help-49

[candid gust]

Thats what the teacher gave i wrote it down

I just translated it for you

[fallow scarab]

then there's some convention your teacher's using or a diagram you didn't include

[lethal path]

the convention should be that alpha is opposite side a; beta is opposite side b

that's pretty standard

anyay it's just the sine rule actually, form an equation using that

[candid gust]

So how to solve it

[surreal moon]

where did you get $\alpha+\beta+90=180$?

[grand pond]

SWR

[candid gust]

Thats the other task

[surreal moon]

what other task?

[candid gust]

For a normal triangle

[surreal moon]

sorry, but I feel like there's just too much missing information to help you

[candid gust]

Bruh then how does my teacher expect me to do it on my own

When info is missing

[fallow scarab]

Probably just forgot to give you more information.

[candid gust]

Can anyone create a simmilar task and help me solve it?

[fallow scarab]

Hard to know what similar means without more info

Maybe
https://www.khanacademy.org/math/geometry-home/triangle-properties/geometry-triangle-angles/v/triangle-angle-example-2

[midnight plank]

@candid gust Has your question been resolved?

[midnight plank]

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[wispy mortar]

hi

[mighty edge]

hi

[rich yew]

hi

[mighty edge]

can i call you pi?

[wispy mortar]

Q5

[mighty edge]

youre cubed

uhm i cant help u there

[wispy mortar]

i don't mind yepp

[mighty edge]

sorry im stupid im only in second yr

[mighty edge]

ty

[west iron]

do you know what an integral domain is?

[lyric charm]

do you know what a unit is & what a zero-divisor is

[wispy mortar]

i don't understand how to do the mod part in zero divisors

[wispy mortar]

yess

[wispy mortar]

yess

[west iron]

a ring without zero-divisors right

but a ring where every element is a unit is a field

[lyric charm]

hold up

we're doing number 5 right

[west iron]

oh oops

[wispy mortar]

yess the proof

the 6th one will just be a ring of integers with element 2 ig

[lyric charm]

ok so

Z_n aka the integers mod n

do you know how to characterize all the zero-divisors in this ring yes or no

[wispy mortar]

gcd(a,n) != 1

[lyric charm]

ok

so you need to show that for every a ∈ {0, 1, ..., n-1} either gcd(a,n) ≠ 1 or a is a unit

[wispy mortar]

yess

so now in this

i don't get how to show the mod part

i actually used google lens and it gave a weird answer

[lyric charm]

wdym by "show the mod part"

[lyric charm]

!noai

[midnight plank]

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).

[wispy mortar]

thats not correct?

[lyric charm]

idgaf if it's correct or not

dont use ai

[wispy mortar]

it was my last resort after maths stack exchange 😭

[lyric charm]

i mean idk what to say really

right now it sounds like you just confused yourself beyond immediate help

[wispy mortar]

what would be the right approach

[wispy mortar]

the reference books isnt of much help :(

[lyric charm]

so you need to show that for every a ∈ {0, 1, ..., n-1} either gcd(a,n) ≠ 1 or a is a unit
rephrase this as
for every a ∈ {0, 1, ..., n-1}, if gcd(a,n) = 1, then a is a unit

[wispy mortar]

oh wait

do we just have to say that

inverse is there

im going to sleep

thanks for the help

appreciate it

.close

[midnight plank]

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[obsidian glen]

I have to prove this statement

[obsidian glen]

and idk where to begin

I mean what are you asking? it's all written there. For any non empty element in the boolean algebra, there must be a non empty set of atoms below it, so I have to prove that x is the least upper bound of all of those atoms

[west iron]

Well there's two steps to proving a least upper bound

eh maybe not with boolean algebras

Ok, there's a very special element among the a_i that will help us out

[obsidian glen]

and that is?

[west iron]

well, it's <= x

and its something we have a name for

any guesses?

[obsidian glen]

i mean every atom is below x, and if we are looking for elements below x that could be greatest upper bound of x and any element really, so x and the atoms as well?

[west iron]

Oh oops, missed the atom part

Let me rethink

[grand pond]

ginny

ginny

ginny

[west iron]

Hmm

Suppose x was less than or equal to something that broke down into an atom not in that set

or oop

I mean greater than or equal to

I lost my train of thought

[obsidian glen]

no worries

[west iron]

My intuition is that anything not ≤ (a1 v ... v an) will need to have an extra atom

Or maybe that's just rephrasing the problem

[obsidian glen]

hmm

however we can say $y \neq 0 : y \leq a_1 \lor a_2 \dots \lor a_n$ then $y = a_1 \lor a_2 \dots \lor a_n$

nvm this is false lol

[west iron]

Maybe if we go back to this

This is $(x\land a_1)\lor\cdots\lor(x\land a_n)$, right

[grand pond]

Dreyuk

[obsidian glen]

hmm

[obsidian glen]

it simplifies to this

[west iron]

ah right

maybe if i look at $x \lor a_1 \lor \cdots \lor a_n$

[grand pond]

Dreyuk

[west iron]

all i can show is that =x though and that gives the same

[obsidian glen]

we need to show this is somehow = a_1 \dots \lor a_n

[west iron]

yeah ig im being lazy

at least one of the steps is going to need to use the definition of atom

[obsidian glen]

this did

as we said x \land a_i = a_i

[west iron]

oh, i thought that followed from $x = a_i \lor (a_1 \lor \cdots \lor a_n)$

[grand pond]

Dreyuk

[west iron]

or wait

not that lol

from $a_i \leq x$ i mean

[grand pond]

Dreyuk

[obsidian glen]

I mean again that is what we used to show x \land a_i = a_i because a_i is an atom below x

if it were any atom we coouldn't have said that I think?

