help-17

[vocal pagoda]

damn, wrong button. I'll have to check again

<@&286206848099549185>

[vocal pagoda]

Please? I'd like some help.

[alpine merlin]

Whatchu want

[vocal pagoda]

Is this correct: $x=(x_1,...xn), V=\Sigma{i=1}^n V_i\partiali$ and $A=(a{ij}(x)){i, j= \overline{1,n}}$, then $V(A)=\Sigma{i=1}^n V_i(\partial_iA)$

[twin meteor]

Wild123
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[alpine merlin]

um

[vocal pagoda]

I'll correct the stuff. apparently

it doesn't copy paste well

[alpine merlin]

!original

[vocal sleet]

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

[vocal pagoda]

I posted it earlier

[alpine merlin]

can you repost it :3

[vocal pagoda]

Just scroll a little up...and you have it all xd

ok I'll try

[alpine merlin]

oh this is easy

[vocal pagoda]

$\Omega \subset \mathbb{R}^n$ is an open set, $A, B$ differentiable functions, $A : \Omega \rightarrow M{mn}(\mathbb{R}), B : \Omega \rightarrow M{nk}(\mathbb{R}), u : \Omega \rightarrow \mathbb{R}^n$. $\forall x \in \Omega, V \in T_x\mathbb{R}^n$, we define $V(A), V(u)$ as the differential of A and u along the vector field V.
I try to see if $V(AB)=V(A)B(x) +A(x)V(B)$.

[twin meteor]

Wild123

[alpine merlin]

yeah im sorry but youre on your own 🙏

too many greek letters

that i dont know

[vocal pagoda]

alright

I'll open a new room

.close

[vocal sleet]

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[vocal sleet]

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[vital wave]

i got [-2 -4 0 1] but my book says the answer is [3 5 1 0 ], [-2 -4 0 1]

can someone help me

[boreal remnant]

define 'null space'

[tawny salmon]

show your work

[vital wave]

this is RREF

there is a point where the vector they havve is similar but not the same.... before it is reduced

the -3 -5 1 but they haeve an answer as 3 5 1 0

i have no idea where they got that answer from, do you?

[tawny salmon]

hmm it should only be {-2 -4 0 1] like you said

[vital wave]

this is their answer

<@&286206848099549185>

@alpine merlin i did

but whyis this other vector in the book's answer?

[alpine merlin]

idk

what is it asking you to do

[vital wave]

#help-17 message

[rugged orchid]

What do you do at this point

[vital wave]

i wrote it in parametric vector form

then i got my answer..

[rugged orchid]

What did you get

Show your work

[vital wave]

i got -2 -4 0 1

that is the same answer as the book

but they have the other vector in there also

[rugged orchid]

Show your work

[vital wave]

where is that coming from

no i got the right answer

[rugged orchid]

…

Clearly not if you’re missing 1 vector

[vital wave]

i used a matrix calculator to get RREF what work would you loike to see

which part

[rugged orchid]

From this point onwards

[vital wave]

It’s number 9

this is their answer

[rugged orchid]

,w row reduce ((1,0,-3,2);(0,1,-5,4);(3,-2,-1,-2))

[twin meteor]

Results provided by WolframAlpha

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

Requested by frosst

[rugged orchid]

They’re wrong I think

[vital wave]

this is like the 10th time they have wrong answers in this book

[rugged orchid]

Ok sorry for saying you were wrong

[vital wave]

lol aright

[rugged orchid]

This matrix takes vectors of ℝ⁴ by having 4 columns

There’s 3 pivots in the rref

4-3=1

The kernel has dimension 1

So there should only be 1 basis vector

(This is rank nullity theorem)

[vital wave]

my university strikes again!

always using the most up to date bs

ok what about this one...

i did RREF

has 0 vector

so x3, x5 are free?

[rugged orchid]

There’s 3 pivots

2 basis vectors of the kernel

[vital wave]

okk cool

Got this

[rugged orchid]

Could be

[vital wave]

good enough for me right now

im gona close this and make another one

.close ty 4 help

[vocal sleet]

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[wraith magnet]

how do I solve a differential equation when both sides are polynomials (of y and x respectively)?

[wraith magnet]

the problem has

I have gotten it down to

hold on

1/2p^2 + p = 1/3t^3 - t + C

I don't know where to go from here

could I complete the square with P or something?

omg yeah it would work!

[dull bear]

(to clarify, you wanna get p as a function of t only?)

[wraith magnet]

I see it now

[wraith magnet]

general solution

ap calc ab unit 7

[dull bear]

Cool, are you given like any initial conditons or anything?

[wraith magnet]

no

just general

well I think I figured it out so bye and thanks :3

.close

[vocal sleet]

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[winged sigil]

Im having a hard time trying to figure out how to set up this problem to find the area bounded by this region

[winged sigil]

what do I do about the square root

tag me for a response

[deft oar]

you need two integrals

one that's sqrt-(-sqrt)

one that's sqrt - linear

[sharp lynx]

or just integrate with respect to y instead

[deft oar]

👺

I think you should do it in polar

[sharp lynx]

https://tenor.com/view/cat-skate-board-gif-778367011615093516

[vocal sleet]

@winged sigil Has your question been resolved?

[winged sigil]

you subtract the bottom function from the top function so in this case is it like intergral 2 to 26 of sqrt -(-sqrt) + intergral 2 to 26 sqrt - linear?

[vocal sleet]

@winged sigil Has your question been resolved?

[faint pumice]

do you know how to do it in terms of y

that would be easier

[winged sigil]

like y= root(x-1) and y = 6-x/-4?

