help-10

[unreal tangle]

with that i can say it is an inflexion point ?

[trim portal]

red is f(x), blue is f''(x)

so at 1, the blue f''(x) is positive, so the red f(x) is gonna be convex
at 3, the blue f''(x) is negative, so the red f(x) is gonna be concave (as we can see)

and at 2 is where f''(x) = 0 and where its sign changes. So that's the inflection point, that's where the function went from convex to concave

[trim portal]

yep

ideally try to add some commentary on why what you did works

[unreal tangle]

last question sir
In the question he already told me to prove i(2, f(2)) is an inflexion point first i will say that f''(2) = 0 but what will i do with f(2) ?

[trim portal]

(2, f(2)) is just the point on the graph of the function

2 is the x coordinate and f(2) is the y coordinate

it's just the point with x = 2 on the graph of f, just written a bit differently

you dont need to do anything with f(2)

[unreal tangle]

oh okay i understand now

a bunch of thanks mate 💐💐💐💐💐

[trim portal]

np

[unreal tangle]

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[obtuse pebble]

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[delicate spire]

4

[delicate spire]

My reasoning so far is yes, because if you’re at an odd degree vertex you’ll travel to an even degree vertex (unless it’s just 2 vertices then the path is finished), and this keeps going until you have to reach the other odd vertex

I guess I would say in simpler terms if you reach a point in which you can’t continue the path, you must have reached a vertex with odd degree

<@&286206848099549185>

[polar fossil]

hmm

that seems reasonably solid

why do you have to end up at the other one, why can't you go in a loop or something

[delicate spire]

We have to prove a path exists, and since there’s another vertice with odd degree in the graph, any loop will eventually stop

And if you have a walk that loops until it can’t anymore, there exists a path that exits the loop on the first go-around

[polar fossil]

i don't understand what this is saying

[delicate spire]

Let me try to explain it better

[polar fossil]

my head currently contains the following image

[glad vapor]

hmm

do you know the result that if a graph has all vertices except two of even order then it has an Eulerian path

[delicate spire]

No

I think that’s what I’m trying to prove here

[glad vapor]

[delicate spire]

I’m saying you’d cut the loop out entirely

If there’s a walk where you go ABAC, there exists a path AC

[glad vapor]

well actually eulerian is overkill here

Let u and v be the vertices with odd degree and every other vertex have even degree

suppose u and v belong to separate connected components, say G1 and G2, then ...

[delicate spire]

Let me explain my logic rn: if you’re at a vertex with even degree there are an odd number of new vertices to which you can travel

If you keep following said new vertices, you will eventually reach the other vertex with odd degree

[polar fossil]

if you keep going to new vertices you'll eventually reach the other vertex with odd degree
false

[delicate spire]

2 vertices with odd degree connected by a single edge…

[polar fossil]

actually

or wait you want to start at one of the odd vertices

[delicate spire]

There’s no required starting point, you need to show a path exists at all

And there’s still a path from odd to odd in your graph

[polar fossil]

there is, yes. the theorem is true. i just don't think i agree with your proof for it

[delicate spire]

Ok I should def specify an odd vertex as the starting point

[polar fossil]

you're saying "start at one odd vertex, and always go to a vertex you've never been to before"?

[delicate spire]

Yes

[polar fossil]

you'll get yourself trapped

[delicate spire]

eventually you’ll either reach an end point or a loop, and a loop requires at least 1 vertex with odd degree

[delicate spire]

There’s still a path though, I’m not claiming that all walks that visit new vertices are paths

[glad vapor]

I think if you keep refining this idea you will eventually arrive at a proof of the Eulerian result

but you need a lot of work to make it rigorous and its full strength is not needed for this exercise

there is a simpler proof by contradiction

[delicate spire]

Saying if there isn’t a path from odd to odd there aren’t exactly 2 vertices with odd degree?

[polar fossil]

yea

okay yeah that is easier

[delicate spire]

It makes sense I’m trying to figure out how to prove that

let me think for a bit

[polar fossil]

or even - suppose there isn't a path from odd1 to odd2, and that there are exactly 2 vertices with odd degree. now dig in the dirt until you find a contradiction

[obtuse pebble]

@delicate spire Has your question been resolved?

[delicate spire]

I’ll keep working at it later and post my proof when done

[obtuse pebble]

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[ember frost]

hi i kinda don't get what they mean by "labeling the last row as HI". do they mean just put the last two values in the last row regardless of their stem and label it as HI? since it doesn't really make sense to have HI itself as the stem

[ember frost]

what i have rn

[dense yoke]

the two largest values are punted to the HI row, yes

[ember frost]

like this?

[kind hawk]

surely 10.41 and 13.44

otherwise thats useless

[dense yoke]

HI doesn't say anything about the integer part of the values, so you should include them in the leaves

[ember frost]

it feels odd to have a clear distinction between stems and leafs for the remaining rows, and then you just toss everything in HI

[dense yoke]

because HI here means hi(gh-ended outliers)

at least that's what I'm sensing here

[ember frost]

hmm

alright i see, thanks

.close

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[broken otter]

determine the 0 if it exists, of each exponential function

[broken otter]

0 = (2/5)^(x+2) - 625/16

625/16 = (2/5)^(x+2)

im stuck here

[tacit fractal]

you can write

625/16

[royal basin]

write 625 as a power of 5 and 16 as a power of 2

[tacit fractal]

as 16/625 whole to the power of -1

then you can simplify it further

[broken otter]

ohh

[broken otter]

oh yeah u can do taht too

5^4/2^4

5/2?

[lucid bane]

Express 625/16 as powers

[lucid bane]

Yep

[winter cargo]

did you cancel the ^4 s..

[broken otter]

yeah

[lucid bane]

Dont cancel

[broken otter]

oh

[lucid bane]

You get (5/2)⁴

Then

[broken otter]

so 5^4/2^4 = (2/5)^(x+2)

[royal basin]

$\frac{a^n}{b^n} \neq \frac{a}{b}$ generally.

[warm shale]

Ann

[lucid bane]

Reverse the fraction and you'll get to the power of -1

[broken otter]

oh alright

[broken otter]

(2/5)^(-4) = (2/5)^(x+2)

[royal basin]

brackets

[lucid bane]

Yep

[broken otter]

ok yeah that makes sense

[royal basin]

brackets (x2)

[lucid bane]

Don't forget the bracekts

[broken otter]

like that?

