help-41

[stoic valley]

No need!

you already di this part!

Now you were going on the correct track here

Just do the same thing for the denominator asw!!

same logic as you aplplied here but with a better result

It wont give you a form like we got before, aka inf * 0 or 0/0

[rapid plaza]

oh

I already did it again lmao

is this on the right track aswell?

[stoic valley]

Lol thats fine!

[stoic valley]

Yeah you are!!

[rapid plaza]

then multply top and bottom by 1/x to put it in the form a/x right

[stoic valley]

Yeah!

Do lmk what you get in the numerator and denominator after putting all such terms with a/x or a/(x^2) form as 0!!

[rapid plaza]

3/2

[stoic valley]

Thats correct good job!!

[rapid plaza]

but what does that mean

horizontal asym?

[stoic valley]

oh you mean graphically

[stoic valley]

Yeah

[rapid plaza]

how I kinda guessed that

my teacher never went over the properties of limits

so like all I know about them is using them for endbehavior at VA

[stoic valley]

wait wdym

va?

[rapid plaza]

im lost I only knew to plug in x as 0 because of this

[stoic valley]

x as 0??

[rapid plaza]

or sorry the lim of 1/x

as x aprouachs inf

[stoic valley]

Ah i see

[jagged zephyr]

Ok guys what happen where we all stuck

[stoic valley]

Sometimes we may not get the best teacher but again it also plays a role on ones resolve to learn

[stoic valley]

on philosophy

[jagged zephyr]

Ic no one taught me that i shoud learn

[stoic valley]

You think more than what you think you thought

[rapid plaza]

so the reason why we can set it to 0 is because when were taking a lim were taking the num were apporaching

[jagged zephyr]

Brain is not braining that what u wanna say?

[rapid plaza]

thats what I thought and is why I mentioned the VA thing cuz thats what we used lims for so far

like setting x as a num extremly close to the asym to find endbehavior

[stoic valley]

Yeah thats what x tending to infinity means!

[rapid plaza]

ohhhhhh wait I thinks its all clicking

okok I think I got it

[stoic valley]

Glad to hear!!

Any more?

[rapid plaza]

tysm guys 🙏

.close

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[sacred sphinx]

Hello

[sacred sphinx]

I need help with this

$\sum_{n=1}^{\infty} \frac{(2n)! + 2^n}{(n!)^2 \cdot 5^n + \sqrt{n!}}$

[grizzled pagoda]

KevinSnow

[sacred sphinx]

I cant know if its convergant or not

[signal furnace]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[round rune]

which of the terms in the denominator is largest?

for large n

think about the numerator too

[sacred sphinx]

nvm I solved it

ty!

.close

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[small copper]

You are welcome, have a nice day!

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[worldly merlin]

could someone help me derive d<p>/dt ?

[amber wasp]

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

[worldly merlin]

2

[shy kraken]

whats ur problem

?

[worldly merlin]

I’m struggling to get to the part where stuff will cancel to give me the answer. I don’t know if I’m making some stupid little mistakes or if I’m missing something

My chemistry teacher gave me an introduction to quantum mechanics pdf but I don’t think I’m ready for it. My teachers don’t know what to do so I was wondering you guys could help me out?

[shy kraken]

idk wtf i am even looking az

[worldly merlin]

It’s deriving ehrenfest’s theorem

[pallid canopy]

you haven't defined anything in here so it's impossible to tell what depends on t and what doesn't

[pallid canopy]

you should really just read the pdf

[pallid canopy]

there's plenty of proofs of this theorem online

https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/03%3A_Fundamentals_of_Quantum_Mechanics/3.04%3A_Ehrenfest's_Theorem

https://farside.ph.utexas.edu/teaching/qmech/Quantum/node36.html

[worldly merlin]

Ok thanks. I’ve been reading the pdf but at the end of each chapter there are questions to check your understanding. I’ll look into those links

[pallid canopy]

https://physics.stackexchange.com/questions/698849/ehrenfest-theorem-proof

[worldly merlin]

.close

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[small copper]

Have a nice day!

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[lime ingot]

Yo, I still have a question unanswered regarding directional derivatives , why is that if we need a derivative in a specific direction of a unit vector we multiply the partial wrt x with the first component of the unit vector and the derivative wrt y with the second component of the unit vector ?

[pallid canopy]

read

where this is g(z)

[lime ingot]

Yeah sorry that’s not helping I need someone to explain the geometric intuition for me

[pallid canopy]

"why" is the proof

https://math.stackexchange.com/questions/2066558/directional-derivatives-geometric-intuition

https://mathweb.ucsd.edu/~ashenk/Section14_5.pdf

[amber wasp]

@lime ingot Has your question been resolved?

[grizzled pagoda]

Azyrashacorki

[full elk]

See this

[lime ingot]

I did but it didn’t help

I just need someone to explain to me the intuition I just work best when talkin about it

Like just the intuition behind the multiplication part that’s it

[full elk]

The point is that this is the intuition

It’s that way because if you write down what you want it to mean to take a directional derivative, you end up with the dot product

[lime ingot]

I know but why tho

Why is it that if you take two partial derivatives and combine them while multiplying each one with one component of the unit vector work

[full elk]

Because the partial in x gives you how f changes in the x direction and similarly with the partial in y.
If you go in some other direction (a,b), then the change in f will be some amount in the x direction proportional to a + some amount in the y direction proportional to b, and those amounts are specifically the partials.

[pseudo crescent]

if you're asking about why multiplication specifically, i'd think about it like this:

as azyrashacorki said, partial derivative in x gives you how f changes in the x direction. Say that it it's = 2 and so it's 2 f-units / 1 x-unit. Now say that the component of your direction vector in the x-direction is 1/2. So it says that we move 1/2 x-unit. Now if f changes by 2 f-units per 1 x-unit and we move 1/2 x-unit, that means that the total change in f is just 2 * 1/2 = 1 f-unit. Notice the multiplication

[lime ingot]

I got lost at:

“If you go in some other direction (a,b), then the change in f will be some amount in the x direction proportional to a + some amount in the y direction proportional to b, and those amounts are specifically the partials.”

