help-39

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

mh

so

.close

[pearl pond]

Channel closed

Closed by @twilit cedar

Use .reopen if this was a mistake.

[midnight haven]

gonna try it myself

[honest solstice]

Which question are you even doing

We dont read minds

[pearl pond]

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[calm chasm]

I know this can be solved easily using u-sub , But i just want to know what's wrong with this approach

[muted shale]

$\int \cos^3(x)\sin(x)dx$

[calm chasm]

3

cos(x)^3

sorry bad handwriting

[jolly parrot]

Flappie

[muted shale]

does it converge?

,w int cos^3sindx

[jolly parrot]

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Requested by flappie.

[muted shale]

i dont think there is a solution

:)

[calm chasm]

for the intergal ?

the intergal is solved using u-sub

and it will be -cos(x)^4/4 + c

[muted shale]

take u=cos(x)

[calm chasm]

yes i know how it goes

but what im asking

[calm chasm]

what is wrong with this ?

[muted shale]

is it not the same thing?

you can rewrite the cos(4x) and cos(2x)

[calm chasm]

here is the intersting part

wait srry

[muted shale]

yeah

that looks fine

whats the problem?

[calm chasm]

but why arent they the same exact thing ?

after some experminting

[muted shale]

owo, whats this?

[calm chasm]

there is a missing

-3/32

[calm chasm]

mmm

but for a fucntion where c=0

[muted shale]

what happens if you add -3/32 on the red line

[calm chasm]

what is this

[calm chasm]

wait let me show u

identical

exactly

the same

what do u think about that ? @muted shale

[muted shale]

yeah

that makes sense

[calm chasm]

why didnt the constant show up in my apprach

approach *

because like , i dont think thats the c

because the red line is exactly equal to -cos(x)^4/4

@muted shale

[muted shale]

in the $\int \cos^3(x)\sin(x)dx = -\frac{\cos^4(x)}{4} + C$ approach there is also a + C

[jolly parrot]

Flappie

[calm chasm]

yes but in f(x)= cos(x)^3sin(x) , c=0

they should be the exact same thing

with out adding the constant

right ?

[muted shale]

im not sure what you mean

[calm chasm]

im saying

assume c = 0

so f(x)=cos(x)^3sin(x)

so if we take the intergal using the 2 approaches

shouldnt they be the same exact thing

[muted shale]

you cant just do this

if you integrate, you will always get a +C

[calm chasm]

yes because we are intergrating an expression

but here im talking about a function

where i already know

that c=0

[muted shale]

but you dont know the C after integration

if you take your other solution

and take the derivative

you will get the same result

as when you take the derivative of -cos^4(x)/4

[calm chasm]

we put +c because of the possibility that the function had a constant and it vanished because we differentiated it

right ?

[calm chasm]

true

[muted shale]

one moment

Let $f(x)$ be some function, consider $g(x)=f(x)+A$, with A some nonzero constant. Then, $g'(x)=[f(x)+A]'=f'(x)+[A]'=f'(x)$. So, the derivative of g(x) is equal to the derivative of f(x), even though they are not equal.

[jolly parrot]

Flappie

[calm chasm]

okay

now i get it

@muted shale ty

so the -3/32

vanished

when we diffrentiated it

[muted shale]

exactly

[calm chasm]

and gave us the same

[muted shale]

you can always add a constant to a function

[calm chasm]

cos(x)^3sin(x)

so my apprach wasnt really wrong

but the constant just vanished

[muted shale]

and when you differentiate it, you use the property of differentiation that [f(x)+g(x)]=[f(x)]'+[g(x)]'

[calm chasm]

yep

much thanks

.close

[pearl pond]

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[pearl pond]

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[wraith vault]

Hello I would like to know if there is a name for this 3D shape, I did 2D of it following this desmos graph, and then fudged it into 3D so from both right and left it looks like an infinity loop, but I'm not sure if it has a name so I can investigate further, ty
https://www.desmos.com/calculator/pu73klljjp

[quiet tendon]

it looks like a hyperbolic paraboloid

and the side view looks like a lemniscate

[summer imp]

Forbidden pringle

[pearl pond]

@wraith vault Has your question been resolved?

[pearl pond]

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[pearl pond]

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[mild tartan]

help

[mild tartan]

heyshatwt

pease please solve this!!!

please

i need so much help!!

ignore 31

i just need help on 32

please use the lift equation for this

@everyone

i hav my exam in 12 minuets

[naive dirge]

What

Who cares about exam?

We do it for fun

Math is a habit

[mild tartan]

ik

but please it just doesent make sense

i have studied

[naive dirge]

I might not be able to answer it within 12 minutes

[mild tartan]

i like eminem btw

[naive dirge]

The question is lengthy

[mild tartan]

i tried iusing chat gpt and copilot and gemini and nothing worked 😦

[naive dirge]

I suggest you to skip it and focus on other simple questions

[mild tartan]

but this is the main uestn

I LOVE ohio

@everyone

sorry

that was my brother

he took my computer

I understand

i will focus on the other questions

sorry

my borhter took my computer

.close

[pearl pond]

Channel closed

Closed by @mild tartan

Use .reopen if this was a mistake.

