help-49

[midnight plank]

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[midnight plank]

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[sand flume]

https://physics.stackexchange.com/questions/314087/why-is-normal-force-zero-at-the-top-of-a-circle . For this I see that we can decrease the value of f by decreasing the inputs. And then if we decrease the inputs x,y the gradient would get smaller. However, The gradient we start off with is $-\nabla f(x,y)$ since we want it to maintain our constraint and keep the object not going through the radius. Now if we wanted this gradient to be positive we would need to change the inputs such that the gradient itself is negative so that after applying the negative it becomes positive. So this can only happen if both inputs are in the 3rd quardrant. But the object could have falllen off the radius if both inputs are negative. And for the second part how does making the radius bigger affect if the gradient will be positive. We already showed it is possible with the current radius.

[grand pond]

BigBen

[midnight plank]

@sand flume Has your question been resolved?

[sand flume]

<@&286206848099549185>

[lilac finch]

ts physics

[midnight plank]

@sand flume Has your question been resolved?

[midnight plank]

@sand flume Has your question been resolved?

[bright shoal]

!status

[midnight plank]

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

[sand flume]

my status is that I don't see how the object wont fall off when the gradient is positive and I don't see how radius plays a role

[violet storm]

so the car has an initial velocity but no acceleration right, so the only force is gravity, and well the normal force of the track

[twilit jetty]

@sand flume you here?

[sand flume]

Yes

[twilit jetty]

you said you dont know how radius plays a role
can you intuitively grasp how a large enough radius would mean the car never reaches the top?

[sand flume]

It doesn't have enough velocity

[twilit jetty]

velocity is a vector, but "doesnt have enough" only describes magnitudes, you mean speed (or specify its the initial velocity, the direction of which is predetermined)

[sand flume]

Ok

[twilit jetty]

also, youre not allowed to modify the initial velocity

youre only allowed to modify the radius

so you can "speed" the car up by shrinking the radius

you can "slow" the car down by growing the radius

does that make sense?

its a yes or no question, its just a step you have to answer before I can continue

[sand flume]

No

[twilit jetty]

Ill let you in on a secret, neither answer would have changed what I was going to say next

you need to be faster with your responses

lets say we have a scenario like this

[sand flume]

Sorry

Ok

[twilit jetty]

now we would expect the car did not have enough velocity to start with,

but lets try and reframe this to "the radius is too large"

first is the shape of the path the car has taken

I drew the bounce for dramatic effect, but really we only need to focus on two parts of it:

[sand flume]

Ok

[twilit jetty]

the part where it's on the track
the part where it's off the track

on the track, the path is a circle
off the track, the path is a parabola

does this make sense

[sand flume]

Yes

[twilit jetty]

now consider the centripetal force when the car is on the track

we know the car has to slow down on the track
however, the track is frictionless, so what slows the car down?

[sand flume]

R

[twilit jetty]

...R?

[sand flume]

radius

[twilit jetty]

thats a strange way of thinking

remember, we are doing a physics problem

presumably a force or acceleration would slow the car down instead

[sand flume]

Gravity sorry

[twilit jetty]

pay closer attention to your causes and effects

you just said earlier you didnt fully crasp how r (which is lowercase, not uppercase) would do the same things as changing the initial velocity

dont attribute things to r for now then, we can prove that it will have similar effects

we can have r be fixed

then vary the initial velocity

we'll start with that

[sand flume]

Greater the velocity the higher it will go

Speed

[twilit jetty]

good to know, but lets try and get some numbers down

since gravity is the only force slowing the car down, we can consider potential energy

until the car hits the ground, we can assume the car and the track are a closed system, with PE being how much energy gets transferred to the car's height instead of the car's speed

[sand flume]

Ok

[twilit jetty]

now at point 1, what energy does the car have?

[sand flume]

All kinetic

[twilit jetty]

yep

now we know the car eventually leaves the track

at point 2, what is the centripetal force?

[sand flume]

0

[twilit jetty]

thats good

now centripetal force is mv^2/r

since the centripetal force of the car is 0, mv^2/r no longer applies

now lets look at a moment just before that

what do you think the velocity of the car would be?

[sand flume]

Something very small

[twilit jetty]

interesting

so at point 2,

does the car have any KE?

[sand flume]

No

[violet storm]

but it's still moving up in the picture!

[twilit jetty]

it sure is

its a trick question

[twilit jetty]

remember that when the centripetal force no longer applies, neither does this law

there might still be some leftover KE

hold on

what am I on about

I gotta be sure about this

[sand flume]

Wait how can that be? If there is no centripetal force no velocity

[twilit jetty]

the velocity is definitely continuous, so thats gotta be the case

let me redraw the picture to be less crazy

[violet storm]

good point 🤔

[sand flume]

?

[twilit jetty]

I dont think I can really model the car now

Im sure the parabola path is still correct, because the velocity still has to be continuous, and gravity cant affect the horizontal speed

none of these need to be considered if the car's point 2 happens to be at the top of the loop:

lets look there then instead

[sand flume]

Ok

[twilit jetty]

the key idea here is that, as before, we know the path eventually transitions to a parabola

theres no upward velocity this time to worry about, any possible acceleration that wouldve pushed it up would be counteracted by the normal force of the track

this means the velocity is purely horizontal

[sand flume]

Wait how is there normal force? Because the whole reason I why trying to read that stack exchange post is to understand why normal force is 0 at the top of tracks in circular motion

[twilit jetty]

you can see we are considering three different paths

one of them has a normal force

[sand flume]

Ok

[twilit jetty]

now before we get to that,

I want to make sure you can get the geometry here

when the ball reaches the top of the track, theres three possible kinds of parabolas the ball might take

ball?

now see here that the parabola either is always above the circle, or intersects with it

[sand flume]

Ok

[twilit jetty]

however, in the middle, the parabola also does not intersct the circle either

if I were to change how much the parabola slopes down any further, it would intersect

[sand flume]

Ok

[twilit jetty]

the topmost parabola is the second picture

anything below that would be the third picture

[twilit jetty]

now the ball can take any of these three cases I showed

[sand flume]

Ok

[twilit jetty]

but notice that, in the second case, or 'boundary' where the ball would do this, the ball would still remain on the track

do you see why thats the case?

