help-33

[lyric kelp]

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[gilded ocean]

need help on 2 and 3

[gilded ocean]

this is what I have for #1

[main plank]

no

[gilded ocean]

this is intro to linear algebra

[main plank]

your cooked

[marsh citrus]

@gilded ocean Has your question been resolved?

[main plank]

chat

😂

@gilded ocean

[gilded ocean]

<@&286206848099549185>

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[marsh citrus]

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[cerulean chasm]

I didn't understand the bit after a²>0

[gilded cedar]

dekho

jo since ke baad 2nd step h

usme wo eqn ya to ek baar touch kregi axis ko

ya kbhi nhi kregi

[proud ice]

[gilded cedar]

to discrimant ya to 0 ho to 1 baar touch

aur D<0 taki ek baar bhi na kre

[proud ice]

@gilded cedar this is an English speaking server

[gilded cedar]

i see many ppl suggesting in hebrew french

i just used hindi so its easier for him to understand

[proud ice]

The users here are overwhelmingly english-only. If someone else shares a language with you, sure. But it's not recommended to assume so

[gilded cedar]

its an indian book

[proud ice]

@cerulean chasm are you fine with this?

[cerulean chasm]

Yes I am

@gilded cedar please explain after that

[gilded cedar]

whi kro bs y ki eqn me

D<=0

[cerulean chasm]

A²<0 kaise aaya

[gilded cedar]

0 hai

[cerulean chasm]

Han wohi kaisa ayaa

[gilded cedar]

wo bs parabola ka mouth btane ke liye

kisi bhi real quantity ka square 0 se to bda hoga hi

[cerulean chasm]

Uske baad wale step mein quadratic ka D lesser than zero kyu hua

[gilded cedar]

kisi quadratic eqn ko

agr >=0 hona h

ek sec diagram dikhata

[cerulean chasm]

Toh d=0 ya d>0 kyu nhi kra

[gilded cedar]

greater than me 2 roots aate

[cerulean chasm]

Are bhai

[cerulean chasm]

Toh kya hua

[gilded cedar]

are diya to h ki jo y me quadratic h

usko >=0 hona h

[gilded cedar]

isme niche wala part eqn ka neg nhi ho jayega

[cerulean chasm]

Zero nhi ho skta?

[gilded cedar]

0 ho skta h

but usse km nhi

[gilded cedar]

isme neg ho ja rha

[cerulean chasm]

Toh hamne sirf lesser than 0 wala hi kyu kiya

=0 wala kyu nhi

Arey hn liya h

Sorry

[gilded cedar]

hn

np

[cerulean chasm]

@gilded cedar thanks bhai

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[cerulean chasm]

.reopen

[marsh citrus]

✅

[cerulean chasm]

@gilded cedar agar x real h toh roots exist krne chahiye na

[gilded cedar]

hn tbhi to wo D>=0 se

Y wali eqn bni

[gilded cedar]

Fir wo eqn >=0 honi thi

To D<=0 kiya

[cerulean chasm]

Jo y ke terms mein eqn thi woh?

[gilded cedar]

Hn

[cerulean chasm]

Okay

[marsh citrus]

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[narrow depot]

this is the solution to the question

i don't understand how they found Q(x). i understand that p(x) = (x-3)^3 times Q(x)

but idk how to find Q(x)

[marsh citrus]

@narrow depot Has your question been resolved?

[narrow depot]

no

[silk sable]

One moment.

[narrow depot]

oh wsorry lol

[silk sable]

What have you tried?

Q(x) is the quotient of f(x)/(x-3)^3.

[narrow depot]

i honestly didn't know that was a method lol

ive never done that before

[silk sable]

You should have learned about this when learning about Polynomial Long Division.

[narrow depot]

i learnt about the remainder theorem, the factor theorem and there was some rule about monic polynomials

that allows you to do things (but i don't remember exactly what)

and i think the answer did something about a monic polynomial siince it said "coefficient of x^4 is 1"

[silk sable]

Anyways, because (x-3)^2 is a root, then (x-3)^3 is also a root of P(x). Expand (x-3)^3 and you will have a third degree polynomial. Divide P(x) by that expanded polynomial and you should get a linear factor as a result with no remainder.

That will be your four linear factors of P(x).

[narrow depot]

that is what i tried but im very confused with what happened

i'll send you what i did

the remainder doesn't equal to 0

[silk sable]

,rotate ccw

[silk sable]

Why did you stop at -9x^3?

[narrow depot]

OBHHH

OK THATS PROBABKY WHY

It keeps coming out as -54x for some reason

[silk sable]

-27x should become addition which results in 54x.

[narrow depot]

OH MY GOODNESS

ok that makes so much sense

THANK U SO MUCHH

[silk sable]

yw

[narrow depot]

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[ionic owl]

why is y not positive? wouldn't -5pi/6 be in the 2nd quadrant?