[lucid bane]

No

[royal basin]

missing one pair still

[lucid bane]

Brackets the other side

Then you'll get the rest

[broken otter]

oh alrl ioke that

yeah

-4 = x+2

x = -6

[lucid bane]

Yep

[broken otter]

alright yeah i see ima revise this one cause u can reverse the exponents and stuff

its important ty

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[plush folio]

Solve the diophantine equation x^2 + y^2 + 2 = 3xy

[junior raptor]

!status

[obtuse pebble]

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

[plush folio]

I have solved this question by showing that the discriminant of the quadratic in y cannot be a perfect square using modular arithmetic (mod 5), but I would like to do this another way.

<@&286206848099549185>

[undone flare]

Hi

[plush folio]

Hi

[velvet spoke]

x = ±sqrt(2), y = ±sqrt(2), x/y = 1

[fringe mango]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lapis thunder]

heya

[obtuse pebble]

@plush folio Has your question been resolved?

[lapis thunder]

For any integers x and y, the expression obtained while solving the equation leads to a condition that cannot be satisfied by any integer. Therefore, the Diophantine equation has no solution in integers.

@plush folio

[obtuse pebble]

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[rapid osprey]

Prove that if f ∈ C2([a, b]) has at least three distinct zeros in [a, b], then the equation
f (x) + f ′′(x) = 2f ′(x) has at least one root in [a, b].
(C2[a, b] denotes the set of all twice differentiable functions from [a, b] to R)

[rapid osprey]

no idea

[hardy widget]

,texsp ||$g(x)=e^{-x} f(x)$||

[warm shale]

Civil Service Pigeon

[rapid osprey]

this is the whole solution

pretty much

$g'(x) = e^{-x}f'(x) - e^{-x}f(x)$ and $g''(x) = e^{-x}f''(x) - 2e^{-x}f'(x) + e^{-x}f(x)$

[warm shale]

CherryMan

[rapid osprey]

g has three roots

so g' has two roots

so g'' has one root

which gives our condition

[rapid osprey]

how did you get g

[hardy widget]

exponential is the magic way to flip signs cause that's basically all it does when you differentiate it

[rapid osprey]

so wait you guessed the form of g? or did you find antiderivatives

$f(x) - f'(x) = f'(x) - f''(x)$

[warm shale]

CherryMan

[rapid osprey]

$e^{-x}f(x) - e^{-x}f'(x) = e^{-x}f'(x) - e^{-x}f''(x)$

[hardy widget]

I mean I just gave you a first instinct (which is based on what I said above)

I probably should've checked that it worked ngl

[rapid osprey]

$(e^{-x}f(x))' = (e^{-x}f'(x))'$

[hardy widget]

but I was lazy

[warm shale]

CherryMan

[rapid osprey]

i think i get it

[hardy widget]

[warm shale]

CherryMan

[tardy epoch]

This was a fun problem

[hardy widget]

[rapid osprey]

i dont know where im going wrong but this is giving me f(x) = f'(x)

[tardy epoch]

Don't think I could have gotten it without the hint

[rapid osprey]

same

[wild swallow]

a solution to y'' - 2y' + y = 0 is e^x

so it's natural to try y of the form e^x u for some u

you'll arrive the answer more naturally this way

[rapid osprey]

right

[wild swallow]

it's a classic reduction of order technique when solving odes

[rapid osprey]

whats an ode

[tardy epoch]

Blud

[wild swallow]

like a poem

[rapid osprey]

oh like an ode

ode to calculus

[hardy widget]

[rapid osprey]

ok thanks

.solved

[obtuse pebble]

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[rapid osprey]

$(e^{-x} f(x) - e^{-x}f'(x))' = 0 => (e^{-x}f(x))''=0$

i guess this is the natural way

[warm shale]

CherryMan

[compact shadow]

$P(D)(e^{\lambda x} f(x))=e^{\lambda x}P(D+\lambda I)(f(x))$ might be useful in the future, where $P$ is any polynomial, and $D, I$ are taking derivative, taking no action (identity) respectively

[warm shale]

Cogwheels of the mind

[rapid osprey]

what is P(D+ lamda I) supposed to mean

[obtuse pebble]

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[latent quiver]

Whats your question

[spice cosmos]

How can I solve this task?

[latent quiver]

First establish a coordinate system

Where would you place your origin or (0,0)

[spice cosmos]

• x = 0 → directly below the net
• y = 0 → ground level

[latent quiver]

Yeah ok

If the total court length is 24, how long is each half

[spice cosmos]

• Origin: (0, 0) at the net on the ground
• Vertex (highest point): (0, 1.3)
• So the parabola has the form:

y = a x^2 + 1.3

(because the vertex is at x=0)

12 m

[latent quiver]

Yeah

Seems like you're doing pretty well on your own. What do you need help with?

[spice cosmos]

• Net is in the middle → 12 m to each baseline
• Ball lands 0.5 m before the baseline

So the landing point is:

x = 12 - 0.5 = 11.5

At the ground:
y = 0

➡️ Point: (11.5, 0)

I was confused

[latent quiver]

...are you just using gpt?

[spice cosmos]

Always

[hybrid cosmos]

then what would be your confusion if you're using GPT to blaze through the question?

[spice cosmos]

I still had to ask because the steps were confusing me.

I really appreciate it. Thank you.

[hybrid cosmos]

how would you have approached the task without GPT? perhaps thinking about it and voicing your confusion that way may be more productive.

[spice cosmos]

Yes, because I wouldn't even start with the questions.

[hybrid cosmos]

then have you actually... studied the material in question?

[spice cosmos]

It's a very difficult topic.

[spice cosmos]

In question ?

Yes but only one day

[hybrid cosmos]

I'm not going to police your studying habits, but difficult or not, slamming everything into GPT will probably yield negative results, on top of making you unable to diagnose GPT's output. like right now.

also, you have not stated what your confusion actually is.

[spice cosmos]

I have the basic idea myself, but do you know what the problem is?

Wait, let me explain

[hybrid cosmos]

I will repeat: you have not stated what you're confused about. all you've placed here is a solution from GPT and that you're confused.

[thick oracle]

Die Leute hier hassen die KI

Aus irgend einem Grund

[hybrid cosmos]

I presume that means you want to take over?