[sand flume]

Yes because the two graphs are extremely close

[twilit jetty]

these graphs are close too, you dont mean "close" here

[sand flume]

Or they are the same for a bit of time

[twilit jetty]

its more that the parabola path would rise above the circle

because of that, if the car takes that path,

it would go through the track

the track's normal force isnt going to allow that

since the car is going into the track, that would mean theres a force from the car to the track

the track in turn pushes back on the car, this normal force is called centripetal force

thats why it always faces inwards

[twilit jetty]

but its not like centripetal force can do anything other than push straight down, just like gravity does

it doesnt stop the velocity of the car from pointing it into the track again, not enough that is

as a result, the car remains on the track

[sand flume]

Wait didn't we see that on picture two it is on the track?

[twilit jetty]

look at this part of it

the part where x < 0

the parabola is above the circle

all parabolas as drawn already intersect with the circle at (0, 1)

[twilit jetty]

in the third case, it would mean it starts below the circle then intersects it

[sand flume]

I don't understand this then

[twilit jetty]

is the parabola ever below the circle in the second case?

[sand flume]

No

[twilit jetty]

when the car eventually loses grip with the track,

it should take the parabolic path instead of the circular one, right?

[sand flume]

Yes

[twilit jetty]

so the car has to go take the parabolic path and... go through the track and outside it?

[sand flume]

How can the car go through the track if the parabola intersects it only at a point

[twilit jetty]

you do know what going through means here right

here, the car remains in the inner hole
everywhere shaded is outside the track

even if you assume the car has a nonzero radius,

that parabola is still going to go right through the circle

any path that goes from the inside to the outside goes through the circle

so both parabolas here go through the track

[sand flume]

I don't see that if the red circle is the track. The blue parabola is the path of the car

[twilit jetty]

look at this

see the path of the car beforehand?

the right half of the red circle is part of the path of the car

following that, the left half of the parabola is the other part of the path of the car

of course it would look something like:

this would be an example of case 3 where the car ends up falling off early

but give more velocity to the car, and this is case 1, where the car would end up going through the track

it eventually intersects

now consider a case like this

do you think this is case 1 or case 2?

as a hint, Ive drawn a black circle on top of the red dashed path

[sand flume]

Case 2

[twilit jetty]

yep

does case 2 intersect with the track?

[sand flume]

Yes

[twilit jetty]

so in case 2,

the actual path of the car,

would following the path of the parabola mean the car goes through the track?

[sand flume]

Yes

[twilit jetty]

in a realistic case 2,

which path does the car take, the green or the blue one?

[sand flume]

Blue

[twilit jetty]

youre saying in a realistic case 2, the car would go through the track?

[sand flume]

Look from what I see is that dotted circle makes the path the car needs to travel to be in the track. If it goes on the blue it exists it

[twilit jetty]

...

do you know where the blue path came from?

[sand flume]

It's the middle one here

[twilit jetty]

:) no it isnt

thats case 3

you can see it falls down far more than it should

case 2 is really hard to draw, so I had to get the desmos graphing calculator to draw it

also, any one of those parabolas are blue paths, you know that

[twilit jetty]

what happened to these three images earlier?

the moment the car switches to the blue path, what is the relationship between the car and the track?

[sand flume]

The car won't be on the track?

[twilit jetty]

youre mistaking velocity and acceleration now

the car is still very much on the track, so much its trying to going right through it

we would have to remove a lot of velocity for the hypothetical case 3 where the car falls right off

the blue path happens the moment the car loses grip with the track, as I said earlier

that happens if for example gravity has removed so much velocity from the car that there isnt enough centripetal acceleration to keep the car pushed up on the track

[twilit jetty]

but as seen in the picture, a lack of a push still allows the car to remain on the track anyway

heres another example of a potential case 3

if this case is correct,

the car first begins on the red path

then at the blue path, it loses grip and begins to fall off the track

it then hits the wall and stays on the track again

maybe it wouldve bounced down the track, maybe it wouldve stayed on it after that

but we know for sure it isnt going to go through it

if the case 3 is close enough to a case 2, the car never slips out of the side of the loop

regardless, that case 3 would still be a car that begins and ends... on the track

we usually dont consider this because we're only focused on this point here

but its a undeniable possibility

if youve driven a car before and theres a dent in the road that makes you feel like falling, youd remember the moment you know the dent ends by the car forcibly being pushed up

case 3 here would be a similar example of that

case 2 and case 1 are examples where the car always remains on the track at all times

even if there isnt a force to keep it on the track

there still is enough velocity for it to head into more track that would then keep it on the track

[sand flume]

So your saying that the car follows the blue position till it hits green then goes back to green?

