help-42

[full quarry]

wat does a 9 point circle mean?

[frozen rampart]

@kindred folio comeon youre here arent you hs

[frozen rampart]

circle of the three midpoints

Of a triangle

And the perpendicular feet

And the midpoints of H to each side of the triangle

Naijie would say 'trivial'

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[frozen rampart]

.reopen

[calm coral]

✅

[frozen rampart]

k phew it works

weird

[frozen rampart]

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[frozen rampart]

.reopen

[calm coral]

✅

[frozen rampart]

the qn btw

[tough prism]

have u tried asking in evan chen discord

they are quite the geo gods over there

[frozen rampart]

ah

its really inactive

i think im almost done btw

[tough prism]

if u post the problem im sure someone will help u

[frozen rampart]

im almost done ty tho

all i have to 证 is MDQ similar to XHA

And were done

[tough prism]

ya u orz i can never understand the geo problems with like more than 3 shapes lmao

[frozen rampart]

nah is ok youll get better

Okay am done

Very good

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[viral badge]

hi i have another question

for this one, i don't really understand the function

(2(n+1)(3n+2))/ n^2

how exactly should/could I be using this function to solve this problem?

any help is appreciated !

[nimble harbor]

so

they give u the approximation of the integral with n subintervals

in a riemann sum

how many intervals do u want to have?

is that ado pfp

[viral badge]

umm i dont really know if im being honest

[viral badge]

yes

[nimble harbor]

infinite subintervals, rught?

[viral badge]

yes

[nimble harbor]

so

take limit as ${n \to \infty}$ and u should get the answer

[potent lotus]

k

[viral badge]

but isnt lim n --> infinity like

oh i need to expand first

wait is it just 6

[nimble harbor]

yes

[viral badge]

oh

so is it like that regardless of the definite interval

[nimble harbor]

yes

lim n-> infity is regardless of the interval

[viral badge]

oh okay

also is this equation

is this in a type of format i should know

[nimble harbor]

nah

it is some arbritary estimation they give u

[viral badge]

oh okay

ok thank u so much !!

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[blazing coyote]

Let $S$ be a set . Prove that the law of composition $ab=a$ for all $a$ and $b$ in $S$ is associative. For which sets does this law have an inverse

[potent lotus]

What a wonderful world !

[blazing coyote]

so I have to show (ab)c=a(bc)

[plucky walrus]

That's trivially true.

(ab) * c = a * b * c = (b * c) * a

[blazing coyote]

this feels off

ac ≠ab

[stable cobalt]

you should apply the law of composition one more time

that should do it

[blazing coyote]

write

*right

just realised

cool

now when does it have an inverse

hmm

ab=a

We start by post multiplying across by b^{-1}a^{-1}

so we have 1= ab^{-1} a^{-1}. \implies 1= ab^{-1} \implies a=1

so a must be 1 for it to have an inverse?

[stable cobalt]

not necessarily
the law has an inverse if ab=ba=1 where 1 would be the identity element
however by defining the law this way, right inverses are not possible when there are 2 or more distinct elements as ab=a and ba=b

[blazing coyote]

ah yes

I forgot that both right inverses and left inverses must exist

Thanks

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[edgy leaf]

I have two questions:

1. can a group consist of 3 arguments? a set and two operations so (S, *, +) for example
2. if yes then can a vector space be represented as a group then with (V, +, *F) where *F is just my notation for scalar multiplying with a field F

[tawdry flame]

No

[glass heart]

a group is just a set with one operation

[edgy leaf]

i see

[tawdry flame]

But i mean you could say

Something like

Vector space with addition is abelian group

[edgy leaf]

yes

why cant we say that for scalar multiplication

[rapid cliff]

yes. all rings (R,+,*) are abelian groups (R,+) endowed with a multiplicative operation *

[tawdry flame]

But it isnt a group

Its a ring

[rapid cliff]

and a square isn't a rectangle?

[edgy leaf]

i havent gotten to rings yet

[tawdry flame]

And a group is onlky if you restrict your attention to addition

[tawdry flame]

Well

[safe oriole]

but she wants to know about groups

[tawdry flame]

Because this clearly isnt a group

[edgy leaf]

okay what about (V, *F) using my notation for the *F

oh is it because *F isnt defined as an operation

[tawdry flame]

Yes

[rapid cliff]

for any ring R, it holds the axioms of a group and is therefore a group

[edgy leaf]

so when we use (V,X), we always mean X is an operation acting on set V and not on any other set

[safe oriole]

really?

[tawdry flame]

No

A group has one operation

Ring has two

It is a group with addition alone

But that removes part of structure

[safe oriole]

clearly

[rapid cliff]

Consider the ring of real numbers under canonical addition and multiplication, (R,+,*).

• R has additive identity 0
• R is closed under addition
• R has additive associativity: forall a,b,c in R. (a + b) + c = a + (b + c)
• R has additive inverses: forall r in R. exists -r in R. r + (-r) = 0
• R has additive commitativity: forall a,b in R. a + b = b + a
  R is therefore an abelian group under addition.

[tawdry flame]

Under addition

Not as a ring

RIng is a triple (R,+,*)

Group is a pair (G,*)

[rapid cliff]

Technically, the ring R is the tuple (R,+,-,*,/,1,0)

[tawdry flame]

/?

[rapid cliff]

because inverses are a function on the set

[tawdry flame]

What

Inverses are function?

That is redundant

[rapid cliff]

Yes. It's an involution

[tawdry flame]

You dont need 1, 0, /

Because thats evident from the operation

[rapid cliff]

there may be many elements of the set satisfying the identity axiom, we need to specify which we want to identify as the unit

[tawdry flame]

Or -

[tawdry flame]

No?

Are you crazy?

Only one element can

[rapid cliff]

I think this is only true in abelian groups?

e1 = e1 + e2 = e2 + e1 = e2
             ^
       commutativity required here

[rapid cliff]

Let me lessen my "yes" earlier. Any ring does have a group embedded in it (see category theory).
I'm now too scared to comment if it is a group, or if it just has group-ness.
Like if a square is a rectangle, or if it just has rectangle-ness

[edgy leaf]

okay thank you!