[faint pumice]

yes

and then you integrate right-left

so linear - quadratic

[faint pumice]

wait hang on

actually looks like youre gonna have to do 2 integrals regardless 😭

wait

nh

[winged sigil]

Im still a little confused on how Im setting this up

[faint pumice]

sorry

[faint pumice]

this would be in terms of x

in terms of y it would be y^2 +1 and 4y+6

using this, you could do 1 integral

[winged sigil]

these are the equations I was given:
y^2 = x-1 and x-4y = 6

[faint pumice]

yes correct

lemme show you

[faint pumice]

have you integrated with respect to y?

[winged sigil]

I don't think your in terms of y is right

I think it's + 1

[faint pumice]

oh yes

sorry i mean y^2 +1

do you know what to do from here though ?

[winged sigil]

Do I intergrate from 1 to 26 with the top function being x = y^2 +1?

[faint pumice]

when integrating with y its a little different, you do the right most function - left most function

so instead it would be linear - quadratic

but yes, the bounds would be 1 & 26

do you know if your teacher wants it integrated in x or y ?

[winged sigil]

Whats the reasoning behind this. I just want to know so I understand it better. And would the bounds not change?

[faint pumice]

okay wait

i am sorry the bounds would be 1 & 5

wait

no

fuck sorry

[winged sigil]

It doesn't matter as long as you get the right answer. I think she went over something like this before but I forgot how to do it

[faint pumice]

okay

cause i know sometimes the answer can be different depending on how you integrate

anyway if youre integrating in y, you need to use the y values of intersection

in this case, -1 & 5

[faint pumice]

the same reasoning with x

i can try to make a visual rq

[winged sigil]

ok

[faint pumice]

okay

this is in terms of X

you subtract the top function from the bottom function to find the region

because that is where the areas do not overlap

and its based on area under the curve going to the x-axis , as you see

however, with y

it is this

the area under the curve goes towards the y axis

in this case, the right most function has a greater area

does this make sense?

[winged sigil]

whos area is the purple the linear function?

[faint pumice]

purple is the quadratic

linear is blue

here does this help

blue cant be the quad but it goes "above" the function (atl in this orientation)

[winged sigil]

Oh it makes sense now lol thats the area under the curves

so it would be linear minus qudratic

[faint pumice]

correct

the area when integrating in y goes from the left of the curve to the y axis

[winged sigil]

So it in terms of y your intergrating along the y axis and in terms of x along the x axis?

[faint pumice]

yes, correct

so you use the respective bounds for each. y integration uses y-value bounds

x integration uses x-value bounds

[winged sigil]

So would the intergral look something like this?

[faint pumice]

(4y+6) - (y^2+1)

but yes

[winged sigil]

oh yea

I think I understand it now. Its just hard remembering all the different methods especially with this volume of a solid thing now

[faint pumice]

yeah i understand, only thing that helps is practice

what calc are you in ?

[winged sigil]

2

[faint pumice]

same bruh

it does not get better

😹

[winged sigil]

[faint pumice]

my uni didnt accept my apcalc bc score

so now i gotta retake it

and im absolutely smoked

[winged sigil]

I didn't even take calc and I regret it now. I took AP stat instead. If I knew it would've let me skip this stuff I would've tried to take it and take it seriously

[faint pumice]

i only took calc cause i have a hatred for stats bahaha

[winged sigil]

Funny thing is I didnt even take the AP stat test. I was just taking AP classes to boost my gpa I really wasnt knowledgible about the credits and stuff

[faint pumice]

it helps that i have a baseline but damn

its a different ballpark for calc 2

[winged sigil]

I think i was traumatized from pre calc tbh

[faint pumice]

i dont even remember precalc 😂

its very funny that i cannot remember very basic math things

i was struggling helping someone with limits earlier, people who say limits are easy are fake news

[winged sigil]

tbh me personally I think pre calc was my hardest my class. Just because you had to learn so many fundamentals

[winged sigil]

[faint pumice]

have you taken a trig course, just trig ?

[winged sigil]

I thought calc 1 was really easy though and I did very well but now calc 2 is like getting hit by a truck

[faint pumice]

trig is tough for me personally

there is so much to remember

and so many identities

thats why i sorta struggle in calc 2 because there is so much trig you should know lol

[winged sigil]

well its included in pre calc. but my advisor did recommend me to take a separate trig course which I did but it really didnt help that much

[faint pumice]

brightside is that calc 3 is easier, or so they say

[winged sigil]

So far I don't think there's been that much trig or atleast its like the ones that you can find out easily

[faint pumice]

how far into your semester are you ?

[winged sigil]

My midterm is on thursday

[faint pumice]

youve not done trig sub yet ?

just wait

[winged sigil]

Yes and no

She told us to watch a video on it but shes not testing us on it I think

[faint pumice]

oh goodluck btw, i just took one last week and absolutely bomed the shit

[winged sigil]

The only stuff I know we have done and been tested on is :
Regular intergrals
Intergral techniques: Partial fractions, By parts, With tables, U sub
Improper intergrals type 1 and 2 like infinity and a denominator going to 0
Arc length
Volume of solids
Some physics thing I don't even remember lol
Area between 2 curves
estimation methods like simpsons rule etc

I might be missing some stuff but I think thats everything we pretty much learned up till now

But its like we skipped past trig subsitution unless thats what intergration with tables is because she was saying something like the tables come from trig subsitution or something

[faint pumice]

ah okay i see

well im wishing you luck for trig subs, i dont think theyre actually that bad

but personally i struggled

[winged sigil]

The only indentities that we kind of used are these

we don't use any of the other ones

[faint pumice]

oh yeah

we use those sometimes

but its mainly like half angle identities and some other random bs

math is so superficial lol

[winged sigil]

But just like the unit circle atleast me personally I realized I was going about it wrong and I should just remember like the concept of it and solve to the find the other values. Even though it takes me like a lot more time to do a problem I think its better than just trying to jam all the information in and then you have to keep calling back to remember it

@faint pumice whats your major though

[faint pumice]

i get that

but there is stuff you just gotta remember because the work is absurd

[faint pumice]

biochemistry haha

hbu

[winged sigil]

oh dam u gotta take them science classes too hell nah

[faint pumice]

chem is on 🔝

you cant even see it bahha

science is fun 🤓

[winged sigil]

I took chemistry during covid in high school and I slept during all the classes during zoom

Once I heard something about a bond I was gone

[faint pumice]

my chem class in hs was abominable, an actual fever dream

[winged sigil]

I still got an A though because I cheated on all the test because the teacher got all of them from online

[faint pumice]

so i get that

whats your major ??

