help-4

[analog schooner]

Its fine

I can read it

Without spaces

$\epsilon_n \to 0$

[rocky lotus]

Dhruv

[analog schooner]

Yes

Wait you're right

Its not 0 × infinity in the first place

I made a mistake

Thank you

Both of you

Yeah

Thank you

I thought it was 0 * infinity instead of 0 ^ inf

That's why I thought it couldn't be ignored

👍

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[river shale]

@versed quarry post your problem here

Nvm, you claimed one

.closed

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[river shale]

No, you’ve already claimed a channel

I don’t have one

I claimed this for you

Nah, it’s alr

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[icy sigil]

I am trying to make a vector that points to the direction that the point at x=1 is moving to, but it seems very off, could someone help me?

[icy sigil]

i need it for its endpoint to be $(x,i\cdot y, i\cdot z)$ where y and z are coordinates of the beginning of the vector

[rocky lotus]

maxtsg

[icy sigil]

and in this case x is just 1

since desmos doesnt accept imaginary vectors i just swapped the coordinates and flipped a sign since a+bi * i = -b+ai

but it didnt work i think

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[lavish tinsel]

So the test says it should be impossible, but I think x²=3-5/3x⁴-2x⁵
x³=3+2x¹-2x⁵
Am I dumb or is that the real answer?

[charred burrow]

if you have five variables and only four equations, you can't have unique solutions for the system

i.e. there are infinitely many solutions

[lavish tinsel]

Its general solution to the basis to the x² and x³

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[cedar spear]

Hi, I neeed someone to explain me this.

i'm learning riemann sum and a teacher wrote this

If f:[a,b] -> R and P, Q partitions of [a,b], and P ⊆ Q, then

S(inferior)(f,P)=<S(inferior)*(f,Q)=<S(superior)(f, Q)=<S(superior)(f,P)

and he said it was bc if A ⊆ B, then inf(B) =< inf(A) =< Sup(A) =<Sup(B)

it doesnt get on my head bc i think the signs (=<, =>) should be inverted, and I need someone to explain me why does this work this way

Thanks

[ruby sleet]

if A ⊆ B, then elements of A cannot go below more than elements of B could, because any place A reaches, B will reach too

so B is the only one that can potentially reach further down

similarly, no matter how far up A reaches, B will reach too

so B is the only set that can potentially reach further up

as a small example in terms of intervals

A = [-1,1]

B = [-2,2]

A ⊆ B

A is a box contained in B, a bigger box

the bigger box goes further down and further up

no matter how big the small box is, the bigger box must, by definition, be bigger

[cedar spear]

but then what my teacher wrote shouldnt be

s(inf)(f,Q) =<s(inf)(f, P)=<s(superior)(f,P)<=s(superior)(f,Q) ?

[ruby sleet]

oh wait actually

It does work the way it's originally written

when you have a partition made up of more subsets

then each subset is individually smaller

that's how you have to interpret it

if P ⊆ Q, then take one subset in the partition P, call it B

then it's either gonna appear as is in the partition Q

or it's gonna be divided into smaller sets, call them A1,A2,...,An

the part of the inferior and superior sums for P that is linked to B will be length(B)*inf(B) and length(B)*sup(B)

whereas the terms that correspond in the partition Q will be length(A1) * ... + length(A2) * ... + ...

because inf(B) <= inf(A1), inf(B) <= inf(A2), ...

we will have length(A1) * inf(A1) + length(A2) * inf(A2) + ... >= [length(A1) + length(A2) + ...]* inf(B) = length(B) * inf(B)

same thing for superior sums

[cedar spear]

i still dont get it how does it apply for this

[ruby sleet]

I'll try again

[cedar spear]

thanks

[ruby sleet]

btw are partitions for you any kind of sets or are they just points that separate intervals?

[cedar spear]

sets i think

[ruby sleet]

ok

what it means for $P\subseteq Q$ in terms of partitions:

If $B\in P$, then you can find $A_1,A_2,...,A_n\in Q$ such that $B = \bigcup_{k=1}^n A_k$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

I just wrote what it means mathematically

if B is a set in the partition P

[cedar spear]

uhm sorry is that U a summatory?

[ruby sleet]

it's a union

B is exactly the union of some subsets in the bigger partition Q

Q is a "finer" partition than P

[cedar spear]

okay

[ruby sleet]

makes sense right? It should be your definition

[cedar spear]

yeah, never read it that formal but it makes sense

[ruby sleet]

ok

let's focus on the inferior sums for now, superior sums will be analogous afterwards

what's your notation for "length" or volume of a set?

so I can be coherent with what you have

[cedar spear]

how many elements it has in

like A = {1,2,3}, its lenght its 3

[ruby sleet]

uh no, like [a,b] has length b-a

[cedar spear]

oh alright

[ruby sleet]

what's the symbol for that length

[cedar spear]

maybe i got confused cause english isnot my first language

[ruby sleet]

that's cardinality btw

[cedar spear]

and im not usually talking abt math in english so im still not used to this

[ruby sleet]

what symbol do you use for this?

[cedar spear]

i dont think ive ever used a sign for that

[ruby sleet]

really? how did you define inferior and superior sums then xdd

[cedar spear]

wait let me see if i can send you a pic

[ruby sleet]

xddd

ok, but have you ever seen mathematical definition of those

like S*(f,P) = ...

anyways

I'll write $\lambda(C)$ the 'length' of set $C$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

for example $\lambda([a,b]) = b-a$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

are you ok with that notation?

[cedar spear]

on the superior sum is the same as inferior but instead of inf it says sup

[cedar spear]

yes

[ruby sleet]

yes, but here I see this is the particular case of where partitions are intervals

[ruby sleet]

I asked if this was the case

[cedar spear]

i thought sets were intervals

[ruby sleet]

nevertheless, we can continue with the general case

[ruby sleet]

intervals are sets

not all sets are intervals

[cedar spear]

ohh i see

sorry abt that

i will get better at math vocabulaty on english

[ruby sleet]

now, for any partition $R = {x_0,...,x_n}$ of $[a,b]$, the definition of the inferior sum of $f$ on $R$ is $S_{inf}(f,R) = \sum_{k=0}^{n-1}(x_{k+1}-x_k)\inf_{[x_k,x_{k+1}]}(f)$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

good?

