help-39

[earnest warren]

Idk, this problem sheet ( it'a actually an old exam paper) came out before they removed complex number

[dusty jungle]

oh

in that case

[earnest warren]

So maybe complex number is the intended solution

[dusty jungle]

in euler's formula

if you substitute x->-x

then you'll get two equations that let you write both sin x and cos x in terms of e^ix and e^-ix

[earnest warren]

Oh

I see

[vital crescent]

is linear algebra in your curriculum?

[earnest warren]

Yes? Idk, we have 3d geo, cross prod and stuffs but not matrices, matrices was removed with complex number

They love combinatorics now, so they feed us a ton of them instead

[dusty jungle]

bruh is this for olympiads or..?

[earnest warren]

Nahh, this problem is from talented engineer program , qualifying exam

[dusty jungle]

damn.. all this not even for an olympiad? cooked

[vital crescent]

i see i see

i meant more so like linear independence

[earnest warren]

I'm still in hs 😭

[vital crescent]

u could use something called vandermonde matrix (specifically that it has rank n) to solve it

but ig that not taught

[earnest warren]

I lost my chance so hehe

[dusty jungle]

ahh forgetaboutit

[earnest warren]

Okay I see

Thank god I do know a little bit of complex number

I learnt them for the sake of solving combinatorics problem

Using generating func

😭

[dusty jungle]

dissapointing cus complex numbers can be super cool, they arent just used for solving problems quick...

[earnest warren]

Idk, I hope I will like complex number in the future

Cuz idk how I would be an EE major if not

Okay thx guys

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[buoyant cypress]

<@&268886789983436800>

sniped

[pearl pond]

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[iron basin]

i got the condition to be b1 + 7b2 = 13b3

[iron basin]

never mind

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[buoyant cypress]

blud grinding

good night

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[golden wave]

Confused on part b, can someone explain in general how to know whether the binomial will give an overestimate or an underestimate please

[left steppe]

where is C?

[golden wave]

oops mb i meant part b sorry

[left steppe]

can you say like

nvm

[autumn fossil]

i suppose that the ans for a was sth oscillating?

positive + negative + positive + negative?

[dire tapir]

is it not positive + negative + negative?

[autumn fossil]

maybe i dont wanna do the math

[autumn fossil]

there are 4 terms tho

[dire tapir]

yeah i meant positive + negative + negative + negative

[autumn fossil]

oh right

[golden wave]

this was my answer to part a

[autumn fossil]

yeah so as you can see, its positive + negative + negative + negative

[golden wave]

yes

[autumn fossil]

how do you think it'd continue if you actually wrote it till infinity?

just the signs, not the coefficients

[golden wave]

it would all be negative

[autumn fossil]

so P + N + N + N is ................ P + N + N + N + N + N + ...
a) underestimating
b) overestimating
c) cant say

[golden wave]

made a sligiht error here is the correct expansion

[autumn fossil]

yeah, the exact expansion doesnt even matter that much

[autumn fossil]

oh and i think it should be 9/4

[golden wave]

ah yeah oops

[golden wave]

c 😭

[autumn fossil]

and the other 2 are wrong too

[left steppe]

notice how it’s getting mote negative as you do more

[autumn fossil]

not quite

[autumn fossil]

yeah, as you add more and more terms, it keeps decreasing

[left steppe]

what does this tell you 🤔

[autumn fossil]

so what will happen when you add all the infinitely many terms?

[golden wave]

erm im not really sure sorry

[left steppe]

well let’s say i have 2 + -9/2

what does that equal

[golden wave]

so here's my thinkig

[left steppe]

also i think it should be 9/4

[golden wave]

yeh 9/4 i did it ages ago

[left steppe]

mb

[golden wave]

i quickly change it and uplaod correct version 1 sec

[golden wave]

this would be -2.5

or if it was 9/4 its -2.25

[left steppe]

next we do 2+ -9/4 - 81/64

[golden wave]

this would be 65/64

[left steppe]

wait oops

[golden wave]

oh

[left steppe]

mb

[feral sedge]

is x=1 here?

[golden wave]

that would be -225/64

[left steppe]

okay

[left steppe]

first: -2.25
second: -3.516

see how it’s decreasing

[golden wave]

yes

[left steppe]

so what does that tell you

[golden wave]

so going to inifinty it'll become more negative

[left steppe]

yes

and can sqrt(3) be negative

[feral sedge]

i'm confused what's going on

[left steppe]

^

[feral sedge]

to calculate sqrt(3) we should be using x=1/9, as the problem says, not x=1

[left steppe]

i thought the stuff from part A

[golden wave]

oh wait

i think i kinda get

so is it like if we compute our approximation

so sub in x = 1/9

then we would get whatevr that is

then from here clearly since out approximation is taking morw things away from it infinwtely to get closer to the true value

our approximation must be an overestimate?

and also

wanted to ask

what would happen if signs were alternating?

[pearl pond]

@golden wave Has your question been resolved?

[pearl pond]

@golden wave Has your question been resolved?

[golden wave]

😭

.close

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[indigo stone]

Any ideas?

[split void]

Am i tripping or is there an error in the question.

[lofty storm]

express it as 3^3 + 3^2 + (1/2)^3 then apply a^3+b^2+c^3 identity

[indigo stone]

Thats an olympiad question, might not be an error

[lofty storm]

my bad

[ashen ivy]

,w 27 + 9 - 1/8

[jolly parrot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

May 1, 2026, 4:14 AM

[split void]

It is VERY clear that 27+9-1/8 is WAY more than 20 itself

[ashen ivy]

this is what my teacher meant when she said i wont always have a calculator

[ivory basin]

...trolls?

