help-39

[wooden flare]

they are

-2

[west sapphire]

correct

so the planes are parallel

[wooden flare]

oh that simple

would this be something similar then?

[west sapphire]

similar yes

in the "good" case, none of the planes are parallel

what would you expect the intersection to be in that case?

[wooden flare]

no intersections

oh wait

[west sapphire]

not true

[wooden flare]

i read that wrong

[west sapphire]

think geometrically

[wooden flare]

3

[west sapphire]

if two planes intersect, what does their intersection look like?

[wooden flare]

a line

[west sapphire]

right

and if a third plane intersects that line, what do you get?

[wooden flare]

the same line

[west sapphire]

no, try experimenting with like a pencil and a sheet of paper

[wooden flare]

wait how is it not just the same line??

[wraith jacinth]

-# maybe it might help to clarify weather they all intersect in the same place or diffrent places?

[wooden flare]

if it intersects that line, then would the line not be the intersection

[west sapphire]

well what's the intersection of the z axis and the xy-plane?

[wooden flare]

a point

[west sapphire]

yeah, i'm just considering a sort of "standard" case here

[west sapphire]

right

the "good" case is that you get a point

[wooden flare]

OH, you meant it like that

lemme draw up what I thought

[west sapphire]

there are situations where that doesn't happen

for example, what happens if two of the planes are parallel but not identical?

[wooden flare]

if this makes sense

[west sapphire]

yea, you can get that case too

[wooden flare]

the red line is the intersection of all 3 planes

[wooden flare]

they dont intersect

[west sapphire]

right

it turns out that there are really only 4 possibilities

for the intersection

[wooden flare]

none, one, two, or 3

[west sapphire]

either it's a plane, or a line, or a point, or empty

[wooden flare]

ok

so back to the question, how do i show that

it is one of these 4

[west sapphire]

well there are various ways to proceed, probably the most straightforward is to start by finding out how the first two planes intersect

and then see how that result intersects with the third plane

if you know some linear algebra then you can do this methodically using row reduction, for example

but if not, then just do it however you usually do it

[wooden flare]

oh hell naw, not that 😭

i had a headache yesterday doing a 4x4

[west sapphire]

what's the context, is this a linear algebra class?

if so then probably you should do row reduction tbh

[wooden flare]

uh, idk. Unit is 3D vectors and vector calc

[west sapphire]

ah ok

just find the intersection of the first two, however you would normally do it

eliminate one of the variables

[wooden flare]

never hurts to practice the matrix nonsense

[west sapphire]

yep, if you know how to do it then go for it

it shouldn't be too horrible hopefully, only a 3x3

[wooden flare]

yea

so ill just do my thingy (probs gonan be at least 5m if I dont get stuck)

and ill get back to u once ive finished

[west sapphire]

sounds good

if i'm not here i'm sure someone else can check

[wooden flare]

ok

-# ty for the helps (if you do end up leaving b4 i finish)

hmm

for my last row I've got {0 0 0 | 51/4}

idk what to make of it

[west sapphire]

hmm, can you show your work?

i get this ```
M =
1 -5 2 10
1 7 -2 -6
8 5 1 20

rref(M)
ans =
1.0000 0 0.3333 3.3333
0 1.0000 -0.3333 -1.3333
0 0 0 0

[wooden flare]

wha

[west sapphire]

shouldn't the 1 on the right be 10?

[wooden flare]

oh it should be too

i copied it down wrong...

[west sapphire]

ah bummer

luckily it only affects the last column

so you don't have to recompute everything

[wooden flare]

hm?

[west sapphire]

your calculations for the first three columns will still be fine (if you didn't make any mistakes there)

i mean when you change that 1 to 10

[wooden flare]

ill just go through it again

should be better this time

oh what is this

i got row 2 and row 3 to be {0 1 -1/3 | -4/3}

is that a problem?

