is not so satisfying to me tbh
It's just that whether or not the 100th guy gets to sit at his seat is fully determined by whether some previous passenger takes alice's seat or his seat. And from perspective of that some previous passenger, those seats are exactly the same. So he just chooses randomly between them
silk bridge
There are 100 passengers waiting in line to board a plane with a maximum capacit...
If alice sat on nth person's seat everyone below n will sit on their seat, but from nth person if chooses to sit on Alices seat then everyone else can sit on designated seats, else nth person sits on kth seat, continues till last person
charred carbon
It's just that whether or not the 100th guy gets to sit at his seat is fully det...
Oh that's same argument as mine
charred carbon
I think the last person the choose a wrong seat have to choose either Alice's se...
Kinda like this
silk bridge
It's just that whether or not the 100th guy gets to sit at his seat is fully det...
I'm slow typing missed all you guys said bare me if if i repeat
brazen lark
?
grave elm
I think the last person the choose a wrong seat have to choose either Alice's se...
It's similar, but i dont think its necessarily the last person. If the 3rd person chooses alice's seat, then it's already determined that the 100th guy will sit at his seat. If 3rd person chooses 100th guy's seat, it's already determined that he'll sit somewhere else
charred carbon
It's similar, but i dont think its necessarily the last person. If the 3rd perso...
If the third person choose alice's seat then 4rd-99th passenger would sit at their designated seat
So he is the last person to choose a wrong seat
grave elm
but what if 3rd guy chooses 100th guys seat?
charred carbon
but what if 3rd guy chooses 100th guys seat?
beside the 100th guy
I meant
still in between them everyone would be in their seat
charred carbon
yay
grave elm
so the only unsolved question is the metal cubes one, which I have no idea how to do
brazen lark
i know how to slove if theres one but 7 makes me dizzy
charred carbon
so the only unsolved question is the metal cubes one, which I have no idea how t...
you can try that later, I still have some weird questions to do
okay doin hw here is kinda fun, 1 more ig
Consider four real numbers (not necessarily distinct), each with an absolute value not exceeding 1/2, and the sum of any three of the four numbers is an integer. What possible values can the sum of the four numbers take among the following options?
-4/3
2/3
silk bridge
There are 100 passengers waiting in line to board a plane with a maximum capacit...
this would be different if a person can choose seat irrespective if their original seat is available
grave elm
0 is obvious
charred carbon
Consider four real numbers (not necessarily distinct), each with an absolute val...
grave elm
suppose exactly 3 numbers are same, then
x + x + x and x + x + a are both integers, meaning that a and x differ by an integer, requiring x = 1/2 and a = -1/2 or vice versa
suppose exactly 2 and 2 numbers are same, then
x + x + y, y + y + x are both integers, meaning that x - y is also an integer, again requiring them to be 1/2 and -1/2
suppose exactly 2 numbers are same, then
x + x + y, x + x + z, x + y + z must both be integers, requiring y - z to be an integer, so y = +-1/2, z = +-1/2 and hence x = 0 by the last one
none of these can produce 2/3 or 1/3
so they'd have to be all distinct
brazen lark
wowx
charred carbon
so they'd have to be all distinct
yeah I get that
grave elm
x + y + z, x + y + w is an integer would require z - w to be an integer
charred carbon
x + y + z, x + y + w is an integer would require z - w to be an integer
HMMM
charred carbon
i.e. z and w must again be +- 1/2
wait what
grave elm
oh im being stupid here
either z and w are +-1/2 or z = w
oh, which means they're the same, which I already dealt with
brazen lark
yeah thats
grave elm
wait what
z-w = 1 or -1 or 0
if it's 0, they are the same (so not distinct)
if it's 1, by the absolute value condition it must be 1/2 - (-1/2)
if its -1, it must be -1/2 - (1/2)
charred carbon
oh interesting
brazen lark
NICE
grave elm
Surprisingly, the only non-trivial cases which can produce something other than combinations of halves are the trivial ones - where all the numbers are the same
charred carbon
x + y + z, x + y + w is an integer would require z - w to be an integer
Haven't thought of this
Wao that's impressive
brazen lark
Haven't thought of this
me too
grave elm
suppose exactly 3 numbers are same, then
x + x + x and x + x + a are both intege...
You probably dont even need all this casework
i was just expecting the distinct case to be difficult
so i did the casework first, but it probably wasnt needed
charred carbon
you shouldn't
brazen lark
You probably dont even need all this casework
but thats the logical way
charred carbon
We have 40 questions in the span of 60 min
brazen lark
wow
silk bridge
We have 40 questions in the span of 60 min
crazy
grave elm
We have 40 questions in the span of 60 min
damn
silk bridge
1.5 min/
brazen lark
we have 7questions per 100min
grave elm
whats the avg score?
i can imagine spending 5 days on the og question and still not getting it
brazen lark
lol
charred carbon
whats the avg score?
