help-49

[wicked mauve]

fundementally if youre solving for x

you need to use x

well no

but in this case it makes no sense

this is apart of cardanos theorem

use this

have you look up the proof for the theorem

[obtuse idol]

Is r p?

Or is q p and r q?

[wicked mauve]

https://proofwiki.org/wiki/Cardano's_Formula

here

just look at this because it might be different depending on the vid you saw

[obtuse idol]

That looks like a whole lot more work than the method I saw

[wicked mauve]

do you have a video of this method

[obtuse idol]

no

[wicked mauve]

that moment when the cubic formula actually turns out to be a lot of work like i said 800 times

[obtuse idol]

One of the videos I saw was by mathologer

[obtuse idol]

Maybe not as much work as it looks

[wicked mauve]

so they say

[obtuse idol]

There is some stuff I haven’t learned yet I’ll need to learn tho

Like i

[wicked mauve]

like

?

[obtuse idol]

i

[wicked mauve]

i

meaning imaginary

?

[obtuse idol]

Yes

[wicked mauve]

okay yes

[obtuse idol]

And cos

[wicked mauve]

that isnt necessarily relevant

to cubic

unless you have a negative root

[obtuse idol]

I saw it in the link

[wicked mauve]

the cosine is for trig functions

and i

well

is i

[obtuse idol]

What is 0 with a line going through it

[wicked mauve]

theta

its for trig

[obtuse idol]

m

[wicked mauve]

nop

q

[obtuse idol]

ok

I think im going to ignore all of ur advice and continue with what ive been doing originally

[wicked mauve]

https://tenor.com/view/yapping-alien-alien-yapping-levels-just-the-two-of-us-gif-996330645811370798

go ahead

[obtuse idol]

Have u not ever used x=y-b/3a?

[wicked mauve]

no

nor have i used the cubic formula for than 1 time

because its irrelevant and useless

nor do i see that when referring to cardanos method

also not ever is redundant, use never

[obtuse idol]

Ye well,

Using x=y-b/3a get most of it right

[wicked mauve]

mostly correct is still wrong

🤓

[obtuse idol]

I’ll see if I get the same answer for p using the other strat

[wicked mauve]

go ahead

[obtuse idol]

Btw what year are u?

[wicked mauve]

eh

id say 1st

[obtuse idol]

And what are learning

[wicked mauve]

calc 2

[obtuse idol]

m

Very nice

[wicked mauve]

ouais

i suppose

[obtuse idol]

Is this goodbye?

[wicked mauve]

https://tenor.com/view/plankton-spongebob-dontlettheflamedieout-flame-gif-19163324

this gif so dumb

[obtuse idol]

Wym

[wicked mauve]

if you dont need no help

[wicked mauve]

exactly that

[obtuse idol]

I don’t get it

[wicked mauve]

exactly

[obtuse idol]

But I do get it

[wicked mauve]

is rage baiting in your top 3 hobbies

[obtuse idol]

I think I got bored of it

[wicked mauve]

of what

rage baiting

[obtuse idol]

Yes

[wicked mauve]

since when

today ??

in the last 20 mins

[obtuse idol]

Like months ago

[wicked mauve]

turnin a new leaf

🤨

so it was

one of your top 3

[obtuse idol]

Yes

Probably

[wicked mauve]

knew it

what math are you in

[obtuse idol]

Algebra 1

[wicked mauve]

you in uni

or nah

[obtuse idol]

What is uni

[wicked mauve]

UNIVERSITY

[obtuse idol]

no

Not even close

[wicked mauve]

ah gotcha

[obtuse idol]

Wym

[wicked mauve]

i could smell it

nah i just can read people

[obtuse idol]

What exactly made u think that of me

[wicked mauve]

when you said cubic for the win

😂

[obtuse idol]

I don’t get it

[wicked mauve]

doesnt matter

only i have to get it

[obtuse idol]

ok

This has gone way off topic

Should we end it?

[wicked mauve]

go ahead

[obtuse idol]

Alr

[wicked mauve]

fun travels with cubic

[obtuse idol]

I think so

.close

[midnight plank]

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[midnight plank]

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[sweet shuttle]

can someone help me sovle 125^x = 5 rn I have 5^3x = 5 cause 5^3 is 125

[sweet shuttle]

but im stuck here

[lavish venture]

equate exponents

[sweet shuttle]

there is no exponent for the right

[lavish venture]

there is

[night obsidian]

change both to base 5

right

[sweet shuttle]

1 ?

3x = 1 ?

[dusk pier]

Congratulations

[lavish venture]

Congratulations

[dusk pier]

You just won the lottery

[night obsidian]

woo

idk why im wooing i didnt win

[dusk pier]

Now you have 2 options

[night obsidian]

im wooing supportively

[sweet shuttle]

WHAT ?? REALLY

[dusk pier]

I take your math skills or your money

Choose wisel

[sweet shuttle]

money

[dusk pier]

Thank you

[sweet shuttle]

take it

[dusk pier]

Ally day?

Any day?

[sweet shuttle]

all day

[dusk pier]

Damn son

Anyways what's the answer

Needed to cheer you up

YOu seem sad

[sweet shuttle]

x = 1/3

[dusk pier]

Good boy

[sweet shuttle]

hell nah

good boy ?

https://tenor.com/view/rigby-freaky-cat-cat-meme-tongue-out-gif-1492882876567450372

i love u all

.close

[midnight plank]

Channel closed

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[dusk pier]

I don't

[sweet shuttle]

fuck you

[dusk pier]

Haha

ezpz ragebait

clipped

[sweet shuttle]

lmfaooo

[midnight plank]

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[solar moon]

Any hint for it?

