help-49

[flat spire]

$\left(w+vi\right)\left(w-vi\right)=w^{2}-v^{2}i^{2}=w^{2}+v^{2}$

[grand pond]

water beam

[flat spire]

is this correct?

[jagged saffron]

yep

is that real?

[flat spire]

real is when Im(z) = 0 right

[jagged saffron]

indeed

[flat spire]

and i dont see any i's so yeah its real then

[jagged saffron]

true

just be careful with "i dont see any i's"

its true here bc w and v are real

[flat spire]

ohh okay

thank u

[jagged saffron]

but i could be sneaky and define z = 5 + di, where d = i !!

then z is real

even tho theres an i 🙂

[flat spire]

so can i be real?

[jagged saffron]

no i is never real

my point is that its important they told you that w and v were real, they need not be when writing z = w - vi

[flat spire]

ohh okay

[jagged saffron]

its just a small subtle point but it can trip you up down the line

[midnight plank]

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[indigo breach]

.close

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[tall lily]

i have a circle that goes through two points, then i draw the smallest arc to go between these two points (ignore the case where teh arc would be the same length on both sides)

[tall lily]

how would i get the dimensions of the smallest rectangle that can fit around that arc

i know the radius of the circle too

and the equation of the circle

[subtle blaze]

2 points is not enough to define a unique circle

[tall lily]

^

[subtle blaze]

The box you drew isn’t the smallest I don’t think

[next rover]

yeah, this doesn;t make sense, because you already have the circle

[subtle blaze]

If you tilt it a bit it is probably smaller

[tall lily]

ah sorry, the box has to be in this fixed angle

there's no rotation on the rectangle

[feral sedge]

this problem seems heavily paraphrased

[subtle blaze]

The problem is we can’t really give you anything concrete

[tall lily]

i am the one coming up with it :p

[subtle blaze]

Because all you’ve said is I got 2 points on a defined circle

[tall lily]

i have the equation for the circle tho

[subtle blaze]

Which isn’t enough information to say anything concrete

[tall lily]

i dont understand how that's not enough

i have 2 points plus the equation

the equation is all you need to get the circle

[subtle blaze]

The box you’re looking for is gonna be a function of 2 points + the equation of the circle

There’s like 7 different parameters here

[tall lily]

yes and i have all of them

[subtle blaze]

Yeah I’m not deriving a 7 parameter formula for you

[tall lily]

well im not asking you to do it for me, just a direction on how to do it would be helpful

[next rover]

in this case you take the center of the circle and you subtract radius, and you get the ?

or like you get the lowermost point

[tall lily]

you subtract the radius from which axis of the center

[next rover]

the y

[tall lily]

ah yeah that makes sense

and then you get the max of the two y's right

[next rover]

no

ok this doesn't work

[supple tusk]

yo

[next rover]

if both points are in the same quadrant, there's no thrid point to include

[supple tusk]

no one indian??

[next rover]

all indians leave within 20 minutes

if you stay you;ll be the first

[supple tusk]

why ??

[gritty light]

Looking for Indians ?

[supple tusk]

15 done

[gritty light]

Hi

[supple tusk]

I'm Indian

[tall lily]

i think i figured out a general case way of doing it

acutally nvm

it's hard to do general case

[supple tusk]

you tell me

[pure pier]

Im indian lol

Jai hind

[supple tusk]

why indian left in 20 min

my 20 min is done

[dull yoke]

Hello Abhinav Sir

[next rover]

you can stay, just no one else does

@tall lily there's 2 cases i think

if the points are 90° or closer, they are corners

otherwise it's what your picture shows but you need to rotate it so they are in the bottom half

[tall lily]

that makes sense yeah

[next rover]

no tha't wrong nvm

[tall lily]

how come

[next rover]

they aren't corners necessarily

[tall lily]

ah

theyre corners if they have the same y

[next rover]

like here they are close

but there's still the third point like in your picture

[supple tusk]

thx

[tall lily]

yaa
you know for at least of the box edges are touching one of the points

can't generalize with x or y cause the dots can be like the image you provided but rotated 90 degrees

[next rover]

there's like 7 cases

[supple tusk]

someone said rights I have to leave now

I didn't understand anything

[next rover]

same quadrant, 1-2, 1-3, 1-5, 2-3, 2-4, 3-4

idk, i give up

[supple tusk]

bye

[next rover]

bye abhinav

[supple tusk]

u said right ✅️

I can't be more there

[midnight plank]

@tall lily Has your question been resolved?

[tall lily]

pretty sure i got it now

it's nothing to complicated
first i get the smallest rectangle between the two points (by subtracting their x's and y's )

then go through all 4 edges of the circle (edge as in left most part, top most part, right most part, bottom most part)
and if on of the edges is found in between the arc, then extend the rectangle to the edge

to check if the edge is in the arc, i do 4 cases of restrictions depending on quadrant of the points

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[obtuse totem]

everything I get up until here..

you want to show f is bounded tho

every bounded sequence has a convergent subsequence, since you already know the convergent subsequence exists, then can you just conclude its bounded

[obtuse totem]

not sure what the significance of this is

oh

so the not bounded above part is the contradiction

it only shows its bounded below

ok

[obtuse totem]

no?

in this case f(x_nk) is bounded below by k

[novel herald]

The converse isn't true

[tribal temple]

Well k isn't constant, k increases, remember it's the index for the subsequence

So f(x(n(k))) is at least something that goes to infinity, which is exactly why you're not bounded above

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

just writing down some more notes and then i'll get to some questions

[midnight plank]

@obtuse totem Has your question been resolved?

