help-49

[blissful pier]

lol matrix is too big for the discord bot I think?

[languid flame]

ask gpt to fix

[grand pond]

PajamaMamaLlama
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[languid flame]

halfway done with getting all the zeros in first row

[midnight plank]

@languid flame Has your question been resolved?

[midnight plank]

@languid flame Has your question been resolved?

[midnight plank]

@languid flame Has your question been resolved?

[languid mica]

Gauss method 😭

[languid flame]

i just finished column 3

[languid mica]

This is pure punishment having to solve all that

[languid flame]

it took 2 hours just for 3 columns

its so wraps

[languid mica]

[languid mica]

I think bro is getting cooked

Idk what the point of it is though

[languid flame]

bro im fucked

i need it for the ec

[languid mica]

Like they can test if you know Gauss method with 4x4 or 5x5

[languid flame]

i tok my first test and after seeingthe grade i really need this

[languid flame]

that was the homework

[languid mica]

Oh

Put it on computer fr

[languid flame]

yea im doing rn

i cant stay up like 4 more hours to solve ts

[languid mica]

Yea fr this doesnt even test if you know math after some point

[languid flame]

.close

[midnight plank]

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[midnight plank]

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[neat silo]

Is this somewhat related to the nine point circle?

[neat silo]

Let M be the midpoint of BC

How do i prove thay XDZY is cyclic

[civic gazelle]

prove angle XDY = XZY

[neat silo]

Hmm

Maybe this is a useful fact to prove it

The circumcenter of ABC is the orthocenter of its medial triangle

[civic gazelle]

nah you don't even need that

[neat silo]

Nvm no need

Yea

[civic gazelle]

notice that once you got XDZY cyclic, you also got XY//BC which means XDZY is an isoceles trapezoid

[neat silo]

I have another one

Since D lies on the nine point circle of ABC it must be the feet of perpendicular or median

But since its not the midpoint it has to be the feet of perpendicular

[civic gazelle]

then you can derive equal sides which leads to XY being the perpendicular bisector of AD

[hard shard]

maybe its naive to say but cant you just backwards construct it by reflecting A across XY

[neat silo]

Can i use the phantom points here then prove the points are congruent?

By angle chase XZY = ABC = XDY so XDZY cyclic then the conclusion follows

Yea phantom point also works

Reflect A wrt XY call it A'. Then prove A' cong D

Alr thanks all

[civic gazelle]

It doesnt ensure that A' and Z are the only points on BC that satisfies that property tho

[neat silo]

Oh off

Oh oof

[wet blade]

@toastaman needs help

[civic gazelle]

!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).

[civic gazelle]

so yea the cyclic quad way is prob the best way

and since it involves the 3 midpoints you can use the nine-point circle argument to prove

just for quick finishing

[neat silo]

I got another way

Since its an isosceles trapezium

AX = XB = XD which means ADB = 90

[civic gazelle]

ye

there are many ways to end this problem after proving the cyclic quad

[neat silo]

The isosceles trapezium one is beautiful

[neat silo]

Yes lol

Alr thanks all

.close

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[minor grove]

Can ts be middle term splitted?

[desert siren]

No

[civic gazelle]

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[civic gazelle]

determinant is ugly so no

[minor grove]

Ok one more

Can anyone send a pic of someone solving ts

Ik I am a dumbass

[civic gazelle]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[desert siren]

Hint: $\frac{a}{(\frac{b}{c})} = a \cdot \frac{c}{b}$

[desert siren]

It's !noans, for future reference

[civic gazelle]

ah I forgot

sry

[minor grove]

Bro

Can anyone please send the pic

[swift osprey]

i can send

[desert siren]

I've already given you a hint, if you insist on getting direct solutions, i will have to ping the mods. Also stop opening multiple channels when your question has already been answered.

[swift osprey]

im pretty useless

[desert siren]

No you cannot.

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[swift osprey]

ok sorry i didnt

know

[civic gazelle]

let's send memes

[swift osprey]

hint

[desert siren]

Did you try this yet

Or do you just refuse to learn

[swift osprey]

$\frac{a/b}{c/d} = \frac{ad/bc}

shoot im trash at latex

[desert siren]

Another $ at the end

[swift osprey]

$\frac{a/b}{c/d} = \frac{ad}{bc}$

[grand pond]

Goofy

[desert siren]

@minor grove

[magic panther]

That’s crazy

[midnight plank]

@minor grove Has your question been resolved?

[midnight plank]

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[inland patio]

I have a basic question. If I have a function that is continuous and I extend it by a single point, does it suffice to prove continuity of the extension by simply proving continuity at the extended point?

[hoary pulsar]

Yeah, so long as that doesn't somehow break the continuity of other points

[inland patio]

Ok. This confuses me to a great extent. How does one "prove" this? I thought that a function is continuous iff continuous at each point. Does this statement not apply here?

[hoary pulsar]

Sorry did the second thing I said there confuse you? Because I don't think that's actually possible (unless you're doing some strange topology stuff maybe, idk)

If you had a function that was continuous on [a,b), and you safely extended it to [a,b], so long as the value at b is continuous with the rest of the function, the extensions is fully continuous

By safely, I mean like how you can't really extend 1/x from (0,1] to [0,1]

[main current]

Are you only considering functions R -> R?

[fallen sparrow]

what if you define it piecewise

yeah no that wouldn't work anyways

[inland patio]

Your statement is fine. 🙂 But I think my own original statement confused me. However, your phrase "...so long as the value at b is continuous with the rest of the function." How does one formalize this?

[hoary pulsar]

Limits

[main current]

The point we're adding is a continuous point

[inland patio]

Yes, more or less. A function X from some topological space to C, the complex field.

[hoary pulsar]

Oh we're talking about topology?

[inland patio]

Yeah, this appears in a topology section.

[hoary pulsar]

I thought this was a calc 1 question and was treating it like one lmao

[fallen sparrow]

take sinx/x for example
design a function such that it is sinx/x for all real except 0 and then define it as 1 for x = 0

oh shoot

my bad this is topology

[main current]

General topological space? I am eager to say this is not true if the space is not Hausdorff but I'll think on it

[inland patio]

Yes, I think you're right. Something like that.

[inland patio]

To be more precise, the continuous function we are extending, we are extending it to X's one-point compaction, where X is assumed to be LCH -- locally compact Hausdorff.

[hoary pulsar]

Yeah, doesn't the double-zero reals break a ton of continuous stuff around 0?

Hmm...

[inland patio]

Oh, you mean the space X = (-1,1) U {0'} where 0' is not an element of (-1,1)?