[winged sigil]

Computer Science

Whole bunch of math, theories, and algorithms

[faint pumice]

compsci scares me

i dont understand none of it

youve gotta take calc 3 then right ?

[winged sigil]

I warn a lot of people about it. If you don't have a passion for it or your not interested do not take it

[winged sigil]

yea either that or linear algebra I think

I still have to take I think differiential equations Im not sure

[faint pumice]

ive heard that diffeq aint terrible

i believe i have to take linear algebra too tho

not shore

[winged sigil]

Yea tbh I think the hardest thing is going to be getting through calc 2

Compsci is basically like here's a 1 million people find the fastest and most efficient way to fit them into this small house

[faint pumice]

ngl the typo made me laugh

but yeah calc 2

[winged sigil]

[faint pumice]

its gonna be tough

im gonna lock in though

cause after that exam

i literally am to believe my grade is boutta go from 93 to 65

dead ass

[winged sigil]

I really want an A but tbh the way its looking we might be shooting for a B lol

[winged sigil]

I pray for u

[faint pumice]

the thing is tho

a B is great for calc 2

ik many people who have taken it at least 2 times

[winged sigil]

Dam

[faint pumice]

i believe you can get an A though

[winged sigil]

Idk man. I lowkey still don't even know how to intergrate using partial fractions

I never understood that crap

[faint pumice]

partial fractions are bs

yk that was actually on my exam, and i spent HOURS practicing it

only for a single question to be on partial decomp lol

[winged sigil]

[faint pumice]

almost crashed out mid exam

[winged sigil]

Like it looks like it shouldnt even be that complicated im just slow

And than sometimes Im trying so hard to remember how to use some of these formulas I forget how to do basic algebra

[faint pumice]

sounds about right bahahaha

look

hang on

@winged sigil

27 mins is insane

i had to do this like 8 times before i understood it

they kept giving me this exact question too

[winged sigil]

lol looking at that now I don't even think I would know where to start

[faint pumice]

its so butt

when my prof went over a similar question

[winged sigil]

Do you use webassign too or something different

[faint pumice]

literally all that was going through my mind is that whoever made this stuff up was fr pullin it out of their ass

[faint pumice]

nah

knewton alta

[winged sigil]

I feel like webasiggn is the worst math software ever created

[faint pumice]

ive not heard of it before now

knewton alta is quite nice, but sometimes i go a lil brain dead and cant seem to understand the steps they skip

[winged sigil]

Its good and bad because the questions will be so hard and out of pocket sometimes but if you figure out how to do them its easy when theres an exam because the questions the teachers give arent nearly as hard

[faint pumice]

and chatgpt cant even come in clutch cause its wrong 99% of the time i use it

[faint pumice]

thats what i prefer

[winged sigil]

Lol u gotta know how to instruct it properly. But it gets kind of tricky when your asking it to answer estimation questions sometimes

[faint pumice]

no i swear

this does not work

i have to ask it bit by bit cause it will get it completly wrong when i ask a full question

itll forget stuff moving from one line to another bahaha

[marsh jungle]

or just learn it tbh

its less work in the long run

[faint pumice]

breuh

i do learn it

i was referring to when my homework software skips steps

i need to see the steps sometimes

[winged sigil]

I will be doing a problem over and over again come to find out Im just missing like a step or something and chat gpt is good because its there to help explain the thing to you vs u just sitting there not having anything

[marsh jungle]

ah

[winged sigil]

And when its all said and done Im turning something in vs nothing if I can't figure it out in the end.

[faint pumice]

sometimes chatgpt is confusing as hell too lol

but i think thats when its also wrong

[winged sigil]

You can ask it follow up questions like explain further how you got step .... for example.

[faint pumice]

i do that yes but

i deadass dont understand what its saying sometimes

[faint pumice]

so for one of these questions i couldnt understand a step it was talking about

and had to ask 3 different times to re-explain lol

[winged sigil]

@faint pumice u speak french?

[faint pumice]

un peu

bahaha

[winged sigil]

My french use to be pretty good but now its down to basic vocabulary and I won't really be able to understand someone if they are speaking to fast

[faint pumice]

better than me, im still learning. my listening is horrible

whered you learn ?

[winged sigil]

I went to a french/english elementary school and everything was in french until 5th grade when we started learning english cuz u have to know once u get to middle school

but like no one spoke french outside of school tho everyone pretty much knew english already

[faint pumice]

you from canada ?

[winged sigil]

except for like proper grammar, spelling etc

[winged sigil]

Nah Im from the US maryland

[faint pumice]

oh thats interesting

makes sense cause its close of french canada but i wouldnt expect the schools to be french-english

[winged sigil]

Theres a lot of like french/english schools that you can just enroll in

@faint pumice r u from canada or the Us?

[faint pumice]

the US 🦅

haha

[vocal sleet]

@winged sigil Has your question been resolved?

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[native falcon]

could someone explain to me why the tangent lines are there? im confused how graphing the tangents of dy/dx would work

[native falcon]

is it just because thats where the graph has a slope of -2?