[old marsh]

@mild tartan Sorry man no excuse…looks like I have to report you to the discord mods

Ok reported

[mild tartan]

sorry

im sorry

bro my brother

[pearl pond]

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[sharp smelt]

Just wanted to be sure, (1,2,3)(4,5,6)(7,8,9)(10,11,12) would be a valid example, right

[sharp smelt]

oh, hello catbit

yeah,I missed the last part

so say (1,-1,0)(2,0,-2),(-4,0,-4)(3,-3,0)?

thanks

[merry carbon]

Those'll do

[sharp smelt]

thanks

.close

[pearl pond]

Channel closed

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[merry carbon]

Oh wait-

[merry carbon]

.careful of the third one tho, only one of those two entries should be negative in there

.so that you satisfy the whole x + y + z = 0 condition and all, but otherwise

[pearl pond]

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[warped crescent]

I had a question

[warped crescent]

https://www.desmos.com/calculator/a7zk73p4yp

I expiremented with using two inverse functions and a line of symettry (y = x) to make a triange

realizing it makes a isosolece right triangle

was wondeeing why

[vocal lark]

because the inverse transformation swaps x and y coordinates

[warped crescent]

and that explains congruency right

[vocal lark]

So if P1 is on the original function,
P2 is the inverse of P1,
and you just draw straight lines to the line of symmetry (y=x) and mark P3 on that line,
then connect the points in a triangle,
then yes, this is what you get.

[warped crescent]

im going to let that sink in

[vocal lark]

start over

start with ANY other point on the green function

[warped crescent]

i can change the value of c

and the triangle is still isosolece right

triangle

interesting

i created equations that model the length

of the two sides

they end up being essentially the same thing

[vocal lark]

yes.

[warped crescent]

|c - f(c)|
|f(c) - c| -> |-(c-f(c))| -> |c-f(c)|

that brings another observation for me

a very intersting observation

[vocal lark]

hang on a sec...

[warped crescent]

we can model the distance between the two points (c, f(c)) and (f(c), c)

[vocal lark]

drawing something in desmos

[warped crescent]

using the distance formula

[warped crescent]

sure no problem

[vocal lark]

https://www.desmos.com/calculator/jljzjvylar

[warped crescent]

interesting

your graph also accounts for negative values

[vocal lark]

refresh it

I just resaved it

[warped crescent]

yea i see

i also was thinking of something else

if you could find a function to model the distance between the two points

most likely the distance formula

you can integrate that function to find the area between the two curves

perhaps that would have some applications in math?

find the area between a shape with symmetry about some line?

[vocal lark]

sure

[vocal lark]

yeah, that comes up a lot in Calculus

[warped crescent]

yea i just finished calculus

[vocal lark]

refresh one more time

ah...it breaks on the negative values

oh well

[warped crescent]

its no worries

i understand now

[vocal lark]

fixed!

[vocal lark]

cool

[pearl pond]

@warped crescent Has your question been resolved?

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[warped crescent]

the volume of a sphere is 4/3pir^3, where did i mess up in my derivation?

[summer imp]

Your expression for the half circle is wrong if you're going to center it at (r,0).

[vocal lark]

you only did half of it

[warped crescent]

yea

[summer imp]

Either keep x^2 + y^2 = r^2 but integrate from -r to r

[vocal lark]

^

other weirdness here

[summer imp]

Or consider (x-r)^2 + y^2 = r^2

[warped crescent]

0 to 2r wont work?

[vocal lark]

It is useful to reposition some figures so they are centered about the origin

[warped crescent]

if it was centered at the origin

[summer imp]

It can work. The issue is that x^2 + y^2 = r^2 describes a circle centered at the origin, and this one isn't

[warped crescent]

oh

i see

[vocal lark]

if it were centered at the origin (as I did it), I would just work from 0 to r and double my final result.

[warped crescent]

so in my case i would have (x-r)^2+y^2=r^2

where im trying to use 0 to 2r

[summer imp]

If you want to do it like that, yes.

[warped crescent]

ok

i see

[summer imp]

But what Melvin pointed out is true. Especially when it comes to very symmetric shapes like this. Oftentimes you'll save some work by centering everything and putting the symmetry to use.

[vocal lark]

yeah, really most useful for spheres and cylinders tbh

but also plane figures

[warped crescent]

so you would say centering at the origin is better

[vocal lark]

that have symmetry

^

[warped crescent]

alr

[vocal lark]

but also consider

that moving the figure will at least make the lower bound 0

[warped crescent]

moving the figure will make the lower bound 0 regardless

of the radius

no

let me think

[vocal lark]

(here we used symmetry and doubled the result though, when we move the figure to the origin, since we are only evaluating half of the figure)

[warped crescent]

moving the figure as in the circle horizontally right

[vocal lark]

correct

lower bound is technically -r, upper is r

but because of symmetry, we can just work from 0 to r.

and double the final result, since we are going to get a hemisphere.

[warped crescent]

ohhhhh

ok

im figuring this out for the first time, this isnt rlly taught in schools

[vocal lark]

don't worry, you would have got this sooner or later

[warped crescent]

alr thx

for the help

really appreciate it

[vocal lark]

np

[pearl pond]

@warped crescent Has your question been resolved?