[midnight plank]

@solar moon Has your question been resolved?

[midnight plank]

@solar moon Has your question been resolved?

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[low marlin]

I can't exactly figure out how 6. d.) works. I've solved the other ones . So far i tried doing it by cases by doing 4!*4! + 5!*3! and then i tried doing 7! - 6!x2! but neither of them are right.

[misty bronze]

the odd numbers are 1, 3, 5, 7
how many groups of 3 can you make from these

[low marlin]

24?

[misty bronze]

no
have you learned combinations ?

[low marlin]

Meant 4

Thought u mean perms or smtn

[misty bronze]

yes

let's say i pick 1,3,5

i take these numbers and put them in a group, so i'll consider them as one entity

(135), 2, 4, 6, 7, 8

how many arrangements can you do of this

[low marlin]

(6-1)! = 5! =120

[misty bronze]

yea
but we know that we can make 4 groups

so you multiply this by

[low marlin]

4 I'm assuming

[misty bronze]

yes

this is one case

where 3 odd numbers are together

if all 4 are together

we can take (1357) as a group

and arrange them

and then u add both the cases

[low marlin]

Doesn't this have to get multiplied by like 3! Tho?

[misty bronze]

why do you think so

[low marlin]

To order the odd numbers

[misty bronze]

oh yea true

4*3!*5!

[low marlin]

And this would be like 1! * 4! * 4! right

[misty bronze]

yes

[low marlin]

Well I added them up and I got 3456

But the answer says it's 2304

[misty bronze]

oh you also need to remove the cases where 7 is next to (135) in the first case

[low marlin]

Ohhh

[misty bronze]

there are 6 spots in the circle
we fix one and give it to the group
5 spots remain , 2 cannot be used by 7
there are 3 spots that can be used by 7
so 4 x 3! x 3 x 4! = 1728

this should be correct now

[misty bronze]

the 4! is arranging the even numbers

[low marlin]

Yep that's it

Thanks appreciate it

[misty bronze]

[low marlin]

.close

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[dark locust]

im sorry my channel closedd 😭 is this correctt? @slender walrus

[dark locust]

@violet dune 🥲

[violet dune]

yes that looks right

i saw your older message but couldnt react to anything or text lol

[slender walrus]

shouldn't be writing the Lim after already having applied the h->0

[violet dune]

you did get the right derivative though

[dark locust]

THANK UU GUYSS

❤️

[violet dune]

ofc

[midnight plank]

@dark locust Has your question been resolved?

[midnight plank]

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[drowsy notch]

Calculate the area of an equilateral triangle in terms of x

[drowsy notch]

Now my answer here was A = 1/2 x(xsin60) but it was wrong.

What do?

[lyric charm]

but it was wrong
says who?

[drowsy notch]

The one that checked my paper

[sharp coral]

x being the side length?

[lyric charm]

did you just get a big fat 0 on the problem or did you get more specific feedback

.... this was an "or" question

you dont answer those with "yes"

which is it

[drowsy notch]

yeah

[lyric charm]

$A = \frac12 \cdot x \cdot x \cdot \sin(60\dg)$ is correct but unsimplified, btw.

[grand pond]

Ann

[drowsy notch]

Big fat X

[lyric charm]

do you have the question paper with you so we can look at the specific instructions written on it

[drowsy notch]

No. 2

[lyric charm]

so that's it.

but then, again, maybe they expected you to simplify it

and/or know the value of sin(60°) and substitute it in

[drowsy notch]

What woild be the simplified answer then? (x^2 sqrt3)/4 ?

[lyric charm]

yes

[drowsy notch]

Zam

Maybe that was it

Thanks

[sharp coral]

also possible that they expected you to use a different method (although imo they should give full points for any correct method)

[drowsy notch]

They should So i can get a higher score

.close

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[solemn solar]

hey, i got a majority of the question right, but for S (state) i dont know how the textbook got 18-2x? i got 2y since i placed a pronumeral for O (overseas), can someone help?

[midnight plank]

@solemn solar Has your question been resolved?

[shrewd tusk]

It mentions 22 in their own state

so people who were only in their state will be 22-2x-2-2=18-2x

[midnight plank]

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[inland patio]

In my linear algebra book they gave some applications to composition of linear maps, and I found the following exercise. \

Exercise. For an incidence matrix $A$ with related matrix $B$ defined by $B_{ij}=1$ if $i$ is related to $j$ and $j$ is related to $i$, and $B_{ij}=0$ otherwise, prove that $i$ belongs to a clique if and only if $(B^3)_{ii}>0$. \

An incidence matrix is a square matrix with $0$s on the diagonal, and $0$s and $1$s elsewhere. A clique is a maximal collection of three or more people with the property that any two can send to each other. I don't really know where to start with this exercise. I know no graph theory and the book hasn't presented any. I notice it is an iff statement, so there are two implications I need to prove. Also, $B$ is symmetric.

[grand pond]

psie

[lyric charm]

well

this is specifically linalg as applied to graph theory right

try to draw up some graph on like 5 nodes that's neither edgeless nor complete
write down its incidence matrix

and then look at its square first

see if you can find meaning in the entries of B^2

[inland patio]

yes, it's an example from graph theory, but I think one should be able to solve it without any knowledge of this subject (which I don't have 🙂 ). Anyway, the meaning of A^2, the square of an incidence matrix is the number of ways in which people can send to each other in at most two stages. The interpretation of (A+A^2+...+A^m)_ij is the number of ways in which person i can send to person j in at most m stages.