Hmm, so there're Math-Science-Comprehending session
Average is 50-60/100
like 75% taker
grave elm
and are most of the questions easier than the ones u sent?
silk bridge
Hmm, so there're Math-Science-Comprehending session
this feels vietnamese
brazen lark
lol he is
charred carbon
and are most of the questions easier than the ones u sent?
yeah, there're around 5-6 hard one
others are mid
Not so easy tho
brazen lark
Average is 50-60/100
our test too
they are all hard
charred carbon
our test too
they are all hard
the time is what make it hard here
silk bridge
avg asian shit 😢
charred carbon
And the precision
brazen lark
the time is what make it hard here
also a lot of creativity to solve
you know my questions
silk bridge
our test too
they are all hard
your are olym type 15-30 min/q
charred carbon
you know my questions
We still have some hard geo questions
brazen lark
yeah ill send you the pdf
silk bridge
what is your exam called again @charred carbon
charred carbon
what is your exam called again <@1265924904969043979>
the TSA
silk bridge
imma check some past year papers if available
charred carbon
I don't think there's any in english
silk bridge
I don't think there's any in english
fine, i'll just have look
btw are all q solved
for this channel
charred carbon
This's a not so creative question from my test
Consider three spheres (S1), (S2), (S3) with radii R_1 = 11, R_2 = 13, R_3 = 19, each externally tangent to the others. A plane (P) intersects the spheres in three circles with centers A, B, C, respectively, and the centers of the three spheres all lie on the same side of (P). It is also known that the three circles of intersection have the same radius.
Given that AB² = 560, the value of AC² is .....................
brazen lark
hmm thats interesting
charred carbon
Given a regular quadrilateral pyramid S.ABCD with a base edge of 24 and a height of 20. Let G1, G2, G3, and G4 be the centroids of triangles SAB, SBC, SCD, and SDA, respectively, and O be the center of the square ABCD. The volume of the pyramid O.G1G2G3G4 is ....................
Another one
charred carbon
That's interesting
charred carbon
Oh that's not many problems I see
brazen lark
but it needs a lot of creativity and fast thinking
brazen lark
ill be back in 15 minutes
silk bridge
Consider three spheres (S1), (S2), (S3) with radii R_1 = 11, R_2 = 13, R_3 = 19,...
here perpendicular to centre of intersection cicrle radii of sphere and radii of circle form a right triangle
charred carbon
you can try that out, imma get some sleep
It's bad that on average I sleep 5 hours/day or even less
Gotta fix it
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silk bridge
It's bad that on average I sleep 5 hours/day or even less
gn
i suck at 3d
but distance b/w center of S1 and S2 = R1 + R2
now we relate that to AB
idk how
charred carbon
i suck at 3d
you're not alone mate
silk bridge
AB is r^2
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✅ Original question: #help-28 message
charred carbon
Here letmme show you something
silk bridge
🫂
brazen lark
what do i need to get
silk bridge
this is for ?
brazen lark
i love geometry
charred carbon
Just an example kek
silk bridge
oh
charred carbon
this is for ?
for how 3d geo here look like
silk bridge
😢
silk bridge
for how 3d geo here look like
we don't learn 3d for same reason
charred carbon
Oh that's not a hard problem , the hard one has even more lines than that
My nightmare back when I prepared for comp
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brazen lark
I don't understand why people love geo
its beautiful
silk bridge
i have seen this
charred carbon
yeah
silk bridge
i dont remember
charred carbon
I'm fionna
silk bridge
who fionna
silk bridge
ohk
silk bridge
right
charred carbon
its beautiful tho
brazen lark
ouch
silk bridge
silk bridge
<@669797638526468106>
wait fionna is girl
do you have multiple personalities or something (light joke)
brazen lark
lol
brazen lark
what
silk bridge
Someone told me to be a girl on that account
wut
silk bridge
Someone told me to be a girl on that account
lance type shit
charred carbon
Not lance lmaooo
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opal wolf
If (xn) and (yn) be sequences of real numbers that converge to x and y respectively.
Show that the sequence (cX) converge to cx , for some c ∈R
errant tiger
are u applying epsilon delta?
errant tiger
ur method is correct but u need to do it more neatly ig
first u need to construct a delta and then find the epsilon
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opal wolf
first u need to construct a delta and then find the epsilon
It was just the idea of proof
I have to make it formal
.close
broken rover
Let n be a natural number such that n+1 is divisible by 24, and let d be any positive divisor of n. Prove that d^2 - 1 is divisible by 24 and that the sum of all positive divisors of n is also divisible by 24.
broken rover
i call k=n+1/d and replace n with kd but im js stuck there, i think mod can be used in some ways but idk how :<
grave elm
i call k=n+1/d and replace n with kd but im js stuck there, i think mod can be u...
maybe we could try rephrasing it all in terms of mod
n+1 is divisible by 24 means n = 23 = -1 (mod 24)
to prove that d^2 - 1 is divisible by 24, we need to prove d^2 = 1 (mod 24)
for that, I'd consider mod 3 and mod 8 separately
fallen scroll
agh got sniped 
charred carbon
-# 
Celine is backkk
broken rover
for that, I'd consider mod 3 and mod 8 separately
um ok let me do it :3 thanks
broken rover
for that, I'd consider mod 3 and mod 8 separately
ooki have proven the first one, the second one seems a bit complicated and general? is there a way to show how many positive divisors of n 😅
grave elm
ooki have proven the first one, the second one seems a bit complicated and gener...
