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[wicked gazelle]

okai okai

[glad iron]

The 1st line in yellow after Let (x,y,) be the foot of the perpendicular is the formula

[wicked gazelle]

okai so how am i supposed to apply this

[glad iron]

Do you not know the standard form of a line?

[wicked gazelle]

i know nothing man i am doing from scratch

[glad iron]

Standard equation of a 3d line is ((x-x1)/a)=((y-y1)/b)=((z-z1)/c)

Compare points here with the form and substitute in the above formula

[wicked gazelle]

alr alr let me see if i get the answer

[knotty grotto]

solved with system of linear equations

linear like the highest exponential degree present is 1
0.12b = 0.10a + 1400 ; linear
0.12b^2 = 0.10a^3 + 1400 ; not linear, is cubic, highest degree is 3

[wicked gazelle]

ohh okai got it

.close

[lone heart]

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[jaunty olive]

What have you tried

[graceful halo]

I have tried like 14.5 - 6/9 but I don’t think that makes sense 😭😭

[jaunty olive]

What is your reasoning for doing 6/9?

[graceful halo]

Honestly I don’t know😭 originally I was thinking of doing this:

But I just had an exam and I honestly cannot think straight at all

[jaunty olive]

You're correct!

You have probably made a calculation mistake somewhere

[graceful halo]

Ohh okay

I’ll recalculate it

[jaunty olive]

Okay

[alpine sable]

Hey guys

[opal jolt]

start by computing the total marks from all the exams

then remove the mark from the one to be revised fromn the total

compute the mean of the remaining ones, remembering that there's one less exam for the new total mark

[jaunty olive]

What she has done is also correct.

[opal jolt]

this one? that would assume that the mean doesnt change, which it does

[graceful halo]

Is the new mean 14.8

[jaunty olive]

? No?

[jaunty olive]

Yes!

[graceful halo]

Or unrounded is 14.8667

[jaunty olive]

Yep

[jaunty olive]

She's taken new mean as x

[graceful halo]

Thank you for the help!

[alpine sable]

Anyone

??

[jaunty olive]

!occupied

[lone heart]

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

[jaunty olive]

!done

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[vast bobcat]

for 3cii, so far i've said that p3 and p2 intersect as because they are non-parallel, they have to intersect. is this correct?

[vast bobcat]

or can you also argue that it depends where p3 is positioned because it could be small enough that it doesn't touch p2

im just finding trouble with drawing c ii)

[vale wigeon]

yes it is correct

any plane parallel to P1 will be nonparallel to P2 and so intersect it

[vast bobcat]

how do you represent that visually?

[vast bobcat]

this sort of idea?

[vale wigeon]

mmm

idk about this

im thinking more like

[vast bobcat]

welp

[vale wigeon]

imagine taking the shape of the symbol ≠

and extruding it vertically

[vast bobcat]

oh

[vale wigeon]

each of the 3 lines in there becomes a place

[vale wigeon]

ish

[vast bobcat]

wait how do i draw it so it looks acceptable to write up

[vale wigeon]

eh. good enough for me tbh

[vast bobcat]

should i draw in the arrows and right angle degree

oh something like this?

[vale wigeon]

yeah

[vast bobcat]

oh this is interesting

thank you

.close

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[wicked gazelle]

How is example 1 not transitive?

[jaunty olive]

(1,0) and (0,3) are there

But no (1,3)

[vale wigeon]

missing brackets but correct points

[jaunty olive]

Yes sorry about that

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@wicked gazelle Has your question been resolved?

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[wicked gazelle]

Thanks

[wicked gazelle]

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[vital blaze]

so i hav to find the bounds for the triple integrals of the following function

z^2+x^2+y^2=9

bounded by the cone z=root(x^2+y^2)

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@vital blaze Has your question been resolved?

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@vital blaze Has your question been resolved?

[last spire]

Have you tried cylindrical coordinates?

[vital blaze]

Ohhh hi

Well i was just looking for the bounds

That is the hard part

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[wispy jay]

can someone explain this in normal english please

[wispy jay]

does that not mean that the lowest heart rate before is 52 😭 ?

[woeful zenith]

Uh

This doesn't look right at all

[lavish cave]

yeah it's a stem-and-leaf diagram

but then the formatting is super messed up

[wispy jay]

wym

[woeful zenith]

most cursed stem and leaf I've ever seen

[woeful zenith]

It's not formatted correctly

[wispy jay]

😭 ikr

looks weird af

thats why i need help gng

😭 🙏

[lone heart]

@wispy jay Has your question been resolved?

[wispy jay]

dk if anyone can help but any help would be appreciated

[lone heart]

@wispy jay Has your question been resolved?

[lone heart]

@wispy jay Has your question been resolved?

[lone heart]

@wispy jay Has your question been resolved?

[wispy jay]

ts on my test tmr, lmk if anyone can help if not all g

[elfin pike]

R u sure u have the right paper cuz there should be numbers in the fitness after the programme🤔

[lone heart]

@wispy jay Has your question been resolved?

[jolly notch]

i figured it out

for the first row of values notice the middle 5 has values increasing on either side

meaning that's the number in the stem

beyond that just place each next number (6, 7, 8...) in the stem and properly format it

wait what the hell is that 4

oh gosh

oh the 4 probably belonds to the first row

this is all speculation

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[faint thistle]

Question says calculate derivative of x^a*y^b=(x+y)^(a+b), now, I don't really know a method to do a 2 variable derivative, the answer is y/x but I want to know how to reach it, it doesn't seem I can use any of my known methods to reach an answer.

[lavish cave]

implicit differentiation, i.e the chain rule

so $\frac{d}{dx} (y^b) = by^{b - 1} \cdot \frac{dy}{dx}$, due to the chain rule

[ocean seal]

south

[lavish cave]

y is a function of x

[faint thistle]

So it would be axby'y=(a+b)(x+y)(x'+y')?

