help-13

[keen nova]

can the unit cube be from , say, -0.4 to 0.6?

(in all axes)

i see some definitions defining it as always [-0.5, -0.5, -0.5] to [0.5, 0.5, 0.5]

in other words, does it matter if the center of the unit cube is precisely in the origin?

<@&286206848099549185>

[vagrant pewter]

I mean it’s kind of a semantic question I guess I would say that typically the unit cube would refer to the unit cube centered at the origin

But you can totally apply a translation to it

It’s changes the equation of the shape though of course

[keen nova]

hmm i see

yeah, i guess i will have to ask my professor about this. He specifically says that we should ensure the barycenter of the shape coincides with the origin, but i am not sure if then further scaling changes that property

as in, after the translation it should coincide, but after scaling he doesnt explicitly say so

[vagrant pewter]

Yeah I would definitely ask them because the cube with side lengths 1 centred at (.1,.1,.1) could be considered a unit cube but it probably isn’t “the” unit cube

Barycentre is the spatial average correct?

Or I guess the “center of mass”

[keen nova]

yeah

but even barycenter seems a bit ambigous to me

because depending on the structure of the shape it can be different

[vagrant pewter]

Well if the scaling is done uniformly in all directions it would be barycentre preserving

[keen nova]

we learned about a weighted area barycenter

[vagrant pewter]

Do you solve for the barycenter computationally or algebraically?

Algebraically I guess you would need some triple integral of some kind

[keen nova]

thankfully cmoputationally ;p

it's kind of shitty my partner did it and he obv used chatgpt. I had to revise the code and then i asked chatgpt different ways to get the barycenter and it gave me literally the exact same code as he wrote

but if you're interested i can show u

[vagrant pewter]

Well the code will most likely just be using numerical methods to compute an integral I guess

[keen nova]

nah, mostly linear algebra i think

or i guess kind of integral

[keen nova]

so i am actualy using a scale operation from a library called Open3d

and this takes a scaling-factor as well as a center

[vagrant pewter]

So it scales it about a particular centre then and the centre you input should most likely be preserved by the scaling

[keen nova]

i am giving it the barycenter as the center and then the scaling-factor is computed as the max-extent of the bounding-box of the shape

the issue might honestly be related to floating point precision

[vagrant pewter]

Max extent I assume is the longest distance from the centre to the perimeter

[keen nova]

because its always about 0.1 or max 0.2 off

[keen nova]

nope, max extent is, if you imagine a box surrounding an object, the max length of any axis

[vagrant pewter]

The orientation of the object must be important too then

You could rotate it in such a way that the scaling factor could be increased and the object would still sit in the cube

[keen nova]

hmm, not sure what u mean

anyhow, i just verified that the barycenter in fact does not change with my scaling

so somehow, i scale with respect to the barycenter with a certain scaling-factor, yet it very frequently goes , say, -0.4 in neg. X and 0.6 in pos. X

i think it's just because the actual barycenter is never exactly [0,0,0]

but i didn't think it would be off enough for it to be even noticeable

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[stark grove]

I’m trying to figure out 26, the second photo is an example of what we’re supposed to do. The third photo is an equation someone gave me when they were trying to help. It looks correct, but I asked my friend who’s rlly good at math and he got a completely different answer.

[glacial field]

The equation I provided is correct

Solving for k from it should be straightforward

[stark grove]

yes but my friend got a different answer when I solved it

I got -1.75 and he got -5.75

[glacial field]

Oh sorry I meant to put plus

[stark grove]

omg I didn’t even notice that

ok lemme try re solving

[plucky owl]

Why did this become minus?

[glacial field]

I made an error

[stark grove]

it’s ok lol I’m gonna redo it rq

I got 1,75

uuuuugghghghghhghgg

[plucky owl]

Show the work

[stark grove]

maybe he got it wrong but I highly doubt it

[glacial field]

You wrote +15

Instead of -15

[stark grove]

AAAA

I didn’t notice that AGAIN

ok let me redo it again..

[plucky owl]

You weren't suppose to change both minus signs to plus

[glacial field]

She gets that

[stark grove]

I got -5.75!!!!

lemme try the next one and then I’ll close if I get the right answer

[glacial field]

Good job

[stark grove]

ok I got 2

[stark grove]

and ty!

[plucky owl]

What's the problem?

[glacial field]

Yes it is 2

[plucky owl]

Fyi, you can plug in your answer back in to check your work, because it's an equation, the left would equal the right side

[stark grove]

ooh yes I forgot about that

I hope my teacher doesn’t make me redo it since I didn’t do it her way

tysm for the help tho!! I rlly appreciate it

.close

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[bold pond]

how is this possible? the inner radius and outter radius ave the same equation.. won't that cancel out the volume?

[remote trail]

doesn't it make more sense to use washer method

oh nvm

[bold pond]

probbly

[remote trail]

im trolling

[bold pond]

not

nm

ok

ha

i was just like arigt

[remote trail]

nah shell is better

that's my bad

[bold pond]

ight

but like don't i need to do outer - inner?

[remote trail]

for the shell method?

[bold pond]

ya

for this problem

[remote trail]

well you want to find r(x) and h(x) for the cylinder

radius and height

the radius will just be x-1 if you can see

[bold pond]

yea

[remote trail]

since you rotate around x = 1

[bold pond]

for the outter is 1 - (4x - x^2)

[remote trail]

and then the height will be ((4x-x^2)-3)

[bold pond]

what why?