[west iron]

$x \leq y$ iff $x = x \land y$ isn't generally true?

[grand pond]

Dreyuk

[obsidian glen]

it is

[west iron]

Probably we'll also need to use negations somewhere since I can think of a counterexample if we don't have them

gl 😭

[midnight plank]

@obsidian glen Has your question been resolved?

[west iron]

Ok, suppose x is bigger than a1 or ... or an

Break down x and not (a1 or ... or an) into atoms?

[midnight plank]

@obsidian glen Has your question been resolved?

[obsidian glen]

and?

OH

[obsidian glen]

this should be 0 right?

[west iron]

Well its 0 iff x ≤ (a1 or ... or an)

That's what you're trying to prove, so assume FTSOC that it's not

[obsidian glen]

yeah I might have got it

if it is not 0 it has an atom below it, which means that atom is below x and the complement of that other term. which means this atoms comes from the set a_i, while this atom being below that other term (complement of lub of these atoms) means it is NOT below the gub of these atoms, which means there is no atom in the set which is which is above this atom but that can't be true since this term is the gub of these atoms and the new atom is in the set of atoms

this is too vague(I can write it out but I thik i got it) yeah there's a contradiction

thanks for your time

.close

[midnight plank]

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[last slate]

hi

[last slate]

how do u find a set of vectors orthogonal to some vector

do you just form the plane equation?

[rose trout]

Well the set of vectors orthogonal to some vector is the plane with that vector as a normal vector.

Given the normal vector the plane equation is basically read from its components so

[deep heart]

What the heck is the quadratic formula my 7th grade math teacher just asked me what it is and none of my other teachers taught me.

[rose trout]

This channel is already occupied, but I'm sure you can open your own and people will help you!
It's also on Google.

[deep heart]

Bro literally just said "You'll learn this next year"

[last slate]

do you need a set notation

they’re asking for “the set”

[rose trout]

I mean yes you could write it as ${ (x,y,z) \in \R^3: Ax + By+ Cz = 0}$ that's probably better.

[grand pond]

Azyrashacorki

[stone pine]

Conceptually if you're trying to find all vectors $v$ orthogonal to $w$ its

${v\in\mathbb{R}^n : v \cdot w=0}$

and remember that $v \perp w$ if and only if $v\cdot w=0$

[grand pond]

flynger

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[grand pond]

Renato

[tidal turret]

I need some help with b)

w ∈ G40 <=> w = e^{2k.i.pi/40}

w^40 = 1

1 + w^2 = -w^8 (1 + w^2)

w^3 = e^{6k.i.pi/40} = e^{3.k.i.pi/20}

[stone pine]

Can't you divide both sides if $\omega^2 \neq -1$

[grand pond]

flynger

[stone pine]

or am i dumb

[tidal turret]

1 + w^2 = -w^8 (1 + w^2)
1 + w^2 + w^8 (1 + w^2) = 0

[stone pine]

yeah you can also do it that way

and factor

its cleaner

[tidal turret]

(1+w^2)(1+w^8) = 0

1. w^2 = -1
2. w^8 = -1

(1+w^2)(1+w^8) = 0
(w-i)(w+i)(1+w^8) = 0

w^8 = -1 and w^40 = 1
(w^8)^5 = w^40

(-1)^5 = -1 = w^40

which is absurd

only possible w ∈ G40 such that that equation is verified is w = i and w = -i

.solved

[midnight plank]

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[midnight plank]

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[near geyser]

#help-44｜stanley-🌲-v2-dans message

[mortal falcon]

$\frac{1}{2}\left(\frac{1}{2n-1}+\frac{1}{2n+1} - \frac{1}{n} \right)$

[grand pond]

snowflake

[mortal falcon]

$\frac{1}{2} \sum_{n=1}^\infty [\frac{1}{2n-1}+\frac{1}{2n+1} - \frac{1}{n}]$

[grand pond]

snowflake

[sullen bolt]

telescoping?

[mortal falcon]

no, but it's not too far off

you get something like

[sullen bolt]

[mortal falcon]

$\frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^n}{n}$

[grand pond]

snowflake

[mortal falcon]

it feels a bit illegal to reorder the terms around

the original series converges absolutely, but the individual ones being rearranged do not :/

[stone pine]

i would combine the first two fractions at first glance

and then prob combine the 3rd one in too

i think it becomes

[mortal falcon]

well that just gives the original series

[stone pine]

wait what are we trying to do

[mortal falcon]

$\sum_{n=1}^\infty \frac{1}{2n(2n-1)(2n+1)}$

[grand pond]

snowflake

[stone pine]

find its value? determine convergence?

[mortal falcon]

find the value

i assume

[modern sapphire]

it strongly feels like two of them telescope and the leftover one can be rewritten as a riemann sum

[mortal falcon]

convergence is kind of trivial

[stone pine]

it is

okay

[mortal falcon]

i dont see how they would telescope, we have 2 positive terms for odd denoms, and a negative term for all denoms

a riemann sum seems promising

[stone pine]

b-a = 1

factor out the 1/2

[mortal falcon]

revisiting this

$\frac{1}{2} \sum_{n=1}^\infty [\frac{1}{2n-1}+\frac{1}{2n+1} - \frac{1}{n}] = \frac{1}{2} [\sum_{n=1}^\infty [\frac{1}{2n-1}-\frac{1}{2n}] + \sum_{n=1}^\infty [\frac{1}{2n+1}-\frac{1}{2n}]]$

[grand pond]

snowflake

[mortal falcon]

those sums look a lot cleaner, i think they might be standard results

variants of basel?