[hoary pulsar]

Yeah, where every open set that contains 0 also contains 0'

I remember that being an example of a topology with some problems

So this is honestly more topology than I think I can review at 1am

[inland patio]

Ok, well, thanks for the help nevertheless.

[hoary pulsar]

Can you think of any reason why adding a point at infinity (which looks to be what you're doing) that is continuous would break the continuity of other points?

[hoary pulsar]

This honestly seems to be very similar to what I said back here

A more general topological notion of it sure, but the same general vibe

[inland patio]

I can't, but only if I prove it, I'm convinced. 🙂

And I don't really know how to show that it suffices to prove continuity at infinity.

[hoary pulsar]

Hmmmmmmmm

If f:X->C is our function, X' the compactification, and F:X'->C our function extension, can you prove that if for some point b other than the one at infinity, that if F(b) isn't continuous then f(b) isn't either?

[main current]

I think I got it. Take the real line, we're going to add another origin.

f(origin2) ≠ f(0)
f is continuous at origin 2

There exists an open set containing f(O2), but not containing f(0).

Therefore the inverse image is an open set containing O2 but not 0, a contradiction.

[hard umbra]

clearly it's sufficient that there exists a neighbourhood of every other point which doesn't contain the new point

[hoary pulsar]

clearly

[inland patio]

I will try. We would need to find a neighborhood U of F(b) such that F^{-1}(U) is not a neighborhood of b. I think this is what it means for F not to be continuous at b. But since F is not given explicitly, I don't know how the preimage of U is mapped.

[main current]

Ah wait, but that's breaking continuity of the original function too, isn't it

[inland patio]

yeah it is

[inland patio]

Why would that be sufficient? Such a neighborhood would perhaps be X itself?

[hoary pulsar]

Hmm

[hard umbra]

is X open?

[inland patio]

Yes, in the one-point compactification topology it is.

[hard umbra]

continuity at a single point can be checked on a neighbourhood

[hoary pulsar]

Well thinking proof-by-contradiction-ey, imagine you've got a nbhd U of F(b) s.t F^{-1}(U) isn't a nbhd of b, what does that tell you about f^{-1}(U)?

[hard umbra]

so if you have a neighbourhood that excludes the new point then you're always fine

[main current]

Okay so I think I can fix my example, only by setting f(O2) = f(0), right? Or is there any other way?

And as such, we are forced to make this extension continuous anyway?

[inland patio]

If U doesn't contain the point that infinity is mapped to, then F^{-1}(U)=f^{-1}(U), right?

[hoary pulsar]

Yup, so then f^{-1}(U) wouldn't be a nbhd of b either, making f non-continuous

There's still the case of U containing F(inf), but I think handling that shouldn't be too bad

[inland patio]

So we need to be able to find a neighborhood that does not contain the image point of inf (not the other way around; a neighborhood of each point other than infinity).

[hoary pulsar]

No I think it's fine if we have a neighbourhood that includes F(inf)

We're trying to show whether or not making F continuous at infinity is enough to make F everywhere right?

[inland patio]

Right.

[hoary pulsar]

So when trying to show that it's insufficient, we should be assuming F is continuous at infinity

One moment let me lay this out a bit

[inland patio]

Yes, that would be great. We are proving F continuous at infinity implies F continuous everywhere, and we are proceeding by proof-by-contradiction.

[inland patio]

So I'm still thinking on this one.

[hoary pulsar]

I'm trying to suggest a proof by contradiction, where f is continuous everywhere and the extension F is continuous at infinity, and our contradictory assumption is that F isn't continuous anymore

So we want to show that F being continuous at infinity but not continuous everywhere implies that f isn't continuous everywhere

[inland patio]

Ok, makes kind of sense. I follow I think.

[hoary pulsar]

Okay I'm definitely on the verge of passing out here, I'm forgetting what the next things I was gonna say are

But yeah, if the preimage of an open set in X' isn't open, then it shouldn't be open in X either

[inland patio]

I think all that is left to show is that if we've got a nbhd U of F(b) s.t F^{-1}(U) isn't a nbhd of b (here U is a subset of C, or R), we need to show f^{-1}(U) isn't a nbhd of b. The case where U contains F(inf) is immediate since f^{-1}(U)=F^{-1}(U) then. But the case when U contains F(inf) remains.

[hoary pulsar]

Hmm.... okay so if F(inf) =/= F(b) then because of Hausdorfness we can find a U only containing F(b)

So we've gotta care about when F(inf) = F(b)

Can you show that if F^{-1}(U) isn't open, then f^{-1}(U) = F^{-1}(U) - {inf} isn't open either?

[inland patio]

Ok, let me ask something slightly related. If U is a neighborhood of F(oo) such that F^{-1}(U) isn't a neighborhood of oo, what does that mean?

A neighborhood of p is defined to be any set such that its interior (an open set) contains p. So I guess it means F^{-1}(U)'s interior does not contain oo, so its interior is a subset of X. I'm a bit lost now.

[hoary pulsar]

That's gotta do with how the compactification is done

[inland patio]

I think I really need to understand the negation of a neighborhood. I don't think I got it right.

[midnight plank]

@inland patio Has your question been resolved?

[inland patio]

The assumption is that $f$ is continuous on $X$ and that $F$ is continuous at $\infty$. Then suppose on the contrary, $F$ is discontinuous at some point $b\neq\infty$. This means that there exists a neighborhood $U$ of $F(b)$ such that $F^{-1}(U)$ isn't a neighborhood of $b$. If $U$ does not contain $F(\infty)$, then $f^{-1}(U)=F^{-1}(U)$ isn't a neighborhood of $b$ either. Impossible. \

If $U$ is a neighborhood of $F(b)$ that contains $F(\infty)$, then $\infty\in F^{-1}(U)$ but it isn't a neighborhood of $b$. Here I'm stuck.

[grand pond]

psie

[hoary pulsar]

I think... if V isn't a neighborhood of b, then wouldn't V\{inf} also not be a neighborhood of b?

[inland patio]

Well, I think you're right. The tricky thing is translating that to f, since one needs to show f^{-1}(U) is not a neighborhood of b != inf for some neighborhood U of f(b).

[hoary pulsar]

Well if U contains F(inf), then f^{-1}(U) = F^{-1}(U)\{inf}

[inland patio]

Ok. A final question perhaps. Where in all this did we use Hausdorff-ness, if at all? 😅

[hoary pulsar]

Hmm... could a Compactified non-hausdorff space have a point that always shows up in open sets with infinity?