[woeful igloo]

Yea, you can see that the slope of tangent is gonna be negative in only these sections of the curve, and if you eyeball it, then its reasonable to see that the central section doesnt have slope of tangent smaller than -1

[native falcon]

ohh okay!! would it still be the case if it asked dx/dy=-2? sorry implicit differentiation kinda confuses me 😭

[woeful igloo]

Once you identify which section is reasonably good enough for this, you draw those 2 lines

dx/dy is 1/(dy/dx)

so you can make suitable changes to guess the correct section

changes = algebraic manipulations

[native falcon]

ahh gotcha 🫡 so a slope of -1/2 on the graph somewhere?

[woeful igloo]

yep

[native falcon]

hypothetically

awesome thank you!!

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[silver lynx]

any ideas on this limit?

[tight hornet]

i think i should be e^n factorial but let me check once

[silver lynx]

it's + not -, my apologies

[paper depot]

take the log

see what comes of that

in fact you will find that the logarithm of this limit will end up being the derivative of f(x) = log(1 + sum sin(kx)) at x=0

(not l'hop)

[silver lynx]

uhh log of a limit? I'm not sure I know how that works(I'm really bad sorry💀)

was it some sort of formula?

[tight hornet]

yea there is a method, u put the limit on e^lim then log

[tight hornet]

sry, nvm this

this is the method, i dont remember the proof, but the limit stays the same if u do this

So, by using this method u can do that

[hybrid flicker]

because ln(f(x)) is equivalent to f(x)-1 as x->a

[silver lynx]

ohhhh I see

I might have seen that before

okay I'll try with that and see how it goes

thanks!

[tight hornet]

oh yea, thanks

[tight hornet]

mhm i think u should get e^n!, by using sinx/x = 1 (when x is tending 0)

[silver lynx]

hmm but what if I do try to use l'hopital?

[tight hornet]

that would solve it too

[silver lynx]

something like that?

[tight hornet]

yea, all the sins change to cos and the x in the denominator will become 1 giving e^n!

[tight hornet]

whats that power?

[silver lynx]

uhh

well cos 0 was 1 right

[tight hornet]

yea

[silver lynx]

and if I derivate each term

[tight hornet]

yea by chain rule u would get 1 + 2+ 3+ 4+ 5..

[silver lynx]

when doing lhopital

yea

[tight hornet]

mhm = n!

[silver lynx]

so that's n(n+1)/2?

[tight hornet]

oh yea

[silver lynx]

it's not ×

[tight hornet]

sryy

my bad

[silver lynx]

ahh it's okay

[tight hornet]

right right

u are correct

[silver lynx]

I'm just confused 24/ hahah

tha k u sm for the idea

I hope u have a wonderful day😭

[tight hornet]

its ok no prob

[tight hornet]

u too

[silver lynx]

.close

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[keen frost]

convergence test

[keen frost]

im using 1/n^2 as a comparison test

so if I'm taking the limit, the answer will be convergence, which is 2. Is it correct?

[winter hawk]

yes

[keen frost]

.close

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[atomic pine]

convergence is correct idk what you meant by 2

[winter hawk]

LCT 2

[atomic pine]

oh i didn't know theres numbering on that

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[weary roost]

i think the answer to this is (5/6)^3, as p(not getting six) = 5/6

[tight hornet]

u r correct

[weary roost]

okay thanks

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[upbeat flax]

Greetings folks and good evening!

The problem: Find the exact length of the curve y = ln(1 - x^2), 0 <= x <= 1/2

Formula used: L = (a to b) sqrt( 1 + [ f'(x) ]^2) dx

What I am having trouble with:

I am currently stuck at figuring out where to go next.

After getting the derivative of f(x), removing the square root through computation, I am at:

(0 to 1/2) (1 + x^2) / (1 - x^2) dx

I do not know where to proceed from here. Do I start integrating this? Or am I still missing something at this step? Any help is appreciated 😄

[gentle sleet]

can you show your work to get to that integral?

[upbeat flax]

Best I can do is give a photo shot. Give me 2 minutes!

[gentle sleet]

yeah, photo is fine

[gentle sleet]

looks good so far,
you can add and subtract 1 from the numerator to split that fraction

oh wait, that last step doesn't seem right

[upbeat flax]

Yeah the last step is where I'm iffy.

I can leave that and keep to the 2nd-to-last step.

Before I put a negative in front of the integral.

[gentle sleet]

instead of that last step, factor a -1 from the denominator

[upbeat flax]

Only from the denominator? o>o

[gentle sleet]

yeah like $\int \frac{x^2+1}{-(x^2-1)}\dd x = -\int \frac{x^2+1}{x^2-1}\dd x$

[patent nymph]

dx

[twin meteor]

Sepdron

[gentle sleet]

then subtract 1 and add 1 to the numerator to split the fraction

[upbeat flax]

Just to be clear, if we factor a -1 from the denominator, wouldn't it make 1 - x^2 into x^2 + 1?

[gentle sleet]

no, if you distribute -(x^2-1) back, you'll get -x^2+1 = 1-x^2

if it turned into -(x^2+1), distributing it back it becomes -x^2-1

[upbeat flax]

Dah right.

1 is positive before factoring the -1.

Ok, so really quickly, am I going to be using partial fractions at this point?

[gentle sleet]

you can but it's a bit overkill

[upbeat flax]

Really?

[gentle sleet]

we're going to achieve the same thing, but simpler

[fossil pier]

hey guys great work

also never give up

❤️

[upbeat flax]

Thank you professor frog

[fossil pier]

plus keep trying

[upbeat flax]

Gotcha, so I am wondering how you concluded on adding and subtracting 1 from the numerator to split the fraction

[fossil pier]

one day you could be a sigma

one day

anyways

[upbeat flax]

[fossil pier]

im just here to support you guys

[gentle sleet]

here's what I mean\
$-\int \frac{x^2+1}{x^2-1}\dd x$\
$-\int \frac{x^2-1+1+1}{x^2-1}\dd x$\
$-\int 1+ \frac{2}{x^2-1}\dd x$\

[twin meteor]

Sepdron

[upbeat flax]

Ok got it, but what rule or why did you do this? o>o

I can see that once you split the fractions, you have one that cancels into 1, and then you just add the two +1s together and have the same denominator. I am wondering how you figured that next step from where I was before.