[pearl pond]

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

how do i go on about doing this

[muted shale]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vestal pelican]

1

[muted shale]

do you know the chain rule

[vestal pelican]

yes

[muted shale]

try that

[vestal pelican]

where

[muted shale]

the chain rule states that, $[g(f(x))]'=g'(f(x))\cdot f'(x)$

[jolly parrot]

Flappie

[vestal pelican]

ik that im saying where do i apply it

[muted shale]

here, we have f(x,y) and g=z

so we get $[z(f(x,y))]'=z'(f(x,y))\cdot f'(x,y)$

[jolly parrot]

Flappie

[vestal pelican]

do i plugin x and y there?

[muted shale]

not quite

note that, z' and f' are matrices of size 1x2 and 2x1

so we get, $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

[jolly parrot]

Flappie

[muted shale]

$\frac{\partial x}{\partial s}$ and $\frac{\partial y}{\partial s}$ you can calculate, since you know x and y in terms of s

[jolly parrot]

Flappie

[vestal pelican]

hey im kinda stuck at this

[muted shale]

that is the chain rule

[vestal pelican]

yes i know but is it applied there

?

[muted shale]

wdym is it applied there?

[vestal pelican]

is the chain rule done there

[muted shale]

it is the chain rule

[vestal pelican]

ok

[vestal pelican]

how did we get this

[muted shale]

that is what you get from the chain rule, if you write it out

f(x,y) has two variables

so its derivative is a matrix

$f'(x,y) = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{pmatrix}$

[jolly parrot]

Flappie

[vestal pelican]

wasn't z also a part of the equation?

[muted shale]

$z'(f(x,y)) = \begin{pmatrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y}\end{pmatrix}$

[jolly parrot]

Flappie

[vestal pelican]

why is there no f

tbh, i need someone else to help me with this, no offence to you and it's not you, but I need someone who can explain it so i understand

<@&286206848099549185>

[muted shale]

wait, im confusing myself

😅

haha

sorry

[muted shale]

the important one is this

im trying to show the derivation, but i forgot part of it, which makes it hard

[vestal pelican]

and another thing, i cant be having to wait for hours to finish an exercise sorry but i just dont have the time to waste rn

[muted shale]

!volunteer

[pearl pond]

Helpers are just people volunteering their time to help you. Be polite and patient.

[vestal pelican]

yes i know, but if u dont know how to solve some parts it'll be harder for me to understand

[muted shale]

all you need to know about the chain rule is that $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

[jolly parrot]

Flappie

[vestal pelican]

where did "s" come in play

[muted shale]

and similarly for r, $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$

[jolly parrot]

Flappie

[muted shale]

thats where we differentiate with respect to

right here

we want $\frac{\partial z}{\partial s}$

[jolly parrot]

Flappie

[muted shale]

and then we take $\frac{\partial}{\partial r}$ of that

[jolly parrot]

Flappie

[muted shale]

$\frac{\partial^2 z}{\partial r\partial s}=\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial s}\right)$

[jolly parrot]

Flappie

[vestal pelican]

are we finding for r and s?

[muted shale]

is that not what is being asked?

[vestal pelican]

ok

wait

ok so we found those

what else

[muted shale]

wdym what else?

thats it

[vestal pelican]

x=r cos(s) ln (r)

y=sin(s) ln(r)

[muted shale]

yeah, so plug it in and calculate

[vestal pelican]

ok

where do i plug it at

[muted shale]

plug this

[muted shale]

into this

[vestal pelican]

how do i do that

[muted shale]

$\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\right)$

[jolly parrot]

Flappie

[muted shale]

$=\frac{\partial}{\partial r}\left(-\frac{\partial z}{\partial x}r\sin(s)\ln(r)+\frac{\partial z}{\partial y}r\cos(s)\ln(r)\right)$

[vestal pelican]

what did we do to this to have that answer

[jolly parrot]

Flappie

[vestal pelican]

how did we get (∂/∂r)(∂z/∂s)

[muted shale]

this is the question

thats whats being asked of us

[vestal pelican]

nono

[vestal pelican]

yes ik

[vestal pelican]

but from that u got this

[muted shale]

yes?

[vestal pelican]

how

what did u do to it

[muted shale]

well, do you know $f''(x) = \frac{d^2f(x)}{dx^2}$?

[jolly parrot]

Flappie

[vestal pelican]

is this some sort of differentiation rule

[muted shale]

what?

[vestal pelican]

idk

[muted shale]

how do you denote taking a derivative multiple times

@vestal pelican

[vestal pelican]

Hmm

Idk

[muted shale]

how do you denote taking a derivative once

[vestal pelican]

Wdym

[muted shale]

like, f'(x)

how do you write that

[vestal pelican]

How do I derive that?

[muted shale]

no

how do you write it

[vestal pelican]

Wsym

Wdym*

[muted shale]

if you take the derivative of f(x)

how do you write that

$f'(x)$ like so?

[jolly parrot]

Flappie

[vestal pelican]

Derivative of f(x) isn't f*1?