[split void]

either this or you haven't sent the entire question.

[ivory basin]

<@&268886789983436800> hi take a look please

[dim linden]

none of the answers make sense

[earnest warren]

Is there anything else add to the context?

[indigo stone]

What will we get here?

[lofty storm]

bro it was a joke

[split void]

Well it wont make any difference.

[split void]

Please dont joke in help channels

[indigo stone]

Ok

[indigo stone]

Thats all

[split void]

Well I suspect that the question is incorrect.

[modern talon]

maybe the question meant 18 lol

[split void]

It won't do much good to hypothesize on the options, I'd rahter have OP consult his teacher.

[indigo stone]

Maybe theres a typo, but im not sure

[split void]

@indigo stone can you consult your teacher on this one?

[light helm]

Question is borked.
Skip and move on.

[indigo stone]

Thats 12 grade olympiad question

[split void]

Well its obviously incorrect. Close the channel and move on to the next question.

[indigo stone]

Ok)))

Thx anyways

[ashen ivy]

mistakes can happen in any contest
personally i would take some time to write out why this is impossible, it might get you some bonus points, but other than that .close

[dim linden]

reading the homepage of the olympiad website, its full of ai slop

[lofty storm]

is this some aptitude olympiad

[indigo stone]

Not really, most problems are pretty easy

[smoky musk]

Can I look at the entire sheet, including all the questions?

[earnest warren]

Maybe I'm too dumb

But yeah there's nothing else

[lofty storm]

if u use their powers though...
3^3 + 3^2 - (1/2)^3

the powers in sum gives 3+2-3 = 2... hence i asked if it is apttitude based.

[earnest warren]

It does not say anything so

[split void]

Well the question is incorrect; I'd love to say that this is really not leading anywhere, OP, you may as well close the channel.

[earnest warren]

Who knows

[indigo stone]

Alr, maybe theres a typo indeed

Thanks

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[visual ravine]

I want to show that in a field for any two elements a, b: a b = 0 iff a = 0 or b = 0

Somehow I struggle with these types of proofs. Have no real idea of what to do and why to do it. Can someone provide a more intuitive way of looking at this while still only argueing with the basic axioms?

[steep saddle]

is that a times b

[visual ravine]

yes

[steep saddle]

a * a^-1 = 1

[visual ravine]

ok I see what you are doing

the thing is

thanks first

the thing is

for some reason, I dont have a clear image of a field in my head

and it bothers me

maybe its just a matter of exercise

anyways, I guess I just have to try more

thanks !

.close

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[pearl pond]

Please don't occupy multiple help channels.

[buoyant cypress]

<@&268886789983436800>

[pearl pond]

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[buoyant cypress]

sniped again

mods are fast

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[pearl pond]

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[valid nova]

<@&268886789983436800>

[pearl pond]

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[royal galleon]

If I am trying to see what the upper bound of the fourth derivative of arctanx is on an interval than is there a better way than taking five derivatives to see its max?

[royal galleon]

The reasoning is for this question I have that difference in the inequality is simply the error of the taylor series for n = 3

so I want to say it is bounded by some constant

even then it doesn't satisfy the inequality which is weird

I on desmos the upper bound seems to be 4.7 but then if you have $\frac{4.7}{3!}\int_0^r(r-t)^3dt>\frac{7}{100}$

[jolly parrot]

BigBen

[royal galleon]

.close

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[ebon skiff]

[pearl pond]

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[daring bay]

how is this the same as x^-3 + x ^-2

[autumn fossil]

split it to 1/x^3 + x/x^3

[daring bay]

yeah so then we have 1/x^3 + x^-2

[autumn fossil]

and x^3 is x^-3

[daring bay]

oh

wait how do we get to the x^-2

again

x/x^3

[autumn fossil]

yeah, that is 1/x^2

you can cancel one x

and that is x^-2

[daring bay]

oh right

yeah

ty

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[daring bay]

hello can someone explain me the thinking behind this

[daring bay]

so they tried finding the point with the x axis

but im confused how he got to it

[verbal whale]

Points on the x-axis have y = 0 and z = 0

[past perch]

in 3d, any point on the x-axis is (x 0 0), similar to how in 2d any point on the x-axis is (x, 0)

you find the value of your parameter from known values on the point

[daring bay]

hmmm yeah

[past perch]

and use that parameter value to find your 3rd value

[daring bay]

oh so you find the value for when y and z is 0

basically

[past perch]

the value of t

yes

and then you use that t in your parametric equation for x

[daring bay]

and then put that into the line

[past perch]

to find your x

[daring bay]

but now

can a line have more than 1 points with an axis?

it can right

[past perch]

a straight line may only pass through a point once

given it is an ordinary straight line

[daring bay]

like it could go through x and y axis

[past perch]

yes but the point on the y-axis may be different
are you trying to imply that the line meets the x and y axes both at the same point?

[daring bay]

no like

a line can meet the x axis and the y axis right

or the y and z axis

[past perch]

yes it can

[daring bay]

then it would be on the plane too right

[past perch]

in the corresponding plane yes

[daring bay]

hm but that wouldnt change the process of finding that point right

like iin the picture i sent

same way?

[past perch]

yep always find your parameter using your known values

and then use that parameter to find your unknowns

[daring bay]

alright okay

ty

[past perch]

np <3

[pearl pond]

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

[daring bay]

wait if i get 2 different values for r

would that mean that there is no point that meets x axis?

[past perch]

yes

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

also if i wanna check if a specific point is on my line then how do ido that

[past perch]

plug all three values into corresponding parametric equations, find your parameters, if they match for all 3 then it's on the line

[pearl pond]

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[plucky shuttle]

how do i find the asymptotes of this function?