[gentle sleet]

it's a pretty common thing to try to reduce the fraction if the top has equal or higher power of x

[upbeat flax]

Ok!! What formula do you use to do that?

[gentle sleet]

I'm not sure what you mean

[upbeat flax]

I do not know how I would deduce what you deduced.

Because I do not know how you would reduce a fraction if the top has equal or higher power of X.

[gentle sleet]

in this case, you can do this trick
but you can also do it using polynomial long division/synthetic division

[upbeat flax]

But again, I am lost for context for the quick trick you did.

All I can assume that you did is:

Because in that fraction, because the two Xs are equal power (numerator can have a greater X power to apply this trick), you added both numerator and denominator constants only to the numerator.

What I got is this from you:

Apply when: The X power of the numerator of a fraction is equal to or greater than the denominator's X power.

What to do: ???

[gentle sleet]

ok so the goal is to reduce the power of x in the numerator\
since in this case the top and bottom are very similar, x^2+1 and x^2-1
it'd be nice if the top can be x^2-1, since then they'll cancel
but I can't just add a -1 there, I have to compensate with a +1

idrk how to explain it, ig it's just a pattern that you recognize

[gentle sleet]

also after this you'd have to do partial fractions, I thought 1/(x^2-1) was a known derivative

[upbeat flax]

So that is what I am doing right now

First long division, and then on the remainder, do partial fractions

I got 1 + (2 / x^2 - 1)

x^2 - 1 would be (x-1)(x+1)

[gentle sleet]

ig that trick is just a shortcut for that long division

[upbeat flax]

AGH YEP.

I just feel like a big dumb for not knowing

Thank goodness I'm refreshing again on a lesson that's supposed to be one of my tools in my toolbox.

Alrighty, I got it.

The missing step was to refer to Integration by Partial Fraction. First step: Long Division if numerator has a greater or equal X power than denominator. Second step: Do the partial fractions. Compute from there, and it's solved.

Thank you for your guidance Sep!

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[upbeat flax]

Myself.

The person assisting me, Sepdron, implemented a technique that I was not sure how he was doing. I referred back to my notes and figured out that I needed to long divide.

[paper depot]

second line, you tried to get the fractions to have a common denominator but you messed this up on both of them by the looks of it

don't call me "dude".

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[paper depot]

<@&268886789983436800> transphobe

[mild flower]

it's my world

and you're leaving it

[paper depot]

haylsune miku

[vocal sleet]

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[spring sleet]

Hii

[spring sleet]

How are u all?

[vast shale]

Not really a place to ask such kinds of questions

[spring sleet]

Oh ok sorry

[vast shale]

U can use #discussion or #chill instead

this is for actual questions

[spring sleet]

okk

thanks

[vast shale]

Do close the channel by typing .close

@spring sleet

[oak magnet]

.close

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[hollow quail]

i've got this limit and im trying to figure out what number y tends to

[next bay]

Have you learnt l'hopital rule?

[hollow quail]

we don't learn that at school

[next bay]

Like is that name familiar

Oh

[hollow quail]

i have heard of it from exercises i've seen online but they don't teach us that at school

[paper depot]

so you're looking at $\lim_{x \to 0^+} \frac{\ln(1+x)}{x}$ now, yes?

[twin meteor]

Ann

[hollow quail]

yep, thats it

[next bay]

Been a long time since ive done a limit -w-

[paper depot]

have you learned about derivatives in general?

[hollow quail]

i have yes

[paper depot]

ok this is the derivative of ln(t) at t=1

literally by the limit defn

[hollow quail]

so I dont have to change the variable?

i was planning on doing y = ln(x+1) to then use the lim when x tends to 0 = e^x -1 / x

which is 1

[next bay]

Oh i get it

[paper depot]

brackets

but you could do that as well, only your limit would become y/(e^y - 1) as y -> 0

and then it would be the reciprocal of the derivative of e^t at 0

no, you don't have to change the variable here

[hollow quail]

yeah, exactly

alright, another question

if i were to change the variable how would I see where y tends to?

if I were to do y = ln(x+1)

[next bay]

$\frac{d\ln(t)}{dt} = \lim_{x \to 0} \frac{\ln(t+x)-\ln(t)}{x}$

Ughh

At $t$=1 is the limit we want to solve

[twin meteor]

not a rich man

[next bay]

Ah thats smart

[paper depot]

\ln

[twin meteor]

not a rich man

[next bay]

@hollow quail ?

[hollow quail]

it works yeah, thing is since I wasn't taught that way I cant use it on tests

which is extremely stupid

[next bay]

That sucks

Gl though

[hollow quail]

do you know how to do this?

[next bay]

Were you only taught in school subtitution?

[hollow quail]

yep

I was taught derivatives but we've never used them on something like this

[next bay]

Try Let $1+x=e^{t}$

[twin meteor]

not a rich man

[hollow quail]

whats "let"?

[next bay]

Just letting some vairbake equal to a different one

[hollow quail]

oh yeah sorry, thats what I did

but I can't solve it further without knowing what y tends to

[next bay]

Was the question explictely asked to be solved using subtitution?

[next bay]

You could tell your professor you solved it like this

[hollow quail]

it wasn't no but in the solutions they use substitution

exactly like Idid

expect they say Y tends to 0+

[next bay]

That isnt necessarily a way you havent been taught unless you havent been taught derivative yet

[hollow quail]

which I don't understand why

[next bay]

Well when you subtiute in a limit you have to also change what the limit tends to

[hollow quail]

yeah, thats what im asking

why does y tend to 0+

and not +inf

or something of the sort

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

i just replace x with 0 in y = ln(x+1) and it gives me 0

is this it?