[solar stump]

Denn sie halluciniert

[thick oracle]

Besser du versuchst alleine zu arbeiten

[solar stump]

"The people here hate AI for some reason"

[solar stump]

"better that you try to work on your own"

[spice cosmos]

Chat GBT solves the tasks in its own way, but not the way we learned. For example, we calculate the zero point and vertex, but in this task, it does it differently somehow.

[hybrid cosmos]

well, I don't necessarily hate AI. but if you are to use it, make sure you know when it's wrong.

[solar stump]

well, we have !noai for a reason:

!noai

[obtuse pebble]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[hybrid cosmos]

either way, I'll take my leave. sorry for intruding.

[solar stump]

No one's suggesting it's inherently bad, but that it's often misused

[thick oracle]

Yeah, just blindly copying the work is not benefitial

[solar stump]

nuu you're fine lol, I'm in the middle of irl teaching lmao

[spice cosmos]

Bist du deutsch

Zeig mal wie das geht

[thick oracle]

😭

So what was your question

Sorey

[thick oracle]

Ok

Also wir kennen den Scheitel

Und den Punkt 11.5 | 0

Daraus basteln wir uns die Parabelgleichung

[sudden saffron]

Machine must be placed 5.52m

Player hits the ball at height 1.26m

[spice cosmos]

Aber woher weißt du das der Scheitel 11,5/0 ist

[thick oracle]

Steht in der Aufgabenstellung

Dass der höchste Punkt genau übers Netz ist

[spice cosmos]

Hast du Zeit in mein channel zu kommen

Und mir zu erklären

[thick oracle]

?

Wieso nicht hier

[solar stump]

Das hier ist DOCH dein Channel

[obtuse pebble]

@spice cosmos Has your question been resolved?

[obtuse pebble]

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[soft scroll]

the degree would be 60 correct?

[hardy widget]

Yeah

[soft scroll]

im assuming theta is the inner angle here

ok

.close

[obtuse pebble]

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[soft scroll]

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[soft scroll]

sin(theta) = y right?

okey dokey

.close

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[soft scroll]

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[soft scroll]

why does cos(0) = 1

because wouldnt 0 degrees just be a flat line

why would that tell you about its orientation

[hollow dew]

What is the x coordinate of the point on the line intersecting the unit circle

[soft scroll]

well its not stated

or is it

[hollow dew]

What is the unit circle

[soft scroll]

a circle with a radius of 1

[hollow dew]

centered at the origin

[soft scroll]

yes

[hollow dew]

So what is the x coordinate if the y coordinate is 0

[soft scroll]

how do we know the y coordinate is 0 though

thats more of my question

[hollow dew]

It's a flat line

The ray starts at the origin

[soft scroll]

so its always straight with x if its 0 degrees

[hollow dew]

Yeah the convention is 0 is straight right

[soft scroll]

im just confused because if you were to rotate the ray

it would still be 0 degrees

[hollow dew]

How would it be 0 degrees

[soft scroll]

straight line right

im trying to pick up the logic for trig here its been a minute lol

[hollow dew]

It's the angle from the ray facing straight right

[hollow dew]

Look at this diagram

[soft scroll]

'

ok

[hollow dew]

Notice that the ray corresponding to 30 deg is 30 deg from the ray going straight right

[soft scroll]

yes

[hollow dew]

Same for 45 and 60

So 0 would be 0 degrees from the ray going straight right, which is just the ray going straight right

[soft scroll]

ok so thats always the anchor point

im getting it

.close

[obtuse pebble]

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[serene drum]

Hey how would i go about graphing f(x) = 1+2sin(1/2*x) ? im having trouble graphing trig functions

[hearty monolith]

desmos lwk

[stable hill]

im assuming they meant without tech

[knotty crow]

Are you familiar with basic transformations for functions?

1st, 7th and 8th row

[stable hill]

it would also be a good idea to try to plot some points that you know

like the ones at x values of x=0, pi, etc.

[serene drum]

thx guys 🙏

.close

[obtuse pebble]

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[cloud echo]

yo

[cloud echo]

can someone guide me through this ?

This is the definition i've come up with

how would i write this more rigorously and how would i explain it more rigorously if i was tlaking to someone ?

<@&286206848099549185>

oh oops 15 mins i forgot

my bad

[obtuse pebble]

@cloud echo Has your question been resolved?

[lapis thunder]

Expectation is the Riemann–Stieltjes integral with respect to the distribution function; for continuous distributions it becomes an integral, and for discrete distributions it becomes a sum over jumps.

[obtuse pebble]

@cloud echo Has your question been resolved?

[obtuse pebble]

@cloud echo Has your question been resolved?

[obtuse pebble]

@cloud echo Has your question been resolved?

[obtuse pebble]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[obtuse pebble]

@cloud echo Has your question been resolved?

[brazen gorge]

the definition on wikipedia is pretty good and got me a pass lol
https://en.wikipedia.org/wiki/Riemann–Stieltjes_integral#Formal_definition

R is sufficient, no need for R^n

[obtuse pebble]

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[twilit quail]

Please help

[latent quiver]

Hint: consider the derivative of log(x)

[twilit quail]

I did it and I got t but I don't get the answer when I put in the values of e and 1

[latent quiver]

Maybe show your work. It'd help us trace the issue better

[twilit quail]

Ok

I am sorry It's taking a little time to upload

[glacial forge]

You are integrating the function of t with the same limits which is wrong
When you are substituting logx=t then you must change the limits too
X=e means loge=t=1
X=1 means log1=t=0

[twilit quail]

Oh yaa.. completely forgot abt it... thx

[indigo niche]

You could also go back to the variable x

[twilit quail]

Yeah right

Thank you

[royal basin]

easier to recalculate the bounds every time imo

[twilit quail]

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[slim ibex]

in the solution i need to take s = 490

idk why

[spice inlet]

what is s

[rose scroll]

coz 490 is half of 980??

[slim ibex]

displacement

[slim ibex]

yeah but doesnt it have a displacement of 980

[spice inlet]

if u mean height its because the question is how long particle takes to travel half the vertical displacement

[rose scroll]

no, you are only asked about half the displacement

[spice inlet]

VERTICAL displacement

[slim ibex]

oh whoops

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

In this textbook, it says that:

The buying rate of 1 USD = 132 X
The selling rate of 1 USD = 133 X

The questions asks if Joe exchanges his 332500 X currency to USD, how much will he get?