[pearl pond]

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[plucky shuttle]

This was during class, but i dont understand when it got to the limit part

i dont get why only for the term on the right, and not for the (x-5/3)^3

[sinful dagger]

is this ur homework?

[plucky shuttle]

yes

[sinful dagger]

oh ok

[plucky shuttle]

<@&286206848099549185>

[summer imp]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[summer imp]

The idea is that the factor on the right becomes essentially 1 for large x, so you can assume that the function will slowly look like sqrt((x-5/3)^3)

So you're not really taking the actual limit.

[plucky shuttle]

i see, im just stuck on how to find asymptotes in general because normally in poly/poly, i can just look at the degree or do long division

i dont really get how you can find asymptote using limits

[pearl pond]

@plucky shuttle Has your question been resolved?

[glad whale]

Is there another problem that you could look at? Becuase for the problem that you have above, you, in a way, just simplify the limit to find the asymptote.

[pearl pond]

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[inland laurel]

How to find solution of this equation without desmos. By either solving or graph sketching

3^|x| *(2-|x|) = 1

[bitter herald]

I think exact solution aint gonna be pretty

[wraith jacinth]

-# i am trying to think of a way to do this as simply as possilbe but i cant really find a good simple way tbh

[bitter herald]

But you can set u = |x| and try to graph 2 - u and 3^-u

Their intersection yields the result

Well

Approximate to how detailed your graoh is

[past perch]

you might be able to design an iterative formula

yeah you're certainly able to design an iterative formula

[inland laurel]

I don't know how to graph it

[bitter herald]

Anything works

u = 2 - 3^-u

Rinse and repeat

[past perch]

yep

[bitter herald]

Although its probably better to graph it dirst so it converges faster

[bitter herald]

2 - u is a line. Do you know how to graph linear equations?

[inland laurel]

Yes

[bitter herald]

Do so

[past perch]

or use the fact that for 3^u (2-u) = 1, since 3^u must be greater than 0, we can impose that 2 - u > 0 hence u lies between 0 and 2

[bitter herald]

True

[honest oyster]

Oh smart

[past perch]

(for OP, the lower bound of 0 comes from the fact that since u = |x|, u must be >= 0)

[inland laurel]

0&1 => 2 roots

[past perch]

more precisely because im pedantic

$u_{n+1} = 2 - 3^{-u_n}$

[jolly parrot]

sam beam

[past perch]

wdym by this

[inland laurel]

Nothing ignore it

[potent flume]

hmmmm, this one is hard

[worthy pine]

Plutonium

[wraith jacinth]

-# hi!

[pearl pond]

@inland laurel Has your question been resolved?

[wraith jacinth]

dont forget to close the channel if ur done!!

[quiet elbow]

Hi plutoooooo!!!!

[wraith jacinth]

Hi tangeerr!! how are you doing?

[pearl pond]

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[quiet elbow]

Good!

I’ll be asking for math help soooonnnn!!!!!

[pearl pond]

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[still dirge]

help pls 🙂

[wraith jacinth]

!status

[pearl pond]

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[wooden flare]

is it not just power rule from here (after simplifying integrand) or am i missing something?

is the top bound meant to be -1 as well as the bottom?

[still dirge]

Hello and thank you for your reply

I restarted the problem and I am stuck again

here

[wooden flare]

so your first integrating with respect to y, so pulling out the x term is a good start

since in the y world, 2x is just constant

[still dirge]

okay, so it stays like that?

[wooden flare]

hmm

i think the way its been written is misleading

i believe it should be more like this:

[still dirge]

here's my proffessor's solution:

I don't even know what's going on here

TBH

where did this come form

[wooden flare]

$\int_{-1}^{2} \int_{-1}^{1} 2x+ydydx \implies \int_{1}^{2} [2xy + \frac{1}{2} y^2] \bigg|_{y=-1}^{y=1} dx$

<@&268886789983436800>

that is terrible

okay so we first integrate the inner function right

2x is constant, and $\int 2x dy = 2xy$

[jolly parrot]

KB

[wooden flare]

and $\int ydy = \frac{1}{2} y^2$

[jolly parrot]

KB

[wooden flare]

then we evaluate these at the given bounds

[still dirge]

okay, let me try now

[wooden flare]

remembering we are subbing y not x (x comes in next integral)

<@&268886789983436800>

twice in like 2 minutes wild stuff

i've got my solution written out which has more in-between working out in case you wanted to have a look at that (after this)

[wooden flare]

hmm almost

ah ic

the 2x evaluated at -1 to 1

is meant to be 2xy and we sub in -1 to 1 in place of y

[still dirge]

why is 2x meant to be 2xy

[wooden flare]

if i have $\int axdx$ this becomes $a\int xdx$ right? by extension, if i had $\ \int 2xdy$ 2x is a constant in the y world so i can bring it out the integral thus obtaining: $\ 2x\int dy$ which is really $2x\int 1dy$ which is equal to $2x [y]$

[jolly parrot]

KB

[wooden flare]

which is why we get 2xy and not 2x

[still dirge]

ah, okay

so uh, do you have some cheatsheet for integration rules? all I have is this:

[wooden flare]

im not familar with ch, sh, and tg etc.

[still dirge]

I don't think we'll do those

[wooden flare]

are they non elementary functions?

[still dirge]

is this table enough?

[still dirge]

I don't know what you mean

[wooden flare]

hmm

its probably just the notation you use

since this is in another lenguage

ah yes, German?