[muted shale]

what?

[vestal pelican]

You mean to show you're deriving f(x)

?

[muted shale]

nevermind

$\frac{d^3f(x)}{dx^3}$ do you know what this means?

[jolly parrot]

Flappie

[vestal pelican]

no

[muted shale]

do you knwo what $\frac{df(x)}{dx}$ means?

[jolly parrot]

Flappie

[vestal pelican]

no

derivative of f(x) wrt x?

[muted shale]

yes

[vestal pelican]

ok

[muted shale]

so then what is $\frac{d^2f(x)}{dx^2}$?

[jolly parrot]

Flappie

[vestal pelican]

second derivative

[muted shale]

exactly

so then do you agree that $\frac{d^2f(x)}{dx^2} = \frac{d}{dx}\left(\frac{df(x)}{dx}\right)$?

[jolly parrot]

Flappie

[vestal pelican]

wouldnt the bottom be d^2 and x^2

[muted shale]

bottom where?

[vestal pelican]

v

dx*dx

[muted shale]

i split it

[vestal pelican]

yes but here it says dx^2

[muted shale]

i have the derivative $\frac{df(x)}{dx}$ and then take the derivative again

[jolly parrot]

Flappie

[muted shale]

taking the derivative means i do $\frac{d}{dx}$ to whatever i want to differentiate

[jolly parrot]

Flappie

[vestal pelican]

if u were to combine this wont you get
((d^2)f(x))/(d^2)x^2?

[muted shale]

ah, i see where its going wrong

$\frac{d^2f(x)}{(dx)^2} = \frac{d}{(dx)}\left(\frac{df(x)}{(dx)}\right)$

[jolly parrot]

Flappie

[muted shale]

does it make more sense like this?

[vestal pelican]

yes

[muted shale]

so then, going back to our partial derivative

$\frac{\partial^2 z}{(\partial r)(\partial s)}=\frac{\partial}{(\partial r)}\left(\frac{\partial z}{(\partial s)}\right)$

[jolly parrot]

Flappie

[muted shale]

does this make sense?

[vestal pelican]

yes

you could have said it was the factorized form btw

[muted shale]

factorised form?

[vestal pelican]

isnt it factorised?

[muted shale]

yeah ig, but you almost never see it like the RHS

[vestal pelican]

anyways

now

[muted shale]

you would almost always write it as $\frac{\partial^2 z}{\partial r\partial s}$

[jolly parrot]

Flappie

[vestal pelican]

we plugin x and y

or r and s?

[muted shale]

first we look at $\frac{\partial z}{\partial s}$

[jolly parrot]

Flappie

[vestal pelican]

ok

[muted shale]

which, by the chain rule is $\frac{\partial z}{\partial x}\frac{\partial x}{\partial s} +\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

[jolly parrot]

Flappie

[muted shale]

x and y are given, and we can calculate their derivatives w.r.t. s

[vestal pelican]

ok

ohhh wait i think i got it

here

how did we get "-"

[muted shale]

look at x

x=rcos(s)ln(r)

what is the derivative of x w.r.t. s

[vestal pelican]

not sure

[muted shale]

what would you guess that it is

[vestal pelican]

idk how to do derivatives like that i find it hard

[muted shale]

okay

imagine you have this

x=4*cos(s)*ln(4)

then what is $\frac{dx}{ds}$?

[jolly parrot]

Flappie

[vestal pelican]

wait

this is product rule

[muted shale]

thats not what im asking for

[vestal pelican]

ik

[muted shale]

if $x=4\cos(s)\ln(4)$ then what is $\frac{dx}{ds}$?

[jolly parrot]

Flappie

[vestal pelican]

4cos1 ln4

do we apply differentiation rule?

[muted shale]

no

[vestal pelican]

ik

[muted shale]

what is $\frac{d}{dx}\cos(x)$?

[jolly parrot]

Flappie

[vestal pelican]

-sinx

[muted shale]

exactly

[muted shale]

so then isnt this $-4\sin(s)\ln(4)$?

[jolly parrot]

Flappie

[vestal pelican]

sure

[muted shale]

if you take a derivative w.r.t. a certain variable, you consider the rest to be constants

[vestal pelican]

ik that

[muted shale]

so then what are you confused about?

[vestal pelican]

if this is with respect to "s" why isnt r treated as a constant

[muted shale]

what makes you think its not?

[vestal pelican]

because instead of "c" it is r

[muted shale]

might be because of my sleep deprivation, but i dont see what you mean

[vestal pelican]

nvm

anyways

is there anything after this

[muted shale]

yes

we still need to diff' w.r.t. r

[vestal pelican]

do we use any differentiation rule?