[midnight plank]

@near geyser Has your question been resolved?

[mortal falcon]

okay wait this is literally the taylor expansion for ln(1 + x) at x = 1

there's no issues with reorderings either

$\sum_{n=1}^\infty [\frac{1}{2n-1}-\frac{1}{2n}] = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$

[grand pond]

snowflake

[mortal falcon]

and $ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$

[grand pond]

snowflake

[mortal falcon]

so the first sum is just ln(2)

the second sum starts at 1/3, so that sum is ln(2) - (1 - 1/2) = ln(2) - 1/2

[near geyser]

Let me open it

[mortal falcon]

overall, we get

1/2 (ln(2) + ln(2) - 1/2) = ln(2) - 1/4

[near geyser]

Damn awesome

Let me write it and try myself

[midnight plank]

@near geyser Has your question been resolved?

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[grand pond]

Renato

[tidal turret]

a^p = a (mod p)

gcd(a,p) = 1 => a^{p-1} = 1 (mod p)

36^{p^2} = (36^p)^p = 36^p = 36 (mod p)

36^{p^3 - p^2} = 36^{p^2(p - 1)} = (36^{p^2})^{p-1} = 36^{p-1} (mod p)

this is where I get stuck

so if gcd(36,p) ≠ 1 then p = 3 or p = 2

p = 3 doesn't work but p = 2 works

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[proud abyss]

how do we start with this one?

[subtle blaze]

try a smaller grid

try do a 2x2 first

[proud abyss]

do we have to count squares and rectangles separately?

[subtle blaze]

i think they just want the total of them added together

[proud abyss]

on a 2x2 grid, it would be 9, right?

[quartz patio]

HLELEPWELPWLEPLEPLE

[proud abyss]

the rectangles

squares would be 5

[proud abyss]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[subtle blaze]

sure

can you keep track of how you counted them?

[stone pine]

what determines a rectangle/square on that grid uniquely?

[subtle blaze]

-# it'll be useful when the grid gets better

[stone pine]

Can you enumerate them?

If you figure out how to enumerate them then you can use combinatorics

[proud abyss]

i do not think i can think of a way to do that

[bold peak]

Not very clear from the wording, might as well do both ig

[stone pine]

Hint: how would you store a rectangle with the minimum possible information in say computer science

what do you need to know

[bold peak]

-# Think matrices

[proud abyss]

its length and width?

[stone pine]

for example for a circle you just need the center and radius

[proud abyss]

hi xavier

[bold peak]

Hiiiii been a while

[stone pine]

those two uniquely determine each circle

what does in your case of a rectangle?

[proud abyss]

two distinct x-coordinates and two distinct y-coordinates?

like

[bold peak]

Do you need all four

[proud abyss]

the coordinates of two opposite corners

[stone pine]

Yes

thats correct

[bold peak]

You can omit one point

Makes this problem easier

[stone pine]

You need 4 values

i think

one point for say the top left

wait

[bold peak]

For an axis aligned rectangle idts

Three should work

We can use the fact that it's axis aligned to our advantage

Think of it this way; you need the top left point and the length and width

[stone pine]

thats four values...

[bold peak]

Which is the same as having three of the points

[bold peak]

wat

[stone pine]

a point is two values

two points is 4

[bold peak]

In that case having four points is eight values

Which is what you were arguing earlier

[stone pine]

It isnt

[wise flicker]

There would be identical cases

[stone pine]

He already got it

use opposite corners

[bold peak]

She*

[stone pine]

ok sorry

point is

They're correct

[bold peak]

Oh oops, I missed that message from her lmao

[stone pine]

now just use combinatorics

[bold peak]

This entire discussion was pointless

My apologies

I thought you said yes to her saying she needed four points

[stone pine]

yes four values is four regardless of how you tuple them

its fine

happens

[wise flicker]

Elaborate

Every four values won't be rectangle

[stone pine]

A value is a number

what i mean

Not points

[wise flicker]

K

Where is op

@proud abyss

[bold peak]

-# Done with our bullshit

[proud abyss]

so for rectangles

would it be (m+1)c2.(n+1)c2?

or no?

[wise flicker]

m?

[stone pine]

that works

except

[proud abyss]

if we are taking an mxn grid

[stone pine]

there is an edge case

[proud abyss]

and that is?

[stone pine]

wait

nvm

no edge case

[proud abyss]

what were you thinking of?

[stone pine]

Something else

it doesnt apply

[proud abyss]

all right

so for the squares

if we know the top-left corner and the length of its side, it works

[stone pine]

yes

[proud abyss]

a square can be 1x1, 2x2, ..., kxk; we now need to figure out how many of them are there

[stone pine]

yeah

i was thinking that too

find how many kxk are there

and sum k

[proud abyss]

i don't know how to do this

ah no

wait

(m − k + 1)(n − k + 1)?

[stone pine]

yeah

except your m is n i think

for this question specifically

[proud abyss]

all right, thank you for the help!

.close

[midnight plank]

Channel closed

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Use .reopen if this was a mistake.

[proud abyss]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[proud abyss]

wait i have another question

so what if we had to count triangles in a pattern? is there a way for that too, like an explicit formula? figuring it out for rectangles and squares was relatively easy but what about triangles?

[wise flicker]

How would you do that for a grid?

Can take diagonals

[proud abyss]

the first thing that comes to my mind is to break rectangles/squares into two parts

not sure if that is the correct way to go on about it

[stone pine]

interesting

but i think there can be triangles on the edge

[proud abyss]

but cannot three points be collinear in lots of ways?