[west sapphire]

no, but it can be simplified further

if rows 2 and 3 are equal then you can subtract row 2 from row 3

[wooden flare]

kk

oh so row3 should just be 0 0 0 | 0

hmm now the "hard" part imo

getting rid of the numbers in the top right of the matrix

[west sapphire]

yea can you show what you have so far?

[wooden flare]

yep 1sec

<@&268886789983436800>

[west sapphire]

so far so good

[wooden flare]

im pretty sure this is the answer

since this is what u have

[west sapphire]

yep it matches what i got numerically

[wooden flare]

so...

parameterization time

[west sapphire]

yep!

[wooden flare]

let z = lambda

ok done

so uh, what does my answer mean...

[west sapphire]

show what you got

[wooden flare]

kk

[west sapphire]

well you have one freely varying parameter, lambda

what kind of object is that gonna give you when you let lambda vary over all real numbers?

[wooden flare]

a line

[west sapphire]

yep

[wooden flare]

oh and thats our answer

[west sapphire]

yeah, depending on what details they want

like do they want a full parameteric equation for the line, or just "a line" as the answer?

btw, this means that you had exactly this sort of situation

[west sapphire]

ah nice

[wooden flare]

idk how they get that ima be real

[west sapphire]

you can get that form by multiplying your equations by 3

and rearranging

(and renaming lambda to t)

[wooden flare]

vector eq of a line is of the form: p = a + t(AB)

[west sapphire]

actually theirs isn't the same as yours hmm

oh, it should be the same, nvm

their direction vector is the same as yours

they just used a different anchor point

[wooden flare]

mhm

so uh, how do I get the line eq.

[west sapphire]

if you plug in lambda = 1 you get their point (3,-1,1)

so their t is equal to (lambda - 1)

well just taking yours:

multiply everything by 3, you get:

3x = 10 - lambda
3y = -4 + lambda
3z = 3 lambda

[wooden flare]

mhm

[west sapphire]

the right hand side in vector form then becomes
(10, -4, 0) + lambda (-1, 1, 3)

so that's one valid parameteric form

any scalar multiple of this also works

you can also replace (10, -4, 0) with any other point on the line

for example (3, -1, 1) as in their answer

[wooden flare]

ooh ok

[wooden flare]

but this is valid

[west sapphire]

yep assuming i didn't make any arithmetic mistakes, check my work 😆

it is 3am here haha

[wooden flare]

no it looks correct

[west sapphire]

ok cool

[wooden flare]

that makes sense

just annoying with matricies n what not lol

[west sapphire]

yeah, the usual grunge, pretty much unavoidable with this kind of question

when possible, always double check with software

so easy to make mistakes doing it by hand

[wooden flare]

yea

ty for the help on these problems

[west sapphire]

yw!

[wooden flare]

we start vector calc tomorrow, would you recommend pre-reading / solving it?

to get an understanding of it

[west sapphire]

if you have the time and energy, sure
but being well rested is also good

[wooden flare]

ive got at least 4h till i sleep

[west sapphire]

doing at least a skim ahead of time should be useful if you are able to

[wooden flare]

mmhm, bc its just parameterization for first chapter

stuff like ths

[west sapphire]

yeah, fairly straightforward but like in this problem, kinda tedious

definitely useful to get some practice

[wooden flare]

or i think its parameterization

[west sapphire]

yep that's what it looks like

[wooden flare]

these look horrid

[west sapphire]

definitely gonna be some algebra in some of these

"horrid" describes a lot of multivariable calculus haha

[wooden flare]

like... idek what t^2 i +t^4 j even looks like

[west sapphire]

the concepts are fine but the calculations can be disgusting

well if you had x = t^2 and y = t^4 , then x^2 = y

and since x = t^2, x is always >= 0

[wooden flare]

ohh

thats what it means

gotcha

[west sapphire]

so it's part of a parabola

[wooden flare]

(x)i + (y)j + (z)k

so if its like e^2t i its x = e^2t

[west sapphire]

yep, the i,j,k is another way of expressing the components of a vector
the vector (x,y,z) is the same as (x)i + (y)j + (z)k as you said

[wooden flare]

mhm ok

it cant be that bad then...