[still dirge]

ah, no, just no

[wooden flare]

hmm nvm then, i've seen others use tg(x) for tan(x)

[remote aurora]

sinh(x), cosh(x) and tan(x)

[wooden flare]

oh interesting

i've never seen that notation being used for those

[still dirge]

So now that I know this rule I will try to solve this problem, in both ways. Be right back. Thank you.

[wooden flare]

as far as integrals go this should be enough, and I assume your comfortable with polynomial long division, partial fractions, u-sub and trig-sub etc. (just basic techniques of integration)

[still dirge]

messed up again somewhere...

[still dirge]

we just started these types of problems(double integration) and I did pass single integration last year... so... I think I know those

[wooden flare]

mhm thats good

[wooden flare]

as far as this problem goes, i believe the mistake is ^ ^

[still dirge]

why did you circle those? you said 2x becomes 2xy

[wooden flare]

mhm

but you replaced the int 1 dy with int y dy

[still dirge]

ok, I'll try again 🙂

ok great, now for x first, brb

[wooden flare]

ill quickly go through this 2m

mhm checks out

did you have another problem/question?

[pearl pond]

@still dirge Has your question been resolved?

[pearl pond]

Channel closed

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[quasi kernel]

Is this how it’s done pls

[light helm]

Why are both your acute angles 22

[wraith jacinth]

these angels are impossible..

[wraith jacinth]

also u need atleast 1 side to know how much all other sides are

[quasi kernel]

Wdym

[wraith jacinth]

what are u trying to solve for here?

[quasi kernel]

Question c5

[wraith jacinth]

oh-

[wraith jacinth]

this is not how its done-

[quasi kernel]

Aw

[wraith jacinth]

have u seen the unit circle before

by any chance?

[quasi kernel]

Kind of

[wraith jacinth]

okay so question what does cosine really mean? like when i tell you the world (cosine) what do uudnertand?

[quasi kernel]

Adjacent over hyp

[wraith jacinth]

okay in the traingle you drew which side is adjact and which is hyptonouse?

[quasi kernel]

Depends on the angle

[wraith jacinth]

(also i am gonna try to explain it to you throguh ur method first before giving you the formula)

((trig identites solve this very fast so if u want a quick solution ican just tell uthat but i thnk expaning this will be better))

[quasi kernel]

Sure

[wraith jacinth]

okay so here is the issue with the traingle u drew-

u drew on side as angle 22*

and u drew naother as 90

that means the last angle should be?

[quasi kernel]

70

[wraith jacinth]

why 70?

(although good job ur close)

[quasi kernel]

90+20

[wraith jacinth]

(but its 22 remeber)

[quasi kernel]

Answer minus 180

Ohhhhh

Mb

[quasi kernel]

68

[wraith jacinth]

now this look like it makes alot more sense right?

[quasi kernel]

Yes pls

[wraith jacinth]

(also for future refrence so udont fall into this always imagne you making the angle smaller until its 0 and niotice how the Side opposite to the angle is influnced by it)

https://tenor.com/view/triangles-sum-angles-geometry-geogebra-gif-19130014

[quasi kernel]

Wdym the side opposite the angle is influenced by it

[wraith jacinth]

here is an example-

[wraith jacinth]

wait let me show you something

[quasi kernel]

Ooo

I get it now

[quasi kernel]

It’s gotten longer bc the angle got bigger

[wraith jacinth]

YESSS!! well donee!!

now notice that the other angle! gets smaller and smaller-

right this is how stuff like sin and cosine works-

because we know that angels add up to 180

and that one of them is always constant

the two other angels have to share the 90 between them-

and because we know that the angles are related to the side length of it

we can simply say that the ratio of the side OPPOSITE to the angle we have right now Compared to The side opposite to the angle (90) which is longest side
Givus us a ratio of how much should go to that side

to expalin this better think the side opposite to 90 As kind of llike the big siblings of two twins
the big sibling always eats a full meal

and the two younger twins always have to share 1 meal!

so if wanted to desribethe ratio of how much food a twin got

[quasi kernel]

Can I go for gum, I feel like vomitting

[wraith jacinth]

yes are you okay-

please do go

[quasi kernel]

Back

[quasi kernel]

Yeah I am

Tyfa

[wraith jacinth]

i would say (how much food the twin ate) / (the total) now because of our scenario the total (which is the two twins added up) is also how much the big sibling ate so
to make it easier we just say twin A ate this much (angle) / how much big sibling ate (total)

and because of that we get sine which is

opposite (side to the angle which describes how much we ate) / hyptonuse and we dont need to specift how much cause hyptonus always has an angle of 90

does that make sense now where it comes from?

[quasi kernel]

Kind of

[wraith jacinth]

for now all you need to know that sin desctrive the ratio of how much that side is worth the total

this is why sin(90) is 1

because if u have two angles of 90 in a traingle it just becomes two straight equal lines
so that means that in this situation one side is as equal as the hyptonuse

here is an example

look at how when we make the bottom left angle as big as possible-
it the side start to get bigger and bigger until (the bottom side is pretty much nothing
and the two standing sides are equal

[wraith jacinth]

if utake a lookat the unit circle thisis what it describes

[quasi kernel]

I’m sorry but I don’t understand

[wraith jacinth]

okay look at these traingels-

notice how as the angle inthe center gets bigger they get taller?

[wraith jacinth]

the ones i highleited

[quasi kernel]

Yes

But which one is the angle in the center

Is it the top one?

[wraith jacinth]

these thtree angels respectively acording to each traingle

[quasi kernel]

Ooo

[wraith jacinth]

so now i want you to image what happens if we increase the angel all the way to 90 degrees what do uthink the traingle willbecome?

[quasi kernel]

Very big

[wraith jacinth]

thats right!

but notice the width of the traingle

what happens to it when we increase the angel?

