help-27

[karmic mesa]

you see, the angle BAC = CBD + DCB

[raw cedar]

Yeah.

@dapper fable Do you understand it better?

[devout snow]

@dapper fable Has your question been resolved?

[devout snow]

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[restive river]

I have a doubt , regarding algebra. I have no correct working

Q36

[raw cedar]

,rotate

[restive river]

If it is blurry then i can send a clearer version

[raw cedar]

!status

[devout snow]

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

[restive river]

1

Wait no 2

[restive river]

Clearer question

[raw cedar]

Can you send what have you tried?

[rain summit]

what's $k$?

[woven radish]

1 divided by 0 equals Infinity

[restive river]

It was incorrect

[restive river]

I dont know

Im assuming we have to use the formula for framing a quadratic polynomial

k{ x² + (sum of zeroes)x + product)}

Is this formula correct

[raw cedar]

x^2-

[restive river]

Oh yeah

[restive river]

Sorry small mistake

[restive river]

But otherwise we have to use this formula?

[raw cedar]

Don't worry.

[pure stone]

the question is wrong

[restive river]

I dont think so

[pure stone]

the sum of the squares of the zeros is actually 56

[raw cedar]

Yeah, we need the k inside the equation.

[pure stone]

better to ignore this question

[restive river]

Oh

But ignoring the values

What way do I have to solve

AI told me its 12 but i dont trust AI

Especially not for math

[gloomy aurora]

!noai lmao

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[restive river]

I dont trust AI for math

Or anything

[pure stone]

as stated before, the problem doesnt give you what k is and the given statement is actually wrong

skip the problem and move on

if you dont have any other questions then i'll close it

[uncut crow]

the irrelevant !noai usage counter increases again

[restive river]

Is there any way to solve it without valjes

[pure stone]

nope

[restive river]

I saw a way in which you assume c= k

Can we do that?

[pure stone]

there might've been a problem during the book's publishing

[restive river]

can we assume c= k

K is constant and c is constant

[pure stone]

c in this problem is already give which is 4 (if i understand what you're meaning here) and the problem asks us to find k

[pure stone]

however the given statement in the problem is wrong because of this

[restive river]

Oh

[pure stone]

any other questions?

[restive river]

So it is absolutely wrong?

[pure stone]

yeah something wrong happened during the book's editing

[restive river]

.close

[devout snow]

Channel closed

Closed by @main axle

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[restive river]

Ty sorry for wasting your time

[pure stone]

nah it's okay

[devout snow]

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[frozen solar]

question 1 and 4

[frozen solar]

in first is there a way to check transitive without taking example of exception?

and in 2nd i got 2^6

[frozen solar]

oh wait not 2^6 but 2^15

[glass canopy]

Please send working

[frozen solar]

wait a sec lemme do it neatly

[lunar harbor]

In the future, please use the ,rotate command.

,rotate

[lunar harbor]

1. Suppose $(x,y) \in R$ and $(y,z) \in R$. Then, $x \in {y, 3y}$ and $y \in {z, 3x}$. This means that $x \in {z, 3z, 3z, 9z}={z, 3z, 9z}$. Since $x$ can be $9z$, the condition for transitivity fails.

[woven radish]

Civil Service Pigeon

[lunar harbor]

Also, in the future, please try to keep it to one question per channel.

Keeps this easier on the organisational side.

Anyway I'll let you two do #2 together

[frozen solar]

ah ok I'll be careful next time

[frozen solar]

this my working for question 4

[frozen solar]

this also works for natural numbers as well right?

[rare kernel]

did u get number 1?

[frozen solar]

I'm trying to understand what he wrote

why does x belong {y,3y}?

[rare kernel]

divide whole equation by y^2 , take x/y = t and then factorize

[frozen solar]

wait a sec

[frozen solar]

ah got it

yeah understood the 1st question

[frozen solar]

@rare kernel can you tell me where I went wrong here

[rare kernel]

1 sec havent read the question

[frozen solar]

kk np

i think i got it right but just want to double check

realised my mistake the moment I said I got 2^6

[rare kernel]

is 2^15 not correct?

[frozen solar]

that's the and given in the textbook

[rare kernel]

yeah its correct only

[frozen solar]

did I make a mistake in my solution?

[rare kernel]

u just wanted to confirm the ans?

nope

[frozen solar]

initially I had 2^6

but I realised mistake

[rare kernel]

i see

well u got it now?

[frozen solar]

yeah

thanks a lot

[rare kernel]

alr 👍 nw

[frozen solar]

.close

[devout snow]

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[devout snow]

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[cerulean ruin]

I am a little stuck on how to approach this

or is it obvious

I think my understanding of this problem is correct right for example this would the transformed graph look like

and the order will be something like

[0, 2, 4, 3, 1]
(1, 2, 2, 3, 4) - degree of undirected graph
(1, 2, 2, 1, 0) - out-degree of a graph

so right now im trying to see how i can use a vertex's outdegree to determine the relative position of verticies in sorted order

evidently if a outdegree = 0, then the vertex is either isolated or its the vertex with the biggest degree

if the outdegree is 1, than either its the second highest degree or it could be a vertex with exactly 1 degree

and so on

hmm

i think that might be a pattern

im actually kind of confused in this case wouldnt the outdegree of 0 $ > \sqrt{4}$?

[raw cedar]

We can have 2 cases, right?

[raw cedar]

Yeah, it's correct.

[cerulean ruin]

but then why is the maximum outdegree O(sqrt(M))

wouldnt it just be O(M) graphs with all verticies connected to each other

because then the maximum outdegree is just M-1

actually i am a little bit confused by the tie breaking process

[raw cedar]

Think about, in a complete graph with n$vertices, the maximum output degree is n-1, but since the total number of edges is $m \approx \frac{n^2}{2}$, then that degree $n- 1$ is still proportional to $\sqrt{m}$.