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

arghhhh

these few days couldnt concentrate for some reason

I always end up taking such a long break

what is the difference between converging pointwise and just limit definitions..

rn its looking similar.. ok thats a bad question..

ok ignore, this I'll write them all out later to distinguish differences

[tribal temple]

As in pointwise vs uniform convergence of functions?

[obtuse totem]

uniform convergence is diff from uniform continous right?

yea

[obtuse totem]

yea

they are the same?

heres an ex, i'll look over a few, that should clear it up

[tribal temple]

They aren't: similar to continuity vs uniform continuity, uniform function convergence doesn't [have the integer n] depend on single points and it works for all valid chocies of x - with said integer only depending on epsilon

One example is that you can have a sequence of continuous (usual sense) functions converge pointwise to a function that isn't cts

However replace pointwise with uniform and that limit function must be cts

[obtuse totem]

so if you have uniform conv, that implies its pointwise

but pointwise to uniform is not always the case

ahh okok

thank you so much! that clears it up alot

[obtuse totem]

https://youtu.be/GsORKmBCLuI?si=EpqhUML60mAnZill

this vid is really good

and also, cauchys thm only applies for uniform convergence I think, because uniform convergence is for all x in S, whereas pointwise is just some x_0 in S

for functions I mean

for regular sequence (xn), the sequence is convergent iff cauchy

[obtuse totem]

4 more pgs to read over then i'll have work on some problems (questions for sure)

[obtuse totem]

corresponding answers/hints

for pointwise convergence, you have the def: for all eps>0, there is an N in natural nums, where N>n, |f_n(x)-f(x)| < eps, for some x_0 in S

I have to show for some x in S, not all because all would be uniform..

but this (not the textbook im using) says otherwise?

[obtuse totem]

so for pointwise, its also all x in the set, now Im confused because visually I know the difference between pointwise and uniform, I know that fn-> f must be continuous for uniform (as in continuity carries over to f) and doesnt have to for pointwise, but other than that, they both have the same definitions?

wait

[obtuse totem]

pointwise: for all eps >0, for each x in graph, there exists N s.t |fn(x)-f(x)| < eps for all n>=N
N can depend on eps and x

uniform: for all eps >0, there exists N s.t |fn(x)-f(x)| <eps for all n>=N, for all x
N does not depend on x, so it must be same for all x

[tribal temple]

Yep, that's it, for pointwise that N is allowed to depend on x, you're given eps and x first then start building based on those
Whereas for uniform [function convergence], you're only given eps first, you don't have x, and from there you have to build your N such that then, whatever x you then choose, it works

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

back from a break

[obtuse totem]

lmao I feel like Im putting in so much work

just to get a mediocre mark at the end

it was a fun learning process tho

I have an exam soon on it and one of my friend told me he just started studying

man... I guess some people built different🥲

oh nooo

okok this is why you absolutely cannot listen to music while doing ra

ok quick rant and im back to work

but

oh god so, I wrote there exists N>n st |fn(x)-f(x)| < eps for all n> N

ohmy goodness💀

😭 😭

[tribal temple]

what music are you even listening to-

[tribal temple]

Happens

[obtuse totem]

Mr.Brightside, a head banger song, the one that sort of makes you dance around the room LMAO I was half singing half writing proof

good thing I didnt start writing down song lyrics🤡

yea one song played, and none will be played in the future

[tribal temple]

would be the typa thing I would end up doing

[obtuse totem]

question

for this question, it seems like any N would work??

[tribal temple]

How do you mean here? The right bit, the absolute value part

[obtuse totem]

absolute value part?

well thats |fn(x)-f(x)|

so its given f(x)=0

and when I sub in x=1 for the function fn(x), it goes to 0

[tribal temple]

It does go to zero sure - though are you trying to show that on the right? For that, you'd want to show that [convergence stuff here] |1/e^n| < eps

So then for some given eps, you wanna find some N such that 1/e^n < eps for all n >= N of course

But then for other choices of x, you wanna show similar too...

[it may be worth noting that each f(n) is a positive function on the interval they give you, and you may wanna think about other properties it possesses, maybe try plotting them for choices of n and see...]

[obtuse totem]

not sure how I could isolate for n…

i feel like I can do smth here but not sure

[tribal temple]

You very likely can though as per before...

[tribal temple]

...did you try plotting them?

(remember that exponentials are "more powerful" than polynomials)

[obtuse totem]

y-intercept is 1?

idk what else..

...idk..

[tribal temple]

Did you plot it properly?

[tribal temple]

Try like this one https://www.desmos.com/calculator/75huvwxk2a

[obtuse totem]

I thought you can just plug in arbitrary values for x.. with restriction that x in [0, inf)

oh

righttt

but n tends to inf

ahh

aha

[tribal temple]

Yep, and if you notice something about the graph, no matter which positive x you choose...

[obtuse totem]

y=0..

[tribal temple]

I mean...

I was more trying to hint at something here

[obtuse totem]

oh

oops

mmh idk still

[tribal temple]

Well, notice how no matter what n you choose, there's that maximum point you have...?

[tribal temple]

...and if you take a look at the hint they give you for the question

[obtuse totem]

ohhh

😃 my bad

so i can just say 1/e^n < eps

and solve for n

[tribal temple]

Yep also notice how here we don't even need to consider the points x, and that it will work for any choice of x in the interval

...that should ring some bells about a conversation we had earlier

[obtuse totem]

so choice of N= 1/ln eps would work for all x in [0,inf)

so this is also uniform convergence

[tribal temple]

Careful of the algebra

[obtuse totem]

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhh

omg

give me

oneee second

N=-ln eps

[tribal temple]

Yep

[tribal temple]

And yea, the convergence is uniform

[obtuse totem]

this is for 2

working on this one now

uniform continous means that: for all eps >0, there exists delta>0 s.t |x-y| <delta implies |f(x)-f(y)| <eps for all x,y in interval

if f is continous on R, is it ok to say it is continous on some arbitrary interval [a,b] where x_0 in that interval (for |x-x_0| < delta...)