[stone pine]

not a half of any rectangle

[slow thorn]

like this?

[wise flicker]

It is but can you only break into two?

[stone pine]

I think they mean on a n x n grid still

[wise flicker]

Use bijection

[proud abyss]

no, more like in a gird

grid*

[proud abyss]

fair point

[proud abyss]

this also seems messy

[wise flicker]

Take 3 vertices

[stone pine]

but maybe you can overcount as rectangles are part of other rectangles?

idk seems a lot harder and i have better things to do

[proud abyss]

fair enough, i am sorry

[wise flicker]

Its ok

[proud abyss]

i will close the channel now, thanks to all who helped

.close

[midnight plank]

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[midnight plank]

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[onyx tide]

Is it correct?

[wise flicker]

yes

[onyx tide]

Really?

[slender walrus]

which region are you in?
usually ] and [ would be used in interval notation
, not > and <

[wise flicker]

That does represent inclusive

[slender walrus]

perhaps? though not in the regions i'm familiar with

which is why i'm asking for clarification

since square brackets are used pretty much everywhere i've seen and in books

< > are used when stuff like vectors are involved

[midnight plank]

@onyx tide Has your question been resolved?

[onyx tide]

It’s okay

It’s used > instead of ]

.close

[midnight plank]

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[elfin fox]

I'm really confused about this step. How do they get rid of the reciprocal?

[heavy valley]

It is using multinomial theorem

[runic hamlet]

the inverse of $f=a_0+a_1x+a_2x^2+a_3x^3+\ldots$ is $g=b_0+b_1x+b_2x^2+b_3x^3+\ldots$ such that $fg=1$. so if you multiply out $fg$ then all terms except the first have to cancel out. in that way you can somewhat easily compute the first few terms of $g$

[grand pond]

Denascite

[elfin fox]

Fair enough thanks

.close

[midnight plank]

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[proud abyss]

the answer is no, right? because v6 in the second graph has a degree four but no vertex in the first graph does, too? also, is there an easier way to check if two graphs are isomorphic or not, especially when we are working with more complicated graphs?

[proud abyss]

the answer sheet says they're isomorphic

[runic hamlet]

answer sheet fucked up

checking whether graphs are isomorphic is a hard problem

[hybrid widget]

for your third question, if one graph has a cycle and the other doesn't, they cannot be isomorphic

[proud abyss]

okay, but that is one case

[runic hamlet]

number of vertices with certain degrees, number of cycles, number of triangles, ... can help

but they will never fully answer the question

[proud abyss]

so you just kinda have to check each vertex one-by-one?

[hybrid widget]

yes. that's the safest way

and the general method

[proud abyss]

a question, when you said this

do self-loops/multiple edges not count? say we have a 'square', each vertex will be of degree two; what if we have another graph with multiple edges between two vertices?

or is that not allowed?

i am not sure if i worded that correctly

[midnight plank]

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[twilit field]

P =$\sum_{i=600}^{1000} \binom{1000}{i} \left( \frac{1}{2} \right)^i \left( \frac{1}{2} \right)^{1000 -i}$

Just wondering if there's any other way

[grand pond]

wai

[dreamy lichen]

There is a possibility that they want u to assume approximately normal distribution btw

[twilit field]

ah

[dreamy lichen]

that'd prolly give u the probability to a decent accuracy

[twilit field]

yea, was wondering why this was a question near the end of the book

[twilit field]

what's the name of this theorm, not sure if we';ve done it yet

[dreamy lichen]

CLT

central limit theorem

maybe there is some weaker version just for binomial, but idk

[twilit field]

ah I think we do that tomorrow

nvm then

thanks

[cerulean oyster]

For this particular case its obvious it satisfies
But for Bin to Normal approximation you usually want to have np, nq > 5

[wispy mortar]

ye using mean and variance?

[twilit field]

another question

I think I should use this here

[dreamy lichen]

btw if u only want to calculate the prev question, just do this. CLT only proves that it becomes approximately normal for large n

[twilit field]

$E[Z] = E[X^2Y^3]+E[X^2]-E[X]-E[Y^3]$

[grand pond]

wai

[twilit field]

now to determine if X, Y are independent

so like $\int_{0}^{2x} 12xdy$ is the marginal of $Y$>?

[grand pond]

wai

[twilit field]

which is $24x^2$

[grand pond]

wai

[twilit field]

Not sure what the marginal for X is

it's just 12xy?

this feels off

.close

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[edgy crater]

is this how u would prove this, and if so is there some kind of name for this proof topic / method

[lyric charm]

your handwriting is very meh

so you're saying that you look at the smallest arc between two adjacent students and saying that those two will throw their balls to each other?

ig that's a proof by construction lol

[lethal path]

you have the right idea but you need to be less handwavy when you explain

[wise flicker]

there can be lesser distance b/w a 3rd student than one of the two

[edgy crater]

sorry

[slow crest]

can sm1 pls help me

[cerulean oyster]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[edgy crater]

the circle / arc was just to get an idea of the picture

[lethal path]

as in, "let d(X, Y) be the shortest distance from player X to Y"
"for any X, Y, we have d(X, Y) = d(Y, X)" (why?)

well actually the distance would just be the straight-line distance

[lyric charm]

i think ops idea was to look at ALL the pairwise distances and consider the shortest one among them

[edgy crater]

yeah

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[empty thistle]

hi, how would i prove that the right side is really equal to zero? do i have to prove it for even and odd numbers separately ? (binomial theorem)