[cerulean ruin]

wait is m representing # of edges or verticies

[raw cedar]

M is the number of edges and n is the number of vertices.

[cerulean ruin]

oops my bad

[raw cedar]

Don't worry. 🙂

[cerulean ruin]

hmm alright now the problem makes sense

now i just need to figure out how to approahc it

[raw cedar]

Right!

[cerulean ruin]

so totally there are m directed graphs in the resulting figure
there will alwyas be a vertex with an outdegree of 0, this will always be the biggest vertex in our "sorted list"

hmm this feels tricky

[raw cedar]

The key is not to look at the entire graph at once, but to focus on a single vertex.

[cerulean ruin]

hmm

[cerulean ruin]

and this is any arbitrary vertex

or should i look at verticies one at a time

[raw cedar]

The objective is precisely to demonstrate that the property holds for any arbitrary vertex v, right? So analyzing just one of them is the right way to go.

[raw cedar]

Do you understand it better?

[cerulean ruin]

I think I understand what I'm supposed to do but its a little tricky

like im going to say if some vertex has k outdegree edges

try to figure something about its position in sorted order

[raw cedar]

Exactly!

[raw cedar]

Don't get obsessed with the position number, think about the fact that everyone who comes after v has degrees so high that they spend the total edge budget very quickly. 😉

[cerulean ruin]

wait

if a vertex has k outgoing edges, there there has to be k verticies who have a higher degree

[raw cedar]

Exaclty!

[cerulean ruin]

ok hmm

imma think about it

i see what you mean though

[raw cedar]

Think it.

[cerulean ruin]

hey

i think i have a weird argument

I don't know if this works

[raw cedar]

Tell me.

[cerulean ruin]

Denote $v_i$ as the vertex with $k$ out-degree edges. There are exactly $k$ verticies who have a higher degree than $v_i$, lets label them as $v_{i+1}, v_{i+2}, \dots, v_{i+k}$. Each vertex from $v_i \dots v_{i + k}$ has at $\geq k$ degree, and there are $k$ of them. I think there is some kind of lemma that tells you the total degree in a graph $(2m)$. So $k^2 < 2m \implies k < \sqrt{2m} = O(\sqrt{m})$

[woven radish]

toast

[raw cedar]

Exactly! It's correct.

[cerulean ruin]

tysm, is there a way to work around this without using the handshake lemma

[raw cedar]

Yeah, it's better.

[raw cedar]

Do you have more questions?

[cerulean ruin]

could we argue that since there are k+1 veritices with at least k edges, then $\dfrac{(k+1) \cdot k}{2} \leq m$

[woven radish]

toast

[cerulean ruin]

since its an unidrected graph, just divide 2 to get unique edges

[raw cedar]

Correct!

[cerulean ruin]

kk

i think this makes alot more sense

imma write it down tysm

i think there is one more problem but i will send after i transcribe this down

[raw cedar]

Good, now you have the a and you need the b, right?

What ideas you have?

[devout snow]

@cerulean ruin Has your question been resolved?

[cerulean ruin]

oh s

wait

ill work on this soon

.close

[devout snow]

Channel closed

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[cerulean ruin]

ill close the channel

[devout snow]

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[cerulean ruin]

okie

So i guess we have to count number of triangles in a graph

and devise some algorithmn for it

so the original nbhd intersection algorithmn is like

for each (u, v) in G we look at the neighbors of u and v, then we take $N(u) \cap N(v)$. This will let us know how many triangles are associated with each edge, but each traingle is enumerated 3 times because each edeg of the triangle is checked

[woven radish]

toast

[raw cedar]

Exactly!

[cerulean ruin]

so i guess our difference is that we actually perform it on the directed graph

and i actually think because its a directed graph, we don't overcount

[raw cedar]

You are correct. 🙂

[cerulean ruin]

now i need to explain why

well suppose we have a triangle in the undirected graph

when converting it into a directed graph using sorted order, we get edges

u -> v
u -> w
v -> w

we can only get at most 1 triangle from this and thats if we check u -> v's edge

any other triangle will have this pattern, so we never overcount

[raw cedar]

Yeah, you can.

[cerulean ruin]

hmm is this a strong enough answer to the problem

• we just run a neighbor intersection on the directed edges. We only count it there exists an intersection between the neigbors and this will only occur once per triangle, as our resulting directed graph has no cycles.

There are m edges, and we know by part a, each edge as at most sqrt(m) outgoing edges, so our total run time will be $O(m\sqrt{m})$

[woven radish]

toast

[raw cedar]

For me looks really good.

[cerulean ruin]

ty

is an argument for each triangle cycle, there will always be one vertex with 2 outgoing edges, 1 vertex with 1, and 1 with none. The only edge that detects a triangle is the edge between the two vertex with at least 1 outgoing edge, since that is the only way to get to the third shared vertex

i think this might be good too

but tysm

.solved

[devout snow]

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[raw cedar]

You are welcome!

Have a nice day.

[devout snow]

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[green grove]

can someone explain why (iii) is true? because all elements of $\mathcal{P}(\bN)$ are countable but the set itelf isnt, is it not?

[woven radish]

kesjykara

[trail eagle]

What would be the sets you’re taking the union of in that case? Would there be countably many of them?

[green grove]

i see i see

[jaunty mantle]

You can do the classic method with ℚ here

Do you know about the diagonal counting method for ℚ?

[devout snow]

@green grove Has your question been resolved?

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[green grove]

yeah

[devout snow]

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[jaunty mantle]

Yeah just line up the A’s in a row and each A has a column of it’s countable elements

[jaunty mantle]

Then you just use the same method

To enumerate through the union

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[runic dew]

Combinatorics might be my biggest weak point... I have no idea how to sum the possibilities here, do you maybe have to calculate the amount of partitions of 2, 3,... 99 into two?