[lyric charm]

ok right

so

(A^3)_ii is the number of paths of length 3 that begin and end at node i

ie (A^3)_ii > 0 iff node i belongs to a triangle in the graph

[inland patio]

ok👍

[runic hamlet]

just fyi, usually a matrix B like this for a graph is called adjacency matrix and not an incidence matrix

[lyric charm]

also B yeah not A

[inland patio]

so my interpretation was wrong here?

[lyric charm]

you didn't need to look at the sum of the powers

B^m holds the number of paths of length EXACTLY m between each pair of nodes

[runic hamlet]

(to avoid some potential confusion, note that path here does not mean that each edge is unique. its certainly possible to go over the same edge multiple times)

[lyric charm]

you will not wake me up at 3am and make me successfully able to explain which of "path" and "walk" means what

[runic hamlet]

me neither

its not like people are consistent with it anyway

[inland patio]

ok, currently trying to make sense of the meaning of B^2 and B^3, by just using the definition of matrix multiplication. I have yet to learn what this terminology all means 😔

[runic hamlet]

do what Ann said

draw a small graph, write down B and during the multiplication pay attention to which 1*1 terms occur and to which edges they correspond

[inland patio]

ok 👍

.close

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[twilit field]

So firstly, this is $\pdv{(\pdv{z}{s})}{r}$, right

[grand pond]

What a wonderful world !

[twilit field]

if so we first find $\pdv{z}{s} = \pdv{f(x,y)}{x}\cdot (2s) + \pdv{f(x,y)}{y} \cdot (2s)$

[grand pond]

What a wonderful world !

[twilit field]

we then have $\pdv{( \pdv{f(x,y)}{x}\cdot (2r) + \pdv{f(x,y)}{y} \cdot (2s))}{s}$

oof, this is going to be a pain

[grand pond]

What a wonderful world !

[twilit field]

$2r\pdv[2]{f(x,y)}{x}( 2s+2r) + 2r \pdv{f(x,y)}{\partial{y} \partial{x}}(2s+2r) + 2s ( \pdv[2]{f(x,y)}{y})(2s+2r) + 2s \pdv{f(x,y)}{\partial{x} \partial{y}}(2s+2r)$

[grand pond]

What a wonderful world !

[twilit field]

Does this work

Well, I have to fix the LaTeX

[subtle blaze]

Maybe use $\partial_x$ notation

[grand pond]

frosst

[twilit field]

[
2r \frac{\partial^2 f(x,y)}{\partial x^2} (2s+2r) + 2r \frac{\partial^2 f(x,y)}{\partial x \partial y} (2s+2r) + 2s \frac{\partial^2 f(x,y)}{\partial y^2} (2s+2r) + 2s \frac{\partial^2 f(x,y)}{\partial y \partial x} (2s+2r)
]

[grand pond]

What a wonderful world !

[twilit field]

hmm, I suppose I should have

[twilit field]

Is this fine

of course I could further factorise this

.close

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[gusty falcon]

Hi, I found this question online, but isn't this an ambiguous question?

[gusty falcon]

what does "solve for all three variables" mean

[icy juniper]

I think to get a value of those 3 variables

[hard umbra]

i mean the obvious interpretation is that A should be 3 and B should be 4

[gusty falcon]

Yes

That's the obvious interpretation 😭

[icy juniper]

yea

[hard umbra]

but like yeah it's kinda not a very well posed question

[icy juniper]

is that wrong?

[hard umbra]

there's a whole wealth of things that can go wrong here

[gusty falcon]

Well it's not wrong but like uh idk this causes ambiguity, no?

Like?

[icy juniper]

there is?

[gusty falcon]

Well it's fair to see that the question is far from well posed so might as well just analyze all the things that can go wrong

[gusty falcon]

u can define z to be a parameter t

[dim cloak]

hi

[gusty falcon]

z = t

[hard umbra]

well it's only in the "general" case that you need 3 for A

[gusty falcon]

and then well other variables would be in terms of said parameter

that also solves all three variables

and well u don't need 3 linearly independent rows for that

[gusty falcon]

hmmm, what's the "non general" case

[hard umbra]

it's easy to construct 3 distinct equations that have 0, 1, or infinitely many solutions

[gusty falcon]

indeed

that was my issue too and 😭 for me infinitely many solutions is still "solving for all three variables" right?

i guess 0 solutions doesn't work? cuz uhm variables don't have values?

[hard umbra]

that's also another issue yes

[gusty falcon]

is what i said like an example which disproves the "validity" of this question?

[hard umbra]

i mean having no solutions is a "solution" in a sense

like you've solved the problem by classifying what the solutions are

[gusty falcon]

😭 whaa, wait do you thijnk it counts for "solve for all three variables"?

[hard umbra]

i.e. that there are none

well who knows

[gusty falcon]

what's the most obvious issue with the question

is it what i pointed out above with the uh

infinite solutions?

and what fixes can one make without introducing linear algebra

[hard umbra]

well just the fact that it's ambiguous

[gusty falcon]

so no "linearly independent" bla bla

[gusty falcon]

right, but if someone asks why it's ambiguous

what's the response to that?