[cedar spear]

good

[ruby sleet]

so now

take $P\subseteq Q$ partitions of $[a,b]$.

[rocky lotus]

Raphaelisius Maximus MMIII

[cedar spear]

yeah

[ruby sleet]

so for example, if we take x_0, x_1 in P

that might become $x_0,y_0,y_1,...,y_m,x_1$ in $Q$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

so I can divide $B = [x_k,x_{k+1}]$ into $A_1,A_2,...,A_m$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

so while in $S_{inf}(f,P)$ we'll find the term $(x_{k+1}-x_k)\inf_{[x_k,x_{k+1}]}(f)$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

in $S_{inf}(f,Q)$ it will be replaced by the terms $Length(A_1)\inf_{A_1}(f)+Length(A_2)\inf_{A_2}(f) + ...$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

does that make sense?

[cedar spear]

some

[ruby sleet]

ok

we can go back to this if it's still unclear

now, we'll look at those two facts

Fact n°1: $Length(A_1) + Length(A_2) + ... = Length(B) = x_{k+1}-x_k$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

that's just stringing up every "small" subset back together, it makes the big set again

Fact n°2: since $B$ is a bigger set than $A_1,A_2,...$, then $\inf_B (f) \leq \inf_{A_1}(f)$, as well as $\inf_B (f) \leq \inf_{A_2}(f)$, etc.

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

so

$Length(A_1)\inf_{A_1}(f)+Length(A_2)\inf_{A_2}(f) + ... \geq Length(A_1)\inf_{B}(f)+Length(A_2)\inf_{B}(f) + ...$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

that's exactly $Length(A_1)\inf_{A_1}(f)+Length(A_2)\inf_{A_2}(f) + ... \geq Length(B)\inf_{B}(f)$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

so everytime we replace a term in this inferior sum P

[ruby sleet]

by those terms in this inferior sum Q

we effectively get something larger than before

adding up larger stuff will result in a larger total

so $S_{inf}(f,Q)$ is larger than $S_{inf}(f,P)$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

$S_{inf}(f,P)\leq S_{inf}(f,Q)$, what was expected

[rocky lotus]

Raphaelisius Maximus MMIII

[cedar spear]

does this happen bc inf(f,b) =< inf(f,a)?

[ruby sleet]

yes

[bronze moss]

is P,Q here a subset of [a,b]?

[ruby sleet]

in the context I think they're both of the form ${x_0,x_1,...,x_n}$ where $a = x_0 < ... < x_n = b$

[rocky lotus]

Raphaelisius Maximus MMIII

[cedar spear]

i think i understood the reasoning, but i still cant get my head around this

[bronze moss]

here is a graph of the area S_inf(f, P) we are calculating for a function (in red) and a partition P (in green). And here is the graph of the area for a refinement Q: S_\inf(f,Q)

you can sort of see why S_inf(f,Q) would be larger

[cedar spear]

yeah

[bronze moss]

intuitively, refinement -> S_inf and S_sup gets closer to the actual "area" of the function

so it is natural to expect S_inf and S_sup to get closer for Q than for P

[cedar spear]

yes

but i still dont get why S_inf(f,P)=< S_inf(f, Q)

i can only see it the other way around, since Q is larger than P

[bronze moss]

didn't the graph gave you a example where you can see S_int(f,P)=<S_inf(f,Q)?

[cedar spear]

only the = part

[bronze moss]

you can't see why thry might be equal?

cuz the example I gave: purple >green

P={0, 0.5, 1} and Q={0, 0.2, 0.5, 0.8, 1}

[cedar spear]

sorry, i dont get where S_inf(f,P)< S_inf(f, Q)

[bronze moss]

here is the partition Q marked in orange

and the partition P

green area represents S_inf(f,P) and purple represents S_inf(f,Q)

[cedar spear]

oh yes I think I got it now

[bronze moss]

extra values is really giving it an extra chance to take higher values than the current infimum

[cedar spear]

so is S_inf(f,P)< S_inf(f, Q) because A(S_inf(f,P))<A(S_inf(f,Q))?

[bronze moss]

well S_inf(f,P) is precisely the area

ie S_inf(f,P)=A(S_inf(f,P))

because if you check your formula it is "length times height" over all the rectangles in your partition

[cedar spear]

thank u so much

[bronze moss]

np

[cedar spear]

thank u both

[bronze moss]

and you can see why sup is smaller right?

[cedar spear]

i got it now i could kiss ur brains

[cedar spear]

kind of the same thing right_

?}

[bronze moss]

yeah

extra values-> extra chance for the supremum to be smaller

if you get what i mean

[cedar spear]

since Q is more precise, it wont be so far away from the function, so its area wont be so large

THANKS

[bronze moss]

finer the partition, closer to the area

sup is over estimation, and inf is underestimation

[cedar spear]

thank u

if i graduate i will write ur names on there

have a great daay

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[bronze moss]

u too

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[cosmic crest]

integral e^3x times sinx dx

[cosmic crest]

needs IBP I think, wanna use DI method but idk

[coral dragon]

Are you familiar with the horse shoe method? Indeed, that method comes from the DI method.

[cosmic crest]

nah, what is it?

[coral dragon]

So do the first two steps of the DI method, and you'll notice that you kind of end back where you started. That allows you to have a relation with some fudge factor, which is the solution.

[cosmic crest]

huh

I know this

for DI method, you go until 0, until row is integrable, or you end up at the same spot as you started excluding the signs for that row

I don't think you ever end up where you started with this problem

[glass kelp]

It won’t go to 0 here

[cosmic crest]

yeah that part is clear

[glass kelp]

But it’ll repeat

[coral dragon]

Yes, that's what I meant when I said kind of. Your second case is relevant.

[cosmic crest]

how?

e^3x will just continue 1/3e^3x 1/9e^3x etc

so ur never gonna end up back at e^3x

[coral dragon]

But you can factor out the fraction.

[cosmic crest]

ah

okay so ig you end up with

$$
\frac{\sin x}{3} e^{3x} - \frac{\cos x}{9} e^{3x} - - \int \frac{\sin x}{27} e^{3x} , dx
$$

I think

[rocky lotus]

Paul04

[fickle rose]

/27?

[cosmic crest]

is it right and I just needa factor out?

[coral dragon]

Without checking if the math is right, you would have that whole expression equal to the original integral. So combine the two integrals (bring that one to the other side) and you'll have the solution.