[twilit jetty]

if we have a case 3 where the blue parabola isnt sloping down that much, the car is still centered on the road, so itll just hit the loop again

this is opposed to the idea that it will never hit the loop again

it may be possible that the actual path may look instead like this

[sand flume]

But this only happens because we say that it follows the blue path for a certain amount of time and then goes back to the green which is the path it needs to traverse the track

[twilit jetty]

you dont seem to let me finish

[sand flume]

Sorry

[twilit jetty]

here, the car continuously falls down on the road

this is also a hypothetical possibility, as with anything about case 3

[twilit jetty]

maybe it stays on the green path, maybe it doesnt

[sand flume]

Isn't the green path the path of the track?

[twilit jetty]

what, is the green path not on the track now?

Im not a perfect drawer you know

do you need the paths to be dashed?

I dont think you have enough material to even ask a question by this point

given all the recent questions you ask just chalk up to "I dont think these two connecting pieces connect, but I wont elaborate more on it"

what do you even mean by a question like this? a green path has always been the path of a car on the track, it must follow the path of the track

[sand flume]

I was saying that it was the track

[twilit jetty]

as in?

you started your question with "Isn't"

[twilit jetty]

the car didnt have velocity to stay on the track, when it bounces off the road its not like its going to remain on it

[twilit jetty]

itll have to stay on the track eventually

as communicated by the picture, with the 3 blue parabolas followed by the green arc

you sound very lost

we can vc and get this explained faster

you dont seem open to texting the questions

[sand flume]

ok

[twilit jetty]

well?

[midnight plank]

@sand flume Has your question been resolved?

[twilit jetty]

.close we worked it out by making a model of the car's FBD in desmos among other clarifications

[midnight plank]

Channel closed

Closed by @twilit jetty

Use .reopen if this was a mistake.

[midnight plank]

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[twilit field]

I would like help with the first part please

[twilit field]

Let $j+I \in J/I$ and $r+I \in R/I$. Then $(j+I)(r+I)= jr+I$.

[grand pond]

wai

[gaunt jetty]

Does your result live in the set of J/I

[twilit field]

no

[unkempt sluice]

R/I->R/J; x+I |-> x+J is surjective and calculate its kernel

[twilit field]

Please don't just give me the answer 😭

[unkempt sluice]

I thought it was a hint

[gaunt jetty]

Why

J is an ideal of R, is it not

[twilit field]

Oh nuts

right

so rj \in J

[gaunt jetty]

What do you conclude

[twilit field]

I thought of this correctly just 5 minutes ago , and then when I started writing that ida left me 😭

[twilit field]

J/I is a left ideal of R/I

[gaunt jetty]

Just a left ideal?

[twilit field]

Simiilarly a right ideal too

[gaunt jetty]

Ok nice

So

[twilit field]

But I had already made an attempt 😭

lemme think of an iso then

[unkempt sluice]

This kernel should be straightforward, obstacle where

[gaunt jetty]

You gotta show the map is a homomorphism as well

[twilit field]

mhm

thinking of one

[twilit field]

no, no. I'm trying to come up with an isomorphism b/w (R/I)/(J/I)and R/J

[gaunt jetty]

If you calculate the kernel you can just apply the first iso theorem

[twilit field]

Yea, I realised that. But I was wondering if there's another iso here

[spiral rock]

That isomorphism is induced from the surjective homomorphism R/I --> R/J

[twilit field]

That will teach me something too

I guess I'll think about this for a while

thanks!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[floral ruin]

$\overrightarrow{r}= \Lambda \overrightarrow{a} + \mu \overrightarrow{b}$

[grand pond]

Prathmesh

[floral ruin]

and then I did $(\overrightarrow{r}= \Lambda \overrightarrow{a} + \mu \overrightarrow{b}) \cdot \overrightarrow{a}$

[grand pond]

Prathmesh

[floral ruin]

i got $2\Lambda + \mu = 0$

[grand pond]

Prathmesh

[floral ruin]

how do I find another eqn in lambda and mu ?

[late rover]

Using the fact that c is unit vector a.k.a its magnitude is 1

[floral ruin]

ohh

but it will have terms like lambda^2 and mu^2 and lambda times mu

how will I solve it?

[late rover]

ehh, it's solvable, just a bit difficult

I wouldn't go this route tbh

[floral ruin]

then what would you do?

[unkempt sluice]

μ=-2λ
1=(λa-2λb, λa-2λb)=((a,a)+4(b,b)-4(a,b))λ^2

[late rover]

Vector (axb)xa would be parallel to vector c

[floral ruin]

it would be in the plane of c vector. How is it parallel?

[floral ruin]

ohh I got it

[unkempt sluice]

He solved it by a^t (a b) (x,y)^t=((a,a), (a,b))(x,y)^t=0, which obtained him y=-2x

[slow thorn]

fun method: the new vector is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{a} \times \overrightarrow{b}$

[late rover]

wydm? it would be in the same plane as vector a,b and c

[late rover]

That's what I said

[slow thorn]

ah

ok yeah saw rn

[floral ruin]

yea

[floral ruin]

so we dot with both the vectors ?

[late rover]

I love this method so much cuz in many calculators they can find cross product easily

[floral ruin]

?

[late rover]

nope

[floral ruin]

then?

[late rover]

we cross prod vector axb with a

[floral ruin]

and then?

oohhh

[late rover]

and the the resultant vector would be parallel to c

[floral ruin]

I get it now

I am sorry I couldn't visualive it properly

I get it now

this method is more time efficient ig

thanks

[late rover]

took me 30s with a calculator, but you're not allowed to use one ig

[floral ruin]

noo 😭

but how do you calculate cross products in a calculator?

like is it a scientific calculator or sumthhing?

[late rover]

Many calculators, even from Casio, have a mode called vector in which you can type the vector out and it will do the cross prod for ya

[late rover]

yeah

[floral ruin]

ohh

[floral ruin]

which model?