[dry prawn]

it'd be in the 3rd

[ionic owl]

hmm

[rare tundra]

dont worry the just did

[dry prawn]

remember for negative angles your terminal arm rotates clockwise

you're rotating fewer than -pi radians

[ionic owl]

maybe i should've just found the coterminal angle, 7pi/6?

[dry prawn]

if that helps

[ionic owl]

alright, thank u

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[harsh pumice]

Hey just need a little quick help I've done most of the work but for the empty answer f^-1(-31)= I keep getting no solution, like I'm solving and not getting any kind of answer. I'm not sure what I'm doing wrong.

[harsh pumice]

I started with x^4-4=-31

and then tried to solve for x , x^4=-31+4 = -27

but it cant be -27?

[still temple]

helllllloooooooo

[dry prawn]

you haven't fully solved for x

[still temple]

u are at f^-1(31) ?

[dry prawn]

you have x^3 = -27 (not x^4, idk why you keep writing that)

[still temple]

true

[harsh pumice]

ohhhhhhh

so completed its -3

[still temple]

yes

[harsh pumice]

got it thxxx 🙏

and there's nothing wrong with my other ones right

[dry prawn]

a) doesn't look right

[harsh pumice]

hmmm

i just resolved for it i still got 77

@dry prawn

[dry prawn]

how

[harsh pumice]

i subbed -3 in the function then subtracted 4

[dry prawn]

what function?

[harsh pumice]

f(x) = x^4-4

[dry prawn]

for the second time, it's x^3

not x^4

[harsh pumice]

omg

i wrote it in my notebook incorrectly

thats why i keep doing it wrong

thx

okay so its -31

[dry prawn]

indeed

[harsh pumice]

tyy

thxx

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[static quarry]

so far so good

[red tusk]

Power rule?

[marsh citrus]

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[sacred rivet]

Could someone assist me with how to do c)?

For a) I got x = (4-2t, 1-3t, -4+4t, t) with t being the free variable

For b) I got x = (8-2t, -9-3t, 7+4t, t) with t being the free variable

For c) (a) Make a system of equations:

4 - 2t ≥ 0
1 - 3t ≥ 0
-4 + 4t ≥ 0
t ≥ 0

I'm getting that this system has no solutions...?

Similarly for (b)

8 - 2t ≥ 0
-9 - 3t ≥ 0
7 + 4t ≥ 0
t ≥ 0

I'm getting that this system also has no solutions?

Feels like I went wrong somewhere since both my answers for c) are no solutions exist... would appreciate if anyone could confirm and tell me where I went wrong..

[marsh citrus]

@sacred rivet Has your question been resolved?

[sacred rivet]

Any <@&286206848099549185> know this?

[red nimbus]

4 - 2t ≥ 0 -> 2 ≥ t
1 - 3t ≥ 0 -> 1 /3 ≥ t
-4 + 4t ≥ 0 -> t ≥ 1
t ≥ 0

we could basically deduce that there is if t is between 0 and 1/3

[sacred rivet]

Simplifying that gives:
t ≤ 2
t ≤ 1/3
t ≥ 1
t ≥ 0
So, there are no solutions that give x ≥ 0.

[red nimbus]

oh i did a mistake you are right

if we draw a 4 number lines they wont all intersect

[red nimbus]

b) should also have none lol

[sacred rivet]

Hmm

That seems sus tho

lol

Could it be that is simply the answer or I may have messed up part a) or b)?

[red nimbus]

it could be either that

a) is right

b) too

lol

[sacred rivet]

oh

bruh

so my answer is correct?

[red nimbus]

ye

[sacred rivet]

What kind of weird question is this lmfao

[red nimbus]

you ask me

[marsh citrus]

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[marsh citrus]

@sacred rivet Has your question been resolved?

[marsh citrus]

@sacred rivet Has your question been resolved?

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[marsh citrus]

@torpid plover Has your question been resolved?

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[brave spire]

U familiar with the binomial theorem?

[hazy fox]

Yes

[sand fable]

consider (7 - sqrt47)^5 = Q

then, consider quadratic with roots P and Q

i do not recommend use binomial theorem on this question

[hazy fox]

Okay

[brave spire]

o good idea actually

[hazy fox]

Yea if prererable id like to avoid skul

[brave spire]

I was planning on considering odd powers for the fractional part thingy

[hazy fox]

what next

[brave spire]

Find a relation between {P} and Q

[sand fable]

rewrite P(1 - {P})

[brave spire]

And use that to do this

[hazy fox]

As?

P-P²+P[P]?

[sand fable]

yes that is fine

[hazy fox]

alr now the qn is the box

[hazy fox]

Kinda stuck

[brave spire]

I can't see what mqnic is trying to do either

[hazy fox]

Let him cook

[marsh citrus]

@hazy fox Has your question been resolved?

[marsh citrus]

@hazy fox Has your question been resolved?

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[marsh citrus]

Please don't occupy multiple help channels.