[tribal temple]

For this one:
Be careful about making sure that N is actually an integer - not a great idea to say that it's equal, but you can e.g. round your choice up or say that it's an integer that's at least that
but otherwise I think I'm happy that said, do take a look at the hint they gave, that f(n)(x) < 1/n for strictly positive x - you can probably see where this is going...

[obtuse totem]

ah so x-eps/eps(x) + 1

ok

wait..

[tribal temple]

Or/also take the ceiling too, to round it up to an integer

[obtuse totem]

|a-b| <= |a-c|+|c-b| right

idk if i even need to use triangle inequaliity

but I messed up the triangle inequality.. it doesnt work..

but this proof looks right

like the idea

ok ill be back in 30min

[obtuse totem]

ok

im back

but after this question im gonna sleep

@tribal temple

also lmao i thought i was just gonna take 30min lol

[obtuse totem]

somethings wrong with the triangle inequality tho..

what do you think

Its not right but I know I gotta use it right?

[tribal temple]

Hmmm it's tough

[obtuse totem]

oh no

ok

aww man i really thought I was actually finally kind of getting more intuition for proofs

[tribal temple]

mind you with many proofs, it can take a while to think of how to do them

Anywaysss gonna have to sleep while I can

[obtuse totem]

ok

thank you!!

goodnight

lol

yea I have an exam on this soon so you prob wont be hearing from me for a bit

im just gonna take a whole day to relax after

but ye

thank youuuuuuuuuuuu🙂

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[twilit field]

$\lim_{n\rightarrow\infty}\frac{\left(\sqrt{n^3-2n^2+1}+\left(n^4+1\right)^{\frac{1}{3}}\right)}{\left(n^6+6n^5+2\right)^{\frac{1}{4}}-\left(n^7+3n^3+1\right)^{\frac{1}{5}}}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

where do I even start

No L'hopital please, it will send me to l'hopital

I thought of taking the biggest power out of each root

thus obtaining

$\lim_{n\rightarrow\infty}\frac{\left(n^{\frac{3}{2}}\sqrt{1-\frac{2}{n}+\frac{1}{n^3}}+n^{\frac{4}{3}}\left(1+\frac{1}{n^4}\right)^{\frac{1}{3}}\right)}{n^{\frac{6}{4}}\left(1+\frac{6}{n}+\frac{2}{n^6}\right)^{\frac{1}{4}}-n^{\frac{7}{5}}\left(1+\frac{3}{n^5}+\frac{1}{n^7}\right)^{\frac{1}{5}}}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

hmm

I could drop the constant terms as n approachs infty

that would give me

$\lim_{n\rightarrow\infty}\frac{\left(\left(n^3-2n^2\right)^{\frac{1}{2}}+\left(n^4\right)^{\frac{1}{3}}\right)}{\left(n^6+6n^5\right)^{\frac{1}{4}}-\left(n^7+3n^3\right)^{\frac{1}{5}}}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

which isn't much better

any hints

[strong lava]

notice the numerator is dominated by the first term, which is equivalent to n^3/2
Same for the denominator
Meaning you should get a limit of 1

[twilit field]

hmm

got it

thanks

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[vague reef]

Let u and v be two non-parallel unit vectors with u ⊥ v, and let
r(t) = u cos(t) + v sin(t).
Show that the curve r(t) sweeps out the unit circle centered at O in the plane P
defined by u and v

[vague reef]

what condition does r(t) have so that it sweeps unit circle

i have tried to show that $|\vec r|$ is constant where $ |\vec r|^2 = \vec r \cdot \vec r $

yep

how do you do that in latex

[bleak pier]

I don't know

[vague reef]

ok

do you have any idea?

[bleak pier]

No

That's nice

[grand pond]

e=mc^2

[vague reef]

Thx

i am very stuck

[midnight plank]

@vague reef Has your question been resolved?

[midnight plank]

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[edgy onyx]

I'm stuck with this problem:

[edgy onyx]

i found the integral of lnx/x

but that y' is messing me up

[still zinc]

Seperate variables @edgy onyx

Separate*

[edgy onyx]

how

[robust isle]

it's already seperated anyway

[still zinc]

I dont want to say something cursed

But just multiply by dx lol

[edgy onyx]

why

[still zinc]

To get one side for x and the other for y to then integrate both sides

[robust isle]

try a u-sub, ||u=y^2, du will absorb that yy'||

[edgy onyx]

i saw somewhere in the internet which seems to be more clear

basically y' became dy/dx and then they multiplied by dx and removed dx completely

not really sure again why they multiplied by dx

[still zinc]

If you have a differential equation of the form a(x) = b(y) dy/dx then if you know A(x) and B(y) anti derivatives of a and b respectively, you can then write A(x) = B(y), then youd need to know the inverse function of B to get the form of y (also dont foget the additive constant)

[edgy onyx]

sorry but i didnt understand

[still zinc]

If you have a differential equation of the form a(x) = b(y) * dy/dx, where a only depends on x and b only depends on y you separate the variables and calculate the anti derivatives for each independently

[edgy onyx]

i dont think you saw the question i had

[edgy onyx]

someone online

transformed y' to dy/dx

and then multiplied it by dx to remove dx

why did he want to remove dx

[still zinc]

Well mathematically you dont remove dx

Its just a way to skip a few steps

Like a trick

[edgy onyx]

so he only did that so he doesnt have more stuff in his way?