[late rover]

This one and many more, you can even see the vector on the screen denote as mode 5

[floral ruin]

i see

are these calculators very useful for math students?

i always wanted one but feared if I'll get too dependent on calculations

[late rover]

I am to dependent on calculator

[floral ruin]

💀

[late rover]

Tbh, not really? but if the exam allows it and you don' have one you're basically cooked

In my country, we're allowed to use calculator and guess what

There're many books that write about How to use calculator effectively or solve extremely hard problem with calculator

[floral ruin]

ohh i see

whatttt

damnnnn

[fathom onyx]

Well hang on

@floral ruin Based on the formatting of the question you posted... Is this not the JEE?

[floral ruin]

it is

[fathom onyx]

That's a non-calculator paper

[floral ruin]

yea

[fathom onyx]

So you do need to be able to do these things by hand

[floral ruin]

I was just asking 😭

[fathom onyx]

(I have a few choice words about the JEE though tbh)

nah it's alright

[floral ruin]

omg sure sure

go ahead

[fathom onyx]

No, a "few choice words" means I have reservations, criticisms

[floral ruin]

I know cuz even I do

[fathom onyx]

It doesn't actually imply I want to say anything right here

That said, basically it's far too much memorisation and not enough actual maths

It instills an incorrect mindset that "there is always a readily calculable answer"

[floral ruin]

like yeahhhhh finally it feels like I am not alone

[fathom onyx]

It's why I'm glad I don't live in India, because I know I'd have to take that exam inevitably

The latter years of your school stuff, though, sure, a calculator is useful

[floral ruin]

and they don't even go in depth. Like remember if you can and get the marks. Ykw? there is exact DE, clairauts equation in the syllabus but they have memorized us the formulaes like xdy + ydx = d(xy) and dozens of formulas under the name of exact DE

[floral ruin]

yea you are lucky

[floral ruin]

ohh i see. a good reason for me to buy it then.

[fathom onyx]

Here that xdy one is a university-grade weapon of a DE tool

And that'd be part of a bigger question, typically in differential geometry, to my knowledge

...idk why diff geo specifically, but that's where I'd first seen it

[late rover]

I just want to say, mine is completely the opposite to the JEE, we don't have to remember anything

But in the cost of many and many and many hard combinatorics problem

[fathom onyx]

Nah as much as you trash on yours, that one is better imo

It actually requires mathematical thinking, not memorisation

[floral ruin]

ohh. I meant isn't the form of exact DE like dM/dx and dN/dy something which i can't type. But they don't teach it in that way. only formulaes

[slow thorn]

not really needing a calculator in this case
cross products arent that hard to find compared to some other demonic contraptions you have to simplify

[fathom onyx]

dm/dx = dn/dy type stuff?

ew....

[floral ruin]

agreed.

[late rover]

If you was talking about me, well it's not good for my grade

[fathom onyx]

I was

You just need 40 daily hours of practice, is all

[lilac finch]

time dilation

[fathom onyx]

You can't try and expect to use techniques you've never seen before, so you need to work through examples, and also make sure you understand them

[late rover]

yeah, some weird ahh technique keep appear everytime I get a new kind of combinatorics problem

[floral ruin]

I wanna study abroad dawg

[floral ruin]

understanding is what i want. I hate chemistry and I always get cooked in the tests cuz of chemistry.

[fathom onyx]

ref. the original problem, I'd imagine you want to check each given answer for c, such that
- it's some linear combo of a and b
- its dot product with a is 0
- it's a unit length (though with experience I can see that they're already all normalised)

I reckon the second one is easier to do first, and you might knock out some answers

[floral ruin]

my first method was this when I saw this question and each option goes well with the dot product

[fathom onyx]

So then the first one is the main check

[floral ruin]

yea

should I close this channel now?

[lilac finch]

yes

.close

[midnight plank]

@floral ruin Has your question been resolved?

[midnight plank]

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:
• 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.

[lilac finch]

.

..

occupy

[lilac finch]

why

then

k

k

[gaunt jetty]

Hint: riemann sums

[lilac finch]

hm..

[lusty python]

hold up

yo

[unkempt sluice]

For a decreasing function f, f(k+1)<=integration of f on [k, k+1] <=f(k). Now f(x)=x^-(1/2) is monotonically decreasing

[lusty python]

NO

NO

NO

PLEASE

WHY

INTEGRATION

[lilac finch]

k

[unkempt sluice]

The question asked for bound, no?

[lilac finch]

ty

[silk pulsar]

Fuck

[gaunt jetty]

You kinda spoiled the whole thing there lol

[quiet hinge]

You can approximate the sum by two integrals, though we usually leave this for the helpee to work out

[lilac finch]

i just integrated

nvm

[tawdry laurel]

[unkempt sluice]

I assumed he knows the anti derivative of x^-(1/2) already

[lusty python]

ANYTHING BUT CALC

PLEASE

[lilac finch]

coz it's from integration

[lilac finch]

sure if you have a way

imma close for now

.close

[midnight plank]

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[quiet hinge]

The point was for the helpee to figure out that

1/sqrt(m)+1/sqrt(m+1)+...+1/sqrt(n)=1/sqrt(m)*(m+1-m)+1/sqrt(m+1) *(m+2-(m+1))+...

and figure out what function to integrate. They do most probably know the anti derivative

[unkempt sluice]

Shit, yeah I messed up

I will edit the two bounds

[lilac finch]

nvm guys

[unkempt sluice]

^

[lilac finch]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[lilac finch]

new question

ik this would use some identity

nvm i figured

ts just 20Ck *1

.close

[midnight plank]

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[midnight plank]

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[lyric charm]

!status

[midnight plank]

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

[lyric charm]

is that your work or somebody else's

[hoary hamlet]

this is mine

so

[lyric charm]

yours isn't loading for me fsr

[hoary hamlet]

in $\sin(2x) + \cos(2x) = 0$

[lyric charm]

can you reupload this

[grand pond]

keqae

[hoary hamlet]

is that okay

[lyric charm]

the trouble with splitting as cos(2x)(1+tan(2x)) = 0 is that the points where cos(2x)=0 are precisely those where tan(2x) blows up to infty.

also 0 < x < pi not 0 < x < 2pi. if anything, 0 < 2x < 2pi is what you may have meant.