[quasi kernel]

What’s the width pls

Is it the bottom part.?

[wraith jacinth]

yes!!

[quasi kernel]

It becomes smaller

[wraith jacinth]

YESSS!!

so ome really smart poeple along time ago discovered this

know do u know what a hyptonuse is?

[quasi kernel]

The side opposite the right angle

[wraith jacinth]

yes!

[wraith jacinth]

now ntoice something very cool here notice how every hyptonuse here starts from the Center of the circle And touches the edge

that means that all traingels i made have an equal hyponuse!

which is equal to 1

does that make sense?

[quasi kernel]

Yes

[wraith jacinth]

okay so the smart poeple along time ago figures this out

so they decided that since what matters most here is the angle

we can go ahead and write a function

a function that says this hieight is equal to maybe 1/2 Of the hyptonuse

for example

bassicly they mapped this fact and they said HIeght / hyptonuse Gives us that angle!

[wraith jacinth]

^

does this make sense?

[quasi kernel]

Yes pls

[wraith jacinth]

SO this is What sine means! really if u think about it

when you do sign 60

the numer it spits out is bassicly saying

if i was atraingle and my hyptonuse was = 1
the hieght of traingle would be equal to sin (60) whcih si equal to root(3) / 2

[wraith jacinth]

if u look around in this image u can see cordinates
theones on the left descibe the result of cosine and the ones on the right the result of sin

[quasi kernel]

Wha

I don’t under

[wraith jacinth]

oaky oaky

so u understand sine

but not cosine right?

[wraith jacinth]

^

https://tenor.com/view/matematica-gif-7322246

look at thsi gif

[quasi kernel]

Yes

[wraith jacinth]

okay so bassicly what cosine does is that it look at the OTHER angel

here let me show u

lets say we have this traingle

now if i did sin (22) it would give me the side opposite to my angel right (the side to the left)? as we explained earlier

[quasi kernel]

Yes

[wraith jacinth]

but waht if i wanted to get the width instead

also called adjacent angle

well i could DO 180 - (22+90)

and get that other angle

which as u said earlier is 68

[quasi kernel]

Yes

[wraith jacinth]

now if we do sin(68) we would get the width right?

[quasi kernel]

Yes

[wraith jacinth]

OKAY HERE IS THE COOL PART

this is WHAT cosine DOES

[quasi kernel]

It affects both sides except the hyp?

[wraith jacinth]

YESS

OAKY SO LOOKAT THIS

try on ur calculator

to do

sin (68)

and cos (22)

U WILL GET THE EXACT SAME NUMBER

[quasi kernel]

0.93

[quasi kernel]

Oo ur right

[wraith jacinth]

because that is what cosine is because these angels are related

all cosine really does

is that it finds the missing agnle

and gives u the side next to it!

this is why sin = opp/hyp

and cos = adj/hyp

does that make sense?

[quasi kernel]

Yes pls

[wraith jacinth]

okay so now we are in the final stretch

what is tan?

[quasi kernel]

OPP over adj

Just so uk, I have a math class soon

Not in a rude way

[wraith jacinth]

thats okay!

okay so we know that tan is opp over adj

which just means we are comparing sides

so if we know that one of the angels is (22)

how can we get both sides so we can do tan?

should i give a soltuion since u dont have time? @quasi kernel

[quasi kernel]

Do u mean the angles?

[quasi kernel]

Uh maybe

Soemtimes my teacher joins later

[wraith jacinth]

oaky here is the soluton

we know that the side opposite to us is = to sin (22)

and we know the side adjacent to us is = cos(22)

so tan = sin (22)/cos(22)

[quasi kernel]

Yes pls

[quasi kernel]

Oo

[wraith jacinth]

did that make sense?

[quasi kernel]

Kind of

No yes

It did

[wraith jacinth]

now this is just the udnerstanding of it

for future refrence when u need to solve this stuff

use this (image)

[quasi kernel]

Tysm

I’ll save it

Class started

Tysm Pluto

[wraith jacinth]

also search up

[quasi kernel]

[wraith jacinth]

TRIG idntites when u can it might help get all the rules!

this is just the basics!

GOODLUCKK U GOT THISS!

[quasi kernel]

Ty!

.close

[pearl pond]

Channel closed

Closed by @quasi kernel

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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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.

[acoustic tangle]

Is this working correct

[acoustic tangle]

How can the area be 0

[heavy onyx]

it is an odd function

[dawn stirrup]

area above x axis is positive, and below it is negative. for odd functions like these those two happen to be equal

so they cancel each other out

hope that helps

[pearl pond]

@acoustic tangle Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[quasi kernel]

I’m doing smth wrong but idk what

[quasi kernel]

Question 7

[verbal whale]

The issue arises from the painful notation 😬

[quasi kernel]

The wha

[lament dawn]

(small pet peeve but pls notate the multiplication dot fully filled in and if it's in degrees you should write the degree symbol!)

also, the last line is wrong, you included the negative sign inside the brackets

[quasi kernel]

Oh

Tysm

[lament dawn]

np!

[quasi kernel]

.close

[pearl pond]

Channel closed

Closed by @quasi kernel

Use .reopen if this was a mistake.

[acoustic tangle]

Hi

[quasi kernel]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[quasi kernel]

Oh

[acoustic tangle]

Dont you have to take the modulus of areas ?

[verbal whale]

There's no mention of that, so no

[acoustic tangle]

When are you meant to take the modulus I thought area was always positive

[quasi kernel]

How do I abandon this chnnel

[proper nova]

wdym abandon?