[still zinc]

In fact you calculate the antiderivative of B(y(x))dy/dx

With respect to x still

[robust isle]

you could just integrate both sides with respect to x if you want yeah

you'd get exactly the same thing in the end

[edgy onyx]

so i can just integrate both sides with dx?

[modern skiff]

Not sure how to get further than this:

[still zinc]

Yea

[edgy onyx]

oh and another thing can someone teach me the method of substituting with letters like "u" for example

[modern skiff]

usually you're using it under a sqrt or when you see something cancelling out

like the x*du cancels out the 1/x

[edgy onyx]

ohhhhh

[robust isle]

if you're wondering how ppl choose substitutions, it's mostly experience, you see what sticks after some point

[edgy onyx]

why did this guy multiply y with dy and did 2 separate integrals

[robust isle]

and if you just want to know how the technique works, well you can just pick your favorite math youtuber, they prolly have a video on it

it's a topic that's been done to death

[robust isle]

what's the problem with splitting the integral

[edgy onyx]

i just saw organic chemestry tutor and understood it@robust isle

[edgy onyx]

why did he split it and why did he multiply it with dy

[robust isle]

"why did he split it" cause he can

less things to keep track of in each individual integral mostly

[edgy onyx]

he split it just because he can not because its some sort of rule right

[robust isle]

yep

[edgy onyx]

then he multiplied with dy because?

[robust isle]

wdym multiplied by dy

explain yourself a bit more, I'm not completely sure what you're talking about

[edgy onyx]

you can see integral of ydy

[robust isle]

ok

[robust isle]

that's just part of the split

integral of (y + y*sqrt(3+y^2)) wrt y = integral of y wrt y + integral of y*sqrt(3+y^2) wrt y

"dy" : I'm integrating wrt y, think about it that way

[edgy onyx]

is it because dy multiplies both the thing in the parentheses and y?

[robust isle]

if you want yes

[edgy onyx]

so what i said is correct

[robust isle]

I'm not the ultimate guardian of intuition

if that helps you remember how integrals behave, use that

[edgy onyx]

no i mean if the thinking i did means its correct then it means i understood it correct, i dont want to believe i know something then it turns out to be incorrect

[robust isle]

there's a few different ways to interpret the dy in an integral

[edgy onyx]

another thing

[robust isle]

and even then, there are multiple ways of defining integrals, where the notion of d[something] differs between them

[edgy onyx]

why did he do + after ydy

[robust isle]

so it's really a question of how deep in the rabbit hole you want to go

learning is done by successive approximations

you're not teaching special relativity to middle schoolers cause classical mechanics doesn't work with objects at super high speeds

[edgy onyx]

umm

[edgy onyx]

do you know why this happened?

[robust isle]

you're very confused by this split

the integral of the sum of 2 functions is the sum of the integrals

that's what's going on

[edgy onyx]

ohhh

ok

whenever splitting you add +

[robust isle]

or in math terms $$\int (f(x)+g(x))\dd{x} = \int f(x) \dd{x} + \int g(x)\dd{x}$$

[grand pond]

aPlatypus

[robust isle]

for whatever f and g (which you can integrate)

[edgy onyx]

im so confused by this split

[edgy onyx]

why did 1+ get replaced with y

[robust isle]

y(1+sqrt(3+y^2)) = y + ysqrt(3+y^2)

welcome to algebra

[edgy onyx]

its because y*1 is y

ok

after integrating we get:
(ln(x)^2)/2=((y^2)/2)+(1/2)(3+y^2)sqrt3+y^2

[robust isle]

(ln(x)^2)/2=((y^2)/2)+(1/2)(3+y^2)sqrt3+y^2
are you sure about that 1/2

[edgy onyx]

original problem was this, keep that in mind

[robust isle]

ah right

ok yes

[edgy onyx]

now we substitute every x and y with 1

and then find c

right?

[robust isle]

yes

[edgy onyx]

so c is -8

and the finishing thing is (ln(x)^2)/2=((y^2)/2)+(1/2)(3+y^2)(sqrt3+y^2)-8

@robust isle quick question again

you said y(1+sqrt(3+y^2)) = y + ysqrt(3+y^2)

but here its not y+ysqrt(3+y^2)

it is

[midnight plank]

@edgy onyx Has your question been resolved?

[midnight plank]

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[last slate]

any ideas

[odd pond]

well there is a way to calculate 1/2+1/6+1/12+1/20...

[odd pond]

so you can straight up find f and then whatever you want i guess

[last slate]

its 1/n(n+1)

we sum 1/n(n+1) 10 times and subtract it from 8 ?

[odd pond]

wait

[odd pond]

how did you get that?

[last slate]

1/2 + 1/6 + 1/12 .... n times

1/1x2 + 1/2x3 + 1/3x4 .... 1/n(n+1) = n/n+1

[odd pond]

wait so f(n) is connected to the sum?

i am confused

[last slate]

yeah

[odd pond]

isnt it like the sum 1/2+ 1/6+1/20+1/30..... + some f(n)=n/n+1

[odd pond]

oh alr

kind of a weird way to put it

[odd pond]

can you send like the whole exersice

[last slate]

its a question

[odd pond]

not connected to anything else?

[last slate]

thats the solution
but i did not understand anything

[odd pond]

lets see

[odd pond]

alr so we have f(n)=1/n(n+1) right?

[last slate]

yeah

[odd pond]

so like 1/2=f(1)...1/6=f(2) etc right?

[last slate]

ye

[odd pond]

so what is basically gives us is that f(1)+f(2)+f(3)+....f(n)= n/n+1 right?