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[rapid cliff]

Yes, you're right. I'm being crazy here. I was thinking of left-groups.

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[crimson swift]

hi, is there a video explaining how to check derivative and integral using fx991 es plus?

[unreal isle]

that's the silver one correct

[crimson swift]

no, the black one with green buttons

[main marlin]

a quick google search gives this

https://www.youtube.com/watch?v=-sn8po-xV4s

[unreal isle]

in either case, you wanna input your polynomial, use a placeholder value of x, and crosscheck by inputting that x in your derivative

same logic applies w limits in an integral

[crimson swift]

well i found a video of another calculator and its output is a table of
x fx and gx
fx being the calculator and gx being what i think the derivative is

but theres no x button on my calc

[unreal isle]

alpha right bracket

that's the standard

[unreal isle]

that's the ex classwiz

[crimson swift]

thats whats being used in the vid, not what i have

[unreal isle]

this is yours from what i know, correct?

[crimson swift]

ya

[unreal isle]

look at the top left, shift integral button gives d/dx

and alpha right bracket gives x

[crimson swift]

that helps checking one point, but not a table

[unreal isle]

you want a table?

[crimson swift]

ya

[crimson swift]

.

[unreal isle]

go in mode i think it should be 9

mode table

top right

i don't remember the plus i use the classwiz as well

[crimson swift]

its mode 7 for me

[unreal isle]

yeah okay

if u want 2 functions tgt you will have to go in setup (shift mode)

[crimson swift]

but that doesnt compare my derivative to the calc's

[unreal isle]

look for table

[unreal isle]

i don't get what you mean

let me clarify something

[crimson swift]

@unreal isle

[unreal isle]

lets say you have 3x³ - 2x²

do you wish to check your derivative in general i.e. 9x² - 4x

or do you want a table where you input an x and it immediately gives you the dy/dx value

[crimson swift]

do you wish to check your derivative in general i.e. 9x² - 4x

[unreal isle]

alrigjt in that case you dnt need table

lemme type

[crimson swift]

ik how to do that already

[unreal isle]

so e.g. u wanna confirm ur derivative of let's say 3x³ - 6x² + 3x

youd do shift integral button, 3 alpha right bracket cubed .... and so on

right arrow to x = blank, then input a value of x of your choice

then use the same value of x in your worked out derivative. if the values match, congrats you're correct

same holds for integrals, so long as you use a valid value of x you can check by just inputting placeholder limits

[crimson swift]

only one value?

[unreal isle]

to confirm, yes

so for instance you manually worked out your derivative expression to be 9x² - 12x + 3

[crimson swift]

and preferably not 0\1\-1?

[unreal isle]

in your calc youd input d/dx original expression

at let's say x=2

[unreal isle]

depends on whether the expressions gets undefined there

[unreal isle]

and then youd cross check the calc value w the value u get by manual substitution

a neat trick, you can do shift rcl (bottom left of the function keys) to STORE a value. follow these steps

press 2. then equals to. then oress shift rcl and then right bracket. you should see smth like stored to x. now if you do X and press equals, you'll see it's 2.

so if your expression is long you can store a value of x and write in terms of x

[crimson swift]

after pressing right bracket i see ans (right arrow)X

[unreal isle]

perfect now try doing alpha right bracket (i.e. X) and press equals

[crimson swift]

i think itll be easier to write it down

[unreal isle]

that way you just have to change your stored value of x and not have to retype the whole calculation for lets say if u want the value at 3 now

[crimson swift]

oh

[unreal isle]

up to you, but yeah the 991 series cant give expressions, it can give you a value to crosscheck, to answer your primary qs

[unreal isle]

yep so u type it out in terms of x once, then u can change your x, up arrow to go to the expression again and equate again to gey the new value of the expressions at the new x

[crimson swift]

if the domain of the function is spreaded to 2 zones, would you check one value for each one?

[unreal isle]

wdym

could you give an example

[crimson swift]

x>0, -5<x<-1

[unreal isle]

also no to check if your derivative is correct you have to just use one value of x, granted the same value in both places, provided that d/dx is defined at that x

so if your d/dx worked out to have a fraction like 1/x you wouldn't use 0

the domain doesnt matter if you're just checking if you're right

cos the polynomial is the same

[crimson swift]

well theres ln as well

and sqrt

[unreal isle]

"so long as it is defined at that x"

if you're just checking if your derivative is correct for a specific polynomial

you just have to take that polynomial

find the derivative yourself

use the calc to find the value of the derivative at a placeholder x

[crimson swift]

not only polynomials

[unreal isle]

use that same placeholder in your derivative

[crimson swift]

mainly not polynomial

[unreal isle]

any expression in x

[crimson swift]

e^x

lnx

ect

[unreal isle]

any expression in x I'll rephrase that's mb

it's 1am

hazy

😭

[unreal isle]

and crosscheck

[crimson swift]

and if not i can always retake it

🫠

thanks

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[shrewd belfry]

My question is how to prove the following function is quasi-concave. f(x1,x2) = x1 * x2 in R^2++. My linear algebra is shaky and i'm having trouble trying to go from the definition of quasi-concavity to proving this specific function is quasi-concave for the positive quadrant of R^2. If possible I would appreciate it if someone walked me through the standard steps to show this

[calm coral]

@shrewd belfry Has your question been resolved?

[shrewd belfry]

<@&286206848099549185> any lads here know about convex optimization? 🥲

[fringe reef]

what's the definition of a quasi-concave function?