[full cloak]

hi

[runic dew]

Hai!

[lament frigate]

So like how many ways 3 numbers can add to 100?

[full cloak]

um, honestly I have no idea

I would say:

you set 1 number and try to see how many ways 2 numbers add up to 100

[lament frigate]

Hold up

I might have an approach

Let’s think of the extremes

[runic dew]

So when x_1 = 1 and when x_1 = 98?

[full cloak]

so:
If the first number is 1, there are 98 ways because the second number can be any number between 1 to 98
If the first number is 2, there are 97 ways because the second number can be any number between 1 to 97
etc
etc
If the first number is 98, there are 1 way, because the second and third number can only be 1

there we have it

add up 1 to 98 is the answer lol

[lament frigate]

Oh! I have this formula!

$\sum _{n=1}^k n= \frac{(k)(k+1)}{2}$

[woven radish]

YeetusDeletus5

[runic dew]

Ah

[full cloak]

the trick is that: Once you have set the first and second number, the third number is automatically filled in, therefore, setting the first number makes it easier to logic and find a pattern

[lament frigate]

,calc (98)(99)/2

[woven radish]

Result:

4851

[lament frigate]

There’s the answer

[runic dew]

Brilliant, thank you both so much!

There is another part to this question that id like to discuss also

[full cloak]

np

ok

[runic dew]

My first instinct was to divide by 6, but i realised that such an approach was flawed, and we'd need a way to separate X into partitions where all elements are distinct, and partitions where 2 elements are the same

I think there's 48 of these partitions of the form (x, x, y)?

[full cloak]

ok, let's break this question down

let's look at the second half first:

so the question states that if two partitions satisfy that x*(a,b,c)=(d,e,f)
Then essentially, (a,b,c) and (d,e,f) and the same

however

is that actually possible?

because we know:

[lament frigate]

Yes

That’s group theory term, they mean swapping the first 2

[full cloak]

oh ok

then I don't need to break it down even further

we can assume:

for (a,b,c) in set G

a<=b<=c

cause why not?

[lament frigate]

That’s not necessary true

[full cloak]

it doesn't need to be

BUT

that encompasses all possible scenarios

[lament frigate]

That assumption isn’t without loss of generality

(98,1,1)

[full cloak]

(98,1,1) might as well become (1,1,98) to fit our assumption

[runic dew]

But in this case, that partition is "essentially the same" as (1, 1, 98)

[lament frigate]

I see

[full cloak]

because we are only counting each set of 3 numbers exactly once

ok so, lets go over the special cases first, because they are easier, no other preferances

the special cases is where there are 2 numbers that are equal in the set of 3 numbers

and you would be one off, I believe there are 49 such sets

1,1
2,2
3,3
...
49,49

again, if the first 2 numbers are known, the third number is set

[runic dew]

Oh wait i realise i didnt specify what the group action here is

The group acting on X is S_3, where some sigma in S_3 permutates some (x_1, x_2, x_3) in x by (x_sigma(1), x_sigma(2), x_sigma(3))

[runic dew]

I see

[full cloak]

ok so now the more complicated cases, where the 3 numbers are different

lets again use the set 1 number trick:

if a=1

then the second number is between 2 and 49, because if the second number goes past 50, the third number is less than 50, breaks our assumption of a<b<c

therefore there are 48 such sets of (a,b,c)

if a=2

[lament frigate]

3 and 48

[full cloak]

the second number is between 3 and 48

46 such sets of 3

if a=3

the second number is between 4 and 48

45 such sets

[runic dew]

So we do 47 + 46 + ... + 1?

[full cloak]

not this time

[runic dew]

Oh

[lament frigate]

No 47

[full cloak]

because there are both jumps of 2 jumps of 1

now

[lament frigate]

46 to 45 is 1

[full cloak]

you can do the good ol' (count down to 1 method)

[full cloak]

48->46->45

[lament frigate]

2->1

[full cloak]

I think I see a pattern

every 3 numbers, there is a skip

47 is skipped

[lament frigate]

U sure this isn’t 44?

[full cloak]

44 will be skipped

yeah not 44

I am sure

so it goes jump 2->jump 1->jump 2-> jump 1 and continues

[lament frigate]

3,48,49

[full cloak]

therefore every 3 numbers, there will be a skip

[lament frigate]

Oh ok

[full cloak]

thus:

47,44,41,38,35,32...5,2 will be skipped

oh wait actually

[lament frigate]

So the pattern is (50-3n)?

[full cloak]

we have to make sure that a<33 as well

[runic dew]

Am i okay to speak about a method that i came up with that might also work?

[lament frigate]

Go for it

[full cloak]

yeah go ahead

[runic dew]

So given that there are 49 permutations of the form (x x y)

Consider that the order of the orbit of all of these permutations is 3

Therefore, when we consider the original set X, these permutations will account for 49 * 3 = 147 of that 4851

So we can do 4851 - 147 = 4704

[full cloak]

omg they are smart XD

yeah that is a much better solution

[runic dew]

These 4704 are now all partitions of the form (x y z)

So the order of their orbit is 6

So we do 4704/6 = 784

Then just do 784 + 49 = 833

[full cloak]

not 147 but um 49

[runic dew]

OH WAIT YEAH

[full cloak]

right answer tho

[runic dew]

Awe thank you :3

.close

[devout snow]

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[devout snow]

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[quiet copper]

I was messing around in desmos and made r = a(theta) + sec theta pi with a being a slider but I don’t understand why it does what it does and wanna know

[shy osprey]

Can you send us an ss of the graph so obtained along with the specifcs as to where you would want to know?

[quiet copper]

I want to know genuinely every part that goes into this

[shy osprey]

well that would be a bit intense,
Sorry am gonna have to sit this one out then😅

[devout snow]

@quiet copper Has your question been resolved?