[remote trail]

imagine taking cylinders throughout the requested area

[bold pond]

oooooo

omg

bc it's rotating on 1

[remote trail]

the height of those cylinders will be the higher-lower function

[bold pond]

so the outer is like 3 away from it

so it's actualy 3 - (4x - x^2)

and the inner starts at 1 tho so does it even need anything

[remote trail]

some kind of shell like this

you can see that h(x) will be upper-lower function

radius will be x-1

[bold pond]

why do you keep saying x - 1 it's 1 - x

[remote trail]

wdym

you're rotating around x = 1 no?

[bold pond]

oh

so is that only for rotations around y axis?

[remote trail]

you can see that the radius from that line to the curve will be the x position of that curve - 1

or x-1

[bold pond]

no it's 1 - x

[remote trail]

assuming the intercepts I was thinking this

[bold pond]

?

i am pretty sure it is 1 - x and 3 - (4x - x^2)

[remote trail]

are we using the same method

[bold pond]

and why are you squaring 4x

[remote trail]

my fault

misinput

just 4x

[bold pond]

do you write it that way when it's a x rotation then?

i recall putting it beofre the function

maybe it was because it was rotated on y = 3

so i put it as 3 - f(x)

[remote trail]

rotating around x = 1 is rotating in the Y axis no?

[bold pond]

yes i am asking about rotating on X axis

is that why is it confusing me

please confirm my question

[remote trail]

for an X rotation you would use the disk method instead

[bold pond]

it is irrelvant

for now

[remote trail]

or invert the function

[bold pond]

just simple question

if it is rotating on X - axis

do you subtract it like 3 - (x+5)

[remote trail]

well to use shell method for x you would need to invert the functions

[bold pond]

OMG

LISTEN TO WHAT I AM ASKING YOU

IT HAS NOTHING TO DO WITH THAT

IN GENERAL

[remote trail]

you don't seem to understand

[bold pond]

i am specifically asking about how to set up

NO U DONT

when it is rotating on x - axis

vs y - axis

[remote trail]

yes

[bold pond]

do you need to set it up a differnt wayu

[remote trail]

go on

[bold pond]

that is the question

[remote trail]

yes

you need to set it up differently

[bold pond]

I AM JUST TALKING ABOUT THE AXIS OF ROTATION WITH RESPECT TO THE VALUE WHER IT IS ROTATING

FOR EX

if blah blah blah rotates about y = 3 vs x = 3

our disagreement is laying on this

you say it is x - 1

[remote trail]

rotating around x = 3 would be rotating around the Y-axis over that line

[bold pond]

i say it is 1 - x

I am asking

if

it

changes

1 - x

x - 1

does that form change

based on whether it is rotating on the x or y axis

nothing else

not asking abt dy or dx here

just pretend everything else is perfectly setup for the equation

because we have a disagreement about how to set up the rotation part

back into the equation

you say x - 1

i say 1 - x

[remote trail]

ok I kind of understand what you are trying to say but it doesn't make sense

[bold pond]

so all i am asking is

[remote trail]

because if you were rotating around the x-axis you can't rotate around the line x = smth

[bold pond]

dude

just poretend

for a second

[remote trail]

you would be rotating around y = 1 or y = 3

[bold pond]

it would work with a different equation

[remote trail]

I can't pretend if it doesn't make sense

[bold pond]

you have no imagination that is why

[remote trail]

because you wouldn't have to worry about 1-x or x-1

[bold pond]

use a different example

yes that makes a different

difference

[remote trail]

why don't you give an example question

[bold pond]

if you cant see that

i shouldn't have to

[remote trail]

1-x and x-1 are different

[bold pond]

you are just stuck

[remote trail]

but clearly for this question

you would use x-1

[bold pond]

WHY

ok

for your sake

if the equation was setup

as a washer method

[remote trail]

because the line x = 1 shifts the graph one to the right? So to find the correct radius

[bold pond]

rotating on the X AXIS

i dont care

[remote trail]

you have to use (x-1) as the radius

[bold pond]

i a m setting up the quetsion for uy

like u asked me

[remote trail]

alright

[bold pond]

so you have a quesion rotaing on the X - AXIS

at y = 3

now

when you go back to put it into the equation

how are you going to do that

would it be x - 3 or 3 - x?

since it is the X - AXIS

not the Y

[astral flower]

You rotate on the Y axis

[remote trail]

how can you rotate the line x = 3 around the x-axis??????

[bold pond]

y = 3

typo

[remote trail]

look at this line

[bold pond]

do you understand what i am syaying tho

[astral flower]

Imagine spinning your hand on each axis and filling the shape

[bold pond]

you guys are totally missing the point

[remote trail]

^

[bold pond]

all i am asking

is

if you rotate on x - axis lets say y = 3

ok Y = 3!

get it:

would you insert it back in

as 3 - x or x -3 to f(x)

if you are using washer method or disk or whatever doesnt matter

do you get it?

[astral flower]

Integrated with respect to x will be a lot more difficult.