Specifically “proportional to a + some amount in the y direction proportional to b”

[lime ingot]

So you mean say you find a derivative in the x direction at some specific point and you get 2?

Is this kind if like a linear combination?

Or am I going too far

Why does this just seem like linear combination @full elk @pseudo crescent

[humble palm]

it is a linear combination

because the differential at a point is a linear map

think of derivatives as "best linear approximation of the functions"

[lime ingot]

So are you basically scaling the components of the unit vector and combining them into a new vector ?

Oh And is that why it needs to be a unit vector

[humble palm]

in this case you get a scalar because the image of g lies in R

[lime ingot]

Yeah I think I’m wrong

[lime ingot]

Mb I don’t get it tho

[humble palm]

think about the tangent plane of the curve: its slope in the x direction is dg/dx and in the y direction it is dg/dy

[lime ingot]

Sure

I mean wouldn’t that be infinite tangent lines in the x direction and vise versa

[humble palm]

there is only one line in the "pure x" direction

at a given point

[lime ingot]

Yeah at a point yes

But given a function

And being told to find its direction vector you start by computing its general partials

Right

Can we go up from here

[humble palm]

sure

[lime ingot]

So now you have infinite tangent lines going in the x direction

And infinite tangent lines in the y direction

And you’re given a unit vector

[humble palm]

only one tangent line no?

[lime ingot]

At some direction

[lime ingot]

If you’re finding the general partial derivatives without a specific point? No

[humble palm]

yeah without a specific point ok

[lime ingot]

Yeah so say you’re given some point now and a unit vector

So the point will narrow the infinite tangent lines to two lines directly perpendicular to one another

Now the vector part is where I’m lost

Like the direction

[humble palm]

ok so picture the tangent plane. the whole plane equation is dictated by the two slopes in the x and y direction

[lime ingot]

Sure

[humble palm]

so the plane equation is like z = ax + by right

a = dg/dx ; b = dg/dy

so if you take a vector, the formula for the derivative gives the height of the plane at the vector-point-direction

[lime ingot]

We’re thinking of <a,b> as a vector right

[humble palm]

a and b are scalars here I think

[lime ingot]

Wait how did you derive the equation

[humble palm]

but it could work with vectors

[lime ingot]

What’s your normal vector

Here

[humble palm]

also I centered the plane at 0,0,0

[humble palm]

and this height, for a unit vector, is equal to the slope in this direction (because rise over run)

I suggest playing around in desmos if you can

[lime ingot]

Honestly it’s here where you lost me

Are you down to go over that?

[amber wasp]

@lime ingot Has your question been resolved?

[humble palm]

a general non vertical plane equation has that form right

the tangent plane has equation $$z = g(x_0, y_0) + (x - x_0) \frac{\partial g}{\partial x}(x_0, y_0) +(y - y_0) \frac{\partial g}{\partial y}(x_0, y_0)$$

[grizzled pagoda]

bloubbloub

[humble palm]

now for the sake of simplicity, take x0 = y0 = g(x0, y0) = 0

This is the same as translating the 3D space such that the point we are considering lies at the origin

then you get something like $$z =x \frac{\partial g}{\partial x}(0, 0) +y \frac{\partial g}{\partial y}(0, 0)$$

[grizzled pagoda]

bloubbloub

[humble palm]

which is what I did earlier

[amber wasp]

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[wild iris]

how do i do b

the gradient -> inf

so (y-x)/(3y-x) = inf

like how do i solve that

[signal furnace]

inf is not a real number, so the division is undefined

when is division undefined

[wild iris]

div by 0?

so 3y -x = 0?

[signal furnace]

yes

[signal furnace]

yes

[amber wasp]

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[tough mica]

can I get some help with b)?

let me translate but the gist is that they want me to prove this propositions using the ordered field axioms

[amber wasp]

@tough mica Has your question been resolved?

[grizzled pagoda]

Renato

[tough mica]

here is the translated problem

i still need help with b)

i know the ordered field axioms a little bit

but idk how to use them like that

i did a) but was different

[astral relic]

Isnt b) basically an axiom.

[tough mica]

yeah... wtf

[tough mica]

exactly dude...

what is there to prove??

c doesnt change the order in the inequality

my friend just said that and moved on with the exercises

[tough mica]

todo bien gordi?

como va?

[astral relic]

Well, it isnt axiomatic in itself, it can be proven by rearranging the relation

[amber wasp]

@tough mica Has your question been resolved?

[tough mica]

care to elaborate

[astral relic]

Substract bc from both sides

[tough mica]

a <= b

a - bc <= b - bc

what now?

c.(a-bc) <= c.(b-bc)

what??

[tough mica]

what do I start with?

[astral relic]

Nope, start at ac <= bc

[tough mica]

ok

ac - bc <= bc - bc

ac - bc <= 0

[tough mica]

but thats what I want to prove, I cant start with what I am trying to prove

you see what I mean?

[astral relic]

you can reason from (a-b)c <= 0 into proving that the case is true

[tough mica]

i cant start from this, then the proof is circular

[astral relic]

the (a-b)c <= 0 is just a reorganization of the thing you want to prove. Not the proof itself

Since substracting the same element from both sides and then later distribution is axiomatic for an ordered field

Althought you could start from
a =< b
a-b =< 0, a-b = X in F
X =< 0
Xc =< 0

But youd have to also prove that if X =< 0, c > 0 ==> Xc =< 0 > Since this one isnt an axiom itself

[vital tinsel]

@tough mica a <= b implies a = b or a < b

If a = b then ac = bc so ac <= bc.
If a < b then
b - a > 0
The axiom of closure of positives under multiplication gives
(b - a) * c > 0
Using the distributive law
bc - ac > 0
so ac < bc, so ac <= bc as desired.