[novel juniper]

oops

Trying to find the basis of $Im(T)$

[elfin berry]

Veni, vidi, perii

[novel juniper]

so uh, I feel like it should be

[iron marlin]

seems to be likely a conceptual error - bases are generally not unique

[novel juniper]

I know

[iron marlin]

but that aside, all you need to do is find linearly independent vectors in $Im(T)$

[elfin berry]

Element118

[iron marlin]

so you can basically do a bunch of substitution to find vectors until you find enough linearly independent vectors

[marsh citrus]

@novel juniper Has your question been resolved?

[novel juniper]

so I Just brute force it?

[iron marlin]

that's one way, though another way can be rewriting it as a map on $\mathbb{R}^4$ and doing gaussian elimination on that matrix

[elfin berry]

Element118

[novel juniper]

hmm, okay

Thanks

why can't the basis have 4 matrices?

I was thinking $\begin{bmatrix} 1&0 \ 0&0 \end{bmatrix}; \begin{bmatrix} 0&1 \ 0&0 \end{bmatrix}; \begin{bmatrix} 0&0 \ 1&0 \end{bmatrix};\begin{bmatrix} 0&0 \ 0&1 \end{bmatrix} $

[elfin berry]

Veni, vidi, perii

[iron marlin]

some of these vectors aren't in the image

unless you are going to take T of those vectors

then of course the result would span the image

but then you have to check for linear independence

[novel juniper]

That's a very good idea

thanks

yeah, they ain't LI

[novel juniper]

ykw, that works

Thanks !

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[swift bronze]

wait this might seem like a dumb question

[swift bronze]

but how am i supposed to show my work for a limit approaching 3 from right

do i just put in in a table

[austere snow]

you can show that as it approaches 3 it comes to some value, so maybe f(3.01), f(3.0001) etc

what kind of maths class are you in?

[swift bronze]

but what about no calculator

im in calc ab the test is on limits and no calc

[austere snow]

You can try direct substitution, though I'm guessing it's not that easy.

[swift bronze]

yea but what if like its a big polynomial function

[austere snow]

Then you just need to show that the place lets call, 3+delta is such and such value

$\lim_{x\to{3}^+}f(x)$

[elfin berry]

Makenna

[austere snow]

Analysis the function on the right side of x=3

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[limpid zealot]

The curves f : y = ax^2 −3 and g: y = −x
2 +b have the common point S(−2| −1).
(a) Determine the functional term of f and g!
(b) The two curves f and g define a finite surface area.
Sketch this surface area!
(c) The surface area rotates around the y-axis.
Calculate the volume of the resulting solid of revolution!

Its what i have made

But pro

But problem is C

I think i calculated it wrong

From the graph it seems like intervals are -3, 3

[leaden mulch]

The area does go from -3 to 3 but you got the formula for the integral wrong

You aren't going to subtract the integrals

[limpid zealot]

Yeahh i also got bit confused bc it has | … | thing

What wait a min

[leaden mulch]

So from -3 to -1 you have the rotation of f

And from -1 to 3 rotation of g

[limpid zealot]

So i have to divide it and find the volume then add together?

[leaden mulch]

Yeah divide it up in the 2 sections and add up

[limpid zealot]

Thank you!

[leaden mulch]

If you take the functions you derived for f(y) and g(y) you get this

[limpid zealot]

Also how do i know the intervals without the graph

[leaden mulch]

It's always helpful to draw out a graph to see but the intervals here go from the vertices of both parabolas to their intersection

And both of those can be calculated algebraically

[limpid zealot]

I see

[leaden mulch]

Here the intersection is f(x) = g(x) and the vertecies are f(0) and g(0)

[limpid zealot]

Thank you!

Now is it correct?

[marsh citrus]

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[devout pier]

why’s the answer c?

i got b

[eager tinsel]

you messed up with the quadrant

[devout pier]

lou wait

nvm yeah got it lol

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[devout pier]

thanks

.reopen

[marsh citrus]

✅

[devout pier]

wait

why is this b and not c

sike

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[unkempt goblet]

How do u understand these kinds of texts. Is there a math topic for it?

[patent sundial]

probably coordinate geometry

[unkempt goblet]

No I mean some things in a math book goes like "let x is something set of defined"

[cloud field]

proofs?

[unkempt goblet]

Which is hard to understand (or maybe I just have bad reading comprehension)

[patent sundial]

You mean set theory?

[unkempt goblet]

Idk

[valid cape]

this looks like set theory

[patent sundial]

Yeah...