[last slate]

yes

[odd pond]

well f(1)+f(2)+....f(n) is the sum of f(n) from n=1 to n right?

do you know anything about sums, i am not sure how much to explain

lemme know if you dont understand

[last slate]

i know

[odd pond]

alr

so we have that the sum of f(n) from n=1 to n is n/(n+1) right?

[last slate]

yup

[odd pond]

alr for n =10 what do we get?

[last slate]

10/11

[odd pond]

so f(1)+f(2)+....f(10)= 10/11

[odd pond]

for n=4?

[last slate]

4/5

[odd pond]

then f(1)+f(2)+f(3)+f(4)=4/5

subtracting those 2

we get that f(5)+f(6)+....f(10)=10/11-4/5 right?

[last slate]

Right

[odd pond]

the first part is the sum of f(n) from n=5 to 10 which is basically what the questions asks for

[last slate]

but n=10 gives us the f(10) only
we need the whole sum of sequence from f(1) to f(10)

[odd pond]

we have that the sum of f(n) from 1 to n is n/n+1 dont we?

[last slate]

ohh

make sense

[odd pond]

when you put n=10 you put it to the sum as well

so from 1 to 10

[last slate]

yeah i get it

[odd pond]

alr

[last slate]

tysm for helping

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[odd pond]

np

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[real eagle]

Distribution of Sample proportion, Stats and Probability

[real eagle]

need to know how the answer sheet got a 0.225

also this relates to B of the question

[runic python]

It's the upper bound

You want to find the probability of at most 45 people from a sample of 200

45/200 = .225

[real eagle]

yeah I know

[real eagle]

oh ok

P hat

I remember now

thank you

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[drifting jackal]

<@&286206848099549185> I need someone who has a TI-84 Plus graphing calculator.

[hollow cosmos]

i got casio fx-82

is that enough?

well if u want help i can

[drifting jackal]

I'm so sorry, but I don't think that will help. 😭 Thanks for the help though.

I will try.

[hollow cosmos]

Ok then close this

if u want

[drifting jackal]

Close it? Why?

[hollow cosmos]

whats y in question a. Like age?

[drifting jackal]

No. A linear equation typically takes the form y = mx + b, where m and b are constants, x is the independent variable, y is the dependent variable.

Does that make sense?

😬 😅

[grand pond]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

You really shouldn't take it literally :upside_down:. Please type ,help ping, for example!
The full command list may be found using ,list.

Please give me something to evaluate.
See ,help calc for usage details.

Apr 13, 2024, 11:23 PM

[drifting jackal]

.close

[midnight plank]

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[sand night]

,help

[grand pond]

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[midnight plank]

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[midnight plank]

@still nacelle Has your question been resolved?

[dawn dagger]

,,\abs{g(z)} = \abs{\frac{f(z)}{z}} \leq \frac{1}{\abs{z}} \leq \frac{1}{r}?

[grand pond]

𝔸dωn𝓲²s

[dawn dagger]

ah no

fuck

[still nacelle]

the last inequaliy would mean |z| >= r, which is the other way round right?

[dawn dagger]

yes

mb

[still nacelle]

nah all good

the question looks so innocent

[dawn dagger]

😅 😦

[still nacelle]

unfortunately that would be proving |g| >= 1/r not <= 1/r XD, also I don't think the last step works cuz |f| <= 1, not >=

[dawn dagger]

damn I mixed things

but wait

[dawn dagger]

r is between 0 and 1

[dawn dagger]

so that should hold?

nah it doesnt

[still nacelle]

r is the radius of the disc, which is between 0 and 1, if we fix r (for example r=1/2), then z is just a point in the disc, so all we can tell is that |z| <= r = 1/2

in fact |z| < r strictly since it's an open disc

[dawn dagger]

1/|z| >= 1/r then

lol

[still nacelle]

yh

[midnight plank]

@still nacelle Has your question been resolved?

[dawn dagger]

I tried this idk haha i am going off

[dawn dagger]

,, \abs{g(z)} \leq \frac{1}{r} \iff r \cdot \abs{g(z)} = r \cdot \abs{\frac{f(z)}{z}} \stackrel{0<r<1}{\leq} \abs{\frac{f(z)}{z}} \leq 1

[grand pond]

𝔸dωn𝓲²s

[dawn dagger]

noo I assumed |f/z| <= 1

If you proved c) we can use that for b) lol

[midnight plank]

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[dawn dagger]

.reopen

[midnight plank]

✅

[grand pond]

𝔸dωn𝓲²s

[dawn dagger]

@still nacelle

[midnight plank]

@still nacelle Has your question been resolved?

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[simple spire]

Points A, B and C are on the circumference o a circle with centre O and radius 2cm. Given that BC = 2(sqrt(3))cm, find the possible values of BAC

[simple spire]

if BC is the radius, then BAC is 90

if triangle BAC is equilateral, then BAC is 60

[slender walrus]

BC can't be the radius

[simple spire]

oh right

cuz its smaller than 4

[slender walrus]

have you drawn a pic?

[simple spire]

yes

[slender walrus]

show your pic

[simple spire]

i dont have my phone with me

[slender walrus]

reproduce your pic on paint

[slender walrus]

label the centre as O
draw segments OB and OC

[simple spire]

BAC can be any angle up to 90?

[slender walrus]

no

determine <BOC

[simple spire]

how?

[slender walrus]

OC and BC are radii

[simple spire]

so i just give a random angel?

[slender walrus]

no

you have all three sides of a triangle

[simple spire]

oh

[slender walrus]

there are multiple methods of determining the angle from that

[simple spire]

ok ill try

its 120

so BAC is 60

[slender walrus]

yes, but note that this is only one case / one possible angle for BAC

A could also be on the other side of BC (the side that doesn't contain the centre)

[simple spire]

BAC can be 135?

wait no

BAC can be 120?