[shrewd belfry]

for quasi-concave its this but for -f

the image is for quasi-convex

if -f is quasi-convex its quasiconcave

sorry last part its missing, the set has to be convex too

[idle yarrow]

so for ur problem, u have $\text{img}f=(0,\infty)$, meaning u want to prove that each of the sublevel sets
$S_\alpha = {x \in (0,\infty)^2 : f(x)\ge \alpha }$ are convex

[shrewd belfry]

yep

well 2 inputs but yeah

[potent lotus]

Bob the Builder

[shrewd belfry]

yep

[idle yarrow]

for $\alpha > 0$

[potent lotus]

Bob the Builder

[idle yarrow]

but then $f(x_1,x_2) \ge \alpha$ is the same as saying $x_1x_2 \ge \alpha$

[potent lotus]

Bob the Builder

[idle yarrow]

can u picture the set of $(x_1,x_2)$ in $(0,\infty)^2$ that satisfy this?

[potent lotus]

Bob the Builder

[shrewd belfry]

i think so

[idle yarrow]

do u know how to then prove that its convex?

[shrewd belfry]

hmm so using the definition of convexity directly on that set?

[idle yarrow]

yeah u could but that might be a bit messy algebraically

in this case i think it might be easier to realize the set is bounded below by the function $x_2 = \alpha/x_1$ and prove that the function is convex

[potent lotus]

Bob the Builder

[shrewd belfry]

ahhh alright i guess along the peak of the curve

how would i show that though?

[idle yarrow]

do u know abt any other results abt convex functions besides just the definition?

[shrewd belfry]

i mean not at the top of my head theres so many lol

[idle yarrow]

ok well ofc u want to use one to prove that $\alpha/x$ is convex on $(0,\infty)$ quickly

[potent lotus]

Bob the Builder

[shrewd belfry]

i think 1/x is convex on the positive quadrant right?

or on rather for x > 0

[idle yarrow]

yea

ig u can assume that or prove it (what i had in mind is ||by showing the 2nd derivative is negative||)

but thats pretty much it

[shrewd belfry]

ahh alright i see, i guess assuming it works, its pretty standard i think.

Thanks for the help!

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[royal ingot]

Hello

[royal ingot]

Can I have some help with ex 2 question 3 and 4

this is in french btw

[rough blade]

What have you tried so far ?

You can just try with successive values of q i.e 1,2,3 ..etc

Until you get 1 as a remainder

You can't solve 4 unless you solve 3

@royal ingot

[royal ingot]

yeah thank you I got it

it works with 13

14*

nta rajel

[rough blade]

Ok cool , is the other question clear ?

Think of a way to simplify that term

dont forget that $4^{14} \equiv 1 \pmod{29}$

[potent lotus]

<rajel />

[rough blade]

You can notice that 1024 is just a sequence of powers of 4

@royal ingot

[royal ingot]

yeah I figured it out

thank you

[rough blade]

Np, btw where are you from
Your math seems familiar

[royal ingot]

hhh l watani

sm

if you know what I mean

[rough blade]

Tunisia ?

[royal ingot]

oh no close

Morocco

[rough blade]

Oh

I see

[royal ingot]

yeah I also thought you were from there lol

I forgot also algeria and tunisia study like us

[rough blade]

Very close

It's Mauritania tho

!done

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[royal ingot]

oh ok ur next to me

.close

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[ancient cipher]

dont rlly understand where to start here. my oriuginal thought was to use the idea of that if abs(an) converges, then an converges, proving that answer choice A was right but i guess that was wrong :(

[pallid halo]

well you know the radius of convergence is at least what, and at most what?

based on the given info

[ancient cipher]

at most 2 right ?

i dont rlly know

[pallid halo]

well it converges at x=7

how far is that from x=4?

[ancient cipher]

ohhh i did nott think at all about it being ceneterd at x=4

so radius of convergence at x=3

[pallid halo]

yea that's the key here

right, radius of convergence is at least 3

[ancient cipher]

ok that makes sense

thank you

[pallid halo]

now what does divergence at x=9 tell you

[ancient cipher]

that its not convergent at x=-1

[pallid halo]

no, you can't be sure of that

[ancient cipher]

ohhh yea

[pallid halo]

the radius of convergence could be exactly 5

[ancient cipher]

its like

indterminate rightt ?

[pallid halo]

and then it might converge or diverge at the endpoints

[ancient cipher]

at x=-1

[pallid halo]

yep

[ancient cipher]

bc i cant test forr it

[pallid halo]

you can't say for sure about x=-1

[ancient cipher]

ohh ok

[pallid halo]

how about x=1

what can you say there

[ancient cipher]

in tthe radius of convergence so its convergent there

[pallid halo]

what if the radius of convergence is exactly 3

[ancient cipher]

wait no its not

bc i said

[pallid halo]

then it could converge at 7 but diverge at 1

[ancient cipher]

radius is only guarantted to be 3

so we cant say atuff abt x=1

[pallid halo]

right you don't know

it might converge there or it might diverge

so how about x=2

[ancient cipher]

in the guaranteed radius of 1 to 7 so it must be convergent there

[pallid halo]

yep good

you know the radius is at least 3, and the distance from 2 to 4 is only 2 so it's inside the interval of convergence for sure

so what's the right answer?

[ancient cipher]

B

[pallid halo]

yep correct

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[molten lichen]

I have no idea how to do this someone pls help 🙏🏻

[leaden thunder]

,tex .maclaurin

[potent lotus]

riemann

[leaden thunder]

Use the first one with -x^2/3 instead of x

[molten lichen]

Right, but what would x and x^2 become?

Would it just be -x^2/3 for rhe second term and -x^4/3 for the third?

[leaden thunder]

Wherever you see an x here

Replace it with -x^2/3

[ancient thistle]

maclaurin/exp

[molten lichen]

So what about the summation notation?

[leaden thunder]

What about it

[molten lichen]

How is it written for this function

[leaden thunder]

Doesn't change if you do the substitution correctly

[molten lichen]

(-1) x^(2/3 • k) /k! ?