[trail eagle]

More or less your r value increases somewhat linearly with theta (giving you a vague spiral effect) and whenever cos(theta pi) = 0, r blows up to infinity which creates the big rays emanating from the origin

[quiet copper]

Ohhh thank you

I see

[devout snow]

@quiet copper Has your question been resolved?

[devout snow]

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[vast ledge]

how to do integral from o to pi of 2sinx+sin2x dx

[winter patrol]

what have you tried?

[vast ledge]

i tried usub

but idk what to pick for u

[winter patrol]

integrate each term individually

[vast ledge]

ok

is it -2cosx-1/2cos2x

oop

ok

[winter patrol]

Yes

Sry misread the question

Don't need C for definite Integral

[vast ledge]

oop

ok

[winter patrol]

then evaluate the bounds

[vast ledge]

ok

i got 4

for thsi question i got that integral is 0 to 3 of sqrt x^3/4-x

how do i do that integral

im kinda lost

[winter patrol]

they tell you to use a graphing utility for it

[vast ledge]

o

i didnt read the insrtuctions

i see

[winter patrol]

because it's hard to do by hand

[vast ledge]

oo

ok

ty

how do i do 57

[winter patrol]

integrate normally then sub

[devout snow]

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[tepid isle]

Could I receive help/answers for these problems? This is for business and economic stat midterm tmro.

[devout snow]

@tepid isle Has your question been resolved?

[raw cedar]

What have you tried or what ideas you have?

[devout snow]

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[tepid isle]

no

[tepid isle]

.reopen

[tepid isle]

I just need an answer key

[raw cedar]

We can’t give you the answer but we can help you.

[devout snow]

@tepid isle Has your question been resolved?

[devout snow]

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[tepid isle]

.reopen

[devout snow]

✅ Original question: #help-27 message

[tepid isle]

would u be able to walk me through 4-6

[ripe grove]

Binomial theorem

[tepid isle]

yes wht about it

[ripe grove]

That's what you need to apply to solve 4

Cuz you have a probability of an event occuring so naturally you know the probability that it doesn't occur

And you know for how many trials you wish for it to occur out of a total number of trials

[tepid isle]

wait are you not able to solve other peoples problems on this server

[ripe grove]

yeah...

[tepid isle]

shoot would you want to dm me or no

[ripe grove]

It's better to get help here there's a lot of smarter people who can answer your questions

[tepid isle]

Yea im on here mostly bc I didnt get an answer key with this sadly and I have a midterm and am struggling to see what answers r correct

[ripe grove]

give me your answers for 4 and I'll tell you

[cloud bough]

can people here help with math questions?

[ripe grove]

Yes! See #❓how-to-get-help

[cloud bough]

ok

is 837.11 cm^2 correct for this

[small geyser]

please get your own channel, as this one is occupied at the moment. #help-44｜stanley-🌲-v2-dans is available right now.

[tepid isle]

I used deepseek as reference

and used the probability formula

[ripe grove]

That's the binomial theorem I was yapping abt fr

[tepid isle]

Yea I know but idk how to make it more simple to remember

[ripe grove]

I mean there's nCr events for which we want p and don't want 1-p

Easiest way to remember is to learn how we derived this formula

[tepid isle]

so how did we derive the formula

[ripe grove]

Best to read a textbook

https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/09%3A_More_Probability/9.01%3A_Binomial_Probability

[tepid isle]

anyone dm me if you would like to help in anyway

[devout snow]

@tepid isle Has your question been resolved?

[ripe grove]

when u get answers i'll mark them

since the issue is not having an answer key

[small geyser]

I'll be around for answer checking as well, if Nyxzore is away and I happen to be here.

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[rain summit]

<@&268886789983436800>

[devout snow]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[normal geode]

hehehe

[rain summit]

you got any questions?

[normal geode]

lol , no bro

[rain summit]

please close the channel or ill report this as a troll

this channel is only for math-related questions that you needed help with

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[normal geode]

close

[rain summit]

it's .close

[normal geode]

yeah bro i got it

[thick lotus]

hi

[normal geode]

aaaaaaaa

[rain summit]

hi

[rain summit]

and please close the channel

[thick lotus]

hello i new

[rain summit]

hello

[devout snow]

@normal geode Has your question been resolved?

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[smoky gyro]

hi so im not sure how to take the grad of x^T*Ax with the transpose there

[vast zenith]

one way is to write it out with Einstein notation

$x^T A x = x_{i} A_{ij} x_{j}$

[woven radish]

plantsyavi

[vast zenith]

this can be differentiated somewhat easily using product rule. the k:th element (gradient with respect to k:th x) is:

$\frac{d}{dx_k}(x_i A_{ij} x_j) = {\delta}{ik} A{ij} x_j + x_i A_{ij} {\delta}_{jk}$

[woven radish]

plantsyavi

[smoky gyro]

ohh im unfamiliar with this. whats delta here?

[vast zenith]

kroneckers delta is super neat! it is equal to 1 if the indices match otherwise 0

e.g in simpler terms if you have a variable x_i and another variable x_j

$\frac{d}{dx_i} (x_j) $

[arctic yarrow]

Another way to do it that leads to the same conclusion is to differentiate $x^TAx = <Ax,x>$

[woven radish]

Lin Xia

[vast zenith]

this is probably better if you’re unfamiliar with Einstein notation

[smoky gyro]

hmm okay

[smoky gyro]

so with this if i let f(x) = <Ax, x>

am i finding the directional derivative in some direction h is that how i would use this?

[arctic yarrow]

yes you can do that

[smoky gyro]

is there anther way u would do it or

[arctic yarrow]

you can take the differential of the function aswell

but i dont know what your supposed to do exactly

So do it with the tools you are supposed to use

[smoky gyro]

ah okay thanks

[arctic yarrow]

ping if you are stuck

[smoky gyro]

i got df = (dx)^T Ax + x^T Adx so far after using product but what do i do with dx transposed

[arctic yarrow]

use symmetry of A

[smoky gyro]

do i like apply the transpose (dx)^T to Ax or something

[arctic yarrow]

apply to the whole thing

hum wait a minute

i think it's easier to use the scalar product expression hhere

Do you know what the differential of the scalar product is ?