[amber wasp]

@tough mica Has your question been resolved?

[barren nova]

simple proof : if a <= b and c > 0, divide the cases
Case 1: a < b

then by the order-compatibility axioms of the ordered field, it follows that ac < bc.
Case 2: a = b
then clearly ac = bc.
Adding up the results, we have that ac <= bc

[vital tinsel]

renato what did you disagree with in this?

[amber wasp]

@tough mica Has your question been resolved?

[left arch]

-# what degree are you doing renato? (just curious)

[amber wasp]

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[abstract glen]

I literally cannot figure this out

I tried using the point distance formula of modulus Ax+By+C/(sqrt a^2+b^2)

but at one point it gets too complex and I can continue no further because simplification makes no sense

[stuck vault]

!show

[amber wasp]

Show your work, and if possible, explain where you are stuck.

[jagged zephyr]

Can u show me where it get complex

[abstract glen]

okok wait

[stoic valley]

<@&268886789983436800>

[abstract glen]

(might take a min, im recharging my phone to take the pic) 😭

[stoic valley]

K sure take your time!

[stoic valley]

Just a quick point out, here it should be A and B in denominator right?

maybe you went wrong there?

[abstract glen]

oh nonon, forgot to capitalise

[stoic valley]

Ah okay

[stuck vault]

So the 2 points make a horizontal line (lie on the x axis)

This means that $\theta$ should be limited to some specific values so that the line given has to be a vertical

[grizzled pagoda]

1 divided by 0 equals Infinity

[stoic valley]

What??

[jagged zephyr]

Wait gang he showing his work

[stoic valley]

I think 1/0 may have misinterpreted the Qsn

[jagged zephyr]

??

1/0 what u mean ?

[stoic valley]

.

this guy

[abstract glen]

guys funny thing, i was redoing mah working and missed one trig identity application, it simplfiied to b^2 😭

imma send it now

phone charged

[stoic valley]

Ah okay!

[jagged zephyr]

I didn't understand what you said earlier about 1/0 man my english weak did u said anything about me ?

[abstract glen]

wasnt so hard in reterospect

my previous working was just shabby so i think i missed the identity

[stoic valley]

No not you, there is a guy here whos named 1 divided by 0

[jagged zephyr]

Oh kk

[abstract glen]

wow ok nvm idk why i missed this

[jagged zephyr]

That nice 👍

[abstract glen]

tysm yall 😭

.close

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[jagged zephyr]

No one did anything you did all work 👍

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[stable cloud]

I feel like I am doing this one wrong

[stuck vault]

,rccw

[crisp stratus]

the last step is wrong, you can't factor out sin(x) like that

but there's a better way to solve this

[stuck vault]

Until here i asusme it's correct because i don't know the identities

[crisp stratus]

use the second formula

[stable cloud]

Ohh

[small copper]

Try it: $$\cos(2x) = 2\cos^2(x) - 1$$

[grizzled pagoda]

Ga³¹Br³⁵I⁵³9000✞

[stuck vault]

But idk how you got to this

[crisp stratus]

you get a quadratic in cos(x)

[stable cloud]

If I add cos(x) to cos^2(x) does that equal cos(x)?

I don’t know how to simplify this more

I made sin^2x into 1-cos^2x but I don’t know how to combine the like terms

[small copper]

$\cos^2(x) + \cos(x)$ is not equal to $\cos(x)$

[grizzled pagoda]

Ga³¹Br³⁵I⁵³9000✞

[small copper]

,rotate

[turbid hatch]

,rccw

sniped

[stable cloud]

?

[small copper]

You need factorise.

[stable cloud]

Why am I getting two different answers based on the method I used? I don’t know which is correct

[small copper]

The problem is that your factoring at the checkout is wrong.

[small copper]

If you multiply $(\cos x + 1)(\cos x - 1)$ it gives you $\cos^2 x - 1$, and you are missing that 2 at the beginning and the $+\cos x$ in the middle.

[grizzled pagoda]

Ga³¹Br³⁵I⁵³9000✞

[stable cloud]

What wdym ag the checkout

[small copper]

In the box.

[stable cloud]

So the other one is correct? So cosx equals 0 twice?

[small copper]

Remember you have to set each parenthesis of the factorization to zero:

[stable cloud]

Wdym

aren’t they already set to zero?

Or 1?

I don’t really know what you mean by set them to zero

[small copper]

$(2\cos x - 1) = 0$ and $(\cos x + 1) = 0$

[grizzled pagoda]

Ga³¹Br³⁵I⁵³9000✞

[stable cloud]

I’m confused

Isn’t that the last step after finding the gcf?

And we wouldn’t do it with the equation from the previous step?

Like wouldn’t I set cos^2x to zero

[small copper]

You can't match single terms like cos^2 x because there are other numbers adding and subtracting. It only works when they multiply.

[stable cloud]

So can I just set two cosx to zero?

[small copper]

You can't separate the 2 from the rest.

[stuck flint]

Last third step is wrong

How do you write that

[stable cloud]

I think I was able to do it when I did it in the box

[stable cloud]

I’m not sure which is the third step

Isn’t the third step just pyth identities ?

I’m confused

I just distributed the -

[small copper]

In cos=2/3 remember that the cos is negative and the other the cos is 1.

[stable cloud]

Oh so it would be equal to pi? Also what 2/3 are you talking about?

[small copper]

Yeah!

[small copper]

-1/2 not 1/2.

[amber wasp]

@stable cloud Has your question been resolved?

[amber wasp]

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[charred mantle]

Hi

[charred mantle]

If x = y

Then y = x ?