And also some coordinate geo

[patent sundial]

Hey! I am working so hard, but still not getting the helpful role ☠️

[valid cape]

you just joined 3 days ago

keep helping and you are rewarded

[unkempt goblet]

I'll send another ss if I find another text of those

[patent sundial]

🤞

[valid cape]

there isnt much to it except you will understand it when you learn about the topic

[valid cape]

this is just calculus

[unkempt goblet]

Ik it's calculus but

What is the text even saying

[unkempt goblet]

Cuz usually I just resort to videos if it's like that

[valid cape]

yeah this is 100% reading comprehension issue

[unkempt goblet]

But I think I'm missing something out if I don't understand the texts

[unkempt goblet]

Rip

[valid cape]

read more

or carefully analyze the text

[naive stump]

i think it is just matter of building up knowledge

a lot of things wont make sense if u dont have the foundation for what is before it

cant really except to run before learning how to walk

[unkempt goblet]

Hm idk but that's possible

But I think it's more of a reading comprehension problem since I understand the concept when watching a video of it

[mellow crag]

i think it comes over time

like the first time you come across dense blocks of maths it can feel a bit like wtf is going on here

but if you just watch more maths videos, read more maths etc. you'll get a sense of normally what they're trying to do in the proof and you'll get used to it more

also some ppl are just slow readers - it's fine if it takes you multiple readings to familiarise urself with the topic

[mellow crag]

i think maybe if you haven't done like an introduction to uni mathematics course then it may be helpful to do one to become more familiar with stuff like this

https://courses.maths.ox.ac.uk/course/view.php?id=4938

working through this may help

[unkempt goblet]

OK thanks for the advice

I'll see those through

.close

[marsh citrus]

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[cloud field]

<@&268886789983436800> ads

[marsh citrus]

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[native tapir]

Yoo

[native tapir]

Hiw would i do this q?

Question 2

[dusky viper]

factorise

[spark otter]

w + iw = 1 + 7i can be factored

[silk sable]

System of Equations

[native tapir]

um.. how

[dusky viper]

um

look lhs

[native tapir]

oh

w(1+i) = 1+ 7i?

[dusky viper]

[native tapir]

what can i do with dis tho

[dusky viper]

now find w

[native tapir]

1+7i/1+i?

[dusky viper]

nah gotta realise the denom

[native tapir]

oh

[spark otter]

yes, w = (1+7i)/(1+i), and now rationalize this

[native tapir]

wait so i dont have to turn w into a+bi

[dusky viper]

[native tapir]

.close

[marsh citrus]

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[flint edge]

i need help solving those

[flint edge]

so first of all number a do i take the bracket down and turn it up and reverse the power

[marsh citrus]

@flint edge Has your question been resolved?

[still temple]

which question are you talking about

oh a

for a: multiply and divide the function and then simplify and then substitute 3e^-x + 4

for b: substitute the entire expression under root

for c: its really easy just do it
for d: first of all use trigonometry formulae to simplify then integrate

[marsh citrus]

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[strong plover]

I need help with solving the area of the trapezoid, I tried to work it out and I got 16cm^2 but my program says its wrong, what am I doing wrong?

[paper raptor]

for shapes like these you can break it down into the elementary shapes

so the total area of this shape = area of parallelogram + area of triangle

[strong plover]

alright, But im still confused on how to work out the areas of the shapes

[paper raptor]

for this parallelogram do you know the base?

[strong plover]

isnt the formula for the parellergram bxh?

[paper raptor]

yep

[strong plover]

and then for triangle its bxh/2?

[paper raptor]

[strong plover]

oh alright :))

so I work out the areas of the shapes then add them together?

[paper raptor]

[strong plover]

is the measurements for the triangle 3x4?

[paper raptor]

[strong plover]

ohhh

[paper raptor]

good insight

[strong plover]

so would it be 20+12?

[paper raptor]

I think you know how to do this lol

[strong plover]

oh-

OH i get it now

[paper raptor]

ahh love when that happens

[strong plover]

so it would be 32?

[paper raptor]

[strong plover]

it still says its wrong?

im not really sure anymore-

hello?

hellloooooo

you there

[chilly spoke]

you got the triangle area wrong

[strong plover]

ohhh

i forgot to divide it

is that right?

[chilly spoke]

yea

[strong plover]

ah i see

tysm

.close

[marsh citrus]

Channel closed

Closed by @strong plover

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[paper raptor]

sorry had to do something

but yeah I didn't catch that my fault

[marsh citrus]

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[balmy root]

Help

[balmy root]

I need some info about integration

And differentiation

[grand radish]

ask

[balmy root]

Sooo

How can I start learning calculus

Like nicely

[lethal bridge]

define nicely

[balmy root]

Like

Form 0

*from

[grand radish]

pre calculus is important

[balmy root]

Straight up rock bottom

[balmy root]

What is that

[paper raptor]

there are many prereqs to knowing calc

[grand radish]

trigonometry

some algebra

[balmy root]

K I know that Lil bit

Like what about it the formulas?

[paper raptor]

calculus exists due to the limitations of algebra especially when it comes to infinity

[balmy root]

Hmm

So algebra

[paper raptor]

you should have a solid foundation in algebra

[balmy root]

K

[paper raptor]

polynomials, trig functions, geometric algebra

parametric functions

[grand radish]

ability to derive formulas and also knowing how to manipulate questions to get the desired answer

[paper raptor]

etc.

