help-49

[scenic wyvern]

gotta wait for the dl then i suppose

[compact copper]

Alright, pic it is @safe notch

[safe notch]

in 30 minutes i gotta go eat and go collage

[safe notch]

hmm

[compact copper]

Dude I'm attending a meeting and simplifying SHM at the same time

[safe notch]

ohhh mai god

[scenic wyvern]

i'm in office too

[safe notch]

damn

am i being annoying

😭

[compact copper]

Nah you're fine

[safe notch]

yall are so busy

[compact copper]

We wouldn't be here if we couldn't help

[safe notch]

thanks guys for ur help

[compact copper]

Just trying to put emphasis on time utilisation

Try to get a pic

[safe notch]

all right gotta go to mom for her mobile

if she may

wait

[compact copper]

Yeah sure

[scenic wyvern]

my speeds are conspiring against this man

[compact copper]

lol don't tell me you found it online

[scenic wyvern]

i won't discuss it publicly lol

[safe notch]

what

[compact copper]

I meant the book

[scenic wyvern]

ye i know, but yea i did. not gonna say anything more than that

[safe notch]

oh so its better

should i click the photo

??

[compact copper]

ahhh okay

[scenic wyvern]

unless you wanna wait another 25 minutes, do it

[compact copper]

if you can

[safe notch]

wait

ill send in msenger

and copy to paste here

[compact copper]

Yeah sure

[compact copper]

What part?

[scenic wyvern]

,rccw

[scenic wyvern]

now i wish i had my copy of Sears and Zemansky on hand

[compact copper]

that's not clear

[safe notch]

page 34 see the case i

it says mean position at x= o

at pg 35

[scenic wyvern]

you sent this picture twice

[safe notch]

i know

just had to order

them

[safe notch]

special cases case i

it says the mean position at y=0

aghhhh and i dont get why it took

y as the displacement not x

idk man so much things

here

[compact copper]

Ehhhh, y = 0 or x = 0, both reference to the Origin

[scenic wyvern]

let's say you have circular motion

[safe notch]

k

[scenic wyvern]

now, let's say you're standing to the side of the circle

the component of motion you'll see is different depending on which side you stand on

[safe notch]

yea

[scenic wyvern]

if you stand on the red side, you see the y-component.
likewise, if you stand on the green side, you see the x-component

so whether y or x as the displacement depends on where you're looking at the object from

but in the end they describe the same motion really

just from different points of view

[safe notch]

wtf

wait

[compact copper]

Exactly, and them being zero is them being at the origin

[scenic wyvern]

have you ever been to a carnival

[compact copper]

You need help with that statement?

[safe notch]

yeah

[compact copper]

Answer this first

[safe notch]

i need

[safe notch]

been years i havent been but yea few times

[scenic wyvern]

then have you seen games where they put something like a plastic or rubber duck that moves around in a circle

[safe notch]

yeah

[scenic wyvern]

then players would stand to one side and try to hit the duck

[safe notch]

a horse

[safe notch]

wwhat

no

no social experience aswell smh

[compact copper]

Even I'm unfamiliar with that, we ride the horses

[scenic wyvern]

i suppose you're talking about a carousel

that works

imagine a carousel with one horse only then

[rough plover]

hopefully this makes sense for the x component

[safe notch]

wait

am i seeing things

[compact copper]

Okay wait

[safe notch]

the point particle has circular motion but its projection performs simple harmonic motion

[scenic wyvern]

exactly.

[safe notch]

in the horizontal line

[compact copper]

Imagine you stand with your eyes at the axis, so you eyes are at the same axis as the plane

[scenic wyvern]

side-on circular motion is SHM!

assuming constant angular velocity

[compact copper]

hence theta = wt

which you mentioned earlier @safe notch

[rough plover]

should help connecting to sinusoids

[safe notch]

wth

[scenic wyvern]

i gave him a vid on that earlier

not sure if he watched it

[rough plover]

oh okay great

[compact copper]

Look at one colour first

[safe notch]

oh

oh

horizontal and verticle shms

[compact copper]

-A is the A' you mentioned earlier

[safe notch]

yeah

[compact copper]

Yes

[safe notch]

makes sense

lemme get this st

isnt mean position always the origin

[compact copper]

yes

[safe notch]

so why the hell book is yellling its x= o an y=o

[compact copper]

unless you're using a cosine wave

[safe notch]

im loosing it

[scenic wyvern]

well...

what's another name for the point (0, 0)...?

[compact copper]

They're considering both vertical and horizontal components

[safe notch]

origin

[scenic wyvern]

congratulations, you just answered your own question

[safe notch]

waaaaaaaait a secc

[compact copper]

Assuming it's a sine wave though

which it is in our analysis

let's not get into cosine wave just yet

[safe notch]

aghhh god guys im gatting late for school can we get on this at exactly 10 hours

[scenic wyvern]

don't think i will be here at that time, but do what you need to

[compact copper]

Well my schedule fluctuates, ping me IG

[safe notch]

i am sorry im leaving 😭 😭

aghhhh

thanks for the help until this moment

[compact copper]

Yeah no problem, focus on your studies

[safe notch]

thank u

have a great day

both

[compact copper]

Anytime🫡

[safe notch]

.close

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[tidal turret]

help

[drifting dust]

can you translate?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

P is the power set of x, R is a relation on P we need to prove is an equivalence relation

[tidal turret]

somehow we have to use iii)

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[lucid delta]

can i replace cos(90-a) with sin(a)?