[winter badge]

yes

[tiny juniper]

Yes.

[small copper]

Correct!

[arctic lagoon]

Yup

[small copper]

Do you have more questions?

[vernal bobcat]

It’s actually one of the 3 main properties of the = sign

[charred mantle]

Ok thanks

No more ques as of now

[small copper]

!done

[amber wasp]

If you are done with this channel, please mark your problem as solved by typing .close

[small copper]

You are welcome!

[charred mantle]

One more

If 2+3 = 5

Is 5-3 = 2 ?

[native relic]

yes

[charred mantle]

Ok thanks

And how can I adopt the math tag

[native relic]

go to settings

[charred mantle]

Ok

[small copper]

Click on adop tag or in configuration.

[stray crow]

Don't troll

[zealous nymph]

does kinda seem like trolling cuz they have the undergrad role

but maybe they just didnt select the correct one

[stray crow]

I have not met a single person in my life who doesn't know that x = y means y = x. Especially not an Undergraduate

And look at #help-2 if you aren't convinced

[zealous nymph]

looks like the message got deleted

[amber wasp]

@charred mantle Has your question been resolved?

[spiral zealot]

.close

[amber wasp]

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[pseudo crescent]

@steel sandal (whoops- didnt mean to reply ping, sorry)

For others - the original question was why complex number are represented on a 2D plane

[pseudo crescent]

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

Our 2 basic operations are addition and multiplication, so lets look at what those do to a real number line first

Addition basically shifts the number line. E.g. adding 1 to every number makes it shift right by one unit (0 moves to where 1 was, 1 moves to where 2 was, ...) you can drag the 0-point in the graph I sent and see for yourself

once you come back, lmk @steel sandal

[amber wasp]

@pseudo crescent Has your question been resolved?

[pseudo crescent]

ill just try to explain the rest and you can read it later

multiplication stretches the number line

now the natural thing to ask would be if there is a number system, that would let us do similar stuff on a 2D plane - now that we know that the real number system lets us do some interesting geometry on 1D line

So we add another axis and a unit corresponding to it - lets call it i (note that we dont know whether i^2 = -1 yet, we just make up a new unit and give it a name i)

now what happens when we add e.g. 2 + 1i?

||The whole grid shifts 2 units right and 1 unit up, a + bi + (2 + 1i) will be (a+2) + (b+1)i, so every point moves 2 units to the right and 1 unit up||

||So addition basically shits the grid again, adding real numbers creates left and right shift and imaginary numbers create up and down shift||

Now what should happen if we multiply it by something?

We certainly want the following 2 things to hold:
0 * x = 0, so 0 will not move, it will stay in place
1 * x = x, so 1 will move to whatever x we are multiplying by

lets first look at what multiplying by omething purely real should do

for the 1D real number line, it simply stretched it

so we should expect some kind of stretch in here as well

for example 2 * 2 should still be 4, so the effect on the horizontal direction should again be stretching

and regarding the vertical direction, we will want e.g. 2 * 2i to be 4i, so multiplying imaginary stuff by real numbers should also stretch it in the vertical direction. So we are stretching it both directions

Now both addition and multiplication by reals make the grid to kind of preserve its shape

it only glides around at gets bigger / smaller

so we will want multiplication by other numbers - such as 1 + 2i to do the same

but since 0 has to go to 0 and 1 has to go to 1 + 2i, there is really only one way to do it

we just slide 1 to 1 + 2i and make the grid such that it preserves its shape

so it kind of forces it to rotate

so addition of this new number system does shifting left, right and up, down

multiplication does stretching and rotating

also notice that if we multiply by this new unit i, it basically causes a 90° rotation

but what happens if we do it twice?

it will be a 90° rotation, followed by a 90° rotation - a 180° rotation, or a "flip"

but -1 also does that

so multiplying by i twice results in the same effect as multiplying by -1

i * i = -1

so this new i unit must be the square root of -1

So to sum it all up:

1. we looked at how the real number line behaves geometrically under addition and multiplication

2. we tried to extend the pattern and find a 2D number system that does the same

3. this number system turns out to be the complex number system

so to answer your question, why are complex numbers plotted on a 2D plane, one of the reasons is that its just really, really useful

so useful that I just used that alone to rediscover complex numbers

there are 2 main reasons why complex numbers are useful in math:

1. with complex numbers, we can finally solve / factor all polynomials
2. complex numbers have a really nice geometric interpretation in the 2D plane

usually the motivation for introducing the complex numbers is the first reason, but I honestly think that the second reason could also be used to "discover" complex numbers

and by discovering them this way, your question is answered automatically

soo that's all, you can ping me anywhere or DM me if youve any questions

.close

[amber wasp]

Channel closed

Closed by @pseudo crescent

Use .reopen if this was a mistake.

[pseudo crescent]

oh and btw if you wanna check the graph:
SR controls addition / multiplication in reals
SC controls addition / multiplication in complex

and you can click the folder icons to hide / show the real and compelx graphs

[steel sandal]

wow man i need a cup of tea to read thos

this

hold on

[amber wasp]

Available help channel!

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[charred mantle]

Why was I timed out ?

[stoic valley]

Dm @dense tiger

[charred mantle]

?

[stoic valley]

Not here

[charred mantle]

Ok sorry

[stoic valley]

. close

[charred mantle]

Which channel

Can I go to

[stoic valley]

for?

[charred mantle]

I want to ask something

[stoic valley]

Ask here then!

[charred mantle]

And also

[stoic valley]

Its a mth doubt?

[charred mantle]

Ok

Yes

[amber wasp]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[charred mantle]

What are the factors to 2x^2 + 3x + 91^3 ?

[stoic valley]

You sure thats right?

[stray crow]

Because you are trolling.

[charred mantle]

My friends asked me this

Idk what to answer

Is the question wrong ?