[balmy root]

I know poly butwhta is trug functions

[balmy root]

Hmmmm

[grand radish]

practive trigonometry

[balmy root]

Can u give me some basic equations I can practice to improve

Like 3-4

[grand radish]

uhh

integrals?

[balmy root]

Questions of this

I want to practice

I want to know what type of it

[grand radish]

you should find a book

for trig

[balmy root]

Can us uggest

*suggest

[paper raptor]

that or there are resources online

[balmy root]

Hmmm

[grand radish]

I will try to find

[balmy root]

Thx for the help

I will practice regularly

And improve

Love y'all thx pajama and alaska

[grand radish]

beginner or

[balmy root]

Begnner

[grand radish]

do you have atleast some knowledge

alr

open stax

https://openstax.org/books/precalculus-2e/pages/1-introduction-to-functions

good intro

@balmy root try khan academy

for lectures

I only had S.L Loney's Trigonometry but not if you are a complete beginner

[marsh citrus]

@balmy root Has your question been resolved?

[marsh citrus]

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[still temple]

f(x)= logx/x find the point of local maximum of f(x)?

[still temple]

can anyone help me in this question?

@night mica

[night mica]

the local maximum points have to be such that the evaluation of the first derivative will vanish

[still temple]

u mean substitute f'(x)=0

[night mica]

impose*

[still temple]

?????????

[still temple]

f'(x)= (1-logx)/x^2

i can't understand can u help by how to proceed

with f(x)

[night mica]

as I said you have to impose f'(x)=0

[still temple]

ok

x = e

when i substitute x = e in f ''(x) is +ve

wait

i got it

.close

[marsh citrus]

Channel closed

Closed by @cloud sigil

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[marsh citrus]

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[upbeat helm]

I think it is a pretty easy ex but I am stuck on 2 b and it is the beginning of the year and I still didn't recover everything I learnt

So could anyone help me out on how to answer this question and the questions after

[marsh citrus]

@upbeat helm Has your question been resolved?

[upbeat helm]

<@&286206848099549185>

[grave dove]

to study the relative position between (C) and (d), we study the sign of f(x) - (x-1)

[upbeat helm]

I did that

[grave dove]

in intervals where f(x) - (x-1) < 0 => f(x) < (x-1) => (C) below (d)

in intervals where f(x) - (x-1) > 0 => f(x) > (x-1) => (C) above (d)

where f(x) - (x-1) = 0 => f(x) = (x-1) => (C) intersects (d)

[upbeat helm]

so according to the question

If I were to draw a table of sign the sign between -infinity and 1 (which is the condition) is minus?

[patent sundial]

And for the part a, you have to find the limit of f(x) when x - > +inf and when x - > -inf

And then you have to compare it with the line y=x-1

[upbeat helm]

I don't get it

[patent sundial]

I mean, you have to prove that the line is asymptotic to the function. And asymptote means, that the value will approach the value of the line, but it won't exactly be the same.

[grave dove]

yes, minus

[upbeat helm]

Which part is this?

[patent sundial]

So, if you find the limit of the function f(x) when x approaches to infinity, you can compare it with y=x-1

[patent sundial]

2 a

[upbeat helm]

I am done with that

[patent sundial]

Oh sorry

[upbeat helm]

I wanted 2 b and the ones after

[upbeat helm]

Np

[patent sundial]

I though you asked for the whole part 2..

[upbeat helm]

no

[patent sundial]

Which part do you want me to explain?

2 b?

Do you get what Emily is saying?

[upbeat helm]

I think

But I didn't realize why did I put minus

[grave dove]

what is f(x) - (x-1) ?

[upbeat helm]

2x-4/(x-1)squared

[grave dove]

yes

when studying the sign of that, whose sign matters?

the top's or bottom's?

[upbeat helm]

That what I don't remember

Ig both

[grave dove]

the bottom is (x-1)² > 0

[upbeat helm]

Like we multiply the signs

[grave dove]

yes, if it was (x-1)

[upbeat helm]

uhh yes

[grave dove]

but its (x-1)²

wont play a role in sign determination here

only in this example

[upbeat helm]

so it is positive

[grave dove]

yeah, and what is positive doesnt really play a role in sign determination

so the sign of f(x) - (x-1) is fully determined by the sign of the top

2x-4

[patent sundial]

top = numinator...

Just for clarification

[grave dove]

i know, of course 😛 😛

its numerator btw, lol

[upbeat helm]

Alright

Now

The second part of the table

Should be in postive

Right?

[grave dove]

show me the table!

[upbeat helm]

so should be something like this where minus is F(x) under D and plus is above

srry for bad writing but it's my first day lol

[grave dove]

its okay

,rccw

[grave dove]

but we said that (2x-4) is what determines the sign

[upbeat helm]

Yeah

[grave dove]

and not the bottom

does the sign switch at x=1?

[upbeat helm]

it should

Right?

[grave dove]

no :p

[upbeat helm]

Uhh

When we have

[grave dove]

substitute by a value less than 1

[upbeat helm]

One root

[grave dove]

and by 1

[upbeat helm]

It is same same

I get -12

For 0.5

[grave dove]

and for x=1 ?