If so, I think this might break

but idk

[inland patio]

Ok, thank you very much for your help and patience. 🙏

[inland patio]

I still don't understand what you were alluding to here.

so if you have a neighbourhood that excludes the new point then you're always fine

A function f is continuous at a point p if for every neighborhood V of f(x) there is a neighborhood U of p such that f(U) subset V. Why would the existence of a neighborhood that excludes the new point make the extension continuous everywhere?

[hard umbra]

well first, you already assume continuity at the new point

so it remains to check continuity at all the points you already had

then you just always intersect U with the neighbourhood exclusing the new point

and thats guaranteed to be a subset of V since U doesn't contain the new point

the new point cant ruin continuity because you can always exclude it from consideration

[inland patio]

that makes sense actually, thank you 🙂 And the existence of a neighborhood that excludes the new point is guaranteed by some sort of T1 assumption (in this case, I actually assume something stronger, that X is locally compact Hausdorff, which implies that the one-point compactification is compact Hausdorff).

[hard umbra]

yeah seems fine

[inland patio]

A final question perhaps. How do we choose U to begin with though? The definition of continuity says that for any neighborhood V of f(x), we need to find some neighborhood U of x.

[hard umbra]

the function prior to extension has a U

if you want to be more precise you can probably refer to the subspace topology on X

f|X is continuous, and for it we can find a U = U' cap X, so take U'

[inland patio]

.close

[midnight plank]

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[worn light]

hi

[worn light]

listen

i graduated highschool in june 2022 only VERY recently have i decided to go back to school im attending the university of washington now but between those years ive obviously have not done any math, i was wondering if doing khan academy would suffice to regain knowledge

or is khan academy not good enough

like algebra 1, algebra 2, geometry, trig, precalc, calc series

[rare maple]

It is fine for an overview of topics

[worn light]

so

lets say i did everything from khan

would that be good enough for university level?

like if i had to take a placement exam for math

and i did every single one of them from khan academy

[grim vector]

depends what are the programs your university is teaching

[worn light]

what do you gusy think

well

Udub has alot of stem related majors

but for like a math placement exam is that like good enough?

or what do you think

do you think i should look at other resources for calculus?

im not reading a textbook

i dont have time for that

instead of khan

[grim vector]

if you have to take a placement exam, try to seek for the past ones

[worn light]

please

youre not answering the question

tis a yes or no question

do you think khan is enough

[glass lantern]

no

it's a start

[worn light]

what do i do then?

lets say i do everything

algebra 1 algebra 2 trig geometry precalc calc series everything from khan academy

now what

its not enough so what do i do then whats the next step

[desert siren]

#study-discussion , #book-recommendations

This is not the place

Also, welcome to mathcord

[flat spire]

relax

theyre just asking a question

[desert siren]

And i'm redirecting them

[flat spire]

take a chill pill vroski

[desert siren]

Closing this since you asked there

.close

[midnight plank]

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[signal holly]

is c correct?

[signal holly]

d is correct, I got it. but c I don't know

[high ocean]

Have you tried any workings for C?

[signal holly]

yeah

it is basically sec^2 (x) =2

but I'm wondering if any more solution is there

[high ocean]

Well, you can use cosec^2(theta) = 1 + cot^2(theta) and substitute that into the equation.

From there, finding theta is trivial

[signal holly]

no

3pi/4 is solution... I just checked

there are more solutions, how to get them?

[hollow grove]

yea, 3pi is sol

[signal holly]

that just gives cot^2 (x)=1, same

umm....

[high ocean]

<@&268886789983436800>

[signal holly]

<@&268886789983436800>

[warm hazel]

Donezo

[high ocean]

so try taking the root of that

cot(x) = +- 1

[signal holly]

I guess Trig triggered his PTSD

LOL

[signal holly]

n(pi) +- pi/4

ok got it...

[magic panther]

Trig isn’t even half as bad as integration 😔

[high ocean]

Yes. to find all solutions , simply plug n as any integer

for example, 3pi/4 = pi - pi/4 (n = 1)

[signal holly]

Tru Vro 😢

[signal holly]

got it

.close

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[gaunt nimbus]

<@&268886789983436800>

Get em

[midnight plank]

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[desert siren]

💀

DM Modmail

Or is he already banned

Ok yh hes banned

block

[gaunt nimbus]

Why would anyone do that😭

[desert siren]

Lots of idiots out there

Delete that also kek, even tho its in dm to you dont post it here

Got kids on the server

[midnight plank]

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[river citrus]

hi guys im new here, could anyone help me with this? im not sure if im right
(sorry, its in italian, heres translation: "Complete the following equalities, deducing from the graph the value of the following limits, if they exist.")

[lyric charm]

a is correct
b and c are switched around
d and e are wrong
f and g are correct

[river citrus]

thank youu

[desert siren]

Welcome to the server, Dudika!

[river citrus]

[lyric charm]

for b and c note that $x \to a^-$ means $x$ goes to $a$ \textbf{from the left} no matter if $a$ is positive or negative itself

[grand pond]

Ann

[lyric charm]

so for x -> -1^- you approach -1 from the left and you end up on the upper part of that curve that shoots off to +∞

[river citrus]

ohhhhh

.close

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[willow lotus]

Why am I getting different values for c and d? All the C(z) are mathematically equivalent.

[grim vector]

They are all at a multiplication by 1 close

[willow lotus]

Yes, that's what I said, so why we have different values for c and d?

if the C(z)'s are all the same

[grim vector]

Not the same, equivalent

[willow lotus]

yea

why do we have different values

[civic gazelle]

it's like having 4a/b and 8a/2b

different values for coefficients but they're equivalent

[grim vector]

Different values do implies that they're not the same but not that they aren't equivalent

Same != equivalent

[midnight plank]

@willow lotus Has your question been resolved?

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[woeful turret]

since its differentiable it will be continuous

so i got one condition by putting x=1

and then i differentiated f(x)

and i got second condition by putting x=1 again

[desert siren]

yes correct

[woeful turret]

but like i was just putting the continuity condition in the derivative of the function

is the actual logic left hand derivative = right hand derivative?

[desert siren]

Yes

You approach the function from the right

you get RHD

and if u approach from x<1

You get LHD

so when you say a function is differentiable at a point, it means LHD=RHD

[woeful turret]

ok so i dont need to use first principle or something to find LHD and RHD?

[desert siren]

If they've given the func is diffb already then no need

since LHD is necessarily equal to RHD

If they're asking if the func is diffb then u check LHD RHD by the limit defn

[woeful turret]

ohh ok got it

thank u

.close

[midnight plank]

Channel closed

Closed by @woeful turret

Use .reopen if this was a mistake.

[desert siren]

[woeful turret]

btw did u give jee?