[slender walrus]

yes

[simple spire]

those are the only 2 possible answers?

[slender walrus]

yes

[simple spire]

thanks

.close

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[red helm]

How do I get second solution of - 2

In question 1

[latent elk]

45 deg could be on both side of y=3x

So if u find the first slope; then u can find the second one by get the slope that peperdiculur the the first slope

U can get the perpendicular using this

[latent elk]

Your first slope of 0.5, then the slope that perpendicular to its will be -1/0.5 = -2

@red helm

[midnight plank]

@red helm Has your question been resolved?

[red helm]

Ohhh

[red helm]

Ty I get it

[latent elk]

👍

[red helm]

I rlly appreciate it 🙏

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[acoustic owl]

How can I eat the antiderivitve of tan^2(x)

[digital crag]

use pythagorean identity to rewrite it

do u know what to do from there

[digital crag]

yup

[acoustic owl]

Not sure how to take the antiderivite of that either

[digital crag]

sec^2(x)

[acoustic owl]

Yeah

[digital crag]

what derivative will get u sec^2

[acoustic owl]

Tan(x)

Right?

[digital crag]

ye

so tan(x)-x

[acoustic owl]

Okay

[digital crag]

and then plug in ur thing

[acoustic owl]

so tan(x)-x

1-pi/4

?

[digital crag]

yea

[acoustic owl]

Alr thanks you

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[versed elbow]

This is gaussian from CLT right?

but then i also know that as n-> infty a binomial approximates to poisson so i'm a bit confused

[edgy herald]

@red helm

dud u

dr du

[midnight plank]

@versed elbow Has your question been resolved?

[red helm]

Yeah 😭😭

[midnight plank]

@versed elbow Has your question been resolved?

[versed elbow]

.close

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[last slate]

Hello , how to find the area without using determinant ?

[last slate]

and without 1/2 | x^2 ( y3-y2 ....

[slender walrus]

distance formula, herons

[narrow fog]

no need heron

its very long

there is rule when they gave lik this u write it down lik this

give me moment

[last slate]

using 1/2(median)(altitude)

[narrow fog]

yea

[slender walrus]

that doesn't look like a valid way to get area

[last slate]

why ?

[weary radish]

If it's an isosceles/equilateral triangle ..sure

[slender walrus]

where did you see that expression?

[last slate]

why need to be isosceles/equilateral ?

[last slate]

its a question

[slender walrus]

show the question and/or where it said

using 1/2(median)(altitude)

[weary radish]

Btw did u mean base instead of median ?

[narrow fog]

like this

[weary radish]

That's determinant.

[slender walrus]

that's shoelace / determinants which is what they said they didn't want to use

[narrow fog]

didnt saw sory

[last slate]

no why ?

we can find the median

[slender walrus]

first show where exactly you're seeing that formula

because its highly sus

[last slate]

If we find the equation of the baseline BC and then determine the perpendicular distance from point A to BC is this considered a median ?

[slender walrus]

no

that would be the relative altitide to base BC

[last slate]

oh

[weary radish]

Median is line joining A to midpoint of BC.

[last slate]

so we find the midpoint of bc and then use the prep distance formula ?

[slender walrus]

no

[weary radish]

Why?

[slender walrus]

midpoint of BC not needed/relevant

perp distance formula uses
an external point A
and a line BC

[last slate]

i mean distance formula *

[slender walrus]

then no

that doesn't help you

[weary radish]

This can be either solved by heron or determinant. Anyother method is way too long or isn't gonna give any ans..

[slender walrus]

medians aren't necessarily perpendicular

and if it isn't then it won't help if trying to apply the basic area formula

[last slate]

so Is this question unsolvable without determinants or the Heron formula?

[weary radish]

You can first check if it's equilateral or isosceles and try apply your formula 1/2(base)(altitude)
Else by determinant or heron @last slate

[slender walrus]

you can use the perp distance formula

to get the relative altitude to the base you're using

as i mentioned

[last slate]

is 1/2(base)(altitude) same as 1/2 median x altitude ?

[slender walrus]

no

hence why i requested on multiple equations where exactly you're getting 1/2 median x altitude

median and base aren't the same thing

[last slate]

cant we just find bc length using distance formula ( base )
and altitude using prep distance
?

[slender walrus]

yes

i've said that twice already

note how median isn't mentioned once in that description

[last slate]

Is that true ?

[slender walrus]

2^2 + 1 isn't 3

[last slate]

its 5

[slender walrus]

yes

[last slate]

i just write it like that

[slender walrus]

i'm referring to your work in the altitude btw

you simplification of
$$\frac{3}{\boxed{\red{?}}} =\sqrt{3}$$
implied $\red{?} = \sqrt{3}$

[grand pond]

ℝαμΩℕωⅤ

[slender walrus]

(which is incorrect as that should be sqrt(5) in the denom)

[last slate]

ohh

i get it

is everthing else ok ?

[slender walrus]

update all your work after that

[last slate]

aight
Tysm for helping

.close

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[midnight plank]

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[novel cobalt]

any idea how to solve this integral?

[novel cobalt]

tried to substitute u = what is inside the scope, couldnt get rid of x

😔

[midnight plank]

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[novel cobalt]

.close

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[last slate]

how to find the answer?

[last slate]

nvm i alr solved it

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[cerulean gust]

how to brute force a double integral

[cerulean gust]

.close

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[sturdy crystal]

can someone explain that how did they define the constraints

<@&286206848099549185>

[last slate]

i cant read the texts

maybe a little bit but perhaps you can provide a clearer picture

[sturdy crystal]

<@&286206848099549185>

[midnight plank]

@sturdy crystal Has your question been resolved?