[next bay]

Let $y=\ln(1+x)$ , then take $\lim_{x \to 0^{+}}

[twin meteor]

not a rich man
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[next bay]

ln(1)

Which is 0

[hollow quail]

yeah

thats it

thanks

how do I put solved again?

[next bay]

Mb for not understanding it correctly

[hollow quail]

yeah its fine, just glad I was able to understand it now

[next bay]

.close

[hollow quail]

ah alrihgt

[next bay]

Byebye

[hollow quail]

see ya!

.close

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[umbral gyro]

Show that for any a, b, c integers, the system of equations (in pic) only has the trivial solution x=y=z=0

[umbral gyro]

I have tried adding the equations together and got something along the lines of x+y+z=0 then a+b+c = 1/2 but I don't feel like thats conclusive

[vocal sleet]

@umbral gyro Has your question been resolved?

[umbral gyro]

<@&286206848099549185>

[fleet ember]

I think you meant different rather than equal, also, you proved that x+y+z couldn't be different from 0: a+b+c has to be an integer since it is the sum of 3 integers

you can continue from there

you know that -z = x+y etc

[vocal sleet]

@umbral gyro Has your question been resolved?

[umbral gyro]

yeah that was what i proved in the first place, my bad for not explaining it properly
but i dont get this part "I think you meant different rather than equal"

[fleet ember]

assume x+y+z =/= 0, then you can simplify x+y+z = 2(a+b+c)(x+y+z)
then a+b+c = 1/2 which is impossible bc a, b and c are integers
so x+y+z has to be 0
that's not quite x = y = z = 0, but it's already a good step

[umbral gyro]

also the author's proposed solution says that the determinant of the system must be different than 0 (in order to use cramer's rule), then says the determinant of the system is an odd number so it has only the unique solution (x,y,z) = (0,0,0)

[umbral gyro]

yeah thats what i did

[fleet ember]

oh I assumed you didn't know the determinant
you can compute it with cramer's rule quickly and conclude yeah

[umbral gyro]

but in class I was told that a linear system of eq that has the results (i dont know what they're called in english) equal to 0 can have other solutions other than 0,0,0

[umbral gyro]

oh okay, i thought it was more of a theory-oriented question, not simply computing the result

[fleet ember]

if you calculcate the determinant, you find (2a-1)^3 + 8b^3 + 8c^3 - 12(2a-1)bc if I didn't make mistakes from head

[umbral gyro]

i think we got the same result

i wrote it differently

[winter hawk]

computations are just applications of theorems hence proofs as well

[fleet ember]

oh no I made a huge mistake (corrected)

[winter hawk]

you can do computations as a proof as long as you invoke the relevant theorems correctly

[winter hawk]

did u do RHS = coeff matrix

[fleet ember]

no I wrote 2a-1 = 2(a-1)

which didn't make it quite odd

now it works

[winter hawk]

👀

[fleet ember]

I hope

the det is an odd integer so is non zero

[winter hawk]

,w det[[2a-1,2b,2c],[2c,2a-1,2b],[2b,2c,2a-1]]

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by rokettojanpu

[fleet ember]

which makes (0, 0, 0) the unique solution

[umbral gyro]

that makes sense yeah

[umbral gyro]

i've have made a mistake somewhere as i got the det = 8* ( something ) without the -1 part

thanks guys, that answered my question

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[vast shale]

is this actually the answer? because i got arcsin(x/2)+x/sqrt(4-x^2)

[old sigil]

They're the same; tan(arcsin(x/2))=sin(arcsin(x/2))/cos(arcsin(x/2))=(x/2)/sqrt(1-sin^2(arcsin(x/2)))

oh wait, did you get **+**arcsin(1/2x)+x/sqrt(4-x^2), not **-**arcsin(1/2x)+x/sqrt(4-x^2)?

[vocal sleet]

@vast shale Has your question been resolved?

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[vapid rover]

Did I find the area correctly? If not, where did I go wrong?

[marsh jungle]

12 is correct.

[vapid rover]

Alright, thank you so much!

[marsh jungle]

yeah no problem

[vapid rover]

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[vast shale]

i might’ve forgot the -

but that would only make me lose a point

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[hexed onyx]

Question about the term divergence test

[hexed onyx]

The test say if the limit is not equal to 0, then it diverges.

but doesn’t a finite existing limit mean that a series converges??

[fleet ember]

the limit of the term and of the serie are two different things

if you sum a sequence a_n = 1/n²

a_n goes to 0

and the sum of the a_n to pi²/6

[hexed onyx]

What do you mean the limit of the term?

[fleet ember]

lim a_n

[hexed onyx]

Am I not using the series when I use tdt?

[hexed onyx]

I thought it converges when it has any finite limit?

[winter hawk]

nope

if a_n diverges or has nonzero limit then sum a_n diverges

[fleet ember]

limit doesn't mean anything if you don't say limit of what, yeah convergence is having a limit, but a sum having a limit and the sequence having a limit is obviously an entirely different thing

as one of my teachers would say, you're using vocabulary in an accident-inducing way

[winter hawk]

careless is the vocab 😉

[hexed onyx]

Ohh yea I was thinking ab Sn

Alr

[fleet ember]

well, I tried my best to translate the wordplay from his language

[hexed onyx]

Can I add u in case I need help later

[fleet ember]

no, but you can ask for help here and ping helpers (or me)

[vocal sleet]

@hexed onyx Has your question been resolved?

[vocal sleet]

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[sand lodge]

For any function f : R → R, there exists a differentiable function F : R → R with F
′ = f.

[sand lodge]

its just true or false

but i can't think of any counterexamples

[manic flax]

what if its not continuous

[sand lodge]

so if F is differentiable, it must be continuous everywhere?