[leaden thunder]

No

[molten lichen]

Sorry if the answer is evidently wrong i just started learning this an hour ago 😭

[leaden thunder]

-x^2/3 = -(x^2)/3

[leaden thunder]

This first expression is the same thing as the exponent in your question

Should review pemdas while you're at it

[molten lichen]

So i just sub in -(x^2)/3 to the sigma notation as well?

Directly

[pale blade]

the sigma notation is just a way to succintly represent the maclaurin series. they're functionally identical. so yes, if you use the sigma notation, you should sub -(x^2)/3 into it

[molten lichen]

So each term gets multipled by the (x^2)/3 in the series and the sigma notation has the same substituted in for x with no other changes?

[pale blade]

each term does not get "multiplied" by -(x^2)/3, it's just that whenever there's an x in the series, it gets replaced by -(x^2)/3. So the first term, for instance, remains unchanged (because it's 1), the second term becomes [-(x^2)/3]/1!, the third term becomes [-(x^2)/3]^2/2!, and so on. not precisely "multiplied" by, rather "substituted in"

yes for the sigma notation. I mean you can just think of it as expanding the sigma; to demonstrate why the sigma notation = maclaurin series.

disregarding the sub-in for now, sigma just means "add this expression, with increasing values of the index, k". If you expand sum(0 to inf) x^k/k!, you get: x^0/0! + x^1/1! + x^2/2! and so on this is more of an aside to your question, not directly relevant

which is identical to the maclaurin series

[molten lichen]

Oh okay, that makes so much more sense now, thanks!

In this term, [-(x^2)/3]^2/2!, would it be acceptable to write it as [-(x^4)/9]/2!?

[pale blade]

wouldn't it become positive?

[molten lichen]

Yes, i forgot to make it positive when i copy pasted the expression

[pale blade]

but yeah its fine otherwise

[molten lichen]

And to approximate f(0.3) id just plug 0.3 into x for the first 4 terms right

[pale blade]

any reason 4 specifically?

[calm coral]

@molten lichen Has your question been resolved?

[calm coral]

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[deft arrow]

I did not understand the step where they wrote C-{0} is open

[calm coral]

@deft arrow Has your question been resolved?

[velvet osprey]

{0}, a set composed of a finite number of points (one point), is closed

the complement of a closed set is always open

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[odd belfry]

Problem: ABCD trapezoid. BC and AD are the bases. AD = 25, BC = 15, AB = 6, CD = 8. Find the radius of circle which passes through A, B and touches CD.

How to solve this? I was thinking about circle theorems but it didn't work out. Or maybe I couldn't )

Any hints for solving this?

[odd belfry]

My drawing

[odd belfry]

Correction: CM is 6

And angle CDA is not 45 degrees

[odd belfry]

<@&286206848099549185>

[calm coral]

@odd belfry Has your question been resolved?

[calm coral]

@odd belfry Has your question been resolved?

[vale flower]

are there any more info given?

[calm coral]

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[quiet island]

If $A = B \cup C$ is an open set and $B \cap \overline C = \overline B \cap C = \varnothing$, then $B$ and $C$ are open sets

[pure breach]

open as in open interval?

just use \phi

[quiet island]

Open as Open Sets in Topology

[potent lotus]

Halex

[pure breach]

uhh i am an idiot but

shouldnt u mention the metric space

so whether its euclidean or arbitary

[quiet island]

The Real Numbers, I am the idiot not you

[pure breach]

I havent even taken a formal course in real analysis, fym

but anyhow

so this an euclidean space

uhhh open set so ur given that two sets together make up an open set, and the intersection of their complements is a null set

right

[quiet island]

$\overline B$ stands for closure of $B$

[potent lotus]

Halex

[pure breach]

oh shit

nevermind

hmm

yeah okay i can see why this should be true

but i cant help u write a proof

because i cant do proof writing at all

[quiet island]

If I get the idea I think it's okay

[pure breach]

well this is a very poor way of thinking about it

but imagine it in R^2 (for visualization purposes)

clearly B and C have no common elements (or common limit points)

now the the A's boundary (so closure of A - A) should therefore be the union of the boundaries of B and C

since this boundary is not in A

therefore it could not have been in B or C either (as B,C \subset A)

and since they dont share elements we can map each limit point in the boundary of A to a limit point in the boundary of B and C

[teal coral]

I think I can formalize this without alluding to the boundaries

[pure breach]

ofc please do, i have 0 background with this except the definitions, i could be just a mad man rambling

[teal coral]

basically, suppose you're trying to locate an open neighborhood around x in B

the claim is that you can simply use the one you locate in A (i.e. B U C)

suppose for the sake of argument that you cannot do this; this means that you have a neighborhood around x that contains elements from not only B, but also C

surely there's some way to prove that x is in the closure of C now

[pure breach]

btw what i said, is it correct?

[teal coral]

it's basically saying that you have two blobs, and they're separated yeah?

[pure breach]

yeah 😂

thats the photo i made in R^2

[calm coral]

@quiet island Has your question been resolved?

[quiet island]

What

[cedar oyster]

agreed, I think that's wrong. if B is not open, there is some x in B for which every neighborhood of x contains points outside of B, not just some

I think you use T4 axiom and separability here

it's already given that B and C are disjoint, so there ought to be two disjoint open sets, U containing B and V containing C

A itself is open, so each x of B begets a basis element K containing x that's a subset of A

now we know that K does not intersect C (why?), so it must be the case that K lies entirely in B, and hence B is open

...don't think I needed to bother with U and V lol

[calm coral]

@quiet island Has your question been resolved?

[calm coral]

@quiet island Has your question been resolved?

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[prime locust]

Where should I start learning more math on my own after I just finished AP Calculus BC?

[west shard]

is there anything you particularly liked?

[prime locust]

Everything except polar tbh

[west shard]

polar coordinates?

[prime locust]

yeah

[west shard]

I'm not familiar with ap so you'll have to help me out here

[prime locust]

It covers all of Calc 1 and a lot of Calc 2

[west shard]

in a completely different direction, did you do much about vectors?