[stoic valley]

Yeah this expression isnt factorisable acc to me since the discrimanant is not only -ve

[mild saffron]

its factorable in complex plane though

[stoic valley]

But also the fact that the question is incomplete

[charred mantle]

Ok thank you @stoic valley

[rotund crag]

u could say 2x^2+3x+753571 then u can use the discriminant

[stoic valley]

Am guessing he hasnt done that tho

[charred mantle]

Okay!

[mild saffron]

undergraduate math*

his role

[charred mantle]

Thank you guys for the help

[stoic valley]

Haan bhai wo toh koi bhi karle

[stoic valley]

Sure!

[charred mantle]

Sahi baat h

[stray crow]

Obviously undergraduate

[mild saffron]

lmao 😂

[stoic valley]

what was this😭

[mild saffron]

[stoic valley]

This seems valid for a timeout

[charred mantle]

That was a dare from my friends sorry

They asked to send this

Sorry

[stoic valley]

what friends are giving you dares in undergraduate math bro💀

[charred mantle]

Idk tbh

[stoic valley]

Anyway

!done

[charred mantle]

Yes

Ty guys

[stray crow]

Editing won't work

[stoic valley]

Figured

[charred mantle]

!done

[amber wasp]

If you are done with this channel, please mark your problem as solved by typing .close

[charred mantle]

.close

[amber wasp]

Channel closed

Closed by @charred mantle

Use .reopen if this was a mistake.

[amber wasp]

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[acoustic plaza]

I thought the answer is 2 but it's 4... now looking closely, I find 3 asymptotes... 2 vertical and 1 horizontal at y=3... where's the last one?

[dry field]

looks like there's only 3

use cursor to interact with graph

maybe there's one hidden somewhere

[acoustic plaza]

i did but can't find it

[dry field]

bruh

[drowsy pawn]

Do you think they're counting the one on the left as two for some reason?

[stoic valley]

Wait @acoustic plaza does the right side also become an asymptote

[acoustic plaza]

right side

approaching 3

[stoic valley]

Ahh see thats also an asymptote then

That explains 4 asymptotes

[acoustic plaza]

how

i already counted y=3

[stoic valley]

See thats the thing!

Its approaching y=3,smth for completely seperate parts of the grapg

[buoyant blade]

Shouldn’t you move it to the other side and multiply it by 10?

[acoustic plaza]

but I have already counted it once right?

[buoyant blade]

Exactly you can see both are connected

[stoic valley]

Ahm thats not how it works!

[stoic valley]

!occupied

[amber wasp]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[acoustic plaza]

how does it work then

[stoic valley]

you need to count them seperately

[buoyant blade]

[stoic valley]

See i would tell you how my teachers taught me asymptotes, but i feel like it would be better to read your notes

due to regional differences and notation

[acoustic plaza]

why? it's the same thing... it's just that the asymptote is on the other end of the graph

[acoustic plaza]

I don't have any notes regarding this

[acoustic plaza]

could you teach me how you learned it

[stoic valley]

just a second

Wait @acoustic plaza just looked at it carefully does the left part also tend to y = 3

Like as is it just the mirror image of the right part?

[acoustic plaza]

i suppose so

[stoic valley]

Sorry mb then!

[turbid hatch]

could you share the link

[acoustic plaza]

you can't access it...it's private

[amber wasp]

@acoustic plaza Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[magic lily]

i know that this is obviously true but how do i prove it

[magic lily]

like -rigorous- proof

[fathom mist]

ok

[twilit dawn]

Well what have you done so far

[fathom mist]

if one of the angles is 90 then product of all cosines would be zero

then u proceed I think

[magic lily]

i just took an angle as 90 and applied basic stuff i know to know

but how do i prove it without assuming that an angle will be 90

[jagged zephyr]

He has to prove triangle is right angle but he can't write ight that he need proof that triangle is right from equation given

[twilit dawn]

You're proving the opposite directions

[fathom mist]

wa

[magic lily]

gimme a hint
ill try to do it from forward direction

[twilit dawn]

Well you have sin² and cos², there's something you can do immediately

[fathom mist]

ok just solve the given eq for the cosine squares of A B C which will be equal to 1

[jagged zephyr]

It's 10 point question bud 😭

[wet socket]

Convert all in Cos(2x) format I think

[magic lily]

yes
i tried converting em all to sines
so i got sum of squares of sines as 2

[fathom mist]

how is that 10 pts

[jagged zephyr]

There written in corner

[fathom mist]

holy

[wet socket]

It's for an entrance exam in india

8 questions in 2 hours

[fathom mist]

so u gotta prove the triangle cosine equation

and then prove the thing aswell

so u gotta start with A+B+C = pi

[magic lily]

7 actually

[wet socket]

8 no?

[magic lily]

ok

[twilit dawn]

Good. Now you can use the fact that the angles sum up to 180

Write C as 180-(A+B)

[magic lily]

oh ye mb

[jagged zephyr]

Now you on right track 🫡

[fathom mist]

C = pi −(A+B)

[twilit dawn]

And recall how sin works with that

[fathom mist]

apply cosine

we get cosine squares of A, B C + 2cos a cos b cos c

=1

now the eq u are given

use that for the squares to be equal to 1

so

cos A cos B cos C =0

[magic lily]

wait mb , sum of squares of sines is 1

[fathom mist]

that means either A/B/C are pi/2

cuz cos pi/2 is 0

so its a right angled triangle

[magic lily]

slow down a bit pls @fathom mist

i want to

try myself

[fathom mist]

sry

good luck

[fathom mist]

did u solve the given equation just asking

converting everything into cosine and taking it to one side

[magic lily]

ive reached here

yes, that is a makeshift whiteboard

[fathom mist]

ok

so u wanna do it in sine

then convert given eq into sine

cos^2 A is 1- sin^2 A

bruh

would be easier to do it in cosines

but yeah

[magic lily]

sure

ill do it in cosines then

[fathom mist]

multiply sin C

theres just alot of ways to do it

[fathom mist]

lets just continue from here

u got squares of sines =1

[magic lily]