[hoary hamlet]

ahh

thanks

[lyric charm]

but anyway pi/4 and 3pi/4 are extraneous for the reason i outlined just now

[hoary hamlet]

that cleared a lot of things up

[lyric charm]

it would have been a lot easier if you had simply divided by cos(2x) to begin with

[hoary hamlet]

is it safe to rule out the cosx=0 case for any division for tanx?

[hoary hamlet]

yeah but i was scared i was going to miss sols

thats why i always opt for factorisation instead of dividing

[lyric charm]

before division by cos(2x), ask yourself what happens when cos(2x) is zero

then, for your question:

sin(2x) will be ±1 depending on x, but in no way is sin(2x)+cos(2x) going to be 0

actually "before dividing by foo, ask yourself what happens when foo = 0" works for division by ANYTHING

and in some cases the answer is "foo=0 can't happen, so N/A => proceed as normal"

[hoary hamlet]

so for this example what would your thought process be?

[lyric charm]

right here

[hoary hamlet]

because i tried to apply that

and i set cos(2x) to 0

and then sin(2x)=0

[lyric charm]

yeah and then cos(2x) and sin(2x) can't be 0 at the same time now can they

[hoary hamlet]

go tit

got it

thanks

.close

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[west sundial]

Hey guys

could you help me with part c please

[compact lotus]

its x = 3pi/8 and 7pi/8 right? for the previous one?

[west sundial]

Are you talking to me?

[compact lotus]

no no

[west sundial]

Oh 😭

[compact lotus]

for the previous question

im attempting yours, as much as i can, for my level

[west sundial]

thank you, it's an a level question for maths

this is the answer

I just dont get why they used the sin^2y+cos^2y = 1 trig identity

[scenic wyvern]

heads up

channel's not open yet

[west sundial]

oh it has to be open?

[scenic wyvern]

ideally yes

[west sundial]

When will it open

[scenic wyvern]

no idea

[west sundial]

Thank uuu

[scenic wyvern]

#help-12 is currently open though

[west sundial]

Alright thank you hanako

[midnight plank]

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[sonic sonnet]

g is inversely propotional to the square of d bwhen d is halved g is multiplied by a factor n
calculate n

[wanton spade]

what have you tried

[sonic sonnet]

I tried 1/d²=g
d²/2=gn

and I didnt know what to do next

[wanton spade]

how'd you come to the second line

[dusty portal]

D is halved not d^2?

[wanton spade]

also for the first you wouldn't use =

[grand pond]

;(

[sonic sonnet]

I would use the inverse propotion but it's not in my keyboard

[wanton spade]

👍

[sonic sonnet]

so how do I solve it

[dusty portal]

maybe let 1/d^2=kg instead

[wanton spade]

okay what's 1/d² if i substitute d/2

[dusty portal]

feel like that would help with the math a bit

[wanton spade]

instead of d

[sonic sonnet]

didn't help

it become d/2=1/d²×n

[wanton spade]

what's 1/(d/2)²

[sonic sonnet]

oh mb

why would I divide?

[wanton spade]

wdym?

[sonic sonnet]

why would I do this

[wanton spade]

well you have $\frac{1}{d^2}$

[sonic sonnet]

ye wth is ts

ohh

inverse

ik that

[wanton spade]

d is halved, and half of d is d/2

[sonic sonnet]

ye ik

[grand pond]

Green

[sonic sonnet]

ye i stuck here

[wanton spade]

so instead of $\frac{1}{d^2}$ you now have $\frac{1}{\frac{d^2}{2}}$

[grand pond]

Green

[wanton spade]

i mean that's what you did right?

[sonic sonnet]

no

I did 1/d²=g
d/2=gn

mb

[wanton spade]

😭

[sonic sonnet]

it can't be that bad

[wanton spade]

it's supposed to be $\frac{1}{\left\frac{d^}{2})\right ^2}$

[grand pond]

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

[sonic sonnet]

why

[wanton spade]

instead of d you have to square d/2

the relation is that you square any given distance

d/2 represents a distance

[sonic sonnet]

so I square it every time or what

[wanton spade]

so you square it whole

[wanton spade]

what would you get if you simplify this

[sonic sonnet]

1/d²/4

ok

one question

[wanton spade]

which is?

[sonic sonnet]

what is the relation between 1/d² and d/2

[sonic sonnet]

4/d²

4/d²=gn

so n=4?

[wanton spade]

i dont think there is any

[wanton spade]

yep

[sonic sonnet]

so why did we square the d/2

[wanton spade]

if i give you some distance l, gn = 1/l²

in this case, l = d/2

so 1/l²=1/(d/2)²

[sonic sonnet]

so I square when it's inverse?

[wanton spade]

it's specified in the question that g is proportional to the inverse of the square of d, which is why we're squaring d

[sonic sonnet]

ye ik that's why we squared d

but why did we square the 2/d

d/2

[wanton spade]

d represents any given distance

in this case d/2 itself is a distance

[sonic sonnet]

so d is always squared?