[prime locust]

I know how to do vectors yeah

[west shard]

I think it you want to pursue pure maths, it'll be useful to investigate vectors in a lot more general sense

not just pure maths, maths in general

[prime locust]

I'm gonna go into engineering probably

[west shard]

That being said, I don't remember what the book is called but some like "linear algebra done right"

ah ok

do you care about the maths being used in engineering?

[prime locust]

Yeah but I'm interested in all math

I'm just gonna go to school for engineering

[west shard]

In that case, Linear Algebra is probably a good place to start, it kinda melds your brain into thinking about mathematical objects in a more general sense

if you liked calculus, looking at different sorts of Ordinary Differential Equations (ODEs) is a good place to start, they come up a lot when you're doing physicsy stuff

[prime locust]

Do I need Calculus 3 before ODEs?

[west shard]

Good question, depends where you're from

If I remember correctly Calc (insert number here) doesn't have a set syllabus

some places do ODEs in Calc 3 though

[prime locust]

I don't know like multivariable calculus

[west shard]

multivariate calculus is not needed for ODEs

it helps in some places but absolutely not necessary

[prime locust]

Okay cool

I think I'm gonna look into linear algebra first

[west shard]

in fact, if you look through the left hand side in the channels list

scroll to the early university section

and type all the names of the channels into google or Wikipedia and see what you like

they're all pretty much accessible from where you are now

[prime locust]

Haha nice okay

Thanks for the help

[west shard]

oh, groups, rings and fields in advanced are also a possibility

in which case I'd look at groups or fields first

[prime locust]

Before like linear algebra?

[west shard]

you can look at them side by side

the main object in linear algebra (linalg) is called a vector space, and they're objects "vectors" connected to a number system "fields"

so they come hand in hand

one also has groups which can be visualised on a vector space

open your own channel

[green pumice]

how

[west shard]

#help-48

did you read #❓how-to-get-help

[green pumice]

yea

i did one

https://discord.com/channels/268882317391429632/1371594838952378459

[west shard]

then keep it there

[green pumice]

Okay thanks

[west shard]

delete this please

[prime locust]

Do you have any video or book recommendations?

[west shard]

for?

[prime locust]

Linear algebra and fields

[west shard]

"Linear Algebra done right"

I don't know about group theory

I learnt that at uni and through a fatass textbook i don't particularly recommend called Fraleigh

Abstract Algebra

(I'm probably slightly biased)

[prime locust]

lol okay

So you don't recommend that one?

[west shard]

Linear Algebra done right is the name of a good Linalg book

I personal didn't like Fraleigh's Abstract Algebra, it was very heavy

[prime locust]

Okay cool I found it

[prime locust]

Do you think I can just learn through like videos?

[west shard]

depends on your person, I like watching videos and doing problems so if you can find a resource I don't see why not

but you'll want someone else's opinion on that question

I'm no teacher

[prime locust]

Yeah that's fair

I just don't have school for a little while

[swift laurel]

for linear algebra i highly recommend "essence of linear algebra" by 3blue1brown on youtube

it's a high-level overview of the topic with nice animations

[prime locust]

lol he's a classic youtuber in the math scene

[swift laurel]

in general introductory linear algebra courses tends to be split into two types, a more "computational" approach and a more "proof-based" approach, so what videos and textbooks are recommended depend on which type of course you're going for

sometimes people do a proof-based course as a second course after the computational one

[prime locust]

I think I'd probably prefer computational

for now

[swift laurel]

in terms of computational linear algebra lectures, gilbert strang is a famous professor in that regard, and there are several different semesters of recorded lectures of his on youtube

if you are interested in following them, there is also associated coursework (e.g. homework and exams) on mit opencourseware, and he has also wrote a textbook

the textbook i used for computational linear algebra was "elementary linear algebra" by anton

also in terms of videos, khan academy has a few but they seem a bit sparse and it also doesn't have exercises

[prime locust]

Khan academy is overrated ngl

[west shard]

hehe agreed

[swift laurel]

it's recommended a lot mostly because it's more comprehensive than any of the competitors

[west shard]

although I did use it to learn about Laplace transforms

[swift laurel]

but it kind of fizzles out once you get past courses offered in high school

[prime locust]

I'm technically still in high school

[west shard]

presumably you're wanting to learn about maths beyond highschool

[prime locust]

Yeah

[swift laurel]

well once you are done with ap calculus you're sort of moving beyond what high school can offer

unless you want to make a move sideways and learn ap-level stats (which is actually not a bad idea itself)

[prime locust]

I've heard it isn't too hard

[west shard]

yes it is. i have a stats exam tomorrow and I'm not liking it

(that's me getting biased again)

[prime locust]

AP?

[west shard]

I'm at uni

[prime locust]

oh nah

im dumb

[swift laurel]

high school-level stats is pretty easy in terms of calculations (since most of the time your calculator does them for you!), the main difficulty is conceptual

but also statistics shows up everywhere so it's worth knowing to some extent

[west shard]

(unfortunately)

[prime locust]

I know some concepts of statistics just not really how to calculate it

Like standard deviation or smth

Is it wrong of me to have an undergrad math role because AP Calc is technically college credit

[swift laurel]

no it's fine

in ap stats, usually you learn the formula for something, calculate it by hand once to get a feel for it, then press the "calculate this quantity button" on your calculator every time after

[prime locust]

The AP test was today and we aren't doing anything more in class

[swift laurel]

because the main thing is being able to know when to apply a concept and how to interpret it

[prime locust]

Okay cool

I know Bayesian probability as well

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[lethal sparrow]

how come i got this wrong?

[lethal sparrow]

im trying to follow this

[torpid canopy]

What was your equation?

if the wind is 90, then with the wind the plane flies at 240, and against it flies at 60, which is 4 times slower

which doesn't match the given numbers

[lethal sparrow]

i started with 300(150+x) = 200(150-x)

then 75,000+300x=30,000-200x

then 500x=-45,000

[torpid canopy]

not quite

Remember distance = rate * time

Since the plane flew 300 miles with wind and 200 miles against wind in the same amount of time, you want to solve
300/(150 + x) = 200/(150 - x)

[lethal sparrow]

oh so i accidentally multiplied instead of dividing

[torpid canopy]

yes

[lethal sparrow]

ok ty

.close

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[royal dock]

Hey, I am pretty confused about dependent variables and independent variables. Can anyone explain me briefly about it.