[lucid delta]

and sin(90-b) with cos(b)

or does that rule not apply in this case

[lyric charm]

[insert some kind of funny response here about imaginary math-bureaucratic red tape]

yes ofc you can do that.

[lucid delta]

its just that when i plug the numbers into the calculator i get different answers

[scenic wyvern]

hang on. a might not be ok

[lucid delta]

eg

[scenic wyvern]

b is stated to be acute, so b is fine

[jaunty ivy]

is ur desmos set to radians

[scenic wyvern]

but a is not necessarily acute

[cerulean oyster]

check for rad(ians) or deg(rees)

[lucid delta]

oh it might be

how do i change?

[cerulean oyster]

options, top right and this buttons

its called graph settings methinks

[lucid delta]

okay now it works

[cerulean oyster]

yeah, for both desmos or an actual calculator, always make sure youre working with the right measure > one of the classic fumbles

[lucid delta]

arcsin0.45 is ~27 so we're good

[jaunty ivy]

yeah the identity works for all a in R

[scenic wyvern]

hm aight, sorry for the misunderstanding

[lucid delta]

thank you all

appreciate it

.close

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[next bramble]

Do you know the definition of an equivalence relation?

[tidal turret]

equiv rela = sym + refl + transi

[next bramble]

yes

is it symmetric?
is it reflexive?
is it transitive?

[tidal turret]

it is reflexive

I was trying to see symmetry

[next bramble]

to prove it's an equivalence relation, you'll need to prove those three properties

[tidal turret]

x R y => y R x,

[tidal turret]

proving reflexivity is straightforward

0 = 0

[next bramble]

reflexivity: you need to show x^2 - y^2 = 93x - 93y when x=y

[tidal turret]

x R x <=> x^2 - x^2 = 93x - 93x

[next bramble]

yep, that

[tidal turret]

x^2(1-1) = x(93-93)

this holds for any x ∈ A

[next bramble]

For symmetry, I'll give you a hint: can you simplify x^2 - y^2 = 93x - 93y?

[tidal turret]

x^2 - x^2 = 93x - 93x

x^2 - x^2 -93x + 93x = 0

(x^2 - x^2) + (93x - 93x) = 0

this holds for any x ∈ A

[next bramble]

oops, sorry. I meant symmetry

you've shown it's reflexive, which is good 🙂

hint: x^2 - y^2 is a difference of squares

[tidal turret]

x R y => y R x
Assume: x^2 - y^2 = 93x - 93y
x^2 - 93x = y^2 - 93y
multiply by -1 both sides
93x - x^2 = 93y - y^2
y^2 - x^2 = 93y - 93x
=> y R x

[steady mulch]

Heyyy guys

[next bramble]

nope, because x^2 - y^2 = 93x - 93y and y^2 - x^2 = 93y - 93x are different equations.

[steady mulch]

Is anyone in 11th grade?

From India

[next bramble]

so you can't just multiply by -1

[next bramble]

!occupied

[midnight plank]

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).

[next bramble]

can you simplify x^2 - y^2 = 93x - 93y?

[tidal turret]

(x-y)(x+y) = 93(x-y)

[next bramble]

yep. now can cancel the (x-y)s?

you'll need to justify why

[tidal turret]

we are diving by 0?

[next bramble]

are we?

when would we be dividing by zero?

[tidal turret]

suppose the relation consist of R ⊂ {(a,a) | a ∈ A}

[tidal turret]

when x = y

[next bramble]

yep. so if x R y and x = y, isn't that just reflexivity?

which we've already shown?

[tidal turret]

yes but suppose R = {(1,1)} and A = {1}

[next bramble]

we have two cases. either x = y which is reflexivity, or x != y so we can divide by (x-y).

[tidal turret]

this is reflexive and symmetric

[next bramble]

then symmetry and transitivity are hold by vacuous truth.

https://en.wikipedia.org/wiki/Vacuous_truth

there are no two elements x,y in A where x != y, so it's symmetric by default

Just like the relation on A = {} is all three by vacuous truth

[tidal turret]

what I am saying is

∀x,y ∈ A,
transitivity: x R y => y R x

like is possible all pairs in the relation are x = y

transitivity doesn't have a problem with that

[next bramble]

Yep. Let's test that on A = {1}.
All the elements in A are 1, so there's only one we need to check

• 1 R 1 => 1 R 1, so it's symmetric

i did swap the elements over, you just don't notice it because they're the same

[tidal turret]

yes, but going back to our proof of symmetry, division by 0 is impossible here

[next bramble]

indeed. but we don't need to cancel

reflexivity proves symmetry

Suppose S is another relation. I'm not going to tell you what it is, because you don't need to know.
Suppose also that S is reflexive - so x S x.
Now, knowing that, prove x S y => y S x when x = y because there's only one element in the set.

To put it another way, because there's only one element in A, we know
for all x,y in A, we must have x = y.

[tidal turret]

I think it can be expressed in a better way dude

,, x \mathcal{R} y \iff x^2 - y^2 = 93x - 93y \ x \mathcal{R} y \iff \begin{cases} x = y \ \text{or} \ x + y = 93 \end{cases}

[grand pond]

Renato

[next bramble]

ye, that's a good way to put it

[tidal turret]

if you re define it like this, then symmetry is two cases, when x = y which is already proved by reflexiveness and the other case which we care of

[next bramble]

transitivity should also pop out from that

[tidal turret]

x R y <=> (x = y) or (x + y = 93)
Symmetry proof:
there's 2 cases, when x = y this is proved by reflexivity
second case is when x R y which is x + y = 93 and because sum is commutative
y + x = 93 and thus,we conclude y R x

[tidal turret]

∀x,y,z∈A ,xRy, yRz => xRz

mmm

[torpid garnet]

Bonita relación de equivalencia.