My obvious thought process was that he is buying the USD so it'd be 332500/132 which is ~2518USD but the textbook says 2500 USD.

It looks like if you do 332500/133 that will get you 2500 USD but that is the selling price(?). Is the textbook wrong or am I wrong?

[mortal blade]

The selling rate is the rate at which the country (US) sells you USD

The buying rate is the rate at which they buy USD from you

As a rule of thumb, the selling rate is the higher one. And if you have two values for a conversion, you stick with the one that benefits you (the individual) the least

[silent flame]

-# why would US buy USD brain ain't braining

[mortal blade]

Ehhh I don't wanna go deep into the economics of international currencies

[timid silo]

so if Joe buys USD using the currency X he is going to use the buying rate?

[mortal blade]

No, he uses the selling rate

Like I said, he uses the one which benefits him the least

[timid silo]

ah ok

[mortal blade]

Because otherwise, you could set up infinite money glitches (this is an actual thing, it's called arbitrage)

[timid silo]

so the selling rate is the rate that the country uses to sell it's currency to you

[mortal blade]

Currency conversions aren't balanced. Usually if you go from A to B to A, you will lose money

[timid silo]

i.e 133 X

Therefore, I am wrong

[mortal blade]

Ye

[timid silo]

Alright, thanks

[mortal blade]

Although I imagine the book didn't explain it wrll

So not your fault

[timid silo]

uh huh

thanks anyways

.close

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[oblique spruce]

How to find p0'? Is it possible or am I missing some data? Everything in green is known.

[versed stratus]

I think that's sufficient

[brazen gorge]

if you know the angle of rotation of the top block, then it's just half width point from left corner turned by that angle

[latent bison]

hold on

do you know separately the longer sides of both blocks

because they should be different

otherwise this construction is physically impossible

[oblique spruce]

do you know separately the longer sides of both blocks -> yes its the vectors drawn in green

[latent bison]

proof by contradiction:

sps longest side is common and equal to l.

l < the part of green line inside top block < green line = l

[oblique spruce]

I don't know the angle of rotation of top block, I just know the 2 green vectors (and I can extract the angle but it's not the same as the angle between blocks).

[latent bison]

vectors?

i assume position vectors of all the green points?

[oblique spruce]

Yeah

[latent bison]

then yes, BUT p1' is restricted by p0 and p1

[oblique spruce]

Technically, I think I need to rotate p1 around "left corner" by some unknown angle to get p0'.

[oblique spruce]

What do you mean by "restricted"?

[latent bison]

just quickly verify that |p0p1| = |p1p1'|

[oblique spruce]

I need a solution where |p0p1| and |p1p1'| could be of different value

But I know those values...

I hope that make sense

[latent bison]

but they are both the same length as the green line

or are the green lines of different length

[oblique spruce]

not equal length

[latent bison]

OK so let me get this straight

you only know position vectors of p0, p1, p1', and that there exists some configuration where widths of both blocks are same and p0, p1, p1' are midpoints

there is no additional info

[oblique spruce]

Correct: p0, p1, and p1', and width are known

Nothing else

[cloud echo]

glob glob

[cloud echo]

ayo get back to work

[past sand]

Let p_c be the left corner point. Compute the coordinates of p_c, then compute v_1 = p_1' - p_c and v_0 = 2*p_1 - p_c - p_0. Use a dot product to compute the angle between v_1 and v_0, that's the same angle as between (p_0'-p_c) and (p_1-p_c)

[oblique spruce]

Ok that's the kind of big brain I needed, I hope you are right because that's black magic to me right now

[latent bison]

i wrongly assumed the diagram was saying the green lines were same length, time to disappear

[oblique spruce]

If so that's what I needed so thank you @past sand

And @latent bison for trying!!

[past sand]

In red are v_0 and v_1, in orange the vectors between which you want the angle (so (p_0'-p_c) and (p_1-p_c)), you can clearly see the angle between the red lines is the same as the angle between the orange lines

It's just a matter of constructing a similar vector in both boxes

[oblique spruce]

Yep I picture it now thank you so much

[past sand]

@oblique spruce I might have mistakenly assumed the boxes have the same size. They have the same width, but it's not clear whether they have the same length. If they don't, this fails, and you would need to figure out the difference in length of the red lines first.

Since the half-widths are the same, you can form right triangles and find the angles that way

[oblique spruce]

They have the same width, but it's not clear whether they have the same length
Same width but not same length

[oblique spruce]

Hmm ok I'm not sure I fully get that but I'll try to figure it out

[past sand]

It's more annoying than difficult tbh

I'm tired so I'll try going through it but slowly

The only known points of the upper box are p_1' and p_c so we have to use them

[oblique spruce]

I guess that's a better representation

[oblique spruce]

True, and we know width.

[past sand]

I'll use v and v' instead of v_0 and v_1, seems to correspond better with the notation

Let the length of v' = p_1'-p_c be l'

Let w be width/2

The other side of the right triangle in the upper box has length sqrt(l'^2 - w^2)

I guess you just have to solve a system of distance equations to find p_0'

So let (x, y) = p_0'

(x - p_1'.x)^2 + (y - p_1'.y)^2 = l'^2 - w^2

(x - p_c.x)^2 + (y - p_c.y)^2 = w^2

Should be enough to solve for (x, y)

[oblique spruce]

Thanks I'll try it out!!

[obtuse pebble]

@oblique spruce Has your question been resolved?

[oblique spruce]

So basically @past sand we are doing this?

we know 2 of the side length and we know 2 corner positions

I could still get a flipped result of p0 (mirrored by hypotenus) but I can work with that.

[past sand]

yes

You can select the correct solution by comparing the distance to p_1 if you really need

Or just cross product

[obtuse pebble]

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[ember frost]

probably overthinking this, but for part d i only simply do 4^10/5^10 right?

[hybrid cosmos]

skimming through tells me that is correct.

[ember frost]

alright thanks

.close

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[silent flame]

How does the recipient not know the person who told the rumor to him knows it

🥀

[hybrid cosmos]

¯_(ツ)_/¯

[obtuse pebble]

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[hard sandal]

Fallback to this
#help-12 message

[hard sandal]

@timid silo Didn't have success with playing with negating either barycentric or trilinear but I have an idea

So I have the 3 yellow circles on the triangle points. the inner soddy circle (orange) and the outer soddy circle (green) and I want to get the red ones. And was trying to get them by making a triangle between 2 yellow center with green center.