[hard umbra]

well you have to define what "solve" means

and what "distinct equations" means

[gusty falcon]

doesn't distinc tequations just mean

distinct equations 😭

like just 3/4 diff equations

[hard umbra]

x + y = 1 and 2x + 2y = 2 are distinct equations

[gusty falcon]

yep

they might've meant linearly independent lol

[gusty falcon]

but yes okay so something liek this gets you to

[hard umbra]

but they're not really distinct in a sense that would be useful

[gusty falcon]

infinite or no solutions right?

[hard umbra]

sure

[gusty falcon]

hmm fair so i guess the consensus is that

we don't know what "solve" means

as in how that works for

a system with infinite or 0 solutions right?

[hard umbra]

i mean i feel like you're reading too much into it though

[gusty falcon]

😭 wait i forgot to post the other choices

but there's merit in reading too much

give me a second

one of the choices is

"can't be determined"

😭

[hard umbra]

💀

[gusty falcon]

so that's the most obvious answer to choose

but it's not

[hard umbra]

okay well rip

[gusty falcon]

they just decided to go with this

[hard umbra]

i guess you're screwed

[gusty falcon]

[hard umbra]

if the only options were A is greater and B is greater then you can just assume the question setter wasn't thinking very deeply

but clearly many things can go wrong

[gusty falcon]

😭

okay i think the question isn’t worth much time right?

[hard umbra]

probably not

[gusty falcon]

do i just say can’t be determined cuz “solved” isn’t a precise vocabulary here

[hard umbra]

[gusty falcon]

to capture cases where the system isn’t full rank?

[hard umbra]

i can't read the question setters mind

certainly that can happen

[gusty falcon]

no i mean like in a mathematical context

[hard umbra]

yeah

[gusty falcon]

does that mean u agree with what i said above for uh why it’s “can’t be determined” mathematically?

if so, thanks

[hard umbra]

yes

[gusty falcon]

😭 this was much easier than the barycentric question anyway

idk why i was reminded of that

[hard umbra]

in general you could require any number equations to determine a unique solution

[gusty falcon]

wait wouldn’t the answers be 3 and 4 for u pique solutions

unique*

and well distinct meaning

linearly independent

[hard umbra]

no you can give 3 equations and have infinitely many solutions

[gusty falcon]

cuz otherwise there’s no meaning to distinct

[hard umbra]

and then give more

[hard umbra]

okay sure if you insist on this

[gusty falcon]

wait we’re not assuming linear independence when u say that right?

ah fine

yeah you’re right

i agree

[hard umbra]

yeah if you want linear indep then you can get away with 3

and 4 or whatever

[gusty falcon]

i think those 2 were probably what they wanted to say

😭

given the answer being quantity B is bigger

anyway alright that’s settled

thank you once again!

[hard umbra]

[gusty falcon]

.close

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[ornate glacier]

Q35

[lyric charm]

might wanna crop your screenshots so you don't flash us with the names of a few dozen of (presumably) your classmates

[grand pond]

jan Niku

[ornate glacier]

Tbh i havent got a clue

Where to begin

[lyric charm]

write the powers of cos in terms of sin using the Pythagorean identity

[ornate glacier]

In the evaluate part it is not cos^6x but cos^6 (x)

[lyric charm]

we recognized that as a typo just fine

[ornate glacier]

I dont think that we gotta use the evaluate part

[lyric charm]

...

[ornate glacier]

I think we should use the given part and mold it to get that

[lyric charm]

well if you wanna throw my suggestion to the ground i can't stop you lol

[ornate glacier]

Im sorry

How do i write cos^6 (x) in terms of sin

[lyric charm]

do you know how to write cos^2(x) in terms of sine?

[ornate glacier]

Yes

Oh ok

Then i would get cos^6 (x) = (1-sin^2 (x) )^3/2

That looks terrifying

[lyric charm]

^3 not ^(3/2).

[ornate glacier]

Oyes

[lyric charm]

you're going to end up with a 6th degree polynomial with sin(x) as its variable.

[ornate glacier]

Hmmm

[lyric charm]

(which you might as well abbreviate to a single letter like s unless you like having your life be difficult)

[ornate glacier]

Ok

[lyric charm]

you can (and imo should) divide this polynomial by (s^3 + s^2 + s - 1) and find the remainder.

[ornate glacier]

Can you reason this out

Oh ok

But thats 0

And dividing by 0 …

[lyric charm]

mmm nope not quite.

we are not actually dividing any numbers by zero as such.

[lyric charm]

@ornate glacier calculate this (on paper), simplify it as much as you can, and show me what you get from doing that.

[ornate glacier]

Ok

[lyric charm]

also just to be explicit: don't advance any further than that.

[ornate glacier]

What a shortcut solution

How could a normal person think of that

.close

[midnight plank]

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[lyric charm]

wow it's also incredibly poorly typeset

[ornate glacier]

Hmmm

[lyric charm]

but also yes i guess it is kinda out of nowhere.

[midnight plank]

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[midnight plank]

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[meager ore]

Let $A = { a_1, a_2, \ldots, a_n, \ldots}$ a set of positive integers. Show that there exists an subindex $k \in \mathbb{Z}^{+}$ such that $$\gcd(A) = \gcd(a_1, a_2, \ldots, a_k)$$

[grand pond]

Halex

[meager ore]

Cannot use congruences

[runic hamlet]

consider b_n = gcd(a_1,...,a_n)

[meager ore]

Ohh didn't consider to do that, but it means the gcd will vary no?

Like it was a sequence

[runic hamlet]

the b_n form a sequence, yes

[meager ore]

So I 'd have to show b_k is the gcd of the all the a_k terms?