[wanton spade]

yeah

[sonic sonnet]

sry for my dumb questions bro

tysm for helping me

I appreciate it

[wanton spade]

nah dw

you can .close if you're done

[midnight plank]

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[twilit field]

Really silly question. But here all $X_i$ have the same PDF , $f(x, \theta)$, right.

[grand pond]

wai

[midnight plank]

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[runic hamlet]

<@&268886789983436800>

[midnight plank]

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[edgy crater]

would 1 point be enough, being the turning point, since you get (-a/2, 3a^2+b/4), and can find a from x coordiante, then b from y coordinate?

[lyric charm]

no bc you then have to supply the info that it is in fact the turning pt

[edgy crater]

okok ty

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[near geyser]

any difference between this?

[prime hornet]

no, they're equivalent

[near geyser]

Booooooom !!

[tawdry kraken]

Someone was having a debate about this on one of the channels earlier

It comes down to a matter of intention and notation but in general sure

!done

[midnight plank]

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[lethal path]

you should always refer to your book to figure out what exactly they mean by some notation

[near geyser]

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[near geyser]

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[soft stone]

what have you tried so far? did you use the triangle congruences

[near geyser]

@soft stone

[soft stone]

use the given congruences to pin down which angles are equal

[near geyser]

I did but wrong?

[soft stone]

uh

what does ΔBMA ≅ ΔWSL tell you about the angles?

[near geyser]

B=w

M=s

L=a

[soft stone]

now do the same for the other two triangles

[near geyser]

I did and mention in the picture

@soft stone

[soft stone]

if you are assuming angle k as x then s should be x/2 according to the question

instead assuming s as x would be easier then k and b would be 2x

[lyric charm]

$\angle K = 2\angle S$ means $\angle S = \frac12 x$ not $2x$

[grand pond]

Ann

[lyric charm]

seconding this

[near geyser]

Ohh...silly mistake

Thanks ann

What next?

The given sum of angles?

2x+2x+x

5x?

[soft stone]

yes that's right

[near geyser]

4x=180

[soft stone]

no

[near geyser]

x=45

So i got 45×5=225

[soft stone]

5x=180 so x is 36

[near geyser]

Really?

How?

[soft stone]

in triangle bma

[near geyser]

If i comparing angles for a triangle i got only 4x

And they are asking for 5x value

[soft stone]

x+x/2+ angle a = x

[near geyser]

Wait a second

[soft stone]

but angle a = angle l = angle k

so x+x/2+x = 180

(this is assuming k as x)

[near geyser]

I got x=72

[soft stone]

yes

now find m w u and add them

u should get 144

[near geyser]

72×2

144

[soft stone]

yep

[near geyser]

Yes

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[twilit field]

I was thinking $\chi^2(10)$

[grand pond]

wai

[twilit field]

based on this formula

[soft stone]

you’re right to think about the chi square

but i think since the terms are weighted differently, the distribution won’t be a simple χ^2(10)

[twilit field]

I don't follow

like can't I treat 2X_1^2 as Y_1^2+Y_2^2 where Y_1=Y_2

[soft stone]

you can’t treat $2X_1^2$ as $Y_1^2 + Y_2^2$ since $2X_1^2$ is scaled it doesn’t follow

[grand pond]

anflo

[twilit field]

okay, then how do I do this?

[soft stone]

to get the distribution we would typically use moment generating functions (MGFs) to combine the terms properly

[twilit field]

mhm

I'll try that

thanks

[twilit field]

so MGF of the entire statistic?

[soft stone]

yes exactly

[twilit field]

hmm

that's $E[e^{t 2X_1^2+3X_2^2+X_3^2+4X_4^2}]$?

right

[grand pond]

wai

[twilit field]

I can assume all RVs are indepndent, right

[soft stone]

yes

[twilit field]

hmm

I'm sort of confused

so I have $E[e^{2tX_1^2}] E[e^{3tX_2^2}]+ E[e^{X_3^2}]+E[e^{4X_4^2}]$?

[grand pond]

wai

[soft stone]

+?

$E[e^{2tX_1^2}] \cdot E[e^{3tX_2^2}] \cdot E[e^{tX_3^2}] \cdot E[e^{4tX_4^2}]$

[grand pond]

anflo

[twilit field]

oops, missed a few "ts"

[soft stone]

is this what you get

[twilit field]

yes

[soft stone]

yes then you're right

do you know what to do next?

[twilit field]

I'm kind of confused of how to deal with the squares

Like is there any quick way

I could calculuate each one by hand but I'd really rather not

[soft stone]

hmm try using the MGF $(1 - 2t)^{-1/2}$ for each squared term

[grand pond]

anflo

[twilit field]

wait, where did you get that

[soft stone]

substituting the appropriate scaled $t$ (e.g., $4t$ for $2X_1^2$)

[grand pond]

anflo

[soft stone]

the mgf $(1 - 2t)^{-1/2}$ comes from the chi square distribution $\chi^2(1)$

[grand pond]

anflo

[twilit field]

mhm

okay

so $\frac{1}{ \sqrt{(1-4t)(1-6t)(1-t)(1-8t)}}$ is the MGF

[grand pond]

wai

[soft stone]

mhm!

[twilit field]

hmm

okay

I seem to have forgotten how to find the PDF from the MGF sorry 😭

I think we normally just were told to recognize. it

[soft stone]

to get the PDF we would normally take the inverse laplace transform of the mgf..