[lavish cave]

slow down

(cause that's completely wrong)

you need to use the product rule first on $x^a y^b$

[ocean seal]

south

[faint thistle]

oh you mean u*v?

[lavish cave]

so that would be $x^a \frac{d}{dx} (y^b) + \frac{d}{dx} (x^a) (y^b)$

[ocean seal]

south

[lavish cave]

that's what I mean yes

[faint thistle]

If you don't mind me asking how did you separate (x+y)^(a+b)?

[lavish cave]

I didn't talk about that part yet

[faint thistle]

Ok so you just did the left side of the equation right? We are missing (x+y) then

[lavish cave]

yes

okay now for the right hand side

you have $(a + b)(x + y)^{a + b - 1} \cdot \frac{d}{dx} (x + y)$

[ocean seal]

south

[lavish cave]

also remember that d/dx (y) is just dy/dx

[faint thistle]

So if I am understanding this correctly
for the left side you did dy/dx uv = uv' + vu'
And for right side you did u^n=n*u^n-1 u'?

[gritty bramble]

not what is asked but u can also just put a=1 and b=-1 and get the answer directly

you'll need to do that anyways since the options dont include a and b

maybe thats the method they wanted you to use (incase implicit differentiation isnt taught, which will also include putting values of a and b in this case, but after the differentiation, as opposed to putting at start)

[faint thistle]

I mean if it gives the correct answer who am I to judge, but wouldn't it give x/y=1? Or am I doing something wrong?

[gritty bramble]

x/y = 1 is the simplification of the original equation

you're yet to differentiate it

it asked you to find dy/dx

[lavish cave]

yeah like if it's multiple choice don't bother doing it the proper way

MCQs don't actually teach maths in my opinion

they teach how to guess correctly

[gritty bramble]

the a=1 and b=-1 step will have to be done regardless

[faint thistle]

So which would be the strat? Try to equal a side to 0 exponent?

[gritty bramble]

but yeah u can just put at start instead of after the implicit differentiation

makes it easier in mcqs

[lavish cave]

really? how?

[gritty bramble]

you still have a and b in your differentiated equation

what do you do about them

[lavish cave]

I mean the assumption is that dy/dx is constant regardless of what a and b are

[faint thistle]

Sorry for asking this question but how do you differenciate x/y=1? It has 2 variables I don't know sorry 😔

[lavish cave]

x = y, or y = x

so dy/dx = 1..... yeah unfortunately it doesn't work

[gritty bramble]

chain rule

well

[lavish cave]

I mean 1 = y/x here

[gritty bramble]

u still have to do implicit differentiation

just on an easier equation

[lavish cave]

but so is x/y = 1

so you've eliminated options A and B

you can't tell C and D apart

[gritty bramble]

yeah but you have to get rid of them, no?

[faint thistle]

How did you guess we could just do a=1 and b=-1 by the way?

[gritty bramble]

to simplify and reach y/x

[gritty bramble]

not guess

[lavish cave]

experience with MCQs

[gritty bramble]

the equation given is x^ay^b = (x+y)^(a+b)

put x=y=1

1^a1^b = (2)^(a+b)
1 = 2^(a+b)

for this to be equal

a+b = 0

so a=1 and b=-1

[faint thistle]

Woah, then after doing that and getting x/y=1 how do i implicit differenciate?

[gritty bramble]

d/dx (x/y)

you know division rule right

(y.1 - x.(dy/dx))/y^2 = 0

the dy/dx comes from implicit differentiation

[faint thistle]

You mean u/v=uv'-vu'/v^2? division rule?

[gritty bramble]

yes

its -, not +

[faint thistle]

Aight

[gritty bramble]

wait

its vu' - uv'

[faint thistle]

Thx got it

[gritty bramble]

i forgot how to use latex so

now just solve for dy/dx

[faint thistle]

y/x=dy/dx?

[gritty bramble]

yes

but u need dy/dx

so flip it

what

no

how did u get dx/dy = y/x

[faint thistle]

I wrote it wrong 😔

[gritty bramble]

lol, yeah its correct now

[faint thistle]

Kinda cool problem, got anything else like this to try and refine my skills? Thx anyways

[gritty bramble]

solve actual questions

[faint thistle]

Understood, by the way what if we did a=-1? is that wrong or would we still achieve the answer?

[gritty bramble]

if a = -1 then b = 1

because a+b = 0

if u put both -1 then you wouldnt reach anywhere since a and b still remain in the equation

[faint thistle]

Would that affect the answer or would it still be fine if we did a=1 b=-1 or a=-1 and b=1?

[gritty bramble]

the only reason we eliminated a and b is because there were no a and b in the options

[gritty bramble]

both are fine

[faint thistle]

Oh I see, thanks for the MCQ hack

[gritty bramble]

a =2 and b=-2 is fine too

or 3 and -3

as long as a+b=0

anything will work

[faint thistle]

Except 0?

[gritty bramble]

that will work too

wait no

u just get 1=1

true but unhelpful

[jaunty olive]

Not really

I got the answer without doing that

[faint thistle]

Ok

[gritty bramble]

how

[lavish cave]

https://www.desmos.com/calculator/6fk2ijg912

same I got it

[jaunty olive]

Just do an implicit differentiation

It's not thay hard

The a and b terms cancel out

[gritty bramble]

nvm u can

i did the implicit differentiation wrong

[jaunty olive]

[lavish cave]

sub in $\left(x+y\right)^{\left(a+b-1\right)} = \frac{\left(x^{a}y^{b}\right)}{\left(x+y\right)}$ then keep cancelling and then expand a bit later

[ocean seal]

south

[jaunty olive]

Yep

But if this an mcq

And it was an exam

I would probably put a=1 and b=1 as well

Or some other value to make it simple

[lavish cave]

yeah

honestly put in a = 1 and b = 0 next

ah that didn't work

[jaunty olive]

I'd agree that mcqs don't help in progressing math skills, they only help in making you quick

[gritty bramble]

agreed

[faint thistle]

Is this always applicable?