Now, what I realized is that I can circle reverse (based on green) the circles and get the inner soddy circle of those.

So now if I get the inner soddy of those two inner yellows and the green I get the white one

inverting the circles back to inside the green one, I get one of those circles I was after in the first image, dotted in red: #help-10 message

This is nice because this means I can do the recursion of solving multiple inner soddy circles with the previous circles and the new generated ones and then reverse them back.

Like so:

Circle reverses to inside the green, like so:

Now, what I need to find is a way to reverse a circle based on center and radius.

In these images I'm just doing that to the points of the circumference, but if I do it to the actual center point, the resulting position does not match the reversed circle center point >_>

So if I circle reverse the the positions of the yellow circumference and it's center:

The reversed center point is not the reversed circumference center point ._.

So the question is.... Is there a way to directly calculate the reversed center point given the circle's center and radius?

Right now I see the only way to go about it to reverse 3 points from the yellow circumfrence, and find their circumcenter... Bit tedious

[obtuse pebble]

@hard sandal Has your question been resolved?

[hard sandal]

Basically this:

Alright, found this solution:

Where:

• The circle C to be inverted is defined by point P and radius s
• The reference circle R of inversion is defined by point O and radius r
• d is the distance between the the two circles centers
• d' is the new distance between the circle R center and circle C inverted center.

Just use d' to scale the normalized OP vector and I'm good!

Or if I don't normalize OP I can just scale it by

[hard sandal]

Alright, it works but if a contained circle' diameter is greater than the reference circle's radius (first state) this does not work (middle state) and they are no longer mutually tangent, cause that circle will become the containing circle (last state)

[obtuse pebble]

@hard sandal Has your question been resolved?

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[chilly arrow]

Greetings, could someone help me with this one? thanks

Having the function: f(x - 1) = x² -(a + 1)x + 1

find f(a)

[royal basin]

what's troubling you here

if it said to find f(7), would you know what to do?

[chilly arrow]

it says find f(a)

[royal basin]

i know it says to find f(a).

i am asking you this for a reason.

either you tell me what troubles you,
or you answer the question that i asked you.

[chilly arrow]

if the fuction was f(x) it would be easy to get f(a), but the thing that causes me trouble is that my function is f(x-1)

[royal basin]

ok, so your function is given with "x-1" as its input.

what should x be so that x-1 would become a?

[chilly arrow]

could it be a + 1?

[royal basin]

not just could be, but is

put x=a+1 into that equation and rejoice

[chilly arrow]

i will try it, give me a second

thank you very much, i got the answer right away

[royal basin]

wonderful

got any other questions at the moment? if not, you can .close the channel

[chilly arrow]

.close

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[thick shale]

Given two lines xx' and yy' intersecting at O. On Ox, take points B and A such that A is the midpoint of OB. On yy', take points L and M such that O is the midpoint of LM. Let Q be the midpoint of LB and P be the midpoint of BM. Prove that LP and MQ pass through A

[thick shale]

Help me

[frozen wing]

okay

gimme a min

is the question correct?

[obtuse pebble]

@thick shale Has your question been resolved?

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[frozen wing]

i nead heapl

[vale pelican]

you're using conflicting notation for the same thing?

$^nC_k=\binom nk$

[frozen wing]

wdym

[warm shale]

Flip

[frozen wing]

yeah same thing

[vale pelican]

wack

what have you tried?

[frozen wing]

expand the insides

[vale pelican]

the inner sum has a closed form for sure

[frozen wing]

what is the closed form?

[vale pelican]

the inner sum is $\sum_{\ell=k}^{2n}\binom{2n-k}{\ell-k}$ with $n=7$

[warm shale]

Flip

[frozen wing]

yah

[vale pelican]

you can reindex this to look more familiar

[frozen wing]

nobody can be arsed to open the whole summation

[vale pelican]

$=\sum_{\ell=0}^{2n-k}\binom{2n-k}{\ell}$

[warm shale]

Flip

[timid silo]

all terms with r-k<0 and r - k >14-k would be zero

[timid silo]

so yh this

[frozen wing]

oakty

[vale pelican]

there are no zero terms though

yeah, none

[timid silo]

uh wouldnt that depend on k

[vale pelican]

original is $\sum_{k=0}^n\frac{\binom nk}{\binom{2n}k}\sum_{\ell=k}^{2n}\binom{\ell}k\binom{2n}{\ell}$

[warm shale]

Flip

[timid silo]

oh

[vale pelican]

each k is between 0 and n, each ell is between k and 2n

[timid silo]

right then

[vale pelican]

there is a prerequisite theorem that you would've learned from maybe foundations, probability, discrete, or combinatorics

[frozen wing]

i dont think so

[vale pelican]

what is $\sum_{k=0}^n\binom nka^{n-k}b^k$?

[warm shale]

Flip

[vale pelican]

it's a theorem dealing with expressions of this form

probability uses it because you'd want to know the odds of flipping 3 or fewer heads when flipping a coin 9 times

discrete/foundations would take it and prove it inductively, combinatorics appreciates it and in fact it's in the name, \binom{n}{k}

[frozen wing]

it is binomial

[vale pelican]

so you thought wrong, you have indeed seen this

[frozen wing]

i didnt know that with the name you remember it sorry

[vale pelican]

the sum of all binomial coefficients is a specification of the binomial theorem, with a=b=1

[timid silo]

$\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}= (a+b)^n$
a=b=1 \
$\implies \sum_{k=0}^{n} \binom{n}{k} = (1+1)^n$ \
$\implies 2^n$

[warm shale]

firestepper

[obtuse pebble]

@frozen wing Has your question been resolved?