[cosmic crest]

OH I think I remember that from somewhere

lemme send my work 1 sec

[fickle rose]

You should but I got /9 inside the integral?

[cosmic crest]

OH

you gotta use the row 🤦‍♂️

not the diaganol

right?

ignore the stuff on the right

[coral dragon]

For the last step if the DI method isn't zero'd, yes you have to use the row rather than the diagonal.

So you seemed to have done a step too many, sorry soldier.

[cosmic crest]

yeah

so how should I move it over I ended up with 1/9 * the integral

[coral dragon]

Bring the 1/9 to the other side.

[cosmic crest]

yeah

[coral dragon]

Also you should have a 1+1/9 then, no?

[cosmic crest]

uh

[coral dragon]

Since the original integral plus the 1/9 * integral, so that's (1+1/9) * integral, unless that's what you meant.

[cosmic crest]

on the right side?

[coral dragon]

So you moved the 1/9 * integral to the left side, which already had 1 * integral

So that's 1/9 * integral + 1 * integral = (1/9 + 1) * integral

Then you can bring that (1/9 + 1) to the left side that you drew, so now you have just the integral in the right side == the solution.

[cosmic crest]

yeah idk where the +1 is coming from

moving the integral out is just doing 1 * 1/9 no?

[coral dragon]

So you have "... - (1/9) * F(x) = F(x)", which becomes "... = (1/9) * F(x) + F(x)", agreed?

where F(x) is the original integral, for easy notation.

[cosmic crest]

oh ur moving it over

why is 1/9 pos on the right side tho

is my double negative = pos wrong?

[coral dragon]

Well to bring a left negative to the right, you have to add both sides by its positive.

[cosmic crest]

well yeah but the 1/9 is pos unless I screwed smth up

[coral dragon]

a - b = c ==> a = b + c

[cosmic crest]

yeah yeah fs but 1/9 is pos on my end look at the pic

[coral dragon]

Oh I see what you meant,

[cosmic crest]

I went off integral udv = uv - integral vdu

and the row had -sin

so I thought the - from the formula with the - from the sin would make pos

but idk

[coral dragon]

When you do row multiplication, you have + -sin(x) * (1/9) * exp(3x), so it should be negative.

[cosmic crest]

yeah, but what about the - integral part from IBP formula

maybe I'm thinking of this all wrong, sorry if that's the case

[coral dragon]

I'm not sure what you mean by the IBP formula. IBP == initial boundary problem?

[cosmic crest]

integration by parts

it's DI method essentially

ok so ur saying it's neg let's just go with that forget what I said

so we end up with 10/9 integral

then move that over to the left and it's gg?

[coral dragon]

anyways,

Yep

You may need a +C factor, I believe.

[cosmic crest]

right

okay so lemme see ig final answer is

$$
\frac{9}{10} \left( \frac{\sin^3 x e^3x}{3} - \frac{\cos x , e^{3x}}{9} \right)
$$

[coral dragon]

Btw, any particular reason you integrated the exponential instead of the other way? Should yield the same either way.

[rocky lotus]

Paul04

[cosmic crest]

LIATE says trig before exponential

for u

[coral dragon]

What is LIATE? I fear that's another region-dependent concept.

[coral dragon]

Interesting, thank you.

[cosmic crest]

np

is my asnwer correct?

idk if it is

[coral dragon]

$$\frac9{10}\left(\frac13\sin(x)-\frac19\cos(x)\right)\exp(3x)+C$$

[rocky lotus]

Ordinary

[cosmic crest]

oh yeah idk where the cubed came from

nice!

tysm for the help it!

[coral dragon]

Np, and if you ever want to verify Desmos is a quick and dirty way.

[cosmic crest]

wdym?

[coral dragon]

One moment as I verify this actually..

Desmos is a free, light-weight website for graphing.

[spark burrow]

I haven't heard of the DI method actually

Is this how you multiply?

[cosmic crest]

yeah

[coral dragon]

Yep, it's good. It looked weird for me for a moment, sorry about that.

[spark burrow]

Yeah, shortens IBP quite a bit, nice notation

[cosmic crest]

can you send a ss of how you checked

like what u did

[coral dragon]

Nothing special, I just graphed it.

Oops, replied to the wrong message.

[cosmic crest]

oh I see, tysm!

I'll try that next time around. thanks again for the help!

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[gaunt fog]

Help

[gaunt fog]

I dont understand how this became

-1 = i

Mb from handrwiting

[river shale]

$\sqrt-1 = i$ ?

[rocky lotus]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[gaunt fog]

Ye

Like how that happen

[river shale]

It’s a definition

Just like how we decide the definition of sine

[gaunt fog]

Idk whats sine

But says here they square root 17

I mean

19

[coral dragon]

Like how pi = 3.1415...

[gaunt fog]

Ohh i thought Negative - 1 came from the square root of 19

But i remember my teacher squaring the negative number then simplified it

How u do that

19 for example

[gaunt fog]

I honestly forgot, should have taken notes

[river shale]

Let’s make things clear, the determinant only decides the status of a quadratic function’s root.

[turbid summit]

mathematicians wanted solutions to $x^2+b=0, b>0$ so they named $i$ as a number that has the property that $i^2=-1$ 🤔

[north walrus]

$\sqrt{-1}$ \underline{is} $i$

[rocky lotus]

Allen

parabolicinsanity

[turbid summit]

i think the actual definition for it is i^2=-1 rather than i =sqrt(-1)

[coral dragon]

Then i and -i would be ambiguous.

[river shale]

Why

[coral dragon]

but I see what you mean, like when we make adjoints R[i]/<i^2+1>

[coral dragon]

Since if the definition was i^2=1, then -i also satisfies that so then -i = i.

[turbid summit]

eh i'm not sure but that's what i remember reading

anyways

[river shale]

3^2 = 9
(-3)^2 = 9
3 = -3

Doesn’t make sense at all

[turbid summit]

well the square root doesn't give one value

or something

[coral dragon]

There's a difference between satisfying, and being defined (precisely) as heh

[coral dragon]

It's taken to be the principal, of course.

[gaunt fog]

Alright guys thanjs

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[neat crag]

Hey in algebra 2 rn doing logorithms how would i go about solving this question

[coral dragon]

Do you remember what the difference between "f(x)" and "f(x + a) + b" is? (recall shifts)

[neat crag]

yea

[coral dragon]

Could you explain what the difference between what "a" does, versus what "b" does?