[twilit field]

We haven't done laplace transforms ☠️

Unless I learnt it under some other name lemme check rq

def not done this

[soft stone]

you can often recognize the mgf and directly match it to a known distribution

[soft stone]

it can get tricky so its fine

In this case, it looks like the mgf corresponds to a generalized chi square distribution so the pdf follows that structure

[twilit field]

mhm

,w (1-4t)(1-6t)(1-t)(1-8t)

[grand pond]

Results provided by WolframAlpha

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Requested by math_rocks

[twilit field]

how did you get that

[soft stone]

the mgf is a product of terms like $(1 - kt)^{-1/2}$ which is typical for a generalized chi square distribution

[grand pond]

anflo

[soft stone]

coming from a weighted sum of chi square variables

[twilit field]

okay I haven't done generalise chi square so guess I'll look into that

Thanks so much!

I guess this is $\chi^2_4$ then?

[grand pond]

wai

[soft stone]

no it's not exactly $\chi^2_4$

[grand pond]

anflo

[soft stone]

but you can look into generalized chi square distributions for a better understanding of how these weighted terms combine

[twilit field]

Thanks

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[lilac finch]

claim

[lyric charm]

$\sum_{k=0}^n \binom{n}{k} (-1)^k(a-k)(b-k)(c-k)$

this?

[lilac finch]

(-1)^n

[spiral rock]

Open parenthesis

[lyric charm]

ah alternating

(-1)^k though

[lilac finch]

yeah

[grand pond]

Ann

[lilac finch]

ok

[lyric charm]

do you know how to sum $\sum_{k=0}^n \binom{n}{k} k^p$ for $p=0,1,2,3$

[grand pond]

Ann

[lilac finch]

no

[lyric charm]

hm.

difficult

[lilac finch]

oh wait maybe ik

[lyric charm]

for p=0 it is 2^n

[lilac finch]

yeah

[lyric charm]

then you can also do some generating function fuckery

get that (-1)^k by putting x=-1

[lilac finch]

was it differentiating?

[lilac finch]

idk generating function

fok

[lyric charm]

yes

[lilac finch]

hm.. maybe i remember we multiply (1+x)n by something then differentiate

can you tell the method in brief

@lyric charm

...

[hallow basin]

It's just differentiate binomial expansion of (x+1)^n

Do once and you get the case when p=1, multiply both sides by x and differentiate for the second time and you'll get the case p=2

[lilac finch]

oh

ty

so after that?

[hallow basin]

You should have the answer after doing so

[lyric charm]

(a-k)(b-k)(c-k) expands to a cubic in k

[lilac finch]

hmm..

[lyric charm]

if you are ok waiting until i come home then i can show you the idea of my method more closely

cause rn i don't have paper

[lilac finch]

k

so at the end all summations would be 0?

hmm..

so for a p degree polynomial it is always zero for n> p?

maybe

[lyric charm]

oh wait yeah you have a point

this looks like the nth difference of the polynomial (a+t)(b+t)(c+t)

and n>3

so that kills it

[lilac finch]

hmm..

ty

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[spare rover]

if E is euclidian , a,b,c in E such that a=/=b and |a-c|=|b-c| , I know that |c-(a+b)/2|<=|a-c| but how to prove that |c-(a+b)/2| =/= |a-c| ? I see it with a drawing but i can't write it properly, the proof should be short

[frozen talon]

how many times can a line intersect a circle?

[visual tiger]

If E is any euclidian space then geometric arguments will be hard to use

[frozen talon]

sure, just an idea

[visual tiger]

Have you seen cauchy schwarz?

[spare rover]

maximum 2 times ?

[spare rover]

yes we have

[visual tiger]

And so you've also seen its equality case

[spare rover]

OOOHH I think I understand now

i'm writing soemthing

[visual tiger]

be a little more precise as to why lambda = 1, but it's the idea

even if it's one more line to write

[spare rover]

by taking the norm we get |lambda| = 1 , we can't get lambda=-1 because <c-a,c-b> >=0 ?

[visual tiger]

even better... you have lambda >= 0 already?

[spare rover]

yes i wasn't sure if i write it lambda in R or in R+

[visual tiger]

also, there's some small ground work to do to make sure (c-b) is not 0, otherwise you can't write linear dependence that way

but the rest is pretty trivial

[spare rover]

oh your right, if c-b= 0 then |a-c|=|c-b| then a=b=c so c-b=/=0 so we can write this lambda now, thank you for your help 🙂

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[spice pebble]

.reopen

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[jade magnet]

what went wrong in part b

[jade magnet]

740g is definitely nowhere near 1960

i knew i did something wrong by that point

i have my answer to part a if it helps

but i doubt there were any errors that carried themselves over

because i am pretty sure i got full marks on part (a)

60g ≈ 588 sounds about right

[elder bridge]

hello guys im new

[bitter glade]

?

[bitter glade]

Wat

[jade magnet]

uhhh this isnt the right channel

to talk about this

#discussion maybe?

[bitter glade]

?

He just said discriminational word of skin color

[lyric charm]

who, the new one?

lmaooo

[jade magnet]

wtf

[bitter glade]

Um it’s very very famous discriminational word

uk

[jade magnet]

yeah everyone knows what word you're talking about

[tribal temple]

(They've also been removed by the bot too )

[lilac finch]

lmao does the bot timeout or ban?

[tribal temple]

Do what our esteemed friend does and the bot will ban you

[lilac finch]

ok

n

nah i ain't

night i meant

[quiet parcel]

$\sqrt{-1}$

[grand pond]

alex <3

[lilac finch]

what was that

[midnight plank]

@jade magnet Has your question been resolved?