[jaunty olive]

In this case yes

[faint thistle]

Where would it not be applicable?

[jaunty olive]

Where the relation is not given?

[lavish cave]

it comes from $x^a y^b = (x + y)^{a + b}$ so yes

[ocean seal]

south

[lavish cave]

what I did holds for any x, y, a, b

well bx - ay can't be 0 but that's the only exception

[faint thistle]

Oh I understand it now

[jaunty olive]

Was also a bit skeptical since this term came in both RHS and LHS While differentiating

[faint thistle]

You mean while?

[jaunty olive]

Ye

[faint thistle]

Then what would the process be? Do I do derivative u/v then?

Well I'll research it myself, thx for the time

.close

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[alpine sable]

Question A(9,a) and D are on a line l1 : hx - 4y + 3 = 0 and B which is 6 units away from A is on a another one l2 : 3hx - 4y + 9 = 0,both line intersect at C

Show that a=3 and h=1

I tried this question a few times and can't solve it, idk if its a flaw in the question or it is my issue cus even if I got the question wrong, my answer is inconsistent

[lone heart]

@alpine sable Has your question been resolved?

[jolly notch]

im inclined to call xy

there's information missing

was there a graph or something we can go off of

!xy

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[alpine sable]

Oh yeah there was actually

[alpine sable]

Sorry the graph was kinda sketched really small on my hand when my friend gave me the peroblem but here is what the graph looks like

He had his exams awhile ago and he asked me if i could solve that question

I feel like there is missing info but im not sure

[jolly notch]

lemme see if relating B and D in some ways gives this result

because as of right now the problem is either over or under determined

[alpine sable]

How would you relate them though

[jolly notch]

im seeing whether they have the same x coordinate or if theyre perpendicular or what, and i'll assume the question gave that info

[alpine sable]

The question has a few parts so i believe B C and D are just cordinates for the other parts

[jolly notch]

because the question you stated has insufficient relationships

[alpine sable]

I see

[alpine sable]

I see

My friend may have just forgotten

But still cant explain how i got my answers inconsistent

I got 4a=9h+3 for substituting A into line 1

Then i solved the gradient of line one using point A and the y intercept

Which i got 4

[jolly notch]

from a purely graphical standpoint USING the values given by the question to get the graph, we see that the question probably tells you that the x coordinate of A and B are the same

[alpine sable]

Its wrong but when i did the same thing using pythogoras and point B with line 2

And got

4 answers which is scary

[alpine sable]

Ooo

[jolly notch]

however B could be either one of the points that intersects the purple line

we just know it's 6 units away

but judging from the nature of the question

it's a fair assumption

[alpine sable]

It could be the second iwrsection tho

[alpine sable]

Yes

[alpine sable]

I see

[jolly notch]

yeah i think that's the missing puzzle piece

[alpine sable]

Thanks though

I dont know how to identify flawed questions haha

How again

[jolly notch]

||do note that even still you have two different solution sets for h and a||

!done

[lone heart]

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

[alpine sable]

Oh icic tq

[alpine sable]

Yes i know

Thanks

[jolly notch]

||you have to type .close if ur done ofc||

[alpine sable]

Oh only i can do it?

.close

[lone heart]

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[jolly notch]

well technically helpful (green) roles can too but i feel it's kinda rude to close someone else's ticket

[alpine sable]

Ohh icic tq tho XF

[lone heart]

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[tardy shuttle]

i can't figure out what these symbols mean

[mortal trellis]

you mean $\circ$ ?

[ocean seal]

Denascite

[mortal trellis]

composition

[tardy shuttle]

no i mean everything

discord cut the rest of it off

[mortal trellis]

well without context we will not be able to do anything either

[tardy shuttle]

if you click on it you will see the rest

it should be on there.

[mortal trellis]

those are three images of a few symbols each

thats not context

[tardy shuttle]

okay il send the full photo

[mortal trellis]

a translation would also be good

but T, omega, S seem to be some sort of geometric operations

that notation is explained somewhere earlier in the book most likely

[tardy shuttle]

i basically dont understand anything its trying to say my teacher did say it was written in an unorthodox manner but like

i really dont get what it wants

what are they even supposed to represent 😭

[lone heart]

@tardy shuttle Has your question been resolved?

[lone heart]

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[calm rose]

I think i might be bugging but how do u integrate a function like this by recognition

[jaunty olive]

1/(x^2+16)^3/2 ?

[calm rose]

Ye

[jaunty olive]

Put x = 4 tan theta

[calm rose]

It doesn't seem familiar to. Me and doing a revese chain rule

Seems tough

[calm rose]

U cant do it by recognition?

[jaunty olive]

Wdym by recognition

[calm rose]

Some tough paper i got lmao i aint got time to do a usub

[calm rose]

Reverse chain rule

Or like yk how u are familiar with a integral

Urs will lead to a answer but it will take

Abit of time

Tbey jumped the step

[jaunty olive]

I'm very familiar with 1/(x^2+a^2)^3/2

So i do x = a tan theta

Why memorise when you can do it just as easily

[calm rose]

Time pressure

[jaunty olive]

It really doesn't take that much time

[calm rose]

That's the full q

I need to do it in 12mins which may sound alot but idk now i think abt it i got plenty of time but when ur stuck

And then u need to do it qucik

Longday maj

Man

[jaunty olive]

It's literally just integral cos theta after substitution..