[desert siren]

yh

[woeful turret]

which year

[desert siren]

gave this yr

[woeful turret]

ohh

how was advanced

[desert siren]

🥀

[woeful turret]

oh

😔

[desert siren]

i think weve already talked abt this tho

did u forget

[woeful turret]

OH

U ARE EXECUTOR

LMAO

[desert siren]

yes

[woeful turret]

u changed ur user

i didnt know

[desert siren]

yeah it's a pun- nw lel

[rare maple]

he changes it every single day

[desert siren]

I've only changed it twice

[silent zodiac]

Hmmm

[fallen sparrow]

yes

[fallen sparrow]

are you a she or a he 😭

[silent zodiac]

Truly a mains question

[desert siren]

I've put an any pronouns but it doesnt show at top idk

[silent zodiac]

Man this area under curve is killing me

[fallen sparrow]

i see

[fallen sparrow]

not here

#discussion

[silent zodiac]

[desert siren]

and anyway any pronouns role color is 🥀 so i just put blue and pink both

but yes am he

[midnight plank]

Available help channel!

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[twilit field]

I'm trying to prove the derived set of a set is closed

[fallen sparrow]

in cases like x^2sin(1/x) you gotta do definition by limit for derivative i believe

oh sorry

[silent zodiac]

[twilit field]

To do so let $ x \in (A^l)^c$, then as $x$ is not a limit point it follows $\forall \varepsilon>0 (x- \varepsilon , x+\varepsilon)\cap (A^l)^c \setminus {x} \neq \varnothing$

[grand pond]

wai

[twilit field]

As $x \in (A^l)^c$ , it further follows $(x - \varepsilon, x+ \varepsilon) \cap (A^l)^c \neq \varnothing$

[grand pond]

wai

[twilit field]

not sure what do from here

[verbal pumice]

Don't trust me?

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

I do

that's why I'm trying to prove the complement is open

[verbal pumice]

Oh mb

[lyric charm]

is the notation for derived set really ^l and not '

[twilit field]

I mean that's the notation my prof's notes uses

[lyric charm]

ok here is an idea

uh wait no hold on

[verbal pumice]

Does this need to be true for all epsilon

[twilit field]

oh, right, no

[verbal pumice]

honestly this interval notation for an epsnbhd is kind of clanky 😵‍💫

[twilit field]

you prefer $V_{\varepsilon}(x)$?

[woeful turret]

ohh alr alr ty i was having trouble in finding what cases to use that

[grand pond]

wai

[lyric charm]

you can make a proof that generalizes to any metric space here

[verbal pumice]

sure or that B(r; t) thing if you like that

[frozen talon]

or any T1 space

as long as singleton lads are closed it works

[twilit field]

Could we please stick to R here 🙏( I'll worry about metric spaces if I do well in RA)

[lyric charm]

you can pretend everything i say is confined to R only

[twilit field]

I should add all this is self studying, so please excuse any lapses in my knowledge

[lyric charm]

let $G = (A^l)^c$ for convenience. we aim to show $G$ is open. to do so, consider $x \in G$; we aim to draw an open ball in $G$ that contains no limit points of $A$ (i.e. lies entirely inside $G$).

\textbf{key step:} let $\boxed{\ep := \inf_{a \in A \setminus {x}} |a-x|}$.

my claim is that this $\ep > 0$, and that $V_{\ep/2}(x)$ [or perhaps even $V_{\ep}(x)$, if we're adventurous] contains no limit points of $A$.

(the idea here is that our x has to lie outside A -- but by how much? that's what epsilon quantifies)

(with an overthinkable but not too hard proof, you get that the ε-ball around x contains no points of A \ {x})

oh yes actually this is cooking

[grand pond]

Ann

[twilit field]

hmm

[lyric charm]

i think you have been struggling on this one problem for a week at least

[twilit field]

on and off, yes

[lyric charm]

also the obvious extension of this to arbitrary metric spaces is that |a-x| gets replaced with dist(a,x)

[visual panther]

isn't inf |a-x| = 0 and therfore not >0, or am I missing something?

[lyric charm]

yeah you're missing the set over which i take the infimum

[twilit field]

how do we know an infimum exists

[lyric charm]

an infimum always exists, even if it's -∞ sometimes

but these distances are always lowerbounded by 0

[twilit field]

oh right, I thought we were taking the min, my bad

[twilit field]

So essentially what I have to do is prove that's now contained in G

this ball is in G that is

[lyric charm]

yes

[visual panther]

The infimum does not have to be in the set, inf |1/x| = 0, even if there is no x in R, s.t. 1/x = 0

[lyric charm]

i think you're overthinking something or just not seeing it

im not even making a claim epsilon as a point belongs to A

im deliberately obscuring the fact that A and G are subsets of R for wai's purposes

cause they dont need to be

epsilon is a distance

[twilit field]

I'm kind of unsure of how to proceed

[lyric charm]

prove that for any limit point $y \in A^l$ you have $|y-x| \geq \ep$ and thus $y \notin V_{\ep}(x)$

[grand pond]

Ann

[lyric charm]

you may find the sequential characterization of limit points useful here

as well as the continuity of the function a ↦ |a-x|

it's probably a bit unorthodox to have x be a fixed thing here but i cbf to shuffle notation right now

[twilit field]

this feels like a lot

[lyric charm]

it's several small things adding up to a somewhat scary-looking total

[twilit field]

I think I got it then.
\
Let $y \in A^l$ then as $ \varepsilon = \inf_{a \in A \setminus {x}} \abs{a-x}$ it follows that $ \abs{y-x} ≥ \varepsilon$

[grand pond]

wai

[lyric charm]

how does it follow

note that nobody said y ∈ A

y may very well be a point in A \ A^l A^l \ A !

[twilit field]

hmm

[lyric charm]

you cannot make this jump directly.

[twilit field]

yea, noted

[visual tiger]

typo?

[lyric charm]

yes sorry i meant them the other way around

[twilit field]

Could we split this into multiple cases.
1)Let A be unbounded, in which case this follows directly
2) Let A be bounded

[lyric charm]

you give no shits about the boundedness of A

and no it still doesn't follow directly

[twilit field]

Can I think about this and ping you if/when I make progress

[lyric charm]

sure ig

[twilit field]

Thanks

I'll close this for now

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[prime hornet]

is it necessary to choose an explicit epsilon here?

if you negate the definition of a limit point, don't we have the existence of such an epsilon automatically?

or am I wrong?