[manic flax]

yeah obv. but like im talking about f

what if f is not continuous

[sand lodge]

sure

i dont see how it relates tho

if f is not continuous somewhere then it is not differentiable there

[proud terrace]

what about $f(x)=1$ if $x \in \mathbb{Q}$ and 0 if $x \notin \mathbb{Q}$

[twin meteor]

tm

[sand lodge]

but the question doesn't ask about derivative of f

[manic flax]

its also non integrable

wait nvm i g there are funcn which are integrable but not continuous

[winter hawk]

mhm

[sand lodge]

oh but ig if a function whcih is not continuous because somewhere it goes to infinity then it is not integrable over that point right

like 1/x

or am i wrong

[manic flax]

yeah basically any function that cant be integrated fits as a counter example by taking it as f and naturally F does not exist

[sand lodge]

okok

.close

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[native falcon]

could someone explain why the correct answer is b? i tried using the definition of the derivative but this is what i ended up with

[manic flax]

what did you get as f'(3)?

[native falcon]

using power rule isnt it just 12?

[manic flax]

correct

now did you try evaluating B?

[native falcon]

i mean im not surprised if it would work backwards but im not sure how to get there

[bitter pilot]

they use a different definition

[native falcon]

doesnt the definition take the limit as x approaches 0?

o

[bitter pilot]

substitute a = 3+h

same limit

[native falcon]

are they using f(x)-f(a)/x-a?

[bitter pilot]

yes!

as x -> a

[native falcon]

that makes a lot more sense 🫡 i was wondering why i couldnt find my answer

[bitter pilot]

maybe should use something else than x

[native falcon]

ty for the help!! :> i think i got it from here, i completely forgot the other one existed

.close

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[modern geyser]

Does someone know why The Constand isnt added in the this step of integration

[winter hawk]

the integrals are implicitly assumed to come with +C

[modern geyser]

Why?

and their are 3 integrals so how would c cancel

[winter hawk]

theres no canceling of c's

just keep in mind to put +c after any one of these integrals

[modern geyser]

but why are you allowed to do that

arent they diffrent integrals

[winter hawk]

$\int vdu+c_1+\int udv+c_2$

[twin meteor]

ロケットジャンプ

[winter hawk]

just add c1 and c2 to make a new constant say C

$\int vdu+\int udv+C$

[modern geyser]

is that constant the same as uv's consatnt

[twin meteor]

ロケットジャンプ

[winter hawk]

if u have integrals on both sides of an equation, just subtract one side

$\int f+c_1=\int g+c_2$

[twin meteor]

ロケットジャンプ

[modern geyser]

So the C's do cancel

[winter hawk]

$\int f=\int g+C$

[twin meteor]

ロケットジャンプ

[winter hawk]

these are not the same c's

these are just individual rules for combining constants of integration

they can always be added or moved to other sides

in the end u always get a single constant on one side

[modern geyser]

but where is uv

[winter hawk]

$uv=\int vdu+\int udv+C$

[twin meteor]

ロケットジャンプ

[modern geyser]

I understand the other side has a constant but since when does that let you remove uv's constant

[winter hawk]

from here

[modern geyser]

Oh I see

but since when are you allowed to move the constant out of the integral

[winter hawk]

integration constants never are inside the integral...

theyre added on

[modern geyser]

but I thought that the intrgral of 1 was x + c, not x

[winter hawk]

see how c is added on?

think of it as integral 1=x, then u add on c

[modern geyser]

but then why in indefinte integrals do people only add c after doing the integral and not before

[winter hawk]

thats literally how i describe how indefinites are done

[warm prairie]

you add c because when you differentiate you lose information - the constant terms 'disappear' so when you integrate you add a +c to account for those constant terms

when its the integral form there's no need to have a +c

[winter hawk]

u focus on applying all rules

once ur completely done add c

[modern geyser]

so c is accounted for in the integral but is loss when you do the integral so you have to add c again

[winter hawk]

maybe an example would help

$\int(1+2x)dx$

[twin meteor]

ロケットジャンプ

[winter hawk]

use the sum rule

$\int 1dx+\int 2xdx$

[twin meteor]

ロケットジャンプ

[winter hawk]

i can add constants for each integral

$x+c_1+x^2+c_2$

[twin meteor]

ロケットジャンプ

[winter hawk]

but i wont actually do that, its tedious

and the constants can be added to make a single constant

so we dont bother adding a constant til the very end

$$

[modern geyser]

I was just confused why consants were added but the integral wasnt evaluated

[winter hawk]

$x+x^2$

[twin meteor]

ロケットジャンプ

[winter hawk]

NOW we're completely done. we can add c

$x+x^2+C$

[twin meteor]

ロケットジャンプ

[modern geyser]

ok so the integral is just the antiderivative

[winter hawk]

thats exactly the same thing happening in ur pic

thats why most books dont bother writing +C every time

its implied u will write +C at the end anyway

[modern geyser]

ok thank you

[winter hawk]

ur welcome

[modern geyser]

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[prisma parcel]

We're attempting to solve part C, so we have [P(x = 0.21) = \int_{0.21}^{0.21} 1(1-0.5x) , dx = 0]

I'm a bit confused should we have gotten zero there?

[twin meteor]

Ryan + James (TCC)

[prisma parcel]

Oh wait my book tells me

oops

well uh...amended question then, intuitively, why can't we say that that some event will occur with a specific probability then

[grim lotus]

can u summarise the q

[grim lotus]

u can, it's 0

[prisma parcel]

yeah ig that's what I'm confused about, why can we say there's a probability that an event can occur over a range but any specific value is just zero if that makes sense

I can see it mathematically but it feels a bit unintuitive ig

[grim lotus]

yes that is normal

probability 0 does not mean impossible

[prisma parcel]

ohhh yeah

it's just almost never

or whatever the term is

[grim lotus]

right

almost sure / almost never

[prisma parcel]

nod

[prisma parcel]

So I guess this means that we have a "higher chance" of our value being in a range rather than at any given specific value in the distribution?