[grim vector]

Well you just did

If its equal then its neccessarly 0 on the right side

[lethal path]

if you try expanding (x + y)^n using the binomial theorem, then sub in x = 1 and y = -1, you get that

[grim vector]

Since (1-1)^n = 0

[empty thistle]

ohhhh

right

im stupid lol

thanks

.close

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[leaden seal]

Hi, can someone explain to me the bit in red (arc length of fiber DE)

[leaden seal]

what is the equation for arc length

[fallow creek]

arc length = radius * angle (in radians)

the box is pointing specifically to R theta, not the whole circled red part

[leaden seal]

ok

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[lyric scroll]

A tiler suspects that the manufacturer has secretly delivered 3rd-class tiles (30% defective) instead of 1st-class tiles (10% defective). He tests a box containing 50 tiles. What decision rule must be used if the error of classifying a 3rd-class box as 1st-class should be below 10%?

[lyric scroll]

hey so,

i rly dont know how to start

all i know is that we are searching for a value

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[tidal turret]

Compute the remainder of $$\sum_{k=1}^{100} 5^{k!}$$
upon division by (5^{2}\cdot 19).

[grand pond]

Renato

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[last slate]

How I do this

[last slate]

<@&286206848099549185>

I need this speed

[midnight plank]

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@last slate Has your question been resolved?

[last slate]

<@&286206848099549185> I kinda need help

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[bronze mason]

Do you know what rotational symmetry is?

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[twilit field]

Just confused here

$\langle Tx, x \rangle ≥0 \implies -\langle Tx, x \rangle ≤0 \implies \langle -Tx, x \rangle ≥0 \implies -$( As $T,-T$ are positive. So $\langle Tx, x \rangle ≤0$. This tells us $\langle Tx, x \rangle =0$, so $T$ must be $0$.

[grand pond]

wai

[twilit field]

does this work

.close

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[viral dagger]

how did they use ptolemy's inequality to get that?

if im not mistaken it should refer to this

the diagram should look smth like this (ignore the circles, its used for a different solution)

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[graceful drum]

they used it on quad AEDX

and got AD by cosine law

and similarly they did it on ABCX

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[viral dagger]

ohh

ok thank you!

.solbed

.solved

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[viral dagger]

seperate the log_4(5^15)/log_64(5^3965)

if we shift the terms on the top to the left (so they share the same base), we can simplify it

so like from log_5(5^24)/log_5(5^12)=24/12

this follows the formula ((k+1)^2-1)/(k^2-4)=(k^2+2k)/(k^2-4)=k/(k-2)

[viral dagger]

so were left with this bit, times the product of k/(k-2) from k=4 to 63-1=62 (cause we got rid of a pair)

[viral dagger]

this equals 9/793

[viral dagger]

and this bit, from telescoping would equal 61*62/6=1891/3

so i got 279/13

the answer is supposed to be 93/13?? where did i go wrong

[cunning hornet]

isreal

[viral dagger]

i somehow got an extra multiple of 3 somewhere?

oh wait

[latent wadi]

It should be 61*62/6
The division of 3 comes from k = 5

[viral dagger]

yeahh i just realized that

[latent wadi]

Oh well

[viral dagger]

when i tried working it out i wrote the 3, but i thought that was wrong since it would only have a 1*2 for some reason instead of 2*3??

ok ty

.solved man maybe i should just sleep

[midnight plank]

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[leaden seal]

can someone help with this trig question

[midnight plank]

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[olive shuttle]

this is statics

not math

[clever sedge]

Right, math & physics are clearly irreconcilable.

[cedar coral]

!original

[midnight plank]

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.

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[midnight plank]

@shut canyon Has your question been resolved?

[west iron]

Do you need help with 1→C?

[shut canyon]

Sure

this is 1

[west iron]

Ok

[shut canyon]

I don't yet see what is in C

connect 1 to any C you get connections to every object in C?

I know x : 1⟶X, or there is ⊤ : 1⟶Ω which seem similar

[west iron]

It's saying if you have a functor between 1 and C that's the same as choosing an object in C

[shut canyon]

select one

[west iron]

Yeah, choosing a functor is the same as choosing an object

[shut canyon]

sounds awesome

[west iron]

So to prove this, what information do we need to completely determine the functor?

[shut canyon]

choosing a natural transformation is choosing a functor?

[west iron]

nah natural transformations are between functors

don't worry about those yet

[shut canyon]

all the arrows that go to and come from it

Yoneda

[west iron]

No Yoneda

That's only up to some kind of equivalence right

Functors are functions

So we need to know where each element in the domain goes

What should the domain of F:1→C be?

[shut canyon]

Like specify objects in these categories?

maybe 1 has ● for example?

the domain of F : 1 ⟶ C is just 1

the category with one arrow one object

[west iron]

Yeah

But like, a functor is in particular an actual function

Like the kind from Set

[shut canyon]

so one object and one arrow is mapped via structure preserving map to another category

[west iron]

Yeah

So we need to figure out

• where does the object go
• where does the arrow go

[shut canyon]

eg F : Top ⟶ Set is a functor I believe

[west iron]

forgetful functor?

[shut canyon]

I'm not sure

[west iron]

nvm doesn't matter

[shut canyon]

forgetful forgets structure?

[west iron]

Yeah

[shut canyon]

not lossless

not monic maybe

[west iron]

It just maps every algebraic structure to its underlying set and the morphisms to the functions

It's injective, so "lossless" in that sense

[shut canyon]

I'm confused if 1⟶C, ah maybe it is mapped to the identity of objects

[west iron]

If there was an identity of objects, that would be a good candidate

But there's not one, so really it can kinda go anywhere as long as we can fit the morphism in

[shut canyon]

@west iron thank you for your help!