[neat crag]

b is vertical shift and a is horizontal

so

b movees

up and down

a left and right

[coral dragon]

Exactly,

So we want to move the blue curve to the left (by how many units?), and up (by some amount)

[neat crag]

left by 1 units and up by about 4?

so

is that just it?

[coral dragon]

Close, we now need to express g(x) as a function of f(x) (so plugging in for a and b into f(x-a)+b)

[livid abyss]

.

[neat crag]

g(x) = log(x +1 ) +4?

[coral dragon]

Yeah, that's it.

[neat crag]

oh that was simpler than i thought thanks for the help

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[velvet prairie]

whats 1/2-4.6=
I got 6.765

https://cdn.discordapp.com/attachments/626167199958040586/1272293716702527580/what.gif

[coral dragon]

!occupied

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[pine prairie]

I once again need some help with pdes

[pine prairie]

The question is to solve u_x + u_y + u = 0, and how I might solve this via method of characteristics

The problem is that I don’t really get what the goal is for picking a suitable coordinate transform is, and how I might go about picking it

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@pine prairie Has your question been resolved?

[north walrus]

-# i think you can ping helpers if you want

-# it's already 15 mins

-# gl i havent learnt pdes

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@pine prairie Has your question been resolved?

[odd garnet]

the idea for method of characteristics is as follows

we want to pick a suitable coordinate transformation such that our pde turns into an ode

recall by the multivariate chain rule that if we have a coordinate t=t(x, y)

then du/dt = du/dx dx/dt + du/dy dy/dt

here we want that to look like u_x + u_y

so we wanna try smth like dy/dx = 1 if that made sense

[pine prairie]

How do we know this exists

[odd garnet]

i mean if you have a pde of the form function * u_x + function * u_y

[odd garnet]

then it kinda looks like that

[vale dock]

@pine prairie Has your question been resolved?

[rigid trench]

lol long time

[coral dragon]

iirc, you just pick a convenient transformation and it's okay as long as the determinant of the Jacobian is non-zero.

So worked out, this is what I find.

[pine prairie]

I pick a linear transformation?

Why linear?

[coral dragon]

It's what I was taught, the answer is "because it's simple, and it works".

[pine prairie]

As in, does linear always work?

And what does “suitable” mean?

[coral dragon]

As long as the determinant is non-zero. This doesn't guarantee that you'll simplify the problem.

[coral dragon]

It's subjective, and all about simplifying the original problem (like what the guy said earlier about turning the PDE into an ODE, like what I did above).

[pine prairie]

What’s the choose c = -d = 1 step what are we doing there

[coral dragon]

In this case, because the problem you gave was quite linear, it works perfectly. If you did something nonlinear like f(x,y)u_x+u_x^2u_y+u=0, linear wouldn't work.

[coral dragon]

We have the general form of the linear change of basis, so now we "eyeball" it to make it into an ODE.

So I chose those values to cancel out the partial beta terms and make the partial alpha terms add to unity.

[pine prairie]

I mean any non 0 det matrix will be a bijection and hence a valid change of basis

[pine prairie]

Why

[coral dragon]

Indeed, but sometimes one'd accidently choose a too good change of basis that'd "simplify" everything but give a determinant of zero so it'd be worthless and wouldn't give a proper solution. So it's mandatory to check the determinant.

[coral dragon]

Because

It's all about reducing the dimensions, which is doable because it's linear.

You could alternatively choose to cancel alpha and put beta to unity, you will get the same answer. Otherwise you did something wrong.

[coral dragon]

(alpha terms, and beta terms)*

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[wise jasper]

i really need help guys, im having a test tmr
supposedly speaking if i had two planes that were perpendiculared on the same line, would the distance between the two opposite apexes to the two opposite planes equal each other
english is not my first language so this could be a little hard to understand
hopefully you guys can help me

[verbal badger]

!occupied

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[worldly shuttle]

Help

[worldly shuttle]

Drawing the flow chart only

[deep fjord]

@minor socket

[wraith heart]

!noping

[vale dock]

Please do not ping individual helpers unprompted.

[deep fjord]

why did he @everyone then

?

[wraith heart]

doesn't matter go dm him not in a helper's channel

[deep fjord]

why

how’s a helper allowed to do ts

[wraith heart]

stop spamming OP's channel

[deep fjord]

?

he’s not gonna let you hit

[wraith heart]

#help-4 message

[cerulean maple]

uuuuh.... hi

im Jasmine

[wet bramble]

Agreed but don't inconvenience others

[wet bramble]

They aren't a helper,idk their role but shoo.

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[hushed kayak]

waitm thats possible?! i suddenly have a reason to learn lambda calculus

[timber echo]

i actually did do that but someone just somehow guessed a functioning tetration function that works better than mine (bc mine contains a pred function and i dont know how to express pred) but is structured extremely similarly to mine

so we figured out 2 things that work

[hushed kayak]

im only barely know that lambda calculus can write stuff in a weird way, so i dont understand any of what you did
but, does it make tetration easier to solve, or just translate it to a physical form?

[hollow dragon]

can someone explain function

[wraith heart]

https://brilliant.org/wiki/lambda-calculus/

[hushed kayak]

based

[final adder]

do you need help with a question?

[timber echo]

it dont make it easier imo its just fun

if you know how to do anything in lambda you can be handed a huge equasion and solve it without knowing whats happening
bc simplifying/solving it is always the same step

ok soooooo

the text form translates to the diagrams

say i have this function
λG.G

like plus or times
λv.v is a function
doesnt matter what letter it is

then we have the input, lets say its BANANA instead of a number

λv. says "OKAY, find all instances of v, and replace it with the input"
so λv.v banana = banana

THATS LITERALLY IT

@hushed kayak you following?

you can make more complicated equasions to do more complicated things

λF.F is just the simplest

say um
λx.λy.λz. x (y z)
is a times function
so say i want 5x3
λx.λy.λz. x (y z) 5 3
5 being the first input, and 3 as the second

you replace x with 5, and then replace y with 3, and then simplify more but that requires knowing how to express numbers in lambda and i didnt explain that
the drawings make a lot more sense