[lyric charm]

dunno about necessity

but this is what i cooked up

[prime hornet]

fair

x is a limit point if for all epsilon > 0, there is a point a of A st a lies in B_epsilon(x)

so negating that...

x is not a limit point if there's an epsilon > 0 for which no a in A lies in B_epsilon(x)

i.e. we have a nbhd of x that is disjoint from A

[fathom onyx]

limit point -> for any positive epsilon, a ball with diameter epsilon contains some points in the sequence
¬limit point -> there is a positive epsilon such that a ball with this diameter contains no such points in the sequence [but we still have to prove such an epsilon exists]

[prime hornet]

how is it not given automatically?

[fathom onyx]

Because we still have to prove that this is true

[prime hornet]

if we do this via the topological definition (nbhds instead of epsilon balls), is it possible to prove the existence of such a nbhd?

[fathom onyx]

In this case, Ann attempts this by construction, i.e. explicitly drawing such a ball

[prime hornet]

x is a limit point if for all nbhds U of x, there is a point a in A such that a is in U

[fathom onyx]

In theory, yes? but since here we're in a specific type of domain we can use some extra definitions (like epsilon) to our advantage

[prime hornet]

so x is not a limit point if we can find a nbhd U st no a in A is in U

[prime hornet]

I've never actually constructed such a set, I don't think

such a U, that is

[fathom onyx]

Yes - but now you're using a different requirement; which is why this doesn't need you to construct the nbhd - you just need to claim if there is one, then something else holds

[prime hornet]

I don't feel like it's necessary to construct one though
if the definition says that something works for all nbhds, then negating it gets you the existence of a nbhd for free, no?

[fathom onyx]

The epsilon one requires you to construct a ball/an epsilon; then show the statement from the definition of ¬limit point holds

[fathom onyx]

But the aim is to prove that this is true

[prime hornet]

maybe I still don't quite understand; why do we need to construct such a ball?

why isn't the existence of one given by negating the definition of a limit point?

[fathom onyx]

We need to prove some set A is closed

So that is equivalent to proving that not(A) is open

[prime hornet]

sure, so we need to find an open ball around x in the complement of A that's contained in A

[fathom onyx]

okay so; ignoring the above...

[prime hornet]

is it not possible to use the ball given by negating the definition of a limit point for this purpose, without specifying epsilon?

[fathom onyx]

You can't just negate a definition and assume it holds

[prime hornet]

why not?

what's going on when we swap epsilon-balls for open nbhds then?

[fathom onyx]

Suppose I want to prove that some object A has a property P, with a definition D

[fathom onyx]

The definitions themselves are different

[fathom onyx]

It may be that if A has property P, then not(A) has property not(P) [let's assume this to be a fact from the definition]

Then not(A) would satisfy definition not(D)

We would still have to prove this

[prime hornet]

sorry, the letters are overloading in my head now

in our case, what is A?

which object is A, that is?

[fathom onyx]

e.g. a set

[prime hornet]

is A the set of limit points, the open ball around our not limit point, something else?

I'm trying to match things one to one

[fathom onyx]

I'm trying to generalise

But suppose we want to prove that A is closed; this is definitionally the same as "A' is open" - but I would still have to prove this

[prime hornet]

sure, I don't doubt that

I don't see what that has to do with my complaint though

[fathom onyx]

We have two different definitions for openness here

[prime hornet]

I am not worried about openness

okay wait, I think I see what you're getting at

[fathom onyx]

But I mean, there are two definitions, and they're slightly different; so to prove their negations true in a given situation requires different things

[prime hornet]

wait no

I'm still confused

we wish to prove that some set A is closed, meaning we want its complement to be open

to prove its complement is open, we want to fix a point in it, and then identify an open set that's contained in the complement

yes?

A is the set of limit points of another set B

if x is in the complement of A, it is not a limit point of B

therefore, there's an epsilon such that no point of B lies in the epsilon ball around x

[lyric charm]

uh

this is wai's channel

[prime hornet]

yes, but it's closed

[lyric charm]

y'all wanna yap about it over in #「helpers-lounge」 ?

[prime hornet]

sure

I'm just confused why we need to choose epsilon

[midnight plank]

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[woeful turret]

so the doubtful points are 0 and pi

so we need to check LHD and RHD at these points right?

[tulip moon]

yes

[woeful turret]

could u check if i got them right

because i dont want to have gotten the answer by accident

[tulip moon]

i can give it a quick look

[woeful turret]

checking for one case would be enough

this is f'(pi)

sorry if the handwriting is bad lmk if u cant understand anything

[tulip moon]

this is f'(pi)

ok

[woeful turret]

yeah

[tulip moon]

yea tha tlooks right

it is differentiable at pi

[woeful turret]

yeah

im having a little trouble with the LHD

is the only difference that h-->0-?

[quartz hornet]

pi-h

no?

[woeful turret]

cause my teacher gave me a definition where
lim h-->0 (f(x-h) - f(x))/-h

[woeful turret]

no x=pi-h but h will still tend to 0 right

[quartz hornet]

yeah

thats right

effectively saying you are taking a x which just a little little *100 tad smaller from pi

[woeful turret]

my doubt is instead of this can i use the definition
lim h-->0- (f(x+h)-f(x))/h

@quartz hornet

[quartz hornet]

that doesnt make sense tho ,if you are checking at x1 ,and you are aproaching from left you need a x which is infinitesmilly smaller than x1
x+h would imply you are looking at a number biger than x , effectively looking from rhs

(imo)

you can try to check the continuity on gif with this , instead of replacing x=x-h, use x=x+h

that should put things in perspective

[woeful turret]

ok so how do i find the LHD

could u give a formula

[quartz hornet]

what your teacher said would work yeah

but idt this question wants ytou to do alltha

[woeful turret]

it dosent?

[quartz hornet]

you could also just differentiate the function (if ou are quick at it) , and quickly check lhs rhs
basically checking the continuity of the differentiated function at the critical points

[woeful turret]

differentiate that shit 🙏 im good

[quartz hornet]

it looks a tad repetative , so made me think there might be a trick to it , thats all , maybe not tho

[woeful turret]

i am not quick at it

[quartz hornet]

thats fine too lhd rhd should work same

[woeful turret]

the solution given showed a one line answer though, i dont understand how

[quartz hornet]

there was a trick to it , shit

oh yeah.

give me a gun

😭

[woeful turret]

was it obvious

[quartz hornet]

yeah

[woeful turret]

so what if they are repeated roots?

[quartz hornet]

you were able to recognise the issue points right

x=0, x= pi

[woeful turret]

yes

yes

[quartz hornet]

now if a function has repetative roots at these points

they are differentiable at thoese points

[woeful turret]

oh huh

[quartz hornet]

think about it
consider (x-a)^2

diff it
2(x-a)(1)
now if i put x=a from either side i would get zero

making the differentiated function continuous , hence the function differentiable

[woeful turret]

oh ok yeah

makes sense

im just wondering why this was not taught to me

[quartz hornet]

similarly if you try to differentiate the above product you will notice that when you diff e^|x| -1 , (x-1) is untounched , hence it vanishes , and similarly other terms will vanish too at x=0 and x=pi

[quartz hornet]

now you know!