[calm rose]

Also not really memorising tbf ur method is probably the way to go

[jaunty olive]

[calm rose]

I get some formulae ij my formula booklet

I see them do it in one step

So i thought i should be able to do it

[jaunty olive]

That's cuz they've probably done the rough work somewhere else before writing it

[calm rose]

Can you help me with another q btw im kind of confused

[jaunty olive]

Wouldn't really be a formula sheet if it contained 3-4 formula on one page

[jaunty olive]

Yeah

Sure

[regal anchor]

ha decay

[calm rose]

Part d and e but lets even starttof with a b c

While we are at it

Q1, u can write both as modulus argument /polar form

[lunar saddle]

what

[jaunty olive]

Yep

[calm rose]

Argument is not unique

[jaunty olive]

Yes

[calm rose]

Then apply demovirws theorem and u get three roots

B is simple

C ie also simple ngl

D and e is where im confused

That's my solution so far brw

[jaunty olive]

D is similar to a

Just take z - alpha as some other complex number z'

[calm rose]

Hmm

[jaunty olive]

The e^itheta term will remain the same

Only instead of beta it will become beta^(1/3)

[calm rose]

Hmmm

I feel weak in this topic ngl

Thats their solution

[jaunty olive]

???

There's no mention of what alpha or beta is anywhere

[calm rose]

Problem solving ig

[calm rose]

That's the full q

[jaunty olive]

I know

[calm rose]

I think it kind of makes sense

Though

[jaunty olive]

There's still no mention of alpha being equalto z0 anywehre

[calm rose]

Its becusse we are told z1 z2 z3

Are solutions

[jaunty olive]

Oh shi

Mb

Yeah then alpha = z0. Makes sense

[calm rose]

Care to explain

Im still confused slightly

[jaunty olive]

When you drew the diagram for a and b part, what is the triangle y9u got

[calm rose]

Eqyalateral

Equailatetal

Equalateral

Yh icl idk the spelling

But we move

Equilateral

[jaunty olive]

So now shift the origin to the point z0

[calm rose]

Ah its kind of like a graph transformation

[jaunty olive]

Yes

You're keeping everything same, just translating the diagram from origin to z0

[calm rose]

Ive always never been happy with those is there a strong idea on why f(x-2) shfits to right without just considering the graph

Itself

Liek ik it does that just from considering tje graph

[jaunty olive]

The idea is transformation in itself

[calm rose]

Wdym

[jaunty olive]

Idk how to explain, ig you could just plot some points to observ

[calm rose]

Yh ik that

Tjwts hwo i know it

Tbf Just considering tje input would work aswell

Ig

[jaunty olive]

Yeah

[calm rose]

But yk for some questions like

Onesec

Ngl ido how to do this

This one i do know

[jaunty olive]

You know vietas right?

[calm rose]

Vietas or yi u mean like sum of roots ect?

[jaunty olive]

Yes

[calm rose]

Yh ik that

But theres a way to go about it with graph transformations

Esp the second one

[jaunty olive]

Intriguing

[calm rose]

This one

Let x= w +1/w

Make w the subject

[jaunty olive]

That's not really transformation of graph, it's transformation of roots

[calm rose]

Then substitute expression of w into the first equation

Similar concept tho no?

[jaunty olive]

And that's going to be lengthy to do

[jaunty olive]

No here y9u aren't shifting the graph of anything

The graphs are entirely different

[calm rose]

Ah but the roots are changing

[calm rose]

I mean i stil feel like it is some sort of transformation

[jaunty olive]

Of roots

[calm rose]

Sure

[jaunty olive]

Actually what we did earlier can't be called transformation

It's more so translation0

[calm rose]

Thats a type of transformation

[jaunty olive]

Like changing our reference

[jaunty olive]

Not really cuz here everything else remains same except for our reference

[calm rose]

The roots are changing and the graph itself will definitely change

[jaunty olive]

When you transform roots, you're changing the Vertex, length of latus rectum, curvature of the parabola

[calm rose]

The worst bit is i understand vaguely what yojr saying 😭

Ur complicating it

Nog even vaguely just understand it

I just understand

Anyway thanks

.close

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[jaunty olive]

????

[calm rose]

Bro said latus rectum

If i didnf watch some niche video on it i would never of understood

But mabye u had high hopes for me 😂

[lone heart]

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[kind sandal]

Hello 🙂 I am struggling to understand why they can just dot product the 'end points' of each line in their solutions to this question (I underlined in green). I would assume, that to see where the paramaterised lines are orthogonal, you need to take into account the initial points of each line because this together with the end points determines the directions of the lines?

[kind sandal]

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[gray summit]

could someone explain how this formula works

[pallid scarab]

$u^v = e^{v\ln u}$

[ocean seal]

rafilou is not not born in 2003

[pallid scarab]

so chain rule with exp says $\frac{d}{dx}(e^{v\ln u}) = e^{v\ln u}\frac{d}{dx}(v\ln u)$

[ocean seal]

rafilou is not not born in 2003

[pallid scarab]

meaning $\frac{d}{dx}(u^v) = u^v\frac{d}{dx}(v\ln u)$

[ocean seal]

rafilou is not not born in 2003

[pallid scarab]

and now product rule + chain rule (for derivative of ln(u))

[gray summit]

how does $u^v = e^{v\ln u}$

[ocean seal]

Nox

[gray summit]

is there a property or rule that I am missing

I understand the rest of your explanation

[gray isle]

u = e^(ln(u))

exponential and ln function being inverses

[gray summit]

ohhh

alright

thanks you

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[vestal shell]

I have one question about LDA

[frail dagger]

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[vestal shell]

Linear discriminant analysis

This is the plot of the scores

Basically i am trying to distinguish the beens of coffe of free different colors

But at least in my opinion there isnt a clear separation of groups

[tropic valley]

i

[vestal shell]

And each group is normally distrivuted and the variance and covariance is equal for the three groups

And the Linear discriminant functions are statistically significant

So this shouldnt happen right ?