2 actually

mb

now what

[fathom mist]

bruh

cosines or sines

[magic lily]

sines

[fathom mist]

can I have a pic

ok

sines

u got it as 2

[magic lily]

yes

[fathom mist]

now A+B+C = pi

[magic lily]

right

[fathom mist]

u manipulate it to get sin^2 C

[magic lily]

mhmm

[fathom mist]

u can square both sides

[fathom mist]

expand sin^2 A+B

[fathom mist]

sin^2Acos^2B=sin2A(1−sin^2B)

honestly cosine equation would be smaller

sin^2A+sin^2B+sin^2A(1−sin^2B)+(1−sin^2A)sin^2B+2sinAsinBcosAcosB

u get that

2(sin^2A+sin^2B)−2sin^2Asin^2B+2sinAsinBcosAcosB

then this

[magic lily]

is there some other way to prove this

without trig

with vectors

complex numbers

or smm

[fathom mist]

yes vectors

[magic lily]

thank god

[fathom mist]

but did they teach u

vectors

[magic lily]

yes ik vectors

[fathom mist]

ok

but tbh it would be alot easier if u did go for cosines

[magic lily]

they expect us to know everything , even mobius inversion and stuff

[magic lily]

i hate trig

dont you like vectors

they cool asf

[fathom mist]

but the cosine part is easier

[fathom mist]

I do like it

[wet socket]

That was a PIE question btw

[fathom mist]

how do I explain it in discord chat tho

[magic lily]

infact ykw
imma try vectors on my own

[fathom mist]

bruh

[magic lily]

idk

[magic lily]

no disrespect, i js wanna try out on my own

thx

.close

[amber wasp]

Channel closed

Closed by @magic lily

Use .reopen if this was a mistake.

[fathom mist]

I can give u hints

wait

[magic lily]

sure imma js get back after tinkering with it

[fathom mist]

lemme use a text to image converter

brb

ok got it

@magic lily

[magic lily]

ye ik dat

[fathom mist]

we have this equation

we need to prove AB and AC are perpendicular

so their dot is zero

ok yeah I would still do it with cosines and trig

cuz its easier

[astral umbra]

if youre still interested in a trig approach then\
$\cos(2A) + \cos(2B) + \cos(2C) = -1$\
$(\cos(2A) + \cos(2B)) + 2\cos^2(C) = 0$

[turbid hatch]

blud

[astral umbra]

wtf

[turbid hatch]

delete the latex after putting a spoiler

[fathom mist]

lmao

[astral umbra]

🤦‍♂️

[jagged zephyr]

Gang wait bro left that chat why we bullying each other

[magic lily]

im here but

.reopen

[amber wasp]

✅ Original question: #help-41 message

[turbid hatch]

but this is w

[jagged zephyr]

Gang bro came back let bully each other

[fathom mist]

lemme open presentation or sum shi

we get that

and thats it

[jagged zephyr]

Bro left the game buddy he alr solved it and went to advance math

[magic lily]

hmmmm

not really

[jagged zephyr]

Don't let me down buddy 😭 i got high hope from u

[wet socket]

Most efficient approach

Since you know one of them is 90

Start from CosACosBCosC = 0

And work backwards

[jagged zephyr]

When he know one of them is 90 ?

[wet socket]

You need to prove one of them is 90

It's not a holistic approach

But yes

[jagged zephyr]

Ye prove so that means he don't know

[wet socket]

Yes you can do that in rough work

You don't need to show that you did reverse engineering

[magic lily]

thats illegal

[jagged zephyr]

He need 10 point ig he need approach

[wet socket]

Rough work

[wet socket]

Ik gng I've solved this exact paper

[jagged zephyr]

Ig i should step in let me open my tresure of notes

[wet socket]

The cos(2A)+cos(2B) method is the most efficient

[magic lily]

is this not right?

[turbid hatch]

works

[wet socket]

Also make a case for Cos(C)=0

And C= B-A

Three cases

[magic lily]

got this

[magic lily]

but why this

oh

cuz cosine is even

[wet socket]

Even fxn

[jagged zephyr]

Quadrant

[magic lily]

alr

[wet socket]

Yepp

[magic lily]

🫡

yall are fire

[jagged zephyr]

🫡

[wet socket]

B6 💔💔💔

B6 is bad 💔

[magic lily]

isnt 6 js easy

[wet socket]

Nawh

[magic lily]

use mod

[fathom mist]

did he get it

[wet socket]

Yes mod is to understand the question

[magic lily]

ye

[jagged zephyr]

How many questions to solve for exam and what time ?

[wet socket]

The real solving begins after that

[magic lily]

idk man
we just have to prove that 3 distinct pairs exist
cant we just use pigeonhole

[wet socket]

But there's an elegant solution

[fathom mist]

8 questions 2 hrs

[wet socket]

Not exactlyy

[fathom mist]

yes

[magic lily]

how did u solve it

[wet socket]

I mean php is used

[fathom mist]

the conclusion is he has to prove cos A. cos B. cos C =0

[jagged zephyr]

We 5 ppl solved this question in 45min ☠️

[wet socket]

Considered a tic tac toe board

[fathom mist]

so ABC is a right angled triangle

[magic lily]

we should move somewhere else

[wet socket]

It's a 5 mins trick

[fathom mist]

its discord chat

[wet socket]

Yeah

[magic lily]

ggs

thanks

[fathom mist]

yk what imma summarize a solution

[magic lily]

.close

[amber wasp]

Channel closed

Closed by @magic lily

Use .reopen if this was a mistake.