[west iron]

Np!

[midnight plank]

@shut canyon Has your question been resolved?

[shut canyon]

these do look like object, arrow and acomposition actually!

these are the categories 1, 2 and 3

they look like
a : a⟶a
f : 0 ⟶ 1
and the diagram for composition f∘g = h

functor one picks out an object (and it's loop?)
functor two picks out an arrow because it picks out two objects and the relationship between them?
functor three picks out three objects and their relaitonships which is a composite

very cool

🎵

[shut canyon]

2⟶ℕ could select two numbers n,m, and an arrow n⟶m,
eg +1 : (2⟶3) ⟼ 2+1⟶3+1

+1 : (2→3) ↦ 3→4

[shut canyon]

∀C, Object(a) ∧ Object(b) → ∃Morphism(f) ∧ dom(f)=a ∧ cod(f)=b
∀f ∃id (identity), ∀f,g (composition associative)

∀C ∀F (Functor(F) ∧ dom(F)=1 →
∃o (Object(o) ∧ F(●)=o ∧ F(id₁)=idₒ))

Σ o:Obj(C), F(●)=o ∧ F(id₁)=idₒ

Σ o₀,o₁: Obj(C), Σ f: Hom_C(o₀,o₁),
G(●₀)=o₀ ∧ G(●₁)=o₁ ∧ G(●₀→●₁)=f ∧
G(id₁)=idₒ₀ ∧ G(id₂)=idₒ₁

structure functor2 (C : Type) [category C] :=
(o₀ o₁ : C)
(f : o₀ ⟶ o₁)

#check functor2 -- picks o₀,o₁ and f

.close

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[cerulean temple]

Compute
[
\int_C \frac{1}{z^2 + 4},dz,
]
where (C) is the positively oriented circle with center (2i) and radius (1).

[grand pond]

Slowaq

[cerulean temple]

how to do this i have no idea

[surreal moon]

You can do residues if you are familiar with them

[cerulean temple]

unfortunately not yet

[surreal moon]

Then you will want to do just a regular path integral

[cerulean temple]

so should i just use definition to do it without residues?

ah alright then

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[violet tide]

can anyone help me with part d

surely they dont expect me to draw whatever that is

anyone

[fallow scarab]

did you draw a picture for C?

[midnight plank]

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[crimson wyvern]

I can't figure out what I'm doing wrong

[crimson wyvern]

I am integrating the joint pdf from 0 to u

dy_1

[fallow scarab]

show your work

[fallow scarab]

nah. you need to use convolution to get the density

https://stats.libretexts.org/Bookshelves/Probability_Theory/Introductory_Probability_(Grinstead_and_Snell)/07%3A_Sums_of_Random_Variables/7.02%3A_Sums_of_Continuous_Random_Variables

[crimson wyvern]

I don't think we discussed this

because there are several other problems where you aren't just looking at X + Y

[fallow scarab]

i suppose you can just apply the definition of the sum from the start

https://www.youtube.com/watch?v=Glff9dvPVEg

[crimson wyvern]

wait this is just giving the answer I had that was wrong though

oh wait

it's divided by

ok I did everything right I just screwed up the actual formula for an exponential distribution

Probability is so tedious

I hate it

it's so difficult to check where you've gone wrong

[placid spoke]

Alternatively if you've seen the gamma distribution, the sum of exponential RVs are gamma distributed

[midnight plank]

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[green kite]

Can someone explain to me all the important derivatives in Physics (I'm in AP Physics C and we just got up to center of mass)

Here are some I memorized, but I don't know:
F = dw/dx
F = -du/dx

[green kite]

not the basic ones like velocity position and accel.

but like literally every other derivative needed in physics

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[elfin zealot]

can i please get some help on understanding how to find formulas and understand what to plug in

[small jasper]

same as #help-23

[midnight plank]

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[vital marten]

could i get a quick check if i did this right

[small jasper]

good

,w differentiate (integral of sin(t^2) dt from 1 to x^2) with respect to x

[grand pond]

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[small jasper]

you can also check like this in #bots

[vital marten]

thank you very much civil service pigeon

.close

[midnight plank]

Channel closed

Closed by @vital marten

Use .reopen if this was a mistake.

[small jasper]

[solar folio]

guys

is the formula

f=kx

and when do we use m g =k x

[midnight plank]

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[untold crater]

Can someone quickly walk me through the steps of solving 6a? I somewhat understand ambiguous cases but it’s not fully clicking. Please help;(

[small jasper]

Did you draw the diagram?

[untold crater]

Yes 🙂

1 second

This would be the base

Right..

🥹

[small jasper]

btw you should show what you've done so far (so stuff like drawing a diagram) when first asking for help - it gives us more context and saves time.

[small jasper]

wdym by "base"

[untold crater]

like before the whole

shifting thing with the ambiguous cases

[untold crater]

Okay! Sorry about that

[small jasper]

You can draw the triangle however you want

doesn't rlly impact anything

Did you do anything else with this? (Like setting up a relevant equation?)

[untold crater]

oh okok

I don’t understand where to go from there

Now I’m guessing I could solve c with Pythagorean theorem

But

I am supposed to use ambiguous case on this

So that’s what I don’t understand

[small jasper]

what trig equation is commonly associated with the ambiguous case

[untold crater]

Uhm, sine law?

[small jasper]

yup

[untold crater]

okay okay

[untold crater]

but I can’t use sine law here can’t I?