[bold pond]

wow

u are stuck

ok

we have to move on from this

u just are not folowing

its very simple

[remote trail]

I think you are either not explaining what you mean well enough or you're misunderstanding something

[bold pond]

you guys are just not understanding

you cannot think outside of this equation

it has nothign to do with this equation

ok

first of all

[remote trail]

you're asking for help right

[bold pond]

just forget about this equation right now

[remote trail]

how do you expect to get help

[bold pond]

then you can think about waht i am asking you

[astral flower]

If you integrate from x=1 to x=3, you’ll have an unnecessary area included in the final shape

[remote trail]

we are trying to think about what you mean but how you are explaining it atm doesn't make sense.

[bold pond]

well it wont make sense at all because like i just said 4 times

[remote trail]

x-3

[bold pond]

it is not related to this equation

[astral flower]

Aka the box “sliver” under the shape you’re trying to find the area of.

[bold pond]

so if u want to keep referring to this equation you will never be able to answer the question

[remote trail]

this is why you would use x-3

[bold pond]

what are you doing?

you are clearly just trolling at this point

[remote trail]

you can see that the line x=3 is transformed to the central axis in this case

[bold pond]

that is for this equation yes

?

or no

[remote trail]

I'm literally answering your question?

you would use x-3

[astral flower]

This is not trolling. I understand this can frustrating.

[bold pond]

or are you still using the example i had a typo on

@remote trail

listen

[astral flower]

Choosing the right method is important

[remote trail]

y would be the same

[bold pond]

i told you it was a typo

i am talking about y = 3

not x = 3

Y = 3

[astral flower]

If you don’t choose the correct method, you make it much harder for yourself.

[bold pond]

do you understand Y = 3?

is rotated on x axis

like i keep saying

[remote trail]

this is y=3

[bold pond]

ok

[remote trail]

you transform y-3

[bold pond]

yes good it is

well

i did hw the other day and we did 3 - yh

3 - y

and it was correct

so why are you saying it is y - 3?

[remote trail]

it is question dependent

[bold pond]

how

[remote trail]

on whether the function is above or below the axis of rotation

[bold pond]

no that would just make it addition or subtraction

[remote trail]

if the axis of rotation is above the function you would use 3-y

and for the counterexample you would use y-3

you're trying to find the radius from the axis of rotation

[bold pond]

if y = -3 then you add ity

add it

[remote trail]

it's not that simple

[bold pond]

if y = +3 you subtract it

[remote trail]

no

it depends on the location of your function from the axis of rotation

[bold pond]

ok this is my question you are finally understand what i am asking then

so please explain, what it is

you are doing a good job so fart

far*

ok but you are missing something

because

wait maybe

[bold pond]

so if we rotate at y = 3 vs y = 4

you are saying what exactly

since one is above and one below

[remote trail]

you would have to invert the function first if the function is y = f(x) but yes

let me draw out the cylinders for you to show

[bold pond]

nonono

u dont need to

i am just asking about this

plz trust me u dont need to

i am just asking

look

if you rotate at y = 4 then are you saying you make it y +4 ? 4 -y? or y - 4 ?

if you rotate at y = 3 then are you saying you make it y + 3 ? 3 - y? or y - 3?

cause you said it depends whether it is above or below

so i want to know which is which that is ALL Ii want to know right now

that is my simple only question here

[remote trail]

the height for cylinder at y = 4 will be 4-y and at y = 3 y-3

[bold pond]

what did u draw ?

[remote trail]

the red squares

[bold pond]

...

r u being serious

what does that represent

i know u drew red squares

i just sent that picture without the square i mean what are those supposed to be

representing

lol r u trolling me

[remote trail]

those represent the cylinders you would be finding the area for in the shell method-

[bold pond]

i dont think that's right

why is one on the outside

it looks like an inner radius

on top

and the other one looks fine

but u drew them for 2 different equastions no?

[remote trail]

because you're rotating around a line above and below. There's two separate different rotations

[bold pond]

i am literally just asking u

yes or no

y would u only draw the top one inner radius then

[remote trail]

yes. They are drawn for two different lines.

[bold pond]

that is not even in the area of the curves

[remote trail]

I don't understand why you keep mentioning inner and outer radius, can you explain your methodology to me?

[remote trail]

you made the lines-

[bold pond]

you drew the cylindar outside of the area

[remote trail]

which cylinder

the top or the bottom one.

[bold pond]

the top one

[remote trail]

ok then

how would you rotate it

explain to me

[bold pond]

that is your rotation? it is not even making sense

[remote trail]

you rotate around the line y = 4

you made the question

[bold pond]

how do i draw on the picture

[remote trail]

you rotate the cylinder I highlighted around y = 4

[bold pond]

is there a way to draw on it in discord

i can highlight the area for you

[remote trail]

I used screenshot

[bold pond]

ok hold on

[remote trail]

this is the rotation I was talking about

[bold pond]

do you get it?

[remote trail]

that also works ig

that doesn't make my region incorrect

it really depends on which region the question asks for

[bold pond]

im saying it is the inner radius

[remote trail]

but also the rotation around y = 4 would be an infinite rotation

unless you set a bounds

[bold pond]

ok

so the bound is 5

[remote trail]

I would argue my rotation drawn makes more sense since it is an enclosed area but regardless both are correct and dependent on what the question asks

[remote trail]

for this also you would probably use washer method

[bold pond]

ok but my question was only just how do you know when to wirte it back as 1 - y vs y - 1

so if the rotation is above then you write it as 1 - y

and if it is below you just do y - 1..