[runic hamlet]

thats the def

you have to show that for some k you have b_k = gcd(A)

[meager ore]

Could you please help me out to deduce it?

Been struggling with this problem

[runic hamlet]

what are some basic properties that sequences can have and does the sequence b_n have them?

[meager ore]

b_n divides all of the terms in the gcd

[runic hamlet]

in which gcd. you need to be more precise

[meager ore]

And also it can be written as a linear combination of such terms

[meager ore]

gcd(a1, a2,...,an)

b_n | a1
b_n | a2
.
.
.
b_n | an

[runic hamlet]

yes

the property I am interested in is monotonicity

what can you tell me about it

[meager ore]

Monotonocity?

Linear combination?

[runic hamlet]

monotone

increasing or decreasing

[meager ore]

Oh

How can we tell? It could be 1 sometimes

Afaik

[runic hamlet]

lets do an example. lets say A={60, 42, 80, 15, ...}

can you compute the first few b_n ?

[meager ore]

Yes

[meager ore]

For the first three, it's 2

For the first 4, it's 1

So the more terms I have, the more chance of their gcd becoming 1?

[runic hamlet]

well thats just because of the specific numbers I chose

what about the first two b_ns

and can you now tell me something about increasing or decreasing?

[meager ore]

3

[meager ore]

I'd say decreasing

[runic hamlet]

?

[meager ore]

If we chose the set of even numbers, wouldn't we always get 2?

[runic hamlet]

yes

[meager ore]

So not increasing nor decreasing in this case, no matter how much terms we have

I think this sequence shouldn't be increasing tho

[runic hamlet]

yes

its something called a "non-increasing" sequence

which is often also called decreasing

[meager ore]

Ohh yeah heard of those

What can we do with that conclusion?

How does it help us find our k

[runic hamlet]

something important about non-increasing sequences of positive integers is that eventually they are constant

cause you cant get smaller forever

[meager ore]

z_n >= z_{n+1}

[runic hamlet]

yes

[meager ore]

But still don't see where the k takes part in

[meager ore]

How do I show it?

[runic hamlet]

because the sequence is eventually constant that means that there exists a k so that b_k = b_(k+1) = b_(k+2) = ...

I am claiming that this b_k is gcd(A)

[meager ore]

I get the point

How do we construct a formal proof then?

That claim must be proven as well right?

[runic hamlet]

yes

what would happen if b_k wasnt gcd(A) ?

[meager ore]

By way of contradiction?

[runic hamlet]

thats what I am leading up to, yes

[meager ore]

Not sure what it implies

[runic hamlet]

what does gcd(A) mean?

<@&268886789983436800>

[meager ore]

The greatest common divisor of (a1, a2, ..., an,...)

[runic hamlet]

no

[meager ore]

Forgot the ...

[runic hamlet]

and that means?

[meager ore]

That the gcd divides all of those terms?

And can be represented as a linear combination?

[runic hamlet]

lets forget the linear combination bit

so the gcd divides all terms

and its the biggest of all numbers which divide all terms

always these two conditions

so there are two possibilities for b_k to not equal gcd(A)

it could not satisfy the first or the second of these conditions

what happens if it doesnt satisfy the first one

[meager ore]

it would not divide the terms

[runic hamlet]

ok but we already know that it divides all a_1, ..., a_k

so therefore?

[meager ore]

It would not divide terms after a_{k+1}?

[runic hamlet]

it might divide some of them, but there is at least one a_n which it doesnt divide

what can you tell me about b_n ?

[meager ore]

Is it not the gcd?

[runic hamlet]

gcd of what

[midnight plank]

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[solemn solar]

this looks so simple but i dont think ive learnt enough content, find x

[solemn solar]

so far i did x/x+24 = 8/20

then simplified 8/20 to 2/5

then i cross multiplied into 5x = 2x+24

7x = 24

when textbook answer is 16

[tacit rose]

This part is wrong

[solemn solar]

oh

[tacit rose]

It's 2(x + 24) = 5 * x

[peak herald]

you didnt distribute the 2

2x + 48

[solemn solar]

ohhh

ok that mkaes more sens

e

i just thought the 2 would go to the x

thanks guys

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[tacit rose]

No because you have something like a(b + c) = ab + ac, distribution

[midnight plank]

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[frail skiff]

heyoo

[frail skiff]

im struggling with finding the parent function of a conservative vector field in 3d

lemme attach my work so far

i integrated each part of the vector field F with its respective position variable but im not sure what to do with these newfound constants g

[solid iris]

parent=potential?

[frail skiff]

my prof described it as a venn diagram, but im not sure where to go from there

[frail skiff]

del dot parent = vector field

because its conservative

[solid iris]

u mean just del?

[frail skiff]

the del operator itself is just a vector of derivatives

[solid iris]

del dot=divergence

ok u wrote del f=F so f is the parent?