[tidal turret]

Transitivity proof:
Assume: x,y,z ∈ A, such that
xRy and yRz, thus
[(x=y) or (x+y=93)] and [(y=z) or (y+z=93)]
the cases where x=y, y=z have been proved by reflexivity, so it can be simplified to proving this: (x+y=93) and (y+z=93)
both need to happen simultaneously, from subtracting both you get
x - z = 0 => x = z, since we already proved reflexivity, we conclude xRz

[next bramble]

hint: if x=y, x=z, or y=z, this is just reflexivity.

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

it's not antisymmetric

antisymmetry: xRy, yRx => x = y

counterexample, 80 R 13

xRy <=> (x=y) or (x+y = 93)

so, 80 + 13 = 93 => 80 R 13

and . . . 13 + 80 = 93 => 13 R 80

but 13 ≠ 80

for ii) I am thinking 🤔

[tidal turret]

thanks dude, I appreciate it

we need to find the equivalence class for every element in A, mmm

so for example

[1] = {y ∈ A | y ~ 1} ⊂ A

where A = {n ∈ N | n <= 92}

[1] = {y ∈ N | y <= 92 and y ~ 1} ⊂ {n ∈ N | n <= 92}

y ~ 1 => (y = 1) or (y + 1 = 93)

thus, [1] = {1, 92}

now for 2, the equivalence class of 2 is

[2] = {y ∈ N | y <= 92 and y ~ 2} ⊂ {n ∈ N | n <= 92}

y ~ 2 <=> (y=2) or (y + 2 = 93)

so [2] = {2, 91}

[solid iris]

yes

[tidal turret]

how to find them all?

[solid iris]

do you see a pattern? try to find [n] for arbitrary n

[tidal turret]

from math, we know the equivalence class of a representative of a class is equal to the equivalence class of another representative from the same class

what I mean is,
if [♡] = {♡,☆} then [☆] = {♡,☆}, we conclude [♡]=[☆]={♡,☆}

[solid iris]

based on prior work what do you think [3] is?

[tidal turret]

easy, [3] = {90,3}

[solid iris]

what about [n] for arbitrary n?

[tidal turret]

assuming n ∈ N, and n <= 92

y ~ n <=> (y=n) or (n + y = 93)

so [n] = {n, 93-n}

[solid iris]

all correct!

[tidal turret]

and from what I said above, the representative of an equivalence class has same equivalence class as the other representatives

meaning [n] = [93-n] = {n, 93-n}

[solid iris]

what are enough representatives to describe all classes?

[tidal turret]

idk

[fathom onyx]

Well, suppose a and b are distinct numbers

How would you know if [a] = [b]?

[solid iris]

start listing representatives at 1,2,…

when do we start repeating classes?

[tidal turret]

idk?

[solid iris]

recall [n]={n,93-n}

[tidal turret]

a drawing would be appreciated

[fathom onyx]

You really don't need to draw anything here tbh...

[solid iris]

im eating lunch at a cafe so no drawing from me

[fathom onyx]

If we take [1]

What other class is that equal to?

[tidal turret]

[1]=[92]={1,92}

[fathom onyx]

Class?

yh there we go

Similarly for [2]?

[tidal turret]

[2]=[91]={2,91}

[fathom onyx]

Right

We can keep going in this fashion, easily

Up to what class?

[tidal turret]

I dont follow

[fathom onyx]

Okay, then I'll ask the same question for [3]

What other class is that equal to?

[tidal turret]

one goes up the other goes down

[3]=[90]={3,90}

[solid iris]

the ideal is to write all classes without redundancy

[fathom onyx]

Focus on the question I've just asked

[tidal turret]

equivalence class of 90

[fathom onyx]

Right]

[solid iris]

when listing representatives we should ignore those that give redundant classes

[tidal turret]

one goes up the other goes down

[solid iris]

if we have 1 then we exclude 92

[tidal turret]

this pattern, up until what number?

[fathom onyx]

So we can keep going

[solid iris]

you should figure that number out

[fathom onyx]

[1] = [92]
[2] = [91]
[3] = [90]
...
What's the last equivalence class?

[solid iris]

but thats the right question

[tidal turret]

dude this is so hard

[tidal turret]

last equivalence class is [92]=[1] but that's already included so since it's repeated we don't count it

this is a combinatorics question dude

can somebody help me

maybe doing a drawing can help

[solid iris]

maybe itll be easier if you think this way

list all representatives then remove redundant ones

[tidal turret]

yes

something like this dude?

[45] = [48] = {45,48}

[46] = [47] = {46,47}

so the number of equivalence classes is 1 to 46

[solid iris]

yes

[tidal turret]

because then [47] = [46]

[solid iris]

the classes without redundancy are [1],…,[46]

[fathom onyx]

So how many are there?

[tidal turret]

yes, but how many classes are those?

[fathom onyx]

Brother....

[tidal turret]

(46+1)×47/2

[fathom onyx]

Those ARE the classes

Suppose you had a school, with classes numbered 1 to 46

How many classes does the school have?

[tidal turret]

46 classes dude

[fathom onyx]

There we go

[fathom onyx]

What is this for, then?