[west sapphire]

because i is just (1,0,0), j is (0,1,0), k is (0,0,1)

but they'll love to throw both notations at you (and maybe some other ones too)

[wooden flare]

ok, ill see what I can get through

ty for the help!

[west sapphire]

yw, good luck!

[wooden flare]

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[swift spindle]

can someone check this please? the answers use other forms so idk

[pearl pond]

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✅ Original question: #help-39 message

[smoky musk]

I think 13 b can still be simplified further.

[swift spindle]

ok thx

.close

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[smoky musk]

@swift spindle Wait, also you made a mistake in 13c

you forgot b in the denominator

[swift spindle]

wait hwat where

[smoky musk]

At the y=0 line?

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[brisk steeple]

What is the purpose of alpha > beta?

[woven matrix]

so you can know the sign of an

[brisk steeple]

Does it matter the sign of a_n?

[woven matrix]

probably not really, but when you give a name to two roots it's better to know which is which

otherwise a_n isn't properly defined

[brisk steeple]

Ohok.

Thank you.

.close

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[muted orbit]

<@&268886789983436800>

[pearl pond]

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[hoary relic]

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[royal galleon]

how could there be a solution on (-infity,+infinity) if cotx is not continuous on that intervall

[rough forge]

They mean to say that there is a solution y that would technically be defined on IR, but it would of course only work on subintervals, that excludes the poles

[royal galleon]

I don't see how we can go about finding this solution however, since our method relies both P(x) and Q(x) being cont

[rough forge]

All your solutions will be continuous on (0,pi) but there will be one solution that you could extend to a smooth function on IR, that's the point

[royal galleon]

ok Ill think about how to accomplish that

[rough forge]

What is your general solution, first of all?

[pearl pond]

@royal galleon Has your question been resolved?

[rough forge]

\prpl Ok so this rewrites to
[ frac{2b-cos(2x)}{2sin(x)} ]
and you wann figure how to choose $b$ so that essentialy you get rid of the $sin(x)$ in the denominator

[rough forge]

There is a trig identity

||1-cos(2x)=2sin²(x)||

[wild fable]

hm

[pearl pond]

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[royal galleon]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[royal galleon]

ok so we just use cos2x = 1-sin^2x and b = 1/2

[rough forge]

yes

[timber cape]

<@&268886789983436800>

[royal galleon]

thank you. Now I understand that we can use b to expand our solutions to certain domains

.solved

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[timber cape]

<@&268886789983436800>

[sharp delta]

Why are you pinging mods

[timber cape]

there was a nsfw server link person, nothing important for the current conversation.

[sharp delta]

Ok

[pearl pond]

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[royal galleon]

what is the point of all three of these intervals. The solution seems to be the same on all three

[royal galleon]

P(x) and Q(x) are the same and are continuous on all three of them. This means the method we use to find the solutoin won't change and then we will just have the same solution

[calm saddle]

hii

[pearl pond]

@royal galleon Has your question been resolved?

[rough forge]

I think each solution would have its own constant though, like they would be independent of each other. You wouldn't have a continuous solution necessarily, if you were to connect the intervals and extend the solution on the connected domain

[plush bramble]

By constant do you mean initial condition

[rough forge]

The constant after integration

[royal galleon]

but we are integrating from a to x. sure a can be different but wouldn't all the solutoins have a in the same place

[rough forge]

You'd have to use a different a, like a1,a2,a3 because on each interval a is forced to take a value that isn't contained in the other intervals

[royal galleon]

ok so literally the same solution I just rename constants. And what about b? Because we don't know if b will be the same or not

also when looking for a solution I have the integral $\int_a^x \frac{(t-1)(t-3)}{(t-2)^2}dx$ I want to use partial fraction decomposition so we have $(t-1)(t-3) = A(t-2)+B$ . I get A=-1 but that can't be right

[jolly parrot]

BigBen

[rough forge]

You'd need to do long division, partial fraction decomposition requires the numerator to be of lower degree than the denominator

Or not necessarily long division, I think if you expand terms, you can split the fraction simply

[royal galleon]

what about this?