[smoky gyro]

i mean this is just d(uv) = uv + vdu right

[arctic yarrow]

ye

but the transpose expression is a bit hard to manipulate

[smoky gyro]

ye im not sure what to do abt that

[arctic yarrow]

just rewrite $(dx)^TAx = <Ax,dx>$ and use symmetry of A

[woven radish]

Lin Xia

[arctic yarrow]

Seems simpler

well you dont even need symetry of A for this one but you need it for the other one

[smoky gyro]

can we say u dot v = u^T v

and use that

[arctic yarrow]

Yes this is exactly that formula

[smoky gyro]

so df = (Adx) dot x + (Ax) dot dx

and Adx dot x = x^T * Adx
Ax dot dx = x^T * A^T dx

and then A^T = A so

x^T * Adx + x^T Adx

df = 2x^T Adx

right

[arctic yarrow]

perfect

[smoky gyro]

and since grad f = 2Ax

we have df = (grad f)^T dx

so (grad f)^T = 2x^T A and then transpose both sides

grad f = (2x^T * A)^T
grad f = 2Ax

@arctic yarrow hows this

[arctic yarrow]

I suppose you want to say df = 2x^T Adx in your first line

or i didnt understand what you did maybe

[smoky gyro]

i want to solve for grad f

wait

idk what i di actually i think imessed up

df = 2x^T Adx

so df = (grad f)^T dx right

[arctic yarrow]

i think you should start from $df = \grad f \cdot dx$

[woven radish]

Lin Xia

[smoky gyro]

df = 2x^T A dx
df = (grad f)^T dx

2x^T A dx = (grad f)^T dx

and matching coefficient i think

2x^T A = (grad f)^T

then transpose both sides so grad f = 2Ax

[arctic yarrow]

i think its correct

[smoky gyro]

okay thank you

[uncut crow]

water blud…

[smoky gyro]

do u think my works look rigth

[arctic yarrow]

you're welcome

[uncut crow]

i haven’t read anything in ts channel

[smoky gyro]

okay i will close now my doubt is solved

[dapper cape]

W

[smoky gyro]

.solved

[devout snow]

Channel closed

Closed by @smoky gyro

Use .reopen if this was a mistake.

[smoky gyro]

blud......

[uncut crow]

water blud…

[thick lotus]

wlud

[uncut crow]

who is this blud

[thick lotus]

mrx

[devout snow]

Available help channel!

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[winged hearth]

I want the proof of log a to the base b is equals to log a to the base C the whole upon log b to. and also how log a to the base C the whole divided by log b to the base C is equal to log a to the base b

[last parrot]

Write out bro

[winged hearth]

my writing sucks

[coarse flume]

Upon log b to what?

[winged hearth]

1 sec I'll write it

[last parrot]

If you speak like that no one understand

[void fox]

You want a proof of

,tex $log_a b = frac{log b}{log a}$

this?

[winged hearth]

yea

so the first line and
the arrow

[void fox]

well, the second line follows from the first line easily

[winged hearth]

yea idk lol

[void fox]

so you just want to prove the first line

[winged hearth]

yes

[void fox]

just use the first line

[winged hearth]

and why the base b and the number b gets cancelled

idk I'm watching a video

[winged hearth]

that's how I got these questions

[void fox]

,,relax (log_b a)(log_c b) = frac{log a}{log b} cdot frac{log b}{log c} = frac{log a}{log c} = log_c a

[winged hearth]

oooo

[winged hearth]

and the first line?

[void fox]

on second thought

we deduced the second line from the first line, right? but we can also deduce the first line from the second

try to prove the second line instead

using the definition of log

[winged hearth]

we used the first line to prove the second so technically we can't use the second to prove the first?

[coarse flume]

Statement 1 implying statement 2 is indeed not reversible, the opposite isnt always true
However you can actually prove the first from the 2nd in this case (divide both sides by log base c of b)

[winged hearth]

yea I got that

in the second line the " b's cancel out on in base and the argument"

why

oh no

yea mb

got it

rysm

bye

.close 👍

[devout snow]

Channel closed

Closed by @winged hearth

Use .reopen if this was a mistake.

[winged hearth]

this was left

reopen

.reopen

[devout snow]

✅ Original question: #help-27 message

[winged hearth]

.close

[devout snow]

Channel closed

Closed by @winged hearth

Use .reopen if this was a mistake.

[void fox]

proving the first line from the second was what I meant

[lament kraken]

https://tenor.com/view/blackbeard-blackbeard-writing-taking-notes-writing-gif-17583038544116987898

[copper harbor]

.This is exactly why we tell people off for forwarding things into hlounge.

[devout snow]

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[tepid isle]

.reopen

[indigo coral]

Need help

[gloomy aurora]

post the qs

!da2a

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@indigo coral Has your question been resolved?

[raw cedar]

!status

[devout snow]

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

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[vagrant fulcrum]

So..
I have event A and B are two seperate events, and event X
-> P( AB | X) = P( A | X) * P(B | X)
I dont understand it, can someone elaborate the function?

[arctic yarrow]

Do you mean independent events ?

Also you might mean $\mathbb{P}(A\cap B)$ right ?

[woven radish]

Lin Xia

[vagrant fulcrum]

yeah

A and B are independent events

[arctic yarrow]

What you don't understand here ?

[vagrant fulcrum]

why P(AB | X) = P(A | X) * P(B | X)?

[arctic yarrow]

You want to prove it ?