[jagged zephyr]

Where

[fathom mist]

oh nvm he got it

[jagged zephyr]

Don't leave me alone gang i am also member

[fathom mist]

he left u alone dang

[magic lily]

sent u req

[wet socket]

Ye accepted

[magic lily]

alr gng
jee aspirant ?

[jagged zephyr]

Yep

How you know ☠️

[jagged zephyr]

Bro is ....

[turbid hatch]

bruh server filled with indians

[fathom mist]

@magic lily ok so this is first step

second step is to get sigma cos^2 x on A,B,C = 1

and third step

eq 1 and eq 2 we get cos A cos B cos C =0

hence proved

[jagged zephyr]

@fathom mist they went to secret place they not coming back and they left me alone 😭🫡

[fathom mist]

they made a gc

to solve einstein's problem

ggs

einstein failed and we wont

[fathom mist]

wait so yall are in pre university

[jagged zephyr]

They really leave me behide 😭🫡

[fathom mist]

wth im younger

[jagged zephyr]

Just 2 month then I will be in college

[fathom mist]

nice

which major

[jagged zephyr]

?

[turbid hatch]

you doin jee right so prob eng major

[jagged zephyr]

Yep oh that was he asking

Then Science major ig?

@turbid hatch you also jee?

[turbid hatch]

yes

[amber wasp]

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.
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[stoic valley]

Confused as to how to start

Kindly ping me if you get it!

[native relic]

by fundamental theorem of caculus $F'(x) = f(x)$ so the inequality $F'(x) \geq c F(x)$ and then $F'(x) - c F(x) \geq 0$ holds true

[grizzled pagoda]

MxRgD

[native relic]

now multiply by the integrating factor e^{-cx} and see what you can find

@stoic valley

[stoic valley]

Hm did tht

[native relic]

what did you get

[stoic valley]

I got F(x) >= 0

[native relic]

how did you get that?

[stoic valley]

Uh e^-cx F(x) is an increasing function

And we know that F(x) = 0

So i simply wrote e^-cx F(x) >= e^-0 F(0)

And therefore e^-cx F(x)>= 0

[native relic]

Nice

now check for F(0)

[stoic valley]

Check?

As in?

[native relic]

Like use x = 0

[stoic valley]

.

[native relic]

$F(0) = \int_{0}^{0} f(t) dt = 0$

[grizzled pagoda]

MxRgD

[native relic]

i mean this

[stoic valley]

Yeah thats what i did

Thats how i got the conclusion that F(0)= 0

[native relic]

oh

I got confused

Anyways yeah, try to do the same for g(x)

I'm going to sleep soon so can't help with the rest

[stoic valley]

You get g(x) < 0

[stoic valley]

Oh okay tysm tho!!

[amber wasp]

@stoic valley Has your question been resolved?

[stoic valley]

Pls ping me if you text anything

[stoic valley]

<@&286206848099549185>

[signal furnace]

The question states that $f$ is non-positive, so its integral $F(x) \leq 0$. Therefore, for all $x \geq 0$, we have
$$F(x)=0 \implies f(x)=0.$$

[grizzled pagoda]

Civil Service Pigeon

[signal furnace]

anything else relevant in here

[stoic valley]

Right

[signal furnace]

This is true for $x>0$.

[grizzled pagoda]

Civil Service Pigeon

[signal furnace]

The finish is immediate from this

[stoic valley]

Yeah and thats whats given in qsn asw

[stoic valley]

so technically its 1 right?

[signal furnace]

Not "technically", the answer is 1.

[stoic valley]

That is at zero itself

[signal furnace]

mhm

[stoic valley]

thats wrong

[stoic valley]

I did all this exaclty same method and logic and results

But answer says its 0

that is there are no roots

[signal furnace]

Here is an explcit example of f,g that show the answer is 1.

[signal furnace]

You should've specified that then

[signal furnace]

eh people forget edge cases all the time

I'd chalk it up to whoever made the answer key being a muppet

[stoic valley]

I got that after mxrgd helped a little

Thanks tho

I have another

Might be silly again should i still ask?

[signal furnace]

Eh start a new channel since this one is already kinda messy

[stoic valley]

Aight ty

[signal furnace]

[stoic valley]

.close

[amber wasp]

Channel closed

Closed by @stoic valley

Use .reopen if this was a mistake.

[amber wasp]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[cunning jasper]

ok so I got 144, it's apparently 166

[cunning jasper]

what I'm getting here is that I just don't know how to do summation

[cunning jasper]

since p is a prime that isn't odd it's just 2, so x is either 2 or 4 so S is either 104 or 208, so getting the number of positive integers less than both that are coprime is just (104-52-4)+(208-104-8)=48+96=144, which I'm not sure----------

I'm realizing I misread it to coprime

but wait no it's 168 then no?

[pallid canopy]

Did you use totient function anywhere

[cunning jasper]

it was easy enough since the prime factorization iz 2 thingies

no

wait I'm rechecking the question

[pallid canopy]

I don't see where you calculated the totient of 104 or 208

[cunning jasper]

104-52-4=48, it works

and 96 for the other part

yeah why is it 166

it's 166

104-48+208-96=168 I'm not getting where the last 2 disappeared off to

wait 1 is included in bothof these

ah I'm stupid

.close

[amber wasp]

Channel closed

Closed by @cunning jasper

Use .reopen if this was a mistake.

[amber wasp]

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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.
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[gritty shard]

Four friends, Stacy, Michelle, Peter, and Robert, planned a reunion. They left from four different places: Las Vegas, California, New York, and New Jersey.
It is known that:
Michelle was the first to reach the destination and occupied a round table.
The person from New York arrived second and sat opposite Michelle.
Peter came in third and sat to the immediate left of Stacy.
The person from New Jersey arrived before the person from California.

Where did Stacy come from?