[lone heart]

@vestal shell Has your question been resolved?

[lone heart]

@vestal shell Has your question been resolved?

[lone heart]

@vestal shell Has your question been resolved?

[lone heart]

@vestal shell Has your question been resolved?

[vestal shell]

<@&286206848099549185>

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@vestal shell Has your question been resolved?

[lone heart]

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[lucid aurora]

how did my teacher get this??

[lucid aurora]

i thought 11pi/6 was equivalent to pi/3

and tan of that is sqrt3?

[gray isle]

how are you getting pi/3

[lucid aurora]

on the unit circle they're equivalent right?

like for their exact value table

[summer dirge]

tan is positive in the first quadrant, and negative in the fourth quadrant

[lucid aurora]

yea i get that part

[gray isle]

what calculations are you doing to get the value of pi/3

[summer dirge]

and tan(11pi/6) is not equivalent to tan(pi/3) in absolute value

[lucid aurora]

well its 2pi/6

[gray isle]

what calculations are you doing to get the value of 2pi/6

[lucid aurora]

on the unit circle they like follow the same line

wait lemme get a diagram

aren't these all equivalent?

in their cos and sin value?

[gray isle]

no

you took the wrong line

you want the angles made with thex-axis to be the same

[lucid aurora]

oh so is it equivalent to pi/6?

[gray isle]

i wouldn't say equivalent

the reference angle would be pi/6

[lucid aurora]

oh yeah

[gray isle]

and since tan is negative in Q4, the result
= -tan(that reference angle)

[lucid aurora]

yup

how does he get -1/2 though?

tan for pi/6 is root2/3

so wouldn't it be -root2/3?

[gray isle]

tan pi/6 is neither of those

the teacher screwd up their values again

[lucid aurora]

oh

[gray isle]

tan(pi/6) would be 1/sqrt(3)
or sqrt(3)/3

[lucid aurora]

yeah

whoops i have to fix my table

so this is an answer error?

[gray isle]

yes.

[lucid aurora]

okay, thank you so much :D

and thanks for ur patience sorry haha

im not running on much sleep

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[twilit dome]

is there a function to approximate how many valid formulas in the first n godel numbers?

[twilit dome]

thanks in advance:3

by valid i mean well-formed, not correct

<@&286206848099549185>

[lone heart]

@twilit dome Has your question been resolved?

[dire plover]

what encoding are you using?

[twilit dome]

sorry, i thought there was only one?

based on prime factorisation?

but either way, being approximated by a computable function is a fundamental property that can not change throughout different encodings no?

@dire plover

https://en.wikipedia.org/wiki/Gödel_numbering#Gödel's_encoding

[dire plover]

yes, there is such a formula, i thought you were asking for the formula

[twilit dome]

what is it

[dire plover]

depends on the encoding, but anyway, checking whether something is well formed is doable(i mean we have algorithms for this which will always halt), so we can just go over all godel numbers till N check whether each one is corresponding to well formed formula, and adding them up

[twilit dome]

sorry, i meant to phrase it clearer, i meant that is there a function to approximate that number, like defined by operations. for example log log x

[twilit dome]

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[worthy lichen]

Which answer form is equal to my answer

From the options?

Why option A is not correct since it includes 1 and excludes 1 and 4

[vale wigeon]

+1 is not supposed to be included either

you have to cut it out

[fickle latch]

Looks like Set Difference Notation. The subtraction sign indicates that 1 is excluded from the interval on the left.

[vale wigeon]

your answer is equal to (-1, 4) - {1}

[vale wigeon]

you mean "excludes -1 and 4". do not drop minus signs.

[worthy lichen]

Is (-1,4) - {1} equivalent to x belongs to (-1,1) union (1,4)?

[vale wigeon]

erase the words "x belongs to", then yes.

[worthy lichen]

Bruh

Concept isn't clear

[vale wigeon]

you can say either of two things:
\begin{enumerate}
\item $x \in (-1,4)-{1}$ is equivalent to $x \in (-1,1) \cup (1,4)$
\item $(-1,4)-{1}$ is equivalent(/equal) to $(-1,1)\cup (1,4)$
\end{enumerate}
either write $x$ in both or drop $x$ in both but don't mix

[ocean seal]

Ann

[vale wigeon]

did you mean this as a statement ("The concept isn't clear to me.") or as a question ("Bruh, is the concept not clear to you anyway?")?

[worthy lichen]

To me

[vale wigeon]

ok, then let me tell you: there is no concept. it is only a style thing.

[worthy lichen]

I want to learn this trick

[vale wigeon]

what trick?

[worthy lichen]

Style

[vale wigeon]

style is not a trick

[worthy lichen]

Actually the format

[worthy lichen]

Which option match this and why?

.

[vale wigeon]

your answer is equal to (-1, 4) - {1}

i already told you this earlier

we can repeat this 20 more times though if you want

your answer $(-1,1) \cup (1,4)$ is equal to $(-1,4) - {1}$, which is option C.

[ocean seal]

Ann

[worthy lichen]

.close

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[sick tangle]

i need help on this question, why dont they use positive value of cos0

[lunar saddle]

Cos can be negative

oh wait

ok

They want BC to be longest right

[sick tangle]

yes

[lunar saddle]

if the cosine is positive then BC is not the longest

[sick tangle]

ah okay thank u so much:D

[lunar saddle]

becos cosine is negative when the angle is obtuse, then BC is the longest side of the obtuse

[lunar saddle]

np

[sick tangle]

.solved

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[vestal shell]

.reopen

[verbal gyro]

If I have a polynomial of degree 2 which attains the value a at x1 and another polynomial of degree 2 which attains the value a at x1, can we show somehow that they have the same derivative b at x1?

[vale wigeon]

nope.