[sharp whale]

you gotta mention also that numbers and true/false take on something weird for lambda calculus

for lambda calculus, everything should ultimately just be in lambdas, so numbers are really some specific lambdas in disguise

[timber echo]

i dont know what youre tryna say im just tryna give a basic explanation

[sharp whale]

that doesnt give me a nice way to follow up

[timber echo]

idc

this isnt about you

[sharp whale]

thats a bit hostile, you dont have to say things like that

lets not point things towards me this fast

[sharp whale]

so the multiplication from earlier would give you 5 (3 z)
now if the multiplication works out, 5 (3 z) should be 15 z

to see how that happens, you can view numbers as the "number of times you apply a function"

so if you try doing 3 f x, the result is f f f x, or the result of applying f to x three times

[timber echo]

no actually you need the () to specify whats going into what

bc f f f x ≠ f (f (f x))

[sharp whale]

by default, you dont need to, parentheses weirdly go right-to-left

[timber echo]

you expressed function f with 3 inputs

[sharp whale]

so f f f x defaults to f (f (f x)) instead of ((f f) f) x

[timber echo]

youre very wrong

and id appreciate it if you would stop inserting yourself into my affairs

[sharp whale]

nicole this question isnt specifically about you, its about me adding to an explanation on lambda calculus with more details

if someone did this to me, I would assume they are elaborating

[timber echo]

(function) 5 4 6 2
is the function with 4 inputs
the () changes things

[sharp whale]

Ive checked and yea, by default its left-to-right
the post I took this from left the parentheses out, got it mixed with function composition

[timber echo]

its not( 5 of 4 6 2), put into function

[timber echo]

this has nothing to do with fucking left to right

oh my god this is so badic

[sharp whale]

I understand what youre talking about, but I dont see the use of phrasing this and then presumably directing it towards me
I mixed up one thing with another, they shouldnt be mixed up

ok now for true/false

the interesting part here is that true is "take in 2 inputs, then output the first"
and false is "take in 2 inputs, then output the last"
they dont act like the usual true/false

[hushed kayak]

i suddenly dont regret doing computer science

[timber echo]

these are two completely fucking different equasions

[sharp whale]

nicole I am aware that you watched the same video I do

[timber echo]

ive done far more than watch the video

[sharp whale]

Ive made an egregious error, but after this its not necessary to tell me the same information again

I didnt imply youve only watched the video, just that the video is very good

[timber echo]

well i dont see you having acknowledged that before this message and if you did i dont recognize it

im

[sharp whale]

thats alright, we can move on from this

[sharp whale]

the message may have been buried, but its here

function composition acts right-to-left, so I mixed them up, is what this message means

[timber echo]

im sorry its just i see someone doing something that ammounts to 1+5 = 5 and ive had 4 shots of coffee and like no food

[sharp whale]

its alright, it happens to me just as easily

Ive been in your position several times and did not calm down

the important thing for me to see is to think: dont attribute it to malice, attribute it to a typo

in this case its a pretty strong typo and Im honestly not happy with having made a mistake like that

[timber echo]

and the sqrt5 equasion isnt fun anymore bc of you so thers that going on and you intervene to me explaining this telling me im not doing it right while doing this and now im like a pissed twitter black supremacist getting pissed that someone said hi

[sharp whale]

its just that theres effective and ineffective approaches

[timber echo]

im not mad at you over this anymore bc its been cleared up

[sharp whale]

and from my perspective, Ive seen that kind of approach enough to know it wont lead you to anywhere fruitful

[timber echo]

and i have like so much energy and no energy and o7r8o67e5i676o8pguopuo

[

[sharp whale]

Im sorry nicole

number theory questions cannot be answered with continuous algebra, until you get to complex analysis

even then I have no idea how they pull it off

until then, we have modular arithmetic and diagonalizing matrices to justify particular behaviors, enough to answer the question

[timber echo]

idk what that means but okay

@hushed kayak im sorry

[hushed kayak]

continuous algebra?

[sharp whale]

you dont have to get into that, we can focus back on the lambda calculus

[timber echo]

explaining the false and true function is important tho simply bc its confusing and the main resource for lambda does a garbage job at it

its really simple tho

[sharp whale]

yea, have you seen an if statement? it looks like this

[timber echo]

just misleading

[hushed kayak]

im trying to find a formula for continuous tetration, so i probably wont get much out of this, will I?

[sharp whale]

continuous? theres many sources about attempting this

[timber echo]

is it "if a then b"
how does if work

[sharp whale]

its more of "if P, then A, else B"

[timber echo]

i can just give you the 2 equasions we found

[sharp whale]

what a true does is select A
and what a false does is select B

[timber echo]

well tetration funtions

[hushed kayak]

they work on fractional heights?

[sharp whale]

yep, thats the idea

[timber echo]

??

[hushed kayak]

2^^1.5

[timber echo]

idk how to express a fraction in lambda nor if it can be done but probs not

[timber echo]

it would

do that

perfectly

but idk if the expression of fractions is possible

[sharp whale]

and at least, lambda calculus isnt going to help you out for these sort of continuous functions

[timber echo]

yeaaaaa

its like base 1

[sharp whale]

lambda calculus is more for showing that yes, its possible
its just not very convenient

[timber echo]

so the number 30 is huge

[sharp whale]

111111111111111111111111111111

[timber echo]

tetration in lambda is too much paper to be practical

here ill show

[sharp whale]

for reference, division alone is messy

[timber echo]

these are 0 through 5

ignore the things at the bottom this is just my note sheet

[timber echo]

have not seen

[timber echo]

@hushed kayak

this is partioally why lambda isnmt helpful for tetration

[hushed kayak]

still cool

[timber echo]

that and itl take longer to do the steps

[sharp whale]

and also to consider fractions and somehow implement that

[timber echo]

yea idk i think its too objective and solid to have fractions be a thing

but i know not

[sharp whale]

too low-level is the usual phrase

there are other approaches than lambda calculus to create a continuous tetration function
granted, you should also learn lambda calculus too regardless

low-level for "the parts you are given are too basic, too small to make something of this complexity without grouping things together"

theres a research paper on a tetration function, I can try and find it

[timber echo]

i mean we are working with the physsical manifestation of the concept of the number 1 soooo

[timber echo]

i didnt know ppl had solved for one before

i mean makes sense duh

but yea

[sharp whale]