[vagrant fulcrum]

i do understand P(AB) = P(A) * P(B)

[vagrant fulcrum]

sure

[raw cedar]

Ok, what ideas do you have?

[arctic yarrow]

Do you know the formula for conditional probability ?

[vagrant fulcrum]

yeah

P(A | B) = P(AB) / P(B), correct?

[arctic yarrow]

Just to be clear, by $P(AB)$ you mean $P(A\cap B)$ right ?

[woven radish]

Lin Xia

[vagrant fulcrum]

yeah

[raw cedar]

Correct!

[arctic yarrow]

Ok maybe you can use this formula to prove your statement

[vagrant fulcrum]

i just dont understand how does this make sense

P( AB | X) = P( A | X) * P(B | X)

[vagrant fulcrum]

how

[raw cedar]

Think that once you fix context X, events A and B become unknown to each other and behave as if they were independent.

[vagrant fulcrum]

uh huh
then why is it P(A | X) * P(B | X)?

what is the X doing here

and why is it.. like that
like if i wanna seperate it, why doesnt it be P(A | X) * P(B) or smth

[raw cedar]

Because multiplication is the mathematical way of saying that two things have nothing to do with each other, so the probability of both occurring is simply the product of their separate probabilities.

[vagrant fulcrum]

uh huh, i do understand that
Im just confused why.. is X like that

[vagrant fulcrum]

and this too

[karmic mesa]

what's the problem

[vagrant fulcrum]

HOLY SH ITS THE GOA

T

[karmic mesa]

im not

[vagrant fulcrum]

@karmic mesa

[karmic mesa]

im terrible struggling with probability

[karmic mesa]

are A and B independent

[raw cedar]

The X has to be on both sides because it is the mandatory context, right?

[vagrant fulcrum]

yes

[karmic mesa]

that is
use diagram

[vagrant fulcrum]

uh..huh.. wait wha?

[vagrant fulcrum]

wha

ho-

how

[karmic mesa]

uhm idk how to explain this to you

[vagrant fulcrum]

can u draw the diagram?

i have no idea hwo to draw in this case

[karmic mesa]

this

P(A|X) = P(AX)/P(X) same to the other P(B|X)

[vagrant fulcrum]

okay.....?

and A and B are independent..

what about X

[karmic mesa]

uhm let just think conditional probability in another way

think of it just like regular probability

but

[vagrant fulcrum]

!?
wait wait wait is this

dont tell me there are two types of independence

[karmic mesa]

modified space

[karmic mesa]

nah

the given condition is the new space

[vagrant fulcrum]

new..space?

so like

in normal case

P(A) * P(B) works in the

uhh

P(omega)?

[karmic mesa]

yeah

[vagrant fulcrum]

but in this case, the omega is C

so

P(A | C) * P(B | C)?

[karmic mesa]

P(A) is actually P(A|Omega)

[vagrant fulcrum]

OH

[karmic mesa]

yeah

[vagrant fulcrum]

HOLY SH*

THANK YOU

THAT MAKES WAAAAAAAYYYYYY MORE SENSE

.close

[devout snow]

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[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

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[crystal dawn]

<@&268886789983436800> here as well

<@&268886789983436800> here too

oh damn

that's 6 in the span of an hour

gotta be a record

[wicked rover]

a ban auto clears all msgs so maybe just one ping will do it

tho better safe than sorry

[crystal dawn]

ah mb

[devout snow]

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[pure cedar]

can someone please explain why n and m have to be of opposite parity

i mean , according to me
m can be any arbitary value and n should be odd

[polar chasm]

it is quite weird

they are wrong it seems

6^1 + 9^1 = 15, which is divisible by 5. And 1 and 1 have the same parity

[pure cedar]

but the answer comes out same

for some reason

[polar chasm]

their counting looks wrong too

oh, nvm, their counting is okay

[polar chasm]

yeah, seems like it does

[pure cedar]

can you please help me understand the way they have counted their cases

[polar chasm]

the 2C1 picks the parity (odd / even), the 25C1 picks m and 25C1 picks n

[pure cedar]

isnt that where the flaw is...
we dont need to pick a parity imo
since there are only two sets

[polar chasm]

Their counting is fine, what isn't fine is the "m and n has to be opposite parity" line

the flaw was in that line, but coincidentally, it didnt change the result

but its still wrong

[polar chasm]

but if you are counting pairs of m,n with opposite parity, you do need to pick the parity

you can pick 25 odd n and 25 even m, that makes 25*25 odd-even pairs
you can pick 25 even n and 25 odd n, that makes 25*25 even-odd pairs
so in total, there are 2*25*25 pairs with opposite parity

[pure cedar]

right right...

thanks

have a good day

.close

[devout snow]

Channel closed

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[devout snow]

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[cerulean ruin]

back with some more fun @raw cedar 🤣

[raw cedar]

[cerulean ruin]

what does multiple range queries really mean? is it like given a series of ranges, find the # distinct elements

so like

[solemn hatch]

We got a toaster here in help channels

Beware

[upper gazelle]

hi

[cerulean ruin]

[2, 3, 2, 5, 1, 1, 2, 3]

would a multiple range be like M = [0, 3) [4, 7)

[raw cedar]

Yeah, exaclty!

[cerulean ruin]

ah oki

ah and the challenge is getting a run time algorithm for n queries in nlogn

[raw cedar]

Yeah!

[cerulean ruin]

i see

and it also seems like queries can overlap

like [1, 5) [2, 4)

[raw cedar]

Yeah.

The best thing is to sort the queries by their far right (r) and use a Fenwick Tree to count how many items have their next appearance outside the current range as you progress.