[frail skiff]

yessir

[solid iris]

ic

[frail skiff]

i gotta plug in the endpoints/starting points so i can use the FTC

[solid iris]

in most contexts we call f a potential function

i just wanted to clarify

[frail skiff]

ohhh okay mb

yeah my prof said parent function and potential function were interchangeable terms

[solid iris]

u cant use g for unknown function every time

[frail skiff]

uhhh

how come

[solid iris]

they may be different functions

[frail skiff]

but if you derive a function with respect to the main variable, the other two would disappear

so if you integrate that function you have to add an unknown constant that would include those two

[solid iris]

thats not the issue

u already used g for an unknown function

u need to use smth else for another unknown function

[frail skiff]

ohhh i get what you mean

[solid iris]

say h

[frail skiff]

i mean the variable choice is arbitrary i just dont know how to assemble the function

i can just use c1 c2 c3

[solid iris]

https://i.gyazo.com/5a588d4c60f6af344ccab344d915b660.png

[frail skiff]

yessir

[solid iris]

lets call the unknown functions g_1, g_2, g_3

[frail skiff]

ya

[solid iris]

theyre all equal yes?

bc they all equal f

so we must compare all the expressions

g_1 consists of terms only depending on y,z

[frail skiff]

Yessir

[solid iris]

can u tell me those terms

[frail skiff]

X^2 y and e^x sin y

Sorry my formatting might be a bit weird I switched to mobile

[solid iris]

no

g_1 consists of terms only depending on y,z

[frail skiff]

Oh LOL sorry dyslexic

Y^3 z and just z

[solid iris]

ye

$g_1(y,z)=y^3z+z$

[grand pond]

ロケットジャンプ

[solid iris]

g_2 consists of terms only depending on x,z

what are they?

[frail skiff]

There are none that have both x and z so ig just z

[solid iris]

ye

$g_2(x,z)=z$

[grand pond]

ロケットジャンプ

[solid iris]

same logic for g_3, tell me what u get

[frail skiff]

X^2 y and e^x sin y

[solid iris]

$g_3(x,y)=x^2y+e^x\sin y$

[grand pond]

ロケットジャンプ

[frail skiff]

Oh is it as simple as just assembling them into a system of equations and crossing out the duplicates

[solid iris]

yes, just matching terms that each function depends on

now we done

[frail skiff]

Got it appreciate it dude I think I was just overthinking lol

[solid iris]

np man 🙂

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[tacit kelp]

why is it 90.0 - 32.6???

[stiff pollen]

try drawing it

[tacit kelp]

forgot to include this

here is the diagram

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[twilit field]

Let $f$ and $g$ be functions of one real variable and define $F(x,y) = F(x \cdot g(y))$ . Find the first partial derivative of the the first order , expressed in terms of the derivatives of $f$ and $g$

[grand pond]

What a wonderful world !

[twilit field]

so $\pdv{F(x\cdot g(y))}{x} = \pdv{F(x \cdot g(y))}{x \cdot g(y)} \cdot \pdv{x \cdot g(y)}{x}$

[grand pond]

What a wonderful world !

[twilit field]

I dont think I've simplified it enough

[twilit field]

Need to get rid of the circled part somehow

[twilit field]

I mean to start this is $\pdv{ F(x \cdot g(y))}{ (x \cdot g(y))} \cdot g(y)$

[grand pond]

What a wonderful world !

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[twilit field]

<@&286206848099549185>

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[lost ocean]

Can someone help me prove the convergence of $$x_{n+1} = x_{n} - \frac{2(f(x_n))^2}{f(x_n + f(x_n)) - f(x_n − f(x_n))}$$, the newton method with the approximation $$f'(x_n) = \frac{f(x_n + f(x_n)) - f(x_n - f(x_n))}{2(f(x_n))}$$

I'm using this prove as a base: https://fractal.math.unr.edu/~ejolson/701-12/code/hw2sol/hw2sol.pdf

[runic hamlet]

just type -

without the backslash

what are you doing

[lost ocean]

I'm trying

[runic hamlet]

just type a normal -

just like you did with +

[grand pond]

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

[lost ocean]

Some keep not appearing but I guess u get the gist

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[slim portal]

No. 20

A box contains five red marbles, two blue marbles, and one green marble. Two marbles will be randomly drawn from the box.

The probability that the two drawn marbles have different colors is...

(A) 9/25
(B) 2/5
(C) 5/14
(D) 11/26
(E) 17/28

[slim portal]

Please help me guiding

[viral dagger]

id count the probability that you draw the same color then subtract it

[slim portal]

Owh yeah

Compliment

Thanks

.close

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[lyric charm]

complement ≠ compliment

[slim portal]

[midnight plank]

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[twilit field]

The equation $f(\frac{y}{x},\frac{z}{x})=0$ implicitly defines $z$ as a function of $x$ and $y$. Say $z=g(x,y)$.
\
Show that $$x \pdv{g}{x} + y \pdv{g}{y} = g(x,y)$$

[grand pond]

What a wonderful world !

[twilit field]

NO idea

[dawn dagger]

have you tried to attain g_x and g_y?

[twilit field]

how would I do that

The best I can think of is

$

[dawn dagger]

you think of money?

[twilit field]

no

$\pdv{f}{x} \cdot ( \frac{-y}{x^2} - \frac{z}{x^2})=0$

[grand pond]

What a wonderful world !

[dawn dagger]

z is a function of x and y

[twilit field]

yea

Oh god

$\pdv{f}{x} \cdot ( \frac{-y}{x^2} - \frac{z}{x^2} \cdot \pdv{z}{x})=0$

[grand pond]

What a wonderful world !

[dawn dagger]

?

you would need product/quotient rule

[twilit field]

hmm?

I just need the chain rule

[twilit field]

why

[dawn dagger]

What if z = e^(x²+y²)

how do you differentiate z/x then

[twilit field]

hmm

fair

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[ornate umbra]

• I need help with d^2y/dx^2

[midnight plank]

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[trim lance]

how do I solve this? I'm stuck from the first step..

[slender walrus]

express x and y in terms of h

[trim lance]

like tan 20 = h/x like that?

[slender walrus]

yes

but isolate the x

[trim lance]

x = h/tan20?