[lilac finch]

t here is half the t used in a)

@jade magnet you there?

[midnight plank]

@jade magnet Has your question been resolved?

[midnight plank]

Channel closed

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[jade magnet]

ohhhh

[jade magnet]

ok

.solved

[midnight plank]

Channel closed

Closed by @jade magnet

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[midnight plank]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[royal dirge]

Stuck here

1. maybe I have the wrong approach overall?
2. if this is the right approach, how do I find ratio of FP to PC?

[late rover]

Use menelaus' theorem instead

That's like, insta kill this problem

[royal dirge]

Got it, let me try that

Ceva's and Menelaus are closely related anyway, at least according to Wikipedia

Wait, how is this applied here?

which line in the problem is the transversal?

oh or maybe it's like this case

Hmm... I don't see it

[late rover]

In this case, you can apply menelaus' theorem for triangle BCF with D,P,A colinear

[royal dirge]

Ohhhh

I was thinking about the big triangle, not the little ones

[late rover]

There're no collinear line that goes outside the big triangle, so our instinct is to consider the smaller ones inside

[royal dirge]

<@&268886789983436800> (there was a spam post)

Anyways got the question figured out, thank you for the help!

.close

[midnight plank]

Channel closed

Closed by @royal dirge

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[midnight plank]

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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.
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• Be polite and have a nice day!

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[brazen onyx]

Hii I’m trying to prove that this equation is solvable in integers for every positive integers n
[ x^2 + xy + y^2 = 7^n ] without much success 😭😭

[brazen onyx]

If n is 1, we have x = 2 and y = 1 as one of the solutions

[junior flower]

where is the n?

[brazen onyx]

Sorry

😭

[grand pond]

Irene

[runic pebble]

Hmmmmm

So it's possible for n=1

Maybe try for increasing values of n

and see a pattern

[junior flower]

factoring in some way may be helpful

[runic pebble]

Trying to solve for n=2

[brazen onyx]

okay let me try that

[junior flower]

@silent dock come help this nice woman

[silent dock]

Ah it's you

I have been summoned

[brazen onyx]

😭

Person with weird emojis

[silent dock]

[junior flower]

[silent dock]

It's an induction problem

[junior flower]

ali blud is skeptical

so is irene

[brazen onyx]

Sorry

😭

[junior flower]

irene is also sad

[silent dock]

if x_n^2 + x_ny_n + y_n^2 = 7^n works then let x_{n+1} = 2x_n - y_n, y_{n+1} = x_n + 3y_n should do the job

[junior flower]

excellent work pure

[brazen onyx]

[silent dock]

My service here is done

[junior flower]

[silent dock]

I shall surrender to my slumber

[junior flower]

[brazen onyx]

Ooh thank you for he your help!!

Good night

[silent dock]

[brazen onyx]

Noooo

[junior flower]

[brazen onyx]

😭😭

[brazen onyx]

Okay let me give this a try

But how is this motivated?

[junior flower]

if the ‘old’ solution gave you 7^n, you want the ‘new’ solution to give you 7^(n+1)

i’m gonna declare f(x,y) = x^2 + xy + y^2 for my sake

[brazen onyx]

Okayy

[junior flower]

so we have x_n and y_n such that f(x_n, y_n) = 7^n and we want to find a new x and y such that f(x,y) = 7^(n+1) = 7f(x_n, y_n)

[brazen onyx]

Yess that makes sense

[junior flower]

so you should be able to find some x and y that work in terms of x_n and y_n

those will be your x_(n+1) and y_(n+1)

[brazen onyx]

Okay I’m following

[junior flower]

that’s what i would try at least. i cannot say this is how pure did it

i am just doing my best to explain the motivation since he is slumbering

[brazen onyx]

Okayy I’ll think about this for a bit! Thank you slayla

[junior flower]

[junior flower]

also i love that this is your image of pure

[brazen onyx]

😭

All of his emojis and stickers are weirdd

[junior flower]

it’s lots of people’s image

[brazen onyx]

I can see why!

[junior flower]

@tribal temple

[brazen onyx]

He also has fancy and gorgeous latex

Hiii

[junior flower]

yea

he’s so cool

[brazen onyx]

Okayy I don’t really know how he figured out the trick so quickly but I understand the motivation/purpose

[junior flower]

he’s too smart for us

[brazen onyx]

Thank you @junior flower and @silent dock

byeee

[junior flower]

also you are so pretty

[brazen onyx]

Have a good day/night!

[brazen onyx]

Awww thank you!!

I just know you’re gorgeous too!

[junior flower]

[brazen onyx]

.solved

[midnight plank]

Channel closed

Closed by @brazen onyx

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• 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.

[sand flume]

For part c how is it A =0? We need g(x+2) = g(x) . A can only be 0 if g(x) =0 which is when x = 0 and g(2) is not equal to g(0)

[midnight plank]

@sand flume Has your question been resolved?

[crystal rivet]

Sine is periodic with period 2 pi and is equal to 0 at 0, pi, and 2pi

In your case, g would be equal to 0 at 0, 1, and 2

Since it would need to go back to zero once in between as well as at the end points

[sand flume]

how is it equal to 0 at 1 and 2?

[crystal rivet]

Oh sorry, i was thinking in terms of f

It still works though actually

If we see that f being an even function relates it to cosine

The integral of f is then related to sine, and would then follow the pattern i said

In (a), you show its periodic as well, so it would need to go back and hit zero to make a full wavelength , as sine does

[sand flume]

how does part a show its periodic. Don't we need g(x+2) = g(x)

we have g(x+2) - g(2) = g(x)