[vale wigeon]

too late. you can only reopen before your channel releases. you need to get a new one now.

!occupied

[lone heart]

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

[verbal gyro]

I somehow want to show that a basis spline of degree 2 has equal derivatives at the knot

[vale wigeon]

<@&268886789983436800> exam cheater

or troll idk

[verbal gyro]

do i have to open a new channel? hahah

[vale wigeon]

no

it was yours to begin with

[vale wigeon]

mmm might need some more details for that

do you have a formula for the splines you're working with

[verbal gyro]

i think i want to show this relation:

If a knot has multiplicity miu, then the continuity there is C^(degree - miu)

i.e. this

though I am not sure how exactly you can see it

[verbal gyro]

https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/bspline-property.html this formula (Cox and de Boor)

[lone heart]

@verbal gyro Has your question been resolved?

[verbal gyro]

<@&286206848099549185>

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@verbal gyro Has your question been resolved?

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[leaden fiber]

anyone know where i went wrong here?

[leaden fiber]

question was: In triangle ABD, AC = CD = CB. Let vector AC = u and vector BC = v. Use vector method to prove that angleBAD is a right angle

[winged pawn]

doesn't seem so, its the solution 🤔

only the white one though, idk what's being written in green (wrong)

[leaden fiber]

green was just my throught process

is this right?

[lone heart]

@leaden fiber Has your question been resolved?

[leaden fiber]

i mean

arent we told that CB = v vector

so like BC = v?

implying that CB = -v?

[rustic coral]

Read what you said

You’re saying that CB is both v and -v

[leaden fiber]

wait

i meant

isnt

vec CB = - vec BC

so like since BC is v right? then CB is also -v?

[rustic coral]

Yes

[leaden fiber]

So like, vec CB = -v because BC = v

is that wrong?

Idk ig my thought process was a little convoluted

[rustic coral]

Direction matters

Same magnitude doesn’t mean they’re the same vector

[leaden fiber]

Yeah that makes sense

[lone heart]

@leaden fiber Has your question been resolved?

[visual rivet]

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[carmine topaz]

i dont get why they used 4th one for lagrange in part (c)

[scarlet thunder]

Are you asking why they have f^(4)(c)? Or am I misunderstanding?

[alpine sable]

maths and physics is so nice

[carmine topaz]

yea im lowkey confused why they used 4 for n+1 and the terms

[alpine sable]

i like both

[unique dune]

P(x) is a third degree polynomial so u use 4th degree for the error bound

[scarlet thunder]

^

Your Taylor polynomial has degree n=3, as such, your n+1 becomes 3+1=4.

[carmine topaz]

but where did the (1/10)^4 come from?

[scarlet thunder]

Do you know the formula for the Lagrange error bound?

[carmine topaz]

is it the |x-a|^n+1

i get that 1/4! is 1/(n+1)!

[scarlet thunder]

partially, yes.

[carmine topaz]

idk where they getting that

[scarlet thunder]

The lagrange error bound has the form $\frac{f^{(n+1)}(c)}{(n+1)!}|x-c|^{(n+1)}$

[ocean seal]

Spuerst

[carmine topaz]

yea

[unique dune]

|x-c| = |1/10|

[carmine topaz]

where does that number come from

[scarlet thunder]

the question itself.

[scarlet thunder]

if you compare this to the sheet, can you see how they get $|f^{(4)}(c)|\left(\frac{1}{4!}\right)\left(\frac{1}{10}\right)^2$ ? by plugging in the numbers you know and are given?

[carmine topaz]

dawg 😭 so i js straight up use that?

[ocean seal]

Spuerst

[carmine topaz]

yea

[scarlet thunder]

oops, made a mistake in the latex. should be (1/10)^4, not (1/10)^2.

[unique dune]

since a = 0 and x = 1/10

yes

[carmine topaz]

why does mine look same

[unique dune]

but we want to exaggerate the error anyway

[unique dune]

i got it from that site too i think lol

[scarlet thunder]

yes.

[unique dune]

wait nvm

i got it wrong

x-a is correct

[carmine topaz]

where do they got max f value? do i js plug in to get it

[unique dune]

ok so

fourth derivative is 5^4 sin(x), right?

max of sin(x) is 1

so max of that is 5^4

[carmine topaz]

yea

[unique dune]

,calc 5^4

[ocean seal]

Result:

625

[carmine topaz]

oh i get it

so the method is find max f and use the 1/10 they give and set 1/4!

1/10 = |x-a| right

1/4! = 1/(n+1)!

[scarlet thunder]

yes

[carmine topaz]

ok

thanks for helping 🙂

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[upbeat pilot]

Don't know how to start

[vale wigeon]

ok so im just gonna paste the problem with the diagram as only one image so as to not let discord cropping get in the way of readability:

In how many ways can we color the points E-J in the figure below, each with exactly one color, if we have red, green and blue available and must not allow any three collinear points to receive the same color?

@upbeat pilot are you familiar with the inclusion-exclusion principle

[upbeat pilot]

Thank you

[upbeat pilot]

No

[vale wigeon]

sad

[upbeat pilot]

Should I be?

[marsh drum]

if i may

[vale wigeon]

well, i was going to suggest a route with it.

[marsh drum]

isnt there a rlly easy way to do it?

[vale wigeon]

there is?

do share, but don't spoil.

[marsh drum]

yes good idea

how do u do spoilers again?

[vale wigeon]

||these||

[marsh drum]

||like this?||

ty

[vale wigeon]

but like dont dump the entire solution

give a direction in which to go

[marsh drum]

||w.l.o.g let F be blue. this forces IJ and GH to be red and green. this forces E to be blue. IJGH are either red blue blue red or blue red red blue. then just multiply by three for all maps from (1, 2, 3) to itself with no fixed points|| <---- full answer

hint: consider the implications of fixing point F

[kind sparrow]

I only looked at that for all of 2 seconds, but are you assuming that all three colinear points must be different colors.