I remember the paper, theres a picture of them generalizing it to complex numbers

at least I remember the graph

might be this one http://myweb.astate.edu/wpaulsen/tetration2.pdf

not very readable

the general idea is that you can consider a specific tetration function F(z) = b^^z

F(z + 1) would be including one b in the "power tower" so thats the same as b^F(z)

[timber echo]

ive found ppls note sheets online that contain functions
and ive never found a function that gives a smaller output that actually worked to any capacity

i know that they dont usually work with small numbers, im speaking in spite of that

[sharp whale]

using F(z + 1) = b^F(z), they use something called "complex analysis" to work out details about this F(z), which by default will work for most real or complex numbers z

I think I know what this table can tell you

this is from the paper, and from their guess at a continuous tetration function

the upper-right table for example lists 2^^0, 2^^0.1, 2^^0.2, ..., 2^^1

similarly for the 3/2, e, and 10 tables

they admit their approach will not work if the base is < e^(1/e) which is about 1.445

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[willow arch]

can unbiased estimates be decimals

[willow arch]

unbiased estimate of population mean

[stark wedge]

yes of course? why couldn't they

the population mean of whatever quantity can be a decimal even if individual observations of it are only integer-valued

[willow arch]

oh ok tysm i js thot population would mean counting people but we cant half a person lol

[sharp whale]

if you randomly toggled or not toggled a light then looked on it, the average brightness is between dark and lit
you can estimate this average and itll also be between dark and lit
doesnt necessarily mean how bright that average is means anything concrete, youd have to be more abstract about it

[willow arch]

ohh alrights thank u

whats the difference betw sample mean and population mean? or can it be the same thing

[sharp whale]

calculating the mean across the entire population is pretty hard, youd need to ask everyone so that you can get their values to average
because of this, you only choose a random portion of them for a sample, for the easier-to-calculate sample mean

if you did multiple samples, the means you get would cluster around the population mean
this makes the sample mean an unbiased estimator of the population mean: the average sample mean is the population mean

the difference is that one of them is feasible and the other one isnt
the sample mean being an unbiased estimator means the sample mean aims to be the same thing as the population mean

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[willow arch]

okk thank u so much!

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[sharp whale]

np

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[ivory bough]

Maa chudao

[ivory bough]

Parhai kartey bhi ho Tum sab

[north walrus]

?

[next vortex]

uhh, do you have a question?

[vale dock]

@ivory bough Has your question been resolved?

[lucid cedar]

💀

[west coral]

Tameez mein rho chutiye

[west coral]

He is js insulting us

[next vortex]

hm

then I'll close this

.close

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[humble flicker]

<@&268886789983436800>

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[split edge]

is it possible to transform this ODE to a diophantine equation? for instance euler's method time discretiton

[vale dock]

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@split edge Has your question been resolved?

[wild linden]

@split edge why specifically a diophantine equation?

because you have 4n/5 in the exponent, if you can you will be very restricted by what values of n are possible if you insist on integer solutions.

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[raven frigate]

if i have a A=>B with perhaps steps in between and i know for certain A is true and from that following B is also true. in that case is it always true to also say B=>A?

[pine prairie]

if i have a A=>B with perhaps steps in between and i know for certain A is true and from that following B is also true. in that case is it always true to also say B=>A?

What does that mean

Also, if A is always true, then B => A is true regardless of what B is

[raven frigate]

i mean that i have proven that A=>B is true#

[pine prairie]

Draw the truth table for B => A

It’s always true when A is true

[raven frigate]

okay so im right, right? so in that case could i have written <=> all along instead of => for the proof A=>B from the beginning or do i have to prove B=>A even tho we know its true given what we just discussed

[pine prairie]

If A => B and A is always true then A <=> B

But if A is always true and A => B then B is always true

[raven frigate]

so i only use that => arrow when i dont know what i start with is for certain true or even false?

[pine prairie]

A => B is just some operation on 2 statements

[raven frigate]

okay thanks

[ebon glade]

maybe a small rephrasing helps. if we know that A is always true, then any statement C=>A is true, no matter what C is. in particular those statements are kinda irrelevant

[ruby sleet]

"If I have 5 apples, then the color blue is blue"

as an example

[raven frigate]

okay, but are there proofs in math where u show that A<=>B is true? in that case u would only need to show that for certain A is true and B is true in A=>B, right?

[ebon glade]

you show that if A is true then B is true

and if B is true then A is true

if you want to show that A <=> B

you have to show both A => B and B => A

just doing one direction is not enough

[raven frigate]

i cant skip this?

[ebon glade]

no

[wide gale]

in general A=>B does not imply B=>A

so you need to show both

[raven frigate]

but does A=>B imply B=>A if we know for certain both A and B are true?

[ebon glade]

well sure but thats meaningless

why do we care whether B => A

[pine prairie]

Then you’re here

[wide gale]

if both statements are true then yeah but theres no use for that really

[ebon glade]

if we know that both B and A are true

[raven frigate]

okay so there will be no question like that to begin with, i see. thanks yall

[ebon glade]

the color blue is blue implies that 2 is even

and 2 is even implies that the colour blue is blue

but who cares

[pine prairie]

I mean it’s a statement that’s always true

It’s a valid statement

It’s just not often useful

[wide gale]

tautology strikes again

[raven frigate]

thanks!

[turbid valve]

!done, I assume

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[buoyant sluice]

I do not want an answer but some kind of hint on where to start with this question. This question is in the series, sequences, and binomial theorem part of my curriculum.

[stark wedge]

try to figure out the lengths of OB_1, OB_2, ...

[red tulip]

Whats the difference between each arc? Like is there a set difference in distance from the center

Oh nvm

[buoyant sluice]

okay! thank you!