[upbeat helm]

Math error

[grave dove]

we're talking top onlyyyy

[upbeat helm]

Condition

Ohh

Ohh

I got it

[grave dove]

its -2

negative as well

it doesnt switch sign at x=1

find where it switches sign :p

[patent sundial]

OMG EMILY!

YOU JUST BECAME BLUE

[grave dove]

yeah lol, idk how

[patent sundial]

CONGRATS

[grave dove]

Thank you xD

[patent sundial]

I am grinding for that color since 3 days now 😭

[grave dove]

i dont even know how

[upbeat helm]

Grinding on my frozen brain

[patent sundial]

I have heard that there is a bot which sees the progress in these help channels...

[grave dove]

noice

i joined on Jul 19

[patent sundial]

But you got the Active role... 🫢 Which is different from the Helpful role

[patent sundial]

Yeah I saw

[patent sundial]

I dunno what's the difference lol

@upbeat helm so sorry for disturbance lol

Are you clear?

[grave dove]

who knows, maybe it comes right after

but theres also Very Active

[upbeat helm]

I got the point where it changes the sign

But what I do with the condition

[grave dove]

maybe that, then Helpful

[grave dove]

what is ittt

[upbeat helm]

Because ai tried both 1.5 and 0.5

[grave dove]

you dont just try, theres a way

[upbeat helm]

both giving negative values

[upbeat helm]

Ik

[grave dove]

if you want to know where (2x-4) changes sign, put it =0

[upbeat helm]

It gives 2

[grave dove]

2x-4=0
2x = 4
x=2

[upbeat helm]

I noticed

[grave dove]

yess

[upbeat helm]

But I remember we should write a condition when we can

Doesn't x = 1 gives a 0 in dominater

Is not that the reason why we do that

[grave dove]

yes, so we mark that in the table

by a vertical bar

[upbeat helm]

yeah

but we don't change the sign on that

ok I get it

[grave dove]

of course

show me table

[grave dove]

,rccw

[grave dove]

yes, and we mark the 2 by a hollow circle

in the middle of the vertical line

[upbeat helm]

Right

[grave dove]

to indicate that x=2 zeroes f(x) - (x-1)

and below x, write f(x) - (x-1)

and its good

[upbeat helm]

Do you mind completing the ex with me?
I used to Master these but I just need to remember them

Can you?

[grave dove]

but i am going xd

leaving i mean

[upbeat helm]

alright

Thank you

[grave dove]

youll find better than me 😛

lets tag him

@patent sundial

haha

[grave dove]

youre welcome

[patent sundial]

Oh hi

sup

[grave dove]

have a nice day both!

[upbeat helm]

Sup

[patent sundial]

No don't leave 😭😭

[patent sundial]

good

So...

[upbeat helm]

what do I do in 3 a 😭

I got g(x)

And got the roots using calculater

[patent sundial]

OMG tbh, I don't understand these notations anymore 😭

Yeah ask for help... someone else can help better

[upbeat helm]

Alright

[patent sundial]

ping again

(I am sorry for not being helpful 😔 )

[upbeat helm]

Np

You have no excuse imagine doing all this help for only a discord role

.close

[marsh citrus]

Channel closed

Closed by @upbeat helm

Use .reopen if this was a mistake.

[patent sundial]

I just love that color... 😔 🫶

[marsh citrus]

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

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[still temple]

does this evaluate at infinity

[charred flicker]

sheesh what level is this

[still temple]

medium

[charred flicker]

could u help me with mine

[still temple]

nah

[charred flicker]

oki 🖕

[still temple]

btw this is my work on this

oh lim n -> infinity of sum from k =1 to n of 1/k is ininfinity

so that gives me infinity still

so i guess infinity??

<@&286206848099549185>

[charred flicker]

the answer is within ur mind

u just havent actived ur full potential

[still temple]

can you just stop low effort troll?

thanks

[charred flicker]

oki doki

[still temple]

either u can help me, or you watch, 1 out of 2

[charred flicker]

ill watch

.........................

[still temple]

whats your homework about

[charred flicker]

it already got solved

[still temple]

k

[charred flicker]

by someone nicer

what u want to do in future?

engineer?

[still temple]

nah

lecturer

[charred flicker]

ur problem looks hard

[still temple]

its not

i think i did it right

[charred flicker]

ah

u just want the confirmation ig?

[still temple]

yea

[charred flicker]

be confident about urself

i confirm its done right

[still temple]

im not sure

looks like its good

[charred flicker]

yea

its good

[still temple]

u cant know that

[charred flicker]

im 99.999 sure its

good

im so confident

@red nimbus hel;p for this lady

[still temple]

im pretty sure its correct

[red nimbus]

what's popping

[still temple]

1. not a lady
2. dont ping individuals its mean

[red nimbus]

it's mean value theorem so true

[still temple]

?