?

did i get that right

[remote trail]

if you look at the region here

the radius of the line from the x will be x-1

[bold pond]

why

wouldn't you draw it from left to right?

not top to bottom

[remote trail]

I don't have a mouse but here

[bold pond]

we are rotating on the y axis

lol u dont have a mouse

[remote trail]

the radius is x-1

[bold pond]

how are u even drawing then

[remote trail]

trackpad

see that the radius is x-1?

[bold pond]

well it depends what direction are you going

[remote trail]

the height is (4x-x^2)-3

[bold pond]

top to bottom?

well in this case

right to left?

left to right?

[remote trail]

it says rotate around the line x=1

???

[bold pond]

omg

so are u going right to left

[remote trail]

sigh

[bold pond]

or left to right

[remote trail]

the line is to the left of the area

so you rotate the area around the line to the left...

I'll just leave my working there

see if you can figure out what I'm doing at least.

[bold pond]

ok

so that is the inner radius

yea

[remote trail]

if it doesn't make sense I would encourage you to go look over the method

[bold pond]

the outter radius is x -2 then

f(x) - 2

right?

[bold pond]

you are just drawing weird stuff

it is not correct

not totally incorrect but not correct either

so not helpful

but thank you for trying

[remote trail]

nono

I'm pretty sure the method is correct

[bold pond]

if you knew how to drawe it better it would prob make more sense you know what you are tlaking abt i can tell

ya just your drawing is confusing

and i am saying the wrong thing

cause in shell you have radius and height

so i keep saying outer radius and inner radius

but you should be able to understand what i am talkikng abt tbh

maybe you are afraid to be wrong it's ok

when i kept saying outer radius i mean height f(x) my bad

[remote trail]

I'm not afraid to be wrong

[bold pond]

anyways you seem to know your stuff

[remote trail]

the height is outer-inner radius

[bold pond]

but you should get better at helping people

[remote trail]

or (4x-x^2)-3

[bold pond]

try to understand what they are saying better

[remote trail]

and the radius is x-1

[bold pond]

i think you may just be using chat gpt or something

anyways

i am off to bed

[remote trail]

sure. But I want you to also consider the converse

[bold pond]

i hav to go to sleep

[remote trail]

and try to understand how to explain better

[bold pond]

i can tell maybe you are using chat gpt or something

it is a new trend in this server

.close

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[stray prairie]

if 3x+5-7 as swaped is 3x-5x+7x and then 2-12 is equal to 10?

[whole spear]

That's a lot of attitude for someone needing help with highschool math

It is cute though, attitude matches the level

[cedar kiln]

@stray prairie Has your question been resolved?

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[astral flower]

I wish my highschool taught Calc II.

[void sand]

you've opened this channel, but haven't asked a question

do you have a question to ask?

[astral flower]

it was by accident sry

[void sand]

no worries

[astral flower]

.close

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[void sand]

just close it

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[vale garden]

My professor in his lecture notes would like me to recall a fact from a previous class, I hardly remember it. Could someone explain to me why a line can be parameterized in the way described by the first paragraph please?

Thank you!

[vale garden]

It doesn't even feel like a line to me, since x(t) seems to take the form of an x and y coordinate

[autumn fox]

The intuition is that t ∈ [0, 1] is a 'weight' between the points P and Q, where at t = 0 we have that x(t) = (1 - 0)P + 0Q = P and then it increases in the horizontal coordinate and the vertical coordinate at the same speed towards Q (and therefore in a straight line) as t goes from 0 to 1, where we at t = 1 then have x(t) = (1 - 1)P + 1Q = Q.

[vale garden]

Hm, but t could also be 0.5, uneven weights, what would happen in that case?

[autumn fox]

Then you would be at the midpoint of the line between P and Q.

[vale garden]

Oh!!!

I get it

[autumn fox]

[vale garden]

Wait that makes so much sense.

Tysm!

I have another question further down my lecture notes, I'm not sure if I should start a new chat or not, but it is somewhat (not really) related.

I don't get the second to last line, when it says ccw pi/3 radian becomes cos(5pi/6), sin(5pi/6)

That doesn't really make that much sense to me. P is the shape we're rotating and it's composed of line segments.

And I guess we are trying to contruct the new rotated basis vectors by rotating the standard bases.

[cedar kiln]

@vale garden Has your question been resolved?

[vale garden]

<@&286206848099549185>

I get it.

.close

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[errant holly]

oops

[cedar kiln]

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[errant holly]

.close

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[crimson sedge]

Hii

[crimson sedge]

Ok so not really a question abt an specific problem

But I'm struggling with math well not really struggling but I keep getting bs in my tests

Might be lack of studying but idk how to study math

[bleak viper]

practice

[crimson sedge]

Ik that

But what should I practice

[bleak viper]

questions

solve more questions

[crimson sedge]

Cause js practicing what they give me won't help me that much

[slim birch]

how r u practicing exactly...

[crimson sedge]

I must research the logical things really deep to even dare go say I grasp this math subject I'm in algebra 2

[crimson sedge]

Well I do problems from the class

I watch random yt vids:3

[bleak viper]

are there more problems in your book?