[woeful turret]

ohh ok

thanks bro

.close

[midnight plank]

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[midnight plank]

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[magic ingot]

e

[magic ingot]

image losding

for #12

[fallow scarab]

unreadable

take a closer picture

[gaunt nimbus]

Combine fractions

Note x ≠ -1,1

Boom

Done

[ruby burrow]

add the fractions, then use butterfly method and solve for x

[midnight plank]

@magic ingot Has your question been resolved?

[midnight plank]

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[midnight plank]

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[modern pendant]

Can someone help me look over an application question? It’s for a SI/TA position for classical physics

[modern pendant]

The question was the “why do you want to be an SI”

Here’s my response

.close

[midnight plank]

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[midnight plank]

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[narrow apex]

what is the amplitude of this sinusoidal signal in AC?
given channel 1 voltage div(y) = 10v/div?

[lyric charm]

do you mean that the y-axis grid scale is 10V per grid division

[narrow apex]

yes

in this case is nDiv only 1.5-1.8? is the mid line on the third line counting from above (the partial square line)?

[lyric charm]

what the hell is nDiv

im seeing from this clipart-looking picture that the wave spans about 5 divisions vertically

which means its peak and trough are 50V or so apart

[narrow apex]

yeah but whats the amplitude? I think the amplitude is only the peak?

ndiv is the number of squares you gotta count as divisions

since it starts at line 3, id assume the amplitude or peak is around 1,5 ndiv or 15V peak amplitude

do you think this is correct?

[midnight plank]

@narrow apex Has your question been resolved?

[midnight plank]

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[molten bay]

Is this closed or open set?

[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

[molten bay]

At x=0

I got [1,infinity)

[lyric charm]

so you got F = [1, +∞) yes?

[molten bay]

I don't know what should I do next?

How can I connect it with definition of closed?

[lyric charm]

your definition of closed set is?

[molten bay]

Which contains all limit points

[lyric charm]

so

does [1, +∞) contain all of its own limit points yes or no

[molten bay]

I am confused at x=1

i think i have to say YES because 1 is not in our set

[lyric charm]

why does x=1 matter? the original definition of F has x go over [0,1**)**

0 ≤ x < 1

[molten bay]

Yeah

And we are getting [1,infinity)

Closed

.closed

.close

[midnight plank]

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[midnight plank]

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[inland patio]

Consider the topology of uniform convergence on compact sets on $\mathbb{C}^X$, generated by sets of the form $$\left{g\in \mathbb{C}^X:\sup_{x\in K}|g(x)-f(x)|<n^{-1}\right}\quad(n\in\mathbb{N},f\in \mathbb{C}^X,K\subset X\text{ compact}).$$Let's denote these sets by $B_K(f,n^{-1})$. Are these sets, for fixed $f$, a neighborhood base for $f$?\

What we need to check is that each $B_K(f,n^{-1})$ contains $f$ (obvious) and that for open $U\ni f$ in the topology, there exists a compact $K$ and an $m$ such that $B_K(f,m^{-1})\subset U$ (how do I do that?). I am sort of considering the topology when $X$ is locally compact Hausdorff.

[grand pond]

psie

[midnight plank]

@inland patio Has your question been resolved?

[midnight plank]

@inland patio Has your question been resolved?

[inland patio]

<@&286206848099549185>

[inland patio]

.close

[midnight plank]

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[midnight plank]

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[hidden oyster]

Can someone clarify?

[jaunty ivy]

what do you need a clarification with?

[lyric charm]

what part needs clarification?

[hidden oyster]

I am studying linear algebra, but I was surprised with this quadratic equations question in the exercises so I am puzzled and can not make a grasp of it

[jaunty ivy]

once you plug in the points, those quadratics just become simple numbers

bx^2 at x=2 is just b(2^2)=4b

[hidden oyster]

even the matrix notation contain the unknowns not the coefficients, what does this mean?

[hidden oyster]

there are no specific points to plug as you can see the figure is vague

[jaunty ivy]

they are generalizing the solution set

so they pick arbitrary points

that (x1, y1) could be any point

[hidden oyster]

still did not get it!! let's go straight to the point, how should I attack this problem?

here is the solution if it could assist someone with explanation

[midnight plank]

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[midnight plank]

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[tardy bloom]

lim n→ infinity nsin(2 pi e n !)

[small yoke]

what have you tried?

[hushed mauve]

is that thing at the end there multiplication or exponentiation

[gaunt nimbus]

[ \lim_{n \to \infty} n\sin(2\pi e n!)?]

[grand pond]

k

[hushed mauve]

i see pi, e and n all there

[gaunt nimbus]

like this?

[tardy bloom]

Yeah

The factorial is only on n

[gaunt nimbus]

can i use stirling?

[tardy bloom]

What is that?

[gaunt nimbus]

or is it unneccessary?

[gaunt nimbus]

approx

[dense harness]

Use minority criteria

[small yoke]

you can just frame it

[tardy bloom]

The original question was that = kpi
Then find k
Don't think approximation will help here

[tardy bloom]

What is that too?

[gaunt nimbus]

ye doesnt wokr

[dense harness]

On second thought the e breaks stuff

[gaunt nimbus]

wdym

oh

[dense harness]

Probably doesn’t converge?

[gaunt nimbus]

most likely

,w lim_{n to infty} n sin( 2pi e n!)

[grand pond]

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Requested by nuclearinteraction

[gaunt nimbus]

yup

doesnt converge

[tardy bloom]

?
So the question is wrong then?

[dense harness]

Would be a yucky thing to prove I think

[small yoke]

maybe its supposed to ask the limit in 0?

[dense harness]

Hmm that would be just 0 tho

[small yoke]

yeah ik

but atleast there is an answer

[hushed mauve]

by any chance do you have the original question? maybe there is a mistake with interpretation

[gaunt nimbus]

Oh

What is that left thing

[hushed mauve]

is that supposed to be a factorial

[tardy bloom]

Another way of writing factorial

[gaunt nimbus]

Where n is an integer

(Natural to be precise)

[civic gazelle]

that's a factorial?

[tardy bloom]

Yeah

[small yoke]

in which country

[civic gazelle]

this is the old notation

ye it's a factorial

[hushed mauve]

i'm admittedly taking a shot in the dark here, but would it help to consider the taylor series expansion of (en!)?