[slender walrus]

yes

same idea for y

[trim lance]

y = h/tan 18

[slender walrus]

and then apply pythagoras

[trim lance]

ohhhhh

thank youu

wait i still cant get past that step though because i have variable h to figure out

[slender walrus]

applying pythag will give you an (quadratic) equation with only one variable h
solving that will give you its value

[midnight plank]

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[gritty holly]

so for part a i need to take the limit of the sequence right?

[tribal temple]

”Need to”, no

Try it anyway (cause I wanna watch you )

[gritty holly]

here is my attempt

[tribal temple]

Actually misread a and b, got them mixed up

[tribal temple]

Yep I thought a was actually b

[gritty holly]

okay sick so its convergent right becasue its 0

[tribal temple]

The sequence is convergent, and converges to 0, yep

Now, time to do part b)

[gritty holly]

ya... and since i know that the lim goes to zero then thats the same for a series divergence test right

but with the divergence test the 0 tells us that it diverges not converges right?

[tribal temple]

This is why I was wanting you to try it out

[gritty holly]

ya im looking at my notes

[tribal temple]

The limit being 0 of the sequence doesn’t tell us whether the sequence converges or diverges the series of 1/n^2 converges (to pi^2/6), the series of 1/n diverges (cough hint), but both corresponding sequences have limit 0

If the limit of the sequence is not zero, and is anything else (including a non zero limit, not converging, going to an infinity) then the series does not converge

They’re telling you to “Justify your answer using an appropriate series convergence test”, so we really should

Of course, divergence test is inconclusive here, so that’s one gone

[gritty holly]

what is it inconclusive? when the limit=0

[tribal temple]

Yep, inconclusive when the limit is 0, because you could have either [series] convergence or divergence if the [sequence] limit is 0

[tribal temple]

See here for a case of convergence and divergence

[gritty holly]

so if i needed to know wat the sequence converged to i would need to do another test? the answer isnt 0

[tribal temple]

Well, for the sequence, if you know it has a limit of 0, you’d have to do more work and another test

If you got the sequence limit as anything that isn’t 0, then you instantly know the series diverges

[lavish venture]

the sequence converges to 0 but do you mean the series perhaps?

[gritty holly]

i havent started the series yet i was just asking if i needed to know exactly what the sequence sin(1/n) converges to would i need to do another test

[lavish venture]

you do know what the sequence converges to

you just took the limit of sin(1/n) to get 0

the sequence a_n converges to 0

this does not mean the series converges to 0 though

[tribal temple]

(Nor that the series converges at all, remember the harmonic series)

[gritty holly]

the harmonic series is (-1)^n/n right

[lavish venture]

no thats the alternating harmonic i guess you could call it

harmonic series is 1/n

[gritty holly]

oh ya okay

[lavish venture]

out of curiosity, does your intuition tell you this converges or diverges

[gritty holly]

what do u mean by this?

the sequence or the series

[lavish venture]

like just by inspection/feel would you say the series sin(1/n) converges or diverges

without doing any tests

[gritty holly]

well i know that 1/n by itself diverges

[lavish venture]

mhm

so what do you think happens for sin(1/n)

[gritty holly]

and isnt sin(n) also divergent

[lavish venture]

it is

[gritty holly]

so is sin(1/n) also divergent

[lavish venture]

well its not because sin(n) is divergent

[gritty holly]

or does something weird happen that makes two divergent things convergent?

[lavish venture]

but yes

[gritty holly]

does this ever happen?

[lavish venture]

consider [\lim_{x \to 0} \frac{\sin x}{x}]

[grand pond]

knief

[lavish venture]

i mean

1/n times 1/n is 1/n^2 which converges but im not sure i know what you mean exactly

[gritty holly]

honestly neither am i its okay

[lavish venture]

do you remember what this is

[gritty holly]

0/0

so i need to do l'hospital right

[lavish venture]

sure

[gritty holly]

oh no

[lavish venture]

but this is a pretty famous limit

[gritty holly]

is it -cos(x)/x^2

[lavish venture]

that is not how you do lhopitals

[gritty holly]

oh god im going to fail my final

this is tragic

[lavish venture]

[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}]

[grand pond]

knief

[lavish venture]

assuming conditions are satisfied of course

[gritty holly]

so i need to keep the -1/x in the denominator?

[lavish venture]

g(x) = x

f(x) = sinx

[gritty holly]

oh so its just cos(x)

[lavish venture]

and as x -> 0

[gritty holly]

=1

[lavish venture]

why is this useful?

[\lim_{x \to 0} \frac{\sin x}{x} = 1]

[grand pond]

knief

[lavish venture]

can you see how this is relevant for our problem

[gritty holly]

um no

[lavish venture]

do you agree that

[gritty holly]

how do i get sin(x)/x from sin(1/n)

[lavish venture]

[\lim_{x \to \infty} \sin \left(\frac{1}{x}\right) = \lim_{x \to 0} \sin x]

[grand pond]

knief

[gritty holly]

ya

[lavish venture]

and you agree that

[\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to 0} x]

[grand pond]

knief

[gritty holly]

ya

[lavish venture]

ok and you still dont see how this shows up?

what tests do you know

that might involve a limit

[gritty holly]

let me think 🥹

[gritty holly]

the limit comparison test?