[crimson sedge]

Eh barely

They're either extremely easy or not helpful at all

[bleak viper]

then look online for more problems

[crimson sedge]

I got this one book that is more complex although it has an specific style of solving problems it's not like just do whatever u find best

It has some sort of structured pattern haven't really looked at that one but I will if it will make it better

Well I've never truly had to study for survival like I am now

Because K to 8th it was extremely easy

9th I had a course in which I could rely to learn stuff

[slim birch]

do the pset and if u dont understand something about a question write down what u dont understand/what u need

[crimson sedge]

9th for geometry I had to learn from a course aswell

[slim birch]

then finish the pset as best as u can then lookup what u dont know

[crimson sedge]

But now I don't have a course to rely on so it's only me and the books

[crimson sedge]

May I ask how u study or used to for math

[slim birch]

thats still how i study for it

go to lecture and write down stuff i dont get to ask or google myself later

[slim birch]

then do the pset using this ^

[crimson sedge]

Ok we only get 1 day of full review

Our teacher usually just let's us study or follow her with some problems

But I work better independently

Any tips on being accurate while being fast?

[slim birch]

1 day of full review as in 1 full day ur not learning anything new?

[crimson sedge]

Yerp

[slim birch]

oh well thats fine

ur supposed to do extra practice after class on what u learned everyday too

every week i do some practice problems from the week prior to keep the knowledge fresh

[crimson sedge]

Ah I see

[slim birch]

if u leave it all on last day u will be cramming

[crimson sedge]

Yeah ion like that

I've only had to do that once I got full score but ik it ain't the way

That one time was because my bio grade mattered more than my math grade

[slim birch]

its pretty common to have the last lecture be on a friday here and midterm exams testing that same material on the monday after

so yea its even more important to chew a little bit at a time

[crimson sedge]

I feel overwhelmed with the hw assignments

Cause it's like 1 daily and It doesn't really tell me much on how to solve each thing haha

[slim birch]

can u show them?

[crimson sedge]

Uh currently no cause I'm out but later yes

Or maybe I'm just not viewing them from the right perspective like I can get them correctly but they don't tell me much in my perspective

[slim birch]

hm well cant help u too much w/o a concrete example

[cedar kiln]

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[torpid adder]

A preorder relation p on the class of all funcitons with domain list X1, ..., Xn is defined as. p(f, g) =

[torpid adder]

The exercise is: Demonstrate that if there is at least one smallest function, then a smallest function exists that is both injective and surjective.

what I dont get is isnt the funciton f: X1, ..., Xn -> (X1, ..., Xn) where f(x_1, ..., x_n) =(x_1, ..., x_n) a smallest funciton

that is both injective and surjective

so why "if there is at least one smallest function"

[vagrant pewter]

What are Y and Z?

I guess f outputs to Y and g to Z

I mean it would have to

[torpid adder]

yeah, the notes arent very clear

is my solution correct or am i missing something, I dont understand why there has to be the "if there is..."

[vagrant pewter]

I guess because there’s probably some weirdness that could happen if you had trivial domains or something

I’m not super sure about this without some more clarification

Like I’m not sure how the function class is actually defined

When you say domain list like do the functions all have a domain being the Cartesian product?

[torpid adder]

from what i understand, the notes are really bad, the class of functions is just any function f: X1, X2, ..., Xn -> Y

where Y is any set

but the domain is not cartesian product, just many inputs

[vagrant pewter]

That is Cartesian product

[torpid adder]

wdym

i mean yeah, but it is defined to take inputs like that, not their cartesian product

[vagrant pewter]

Like the input to f is the the tuple (x1, x2, …, xn) where the xis are in Xi

[torpid adder]

but either way the function f: X1, ..., Xn -> (X1, ..., Xn) where f(x_1, ..., x_n) =(x_1, ..., x_n) clearly satisfies p(f, g) for any function g

so...

[vagrant pewter]

And Y and Z are arbitrary?

[torpid adder]

yeah

[vagrant pewter]

A function would be smallest if it is related to all other functions right?

[torpid adder]

yes

i can also send u the full notes if u would like, they are like one page in total

and pretty poorly written

[vagrant pewter]

I’m pretty sure that your function just the identity function

So surely for any function there’s is a phi, namely g itself

So yeah it should be the smallest

I think the reason you have the premise with the if is in case all the Xs were empty or some weirdness like that

Or maybe the class could be empty

[torpid adder]

oh

[vagrant pewter]

Something like that i think is probably what’s going on

But yeah that identity function should be all you need to prove the existential quantifier

Looks good to me

[torpid adder]

i mean if the Xs are all empty wouldnt like the function f: empty set -> empty set be smallest

its also bijective

[vagrant pewter]

I mean i guess the empty function does exist so technically that’s true

I have no clue then why the premise is there maybe it’s redundant or maybe we’re missing something

[cedar kiln]

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[spice phoenix]

Im stuck on epsilon/(x-1)

What should I do from here?

[acoustic pivot]

for this kind of limit proof a trick is to instead give a sort of minimum for what delta can be, so your delta would be of the form

$\delta=\min{\text{constant}, \text{thing dependent on }\epsilon}$

or wiat is it minimum

[spice phoenix]

I think it’s minimum

[acoustic pivot]

well something along those lines. Knowing something delta is at most some constant helps you bound one of the factors

[wraith dagger]

WHAT?