[hushed mauve]

i went to your query on wolframalpha, and it seems like if you tell it to assume a discrete limit, there is an answer

[gaunt nimbus]

Interesting

[hushed mauve]

and i think the idea behind it is to split the taylor series expansion of (en!) into an integer and fractional part

would that be enough information to solve?

[tardy bloom]

I am not getting it.
(1+1/1!+1/2!+....1/n!) (n!)

What to do after this

[hushed mauve]

this is the taylor series expansion for en!, right?

[tardy bloom]

I just wrote the expansion of e
And then multiplied it by n!

Did you mean something else?

[hushed mauve]

what if you look at it this way?

can you now split this expansion into an integer part and a fractional part?

hint: compare m to n. when will n!/m! produce an integer? when will it produce a fraction?

[tardy bloom]

Wouldn't it always be an integer since n is bigger than m

[hushed mauve]

m is also growing without bound

basically both n and m are growing without bound, and we aren't fixing n >= m

[hushed mauve]

so let's come back to this question. what do you think?

[tardy bloom]

I am not getting it. Wouldn't it just give
n!/1+ n!/2!+....+1

[hushed mauve]

when m > n, what happens?

[tardy bloom]

How will m be greater than n, n approaches infinity?

[hushed mauve]

m is also approaching infinity

i mean, we are summing to infinity

for any given n, there will always be values of m > n because m is tending to infinity too

[tardy bloom]

Oh

How does it help then

[hushed mauve]

alright

so let's split the series

and let's just consider the fractional part (on the right)

for large n, what does the fractional part approximate to?

[tardy bloom]

1?

[hushed mauve]

so we have 1/(n+1) + 1/(n+1)(n+2) + ...

[tardy bloom]

0

[hushed mauve]

i think the idea behind the question is that the whole series approximates to 1/(n+1)

[tardy bloom]

How does the 1/n+1 come?

[hushed mauve]

because the decay in value from subsequent terms is significant enough

(for large n)

[tardy bloom]

Where does this lead to?

[hushed mauve]

ok so we have split the taylor series of en! into its integer and fractional parts

let's call the integer part I and the fractional part F

[tardy bloom]

Ok, what after that??

[hushed mauve]

so we have

sin(n(2pi)) for integer n is?

[tardy bloom]

0

Ok, so you expand the sin and solve to get k=2?

[hushed mauve]

so you can remove 2pi(I_n) from the sine

you will need to use small angle approximation

[tardy bloom]

I get it

[hushed mauve]

remember the approximation of f_n? substitute it here

then use your standard limits techniques to solve for k

[tardy bloom]

n sin(2pi/k+1) = 2

[hushed mauve]

n+1*

[tardy bloom]

I get the answer but this part still bothers me

[hushed mauve]

you're not wrong to question it, but it is an approximation

if someone else has a better way or can justify it, i'll be happy to hear

but for me, that's my only justification unfortunately

[tardy bloom]

Thanks a lot

[hushed mauve]

glad to have helped!

[midnight plank]

@tardy bloom Has your question been resolved?

[midnight plank]

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[sage zodiac]

How to solve it through substitution???

[zenith snow]

Use anyone of these equations

[lyric charm]

would you know what to do if the numbers were not roots

[zenith snow]

And find the value of one variable

[lyric charm]

it's basically the same shit

[sage zodiac]

Yeah!!

[sudden yacht]

Then what's the trouble?

[civic gazelle]

square root phobia

[sudden yacht]

Roots are just normal numbers

[sage zodiac]

I get it

[lyric charm]

roots are just normal numbers but pedagogically they are a hurdle

[lyric charm]

rhizophobia i guess or something

like... ok, you do need to exercise a bit of extra care when handling these roots

in particular make sure not to confuse $\sqrt{2}x$ with $\sqrt{2x}$ --- one handwriting trick that helps for this is to put a little hook on the end of your root symbols

[grand pond]

Ann

[buoyant linden]

wait can u show an example of the hook

i dont rly know what u mean

[buoyant linden]

ah

thx

[sage zodiac]

I am stuck 😑

[zenith snow]

Show ur work

[zenith snow]

Where did the root over 2 go

[sage zodiac]

Look there is still a sign of root

@lyric charm

[slender walrus]

not over the 2 in the last equation which is the issue being pointed out

[sage zodiac]

Aah I see

Let me fix it

[zenith snow]

But either way u continue in the normal way

Subtract

To make it easier briing root 8 to a simpler form

[lyric charm]

what is this word

[sage zodiac]

Ehh?

[lyric charm]

specifically what is the first letter in there

[lyric charm]

here you don't need it...

anyway

[sage zodiac]

Put

[lyric charm]

the important thing you care about is that this equation reduces to $\paren{\frac{3}{\sqrt{2}} - \sqrt{8}}y = 0$, and one thing you can say for sure about $\frac{3}{\sqrt{2}} - \sqrt{8}$ is that it's not zero

[grand pond]

Ann

[lyric charm]

if it says "put" then what are these bits doing there

[zenith snow]

Cursive

[lyric charm]

💀

[sage zodiac]

That's cursive

Isn't it cool?

[lyric charm]

no it's not

[sage zodiac]

[lyric charm]

you're talking to someone who deliberately taught herself a semi-cursive hand out of hatred for Russian cursive

[sage zodiac]

Back to the question

[lyric charm]

anyway you reacted with ❓ to this. what's unclear?

[sage zodiac]

Your English and explaination

[lyric charm]

you collect like terms and the equation becomes $\paren{\frac{3}{\sqrt{2}} - \sqrt{8}}y = 0$

[grand pond]

Ann

[lyric charm]

understand or no?

[sage zodiac]

Understand

[lyric charm]

$\frac{3}{\sqrt{2}} - \sqrt{8}$ is not zero

[grand pond]

Ann

[lyric charm]

and to say the quiet part out loud: the best next step is to divide by that

[sage zodiac]

So everything becomes 0

[lyric charm]

⚠️

[sage zodiac]

Right?

[lyric charm]

"everything" is not a word you should throw around lightly like that

[sage zodiac]

Y=0

[lyric charm]

yes

lowercase y not uppercase Y, but yes. correct.

[sage zodiac]

Ohk that's all right!!

"Instead of zero" I forgot to write it

[lyric charm]

too vague to answer.

[sage zodiac]

@lyric charm

[sage zodiac]

Ehh??

[lyric charm]

i do not know what you mean by "there is any natural number"

[sage zodiac]

If there is any natural number Instead of zero

[lyric charm]

like on the RHS?

well then you may have to do some simplification magic with the radicals

[sage zodiac]

Yeah

[lyric charm]

eventually you'll come to a point where you need to do something like multiplying by the conjugate to rationalize the denominator

best if you come here with a specific example question though tbh

[sage zodiac]

Oh yess!