[red nimbus]

also if you are not why do you have wrong pronouns

you are genuinely causing a confusion

[still temple]

im sorry

,selfroles

,iamnot

[elfin berry]

Gave you the Talks selfrole.

Removed the she/her role from you.

[still temple]

sorry

[red nimbus]

no it's good

[still temple]

is my work rigt?

right?

like i got infinity as wolfram said but is the solution right

[red nimbus]

let me see

[still temple]

.

[slim turtle]

How'd you get the n^3 in the second step? Shouldn't the k^2 in denominator be divided by n^3 and added 1

[still temple]

this is the only mistake in my work

[still temple]

since the sum is not indexed by n, i can pull it out

its indexed by k

can i not pull it out

[red nimbus]

you can pull it out, but not the way you did

[slim turtle]

Yeah but shouldn't the denominator be (k^2)/(n^3) + 1

[red nimbus]

,, n^3+k^2 = n^3 \left ( 1+\frac{k^2}{n^3} \right )

[elfin berry]

bacc

[still temple]

that does not change anything does it

im pretty sure i pulled it out right

[slim turtle]

And the numerator is also wrong there

It should be k^2 + 2k

[still temple]

oh yes i see the numerator

true

but

[red nimbus]

,, \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=0}^n \frac{k(k+2)}{ \left ( 1+\frac{k^2}{n^3} \right )}

[elfin berry]

bacc

[slim turtle]

You have written only k^2 in denominator

[slim turtle]

This is right

[slim turtle]

..

[still temple]

true

ok so

how do i calc that sum now

[slim turtle]

I think it'll turn into a integral

Lemme try

[still temple]

ok that's just messed up now

the sum

[red nimbus]

um

maybe we could add and subtract rather n³ in the numerator

[still temple]

and that turns into what

[red nimbus]

[ \lim_{n \to \infty} \sum_{k=0}^n \frac{k^2+2k}{n^3+k^2} ]
[ \lim_{n \to \infty} \sum_{k=0}^n \frac{n^3+k^2+2k-n^3}{n^3+k^2} ]
[ \lim_{n \to \infty} \sum_{k=0}^n 1 - \frac{n^3-2k}{n^3+k^2} ]

[elfin berry]

bacc

[still temple]

ok yeah

thats so cool

didnt see that

[red nimbus]

yea but does it help

i am thinking

[still temple]

uhh

well we can split the sum

[red nimbus]

[ \lim_{n \to \infty} \sum_{k=0}^n 1 - \frac{n^3}{n^3+k^2} + \frac{2k}{n^3+k^2} ]

[elfin berry]

bacc

[still temple]

yeah

that

but i didnt mean that

but it still god

good

[red nimbus]

i know what you meant

but the issue is we can only do that if we know that the sum exists

[still temple]

really?

[red nimbus]

yea

[still temple]

i didnt know that

so you are saying i can do $\sum_{k=1}^n a+b = \sum_{k=1}^n a + \sum_{k=1}^n b$

OH NO

what d a do

[red nimbus]

sum_

[elfin berry]

daniel

[still temple]

yeah

[red nimbus]

yes

[still temple]

i can do that

only if that exists

[red nimbus]

yes

[still temple]

oh that's new

i didnt know that

[red nimbus]

otherwise if they dont you might get funny things because things get strange with infinity

[still temple]

hah

[still temple]

im thinking

how we could simplfy this further

what is the value of n inside the sum when k = 1 ?

is it infinity?

i mean not literally, it doesnt reach it

but yk what i mean

[red nimbus]

ah yea i dont think the approach will work

[still temple]

why

[red nimbus]

haha i dont know how to continue

[still temple]

its okay we can think

take your time

[elfin berry]

señor matlåda

[still temple]

whats the original question

[slim turtle]

Maybe we can approximate it if we neglect k^2 in denominator?

[red nimbus]

I think integrating might be the way

but I can already tell it's not infinity

[slim turtle]

Yeah it does converge

[red nimbus]

we could neglect it

[slim turtle]

I think

[red nimbus]

,, \lim_{x\to \infty} \int_1^x \frac{t^2}{x^3+t^2} : \dd t

[elfin berry]

bacc

[red nimbus]

for soem reason this integral yields the same value as the sum

even though it's an approximation and we also integrate

[ \lim_{x\to \infty} \int_1^x \frac{x^3 + t^2 -x^3}{x^3+t^2} : \dd t ]
[ \lim_{x\to \infty} \int_1^x 1-\frac{1}{1+\left ( \frac{t}{x^{\frac{3}{2}}} \right )^2} : \dd t ]

[elfin berry]

bacc

[red nimbus]

That would be arctan basically

[ \lim_{x\to \infty} \left ( t - \arctan( \frac{t}{x^{\frac{3}{2}}} ) \cdot x^{\frac{3}{2}} \big |_1^x \right ) ]