[leaden thunder]

https://youtu.be/afp3jhPNS3g?si=JYixsjiXm3Z4Hy1T

[royal dock]

I understood this video.

But I got another question that why is y is a dependent variable and x is an independent variable? Why can't x be dependent and y be independent?

Why is it so fixed

?

[unreal isle]

if you were to plot time on the y axis and distance on the x axis, and the variable changing is time, then x becomes the dependent variable in that case. it's just a matter of convention since mostly y = f(x)

[royal dock]

Oo

Ok

[royal dock]

But how is x dependent here.

[unreal isle]

time is the factor that you are changing, x is the one that depends on it then

[royal dock]

Time changes according to distance.

[unreal isle]

well no if you go backwards you're not going back in time

gimme a sec to type

[royal dock]

Yeah sure

[unreal isle]

you should look at it from the perspective that okay take the equation s=d/t
it depends on what your experiment was doing.

Case A: you take a ball and you track the distance at a range of different times.
here, the variable YOU'RE changing is time, and the variable that changes as a consequence is disntance

Case B: you take a ball and you track the time it takes for it to cover a certain range of distances. here, the variable YOU'RE changing is distance, and the result that changes is time.

the dependent and independent variables change.

now at the end of the day, you could put any one of the variables on any access, however "most" functions conventionally are given as y in terms of x.

[royal dock]

So, the dependent and independent variables r changeable according to our convenience but in general people conceder x as independent and y as dependent. Right?

[unreal isle]

yep, conventionally

also the dependent and independent variables are not "changeable" per se

that could be misleading

you cant just choose which is which, you have to consider what the data shows

[royal dock]

Oo

Tqsm.

.close

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[calm coral]

@fringe copper Has your question been resolved?

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[bleak thistle]

Hello, over here why isn’t the answer B) ? I thought if f’’(x)= + (concave) then that’s where f’(x) is increasing
[the answer is D)]

[bleak thistle]

@ me when someone answers thanks !

[primal bear]

can you plot f'(x)

[bleak thistle]

no

[shy iris]

May God reward you if help me

[mellow harbor]

!occupied

[calm coral]

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).

[mellow harbor]

also

!noans

[calm coral]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[mellow harbor]

!nosols

[calm coral]

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.

@bleak thistle Has your question been resolved?

[calm coral]

@bleak thistle Has your question been resolved?

[calm coral]

@bleak thistle Has your question been resolved?

[flat orchid]

It's not necessary for f" to be positive for the function to be increasing

f' > 0 is the necessary condition

You can check the above by drawing slopes to the graph and checking where the slopes are positive

[bleak thistle]

No I mean like since f(x) is concave which means f’’(x)= + and since f’’(x) is positive (f’’(x) is the derivative of f’(x)) so f’(x) is increasing at the interval ]-♾️,2]

Is my reasoning wrong?

It’s asking for f’ so

[flat orchid]

Oh no my bad then

You're right

[bleak thistle]

The answer is D) doe so 😭

[flat orchid]

Hmm i also think it should be B

[bleak thistle]

so is the answer key wrong

if it asked for f(x) increasing then the answer would be D)

Right?

[flat orchid]

Yeah

I believe so

[calm coral]

@bleak thistle Has your question been resolved?

[bleak thistle]

<@&286206848099549185>

[flat orchid]

Did you also have other doubts @bleak thistle

[bleak thistle]

Wdym

[flat orchid]

Oh are you just verifying if the answer key they've given is wrong by asking other helpers

[bleak thistle]

yes

[flat orchid]

Nvm then continue

[bleak thistle]

Until one of them answers

So we gotta wait

[flat orchid]

[leaden thunder]

show the answer key to the problem

[bleak thistle]

It’s quesyion 42

It doesn’t show the answer

Just the letter

[leaden thunder]

yea then they're wrong

typo in question or answer key

[bleak thistle]

how though? They’re right to exclude 2 because 2 would be a critical value in f’(x) idk

[bleak thistle]

Yeah probably

[gentle bane]

Well I found something on Math Stack Exchange, was not actually aware that ] ___ [ denotes an open interval... Then it would be a typo.

You learn something new everyday eh.

Yea then it should be B

[bleak thistle]

Okii thank you guys :)

have a great day/evening/night !

.closw

O

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[spice venture]

f(1) + f(2) + f(3) = 3

so the only possibility is 0,1,2

f(1) can be 3 choices

f(2) can be 2 choices and f(3) can be 1 choice

so wont it be 3x2 = 6

[pallid halo]

that's the number of ways to permute 0,1,2 yes

then you have to multiply this by the number of ways to permute the others

[spice venture]

oh shit ;/

so 5!

x 3!

[pallid halo]

yep looks right

[spice venture]

thanks man

[pallid halo]

yw

[spice venture]

idk how that happened to me 😭

.close

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[exotic turtle]

the straight line with the equation y - 2x = 7 is the perpendicular bisector of the line AB, where A is the point with coordinates (j,7), and B is the point with coordinates (6,k)

find the coordinates of the midpoint of the line AB

[bold hatch]

Do you know how to find the slope of a perpendicular line?

[exotic turtle]

the gradient?

would i need to rearrange into y = mx + c

y = 2x + 7

the gradient is 2

[calm coral]

@exotic turtle Has your question been resolved?

[calm coral]

@exotic turtle Has your question been resolved?