[spice phoenix]

Limit proofs is kind of confusing to me, sorry, but I have (x-2)(x-1) < epsilon. Should I be getting the minimum of those? So x=1?

I got to the point where I’m dividing epsilon by (x-1), so that I have (x-2) alone

[acoustic pivot]

so now the idea is to choose a constant that delta can't exceed so that you can put an upperbound on 1/|x-1|. Does that make sense? Then your delta would look like min(constant, epsilon/upperbound)

[cedar kiln]

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[cedar kiln]

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[old fjord]

Hey can someone help with B.

i just cant remember how to flip (x-2)/x

[humble marsh]

flip?

[old fjord]

like how 1/x is x^-1

[humble marsh]

oh

(x - 2)/x is (x - 2) * 1/x, right?

[old fjord]

yeah

[humble marsh]

so (x - 2)/x is (x - 2) * x^-1

[old fjord]

so (x-2) * x^-1

ah

ok

[humble marsh]

yep

[old fjord]

thank you

[humble marsh]

[\frac{\mathrm{stuff}}{\mathrm{other\ stuff}} = (\mathrm{stuff}) \cdot (\mathrm{other\ stuff})^{-1}]

[wraith dagger]

Invariance

[old fjord]

ahh good explanation

just saved that as mathstuff.pdf

haha

[humble marsh]

lol

[old fjord]

.close

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[viral fox]

Hi, I have a linear-algebra problem. How was the red-underlined part formed?

[steep lark]

Can you provide the full image?

[viral fox]

I believe it's a full image, you may please clock to see it expanded please

[crimson delta]

if you foil then you get uv^tuv^t(...)

in the middle the v^tu is a number

so you can move it to the front

[steep lark]

That underlined part is formed as a result of factoring out v^tu from the terms inside the expansion of second term.

[crimson delta]

it should say (1-v^tu)^-1 and not (I-v^tu)^-1

those are numbers, not matrices

[viral fox]

Yes, a little confusing. I suppose the 1 was due to typo, or scan issue. I still don't understand why v^tu should be number?!?! if v, and u are 3x3 matrices, how could v^tu become a number?

[crimson delta]

u and v are vectors

v^tu is a dot product

[viral fox]

DANG!

Ouch, so now it make sense, so in case the yare matrices, then that v^tu would be a mistake I suppose?

[crimson delta]

well the whole thing doesnt work if they are matrices

although iirc there is at least a similar result

[viral fox]

Thank you, much appreciate your time sir. Thank you @steep lark too.
Math bless us all.

.close

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[azure glade]

yo i got a question

[azure glade]

so if i wannt write this inequality using interval notation

x < 3

would it be this

(-inf , 3)

[bleak viper]

yes

[azure glade]

and can you also write it like this
(-inf, 2]

[bleak viper]

no

[crimson sedge]

isnt it ]-inf,3[ ?

[azure glade]

huh

why are ur brackets like that

[crimson sedge]

maybe we dont have the same notations in differents country

[azure glade]

im sure this bracket notation would be the same everywhere

[crimson sedge]

why parenthesis then ?

[mental trail]

]-inf, 3[ and (-inf, 3) mean the same thing

[crimson sedge]

ah okay

[mental trail]

different notations

whatever the notation convention, you'll find that closed brackets [a,b] are common in both

so parentheses, as well as brackets facing outwards, are the same

[azure glade]

never seen that before, i am quite new to this though

is it an american thing

[bleak viper]

makes sense

facing outward as in not including those

[crimson sedge]

yes

[mental trail]

so here the problem is that you're reasoning in terms of integers

you're thinking that x < 3 is the same as x <= 2

but that's only true for integers

[crimson sedge]

so therr is no stuff with )3,2) ?

[azure glade]

oh shit yea

[mental trail]

there are real numbers between 2 and 3

[crimson sedge]

for example

[azure glade]

yea

[mental trail]

parentheses facing outwards don't exist in any convention

[crimson sedge]

i mean u dont write like this ?

[mental trail]

it's either parentheses, or brackets facing outwards

[azure glade]

ive been working with these inequalities so much in stats so i was only focusing on discrete valuies

[crimson sedge]

ok👍

[azure glade]

forgot about continuous ones

[mental trail]

[a,b] <-> [a,b]
[a,b) <-> [a,b[
(a,b] <-> ]a,b]
(a,b) <-> ]a,b[

[cedar kiln]

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[blazing zephyr]

let I be the centeroid of triangle ABC, prove that AIB=90+C/2

[fathom carbon]

anybody have any idea how to solve this? I'm stuck and it will be greatly appreciated

[blazing zephyr]

!occupied

[cedar kiln]

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).

[blazing zephyr]

if it matters, there is (incenter lemma) next to it but i dont see anything on google thats not incircle excircle lemma

[loud gyro]

pretty sure I has to be the incenter and not centroid

[blazing zephyr]

ok wierd then

[loud gyro]

its easy to prove AIB = 90 + C/2 for I being the incenter

[blazing zephyr]

yea ik its "incenter lemma" but still google says its the centeroid

how fo you prove it then

[loud gyro]

prove what? AIB when I is incenter?