[rough forge]

What do you mean by b

[royal galleon]

well for the we have the condition f(a)=b

b could be the same or it could be different. does it not matter though and we still name it b_1,b_2....

[rough forge]

You should name it differently because the condition can be different on each interval

[royal galleon]

wdym? aren't the conditions just that P(x) and Q(x) need to be continuous. If we are naming the inputs a_1,a_2 and so on woudln't we do the same with the outputs?

[rough forge]

I thought you mean IVP by condition?

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[royal galleon]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[royal galleon]

What is Ivp condition?

[rough forge]

Initial value problem

For example, from your family of solutions, you wanna choose the one that addictionally satisfies a constraint called IVP that is f(x0)=y0

[royal galleon]

Yes so that's what I'm saying so each a_n will have a b_n

[rough forge]

yes, that was also what I was saying

[royal galleon]

ok then. I understand now. thanks

.solved

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[worldly jewel]

intermediate value pearoem

[pearl pond]

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[wild fable]

so im trying to understand whats going on here, correct me if im wrong but basically depending on where the characteristic curve lands, the boundary or initial conditions there determine the solution we use?

[wild fable]

<@&268886789983436800>

[pearl pond]

@wild fable Has your question been resolved?

[dreamy seal]

Bro you solved

It

[wild fable]

did you even read my question

[dreamy seal]

Wait I am gonna solve it give me a minute

[wild fable]

im not asking how to solve it

i know how to solve it

[dreamy seal]

Exactly! You got it 100% right

If a characteristic curve hits the x-axis it picks up the initial condition g(x). If it hits the t-axisit picks up the boundary condition h(t)

[hoary relic]

<@&268886789983436800> AI

[dreamy seal]

Nah

You want me to see I splve it

[hoary relic]

No human types like that

You just edited it

[wild fable]

oh word im being trolled

or am i

chat is dmx cooking

[dreamy seal]

Because I don’t know good English I translated

[feral sedge]

I feel like 6 words isn't really a good sample size

[dreamy seal]

That I wrote

[hoary relic]

The message right after it has AI signs

[dreamy seal]

Yes because I translate my paper it was in Greek

[hoary relic]

Namely weird unicode characters that ppl cant find on their kbs normally

[dreamy seal]

You see what I say

[feral sedge]

Oh did they get edited out

fair enough then ig

[hoary relic]

Ye...

[wild fable]

am i being punked

[dreamy seal]

Bro ok listen

I translated my Greek

[wild fable]

ok

👍

[dreamy seal]

Bro you find it If a characteristic curve hits the x-axis if it touch t axis so x=t^2/2 is boundary condition of h(t) sorry for my bad English

[wild fable]

ya

i got it now

.solved

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[cobalt torrent]

givemeyourIPandbank.com

[cobalt torrent]

.close

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[prime pelican]

Does anyone know how to do this

Pls…

[cobalt haven]

I mean, this looks like Uni math

But I'm not sure

[woven matrix]

log it

you can also factor the 2^(something) on left side

[cobalt haven]

Wouldnt that make it more complicated

But idk, I'm a pre uni too

[prime pelican]

This is accelerated algebra 2 im in high school

[woven matrix]

you won't be able to solve it exactly, you can only prove that an unique solution exist

[west sapphire]

!original

[pearl pond]

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.

[prime pelican]

Thats what im trying to

[rich portal]

yeah so could you solve it?