[cerulean ruin]

hm

i see

(a) Define next(i) as the smallest index j > i such that A[j] = A[i], or n if no such index exists. Argue
that an element at position i contributes to the distinct count of a range [l, r) if and only if l ≤ i < r and
next(i) ≥ r.

so for part a, i think its just like

suppose we have some l <= i < r as our first index. There are no indicies j before i such that A[j] == A[i] by definition of i. Now suppose next(i) >= r. Then in the range [l, r) there is only one such instance of i so it contributes to the distinct count

for the <= direction

[raw cedar]

Right, exactly.

[cerulean ruin]

actually

hm

i think i misunderstood

the index we are counting is the last representative index

so if we have multiple duplicates we only consider the last one, so next(i) is either outside of the range, or out of bounds entirely

[hallow pollen]

Yo y’all occupied?

[raw cedar]

The channel, yes.

[raw cedar]

Yeah! You are correct.

[cerulean ruin]

here is a proof i wrote

(a) $(\impliedby)$ Suppose $i$ is bounded within $[l,r )$ and $next(i) \geq r$. Clearly, there are no other indices $j > i$ where $j \in [i+1, r)$, where $A[j] == A[i]$, otherwise $next(i) \geq r$ is a contradicted.
\\
($\implies$). Now suppose an index $i$ contributes to our total distinct count. Assume by way of contradiction $i$ is bounded by $[l, r)$ and $next(i) < r$. $A[i] == A[next(i)]$, and $i, next(i)$ are both in the range. Both elements will be counted, hence a contradiction.

[woven radish]

toast

[raw cedar]

Good, looks correct for me.

[cerulean ruin]

kk tysm

(b) Explain why sorting the queries by their left endpoint allows us to process them efficiently. What
invariant do we maintain as we sweep from right to left (from i = n − 1 down to 0)?

hmm

[raw cedar]

Think about, what ideas you have?

[cerulean ruin]

so wait we are ordering the queries ranges themselves from least to greatest by left endpoint
[1, 3) [1, 4) [2, 6) [2, 8) like so

we are sweeping right to left in the individual query? or do we consider query as a whole

[cerulean ruin]

ok

my idea is like

if we sweep right to left, we have an idea where next(i) could be without having to search for it

because we have al;ready searched for it

idk if that makes sense

actually hmmmm mm m m m m

actually that might work i am not sure

[raw cedar]

Yeah, it will work.

[cerulean ruin]

(b) If we sort by the left point and start sweeping from right to left, we can maintain the index of $next(i)$.

[woven radish]

toast

[cerulean ruin]

im not sure how to formalie tho

ah

i need to really learn segment trees

lamoo

[raw cedar]

Yeah. It’s necessary to do this.

[raw cedar]

All you need to do is describe what information you mark on the tree as you move from right to left.

[devout snow]

@cerulean ruin Has your question been resolved?

[cerulean ruin]

kk

time to start again

[cerulean ruin]

so like

we can set each index to 1 if its the first occurance

and 0 otehrwise

like

[2,3,2,4,1,2,3,4]
[1,1,0,1,1,0,0,0]

but this only works for range queries starting from 0

oh

so we start from the right

so we dont have to re buiild every time

ok so this is where the hashmap comes in, we store the last seen for each index

for example in my example

at index i = 4

we have

[0, 0, 0, 0, 1, 1, 1, 1] HM = {1: 4, 2: 5, 3: 6, 4: 7}

then we have i = 3

well, A[i] = 4 and there exists 4 in the hashmap

set A[HM[4]] = 0

set A[i] = 1

set HM[4] = i

[cerulean ruin]

so is this the idea
we have a list of n range queries
we sort them from least to greatest by lower endpoint
then we start backwards from the last index
every time we get a new index, we set it to 1, and set HM[A[i]] = i (any value before index i should just be 0)

For every query [l, r) (in this case l = i) that starts at index i, simply perform a range sum query from [i, r) and this is our count of distinct elements

@raw cedar i think i got it

[shy osprey]

Yeah i think hes asleep rn

[cerulean ruin]

deserved

bro is a godsend

😭

[shy osprey]

Fr

[devout snow]

@cerulean ruin Has your question been resolved?

[devout snow]

Channel closed

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[quasi nebula]

I dont really know what these h's are, it says its a linear filter, whatever that means

[quasi nebula]

I need to model this system, problem is it wants initial conditions for a bunch of things and i dont know what the equations are, so i dont know how i could set initial conditions

for context here is the full system I need to model

[devout snow]

@quasi nebula Has your question been resolved?

[quasi nebula]

<@&286206848099549185>

[quasi nebula]

.close

[devout snow]

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[faint gorge]

What is your question?

[upper forge]

I just don’t understand how to start these questions, I understand I’m trying to find the asymptotes for the tan but I don’t know what to do with the value in the parentheses

[faint gorge]

Do you remember the approach from last time?

[upper forge]

Oh dang that’s so embarrassing you already helped me with this exact thing 😭

[upper forge]

Nah it still confuses me

[faint gorge]

It's okay

[faint gorge]

If the denominator gets close to 0 then tangens explodes, which means it goes to infinity or -infinity

That's why we do that

[upper forge]

Ya I understand that

[faint gorge]

These examples are all similar, the only thing they differ in is the period, so their asymptotes appear at different times

[upper forge]

Ya because tan is pi/b

.close

[devout snow]

Channel closed

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[devout snow]

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[turbid spoke]

can anyone give me a hint for this problem, i see the homotheties, but idk what to do with them

[last parrot]

Area of big shape and subtract area of triangles

[turbid spoke]

yea i've been trying to find the area of the triangles with side elngths 1, 3, something, and 2, 2, something

but idk how to find those areas

[last parrot]

You got the area of the square triangle?

[turbid spoke]

yea that's 1

and the one with with 3, 2, sqrt5 the area is sqrt5 by heron's

yes i know i found AC is 3sqrt5

okay i'm just stuipd

thanks!

.close

[devout snow]

Channel closed

Closed by @turbid spoke

Use .reopen if this was a mistake.