[blazing zephyr]

yea

[loud gyro]

Just measure the angles of the triangle AIB

I being incenter means AI, BI and CI are angle bisectors

[blazing zephyr]

so like, AIB=180-A/2-B/2=90+(180-A-B)/2=90-C/2

oh

[blazing zephyr]

it also wants to prove EA=EB=EI

but this looks so wrong

what lmao im pretty sure thsts dtraught up not true??

[loud gyro]

what is E?

[blazing zephyr]

i mean F, in the image its a continuation of AI

[loud gyro]

translate says incenter

[blazing zephyr]

translate doesent make sense, google shows you centeroid

i asked my friend and they said its also centeroid so :shrug:

[loud gyro]

Nahh, I just wanted to verify, coz translate got that wrong

so if the rest of the problem is correct

it also asks D is center point of BIC, so it makes sense if it means angle bisector

[blazing zephyr]

so what is D lmao

[loud gyro]

AI and BC intersection

[blazing zephyr]

in a single sentence it contradicts itself 3 times

[blazing zephyr]

but it claims DB=DC=DI

wait am i stupid

[loud gyro]

Is it part of some bigger problem? Coz this makes no sense

[blazing zephyr]

no thats the whole problem

[loud gyro]

then this means D is circumcenter of BIC, which is true if BIC is 90 deg

[blazing zephyr]

my best guess is the wording is just fucked up and they meant like "if DB=DC=DI, prove that D is the centerpoont of BIC"

[loud gyro]

i'd suggest skip this one

[blazing zephyr]

sure

alright thank you

.close

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[daring barn]

what they did in last step to get the value of r

[mental trail]

factor 1-r^3 (difference of cubes)

then cancel 1-r and 27

multiply everything out to have no denominators

-> at most a quadratic

(in fact it's actually a linear equation on r)

[daring barn]

i got quad eqn

but result is something diff

the eqn i get is 3r^2-3r=1/19

[cedar kiln]

@daring barn Has your question been resolved?

[daring barn]

<@&286206848099549185>

[cedar kiln]

@daring barn Has your question been resolved?

[daring barn]

<@&286206848099549185>

[past wave]

I think they just solved for r

1-r^3 = (1-r)(1+r+r^2)

r2 + r + 1 = 19(r2 - 2r + 1)

r2 is r^2

@daring barn

[mystic flame]

What's the problem?

[cedar kiln]

@daring barn Has your question been resolved?

[cedar kiln]

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[static bramble]

I need help finding The second derivative of this function! I am stuck

[fallen moat]

$\frac{1}{\ln{(a)}(x+1)}$ this?

[wraith dagger]

Biscuity

[fallen moat]

where a is a constant, right?

[static bramble]

Yes

[fallen moat]

so, basically you treat ln(a) as a constant, lets say

$\frac1{\ln (a)}\cdot\frac1{x+1}$

[wraith dagger]

Biscuity

[static bramble]

Yes

What I could do is take it back to the original form?

The right side is ln(x+1)

[fallen moat]

so, we have $\dv{}{x}\left(\frac1{\ln (a) (x+1)}\right)=\frac{1}{\ln (a)}\cdot \dv{}{x}\left(\frac{1}{x+1}\right)$

[wraith dagger]

Biscuity

[fallen moat]

sorry typo

[static bramble]

That should absolutly be it !!

The right hand side will be 1/(x+1)^2

Right ?

[fallen moat]

careful

,w d(1/(x+1))/dx

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by biscuityxd

[static bramble]

Ye minus

-1*(x+1)^-2

So you don’t derivate the first term?

How come ?

[fallen moat]

since it's a constant

[static bramble]

Well if it’s a constant don’t you remove it ?

Derivative of 1 is 0

[fallen moat]

$\dv{cf(x)}{x}=c\dv{f(x)}{x}$

[wraith dagger]

Biscuity

[static bramble]

1 is constant ?

[fallen moat]

it's multiplication, not addition

[static bramble]

Oh right

Thanks then!!

[fallen moat]

Cheers!

[static bramble]

Appreciate it!🤩

.free

[fallen moat]

it's .close

[static bramble]

Hahaha

Okey thanks

.close

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[bronze shore]

Hello i need help for this :

Let g : R² -> R be a C^1 (R²) function; define

h : R² -> R

(x,y) I-> g(y,x)

Calculate the partial derivatives of h as a function of those of g.

[dreamy void]

Is there all there is?

I don't see a relation between h and g

[bronze shore]

Yes, that's why I'm blocked.

I don't have any more information

[dreamy void]

I see

You define a function h

that maps to g(y,x)

h(x,y) = g(y,x)

wtf

So basically your partials will be in terms of g

[bronze shore]

There may be a specific property in this case in my course, but I haven't found it.

[dreamy void]

for example if g(x,y) = x+y² then h(x,y) = g(y,x) = y+x²

[bronze shore]

We're not being asked to define a function ...

[cerulean sail]

Also if I rephrase the question a tiny bit:
you have $h = g(u, v)$, and $u = y, v = x$, so

[wraith dagger]

@cerulean sail

[dreamy void]

[cerulean sail]

That was an example of how it works for a specific function, to give you intuition of how you'd go about it-

[bronze shore]

The problem is stating it in itself: calculating a partial derivative is not complicated.

[cerulean sail]

Well then, do you know how to find the partial derivatives of this function, in terms of u, v, x and y then?