[gritty shard]

could someone help me verify? the answer i'm getting is New york

[stuck vault]

!show

[amber wasp]

Show your work, and if possible, explain where you are stuck.

[stuck vault]

But yea, Stacy's from NY

Ya don't really need the 4th clue for that

[amber wasp]

@gritty shard Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

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[amber wasp]

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[zinc gust]

how to find reciprocal of 7 over 2?

[gloomy tide]

just 2 over 7, unless i'm misunderstanding you

[zinc gust]

the question says

"write down the reciprocal of 7/2"

idk how to do the fraction command

[gloomy tide]

yeah so the reciprocal of $\frac{\rsq}{\gsq}$ is $\frac{\gsq}{\rsq}$

[grizzled pagoda]

professional attention seeker

[zinc gust]

let me send a pic

the first question

[spiral zealot]

yeh, so follow what's mentioned above

[zinc gust]

how do i show my work?

[spiral zealot]

no need for additional work

apply that property/definition
and you can jump immediately to the final answer

[zinc gust]

okay let me check the marking scheme for the answer

[spiral zealot]

for 1 mark, the answer is enough

[zinc gust]

yeah thanks

how do i

do probability

[amber wasp]

@zinc gust Has your question been resolved?

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[hallow cipher]

Help on how to find the domain, horizontal asymptote, and range. :D

[amber wasp]

@hallow cipher Has your question been resolved?

[bleak bloom]

domain = everywhere except where it doesnt exist (one value which you can solve when you remember you cant divide by 0)
im not sure what range is sorry
horizontal asymptote = limit in infinity

[amber wasp]

@hallow cipher Has your question been resolved?

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[pastel meteor]

how does this transformation work?

[pastel meteor]

I mean, sure, I can just multiply the bottom one and see that they're equivalent, but is there a law behind how that transformation makes sense?

[hybrid arrow]

thats a identity for the difference of cubes

https://www.youtube.com/watch?v=PGs-mHEKhX0&ab_channel=Mathispower4u

check this out

[amber wasp]

@pastel meteor Has your question been resolved?

[pastel meteor]

why does my intention to understand cubic formula has so many related topics smh , thank you very much

.close

[amber wasp]

Channel closed

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[carmine kettle]

Find the solutions on the interval 0 ≤ θ < 2π . (Give exact answers) 6 cos(2𝜃) = 6cos^2(𝜃) − 2

[carmine kettle]

I don't know how to answer this

[amber wasp]

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[stoic hawk]

why is this D and not E?

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@stoic hawk Has your question been resolved?

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Remember:
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[craggy latch]

im struggling a bit with limits, can someone please check my answers on these?

[amber wasp]

Please don't occupy multiple help channels.

[craggy latch]

my teacher wants answers typed thats why its weird.. not sure why

[craggy latch]

closed my other one btw

[waxen oak]

i can tell you a and e are incorrect, but that really is a horrible graph

as for h, it looks like the line goes above 1... so it's hard to say...

same with x->3^-

seems like it's off from 2...

[craggy latch]

yeah the graph is not good

im confused on how e isnt 0?

[waxen oak]

what's the y value as x approaches 0 from the right?

well.. approximate value..

[craggy latch]

-1

a is confusing, i thought it was 2 since it passes through 2

[amber wasp]

@craggy latch Has your question been resolved?

[craggy latch]

@waxen oak sorry for the ping but i still need a bit for help for this if you are free

[waxen oak]

you're looking for the y value when x=2

since you've identified the limit exists

[craggy latch]

so -1? how does that work if it passes it

[waxen oak]

where are you getting -1 when looking at x=2

[craggy latch]

sorry i was looking at -2, theres no negative signs on the graph so i got mixed up

so thats like 0.3?

[craggy latch]

but then im still consfused since it passes x = 2

[amber wasp]

@craggy latch Has your question been resolved?

[craggy latch]

ill jus ask my teacher tmr thanks tho

.close

[amber wasp]

Channel closed

Closed by @craggy latch

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[misty igloo]

.help

[amber wasp]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[misty igloo]

.help me

[amber wasp]

No command called "me" found.

[misty igloo]

.reopen

asasdaka;

[atomic needle]

Ay

[misty igloo]

i need help

[atomic needle]

@misty igloo first read #❓how-to-get-help

[misty igloo]

ok

[misty igloo]

done

[atomic needle]

Okay, now what do you need help with?

[misty igloo]

The squash club needs to raise funds for renovations to club buildings. 2 people jason and mark want to help by organizing some local teenagers to sell boxes of chocloate. their goal is to reach 20000. So how many boxes of chocolate bars will they need to sell if 1. each box has 20 choko bars in it 2. each chocko bar sells for 2.50 , it also includes a 12% mark up. 3. the club gets to keep the mark up percentage on each box of chocolates sold. 4 the company for the bars will donate 5 for each box they sell. 5. Calculate the number of boxes of chocolate that the club needs to raise to reach 20000

@atomic needle

what i did was 2.50 x 1.12 = 2.80 then i times it with the 20 making 56, after that i added the 5 dollars with 56 making 61 which is the total for 1 box, after that i did 20000 divided by 61, then it shows 327 boxes

but i dont know if its correct

[faint sleet]

thats not right at all

[misty igloo]

what

howwww

[faint sleet]

For starters 2. refers to each box selling for 2.50

[misty igloo]

but

[misty igloo]

i dint say that

i said

each chocolate bar

is 2.50

[faint sleet]

then what you calculated is the total

but you want the amount raised for the club

most of the money goes to the company

[misty igloo]

huh

[faint sleet]

the club only gets 12% of sales + 5 dollars per box sold

[misty igloo]

ohhh

[misty igloo]

wait..

does that mean

the club earns 11 dollars

for each box they sell?

????????????

[faint sleet]

algebra is your friend

watch how easy it becomes by expressing the information given as an equation

[misty igloo]

can i do it

without algebra