[obtuse pebble]

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[narrow hinge]

Hi y'all, I have been struggling with problem for a while,
Could i get any help 🥰

[zenith raft]

hm well a pretty ‘bad case’ is when we take the last n elements the set

[narrow hinge]

If it was d = a + b + c, i could do it by using the Dirichlet principle

but now its d = a + 2b + 3c it seems to be quite
difficult

[zenith raft]

the subset {ceil(999/6), …, 999} is a bad set

so n needs to be pretty large

at least 999-165

[narrow hinge]

i've used the same method as the method of solving the d = a +2b + 3c

but idk i got n = 1250

somehow

[zenith raft]

[narrow hinge]

i dont think i was using the right method lol

[narrow hinge]

Wait, how do you know that this is a bad set

oh

a + 2b + 3c >1000

oooh

[zenith raft]

yea

[zenith raft]

you can probably put something a little lower than ceil(999/6) there if you take a closer look

i was just giving an example

i think if k is the lowest number such that {k, …, 999} is a bad set, then 999 - k + 1 should be pretty close to the answer

if not the answer

and k should be figure-out-able with a little algebra. just find the largest k where k+2, k+1, k, satisfy a + 2b + 3c < 1000

[narrow hinge]

Hey, you got a point!

[zenith raft]

it should be pretty close to ceil(999/6)

[zenith raft]

i dunno how to prove this btw. but i think the tails are kinda ‘worst case sets’ for large n so that’s why i think the answer should lie there

[narrow hinge]

Yeah i get it, thanks for the help anyway

I'm trying to figure out to prove with n = 835

aka 999 - 165 + 1 as you said

[zenith raft]

165 was the k you found from this?

[narrow hinge]

Yep

[zenith raft]

okay

[narrow hinge]

cant make any progress tho

[zenith raft]

my intuition says if you take <= 164 elements from {165, … 999} and replace them with elements from {1, …, 164} it will still be a good set. maybe you can formalize that, idk

that does cover all the size n = 835 subsets

or maybe we can try to upper bound the answer by 835 somehow. since we have already lower bounded it by 835

[narrow hinge]

i do think your intuition is right

but agghhh

[zenith raft]

yea i’m not really sure how to proceed

ts is so tuff

[narrow hinge]

Appreciate your help =)

forgot to mention

its discrete math (combinatorics) and in the topic of Dirichlet Principle

[zenith raft]

i was never any good at combinatorics

[narrow hinge]

me neither

[obtuse pebble]

@narrow hinge Has your question been resolved?

[obtuse pebble]

@narrow hinge Has your question been resolved?

[narrow hinge]

<@&286206848099549185> hayyaaaa sorry for pinging =(

[timid silo]

https://tenor.com/view/how-can-i-help-dr-max-goodwin-ryan-eggold-new-amsterdam-can-i-do-anything-gif-16988221

[narrow hinge]

I'm struggling with da
problem

#help-10 message

[timid silo]

i run when i see combinatorics

by

🥀

[narrow hinge]

rip

🙃

[timid silo]

but does this come under a topic

[narrow hinge]

Yeah, dirichlet principle

[zenith raft]

i have an idea, i may try it later when i have time

[obtuse pebble]

@narrow hinge Has your question been resolved?

[obtuse pebble]

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[fallen igloo]

hello everyone.

i know some math. basic calculus, some trig, matricies and such.

i came across a system called a double pendulum, and it seemed like something cool to code. keep in mind, i'm much better at coding than maths...

so i decided to read this wiki page https://en.wikipedia.org/wiki/Double_pendulum instead of watching a tutorial, to see if i could challenge myself and implement a double pendulum.

i couldn't lol.

could anyone help me understand the equations? i just want to be able to model the motion. the first pendulum is easy, and the angle is just sine of the time. but the second pendulum makes things complicated.

any help whatsoever (even if its just how to approach reading this gargantuan document) would be much appreciated. thanks!

[worn yoke]

the wikipedia has the equations of motion, so a simulation would consist of putting those into an ODE solver

[fallen igloo]

what's an ode solver?

the equations can get pretty complex

[worn yoke]

an ordinary differential equation solver

that's just the lagrangian, the equations of motion should be differential equations

any programming language should have popular differential equations solvers that you can use

[fallen igloo]

well this sounds quite promising

but im not sure i understand all the terminology, for example the lagrangian... i looked at the wiki page but it doesn't help.

are there any resources i could use to learn this stuff?

im currently watching this 3blue1brown video: https://www.youtube.com/watch?v=p_di4Zn4wz4

[worn yoke]

you don't need to know all the lagrangian mechanics for this because they give you the final equations of motion (differential equations) which you can use to simulate.

if you want to learn about the lagrangian mechanics used to derive the equations of motion you can check out books on classical mechanics (e.g. classical mechanics by taylor, introduction to classical mechanics by morin). you will want to be very familiar with newtonian mechanics first (as taught in introductory physics courses, e.g. sears and zemansky or resnick and halliday)

[fallen igloo]

thank you

[obtuse pebble]

@fallen igloo Has your question been resolved?

[worn yoke]

one thing which might help as an intermediate step would be to simulate a single pendulum using a differential equation solver

[obtuse pebble]

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[broken otter]

how can i put 3 log_e (x) + 2 ln(2x) - log_e (x) into one single logarithmic

[broken otter]

multiply x and 2x

and multiply 3 and 2?

[neon vector]

What do you mean by multiply x and 2x

And multiply 3 and 2

$3 \cdot \log_e{x} + 2 \cdot \ln{2x} - \log_e{x}$

[warm shale]

USS-Enterprise

[neon vector]

This?

[broken otter]

nvm

[tardy epoch]

Why do you have both log_e and ln

[broken otter]

?

[neon vector]

log_e == ln

but I assume it's written like this in the example

To "confuse"?

[broken otter]

its written like that

[neon vector]

Yeah

Well first

It'd rewrite them as ln()

or log_e if really want to

[tardy epoch]

Do they mean different things then?