[autumn fossil]

using .close

[quasi kernel]

Someone’s using this

[proper nova]

that's you 🥀

[autumn fossil]

oh you cant leave it to other people

[dawn stirrup]

if the area is below the x axis, its considered negative and yes integrals can yield negative values too

but the area itself isnt negative

[quasi kernel]

Aww

[dawn stirrup]

thers nothing such as negative area, the sign just represents the direction

[proper nova]

!redir please

[pearl pond]

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

[proper nova]

please don't raid other people's channel

[acoustic tangle]

Oh ok

[autumn fossil]

yall should probably move to a new channel @dawn stirrup @acoustic tangle if yall wanna continue for a longer time

[dawn stirrup]

mb i thoguht it was closed?

uh

ig

and reopened too

[proper nova]

also please close because it got reopened

[wraith jacinth]

hello! sup? whats ur q?

[autumn fossil]

yeah, but its not yours rn

[dawn stirrup]

mb twin

[autumn fossil]

so you cant even control it

its easier to use your own channel

ill close this one and yall can make a new one

.close

[pearl pond]

Channel closed

Closed by @autumn fossil

Use .reopen if this was a mistake.

[dawn stirrup]

oki dogi

[proper nova]

maam's living under the rock 💀

[wraith jacinth]

i am hella confused

[dawn stirrup]

[quasi kernel]

Should I open a different channel

[proper nova]

you can always reopen until this thing closes for sure

[autumn fossil]

if you have a question, you can reopen this one

i thought you didnt have one, sorry, mb

[quasi kernel]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[quasi kernel]

I’m confused about the cos part

[proper nova]

@wraith jacinth

[lament dawn]

what about the cos part?

[wraith jacinth]

!status

[pearl pond]

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

[lament dawn]

in any case, sine and cosine have a period of 360deg (meaning cos(theta+k*360deg) = cos(theta) )

[quasi kernel]

I’m confused on whether it’ll become positive

Bc of the rule

[lament dawn]

you can reduce the degrees inside of the cosine function to be one lesser than 360deg, then use the unit circle to deduce the sign and reference angle

[quasi kernel]

Oh

Ty

.close

[pearl pond]

Channel closed

Closed by @quasi kernel

Use .reopen if this was a mistake.

[quasi kernel]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[quasi kernel]

I’m confused bc of the sinx

[bitter herald]

what happened in the denominator

[lament dawn]

hold on, what happened from the second last line to the last

[bitter herald]

why did sin(270) transform into -sin(x)

[quasi kernel]

Oh

-sin90

[bitter herald]

what is sin(90)

additionally what is tan(45)

[quasi kernel]

1 over 1

[lament dawn]

yes

1 - sin(x) all over sin(x) - 1 evaluates to?

[quasi kernel]

-1

[lament dawn]

there we go!

however, you have to exclude a general form for x

[quasi kernel]

I don’t understand

[quasi kernel]

Why this

[lament dawn]

if sin(x) - 1 is in the denominator, we have to exclude some x, or else the expression becomes undefined right?

[quasi kernel]

If we use the triangle will it be 2 over 2.

So 1

[lament dawn]

(tan45deg - sin x)/(sin x - sin90deg)

now, input the actual values of tan45deg and sin90deg

[lament dawn]

? what are you referring to here

[lament dawn]

uh huh

[lament dawn]

right, in any case

[quasi kernel]

Like this?

[lament dawn]

u missed an x, and no, it would be (1 - sin x )* 1/(sin x - 1)

[quasi kernel]

I’m still confused

[lament dawn]

which part confuses you?

[quasi kernel]

The process of getting it to look like what u said

[lament dawn]

yep, sure

[quasi kernel]

Ic

Tysm

[lament dawn]

but! you still need to exclude an x

[quasi kernel]

Ur I also need help w this

But number 10

[lament dawn]

whereas, a general form of x

[quasi kernel]

Wdym

[lament dawn]

since sin x - 1 is in the denominator, we cant have it equal zero, since then the entire expression becomes undefined

therefore, you need to exclude the cases of x where sin x - 1 = 0

[quasi kernel]

Oh

[lament dawn]

sin x - 1 = 0 means we have to exclude sin x = 1, what is the general form of x such that sin x = 1?

[quasi kernel]

I’m not sure

[lament dawn]

well, what is one example of x that gives sin x = 1

[lament dawn]

if you want to go onto this, sure but youd must exclude some general form of x btw!!

[lament dawn]

and please, notate with a degrees symbol if its in degrees (otherwise, it could be radians too), and make sure your signs dont spill out of the fraction and is distinctly separated, or else all of them could be interpreted differently and is very taxing on the eyes

[quasi kernel]

Ok sorry

I’ll do that for, one one

Now one

On

[lament dawn]

cos(360deg - 45deg) = cos 45deg, not -cos 45deg

[quasi kernel]

Oh but why

Ik it’ll be positive bc of the Cartesian plane but there’s a separated negative top

I thought negatives were bigger than positives

[lament dawn]

well, since 360deg - 45deg = 315deg, which lies on the fourth quadrant, which means the cosine of any angle there would be positive

[lament dawn]

i dont understand what youre saying, you might be callously conflating terms here

[quasi kernel]

Wha

[quasi kernel]

Yes but what about the negative taken out

That’s what I meant

[lament dawn]

huh? cos(-315) = cos(360-45), thats correct, theres no negatives taken out

[quasi kernel]

My teacher told me to take out the negative so it’ll be -cos315

[lament dawn]

thats wrong

cos(-x) = cos(x), not -cos(x)

that only works for sine

sin(-x) = -sin(x)

[quasi kernel]

Oh

[quasi kernel]

Then how would u solve this pls

With the -315

Would u add 360?