[marsh drum]

yep, thats part of the problem statement lol

[marsh drum]

.

[upbeat pilot]

Can you explain that, please

[marsh drum]

ok

whats your fav colour

out of red green blue

pick any, it doesnt matter which one u choose

[upbeat pilot]

yellow

[vale wigeon]

hooold on.

[kind sparrow]

I interpret the problem that if you have three points on a line, they can't all simultaneously be the same color, not that they must be colored RGB

[vale wigeon]

ok yeah jessica pointed out the problem

yeah like RRG along a line is legal

[marsh drum]

AH

[kind sparrow]

If you fix F to be a color and make that assumption, you don't need to do any math to know that the options are combinatorically impossible

[marsh drum]

oh that makes it a bit more complicated then my bad

[marsh drum]

how so?

[upbeat pilot]

Yes

[marsh drum]

it doesnt lead to any contradictions

[kind sparrow]

Those numbers are huge and I expect your answer with those assumptions to be small

[upbeat pilot]

(A) 279, (B) 324, (C) 405, (D) 450, (E) 648
These are the options

Guys, I think I have and idea. I know that for the lines HJE and HGF are 108 legal ways to color

Like, H is red, if J is red, there are 2 colors for E, if J is blue, are 3 colors for E and if J is green there are 3 colors for E, so there are 2 * 3 * 3 = 18 ways to do it

The same thing happens to HGF, so 18 * 2 = 36. H can be any color, so 36 * 3 = 108

Now I have simply to discover I possible colors, right?

<@&286206848099549185>

[reef coral]

but thats not an option

[upbeat pilot]

wdym

[reef coral]

108 is not an option in the question you sent back there

[upbeat pilot]

Indeed, that is why I have still to discover I possible colors

I am dumb, I shouldn't have multiplied lol

So the total possible colors for HJE is actually 24 because there are 3^3 - 3

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[rose hinge]

hey guys learning gtraph theory rn. so i get the gist of it; like lets say k = 2. just chooose some arbitrary vertex v1 and that vertex v1 is connected to two other vertices: v2 and someone else. then v2 is connected to v1 and some otyher vetex v3. but im not sure how to turn this into a rigorous proof. could i get a small hint?

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@rose hinge Has your question been resolved?

[rose hinge]

<@&286206848099549185>

[vivid anvil]

let P be the longest path in G, and use the maximality of P

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@rose hinge Has your question been resolved?

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[vast harbor]

how to prove that this angle is $\alpha$

[ocean seal]

<rajel />

[vast harbor]

\begin{align*}
&\text{we are given a circle and given that} \ \
&\alpha = \left(\vec{v'}, \vec{v_0}\right)
\end{align*}

[ocean seal]

<rajel />

[humble totem]

How do you do that

[vast harbor]

do what

[humble totem]

Latex formulas

[tardy stag]

you can do it with just the blue parts i think?

[vast harbor]

ask in #latex-help

[vast harbor]

interesting step

so the right triangle can help ?

[tardy stag]

assuming those angles are all right angles this should be pretty straightforward

[vast harbor]

yep they all right angles , but i still cant see it lol

[tardy stag]

redraw it without all the other stuff

and then figure out what angle M is

[vast harbor]

,, \frac{\pi}{2} - \alpha

[ocean seal]

<rajel />

[tardy stag]

sure yeah

[vast harbor]

the thing is how would i relate it to the one in the left

[tardy stag]

then examine the kite-looking quadrilateral on the left

[vast harbor]

i see it now thx

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[sharp stag]

is that the same as (f^-1)'(a) = 1/f'(f^-1(a))?

[buoyant saddle]

yes because x_0 = f^-1(y_0)

[sharp stag]

so how would I use this rule to Find (f^-1)'(3) when f(x) = x^3 + x +1?

I guess y0 = 3

so 3 = f(x)

[buoyant saddle]

yep

[sharp stag]

okay so I solve 3 = x^3 +x + 1 I'm assuming

then what?

[buoyant saddle]

solve for x

there’s a trivial solution to this

[sharp stag]

x=1

okay so knowing this what info do we have?

[buoyant saddle]

1/f’(1)

you solved for x_0

[sharp stag]

okay so I need derivative ig

3x^2 + 1

then just plug in?

okay so you get 4

we know f’(1)=4

so is the answer just 1/4?

betttttttt

thanks so much!

[buoyant saddle]

you’re welcome

[sharp stag]

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[potent mirage]

the emoji spam bro

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[tacit sierra]

why is the solution of a non homogeneous linear recurrence relation the sum of the general solution of the associated homogeneous relation and a particular solution of the non homogeneous relation? This seems very similar to how non-homogeneous solutions of a linear system are just displacements of the kernel, but I can't make the connection and it's driving me crazy

[raw linden]

Have you seen https://youtu.be/TXg0lajMdn4

[tacit sierra]

I haven’t, from a first skim it seems to talk about ODEs which I have seen compared often to recurrence relations. But I’m not familiar at all with differential equations and their solutions yet

[raw linden]

It’s more about what’s similar in different areas of math

And your linear algebra example is one of the examples

[tacit sierra]

I have studied group theory so I’m familiar with the concept of isomorphisms and homomorphisms. In fact I suspect there must be some such relationship btw linear systems and linear recurrence relations

[tacit sierra]

The video showcases the reason why this is the case for linear systems, which Im familiar with from group theory and the first isomorphism theorem, but I don’t know the homomorphism that would let me apply this same lens to linear recurrence relations

[raw linden]

The ODEs can also be solved by linear recurrence relations , just by trying power series solutions

That’s one way of getting your link

[tacit sierra]

I wish. Apparently the annihilator method and such are quite linked to this, but since I’m completely unaware of ODEs and their theory that approach is a bit too much for me atm

[raw linden]

Keep it in mind for later 🙂

[tacit sierra]

I will, and surely when I study ODEs I will come back to this and understand better. But rn I was asking for help from a viewpoint I could understand, which includes linear algebra, group theory and, if necessary, up to calc 3

[raw linden]

Ok, this is as far as I go. Good luck!