[neon vector]

and then look at the first and last term

[broken otter]

yeah

i did

look what i did

3 ln (x) + 2 ln(2x) - ln (x)

3 ln (x) - ln (x) = 2 ln (x)

2 ln (x) + 2 ln (2x)

[neon vector]

yes

[broken otter]

ln (x)^2 + ln (2x)^2

2x^2 = 4x^2

(2x)(2x)

ln x^2 + ln(4x^2)

ln 4x^4

4 ln 4x

4 log_e 4x

[neon vector]

if this is ln(x^2) and ln((2x)^2) then yes

[neon vector]

(2x)^2 = 4x^2

[broken otter]

i got this

[neon vector]

yes

hang on

[neon vector]

THis is incorrect

We have

$\ln{4x^4}$

[warm shale]

USS-Enterprise

[neon vector]

You used the property $\log{a^b} = b \cdot \log{a}$

[warm shale]

USS-Enterprise

[neon vector]

to get 4*ln(4x)

however

We have 4x^4, not (4x)^4

Only x is to the power of 4

Not the entire 4x

so you can't just pop the ^4 out

[broken otter]

so i cant do 4 log_e 4x

[neon vector]

no

Because that would then equal ln((4x)^4) right

which is ln(256x^4)

[broken otter]

so what do i do after ln 4x^4

[neon vector]

well we can try writing it in this form

Can you rewrite 4x^4 to be something to the power of something else

[broken otter]

uhh

idk how

[neon vector]

What is 4

written as a power

[broken otter]

2^2

[neon vector]

and x^4

[broken otter]

ohh

[neon vector]

can we write it as something squared

[broken otter]

x^2

[neon vector]

exactly

so the whole thing is (2x^2)^2

[broken otter]

2 ln (2x^2)

[neon vector]

exactly

and that's all we can do

[broken otter]

oh alr

[neon vector]

👍

[broken otter]

ty

[neon vector]

any time

[broken otter]

.close

[obtuse pebble]

Channel closed

Closed by @broken otter

Use .reopen if this was a mistake.

[obtuse pebble]

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[oblique igloo]

Can someone please explain this to me, (a +b)^2 = a^2 + 2ab + b^2 makes complete sense

Yet in the second step what are they doing?? Why is the book creating an inequality?

[trail pond]

Hey, I'll help you.

[oblique igloo]

Can someone also explain the next step where it's saying that | a + b | = | a | + | b |

If a = -2, and b = 3 then wouldnt' that be

| -2 +3 | = 2 + 3
1 = 5? Or am I thinking of the absolute value wrong?

[oblique igloo]

Thanks!

[trail pond]

I'll take a quick look and give you the solution, or do you want me to explain?

[oblique igloo]

I just want an explanation to be honest

If you have the time

[trail pond]

Okay, yes I do.

[trail pond]

I'll give you the whole explanation so you understand.

[oblique igloo]

Damn really aprpeciate you taking the time G

[slender plover]

for future reference, helpers shouldn't give helpees the answer and should more or less guide them to the answer also, welcome to the server :)
anyway, continue with the explanation :p

[trail pond]

Ok, Sorry

I'm having trouble sending it using the format my phone is using.

[oblique igloo]

No worries man, if you can't send it it's fine, but if you're able to send it some other way would appreciate it

[daring sorrel]

This is using a less than or equal sign, not equals.

[oblique igloo]

Oh

So that's cause I didn't read it properly mb

But the first part

I'm still confused what it's trying to argue for

[daring sorrel]

The inequality comes from the first part of the image you sent

[trail pond]

I can't send it.

[daring sorrel]

$ab \le |ab| = |a||b|$

[warm shale]

Max Coy

[trail pond]

My phone is sending it in a completely different format.

[daring sorrel]

Did you copy and paste your message from somewhere else?

[oblique igloo]

Ohh okay got it

Makes sense so it's not saying it's actually less than or equal to a^2 + 2 | a | | b | + b^2 rather just a continuation to get what it really equals

It's the same as saying

(a + b)^2 = ( |a| + |b| )^2

[oblique igloo]

Dw G, thanks for atleast trying to help me out

[daring sorrel]

Not quite, depending on the values of a and b it may or may not be equal

[oblique igloo]

Oh cause if a < 0?

or b < 0

Then what is this part of the text trying to say?

[daring sorrel]

If a and b are different signs, then $ab < |a||b|$

[warm shale]

Max Coy

[daring sorrel]

if a and b are the same sign then $ab = |a||b|$

[warm shale]

Max Coy

[oblique igloo]

Is it saying that depending on the sign of a and b THEN it would be less than or equal, or just equal?

if a > 0 and b > 0 then it would be the 3rd line

if a < 0 and b > 0 or other way around then it would be the second line

[daring sorrel]

the 2nd and 3rd line are completely equal

[oblique igloo]

Oh yeah

Hmm

Then why the inqeuality difference is it cause if a and b are differnet signs?

But why does the inqeuality sign change?

I don't get htat...

[daring sorrel]

Do you agree that $ab \le |a||b|$ for all possible values of $a$ or $b$?

[warm shale]

Max Coy

[oblique igloo]

Yes of course

[daring sorrel]

Do you then agree $2ab \le 2|a||b|$

[warm shale]

Max Coy

[oblique igloo]

Yes

[daring sorrel]

Add a^2+b^2 to both sides

$a^2+2ab+b^2 \le a^2 + 2|a||b| + b^2$

[warm shale]

Max Coy

[daring sorrel]

Do you agree with this?

[oblique igloo]

Yes

Make scomplete sense

cause a and b aren't necissarily positive

in the first part

[daring sorrel]

Yeah

[oblique igloo]

so it could potentially be less

[daring sorrel]

The left side of the above inequality is equal to (a+b)^2. Do you agree with this?

$(a+b)^2 = a^2+2ab+b^2$

[warm shale]

Max Coy

[oblique igloo]

Yes

Makes complete sense

[daring sorrel]

and the right side of the inequality is equal to (|a|+|b|)^2

$a^2 + 2|a||b| + b^2 = (|a|+|b|)^2$

[warm shale]

Max Coy

[oblique igloo]

Yes

[daring sorrel]

Ok so you can combine all of that together to get

$(a+b)^2 = a^2+2ab+b^2 \le a^2 + 2|a||b| + b^2 = (|a|+|b|)^2$

[warm shale]

Max Coy

[oblique igloo]

Okay yeah this makes complete sense now thanks

[daring sorrel]

Yeah and now this is just written across multiple lines

[oblique igloo]

Yeah it makes sense now, I htink I was confusing how the author was getting there step by step

But your explanation clarified it so thank you so much!

[daring sorrel]

np

[oblique igloo]

.close

[obtuse pebble]

Channel closed

Closed by @oblique igloo

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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[narrow hinge]

Maths hasnt been fun for a long time

[obtuse pebble]

@narrow hinge Has your question been resolved?

[narrow hinge]

<@&286206848099549185>

[jaunty ocean]

What have you tried?

consider constructing the longest possible sequence of numbers starting with 1 where no pair satisfies the inequality..