[lament dawn]

but havent you solved it? youve written cos(360deg-45deg) which then becomes cos(45deg)

[lament dawn]

ah, well cos(-315deg) just becomes cos(315deg)

since cos(-x) = cos(x)

so now, all you need to do is evaluate cos(315deg)

that youve (almost) correctly done

[dawn stirrup]

u can do that too, because going around full circle doesn't affect it at all, so its equal to cos(-315+360) which becomes cos 45 deg like wjs said

[quasi kernel]

I removed the negative first

U said that’s wrong

[lament dawn]

huh?

[quasi kernel]

I did -cos(315)l

And solved or

It

[lament dawn]

im really confused, but to clarify:
cos(-x) =/= -cos(x), which is why i said cos(-315deg) =/= -cos(315deg), so you cant "take out [a] negative" here
in your original solution, youve written cos(-315deg), in the subsequent line, youve written cos(360deg-45deg), that is correct, given cos(-x) = cos(x)
however, your mistake is in the next line, where cos(360deg-45deg) became -cos(45deg), that is incorrect, since cos(360deg-45deg) = cos(45deg)

[quasi kernel]

Oh ok

Tysm

[lament dawn]

no worries

[quasi kernel]

.close

[pearl pond]

Channel closed

Closed by @quasi kernel

Use .reopen if this was a mistake.

[sterile python]

https://tenor.com/view/reze-dance-chainsaw-man-silly-gif-3948498077150730027

[lament dawn]

STOP STALKING ME 💣 💥

[sterile python]

Potential helpful

[humble ferry]

my fellow hu tao enjoyer

[sterile python]

https://tenor.com/view/reze-reze-arc-reze-chainsaw-man-chainsaw-man-reze-chainsaw-gif-8800238631908387688

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

[long nova]

i dont understand how length of projection of AC onto u makes sense. Like what if u is shorter than AB, then if you project AC onto u, you wouldn't get the value of AB.

[earnest warren]

the projection of AB onto u can also be called component of AB along direction of u, so u here is just the direction

More like projection AB onto a line parallel to u

[long nova]

yeah but what if u is smaller in magnitude

then aren't you projecting a magnitude smaller than AB and thus not AB

[earnest warren]

Uh... a line is infinitely long

mag u doesn't matter here

What we care is the direction of u

component of AB along direction of vector u

[humble ferry]

it works cause projection only cares about direction, not how long u is
the formula divides by |u|, so even if u is short or long it cancels out
so you’re basically projecting AC onto the direction of AB, and since AB ∥ u, that gives the length of AB

the size doesn't matter

[long nova]

oh ok i see

thanks for the help @earnest warren @humble ferry i appreciate it

.close

[pearl pond]

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[pearl pond]

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[quasi kernel]

Can u pls help me

[ivory basin]

Girl why are you doing it that way

It says "show that this value satisfies this equation"

Just plug in the value directly

[quasi kernel]

Why tho

[modern talon]

it's not asking you to find all the solutions, just that 30 is a solution

"can you show that x = 2 is a solution to x + 2 = 4?" sure, 2 + 2 = 4

[quasi kernel]

Oh

If

Ty

[ivory basin]

Thanks blanket

[modern talon]

yw zavier

currently dodging my responsibilities in studying

[ivory basin]

-# and sorry skittles got distracted by other stuff for a moment

Aren't we all

[quasi kernel]

That’s ok

So for this I just input

[ivory basin]

Yeah

Just plug in theta = 30 and show that it works

[quasi kernel]

Tysm

Both of you

I gonna keep this open

[pearl pond]

@quasi kernel Has your question been resolved?

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[sharp smelt]

So firstly the mean

I'm lost completel here

I pressume i use a similar method to finding the MLE of Unif (0,t)?

and why is invariance needed here?

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[bitter herald]

<@&268886789983436800>

[pearl pond]

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[ebon skiff]

[pearl pond]

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[dusky wing]

Given X & Y (red & blue), how would you create an area formula for either shape? (second image is how I generated the top shape.)

[dusky wing]

Top shape is 3 sides, 2 of the length of Y (Blue) & 1 side is the length of X (Red)

Bottom shape circumference is 2Y (Blue) + X (Red)

[fathom pagoda]

do u mean what function has the graph in that shape of the thick balck lines of photo 1?

[dusky wing]

No, what function could create the Area for either shape, given X & Y.

I'll change the question to that

[rough stream]

Depending on which side you pick, go to the circle for the corresponding side, and use that formula

[dusky wing]

What, no the graph is only how the shape is created

[rough stream]

Sure! And x² is how a parabola is created

[dusky wing]

I dont actually know how the function is created, only that it is the intersection between 2 circles, above 0

[dusky wing]

Are you following what I'm saying?

I dont know the circle on either side, I only know X & Y

Is that not enough info to be able to get either shapes area?

Like let me put numbers to it:
Y (Blue) is 18
X (Red) is 4

[honest spear]

the second one is the area of a lens

u should be able to find something to read on it

draw two lines from the centre of the circle to the points where they intersect

and recall area of a segment

[dusky wing]

So do you think you could derive an equation for the area, given X & Y

[honest spear]

whats x and y here

[dusky wing]

X is Red, Y is blue

[dusky wing]

^

[honest spear]

incheresting

feel like its definitely some integral

[burnt thicket]

you can do it with geometry if you know the radii of the circle and then angles they seek out

the problem is finding that from the given info

[honest spear]

well i mean we are just given X and Y no?

[dusky wing]

Yes

[burnt thicket]

yeah

I think you have enough degrees of freedom but I’m not sure how you would do it without some serious bashing

[cinder flower]

[honest spear]

this is tuff

[dusky wing]

If I were to throw Z (radius) of the bottom shape (that would make the perfect circle, not the actual shape) it would be 5.