[tidal turret]

I thought 1 doesn't count

[fathom onyx]

Why?

[tidal turret]

or that we were overcounting

[solid iris]

the number of classes is the number of representatives without redundancy

thats as simple as it gets

[tidal turret]

yes, is tricky dude

[solid iris]

youll get the hang of it

takes time just like anything else 🙂

[tidal turret]

the diagram helped a ton dude

im bad with mental math

[solid iris]

you said dude a million times in the last week

its so american i love it

[tidal turret]

I appreciate the help fellas 🙏

🇺🇸

[solid iris]

np

[tidal turret]

.solved

[midnight plank]

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[lethal owl]

for an infinite sequence of the digits of the positive integers that goes like this:
1234567891011121314151617181920.....
I am given an integer K, such that I take the first K digits of the sequence and discard the rest of the sequence
Is there a way to calculate the sum of the K digits without brute force.
I successfully found the sequence of the sum of the first K digits for that sequence on OEIS but it didn't contain any formula that I may use to find its value. Here it is https://oeis.org/A054632

[midnight plank]

@lethal owl Has your question been resolved?

[runic hamlet]

well there are of course much better ways than just straight up bruteforce

but still not a nice closed formula

[midnight plank]

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[vocal raft]

If you see the series as
00 01 02 03.. 09 10 11 12 ...

then sum of first k digits is given by

[vocal raft]

https://py3.codeskulptor.org/#user310_oIXwFakrOz_4.py
you can see it works

You can also find a formula for the exact series but I changed the starting to make it look clean

[zealous venture]

Very off topic, but pfp source?

[vocal raft]

Xie Yan from Evil God manhua

[stray sapphire]

how to deal with trignometeric function of jee

[jaunty ivy]

!occupied

[midnight plank]

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).

[vocal raft]

.close

[midnight plank]

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[woeful turret]

if only there was a single answer lol

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[chrome swift]

Bob and Sasha are brother and sister. Bob has as many brothers as sisters, and Sasha has twice as many brothers as sisters. How many boys and girls are there in this family?

[chrome swift]

how would u set this as a matrix

i got the answer but i want to see the math way of doing it instead of just picturing it in my head

[lyric charm]

let b be the number of boys including Bob, and g be the number of girls including Sasha.

then there are b-1 boys besides Bob and g-1 girls besides Sasha.

Bob's brothers = ?
Bob's sisters = ?
Sasha's brothers = ?
Sasha's sisters = ?

[lyric charm]

write these down in terms of the variables b and g, then form two equations

[midnight plank]

@chrome swift Has your question been resolved?

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[chrome swift]

preciate it

4 and 3

thanks

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[twilit field]

I'm trying to provide a combinatoral argument for $\binom{n}{r}= \binom{n}{r,n-r}$

[grand pond]

wai

[twilit field]

I was thinking I could just say to select r objects from n objects we can either select r people from n people or divide n people into two groups, r of whom we select, n-r which we don't

does that work

[lavish venture]

yea

[twilit field]

Thanks people

.close

[midnight plank]

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[lucid delta]

how do i plot a quadratic graph on desmos when i'm only given the solutions?

[lucid delta]

cause this is just giving me y=0

this is the question

[scenic wyvern]

form the factors

then multiply them back out (or not)

[lucid delta]

i feel like thatd take too much time given that a is an integer greater than 1

so it wont just be (x-7)(x+3)

[scenic wyvern]

hm

[lyric charm]

it'll be a times that

[lucid delta]

got it

ty all

.close

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[rugged thunder]

Help please

[rugged thunder]

How do I know

[lyric charm]

what's the derivative of a constant

[rugged thunder]

0

[dawn dagger]

Also notice each vector has the form (x,y)' = (1,y')

[rugged thunder]

I don’t understand

[dawn dagger]

The arrows of your slope field are vectors

You have 4 possible solutions which are all constant. Your job is to figure at which value of y (see the vertical axis) do you have the vector (1,0) for all x

[rugged thunder]

Um

[rugged thunder]

We not taught that we are taught to draw the possible slopes following the lines and guess y (x) so I don’t know what this means

[rugged thunder]

So it’s -1 and 2??

[lyric charm]

yes exactly

those are the only places where the slope field is horizontal

[rugged thunder]

Oh ok and since 2 is not option it’s -1

[lyric charm]

yes

[rugged thunder]

Ok thank you

[midnight plank]

@rugged thunder Has your question been resolved?

[midnight plank]

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[twilit field]

There is a one-to-one correspondence between any two right cosets of H in G.

[twilit field]

I don't get this, by definition isn't the coset for a given subgroup unique?