[tacit sierra]

Thanks for your time regardless

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@tacit sierra Has your question been resolved?

[lone heart]

@tacit sierra Has your question been resolved?

[rigid thistle]

a

[lost belfry]

that is true! i had this thought:
one may think of the following "linear operators" on sequences:
$\phi_{i}(a_k) = a_{k+i}$ for every integer $i$. Then A system of recurrences may be seen just the way you see systems of linear equations

[ocean seal]

fastrack_and_backtrack

[lost belfry]

if you have studied differential equation theory, it may ring a few bells too

[tacit sierra]

Unfortunately I haven't. But I've devised the following, if you would care to see if it makes sense to you

[tacit sierra]

it's similar to what you pointed out here

[lost belfry]

i have no problems in seeing 😄

[rigid thistle]

hi yall

[tacit sierra]

let $$ a_n = c_1a_{n-1}+c_2a_{n-2}+p(n) $$ be a non-homogeneous recurrence relation:
let h be the general solution to the associated homogeneous relation
let s be a particular solution of the relation.

If we claim any solution has the form h+s, then we could surmise that h must be the kernel of some homomorphism P by the 1st Iso thm

Hold up ima use tex cause this will be unreadable otherwise

[lost belfry]

im lost in the first line.. recurrence relations define sequences, so a_n would presumably be a sequence, not a recurrence relation

[tacit sierra]

sorry yeah

my bad

let me correct one sec

wth

ugh

[ocean seal]

Jake The Human

[tacit sierra]

$$P:f(n) \mapsto f(n)-c_1f(n-1)+c_2f(n-2)$$

[ocean seal]

Jake The Human

[tacit sierra]

if we consider P to map from the space of all functions of n to all functions of n (though I'm not so sure of this)

[lost belfry]

that homomorphism idea is right!\
Consider the vector space S of functions from $\mathbb{N}\rightarrow \mathbb{R}$. This is precisely the space of all real valued sequences.\
we can define the following family ${\phi_{i}}{i=1}^{\infty}$ of linear operators from $S$ to $S$:\
define $\phi{i}(f(n)) = f(n+i)$, i,e $\phi_{i}$ returns the sequence shifted by $i$ where $i$ is a positive integer.\
now let us note that a recurrence like $f(n) + c_1f(n+1) + c_2f(n+2) = p(n)$, like the one you have written can be rewritten as $(id + c_1\phi_{1} + c_2\phi_2)(f(n)) = p(n)$ so yes, the homogeneous solution $h$ in the kernel of homomorphism $id + c_1\phi_{1} + c_2\phi_2$.

@tacit sierra is this what you were trying to say?

indeed we see that the homogeneous solutions are the kernel of the associated homomorphism

[tacit sierra]

Yes, we agree on our choice of homomorphism, though I am a bit lost on why the need to define the linear operators $\phi_i$

[ocean seal]

Jake The Human

fastrack_and_backtrack

[tacit sierra]

so the homomorphism as you have presented it is a linear combination of linear operators

[lost belfry]

so that we can obtain the solutions to the homogeneous recurrence as the kernel of a linear combination of the phi_is

[tacit sierra]

Why do we want the kernel to be expressed as a LC of phi_is? Maybe to see that it is a vector space wherein vectors are functions acted on by linear operators?

[lost belfry]

the kernel wont be expressed as a linear combination of the phi_is

the kernel is a set, the phi_is are functions

[tacit sierra]

but I thought our kernel was a subset of the functions on n

?

[lost belfry]

im saying that the kernel of the homomorphism $id + c_1\phi_{i} + c_2\phi_2$ is exactly the set of solutions to the homogeneous recurrence $f(n) + c_1f(n+1) + c_2f(n+2) = 0$

[ocean seal]

fastrack_and_backtrack

[tacit sierra]

this homomorphism acting on a function f(n) I suppose. Then the kernel must be some set of functions, and I know (but I can't prove as of yet) that this kernel is a vector space

[lost belfry]

that just follows from linear algebra

the kernel of a linear map is always a subspace

[tacit sierra]

ah yeah true

this is a LT after all

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[candid dune]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[candid dune]

Dont know how to start thos

[potent mirage]

find the length of the diagonal first

that equals side length into root 2

from there u can find the side length and square that to get the area

[candid dune]

Whats the equation

[potent mirage]

ok take a general square of side length a

[candid dune]

Alr

[potent mirage]

can u find the length of the diagonal in terms of a

[candid dune]

Uh no

I dont think thats how i gotta do it tho

[meager dagger]

You've seen how to get the distance between two points right, with the square root

[potent mirage]

using pythagorus theorem??

[candid dune]

Ye

[candid dune]

Whats the equation cause i dont know the english terms

[potent mirage]

what equation

[candid dune]

Pythagoris theorem

[potent mirage]

a^2 + b^2 = c^2??

[meager dagger]

Ok, commence par calculer la distance entre A et C (la longeur de la diagonale AC), tu trouves combien ?

[candid dune]

Do yk thats for triangles right?

[potent mirage]

yes bro when u join the diagonal u get 2 triangles only no?

[candid dune]

Yeah i asked this

I said idk what the equation is

[potent mirage]

ok imma just let the french guy help u

[candid dune]

k

.close

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[meager dagger]

Tu as jamais vu la grosse formule avec la racine carrée ?

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[patent vale]

I made this problem

[patent vale]

it should look more square like i guess

but i didnt make the drawing scale