[muted shale]

product rule and chain rule

[vestal pelican]

ok

[muted shale]

$\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial y}r\cos(s)\ln(r)\right) = \left(\frac{\partial}{\partial r}\frac{\partial z}{\partial y}\right)r\cos(s)\ln(r)+\frac{\partial z}{\partial y}\left(\frac{\partial}{\partial r}r\right)\cos(s)\ln(r)+\frac{\partial z}{\partial y}r\left(\frac{\partial}{\partial r}\cos(s)\right)\ln(r)+\frac{\partial z}{\partial y}r\cos(s)\left(\frac{\partial}{\partial r}\ln(r)\right)$

[jolly parrot]

Flappie

[muted shale]

this is the product rule

on the second term

[rustic gate]

that's quite some latex you're writing there

[muted shale]

ctrl v did a lot of work

[vestal pelican]

umm

how did we get this

[muted shale]

from here

[vestal pelican]

aha

ohhh wait

ok what

this

is = to this?

[muted shale]

no

first of all, i only took the second term

this one

(i didnt want to write out the full thing)

[muted shale]

and this is differentiated again, but w.r.t. r

hence this at the start

[vestal pelican]

how did we get this

[muted shale]

that is the product rule

[vestal pelican]

ohh yeah nvm

[muted shale]

$\frac{d(uv)}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}$

[vestal pelican]

ikik

this is very long bro

[jolly parrot]

Flappie

[muted shale]

i suggest you revise your differentiation rules in single and multiple variables

you struggle a lot with them

[vestal pelican]

its just that there are a lot of variables in play and it becomes confusing

[muted shale]

indeed, so having a strong grasp of the basic differentiation principles will make it easier and faster

[vestal pelican]

so this just need to keep on derivating right?

[pearl pond]

@vestal pelican Has your question been resolved?

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[pearl pond]

Please don't occupy multiple help channels.

[vestal pelican]

which test should i apply

[pearl pond]

@vestal pelican Has your question been resolved?

[midnight haven]

you could use the integral test, I think?

it's decreasing, positive on the domain $[1, \infty)$ and continuous

[jolly parrot]

45

[midnight haven]

I suggest you do $\sum_{n=1}^\infty \frac{n+1}{n^2\sqrt{n}} = \sum_{n=1}^\infty \frac{n+1}{n^{5/2}}$ then divide by the denominator and apply the integral test

[jolly parrot]

45

[vestal pelican]

yeah limit test

[midnight haven]

well, you take the limit as the upper bound of the integral goes to infinity

[vestal pelican]

nono i got u dw

quick question

im doing the integral

[midnight haven]

yeah?

[vestal pelican]

im here

i should use this

right?

[pearl pond]

@vestal pelican Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[pearl pond]

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>.
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• 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.

[sudden cargo]

How to find minimum of trig functions involving reciprocal terms?

[sudden cargo]

Like 14cosec^2 x + 9sin^2 x?

[sharp smelt]

You want to minimse $14cosec^2(x)+9sin^2(x)$?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

I'd suggest $AM \geq GM$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[pearl pond]

@sudden cargo Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sudden cargo]

.reopen

[pearl pond]

✅

[sudden cargo]

yes

[sudden cargo]

I havent studied that

just tell me the other way

[sharp smelt]

That's the only way I know that doesn't require calculus

Do you know how to compute the arithmatic mean of two numbers

and their geometric mean?

[sudden cargo]

Cant we use completing middle term?

[sharp smelt]

I mean you could I guess

[sudden cargo]

I know arithmatic mean but not geometric

[sharp smelt]

the arithmatic mean of two numbers a,b is defined to be $\sqrt{ab}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[midnight haven]

Bruh

[sudden cargo]

bruh what

[midnight haven]

Geometric*

[sharp smelt]

my bad

yes

[midnight haven]

He meant

[sharp smelt]

geometric

[sudden cargo]

oh

okay

good enough

now how do we apply that here

[sharp smelt]

also know that $AM \geq GM$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

okay

$14cosec^2(x)+9sin^2(x)$

[jolly parrot]

Aryan07

[sudden cargo]

yeah now?

[sharp smelt]

so $\frac{14cosec^2(x)+9sin^2(x)}{2} \geq \sqrt{ 14 \cdot 9}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

hmm

[midnight haven]

6root14 🗿

[sudden cargo]

OH

its that simple?

💀💀

wow

[midnight haven]

I told you it's one liner

Earlier lmao

[sudden cargo]

lol

damn you know more useful formulas than me

[midnight haven]

Bruh nah, you would learn it too

[sudden cargo]

so is this applicable to all the minimum value questions for functions involving trigonometry?

[midnight haven]

Anyways bye

[sharp smelt]

There's one more useful inequality you should know

[midnight haven]

HM?

[midnight haven]

??

[sharp smelt]

$QM\geq AM \geq GM \geq HM$

[sudden cargo]

uh oh

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

now whats QM and HM

[sharp smelt]

Quadratic mean

and harmonic mean

[sudden cargo]

formula?

[sharp smelt]

The QM of n numbers is defined to be $\sqrt{\frac{\left(x_1^2+x_2^2+x_3^2.......x_n^2\right)}{n}}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[midnight haven]

Uh quadratic inequality is rarely used tbh in JEE

[sharp smelt]

still useful

especially for Adv

[sudden cargo]

hmm

ig i dont need it now

[midnight haven]

You might see few questions
I haven't seen one until now

Ofc you can learn it

[sudden cargo]

maybe ill learn it later

soo

I wanted to ask

[sharp smelt]

and the harmonic mean of n numbers is defined to be$\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}......\frac{1}{x_n}}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

is AM greater than/equal to GM applicable for all these types of question?