[small jasper]

why not

[untold crater]

Oh wait

Yes I can

[small jasper]

yeah the sine law works on all triangles

[untold crater]

Okay, and then what would I do? Minus theta B & 29* from 180?

To find theta C?

[small jasper]

yes that comes from the angle sum of a triangle

[untold crater]

And then use sine law again

But where does the ambiguous case part come in

[small jasper]

one thing at a time

[untold crater]

okok

I’ll do that rn

[small jasper]

do the sine law on this

or rather

what equation does the sine law give you on this

[untold crater]

Could this be correct

[small jasper]

not necessarily

You had $\sin B=\dots$ and took the inverse sine of both sides, right?

[grand pond]

Civil Service Pigeon

[untold crater]

Yes

[small jasper]

here's the thing

actually

let me ask like this

[untold crater]

Well I did inverse sine for B

okay

[small jasper]

what is the range of the inverse sine function

[untold crater]

Uhm 🥹

The range?

[small jasper]

what are the possible outputs of the inverse sine function

[grand pond]

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[untold crater]

uhm.. idk 😞

[small jasper]

I gave you the graphs above

you should be able to make a guess based off of that

[untold crater]

It seems like they go 1.5 < y < -1.5

[small jasper]

-pi/2 to pi/2

,calc pi/2

[grand pond]

Result:

1.5707963267949

[untold crater]

oh okok

[small jasper]

do you see an issue with this

[untold crater]

But I got 0.7831 for sinB before i did sine inverse, is that not between those numbers?

[small jasper]

I'm not talking about your specific case

I'm talking about in general

[untold crater]

oh ok

IDK?

😞

[small jasper]

think about obtuse angles

[untold crater]

Oh

That it’s between 90 and 180 and that wouldn’t be under 1.57

? Is that what you’re talking about?

[small jasper]

yes

the inverse sine function doesn't output obtuse angles

the inverse sine function (at least directly) only gives you right and acute angles

[untold crater]

Oh so I can’t use sine law to solve c

[small jasper]

how do you think we can get the obtuse solutions

hint: you can get it from the acute one

[untold crater]

Uhm..

Well wouldn’t obtuse solutions have the same sine anyways?

[small jasper]

yes

the idea is that $\sin(\text{angle})=\dots$ should give you an acute angle and an obtuse one

[grand pond]

Civil Service Pigeon

[untold crater]

That’s the idea of the ambiguous cases?

[small jasper]

yes

there's more than one possible angle

if both angles are possible, then it falls under the ambiguous case

since the angle can't be uniquely determined

[untold crater]

So how does this lead to the side of angle swinging closer to the other side

[small jasper]

bigger angle -> the sides that form said angle are "farther apart" from each other

this should be intuitively obvious (draw a few diagrams if you need a visual)

[untold crater]

Also couldn’t I just use cosine law here?

[small jasper]

to compute side C? yes.

[untold crater]

so then what’d be the point of ambiguous cases

[small jasper]

a different value for angle B gives a different value for angle C

since angle C depends on angle B

hello?

[untold crater]

Sorry, I’m trying to solve the problem

Okay here, my friend sent me a picture of what she did. Can you explain to me what she did kind of?

[small jasper]

did you address this?

[untold crater]

Well there’s cosine law, and also trig ratios in different quadrants

That’s what I’m thinking

[small jasper]

I like the different quadrants idea

[untold crater]

But on my note, it says I have to minus 180 from the acute angle

[small jasper]

yes

because sin(pi-x) = sin(x) for all x

[untold crater]

that’d just make the other angle that’s the same sine

[small jasper]

yes

[untold crater]

okay but it’s just not clicking how to graph it.
Like how do you know where to draw the new middle line, and why does it even really matter if in the end she just did sine law with sin99

[small jasper]

you can draw the picture if you want to

it's schematic though anyway so it doesn't really matter

the main point is that you have your acute angle from taking the inverse sine

and then your obtuse angle from doing 180 - acute angle

but for the obtuse angle, you need to test if it's possible

(make sure it doesn't make the sum of the angles exceed 180 deg)

that's it

you can kinda see the idea of acute angle and 180-acute angle from the diagram tbf

so if it helps you visualise what's going on with the symbol pushing then that's great

drawing it for every question seems a bit excessive and inefficient though

depends how much of a factor time pressure is for you

but I digress

[untold crater]

So in the end, all i have to show for is that the obtuse angle is possible, and that it’s 180- the acute angle?

At the end of the day I could’ve solved it without the ambiguous case

[small jasper]

the ambiguous case is the fact that you can have two different angles

that's exactly what is reflected by the acute and obtuse angle cases

[untold crater]

You can have two different angles and still have the same sine

is that it?

[small jasper]

the ambiguous case [relates to the fact that] you can have two different angles [with the same sine]

[untold crater]

okay okay

I get it I guess 😭

thank you very much ☺️

[small jasper]

[midnight plank]

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

[untold crater]

.close

[midnight plank]

Channel closed

Closed by @untold crater

Use .reopen if this was a mistake.

[small jasper]

yeah tldr is that you can have two different angles with the same sine, and the ambiguous case arises when both angles are possible

that's rlly it

[midnight plank]

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[pastel spoke]

Aghhh

[pastel spoke]

Ehhh

[lusty python]

???

[lyric charm]

@pastel spoke do you have a question to ask? y/n

[lyric charm]

translation?

[pastel spoke]

It is basically you doing some manipulation to get the left to the right matrix

Then it asks for a,b,c

Ehhhh

Annoying one

[lyric charm]

ok, so we do operations on the COLUMNS to get from matrix A to matrix B?