[calm sierra]

im inclined to state the general chain rule (the one with matrices) even if its overkill here

[dreamy void]

have you seen the example?

[bronze shore]

[dreamy void]

bet

[bronze shore]

mb

[bronze shore]

∂g/∂y(x,y) = ∂h/∂x(y,x) , ∂g/∂x(x,y) = ∂h/∂y(y,x)

[calm sierra]

actually ill suggest it for real

[bronze shore]

I'll have the correction tomorrow

[calm sierra]

do you know the general chain rule?

[bronze shore]

yes, I saw this a long time ago

[calm sierra]

for functions R^n->R^m?

[bronze shore]

nop

[calm sierra]

do you know the jacobian?

[bronze shore]

nop

[calm sierra]

did you take vector calc?

[bronze shore]

Yes, the basics

[calm sierra]

ok then jacobian should be a familiar idea

if not, https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant#Definition

here ill use Df for the jacobian of f

general chain rule: if $f:\bR^n\to\bR^m$ and $g:\bR^m\to\bR^k$ are $C^1$ then so is $h=g\circ f$, and
[Dh=(Dg\circ f)\cdot Df]
with $\cdot$ being matrix multiplication

[wraith dagger]

RokettoJanpu

[dreamy void]

honestly, charbits approach was the best

[calm sierra]

your problem is notationally convenient. find f so that h=gof, then use the chain rule

[calm sierra]

its overkill for this problem but its good to know in general instead of having 100 specialized chain rules

[bronze shore]

Ok thanks for these answers

[cedar kiln]

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[jade compass]

for all x and for all y if x is greater than or equal to 0 and y is greater than or equal to zero then xy is greater than or equal to zero?

[dusk finch]

Yes, that's what the formula says

[jade compass]

sick

[dusk finch]

are you supposed to prove it? Or just interpret it

[jade compass]

interpret

i was gonna ask why there are two for all

but then i just read it and it clicked

thank you tho

[jade compass]

havent gone into proofs yet

not thrilled for it but we;ll see

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[austere gyro]

I found the marginal probabilities using a contingency table. now what? how do I present it like as a answer?

[empty locust]

in method of variation of parameters can we choose any y1 and y2 value?

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[bitter musk]

Is it possible tho show that fsubscriptn of x = x^n E C subscript b,uc(D) where delta in (0,1) and D = (-delta, delta)

[bitter musk]

Basically want to show that given delta in (0,1) and D = (-delta, +delta), and given Fn(x) = x^n, i want to show that x^n is in the space of bounded and unifromly convegent funcrtions on D

Any ideas on how to do this?

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[crimson sedge]

How can I find the domain?

[dreamy void]

the argument of the square root must be non-negative

[crimson sedge]

I know

let me show a pic

The -π/4 is closed

I just do't know how to fit the |x|

mmmmmmmm I think I know what to do

it's from <-π/2; -π/4] first

and then

[π/4; π/2>

[crimson sedge]

it's 3π/4

not 5π/4

soo A

I think

yeah

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[lament spindle]

Can someone help dumb down what this means cause I cant understand it

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[lament spindle]

<@&286206848099549185>

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@lament spindle Has your question been resolved?

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@lament spindle Has your question been resolved?

[vagrant pewter]

It’s related to the free variable thing from yesterday

If the two statements evaluate to the same truth value no matter the specific predicates then they are called logically equivalent

So for example you have a sentence A or B

This is logically equivalent to the sentence D or E

A better example is the pair of statements “A” and “not (not A)”

These two statements are logically equivalent since you can insert any sentence/predicate in place of A and the two statements will evaluate to the same truth value

Take a look at the Wikipedia page for more examples of equivalent statements

https://en.m.wikipedia.org/wiki/Logical_equivalence

[lament spindle]

ok thank you

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[half swift]

Can somebody explain this to me?

Why wouldn't we set it up as an inequality that has to be more than 0

Since isn't it the denominator of a fraction? \

Also how does the square root work here?

[sly abyss]

because

if we have any number

squared

it will always be positive right

[half swift]

Yeah

[sly abyss]

so any other number added

with always be positive

therefore a positive plus a positive will never be 0

[half swift]

ohhhhhh

[sly abyss]

x^2+3

does not have any real solutions

[clear umbra]

denominators can be negative, so inequality with >0 doesnt make sense

it just cannot be 0

[half swift]

what if the number wasn't add and it was subtract?

do I still set it up the same way?

[sly abyss]

you would set yp

x^2-3=0

x^2=3

and solve for x

you want to find where the denomiator is equal to 0

[half swift]

wait, so if I solve for x wouldn't it be the same as the other time with + or - sqrt3

[sly abyss]

?

[half swift]

like if I set it up like this to solve for x

would domain be like

(-inf, sqrt3) U (sqrt3, inf) ??

[sly abyss]

no

(-inf, -sqrt3) U (-sqrt3,sqrt3) U (sqrt3, inf)

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[half swift]

ohh I see thank you

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[forest scarab]

show that if
cos 𝑥 − sin 𝑥 = 1, then cos^2024(x) + sin^2024(x )= 1

[forest scarab]

How can i solve this

[dire geode]

Solve for cos or sin and plug into the right equation?

[forest scarab]

but they have power 2024