[midnight haven]

Harmonic is basically inverted arithmetic

[sharp smelt]

yeah, by definition

[sharp smelt]

usually

at times you use it with calculus/ other identities

[sudden cargo]

hmm

[midnight haven]

Ig it's for non negative only
Am I right ? @sharp smelt

[sudden cargo]

lets say we have find the range of 1/(3sinx + 4cosx+2)

[sharp smelt]

I think so

[sudden cargo]

oh wait i think thats easy

[sharp smelt]

here DON'T use $AM \geq GM$, try finding the amplitude of the function instead

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

is it [-1/3, 1/7]

[sudden cargo]

whats amplitude?

[midnight haven]

You multiply and divided by 5 to get together sinx and cosx then use the fact that sinx lies from -1,1 then do the transformation

[midnight haven]

Umm idk

[sharp smelt]

,w maxima of 3sin(x)+4cos(x)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by physicsrocks

[sharp smelt]

yup, 5

yes

you're right

[sudden cargo]

this right

[sharp smelt]

yes

[sudden cargo]

hmm

[sharp smelt]

let me verify it

wait

won't it be R

[sudden cargo]

uhhhh

[sharp smelt]

as 3sin(x)+4cos(x)

can be -2

[sudden cargo]

when tho

[sharp smelt]

,w 3sin(x)+4cos(x)=-2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by physicsrocks

[sharp smelt]

here

[sudden cargo]

bruh

[sharp smelt]

,w range of $1/(2+3sin(x)+4cos(x))$

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[sudden cargo]

,w range 1/(3sinx + 4cosx+2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 99r_ust

[sharp smelt]

huh

my bad

sorry

[sudden cargo]

uhh

how does it go till infinity

i got upper limit as 1/7

oh wait

[sharp smelt]

3sin(x)+4cos(x) can be -2

[sudden cargo]

yeah

[sharp smelt]

yes?

[sudden cargo]

1/7 is upper limit

all good

[sharp smelt]

*lower limit

[sudden cargo]

nope

-1/3 is lower limit

hold on

1/5+2

so 1/7 is minimum

and

1/(-5+2) = 1/-3

= -1/3

is upper limit

but how tf

is upper limit smaller than lower limit

[midnight haven]

What the heck is this

[sudden cargo]

3sinx + 4cosx can be [-5,5]

then + 2

so denominator is [-3,7]

[midnight haven]

True ofc

[sudden cargo]

but then if smallest denominator value is -3
we get whole expression as -1/3

and if denominator is highest, lowest value of expression goes to 1/7

idk how is that possible

[sharp smelt]

why isn't it possible ?

[midnight haven]

Yeah it is positive

Possible

[sudden cargo]

according to this, range of expression would be [1/7, -1/3]

[sharp smelt]

no

[midnight haven]

Everything minus this interval

[sharp smelt]

this open interval

[midnight haven]

R -

The thing

[sudden cargo]

im brainded

idk how

,r minima 1/(3sinx + 4cosx+2)

[midnight haven]

The inequality would get reversed as it is in denominator

[sudden cargo]

bruh

,r

[sharp smelt]

*,w

not ,r

[sudden cargo]

,w 1/(3sinx + 4cosx+2)

my ba

d

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 99r_ust

[midnight haven]

Nice

[sudden cargo]

,w range 1/(3sinx + 4cosx+2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 99r_ust

[sudden cargo]

,w maxima 1/(3sinx + 4cosx+2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 99r_ust

[midnight haven]

Yeah this correct

[sudden cargo]

so maxima is -1/3

[midnight haven]

What are ya tryna do

[sudden cargo]

,w minima 1/(3sinx + 4cosx+2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 99r_ust

[sudden cargo]

HOW IS MAXIMA LOWER THAN MINIMAA

[sharp smelt]

these are local minimas and maximas

to understand them you need to understand calculus

[midnight haven]

Yeah better not rush things

[sudden cargo]

should i jump off a bldg

[sharp smelt]

please don't.

[midnight haven]

@sudden cargo you there ?

[sudden cargo]

yes

no

im going to terrac

e

its over

[midnight haven]

Stopp

[sudden cargo]

fk this

[midnight haven]

I will teach you

[sharp smelt]

We're here to help

[midnight haven]

Waitt

[sharp smelt]

please don't do anything to yourself

[sudden cargo]

jumps

[sharp smelt]

hey, you still here ?

we're here if you need help

please don't do anything to yourself

[sudden cargo]

im back

[midnight haven]

It's yer fault for doing suchthing now

[sudden cargo]

typing from heaven rn

jk guys I was in lift so no network coverage for few seconds

now its good

[sharp smelt]

what do you need help with exactly?

like what do you need explained ?