[opaque cove]

Hi! To find the point both lines cross, we will have to find values for j and k… first, we know the slope of the line created by A and B has the inverse slope of the original line, since they are perpendicular… the inverse slope of 2 is -1/2.. we can use the rise over run formula to find j and k: -1/2 = (k-7)/(6-j) … then solving for j we get: j = 2k - 8 … choose some value of k, i chose 1, and get the associated j value, i got -6… then write a formula for the line created by A and B using point-slope form… then set both equations equal to each other and solve for x, and then plug in that x to find the corresponding y value… that’ll be your midpoint! Let me know if this makes sense… I sent some pictures of my work as well:

[exotic turtle]

thank you very much

[opaque cove]

of course! I hope it was helpful in understanding😸

[calm coral]

@exotic turtle Has your question been resolved?

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[surreal ferry]

I tried to recreate it as well as I could:

What's the measure of segment AE?
mBD = 30m
Angle BAC = 20 degrees
Angle DEC = 44,15 degrees

[tranquil wasp]

let me see i wonder if this is even solvable

[velvet osprey]

is that decimal degrees or is it really 45°15' btw

not that it's a big difference but figured i'd ask

also hm i dont think there's enough info actually

[surreal ferry]

Wdym

[tranquil wasp]

does it mean like 44.15

[surreal ferry]

Yeah

44.15 degrees

[velvet osprey]

44 degrees and 15 hundredths of a degree?

or is it 44 degrees and 15 minutes?

anyway whatever it is

[patent tide]

well if AB and DE are perpendicular to BD, doesnt that mean AE is parallel to BD and it should be a square???

[velvet osprey]

and who said angles BAE or AED are right?

OP didn't even draw them as such

@surreal ferry the decimal thing is not nearly as big of an issue

the bigger issue is there's still not enough info

[patent tide]

oh yeah okay i see where i got wrong it's just that AE isnt straight theoretically

[tranquil wasp]

if the triangle was isosceles it would probably define it

[velvet osprey]

are you 110% sure you didn't miss any other data

[surreal ferry]

Yeah

AE isnt straight

[patent tide]

it would if AB and DE were the same length and i thing we could try finding there length to then know the height of A and B

[tranquil wasp]

are you saying its not a line

[surreal ferry]

I meant

It doesnt form a square

[patent tide]

just that it's not parallel to BD

[surreal ferry]

Yeah that

[patent tide]

it has a slope

[tranquil wasp]

do you know anything about triangle CEA

[surreal ferry]

Nope

[tranquil wasp]

There is nothing fully defining this problem so there is no answer

[surreal ferry]

Like this is sort of what I did but idk if it's right 😭

I gave up

[tranquil wasp]

did chatgpt manage to solve it

[surreal ferry]

I forgot what I got LMAO

Wait

That doesntm ake sense

[tranquil wasp]

it just guessed lmao

[surreal ferry]

Yeah

[patent tide]

if we assume but i dont like assuming

[tranquil wasp]

the guess isnt even right lmao

[surreal ferry]

Yeahh..

[tranquil wasp]

its 39.54

if you assume x=15

[surreal ferry]

😭

My final answer was 30.04m

[patent tide]

if we knew c was the midpoint it would be easy to find AB and DE, and then to find AE

[tranquil wasp]

Do you have the actual problem

[surreal ferry]

It was an exam problem so no but thats exactly what the question was

[velvet osprey]

!nogpt

[calm coral]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[velvet osprey]

just sayin'

[tranquil wasp]

yes it was very wrong lol

[patent tide]

i would be curious to see the answer once you get back your exam

[surreal ferry]

Mbb

Was just seeing waht it would try

[velvet osprey]

was this problem on your geometry exam

[surreal ferry]

Yep

[velvet osprey]

so you don't have its statement written in full

[surreal ferry]

Wdym

[velvet osprey]

like you don't have it on paper where you could reread it

[surreal ferry]

No I don't

[velvet osprey]

and you could have forgotten some length or angle or other info

[tranquil wasp]

you can bound the answer between around 30m-88m

[surreal ferry]

I reread it so many times I think I got every info right but idk

Could be that

[patent tide]

can you tell me why?? just curious

[tranquil wasp]

you can tell by the image the slope of AE if it was a line was negative

and assuming the image isnt totally wrong you can use it to bound it

[patent tide]

by saying that the angle between BAE is less than 90 ??

[tranquil wasp]

yea

[patent tide]

yeah okay i see

thanks

[surreal ferry]

Oh well

.close

[calm coral]

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[calm coral]

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[meager sparrow]

[indigo grove]

[meager sparrow]

,, \sum {k=1}{\inf} ((\frac {-3/4})^{k-1})

Welp

Anyways

[indigo grove]

}

[meager sparrow]

Ehhh

You got the idea?

[indigo grove]

put the ) after the whole expression

[indigo grove]

but yeah

geometric series

[meager sparrow]

Imma give up that for now

So

I tried to a/1-r immediately

a=1 and r=-3/4

[indigo grove]

yep

[meager sparrow]

3/4 is less than 1 so the series converges

But how do ik my answer is correct

[indigo grove]

write your answer

$S_n=\frac{a_1}{1-r}$

[potent lotus]

yoboiqimmah

[meager sparrow]

1/1+0.75

[indigo grove]

i prefer keeping it as a fraction

[meager sparrow]

Should be 4/7?

[indigo grove]

yep

[meager sparrow]

Alrighty, thanks

But one more question if i may ask

[indigo grove]

ok

[meager sparrow]

Is there way to do the ones with the form of ar^k?

[indigo grove]

find a1 at first by plugging the initial value for example
5(2/3)^k-1 the first term would be 5(2/3)^0=5
and r is just the difference being 2/3

[meager sparrow]

Isnt the difference 1/3?