$Ha ={ha \mid h\in H}; a \in G$

[grand pond]

wai

[twilit field]

oh, $a$ varies

[grand pond]

wai

[twilit field]

.close

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[chrome swift]

hey

[chrome swift]

i dont get it how am i still wrong

pls just point out my mistake cuz im lost

<@&286206848099549185>

[scenic wyvern]

row 2's leading entry is a 4, not a 1
leading entry in row 2 is further right than that in row 3
column 4 has a confusing pivot situation - if it has a pivot, then the rest of the column isn't 0
two different rows have the same column as their leading entries (2 and 4)

[midnight plank]

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[chrome swift]

.reopen

.open

[chrome swift]

wait

[chrome swift]

im sorry i still don't get what u said

[hushed mauve]

list down the conditions for RREF

[chrome swift]

it's an augmented matrix btw

[chrome swift]

ok

1.) the bottom row should be all zeros (if it can)
2.) there should be a staircase pattern going southeast for every leading entry
3.) everything below a leading entry should be 0
RREF:
4.) all leading entries are 1
5.) column with a leading entry should only have the non-zero leading entry (1) and everything else in the column is 0

lmk if im wrong

[scenic wyvern]

oh-kay

so row 2 broke rule 4

row 3 broke rule 3

column 4 broke rule 5

[chrome swift]

no it's an augmented matrix

[scenic wyvern]

and you forgot one extra rule - each column can only have one leading entry or pivot

[chrome swift]

that's rule 2

[scenic wyvern]

i'm pretty sure the same rules apply to an augmented matrix. and btw, the word augmented is in brackets

[chrome swift]

ah seriouesly?

[scenic wyvern]

in that case, rows 2 and 4 together break rule 2

[chrome swift]

ok that might help

[midnight plover]

[chrome swift]

bruh

ok

this should be good right?

naw nvm

why is this kinda tricky

[chrome swift]

but im not tryna have a pivot on every row

or column

[scenic wyvern]

you have the wonderful option of zeroing out one row at the bottom and then you now have 3 rows and 4 columns to work with

so definitely one of the columns won't have a pivot

[midnight plank]

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[vast loom]

Whats your progress so far?

[tidal turret]

dont know how to start dude

[vast loom]

Well, do you have the definition of a function?

[tidal turret]

@vast loom

[vast loom]

ok so for the first one, can you find anything wrong with it?

or not

maybe its correct

What you know is that a function gives each starting value a unique outcome

maybe the function doesnt obey that?

[tidal turret]

1 R a and 2 R a

this is fake asfk

no?

[vast loom]

no

thats not the problem

[tidal turret]

ok jaja

[vast loom]

the starting values are 1,2,3,4,5

here you have 1 -> a and 2->a which isnt a problem
You would have a problem if lets say for example you have:
4 -> b And 4 -> c

because then 4 would be given two different values which you dont want

[tidal turret]

3 R a, 3 R d

[vast loom]

(it isnt problematic that two different starting values go to the same thing)

[vast loom]

yes thats an issue

so the first isnt a function

[tidal turret]

f(3)=a=d

absurd becase its not a = d

definitely not a function

[vast loom]

yes

now second one

can you find anything wrong? or not

[tidal turret]

its not a function

[vast loom]

why?

[tidal turret]

5 isnt mapped to anything

[vast loom]

yes correct

now onto 3

[tidal turret]

1 R a and 2 R a implies 1 = 2

[vast loom]

no

its the same as before you can have two things that are mapped to the same thing

what you dont want is one thing to be mapped to two different things

[tidal turret]

true!! i keep forgetting hahaha

[vast loom]

You can for example think of the function for all x in N, f(x) = 0

everything is mapped to 0, but thats not a problem

so whats your verdict for the 3rd one then?

[tidal turret]

its a function

[vast loom]

yep

everything is mapped to something, and that thing is unique

now for the 4th

[tidal turret]

oof dude

[vast loom]

lol

take your time its ok

[tidal turret]

it is quite a big jump in difficulty

[vast loom]

it may seem like it, but if you go slowly to verify what you want it should be ok

[tidal turret]

tf you mean?

[vast loom]

so first of all, is everything in A mapped to something?

[tidal turret]

3 = 2b - a

[vast loom]

sure

[tidal turret]

dude yes

3 + a = 2b

[torpid garnet]

Renato si quieres ayuda en español, me dices.

[tidal turret]

i prefer to keep it in english so thr progress is not lost ykwim

[tidal turret]

R includes Q, so I think yes

is bad justification but i am convinced

[vast loom]

hm

sure

so everything is mapped to something

now is that thing unique?

[tidal turret]

1 = 2b - 3

b = 2

[vast loom]

sure

so (1,2) is in R

[tidal turret]

it is unique

[vast loom]

why is it?

[tidal turret]

2 = 2b - 3

b = 5/2

[vast loom]

yes

basically, when you fix your a, you get b

if you know what a is, then you can find b and he is unique

so that's why it works

[tidal turret]

theres only one b for each a

[tidal turret]

yesss

[vast loom]

ok good, so it is a function

now onto 5

[tidal turret]

-1/2 = 2b-3

dude this one aswell

[vast loom]

?

start with the two questions

is everything mapped to something?

[tidal turret]

its not a function

JHAJAJAJA

[vast loom]

why so?

[tidal turret]

JAJAJAJAJAJA

[vast loom]

lol

[tidal turret]

dude

[vast loom]

so funny huh

[tidal turret]

0 R nothing

[vast loom]

huh?

[tidal turret]

a = 2b - 3

[vast loom]

yes

[tidal turret]

right?

[tidal turret]

a in R, b in N

[vast loom]

indeed

[tidal turret]

right?

a = 2b - 3

let a = 0 in R

0 = 2b - 3

[vast loom]

oh thats what you were saying

[tidal turret]

JAJAJAJAJAJ

[vast loom]

yup

[tidal turret]

https://tenor.com/view/yesyesyesyesyes-jojo-gif-8822335

[vast loom]

they all are mapped to something, but that something isnt always in N, so its cooked essentially

well done

now for the last one

[tidal turret]

yes it got cooked

with our counterexample

this exercises are good jajajjaj

[vast loom]

you see its not that complicated if you go slowly :)

[tidal turret]

a+b|5

[vast loom]

you are able to do it

is it a+b | 5 or 5 | a + b?

im not that fluent in spanish, but i think its the second one no?