[sturdy crystal]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[hearty grove]

My brother in Christ, do not spam

[sturdy crystal]

its been FORTY FIVE minuttes

bye bye

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[cedar kraken]

Zero patience

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[hybrid grail]

Can someone help me with the next step

[midnight plank]

@hybrid grail Has your question been resolved?

[midnight plank]

@hybrid grail Has your question been resolved?

[midnight plank]

@hybrid grail Has your question been resolved?

[still nacelle]

you can simplify your last expression to
$6 \int sec^2(x)cosec(x) \ dx$

[grand pond]

lgkoo

[still nacelle]

Then use $sec^2(x) = tan^2(x) + 1$

[grand pond]

lgkoo

[still nacelle]

then simplify and you should get really nice and simple integration

[still nacelle]

@dawn dagger that last step uses part (c), which is proven by using part (b) lol.

[hybrid grail]

how?

1/cos#theta and costheta/sintheta j becomes 1/cos^2theta no?

[still nacelle]

wait no, you're trying to simplify 1/cos^3 and cos/sin, which is just 1/(cos^2 x sin)

and 1/cos^2 = sec^2

and 1/sin = cosec

[hybrid grail]

i'm confused

[still nacelle]

so inside the integral, you got
$\frac{1}{cos^3(x)} \cdot \frac{cos(x)}{sin(x)}$

[grand pond]

lgkoo

[still nacelle]

which simplifies to
$\frac{1}{cos^2(x)} \cdot \frac{1}{sin(x)}$

[grand pond]

lgkoo

[still nacelle]

which is just
$sec^2(x) \cdot cosec(x)$

[grand pond]

lgkoo

[still nacelle]

does that make sense? @hybrid grail

[hybrid grail]

oh yea

so now we just integrate sec which is tan but what about cosec?

[still nacelle]

this is a product, you can't just integrate sec alone

[still nacelle]

use this hint

[hybrid grail]

but the derivative of tan is sec^2 so the integral of sec^2 is tan, no?

[still nacelle]

that is right

so?

you have sec^2 times by cosec

so you gotta think about the consequence of product rule

e.g. integrate x*sin(x)

you can't just integrate sin and integrate x separately

[hybrid grail]

okay, i think ima come back to this one later. i have a few questions left, do u mind helping me with them?

[still nacelle]

feel free to ask the questions here and someone will help

[hybrid grail]

@still nacelle can you take a look ?

[still nacelle]

you made 2 mistakes

firstly, $x^3 = 64 sin^3 \theta$

[grand pond]

lgkoo

[still nacelle]

but that's easy to fix

your second mistake is for your triangle with angle theta, opposite = x and hypotenuse = 4

the adjacent should sqrt(16-x^2), not sqrt(x^2-16)

lastly, I would like to point out that you should be more careful with your limit, after you made substitutions, your limits also change. Since you converted it back to in terms of x at the end, that ended up fine, but I would suggest explicitly writing **x = 2sqrt(3) and ** x = 0 when your integral is in terms of theta or u

but all of these mistakes are easy to fix and you should get the right answer now

[hybrid grail]

It's been a while since I used the triangle, so can you explain when to do x^2-16 and when to do 16-x^2? ik the hyp is x^2+16

but how do u differentiate

[still nacelle]

well you're essentially just rearranging the Pythagoras Theorem $a^2 + b^2 = c^2$. Here you have c is your hypotenuse, and it doesn't really matter which of the remaining sides you pick as your a,b. Let's say then you're given a and c, and you wanna find b, it is just $b^2 = c^2 - a^2$

[grand pond]

lgkoo

[still nacelle]

which is equivalent to $b = \sqrt{c^2-a^2}$

[grand pond]

lgkoo

[hybrid grail]

oh i got it now

i was just freehanding it but i didn't think of the equation

i think this is still wrong

[still nacelle]

also a quick sanity check you can do is that: the hypotenuse length 4 has to be longer than x, so it has to be sqrt(16-x^2) otherwise you're taking the square root of a negative number

[still nacelle]

that looks fine, why do you think it is wrong?

[hybrid grail]

i put it in and it marked it wrong

[still nacelle]

did you simplify the expression?

you can simplify the square roots

[hybrid grail]

even if i submit it unsimplified

it usually marks it right

[still nacelle]

oh I know why

[hybrid grail]

why?

[still nacelle]

cos(0) is not 0

[hybrid grail]

sso 1-1/3?

[still nacelle]

sry I don't mean cos cuz you converted it back

[hybrid grail]

ah yea

thats why

btw

[still nacelle]

yh

[hybrid grail]

are u good with multivariable calculus

[still nacelle]

I'm surprised you understood what I meant

[hybrid grail]

yea lol i realized it as soon as u said it

[hybrid grail]

or linear algebra?

[still nacelle]

uhh I would say I'm not good with anything 🙂

so don't rely on me hehe

[hybrid grail]

are you famlilar with the classes?

[hybrid grail]

ur pretty godo at this stuff though lol

i'm understanding it

[still nacelle]

what kind of classes, do you have an example?

[hybrid grail]

calc 3?

line integrals

[still nacelle]

thanks

[still nacelle]

I only started learning it this year so I wouldn't say I'm good

[hybrid grail]

oh okay

okay ima work on the other question now, i'll lyk if i need help w it

[still nacelle]

As I said don't rely on me, I'm revising for my exams rn so I'll only show up randomly. If you just ask the questions in the help channels someone else can help you too. Also, I'll suggest you close this channel and open a new one when you need help (newer channels show up near the top so more likely to be seen, maybe)

[hybrid grail]

alright, can u help me with this last one?