.close

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[harsh jolt]

How am I supposed to do this question

Never seen something like it

[rough talon]

which number

[opal pendant]

take the derivative of that right expression

[harsh jolt]

Ok you will get $8k^7 -16 k^3$

[opal pendant]

oh wait sorry dont differentiate yet

[rocky lotus]

playboi69

[opal pendant]

take k^4=a

[harsh jolt]

Alr

[opal pendant]

so now u have a quadratic in a

[harsh jolt]

Ye

[safe fulcrum]

RHS is a quadratic in k^4

[harsh jolt]

Ye

[opal pendant]

quadratic eqns have a minimum value

can u find that out

[harsh jolt]

-D/4a

[opal pendant]

yes

[harsh jolt]

It's -1/2 for this one I think

[safe fulcrum]

oh but that quadratic has complex roots

[harsh jolt]

I meant 1/2

[opal pendant]

a^2-4a+5
2a-4 = 0
a=2
4-8+5

how did u get 1/2

[harsh jolt]

Ye you are right

[opal pendant]

do u understand or are u just agreeing

[harsh jolt]

Nope i get ir

It

[opal pendant]

ok so minimum value is 1

can u see what we can tell now

[harsh jolt]

Ye

No not really

[safe fulcrum]

what about statement 2?

[harsh jolt]

Except drawing a rough graph of the quadratic

[opal pendant]

yes you should get a good idea what ur supposed to do based on statement 2

[harsh jolt]

So what am I supposed to do now

[safe fulcrum]

figure out if statement 2 is true or false

[harsh jolt]

Ye but how

I can't exactly draw a rough graph of it

To determine if it's true

[safe fulcrum]

let me remind you of function transformations then

start from sin(x)

then sin(3/2 * x) is a horizontal compression

with scale factor 3/2

[harsh jolt]

Ye but there is y inside cos

[safe fulcrum]

it's the same reasoning for cos

[harsh jolt]

No I meant the variable y

[safe fulcrum]

f(y) = cos(5y/3), y is a dummy variable

f(t) = cos(5t/3)

f(x) = cos(5x/3)

these are all the same function

[harsh jolt]

Ye true but what this all means sin has a different variable

Cos has a different variable

And they are multiplied together

Unless x = y

[safe fulcrum]

the question states that $x, y \in \mathbb R$

so $x, y$ are being treated as the domains of $\sin, \cos$

[rocky lotus]

south

[safe fulcrum]

what's the range of sin(3x/2) then, if you've managed to draw it?

[harsh jolt]

The range will be -1,1 for any sim function

Sin

Unless there is a restriction of domain

[safe fulcrum]

yep!

[harsh jolt]

In our case there is none

[safe fulcrum]

and the exact same is true for cos as well, right?

[harsh jolt]

Ye

[safe fulcrum]

yeah okay then statement 2 is true

[harsh jolt]

Ye

[safe fulcrum]

okay now, what's the absolute maximum of sin(3x/2) * cos(5y/3)?

based on multiplying them together as you said

[harsh jolt]

I have no idea

[safe fulcrum]

well, you know that sin(3x/2) <= 1 for all real x

[harsh jolt]

Yea

[safe fulcrum]

well and you can assume that cos(5y/3) will be positive, since we want the maximum value

so multiplying gives sin(3x/2) cos(5y/3) <= cos(5y/3)

[harsh jolt]

Huh

[safe fulcrum]

multiply both sides by cos(5y/3)

[harsh jolt]

Alr

Ye how will this help us tho

[safe fulcrum]

well then you've already found the max of cos(5y/3)

[safe fulcrum]

the goal is to now check statement 1 ofc

[harsh jolt]

Ye

[safe fulcrum]

it's like if I have p(x) whose range is 0 to 10
and if I have q(x) whose range is from 11 to 100

I can conclude that p(x) and q(x) never intersect

[harsh jolt]

Ye

[safe fulcrum]

you only need to check that 10 < 11 in that situation

[harsh jolt]

Ye just the range

[safe fulcrum]

okay so now

what's the maximum of sin(3x/2) * cos(5y/3)?

[harsh jolt]

Cos 5y/3

[safe fulcrum]

and what's the maximum of that?

[harsh jolt]

Wait one second

Cos 5y/3 <= 1
sin 3x/2 cos 5y/3 <= sin 3x/2

Is also true then

[safe fulcrum]

yes

yeah if that's easier for you, I see now

[harsh jolt]

Ye like then what is the maximum value

Is it cos 5y/3 or sin 3x/2

[safe fulcrum]

it's a number

so it's neither of those

[harsh jolt]

Ic

[safe fulcrum]

sin 3x/2 cos 5y/3 <= sin 3x/2 <= what number?

[harsh jolt]

No like that was the inequality

I got

Because both the statements should be true

[safe fulcrum]

well we previously established sin 3x/2 <= 1

so sin 3x/2 cos 5y/3 <= sin 3x/2 <= 1

[harsh jolt]

E

Ye

Ok so less than 1 in both cases then

Or equal to 1

[safe fulcrum]

okay!

yeah actually if sin(3x/2) = -1 and cos(5y/3) = -1

you also get - 1 * -1 = 1 as the maximum, so

[harsh jolt]

Yee

[safe fulcrum]

the negative numbers also worked out here

okay do you recall what the minimum of k^8 - 4k^4 + 5 was?

[harsh jolt]

Yea won't it cause problems

[harsh jolt]

1

[safe fulcrum]

cool, so we have LHS <= 1
RHS >= 1

what must happen if LHS = RHS

[harsh jolt]

They both are equal to 1

[safe fulcrum]

yes if we had -2 < f(x) < 1 and -3 < g(x) < 1, we'd get max f(x) * g(x) = 6 instead, just to clarify

[safe fulcrum]

yep!

okay so we're nearly done actually

it seems pretty obvious, but the important thing is that one value of x satisfies x^2 - 4x + 5 = 1

[harsh jolt]

-1

[safe fulcrum]

now at the start, we set k^4 = x

[harsh jolt]

I meant 2

[safe fulcrum]

oh yeah we also need that the x-coordinate of the quadratic vertex is 2

so k^4 = 2, that has 2 solutions

k^4 = -1, well that has 0 real solutions, completely different story

[safe fulcrum]

indeed

[harsh jolt]

2 or 4?

[safe fulcrum]

take the 4th root of both sides

that's the same thing as square rooting both sides, and then square rooting again

but you can't square root a negative number in the reals

so like k^2 = -sqrt(2)

that has no solutions

we just have k^2 = sqrt(2)

2 solutions

[harsh jolt]

Ic

[safe fulcrum]

in general the (odd number)th root will give 1 sol

(even number)th root will give 2 sols

[harsh jolt]

Alr

[safe fulcrum]

yeah so answer is D

I found a million YouTube videos on this exact question lmao

[harsh jolt]

Oh damm I've never seen this type of question tho

[harsh jolt]

Ye thats correct

Thanks man

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[tranquil crane]

Is there any chance that someone could explain this theorem and the subsequent proof? I’m confused on the definition of an orbital and how n-m is obtained

[ebon glade]

do you know what the representation of a permutation as a product of disjoint cycles is?