Oh got what you meant

Thanks again

.close

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[indigo grove]

Wording is wrong should be ratio

My mistake

[meager sparrow]

It’s all good

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[timid acorn]

I’m not sure how to visualize this, the first de Morgan’s law

Like “the intersection of the difference” is uh

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[timid acorn]

I’m not sure how to visualize this, the first de Morgan’s law

Like “the intersection of the difference” is uh confusing

[old falcon]

what you drew does not match what you wrote

[timid acorn]

Yeah I know

[old falcon]

i dont get it

by visualize do you mean draw

[timid acorn]

I’m asking about the correct diagram

[old falcon]

oh

[timid acorn]

Yeah

[old falcon]

well, start with A

then remove the parts included in (B U C)

if you aren't sure whats included in (B U C), use the definiton

its everything thats in either B, or in C

does that help?

[timid acorn]

Would this be correct?

[timid acorn]

I’m mainly trying to get the diagrams correct for myself… it’s not attached to a problem

Issues with the definition

[old falcon]

yea, that looks good to me

[old falcon]

no worries

[timid acorn]

Great

[old falcon]

hmm i think you might be missing something here

what happens if $a \in A$ but $a$ isnt in $B$ or $C$?

[potent lotus]

jan Niku

[timid acorn]

Then it’s this

[old falcon]

well, kind of

can i try to describe it in a way that might be confusing

lets say i have two pockets

what happens if i take everything out my pockets thats not in my left pocket

and combine it with everything in my pockets thats not in my right pocket

i am famously bad at coming up with examples and metaphors so dont sweat it if its not helpful

[timid acorn]

Then you have nothing?

[old falcon]

well, you have everything

at least, everything that was in your pockets to begin with

[timid acorn]

Mmmm

[old falcon]

since stuff in your right pocket is excluded in the second, but not the first

and visa versa

its kind of like your problem

you start with everything in A, except whats in B

and combine it with everything in A, except whats in C

well maybe this metaphor isnt helpful or at least as clear as i thought

i honestly think its maybe better when youre just starting

to start with the venn diagram and check each region

[timid acorn]

Yeah probably

[old falcon]

sometimes its helpful to draw multiple

like when its complicated like in this problem

do each difference separately

then do the union on a third

[timid acorn]

A union B means you shade in two circles

[old falcon]

It means everything in A, or in B

[timid acorn]

A intersection B means you shade in the shared region

Yeah

[timid acorn]

But that’s the first problem fuck

[old falcon]

its like this

lemme see if i have paint

i do not

lemme install it one sec

[timid acorn]

Ty

[old falcon]

here is A - B

[timid acorn]

alright

[old falcon]

now the blue is A - C

wait is it just the first problem

sorry mate

[timid acorn]

Lol

[old falcon]

but fun to go through the checkin gprocess, i guess

so fun

[old falcon]

you recognize this

[timid acorn]

Very fun

[old falcon]

its $A - ( B \cap C )$

[potent lotus]

jan Niku

[old falcon]

so then I guess $(A-B) \cup (A-C) \equiv A - (B \cap C)$

[potent lotus]

jan Niku

[old falcon]

but i guess you are doing demorgan

[timid acorn]

Yeah I’m just trying to get a better understanding of these cause

They were just meaningless signs before

[old falcon]

it takes some practice for sure

but itll pay off

[timid acorn]

So it’s just the same thing lol?

[old falcon]

these are, yea

[timid acorn]

Hilarious

[old falcon]

sorry im comin off a 11 hour shift so im not so sharp

[timid acorn]

All good

Thank you

.close

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• Show your work, and if possible, explain where you are stuck.
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[atomic slate]

i'm trying to find the intersection between 2 lines in 3d, given
$L_1$ contains $Q_1(-3, -1, 5)$ and $Q_2(-9, -7, 2)$
$L_2$ contains $P_1(16, 17, 12)$ and has direction vector $d = [3, 2, -1]^T$
i can get the direction of $L_1$ and get both lines to parametric form,
$\begin{bmatrix} -3 & -6s \ -1 & -6s \ 5 & -3s \end{bmatrix}$ and $\begin{bmatrix} 16 & 3t \ 17 & 2t \ 12 & -t \end{bmatrix}$,
and rearrange it to a single matrix,
$\begin{bmatrix} -6s & -3t & 19 \ -6s & -2t & 18 \ -3s & t & 7 \end{bmatrix}$,
but i have no idea what to do with it - the examples in my textbook are extremely sparse and skip over most steps. i assume i need to reduce it in some form, but what to do with it then, to find the intersection point?

[potent lotus]

amber

[atomic slate]

i feel awkard about pinging for help but i really dont understand, <@&286206848099549185>

if anybody can help i'd appreciate a ping as i cannot keep this chat focused

[mental tree]

@atomic slate

[atomic slate]

yes

[mental tree]

So here, I’m not sure about the conversion from those 2 matrices to the simpler 3 x 3

But I have another method that might help

[atomic slate]

ok

[mental tree]

So 1st we find the direction vector of L1

[atomic slate]

yes, thats -6 -6 -3

from Q2 - Q1

[mental tree]

And then we represent the line in symmetric form

And here we can take any point either Q1 or Q2 as x1 y1 z1

And then we replace a b c with their direction

And then from symmetric form

[atomic slate]

where x y and z are what?

[mental tree]

x y z are variables

[atomic slate]

ok

[mental tree]

As you need those variables to represent the line

So now you just switch them to parametric form as shown below in the image

And then you represent both the lines you have in that parametric form

[atomic slate]

so $\frac{x+3}{-6} = \frac{y+1}{-6} = \frac{z-5}{-3}$

[potent lotus]

amber

[mental tree]

Yep precisely

[atomic slate]

i already have parametric form for both lines

$\begin{bmatrix} -3 & -6s \ -1 & -6s \ 5 & -3s \end{bmatrix} = \begin{bmatrix} 16 & 3t \ 17 & 2t \ 12 & -t \end{bmatrix}$

[mental tree]

Okay so now all you have to do is

Equate the x from both sides

[potent lotus]

amber

[mental tree]

And y from both sides

And z from both sides

So you basically get 3 equations in 2 variables

Solve and 2 of them as you need 2 equations to solve 2 variables

[atomic slate]

this is the parametric form i got before coming here, i dont understand what im supposed to do with them and xyz