[last parrot]

You are not brother

[woven radish]

BeautifulSoup

[devout snow]

Available help channel!

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Remember:
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[shy osprey]

.

State the original question as it was given to you!

Okay, well could you show your working?

[ebon coyote]

Because, PCMG, people aren't here to do your shit for you.

[shy osprey]

#discussion

#math-discussion

[shy osprey]

.

[placid rover]

Please communicate in a more academic manner

this was not clear from your original post

[ebon coyote]

<@&268886789983436800> Look, I don't wanna have to jump through loops again; can someone please communicate this to him again

[shy osprey]

Does your query fit any of these criteria?

!status

[devout snow]

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

[shy osprey]

Are you familiar with the method of finding square roots manually?

[placid rover]

list the first 10 squares

[hushed wraith]

whats the closest number to 29 that u know the square root of

[shy osprey]

I would suggest seeing a yt video/ rewieving your notes then!!

...

just seach?

[upper schooner]

Well, you certainly aren't communiucating what you mean clearly at all

[ebon coyote]

...

Like, jeez.

[upper schooner]

Whale?

[ebon coyote]

this isn't even a language barrier at this point, honestly; this is just a lack of reading comprehension

[upper schooner]

What's that supposed to mean

[hushed wraith]

https://tenor.com/view/omg-really-facepalm-gif-19945159

[placid rover]

its not rocket science, we can tell you how in 5 minutes if you actually try communicating normally with us

without trolling

[shy osprey]

Please use the help channels responsibly as there a lot of people who dont really have all day for some stupid trolls!!

[placid rover]

@shy osprey its not that bad

[last parrot]

Dont need to help btw

[upper schooner]

Well then, you can close the channel if you don't have any extra questions

[devout snow]

Channel closed

Closed by @royal laurel

Use .reopen if this was a mistake.

[upper schooner]

And please don't call them a "stupid troll", as much as it may be frustrating, if you think someone's acting in bad faith, you can ping us (as @ebon coyote did ) or use ModMail to detail your concerns

[shy osprey]

Okay sorry!!

[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[heavy pewter]

Hi could i please get help

[heavy pewter]

This is a functions question

im kinda sped

so i need some help

[shy osprey]

Can we split the numerator into something desirable.

[heavy pewter]

what do u mean

[shy osprey]

Wait before going to this Question, let me ask smth else

5/6

[rain summit]

Try to split so that the denominator appears in the numerator

[shy osprey]

cam you convert this into theform you want?

[heavy pewter]

times the bottom by the top

sooo like x-2 * x-1

i meant x+1

[rain summit]

Not really

[heavy pewter]

then how do i go

[rain summit]

.

[heavy pewter]

how do i do that

i used to fail

in mathematics

now im trying

[shy osprey]

can you rewrite x-4 as x-3 - 1?

(just an example)

[heavy pewter]

ohh yeah u can

[shy osprey]

Well did it click for this one then?

[heavy pewter]

sooo x-1+1

x-2+1

[shy osprey]

yeah!!

wait no

[shy osprey]

Can you tell what this is?

[heavy pewter]

a linear line

[shy osprey]

x - 1 +1 is just x right?

[heavy pewter]

yeah its just x

but i wrote it wrong

[shy osprey]

but your numerator isnt x is it

its x -2

[heavy pewter]

yeah??

[shy osprey]

so once again, note that we have to manipulate the numerator in this manner

but not change the value of the numerator !

[heavy pewter]

can u give me an examp[l;e

x-2+(-x-1)

[shy osprey]

okay lets take (x-7)/(x-1)

[heavy pewter]

yep

[shy osprey]

now we want x-1 smwh in the numerator

Since both denominator and numerator are linear, we simply do this
x-1 + (some constant) = x-7

[heavy pewter]

yep?

soooo x-1+(-6)

[shy osprey]

i am now simply trying to find what i can split the numerator into such that it contains th denominator!

[shy osprey]

thats correct!,

[heavy pewter]

and after what do we

sooo

[shy osprey]

Also note that when you simplify this, you get x-7 back which tell us that we didnt change the value of the numerator

[heavy pewter]

sooo x-2 + some constant = x+1

x-2+ 3

now what do i dooo

[shy osprey]

solve it carefully

*write

[heavy pewter]

solve for what x?

[shy osprey]

Nope!

[shy osprey]

what did we do here?

we solved for ??

[heavy pewter]

we solved to get x-7

[shy osprey]

okay lets try a different approach

Scrape all that before

[heavy pewter]

okayyy

[shy osprey]

We need to split x-2 into something which has x+1

[heavy pewter]

if i have an example i can follow its process

[shy osprey]

back to this again,

we tried to rewrite x-7 as something like x-1 + c

[heavy pewter]

c should be -6

[shy osprey]

now can you tell me what would c be if
x-1 + c = x-7
observe that i am applying this as i mentioned before, we need to value of numerator to remains the same even after spilttig/manipulatig

[shy osprey]

good!!

so now what do you have?

[heavy pewter]

x-1-6=x-7

[shy osprey]

(x - 1 -6 )/(x-1)

[shy osprey]

is this correct?

[heavy pewter]

ohhhh yes

correct im guessing

[shy osprey]

Good!

And now do you also see that we brought the denominator in the numerator

[heavy pewter]

yes i see it

sooo that means it should be x+1-3/x+1

[shy osprey]

Yeah!!!

That was quick good job!

[heavy pewter]

sooo now how do i convert tp the form

[shy osprey]

Now lets remember back to fractions,

[shy osprey]

is (a+b)/c

same as a/c + b/c?

[heavy pewter]

yes it us

is

due to brackets

[shy osprey]

Okay now can you applysimilar logic here?

[heavy pewter]

soo

[shy osprey]

yeah!!

thats correct

now then its pretty simplified isnt it?

[heavy pewter]

YOOOO

whatttttt

[native shard]

It's solved

Now