[midnight haven]

|| iam ignoring you now for pulling such a prank||

[sudden cargo]

whats the easiest way to find mimima and minima

[sudden cargo]

[sharp smelt]

depends on the problem

at times inequalities

at times calculus

[sudden cargo]

no give me a 1shot, like a solution to all problems

[sharp smelt]

calculus , but it can get really ugly

[sudden cargo]

whats there in calculus exactly

[sharp smelt]

differential calculus

turning points

monotonicity

integrals

[sudden cargo]

havent studied any of that

maybe basic integration

[midnight haven]

You can calculate area of a rotating ellipse about its axis
Jk this is too much

[sharp smelt]

but really you have to chose the method based on the problem

[sudden cargo]

hmm

[sharp smelt]

there's never one method in maths

[sudden cargo]

value of cos^2 pie/16 + cos^2 3pie/16 ....... till 4 terms

@sharp smelt@midnight haven

did yall jump too 💀💀

[sharp smelt]

no

I'm thinking

[sudden cargo]

ok

[sharp smelt]

$cos^2 (\frac{\pi}{16} ) + cos^2 (\frac{3\pi}{16})+ cos^2 ( \frac {5\pi}{16} + cos^2(\frac{7 \pi}{16}}$$

[sudden cargo]

we were taught formula when when angles are in AP, but it was for cos not cos^2

it is till 7pie/16

[midnight haven]

Till 4 terms ?

[sudden cargo]

4 terms

[midnight haven]

Okay so ap

Ugh ugly lol

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[midnight haven]

Ye

[sudden cargo]

better

[midnight haven]

Value you want alr

[sudden cargo]

yep

[midnight haven]

Use trigonometry circle

[sharp smelt]

I'd maybe use euler's formula

[sudden cargo]

whats that

[sharp smelt]

$cos(\theta) + i sin(\theta)= e^{ i \theta}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

nahhhhhhhhhhhhhhhhhhhhhhh

[midnight haven]

Ye you can use it

[midnight haven]

Lol

[sudden cargo]

tell me some real ways bruh

i dont know complex numbers yet

[sharp smelt]

"real ways" good pun

use transformations

[sudden cargo]

uhh

[midnight haven]

Look for complementary angles
That's what I mean

[sudden cargo]

there isnt any I can find

[midnight haven]

Try to convert one of them to sign

Sin*

[sudden cargo]

maybe pie/16 + 7pie/16 = pie/2

[midnight haven]

Yay

Ypu you got it

So

First and last term

And second and third term

[sharp smelt]

[midnight haven]

2 is the answer boi

[sudden cargo]

bruh

the fk

how

[sharp smelt]

yup

[midnight haven]

😭 you got it too

[sharp smelt]

euler's formula is def overkill here

[midnight haven]

This

[sudden cargo]

OH YEE

I GOT IT

[sudden cargo]

frfr

[sharp smelt]

[midnight haven]

Nah it's normal lmao

[sudden cargo]

next question

[midnight haven]

Let's do Euler one now ?

[midnight haven]

unecessary here

[sharp smelt]

I know

[sudden cargo]

2(sin1 + sin2 + sin3............. sin89)/(2(cos1 + cos2 + cos3 ............. cos44) + 1)

[midnight haven]

look at extreme angles

[sharp smelt]

notice the angles are complementary

[sudden cargo]

yes

but hows that useful

[midnight haven]

89 + 1 = 90

[sudden cargo]

we pretty much have sin1 + sin2 ....... sin44 and cos1 + cos2 ...... cos44. and remaining is sin45 which is 1/roo2

thats in numerator

[midnight haven]

Anddd
89-1/2 = 44

That's it ig

[sudden cargo]

huhhh

,w 2(sin1 + sin2 + sin3............. sin89)/(2(cos1 + cos2 + cos3 ............. cos44) + 1)

[midnight haven]

Since + sind eee

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by 99r_ust

[midnight haven]

Sinc

[sudden cargo]

bruh

do i need to do it for every single pair

[sharp smelt]

Notice a pattern will form

[midnight haven]

I mean if you really wanna learn maths
Then sure

Or you can do for some specific

And follow trend

[sudden cargo]

sin1 + sin2 is 2*sin(3/2)cos(1/2)

[midnight haven]

Alr I gtg bye
Ask why am I here , he is a sweet person

[sharp smelt]

I have to go too, have to study for bitsat

[sudden cargo]

bruh

yall abandoning me

[midnight haven]

Bruh what

[sudden cargo]

[sharp smelt]

but I can stay for this problem I guess

[midnight haven]

You are Indian too

[sudden cargo]

ofc

[sharp smelt]

half this server is Indian

[midnight haven]

Dude why do you ask questions which I usually don't get 😭
What are you preparing forrr?

[sharp smelt]

just ping helpers, someone will come to help you

[midnight haven]

Olympiad?

Vectors thing

[sharp smelt]

My parents want me to write bitsat, but I'm studying for college

[midnight haven]

Engine 1st year stuff it is Ig

[sharp smelt]

that's linear algbera

[midnight haven]

Yeahh makes sense

Yeah I have just seen it from far away, never touched

[sharp smelt]

anyway, we're getting off topic

[sudden cargo]

yes

[sharp smelt]

back to OP's question

[sudden cargo]

2(sin1 + sin2 + sin3............. sin89)/(2(cos1 + cos2 + cos3 ............. cos44) + 1)

[midnight haven]

All the best for yer bitsat

[sharp smelt]

notice that sin(46)=cos(44), and so on

think that will help you

[sudden cargo]

hmm

we get sin1+ sin2..... sin44 + cos1+cos2 ..... cos44 + sin 45

[sharp smelt]

wait

can you TeX the problem?