[lyric charm]

rather than force me to explain things in generality and fail to make you understand

[sage zodiac]

Can you rationalize a denominator with example

Please

I forger

[lyric charm]

https://www.youtube.com/watch?v=5j8a75aHaSE

[sage zodiac]

Ohk!!

Well @lyric charm why there is no vc?

Vc will help to end the doubt more quickly

[lyric charm]

dunno, ask the mods. not me.

• i don't have the energy for voice chat right now anyway

[sage zodiac]

Who's the mode

How can I meet them

[lyric charm]

mods == moderators

if you want to speak to them as a team, you can DM @shadow scaffold

[sage zodiac]

I see

.close

[midnight plank]

Channel closed

Closed by @sage zodiac

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[fallen vault]

I have the function 144x/(x+9)^2 but only in the domain x>=0.
Is there a tangent to the function at x=0 or no?
Because without the domain constraint it does have a tangent there but with it, the function isn't differentiable at x=0.

[lyric charm]

why do you claim it's not differentiable at x=0?

$\lim_{h\to 0} \frac{f(h)-f(0)}{h}$ still exists just fine, does it not??

[grand pond]

Ann

[lyric charm]

it just resolves to the one as $h \to 0^+$ due to the domain restriction, that's all -- but that doesn't stop it from existing.

[grand pond]

Ann

[fallen vault]

I believed that at "endpoints" functions are not differentiable

[lyric charm]

what, like inherently?

[fallen vault]

yea

[lyric charm]

well you're wrong about that, what else can i say

[fallen vault]

good to know ig

So because the right side limit of the derivative exist, a tangent does exist to the function at that point?

[lyric charm]

yeah sure

[midnight plank]

@fallen vault Has your question been resolved?

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[prime garden]

Is this a correct proof of the product rule?

limit h->0 (f(x+h) - f(x))/h = f'(x)
Therefore, f(x+h) = hf'(x) + f(x)
d/dx f(x) g(x) = limit h->0 (f(x+h)g(x+h) - f(x) g(x))/h = ((f'(x)h + f(x))(g'(x)h + g(x)) - f(x) g(x))/h
= (f'(x)g'(x) h^2 + f'(x) g(x) h + f(x) g'(x) - f(x) g(x))/h = h f'(x) g'(x) + f'(x) g(x) + f(x) g'(x) = 0 + f'(x) g(x) + f(x) g'(x)

[fallow scarab]

There are dozens of proofs online you can compare yours to

[lyric charm]

f(x+h) = hf'(x) + f(x)
this equality is not literally true.

it provides good intuition, but for a proof you may need something more precise like little o notation

[prime garden]

Okay, thanks.

CLOSE!

.close

[midnight plank]

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[midnight plank]

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[proud ember]

Hi! If a/b,c/d are Farey Neighbours in F_n, why can't they have more than 1 fraction between them in F_{n+1}? I wanna try justifying this without the fact that |bc-ad|=1? Now if b+d>n+1, they're still gonna be neighbours but what if b+d=n+1?

[proud ember]

Here's what I got so far:
so firstly, if a/b<r/(b+d)<s/(b+d)<c/d
k/(b+d) must also lie between a/b and c/d for all k beteween r and s both inclusive
so I only took the case where s=r+1
and when I did that what I got is a/b<(2d-1)/(2c-1) and 1/(bc-ad)<1/b+1/d

<@&286206848099549185>

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@proud ember Has your question been resolved?

[proud ember]

<@&286206848099549185>

[midnight plank]

@proud ember Has your question been resolved?

[midnight plank]

@proud ember Has your question been resolved?

[midnight plank]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[lime oracle]

hellooo does anyone know why the magnitude and angle of RF are wrong?

[lime oracle]

the table is correct

you dont need to give me answer im so grateful if someone guides me through it 🥺 🙏

<@&286206848099549185> 🙏

is this using the phytagorean theorem and trigo?

[midnight plank]

@lime oracle Has your question been resolved?

[carmine sigil]

@lime oracle the reason why you haven't gotten much in the way of answers is that the information you have provided is incomplete.

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[carmine sigil]

And also

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[lime oracle]

hold onnn, sorry. this part is actually supposed to be done by my groupmate but i am doing it insteaddd so i dont have her solution (I think she used ai 😭)

[lime oracle]

this is only what our teacher gave usss

let me showww

[carmine sigil]

Do you not have a diagram of the way the pulleys are arranged?

[lime oracle]

aaaa wait i just realizeddd

im also watching this video right now, not sure if its the right one that i should watch but https://youtu.be/RxV06Oj5sMc?si=veTlzsGBCy_Lntee

[lime oracle]

i genuinely just saw the diagram man

[carmine sigil]

This seems like a good video to watch

[lime oracle]

i looked for the N part myself 😭

[carmine sigil]

No worries. A difficult part of learning is not knowing which questions to ask.

[lime oracle]

thank you, i just thought it was a picture of our teacher showing us how to assemble the force table

[carmine sigil]

So when I ask, "do you have a diagram?" That can be useful as an answer to a question you didn't even know you had 🙂

[lime oracle]

yesyesss ❤️

this is my current progress

[carmine sigil]

Good luck with the Squid Game btw.

[lime oracle]

its very small LOL

[lime oracle]

AGHHH s1 will always be my fav

[carmine sigil]

No worries.

Follow along with the video

[lime oracle]

Okaaayy

[carmine sigil]

And then, if you still have questions, let me know.

[lime oracle]

okay okay, thank you!!!

[carmine sigil]

Ping me, I'm going to look elsewhere for a minute

[lime oracle]

if im done will it be alright if u check it?

yes sure

[carmine sigil]

Absolutely

May I also make a recommendation, you don't have to do this, but it might help your other students (and yourself)

Once you figure out how to do this, it might help you solidify your understanding if you teach your project partner how to do it.

[lime oracle]

YESS SURE

slr i just saw ur message

i am halfway done btw

@carmine sigil i am done but i am very unsureeee

for the clockwise/counterclockwise part when u subtract 180 degreess

[carmine sigil]

,rcw

[lime oracle]

i followed with the guy in youtube, but for his one of his sign was negative but mine are both positiveee

so i think i followed wrong HAHAHAA

[carmine sigil]

Well, you two were doing slightly different problems

Let me get oriented and see

[lime oracle]

thank youuu 😭 yesyes

also my part of the activity was vector resolution, and it's the same but for mine it's 46.6

[carmine sigil]

So I'm seeing something a little bit unusual, and I was hoping you could clear it up