[grim lotus]

well yes

it has higher probability

are u familiar with the measure theoretic formulation of probability?

[prisma parcel]

Vaguely

Just with the definition of a probability space and measure, no further than that

[grim lotus]

this issue can arise when ur sample space is uncountable

cuz the probability measure is only countable additive

[prisma parcel]

ahhhhh

[prisma parcel]

ex if it's R?

[grim lotus]

uncountable additivity doesn't make sense in this context afaik

like uncountable sums

[grim lotus]

yes

with Lebesgue measure

[prisma parcel]

makes sense

tysm :)

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[hearty delta]

do group actions have closure by definition? Like does $gx \in X \quad \forall g \in G$?

[hearty delta]

im asssuming this is the case

[twin meteor]

BOSS

[grim lotus]

absolutely.

[hearty delta]

And thats not like a proof, its legit just the definition of a group action?

I was asked to prove it for a problem so im curious if i can just write By Def

[cyan talon]

if you have to show a specific thing is a group action, well yeah you gotta actually prove it, it's part of the definition

[hearty delta]

Ah, so above the identity and associativity I have to always prove closure?

[cyan talon]

what def of group action do you have ?

it's prolly implicitly stated somewhere

[hearty delta]

Legit the question is "Prove that group actions meet closure"

Its a prelim so they are usually easy, I jsut wanted to make sure its something I can say

"by definition"

[cyan talon]

pretty odd

but I'm not asking about the question you have to solve, I'm asking about the defn itself

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@hearty delta Has your question been resolved?

[hearty delta]

ah sorry

[grim lotus]

yea u gotta show the actions actually maps into X

G x X --> X

if the action is transitive and free the X is called a G-torsor

which are the fibres of a principal bundle

[hearty delta]

what does this mean

[grim lotus]

$ \pi: E \to X$. $ \pi^{-1}(x)$ is a $G$-torsor for all $x$. There is a free group action $G$ on $E$, whose orbits are the fibres

[twin meteor]

martingale

[hearty delta]

ah

ok

Sorry working it out lmao

Really quickly, are all points in an orbit distinct?

Thet would have to be right

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[agile lava]

For part (ii). Ik that that the max occurs at x=2, but how to find the y value

[marsh jungle]

Do you know how to find out if f(x) is increasing by looking at the graph of f’(x)?

[agile lava]

yeh

[marsh jungle]

So which values do you think it would be increasing on?

Oh sorry you already noted that

Maxima and minima occurs at zeroes of the graph.

[agile lava]

yeh, x=2, what is y though

[marsh jungle]

When the graph goes from positive to negative, its a maximum, and if it goes from negative to positive it as a minimum

So you have to test that and the endpoints as well

[agile lava]

ik

what is the y value

[marsh jungle]

Does it ask for the y-value? I think you could just say it occurs at x=2

[agile lava]

no

maximum value is y

[marsh jungle]

Ah i see

You would have to use integration

From 0 to your max x value

Remember your formula for it

F(2) = f(0) + int from 0 to 2 of f’(x) dx

[agile lava]

integration formula?

what int

[marsh jungle]

Integral

Integral from 0 to 2

[agile lava]

integral of f'(x) is just f(x)

[marsh jungle]

Integral of f’(x) is also represented as the area underneath the graph

Therefore thats also the reason why the area under the graph is given to you

Does that make sense?

[agile lava]

oh ok

so the area A1 = 4

and hence max value = 4

[marsh jungle]

Yes

[agile lava]

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[deft vapor]

for part (e) I'm still debating whether it's

[deft vapor]

$1 - e^{-\rho}$

[twin meteor]

dXₜ = μXₜdt + σXₜdWₜ

[deft vapor]

or

$1 - e^{-(\rho + \beta)}$

[twin meteor]

dXₜ = μXₜdt + σXₜdWₜ

[deft vapor]

<@&286206848099549185>

[alpine merlin]

No idea

[deft vapor]

It's basically finding

$\mathbb{P}(T_{\rho} \leq 1 \mid T_{\rho} < T_{\beta})$

[twin meteor]

dXₜ = μXₜdt + σXₜdWₜ

[deft vapor]

but I don't know if it's with respect to

$Exp(\rho)$

[twin meteor]

dXₜ = μXₜdt + σXₜdWₜ

[deft vapor]

or the joint distribution of the two birds

$Exp(\rho + \beta)$

[twin meteor]

dXₜ = μXₜdt + σXₜdWₜ

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[charred sierra]

This is good but what is this??

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[sour depot]

can someone help draw a diagram I have no idea how to

[marsh jungle]

i cant help too much with this because i dont wanna give you the wrong help. but i can tell you for sure that an angle of depression goes down from the origin

so your 25 degrees would be 25 degrees south of east

same with the 33 degrees

[sour depot]

oki thanks

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[ocean badger]

Hello

[ocean badger]

I would like some help

<@&286206848099549185>

I've tried to find the similar sides

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[fast monolith]

Hello is there something wrong with my process or is this fine

[fast monolith]

Or might just be wrong

Nvm

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[wheat garnet]

how do i start this question, i dont know any process to begin with

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@wheat garnet Has your question been resolved?

[wheat garnet]

<@&286206848099549185>

[kind hemlock]

yeah idk man

[wheat garnet]

vectors are confusing

[kind hemlock]

draw it?

[wheat garnet]

i have the paper version of the question

aswell

[kind hemlock]

why am I tryna help vro

I'm only here for help myself 😭

[wheat garnet]

i just look around and see if i can do it

if i can, i help

in terms of questions.

<@&286206848099549185> sorry for the double ping

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