[tranquil crane]

Yes, I do

[ebon glade]

if I have a cycle (13562), can you write that as a product of 4 transpositions?

[tranquil crane]

(1 2)(1 6)(1 5)(1 3)

[ebon glade]

so for a permutation like (142)(35)(768) on n=8 with m=3 cycles, can you write it with 5 transpositions?

[tranquil crane]

(1 2)(1 4)(3 5)(7 8)(7 6)

Think I've gotten that right? Took me a while to type

[ebon glade]

well the point being that you write each individual cycle as a product of transpositions

for each cycle you need one less transposition than the lenght of the cycle

so if you have m cycles then you need n-m transpositions

which really is just the statement of the theorem

[tranquil crane]

Got it now, I couldn't see the pattern, nor the fact that orbitals is another word for cycles. Hours upon hours of studying and foggy brain! Thank you for your help!

[ebon glade]

orbits*

take a break

breaks are good

[tranquil crane]

There I go again!

Thank you again for your help!

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[harsh jolt]

I figured out the range for this one

$a^2 - 2a\in [-1,\infty)$

[rocky lotus]

playboi69

[harsh jolt]

$cosec^2(\frac{\pi}{2}(a+x))\in [1,\infty)$

[rocky lotus]

playboi69

[odd jackal]

uh

[harsh jolt]

Wait a second

[odd jackal]

it's a^2+2a

[harsh jolt]

Ye

Right

It's still -1

Tho the minimum value

[odd jackal]

yes

[harsh jolt]

So the only value where they can be 0 is when when the quad is -1

And cosec^2 is 1

[odd jackal]

yup

[harsh jolt]

So a = 1 but x should be

[odd jackal]

a=-1 you mean

[harsh jolt]

This aligns with option b but x is not a integer

Where did I go wrong

[harsh jolt]

A= 1 for the quad to give -1

No nvm -1

Ye so where did I go wrong here?

[odd jackal]

,rccw

[odd jackal]

okay so you want to solve $1= cosec^2(\frac{π}{2} + \frac{ πx}{2})$

[rocky lotus]

wai

[odd jackal]

can you silve this

[harsh jolt]

Ye

$sec^{2} (\frac{\pi x }{2})$

[rocky lotus]

playboi69

[harsh jolt]

Ye got it

Thanks

Man

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[random hull]

I have questions regarding this. Not only solving but i want to know the rules like i have some quries in mind

[random hull]

Helllo anyone here

[tidal terrace]

Hello, You just wait for the people

[random hull]

Oh okay

[tidal terrace]

Do you know how integrals work ?

[marble furnace]

Yeah?

[random hull]

Yes i do. But the thing is i ended up with y^3/3 . a root 2.

But this is wrong. Just to be clear i'm first time encountering double integrals and triple. So yes i'm finding it hard

1. Can i just swap the limits like inner limits to outer and outer to inner? and also swap dx, and dy? and then solve? Because there is no way to plot the region because how the function and its limits aare given.

2. is the question wrong?

[hollow shuttle]

i think the question might be wrong

currently the answer depends on y

can you try replacing the upper bound of the inner integral with a and see if that gives you the right answer?

[random hull]

Yes but it wont make any difference in getting that "Pi" in the answer.. it will just simplify it like make it concise

Can i just swap the limits like inner limits to outer and outer to inner? and also swap dx, and dy? and then solve? Because there is no way to plot the region because how the function and its limits aare given.???

Does this work? Or do i need to redefine the limits? like in each case.

[hollow shuttle]

in this case you should be able to swap the integrals yes

but there's something up with the question, you should probably ask whoever gave it to you to confirm if it's correct

[random hull]

Oh but wait doesnt it violate the rules? Like the ai said

[hollow shuttle]

the inner integral doesn't depend on x

so you can just move it out like you would with any other constant

and then move the other integral inside, because it doesn't depend on y

[craggy rock]

Bro

After we integrate the first one

We can move it out as a constant?

Cuz it has no relation or isn't a function of x

[hollow shuttle]

yep

[random hull]

I did that but it doesnt result in answer

[hollow shuttle]

yeah the question is almost definitely wrong

[random hull]

Wait so i can just swap the dx before and put dy behind and integrate the Y^2 respective to x?

[craggy rock]

No

How would you do that?

You can't swap dx and dy😭😭😭😭

[random hull]

Just take 1 as a in multiply of Y^2 and get the integration "x" and the consider y^2 as constant 😭 DONT JUDGE ME I HAVE JUST STARTED LEARNING THIS..

[craggy rock]

I'm not lol

I'm also dumb

😭😭

I'm so sorry i didn't mean anything like that

[random hull]

😭 YOU ARE LITERALLY BENZENE TEACHING ME MATHS. Pls i really want help. and look this is what my teacher said

🙂

Change the order of integration

Yes thats very helpful

[hollow shuttle]

does this make sense?

[craggy rock]

Idk what that is either

[craggy rock]

Oh wait

[hollow shuttle]

every operation in this image is just moving constants in/out

[craggy rock]

Yup

That makes sense

[random hull]

Woah. Wait this kind does lemme try

[craggy rock]

Got it

This works

⭐

[hollow shuttle]

the thing you ended up with here is correct, there's something wrong with either the question or the answer key

[craggy rock]

Exactly

I'm getting that too

I'm confused

Is it possible the answer key is wrong?

To the guy who sent this

[hollow shuttle]

i'm pretty sure the issue is that the "y" in the upper limit of the inner integral should be replaced with something else

[craggy rock]

Some constant limit

Instead of just y

Could've been a typo

[craggy rock]

Mr. Muffin please ask your teacher da

[random hull]

Yesss she provided me some material saying go through this. and then solve. So okay no worries. I will do it Thakn you so much for trying t o help mee

[craggy rock]

Bro

Send me also

[random hull]

Yess

[craggy rock]

Please😭

[random hull]

What material do you want? like what are you specifically studying

[craggy rock]

12th

Whatever you're doing
Because I'm interested to see

[random hull]

But you dont get to do Double integrals in 12th

[craggy rock]

Doesn't matter