[lavish venture]

yep

can you state what the test says

[gritty holly]

okay thank god

[gritty holly]

um i can try to type it or i can send a pic of the notes

[lavish venture]

either

whats the important bit

[gritty holly]

L= lim a_n/b_n and the 0<L<inf then they behave the same

if that makes sense

[lavish venture]

yep

can you identify what a_n and b_n we might be interested in

[gritty holly]

well a_n has to be sin(1/n) and can b_n be sin(n)? or does that not work

[lavish venture]

why?

[gritty holly]

why what?

[lavish venture]

why b_n = sin n

what was the whole point of me showing you sinx / x

[gritty holly]

that the limit was 1, but how can comparing sine to x be legal? dont a and b have to be related

[lavish venture]

idk what you mean legal

[lavish venture]

why would i ask you this?

[gritty holly]

like acceptable

[lavish venture]

and this?

[gritty holly]

but the limit is changing from infinity to 0 in those but in the limit comparison its just infinity

[lavish venture]

yep

so

what should b_n be

[gritty holly]

oh fuck is it 1/n

im over here about to cry this fucking class is so frustrating!!!!

[lavish venture]

you got it though

you need to know this to solve climate change

[gritty holly]

so i have this

[gritty holly]

no to solve climate change we need to reorganize society away from fossil fuels and consumer culture

[lavish venture]

what happened to sin(1/n)????

[lavish venture]

word

and determine the convergence of series

[gritty holly]

they left the chat

[lavish venture]

every problem you do cleans up 100 pounds of microplastics in the ocean

probably

https://tenor.com/view/jolyne-meme-discord-discord-server-dead-chat-disappear-gif-22305219

chartbit has the best reacts

[gritty holly]

okay wait give me one sec my dad called me and i need to pay tuition real quick

[lavish venture]

rough

at least you have in state tuition

[gritty holly]

okay im back

[gritty holly]

ya but its still basically 15k a year

[lavish venture]

better than 60

[gritty holly]

ya thats crazy

[gritty holly]

this is so true🫶

[gritty holly]

so now i can say that this limit is 1 (i fixed the 1/n)

[lavish venture]

which means

[gritty holly]

that they behave the same and are both divergent

[lavish venture]

killing it

try this one

$\sum_{n = 1}^{\infty} \frac{\cos^2 n}{n^2}$

[grand pond]

knief

[gritty holly]

oh no the anxiety spike from this

can i do the divergence test?

[lavish venture]

try it

[gritty holly]

is it 1?

[lavish venture]

no

[tribal temple]

You're banned from using the word "it"

[lavish venture]

i think if i tell you the trick you won't learn

hmm

do you know any inequalities involving sin and cos

[gritty holly]

noo what am i going to do

[tribal temple]

State what you mean as clearly as you can (else you'll get a react, you even admitted you use ambuguity to get away with it )

[gritty holly]

they are both confined between -1 and 1

[lavish venture]

which means what about cos^2

[gritty holly]

ya i should have kept my mouth shut about that... now its biting me in the ass

[lavish venture]

$-1 \leq \cos x \leq 1$

[grand pond]

knief

[gritty holly]

ya i have no idea how squaring it would change it

[lavish venture]

well what do you know about x^2 to begin with

do you know any inequality about x^2

[gritty holly]

its a U shape graph

[lavish venture]

inequality

[gritty holly]

that goes on to infinity

[lavish venture]

and what else

so it has no maximum

does it have a minimum?

[gritty holly]

ya 0

[lavish venture]

^

[gritty holly]

0<x^2<infinity

[lavish venture]

x^2 >= 0

how about that

[gritty holly]

looks great!

[lavish venture]

so what can we say about cos^2 x

[tribal temple]

(on a more serious note though, it does mean we can make sure you understand stuff properly, and of course I'd like to think you know it's safe to try anything and we won't roast you )

[gritty holly]

umm would cos^2 have to be greaterthan 0 too?

[lavish venture]

yes

and whats its maximum value?

[gritty holly]

but cos^2 cant to to infinity can it?

[lavish venture]

it can not

[gritty holly]

would cos^2 still have 1 be the max

[lavish venture]

$0 \leq \cos^2 x \leq 1$

[grand pond]

knief

[lavish venture]

why is this useful?

[gritty holly]

also the amount of times ive had to change "it" to cos^2 already is horrendous

[gritty holly]

maybe to know that cos^2 wont be negative?

[lavish venture]

well we have this yea

lets start with what your intuition tells you

is it going to converge or diverge

[gritty holly]

my intuition is shit!

[lavish venture]

50/50

[gritty holly]

ill say converge becasue p>1

[lavish venture]

how can we use this?

what test should we use to prove this

we know 1/n^2 converges since p > 1

[gritty holly]

ooo umm the direct comparison

[lavish venture]

can you see why this is useful

[gritty holly]

honestly no

[lavish venture]

well whats the goal of direct comparison

like what do we need to show

[gritty holly]

to show that they behave the same

[lavish venture]

but how

not with a limit

so how

refer to your notes if you need to

[gritty holly]

showing that if one series is smaller that something convergent it converges and vice versa for something larger than a divergent series

[lavish venture]

great so how can we use this

[gritty holly]

would it be since cos^2 is between 0 and 1 that its smaller?

[lavish venture]

$\cos^2 n \leq 1 \implies \frac{\cos^2 n}{n^2} \leq \frac{1}{n^2}$

[grand pond]

knief

[gritty holly]

yay okay

[lavish venture]

would you like another one

[gritty holly]

ya

[lavish venture]

$\sum_{n = 2}^{\infty} \frac{1}{(n + (-1)^n)^2}$

[grand pond]

knief

[gritty holly]

then ill keep trying to do my practive problems