[sudden cargo]

idk how to use it

,help

[jolly parrot]

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!

[sharp smelt]

https://approach0.xyz/search/

use this

start by typing dollars

and then type like sin(x)/x or whatever

it will code it for you

[sudden cargo]

lemme try

ignore the 0 at the top

[sharp smelt]

ok

so notice that cos(0)=1

so you have to solve

[sudden cargo]

we dont cos0

we dont get cos0

[sharp smelt]

cos(0)=1

$\frac{2\left(\sin\left(1\right)+\sin\left(2\right).....\sin\left(89\right)+\sin\left(89\right)\right)}{2\left(\cos\left(0\right)+\cos\left(1\right)+\cos\left(2\right)....\cos\left(44\right)\right)-1}$

yes?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sudden cargo]

it starts from cos1 not cos0

[sharp smelt]

what is cos(0)

[sudden cargo]

1

[sharp smelt]

yes

which is why I've written it like this

[sudden cargo]

oh okay

then?

[sharp smelt]

$\frac{2\left(\sin\left(1\right)+\sin\left(2\right)..+\sin\left(45\right)+\cos\left(44\right)+\cos\left(43.\right)..+\cos\left(1\right)\right)}{2\left(\cos\left(0\right)+\cos\left(1\right)+\cos\left(2\right)....\cos\left(44\right)\right)-1}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

now pair common terms in the numerator and use sum to product transformations

[sudden cargo]

you mean sin1 + sin2?

[sharp smelt]

wait, don't do that

https://artofproblemsolving.com/community/c4h537722p3090171

this is more elgant

*elegant

[sudden cargo]

dont know how root2/2 comes

[sharp smelt]

sin(A-B)

[sudden cargo]

thats sinAcosB - cosAsinB

but we dont have any multiplication

[sharp smelt]

yes

please ping helpers, I have to go

[sudden cargo]

@lunar nacelle

<@&286206848099549185>

$\frac{2\left(\sin\left(1\right)+\sin\left(2\right)..+\sin\left(45\right)+\cos\left(44\right)+\cos\left(43.\right)..+\cos\left(1\right)\right)}{2\left(\cos\left(0\right)+\cos\left(1\right)+\cos\left(2\right)....\cos\left(44\right)\right)-1}$

[jolly parrot]

Aryan07

[abstract vine]

bruh stop

[sudden cargo]

nah

help me

[abstract vine]

let me cook

[sudden cargo]

bruh did u get cooked instead?

@abstract vine

[abstract vine]

use the identity: $sin(90-x) = cos(x)$

[jolly parrot]

clonesolopros

[sudden cargo]

yeah then???

[frosty bay]

sin 1 + cos 1 = sin 1 + sin 89 = 2 sin(1+89)/2sin(88/2), so on till
sin 44 + cos 44 = sin 1 + sin 46 = 2 sin (90)/2 sisn(45)/2

[sudden cargo]

thats fked up

imma just ask my math teacher ig

.close

[pearl pond]

Channel closed

Closed by @sudden cargo

Use .reopen if this was a mistake.

[pearl pond]

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.

[midnight haven]

23

[sharp smelt]

Well , what have you tried

[midnight haven]

Nothing

[sharp smelt]

Try analysing it case wise

[midnight haven]

Like?

[sharp smelt]

Or use the definition of floor (x)

[sharp smelt]

Greater than 0 and less than 0

[midnight haven]

Ugh I hate these kind of question
Lemme try

[sharp smelt]

Try using floor(x)= x-fractional part of (x)

[jolly parrot]

Bob l'éponge

[midnight haven]

Oh oh

[glacial sequoia]

take 1.1 and -1.1, 69.420 and -69.420

floor them and see what you can come up with

[midnight haven]

69 I see

[rustic gate]

not even 1.337 and -1.337

smh

[sly dagger]

this is childish behaviour, i expect better from u guys, try a number like 8008.135

[midnight haven]

I got it

Thanks

. close

Fok

.close

[pearl pond]

Channel closed

Closed by @twilit cedar

Use .reopen if this was a mistake.

[pearl pond]

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[reef quiver]

Consider an $n\times d$ data matrix $X = [X_1 \ldots X_n]^T$.

Suppose the data has been whitened, meaning that a linear transformation is applied to $X$ that made the resulting features be uncorrelated and have variance 1.

My text says this leads to the condition

$$X^TX = nI$$

where $n$ is the number of sample points.

But I can't see why this holds. Any help would be appreciated.

[jolly parrot]

shimamooo

[reef quiver]

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[pearl pond]

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.

[dim timber]

just for the circled part here, how does this rearranging work?

[dim timber]

is it because acceleration is equal in one system

?

[vast berry]

the purple one?

[dim timber]

yeah

like why is it (m1+m2)a

i think its because acceleration is trhe same in a system