[pastel spoke]

Yes

by my previous experience with quizzes this types, we should avoid operations involving any unknowns

It is either you get the solutions from the right matrix to the left or the other way around

[lyric charm]

$A=\bmqty{1&2&3&10\2&3&1&13\3&a&2&13} \ B = \bmqty{1&0&b&5\0&2&1&5\1&c&0&9}$

[grand pond]

Ann

[lyric charm]

btw are you sure it says column operations and not row operations?

just wanna make sure so we don't run into language issues

[pastel spoke]

I think it is called Gaussian elimination

[lyric charm]

i think you're not answering my question

[pastel spoke]

It is addressed as this in my book

[lyric charm]

if it's called Gaussian elimination, then i would imagine we are doing stuff with rows.

[pastel spoke]

You are right

列=row

[lyric charm]

ok

[pastel spoke]

Should I spoil you with my answer book which I cannot understand

[lyric charm]

i was going to give suggestions for first steps myself

[pastel spoke]

Okay

[lyric charm]

and the first move seems to be R1 -> R1 - R2 on matrix A

err wait no

no, scratch that

R2 -> R2 - 2*R1

$A_1 =\bmqty{1&2&3&10\0&-1&-5&-7\3&a&2&13}$

[grand pond]

Ann

[pastel spoke]

Actually, this is exactly what I did

[lyric charm]

ok

what'd you do afterwards

[pastel spoke]

But it turns out to be vain attempt. There is the a which dissuades me from continuing it

[lyric charm]

let's see how far we can take it before having to deal with the a...

[pastel spoke]

I can make the unknown disappear

Alright

[pastel spoke]

It may be quite ugly the calculation. But I will do it

[lyric charm]

$R_1\to R_1+2R_2$ and then $R_2\to -R_2$ gives us $$A_2 =\bmqty{1&0&-1&-4\0&1&5&7\3&a&2&13}$$

[grand pond]

Ann

[pastel spoke]

Then we fk the row 3 with row 1.
To make a clean
1
0
0
On the first column

Is that it

[lyric charm]

that's what i was thinking about next

[lyric charm]

$A_3 =\bmqty{1&0&-1&-4\0&1&5&7\0&a&5&25}$

[pastel spoke]

What’s next

[grand pond]

Ann

[lyric charm]

this is what would happen with my route

[pastel spoke]

Sure

Then you have to deal with the a

Which seems impossible to get rid of

[lyric charm]

hmm yeah

i think this is where the ugliness really shows up

[pastel spoke]

I hate it

[lyric charm]

it is not obvious to me how to continue here

[pastel spoke]

Should I spoil you with my answer book

Then you may help me to understand it

[lyric charm]

sure

[lyric charm]

how did they just fill some shit in

[pastel spoke]

Yes

[lyric charm]

sounds like they just pulled the correct answer out of their ass

wtf is up with that

they don't explain how to get there

[pastel spoke]

0 2 1 5 is a row from the matrix B

[lyric charm]

stupid bullshit

oh i guess we just work with all the rows without letters?

that makes a bit more sense

[keen widget]

this is just R2 - 2R1 and then R3+2R2

[pastel spoke]

Why it works

[pastel spoke]

Wdym

[keen widget]

row transformation

[pastel spoke]

.close

[midnight plank]

Channel closed

Closed by @pastel spoke

Use .reopen if this was a mistake.

[pastel spoke]

It is solved

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[craggy lintel]

how do you mathematically denote a 3 dimensional object? I mean say a chips packet for example, you can't literally say "For every particle considered to be in a chips packet"
so how do you denote complex objects like dat?
say we exclude particles and more especially billions of em.
Let's say the chips packet is a continuum of stuff/source.
how do you denote it still? in 3 dimensional space. u can't just say "oh so this chips packet exists in 5m² radius". How precisely can you denote such a thing?
and also, it's not like you find functions of random objects to begin with ~ the universe isn't so neat that tells you functions and u 3 dimensionally integrate them.
suppose u know nothing of the chips packet, and you just can observe it. Tell me a legible way to denote all the areas precisely enough

[craggy lintel]

<@&286206848099549185>

[midnight plank]

@craggy lintel Has your question been resolved?

[fathom onyx]

https://tenor.com/view/cow-spherical-cow-spherical-sphere-yay-gif-24269101

So, technically, one: this is more a question of physics (and perhaps to a lesser extent mathematical modelling)

Two: in a number of contexts, you can in fact claim "for every particle considered to be in a chips packet..."

But three, and chiefly: how you model an object depends on what you're modelling it for

To that end:
https://lmgtfy.app/?q=mathematical+modelling+lecture+notes+pdf

[craggy lintel]

I'm modeling it for depicting a precise position in 3 dimensional space, no other meaning to it, it needs to be precise and mathematically formulated

so how would I do it then?

[fathom onyx]

But for what

Else you're gonna be as vague as "stuck cylindrical object" and we can only infer what your goal is

[quiet hinge]

well for a more theoretical viewpoint, you need to first define a chips packet to be able to denote it in some mathematical manner

[quiet hinge]

when you try to do this you will realise that this is simply not possible

or not practical

[fathom onyx]

Are you challenging me?
-# - Light Yagami

[quiet hinge]

you could for example take every particle making up a chips packet in some moment in time and declare that set of particles (lets call it S) to be the chips packet

then i guess you can simply denote the chips packet mathematically by "S"

[craggy lintel]

I mean all this is cool, but it's not physically depictable/calculatable on what is considered a chips packet at all