[narrow hinge]

Errrrrrr many things but couldnt make progress

alr i gtg, closing this bye

.close

[obtuse pebble]

Channel closed

Closed by @narrow hinge

Use .reopen if this was a mistake.

[obtuse pebble]

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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• 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.

[hybrid cosmos]

<@&268886789983436800>

sigh.

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[unborn knoll]

you sniped

lmao

That was fast asf

[wet wren]

Can someone help me with this question from the Junior Cert.
I included both the question and the marking scheme.

[hybrid cosmos]

hi, might wanna open a new channel. the spam botters got this one.

[old lily]

idk that looks pretty dull

[unborn knoll]

Sorry mate this channel is closed

And it will be locked or recycled

[wet wren]

ohh okk, how do i do that? i just joined

[hybrid cosmos]

send your question into an available channel.

[unborn knoll]

use these

[wet wren]

okkk tyyy

[hybrid cosmos]

and hope a bot doesn't snipe the channel first.

[obtuse pebble]

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

[hybrid cosmos]

!status, please?

[obtuse pebble]

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

[frozen wing]

Strauss

[silk fjord]

I got no answer

[frozen wing]

Ok

[silk fjord]

What do I start

[hybrid cosmos]

did you try and get no answer, or have you not started?

[silk fjord]

Im not sure where to start

[hybrid cosmos]

alright. you have a linear equation and quadratic in two variables.

[frozen wing]

Convert both into single variable

[hybrid cosmos]

you should probably rearrange that linear equation in terms of one of the two variables, then substitute it into the quadratic.

don't bother rearranging that quadratic, it's a death sentence. kinda.

[silk fjord]

Ok jma try

[hybrid cosmos]

btw just a simple reminder that I don't accept random DMs. (continue this conversation in a discussion channel, just thought to mention this since you are here.)

[frozen wing]

Where is discussion channel tho?

[hybrid cosmos]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[hybrid cosmos]

these two.

[frozen wing]

It is showing no access

[hybrid cosmos]

you have the studying role. go to bots and remove it.

[frozen wing]

Ok

Thank you

[hybrid cosmos]

um, what did you get from the linear equation?

[silk fjord]

Wdym

[hybrid cosmos]

what are you substituting?

[silk fjord]

I substituted

[hybrid cosmos]

I don't see exactly what you're substituting.

you should probably write down what you're substituting into the quadratic (x = {something} or y = {something}).

[silk fjord]

Whoops

Tu

Ty

[hybrid cosmos]

squaring before multiplication!

[vestal meadow]

yep

[hybrid cosmos]

oh Roy, wanna take over?

[vestal meadow]

sure if you are going

[vestal meadow]

dont forget to square 2 as well in denominator

[silk fjord]

[hybrid cosmos]

your numerator is also wrong btw.

[vestal meadow]

oh yea xd

[silk fjord]

Is it +x^2

[vestal meadow]

no

its 25 - 10x + x^2

[silk fjord]

As the finl trinomial?

[vestal meadow]

i think so yea

[hybrid cosmos]

<@&268886789983436800>

[vestal meadow]

bruh

[vestal meadow]

its not 25+x^2, its 25 -10x + x^2

dont forget the -10x

[hybrid cosmos]

$(a - b)^2 = a^2 + b^2 \color{red}{- 2ab}$.

[warm shale]

Nicole

[vestal meadow]

because (5-x)^2=25x - 10x + x^2 because of the identity

[silk fjord]

Ohhhhh

I have to restart

[vestal meadow]

yea its sad

[silk fjord]

Ahhh

Okk

FINALLY

Omd thank you so much

[vestal meadow]

the steps are wrong, the answer is right magically

[hybrid cosmos]

don't forget to find the corresponding values of y!

[vestal meadow]

do we just allow him to be wrong or do we point it out

[hybrid cosmos]

point it out, of course, but my eyes are failing me so I'll leave pointing it out to you.

[vestal meadow]

kk

[hybrid cosmos]

I apologize for troubling you though.

[vestal meadow]

nah i like to help ppl too

its just i cannot really write it down on laptop so i had to pull out a paper and my phone

[silk fjord]

Ohhk thanks

.close

[obtuse pebble]

Channel closed

Closed by @silk fjord

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[brittle trail]

So, I am a little confused, because an action is a function $\pi: G \to \text{Aut} (S)$, but if I conjugate $a \in A$ by $gA$, where $g \in G$ is arbitrary, I expect to get $gAag^{-1}A = A$, which is not an element of A, but the entire set... Is there something I'm not getting here?

[warm shale]

Ruby

[shadow dagger]

ok so yes, the definition of G/A is like the cosets like gA

but often, the better way to think about G/A are "elements of G mod A"

(so like Z/5Z are "numbers mod 5", you still think of them as numbers 1,2,3,4,... etc. but like i'd just consider 1 and 6 the same)

so G acts naturally on A by conjugation

so when it says G/A acts on A by conjugation

you wanna show that if 2 elements are the same mod A, then they act the same on A by conjugation

[brittle trail]

Ok, that makes more sense. Thank you :)

[shadow dagger]

nws

[brittle trail]

.close

[obtuse pebble]

Channel closed

Closed by @brittle trail

Use .reopen if this was a mistake.

[obtuse pebble]

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Send your question here to claim the channel.

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

[sonic pumice]

A level vector question.. id really appreciate if someone could explain this to me like im 5 because im not even sure where to start

[frozen wing]

Ok

[sonic pumice]

Thanks

[frozen wing]

Take it as a unit cube

Origin as 000 bla bla ok?

[sonic pumice]

yes

[frozen wing]

Then you know the formula for point dividing in a ratio yeah?

[sonic pumice]

i think so

[frozen wing]

Write it for me

[sonic pumice]

yes

ok so OM is 3/4(OE)

like that?

os u would divide all the co efficient

sorry times

by that

[frozen wing]

Yea but it has a easier to remember version, whatever suits you as long as you can get coordinates of m

Ill put the formula in the end

[sonic pumice]

yes

[frozen wing]

Now we need to work with ap

[sonic pumice]

okok

[frozen wing]

So, let p has coordinates a,b,c such that m divides ap in the given ratio

You can now calculate a,b,c separately

Yeah?

[sonic pumice]

ohh yes

[frozen wing]

Yea