[ \lim_{x\to \infty} \left ( x - \arctan( \frac{x}{x^{\frac{3}{2}}} ) \cdot x^{\frac{3}{2}} - 1 + \arctan( \frac{1}{x^{\frac{3}{2}}} ) \cdot x^{\frac{3}{2}} \right ) ]

[slim turtle]

Maybe we neglect the 2k in the numerator also

(My physics friend would be proud of me rn)

[slim turtle]

That'll give us ( k^2)/(n^3)

[elfin berry]

bacc

[slim turtle]

That can be transformed into integral

[elfin berry]

bacc

[red nimbus]

so t^2/x^3

[slim turtle]

Yeah

[red nimbus]

i just fear it's too big of an approximation but sure

[red nimbus]

idk how to evaluate this but it's 1/3

[slim turtle]

Yeah

[red nimbus]

so basically lim x->oo t^3/(3x^3) from 1 to x so

x^3/(3x^3) - 1/3x^3

it works

lol

you also get 1/3

weird

I understood leaving out 2k but also k² from the denominator

but honestly

we only considered the dominant terms in the numerator and denominator

so that might be it

[slim turtle]

Maybe , I don't even know anymore

It should be approx but it's exact..

[red nimbus]

[ \lim_{n \to \infty} \sum_{k=1}^n \frac{k^2+2k}{n^3+k^2} ]
[ \leq \lim_{n \to \infty} \sum_{k=1}^n \frac{k^2+2k}{n^3} ]
[ \leq \lim_{n \to \infty} \sum_{k=1}^n \frac{2k^2}{n^3} ]
[ \approx \lim_{n \to \infty} \int_1^n \frac{2k^2}{n^3} : \dd k ]

[elfin berry]

bacc

[slim turtle]

Anyway is 1/3 the answer?

[red nimbus]

wait

it could be either way 2/3 or 1/3

the problem with the 1/3 is we need to approximate correctly

if we leave terms out that then that either makes the sum greater or less

leaving out k^2 makes it greater but leaving out 2k makes it less again+

but I can surely write instead of k^2+2k <= k^2+k^2 = 2k^2

[still temple]

is the answer 1/3?

[elfin berry]

señor matlåda

[red nimbus]

.

yea

[still temple]

you also got 1/3?

is it right?

[red nimbus]

yes

[slim turtle]

Yeah but my method is by neglecting a lot of terms so idk if it's even right

[slim turtle]

It is?

[red nimbus]

,w Limit[Sum[(k^2+2k)/(n^3+k^2),{k,1,n}],n->inf]

[elfin berry]

Results provided by WolframAlpha

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Requested by adoniszgks

[red nimbus]

yea i am a bit confused but whatever

at least it converges

not infinity haha

[still temple]

same i ended up with just x2dx

[red nimbus]

my guess is the idea is to leave out non dominant terms

[slim turtle]

We should be happy it does

[slim turtle]

I did the exact

[still temple]

idian?

if yes we prolly learnt it the same way

[red nimbus]

what way

cause in europe you aint doingthis stuff

this unmathematical stuff

[still temple]

kick everything other than equal power

[slim turtle]

Maybe, but we can't say it's exactly equal to the initial limit

[still temple]

no its not

[still temple]

limits are not exact anyways

[slim turtle]

Indian? Yeah

[red nimbus]

wdym by that

[still temple]

yeah

[red nimbus]

so i only consider 1/n³?

[still temple]

ill just write it

[red nimbus]

thanks

i hope you pass JEE

[red nimbus]

you dont know my limit

[slim turtle]

You shouldn't do it in maths, but in other fields they just neglect the terms that make it a hard problem

[red nimbus]

yea

[red nimbus]

physiczzz

oh my

what am i seeing

my eyes

[still temple]

integration as a limit of sum

[red nimbus]

So you have

,, \frac{\frac{k^2}{n^3}+\frac{2k}{n^3}}{1+\frac{k^2}{n^3}}

[elfin berry]

bacc

[red nimbus]

now what

[slim turtle]

That notebook screams Indian to me

[red nimbus]

😂

[still temple]

its record book

[slim turtle]

Notebook

[red nimbus]

For me it was the handwriting

[still temple]

now do you know what integration as a limit of sum is?

if yess apply it

[red nimbus]

no explain it

please

you must be very smart at school

[still temple]

i wish i had the time ,the best i can do is brief

[still temple]

youre right

[slim turtle]

Buddy you're a math guy, trust me you don't wanna know what else we do in physics

[red nimbus]

oh nvm i see it lmao

[still temple]

you dont need an explaination?

[red nimbus]

if you got no time, then no

[slim turtle]

Is that sarcasm I see?

[still temple]

ill give you a smol explaination

[still temple]

its ok im in b1 my ego is huge

idc

[red nimbus]

yes please

cause i didnt get it

seems like wizard lizard math

[slim turtle]

What is b1?

[red nimbus]

yes whats b1

[still temple]

batch 1

in my coaching

[red nimbus]

k²/n³ = 0 😭

you coach wizard hoax math 😭