[still nacelle]

what is $sec^2(x)$?

[grand pond]

lgkoo

[hybrid grail]

oh wait

it's 1/cos^2theta

[still nacelle]

yup

[hybrid grail]

can i do this

[still nacelle]

yes

[hybrid grail]

and how do i integrate that?

do i use the powers of cosine

like

make it 1-sin^2theta

[still nacelle]

that's not gonna be helpful cuz then you'll have to deal with sin^2

tip:

whenever you have to integrate sin^2 or cos^2, usually convert it into cos(2theta) or sin(2theta) using double angle formula

[hybrid grail]

so after i convert it to cos2theta, i can just integrate it to -sin2theta/2?

[still nacelle]

yup

well not the - sign

[hybrid grail]

oh yea

how does this look

[still nacelle]

uhh cos^2 (theta) does not equal 2cos(theta)

[hybrid grail]

what is it?

[still nacelle]

do you know the double angle formula of $\cos^2(\theta)$ in terms of $\cos(2\theta)$ or $\sin(2\theta)$?

[grand pond]

lgkoo

[hybrid grail]

i knew it but i forgot it

[still nacelle]

look it up

[hybrid grail]

is it 2sinthetacosthta

[still nacelle]

that for sin^2

[hybrid grail]

cos^2thetasin^2theta

[still nacelle]

for reference, all you had to google was cosine double angle formula

[hybrid grail]

oh alright, but how do i integrate that

[still nacelle]

well first step choose the right formula for this case, and rearrange for cos^2

[hybrid grail]

wdym?

[still nacelle]

the question is to integrate cos^2

and the trick is to convert that into cos(2theta)

so which of the 3 cosine double angle formula do you use to do so?

[hybrid grail]

wait so we have 2costheta but idk which equation to plug in for it

[still nacelle]

well it's obviously the second one because we only had cos^2 in the question

the other two requires you to have sin^2

[hybrid grail]

oh wait i think i got it

u move the 2 out

and then cos^2theta-1

but thats not sin so do u just integrate it>

[still nacelle]

idk what you mean

write it down or type it pls

[still nacelle]

???

$\cos(2\theta) = 2\cos^2(\theta) -1$

rearrange to make $\cos^2(\theta)$ the subject

[grand pond]

lgkoo

lgkoo

[hybrid grail]

yea thats what i had

and i moved the 2 out

leaving you with cos^2theta-1

[still nacelle]

mate

you wrote cosine square = 2 times cosine

which is wrong

and 2 cosine is not the same as cosine of (2 times theta)

I will just show you

[hybrid grail]

alright

[hybrid grail]

oh okay i gpt it

i think i can do it from here, i don't wanan keep bothering u since u have exams so i'll work on the other questiosns and try to do them

thanks a lot for the help

[still nacelle]

np

[hybrid grail]

.clsoe

.close

[midnight plank]

Channel closed

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[midnight plank]

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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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[half meteor]

Why is it (n-1) ?

[cedar vortex]

bc we consider the "0th" term

[tribal tartan]

they counted x^0/0! as the first term

[cedar vortex]

^^

[half meteor]

So generally if we have a 0th term

Then, the nth term will have to be some (n-1) ?

[sharp coral]

we have two options with an n=0 term:

1. count it as the first term, then the n= 1 term is the second term (what they did here)
2. count it as the "0th term", then the n=1 term is the first term, etc
   neither is necessarily wrong, although you should make it clear which one you use

[half meteor]

hmm ok

I think i get it

Thanks!

.close

[midnight plank]

Channel closed

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[midnight plank]

Available help channel!

Send your question here to claim the channel.

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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vast plover]

I need help

[vast plover]

how does he get 46 units

[midnight plank]

@vast plover Has your question been resolved?

[midnight plank]

Channel closed

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Send your question here to claim the channel.

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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jaunty dagger]

𝐴 = {𝑥 / 𝑥 is a professional soccer team and belongs to Bogotá} 𝐵 = {-4, -2, 0, 2, 4...}

Based on the selected items, the following responses are provided:

According to the selected item, determine by Extension the set given by Comprehension and determine by Comprehension the set given by Extension.

By Extension:
Set given by Comprehension: {𝑥 | 𝑥 is a professional soccer team and belongs to Bogotá}
By Comprehension:
Set given by Extension: It is not possible to define the set by Extension without knowing the specific elements that satisfy the condition given by Comprehension.
Find the cardinality of each of the sets:

Set 𝐴 is undefined in size (a specific list of professional soccer teams belonging to Bogotá is not provided), so its cardinality is unknown.
Set 𝐵 is an infinite set, as the given sequence extends infinitely in both directions. Therefore, its cardinality is infinite.
Identify the types of sets:

Set 𝐴: Infinite (no specific limit on the number of professional soccer teams belonging to Bogotá is specified).
Set 𝐵: Infinite (it is an infinite sequence of integers).

[midnight plank]

Please don't occupy multiple help channels.

[polar mortar]

is there a question here?

[midnight plank]

@jaunty dagger Has your question been resolved?

[midnight plank]

@jaunty dagger Has your question been resolved?

[midnight plank]

@jaunty dagger Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

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[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dull ember]

i need help brah i forgot how to do this shit

[pastel spoke]

*Bruh

[dull ember]

brah

[astral garnet]

breh

[dull ember]

fr tho help

[last slate]

what question

[astral garnet]

find the roots of the first polynomial

then sub the roots into the seccond polynomial and if it equals zero then they share the same roots

[dull ember]

what are the roots?

x^2 + 3x?

oh fuck find the roots

ok god im shit at this

[astral garnet]

yuh