So, Lets say we now have:
X = 4
Y = 18
Z = 5

I would prefer an equation for both shapes WITHOUT Z but it might be absolutely needed

[kind rampart]

are you looking for a general formula for the area of the figure or just the area given the values above

im assuming its the former

[burnt thicket]

well I think the problem can be reduced to finding the angle sought out by a circle on a given arc

[sterile gazelle]

[burnt thicket]

but I’m not sure how to do that even though it feels like I should

[cinder flower]

[dusky wing]

general formula for both shapes, given X & Y variable... Possibly Z

[kind rampart]

if you have the radius and X i think i can make a formula for the bottom figure but the above one is out of my paygrade

[burnt thicket]

yeah I don’t think this is doable. if you cut a given arc in half (i.e. bisect the angle) and then make its corresponding circle’s radius two times larger the arc length is preserved, changing the area of the top figure

you need more info

[dusky wing]

Here I'm gonna make like a better image for the general problem.

[honest spear]

$A = \frac{YR}{2} - \frac{X}{2}\sqrt{R^2 - \frac{X^2}{4}}$

[jolly parrot]

Nyxzore

[honest spear]

but this would need the radius

but its the best i could come up with

[burnt thicket]

yeah that’s probably the best you can do

[honest spear]

I can't think of a way to just use X and Y

[burnt thicket]

pretty sure it’s impossible ^

[honest spear]

i mean you can probably 'prove' its impossible since X=2Rsin(theta) and Y=2R(theta)

then you can solve for theta

as Y/2r = theta

[burnt thicket]

I mean this formula also proves it since you get different areas even if you fix X and Y

[honest spear]

$sin(\frac{Y}{2R})=\frac{X}{2R}$

[jolly parrot]

Nyxzore

[kind rampart]

the Y here is wrong since i assumed it was the entire blue circumference but you can change that if you want

[dusky wing]

Not to say that the values help here IN ANY WAY

[dusky wing]

So, so far we've got this equation for the bottom shape?

[burnt thicket]

top shape is possible if you have the radii for the two intersecting circles

you can make an isoceles triangle in the middle and the two sector-ish parts surround it which isn’t too hard to find

[dusky wing]

I could technically get Any horizontal slice measurement of the lense, however I dont think that'd do us any good

[honest spear]

smth smth

A= X(sqrt(4Z² sin²(Y/(2Z)/2) − X²/4))/2 + Z²((Y/(2Z)) − sin(Y/(2Z)))

is my final submit dk if its fine

left as an excercise to the reader to check my work :/

[kind rampart]

just wanted to point out btw a circle with radius 5 has 10pi ≈ 31.4 units has its circumference which 18 + 18 = 36 > 31.4

[dusky wing]

I'm measuring off of furry textiles, so my accuracy might be off

😭

[pearl pond]

@dusky wing Has your question been resolved?

[burnt thicket]

you have all the formulas you need, no?

was there something else?

[dusky wing]

For the lense?

[burnt thicket]

yes here’s the method

[burnt thicket]

I believe this is the formula but I haven’t verified

it might be two times that?

[dusky wing]

How could it be, theres no Pi

[burnt thicket]

I’ll work it out if you want me to but Y is an arc length so the pi’s are built in

pretty sure it’s ||YR - R^2 * sin(Y/R) + (X/2) * sqrt((2sin(Y/(2R))^2 - (X/2)^2)||
if you see this, feel free to ping me

[pearl pond]

@dusky wing Has your question been resolved?

[dusky wing]

I'll just ask it when im putting it into practice tomorrow

[pearl pond]

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[desert girder]

im unsure whether or not i should take the integral from 1 to 6 in order to find the value of 6

and dont know how to find f(-2)

[pearl pond]

@desert girder Has your question been resolved?

[flat onyx]

did you do a variation table of f ?

[flat onyx]

what can you say about the integral of f' on the interval [-2,0] ?

[pearl pond]

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[wicked edge]

Note that the derivative is 0 for 2 points x = 0.5 and 5.

[wicked edge]

.close

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[sharp smelt]

very silly question, why can't a likelihood function be negative

[sharp smelt]

the PDF doesn't have to be continuous does it

so we can have for instance a function defined as the pdf of N(0,1) , when x≠0, and -1 when x=0?

[smoky gull]

.pin

[inland ivy]

there's never any practical reason to do this

strictly speaking your pdf can be negative on a null set

[sharp smelt]

sure, but the question is theortical right

Which is why I'm pushing the definiton to something that we'd never use, but still valid

[inland ivy]

i'm pretty sure likelihood is only defined for absolutely continuous distributions

at least normally

[inland ivy]

this is just a matter of definitions, how does your book define likelihood and pdfs?

[sharp smelt]

lemme check pdfs once more

[latent blaze]

Hi there, I'm currently trying to make a double pendulum simulation. The motion equations for the pendulum's angles are listed as such on Wikipedia with Theta1 and Theta2 being the first and second pendulums angles respectively. The problem is that they have a circular depdency on each other, and to simulate the pendulum I need to bring the differential equations in some form where I am able to calculate the first second order derivative of say Theta1, but don't know how to do so. Simply solving it like a linear algebraic system doesn't work because when I solve for Theta2DoubleDot and plug that into the first motion law then it just contains Theta1DoubleDot while I'm actually just trying to calculate Theta1DoubleDot. How do you solve for these second order derivatives based only on the angles themselves and their first order derivatives?

[sharp smelt]

okay, I see. I had missd this bit

Thanks!

[sharp smelt]

!occupioed

!occupied