[tidal turret]

latter

[vast loom]

ok that's what i though

so 5 | a + b

any ideas?

[tidal turret]

yes

(a,b)=(3,3)

(3,3) in ZxZ

6 is not divisible by 5

[vast loom]

sure

but that doesnt give anything does it?

[tidal turret]

wdym

[vast loom]

well for example look at the 4th one, (0,0) in NxR but 0 isnt equal to -3

yet we said its a function

(which it is)

do you understand?

[tidal turret]

im stupidddd

[vast loom]

what you said here is that (3,3) isnt in R that defines our function, but we dont really care about that

[vast loom]

nah you're good

happens

so any other ideas maybe?

we want to see if everything is mapped to something, and if that something is unique

[tidal turret]

this is not a function

[vast loom]

and why is that?

[tidal turret]

wait

10 is divisible by 5?

yes

[vast loom]

yes

5, 10, 15, 20, 25, ... are divisible by 5

[tidal turret]

10 R 0, 10 R 10 => 0 = 10, absurd

[vast loom]

yes bro

10 is mapped to more than one thing

(so is every number actually)

[tidal turret]

https://tenor.com/view/yes-yes-yes-yeah-yep-yas-gif-15410358

[tidal turret]

dude, it was hard or no?

[vast loom]

what do you mean? what was hard?

[tidal turret]

exercise

[vast loom]

depends which level you are

for me it wasnt hard no, but if you asked me 10years ago, the answer wouldnt be the same

[tidal turret]

i appreciate it dude

[vast loom]

my pleasure broski

[tidal turret]

.solved

[midnight plank]

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[ebon spear]

I want to learn proofs without reading a textbook is that possible

[midnight plank]

Please don't occupy multiple help channels.

[fallow scarab]

.close

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[gritty holly]

.close

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[lyric charm]

"a intersection b complement" is ambiguous

bc you chose to put it into words it's impossible to tell if you meant $A\cap\overline{B}$ or $\overline{A\cap B}$ (also uppercase A and B not lowercase)

[grand pond]

Ann

[lyric charm]

anyway not sure where your confusion is with the pink highlighted stuff

[midnight plank]

@lunar ridge Has your question been resolved?

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[oblique heath]

is ln5?

[jaunty ivy]

what is the question

that is just some expression

[oblique heath]

solution of the ques

[dreamy lichen]

are u asking why the limit of that as x->0 is ln 5?

[oblique heath]

no

wait

[oblique heath]

is this equal to ln 5

when lim x-->0

[dreamy lichen]

No, but the limit as x -> 0 of that is equal to ln5

[dreamy lichen]

then yes

[oblique heath]

oh

ok

thanks

.close

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[molten bay]

What is this solenoidal?

[lyric charm]

https://en.wikipedia.org/wiki/Solenoidal_vector_field maybe this?

and they meant "solenoidal vector field" but somebody swallowed the word "field"

[molten bay]

But what is that?

Have you ever used it?

Of you have read vector calculus?

[lyric charm]

... the wikipedia article lists a definition in the very very beginning

a vector field is solenoidal iff its divergence is 0

[molten bay]

What is n?

In the given question

.close

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[ruby dust]

here, we need force eqns varying with x

hence integrals must be applied

where

[ruby dust]

the bracket shows cross products

[lyric charm]

last time you posted this question here i tried to ask you where the integral came up, and you instead decided to run for the hills and close the channel rather than explain your reasoning

[ruby dust]

hmm

now comprehend it

[lyric charm]

oh yeah, it was about 5 hours ago that happened

[ruby dust]

my class was also running simultnaeously so yea

[lyric charm]

gonna ask for explanation here on what $d\vec{l}$ means and where this formula comes from. if it has a name, i would like to hear that as well so i can look it up myself.

[grand pond]

Ann

[lyric charm]

also once again, how and where EXACTLY does the integral emerge?

[midnight plank]

@ruby dust Has your question been resolved?

[dusk furnace]

No, because even if there is a force field, nothing is moving along any path in the force field. Integration would give you total work if something did move, but it doesn't. As stated in the problem they're simply measuring the force at each point.
In order to find the force on the paramagnetic ball from the expression for B, I believer you'll need to differentiate the flux density.

[ruby dust]

elaborate

[dusk furnace]

Not unless you can ask a more specific question -- or perhaps answer Ann's clarifying question about what you want to integrate and why.

[midnight plank]

@ruby dust Has your question been resolved?

[ruby dust]

ahem.

will try to do it myself @dusk furnace @lyric charm

will report when i face a issue

.close

[midnight plank]

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[wind oxide]

<@&268886789983436800>

.

[ruby dust]

??

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[jade magnet]

3.i.c

[jade magnet]

for c i get a not nice quartic

and idk how im gonna simplify it

i get

(k - (36/(k-36/(k-36/(k-36/2)))) = 2
(k - (36/(k-36/(k-36/(k-18)))) = 2
(k - (36/(k-36/(k-36/(k-18)))) = 2
k(k-18) - (36/(k-36/(k-36))) = 2k - 36
k(k-18)(k-36) - (36/(k-36)) = (2k-36)(k-36)
k(k-18)(k-36)(k-36) - (36) = (2k-36)(k-36)(k-36)
k(k-18)(k-36)(k-36) - (36) = (2k-36)(k-36)(k-36)
k(k-18)(k-36)(k-36) - (36) = (2k-36)(k-36)(k-36)

[lyric charm]

did you solve the cases of periods 1 and 2

[jade magnet]

yeah

those were easy

[lyric charm]

these will also appear as roots of this quartic

[jade magnet]

oh yeah

true

[lyric charm]

you can put it in std form and factor them out

should be manageable from there

[jade magnet]

wdym

as in (k-36)^2 instead of (k-36)(k-36)

that sort of thing?