do you mean that you proved mod 3 or that u proved the d^2 - 1 question?
broken rover
do you mean that you proved mod 3 or that u proved the d^2 - 1 question?
the d^2-1 one
torpid vault
ooki have proven the first one, the second one seems a bit complicated and gener...
hint, prove that for every divisor d of n, d+(n/d) is divisible by 24
grave elm
and hint #2: Q1 can be useful
broken rover
hint, prove that for every divisor d of n, d+(n/d) is divisible by 24
but how can that be helpful in any ways? i dont really understand why is ther suddenly d+(n/d), it looks quite random to me 😵
grave elm
but how can that be helpful in any ways? i dont really understand why is ther su...
you can pair the divisors into pairs d, n/d
for example
24
1, 24/1
2, 24/2
3, 24/3
4, 24/4
this is the only "symmetry" of divisors, so that's the motivation for trying it
broken rover
this is the only "symmetry" of divisors, so that's the motivation for trying it
sure, ill come back in a minute
broken rover
this is the only "symmetry" of divisors, so that's the motivation for trying it
oh i get it now and also i solved the problem, thanks alot :3
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twin parrot
An old lady went to the chemist shop to buy medicines. She fell short of the total amount required but the kind shopkeeper sold the medicines at a loss of 12% so that she could buy them. If the cost price was ₹2200, find the selling price.
twin parrot
I don't have a phone rn
onyx glen
so you swear your life on the fact that it says "she fell short" and not "she fell Rs. 999 short"...
onyx glen
yeah, i was going to say you should skip.
silk bridge
he has left...
onyx glen
he's not left the server as far as i can see
silk bridge
i mean offline
slate violet
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slate violet
well, let his channel time out then
they left chat without closing it lmao
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twin parrot
nope he's still in the server
I didn't ignore you I really went offline. Why would I do that. Sry if it made u feel uncomfortable 😔
onyx glen
thank you for editing it out as i asked.
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void jay
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void jay
is my approach solid or would you solve it in a faster way
also, i was hesitant regarding the movement of decimals to eventually get to 1+4 = 5 in the exponent
silk bridge
looks good
spiral vigil
also, i was hesitant regarding the movement of decimals to eventually get to 1+4...
i do exactly this with the bouncing and everything
void jay
ok thanks
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opal wolf
I proved in a different manner . Since we had already proved the theorem : if xn≥yn , then limit(xn)≥ limit (yn)
My proof :-
From hypothesis, we have
xn≤yn for all n ∈N
Theorem 3.2.5 implies ,lim(xn)≤lim(yn) Also
yn≤zn for all n ∈N
Theorem 3.2.5 implies ,lim(yn)≤lim(zn).
Since lim(xn)=lim(zn), we have lim(xn)= lim(zn)≤lim (yn)
Combining the inequalities,
lim(zn)≤lim (yn) and lim(yn)≤lim (zn)
We have lim(yn)= lim(zn).
So that lim(xn)=lim(yn)=lim(zn).
QED
fast peak
your proof assumes that lim y_n exists
that is a strong assumption
one strength of the squeeze theorem is that you do not need that extra assumption
the theorem itself shows you that lim y_n exists
opal wolf
that is a strong assumption
Hypothesis....
xn≤yn≤zn
So lim(xn)≤lim(yn)≤lim(zn)
Doesn't it help
fast peak
but lim y_n doesnt have to exist in general
-1 <= (-1)^n <=1 but lim (-1)^n doesnt exist
you have to prove that such a situation doesnt happen here
opal wolf
Hypothesis....
xn≤yn≤zn
So lim(xn)≤lim(yn)≤lim(zn)
I didn't understand how this can't tackle this case
fast peak
you are writing lim(yn)
how do you know that you can write that
how do you know that it is a number
opal wolf
how do you know that you can write that
am i correct
opal wolf
Ok thank u
opal wolf
and then your proof works
What changes in this proof , to make it fully correct
fast peak
they show -eps < y_n-w < eps
which essentially directly shows that y_n converges to w
which is immediately what you want
opal wolf
which essentially directly shows that y_n converges to w
So essentially I have to follow book method only
fast peak
well its not like you have to follow it but essentially all proofs you see have been optimized over decades/centuries and you can rarely improve them
opal wolf
I was just trying to find an alternate proof
Seems like , it diverts to books prove ,...only
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sturdy valve
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sturdy valve
not able to start
fleet radish
Bayes‘ theorem?
silk bridge
classic bayes
hidden vault
yup bayes
sturdy valve
cn u do without that
hidden vault
hidden vault
cn u do without that
yeah but you wd js find the formula itself again
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sand badge
Pls explain 2nd degree 😭
hidden vault
u mean ax^2+bx+c ?
silk bridge
!help
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sand badge
what is 2nd degree then
3x^2 - 7x - 5 = 2x^2 + 4x - 9 like ts