[jade magnet]

even then idk what to do with the extra 36

[lyric charm]

i mean expanded form

[midnight plank]

@jade magnet Has your question been resolved?

[jade magnet]

could you show what you mean

[lyric charm]

ughhhh it's going to be a bit unpleasant i think

gimme a few

[jade magnet]

yea

[lyric charm]

can you tell me what the k values were for constant and period 2 btw

[midnight plank]

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[jade magnet]

for the constant it was 20 iirc

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[lyric charm]

welp now we have to reopen it in my name ig

[jade magnet]

yeah why does this keep timing out

[lyric charm]

you didn't click on bot's prompt

[lyric charm]

right here

anyway this is what i came up with

this factors quite nicely

[midnight plover]

eh thought u need help for the first time

got exited for nothing

[lyric charm]

nope not this time

waiting on @jade magnet to confirm or ask

[jade magnet]

yeah nope

[jade magnet]

oh ok

so k = 0 is obvs a solution then

[lyric charm]

yes

as is the k=20 you mentioned

0 should have come up for period 2 btw

[jade magnet]

wait did it?

oh yeah

true

idk why i got 30 something first time

[lyric charm]

@jade magnet anything else or can i close this now

aight fuck it im closing. @jade magnet when you come back you can ping me in like #precalculus or sth about this if you still need me to yap

[jade magnet]

nope

[lyric charm]

.close

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[vast loom]

are you stuck on a or b?

[midnight plank]

@wanton grove Has your question been resolved?

[vast loom]

for a it is the same k

[midnight plank]

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[viral dagger]

whats the motivation of inducting on 2^m?

[hard shard]

what is the step before the win state?

personally, i would put it in binary and that almost trivializes the problem

[viral dagger]

ohh

[viral dagger]

wait binary?

[hard shard]

yes, that or consider v2(p) and v2(q) and the v2 of the resulting piles

[fathom onyx]

[Another hint for the motivation might have been that 64 = 2^6]

[Not a definitive hint, it doesn't really explain why you should think that to be helpful, but it gives you a road you can try driving into]

[hard shard]

oh we're looking for motivation

i kinda sidetracked myself to solving the problem whoops

[fathom onyx]

I mean the solution is there lol

[hard shard]

yeah

[viral dagger]

how would you even do this in binary

[fathom onyx]

Though putting it into binary, if you can do it properly, is an interesting approach

Well, if you add two numbers in binary, say P + Q

Wherever there's a 0 in Q, the total in that column's just whatever's in P

If there's a 1 in Q, then the total's gonna just be 1 if the column in P is 0, or 10 (where the 1 is pushed left) if the column in P is 1

[Honestly it's a messy explanation even if trivial]

oh no wait, one pile becomes P-Q and the other is 2Q

2Q is simple - you just add a zero to the right of whatever v2(Q) is

For P-Q, ... er...

well shiet idk how to explain that one chief

[viral dagger]

isnt vp(Q) a function for the power of the prime p that still divides Q?

[mortal falcon]

ideally you want each move to be somewhat “constructive”, so hopefully P-Q also increases the v2

[fathom onyx]

(Honestly I've never seen this notation, I thought it was the binary rep. of Q)

[mortal falcon]

that motivates how you structure your moves

[viral dagger]

ooh wait

[mortal falcon]

If you can keep increasing the v2 of every pile, you have to eventually get 2^6

[fathom onyx]

(doesn't this decrease the v2?)

[mortal falcon]

er it shouldn’t if you make good choices

e.g if we do 3 and 7, we get 6 and 4, increasing the v2 of both

[viral dagger]

is it like there are an even number of the things that have the last digit of 1 in the binary rep of the piles, so you can do the move on all of them and v2 of P-Q will increase, so every term is divisible by 2

[mortal falcon]

you keep identifying piles that share their “lowest active bit”, and then your transfer eliminates that bit in both

[viral dagger]

after doing it theres also an even number of things with the second last digit of 1 and you can do them, so every term is divisible by 4

and you can repeat this?

[mortal falcon]

Yes exactly and this is precisely because we know the piles sum to a power of 2

[viral dagger]

ohhhh

[viral dagger]

actually this is like really simmilar to the solution here, this used induction and we used an algorithm of sort?

[mortal falcon]

(Also just to clarify, vp(n) is just the highest exponent x such that p^x divides n, it’s just the power in the prime factorization)

yeah any (good) inductive problem is really just an algorithm in disguise

[fathom onyx]

Which equals the number of digits in base p, minus 1?

[mortal falcon]

moreso the position of the last nonzero digit

[fathom onyx]

oh wait divides

Not is less than

mb

[viral dagger]

12's binary representation is 1100, so v_2(12)=2

[fathom onyx]

yee

[mortal falcon]

vp is just a funny way of saying “we’re a multiple of this power now yay”

[viral dagger]

ok guys thank you very much!

.solved

[midnight plank]

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[polar pulsar]

i dont get how they got thisw

how is it 40x

[pearl hull]

Can you provide the uncut question?

[gaunt nimbus]

a^2 + 2ab + b^2

I assume they are expanding the first bracket rn

Cuz 2(4)(5) =40 :/

[fathom onyx]

yep

[polar pulsar]

so its a perfect square?

[fathom onyx]

it's just from this

= (5x + 4)(5x + 4)
= 5x(5x+4) + 4(5x + 4)

You should probably be able to see the rest of that expansion

[polar pulsar]

i didnt think you did all that. i thought it was 5x^2 and 4^2 and youre done

[fathom onyx]

In general it is NOT the case that $(a+b)^2 = a^2 + b^2$.

[grand pond]

Waes (Wires)

[lusty furnace]

(5x+4)² = 25x² + 2*5*4x +16 -15x - 12 +1

2*5*4x = 40x, so I suppose you paused the video before they had time to put down the -15x.

[fathom onyx]

You should be familiar with this being the expansion

[lusty furnace]

$(ax +b)^2 = a^2x^2 + 2abx + b^2$

[grand pond]

Médicis

[polar pulsar]

when is it the case

[fathom onyx]

(lol get š'd)

When a = b = 0

[lusty furnace]

in $\mathbb{R}$, only when $a = 0$ or $b=0$.

[grand pond]

Médicis

[fathom onyx]

My point is, this isn't a valid expansion of the brackets

[fathom onyx]

Break it up like so, and you're far less likely to make that mistake

[midnight plank]

@polar pulsar Has your question been resolved?

[midnight plank]

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[molten bay]

Limit n tends to 0

e^(1/(nlogn))

[molten bay]

limit ( x ) tends to 0 ( e^{\frac{1}{x \log x}} )

[grand pond]

Andy

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

<@&286206848099549185>

[midnight plank]

@molten bay Has your question been resolved?

[mortal falcon]

You can take the limit of xlogx separately and then plug it in

[quiet pawn]

its 0 from e^(-\infty)

[molten bay]

,w e^(-1/infinity)

[grand pond]

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[molten bay]

@quiet pawn

[quiet pawn]

…e^-\infty

not -1/\infty

[molten bay]

Write it properly

What is this?

,w e^(1/-infinity)

[grand pond]

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[quiet pawn]

bro

oh my god

$e^{-\infty}$

[grand pond]

moon

[mortal falcon]

wait but Andy do you get how to arrive at that

[molten bay]

No

,w limit x tends to 0 1/(xlogx)

[grand pond]

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[molten bay]

@mortal falcon @quiet pawn

[quiet pawn]

what

how did you take the 1/x out

[vocal moon]

The limit should be 0+

Otherwise you're cooked

[quiet pawn]

thats not even the point

[vocal moon]

No I'm just saying

Because it'll be soon needed

[quiet pawn]

1/(xlogx) ≠ 1/x * logx

??????

[vocal moon]

Yeah 😭

[quiet pawn]

how bro felt after pinging us and disappearing

https://tenor.com/view/entourage-entourage-phone-intergalactic-quality-intergalactic-angry-gif-3361600449112277023

[vocal moon]

I'm talm about this, if it's 0+, then the answer is defined

[vocal moon]

Rs

[molten bay]

I see

,w limit x tends to 0 1/(xlogx)

By l'hopital i got -infinity

.close

[midnight plank]

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[midnight plank]

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[lyric charm]

!status

[midnight plank]

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

[oblique heath]

idk what to do after this

[oblique heath]

2

[lyric charm]

f(1) is just 3

you can write your limit as $\exp\paren{\frac13 \lim_{x \to 0} \frac{f(1+x)-f(1)}{x}}$

[grand pond]

Ann

[oblique heath]

hmmmm

[lyric charm]

this form should be familiar

[oblique heath]

legrange's theorem

[lyric charm]

no need for any fancy theorems!

$\lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$ is?

[grand pond]

Ann

[oblique heath]

f'(1)

ahh

now i just have to put the values

thanks!

.clsoe

.close

[midnight plank]

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[twilit field]

A woman has $n$ keys, of which one will open her door
\
If she tries the keys at random discarding those that don't work that's the probability that she will open the door on her kth try

[grand pond]

wai

[lyric charm]

do you mean on or before the k'th try, or do you mean exactly on the k'th try

[twilit field]

exactly

[lyric charm]

so her first k-1 keys need to be wrong and the k'th needs to be right

[twilit field]

yes

[lyric charm]

!status

[midnight plank]

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

[twilit field]

2

I know the size of the sample space is k!

[lyric charm]

mm

using "sample space" to mean "size of the sample space" is a bit eh

[scenic wyvern]

idk if i can give a hint without spoiling the type of distribution this sounds like

[lyric charm]

you dont need to think about distributions here

op is doing extremely basic probability with explicit counting of favorable vs total outcomes

[scenic wyvern]

i see, fair

[lyric charm]

also the total number of permutations of the keys is n! and not k!

[twilit field]

Actually the size of the sample space might be npk and the number of favourible ones would be (k-1)!

[lyric charm]

actually the word "sample space" might, once again, refer to a SET and not to a NUMBER?

[twilit field]

Fixed

[lyric charm]

what are you considering to be a 'sample' here then

if you're claiming that their count is P(n,k)

[twilit field]

I'm thinking the sample space includes the sets of ordered permutations of the k keys

[lyric charm]

there are n keys not k keys

[twilit field]

But we need to select k-1 keys to try

the last key being the one that works