[prime pelican]

No ive been trying do this for a while bc im self studying so i literally dont know what im even doing im tonna be honest but ive been working with logarithms i guess and one of my friends whos in acc algebra 2 sent me this to practice

[woven matrix]

that's more calc practice than algebra practice

[rich portal]

I can agree because theres 'e' and this looks tuff tbh for me

[woven matrix]

you need to see left side - right side as a function, so that you're just looking for when a function equals 0

f(x) = 6^(2x-4) - 2^(x+2) - e^(2x+4)

you're trying to prove f(x) = 0 has a unique sol

which will be true if you can for example show that the function is strictly increasing

starts negative and ends up positive

but it will require to do a full study of the function

[rich portal]

Is this a troll question in any sense?

Man you need so much stuff for an algebraic expression and shes just learning logs

[woven matrix]

well for a highschool student, I do expect it to take at least 30 to 45 min ig

unless you find a smart trick to accelerate things and avoid some calculations

I don't see how it's related to algebra 2 more than calc but it's doable if you did highschool calc

[prime pelican]

Im in 10th grade but im in geometry rn. Im just trying to study algebra 2 and pre calculus on my own so I can test out and take ap calc bc senior year but its also just for fun because i find math interesting despite the fact that im not exactly good at it i want to get better thank you sm for your help

[pearl pond]

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[pearl pond]

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[stoic imp]

any examples? I dont understand the theorem

or if we can chat about this a bit

[pastel umbra]

You can probably work this out from the theorem itself

If I_(n+1) is inside I_n, what can you conclude about a_n and a_(n+1), say?

[stoic imp]

unsure

[pastel umbra]

You have two intervals

And one is inside the other

What does that tell you about the relation between the left-sides of those intervals

(draw this out if you're still lost)

-# And for a prototypical example of the theorem at play, I_n = [-1/n, +1/n] is pretty easy to check

[stoic imp]

how?

[pastel umbra]

You do know how to draw an interval, right?

[stoic imp]

[------[----]-----]

[pastel umbra]

okay

If those four markers represent
a_n, a_(n+1), b_(n+1), b_n
respectively, what must we conclude about a_n and a_(n+1)?

i.e. how they relate to each other?

[pastel umbra]

But even if you don't want to answer that - and tbh it's not a requirement for understanding the theorem - suppose ALL your intervals were like that

[stoic imp]

an ----an+1----bn+1-----bn

[pastel umbra]

i.e. they're all nested inside each other

(if you like, you could stack them like a pyramid, then just collapse them all onto the same number line)

[stoic imp]

unsure

[pastel umbra]

(think "order")

[stoic imp]

i want to say that an is monotonically decreasing

but i pulled that out of my ass

[pastel umbra]

decreasing?

If I go from the nth one to the n+1 'th one, am I going UP or DOWN?

[stoic imp]

is this mathly correct statement

[pastel umbra]

(If I move at all, that is)

-# what you just said is barely even grammatical

If you go from a_n to a_(n+1), are you going to the right on the number line, or to the left?

[stoic imp]

right so an is monotone increasing

[pastel umbra]

Right

-# so why did you say decreasing

[stoic imp]

bn is monotone decreasing .

[pastel umbra]

yes

[pastel umbra]

Okay so imagine this

[stoic imp]

i am in the learning phase, I am allowed to make mistakes

[stoic imp]

but, why?

[pastel umbra]

wdym why

This is literally the theorem

[pastel umbra]

You have a bunch of closed intervals, nested inside each other

That's literally the FIRST requirement

[stoic imp]

why does lim (bn - an ) = 0

@pastel umbra

[pastel umbra]

The SECOND requirement is that the leng-

You can see that I'm typing, you do NOT have to ping me.

[pastel umbra]

The SECOND requirement is that the length of these intervals approaches 0

(one way of writing this is the limit-statement as written there)

The theorem then says that the infinite intersection of these intervals does exist (and namely as a singleton)

[stoic imp]

why lim length = 0

[pastel umbra]

Okay I'm just gonna lay out the rest of this then

[west sapphire]

if the lengths don't go to zero, you can hardly expect the intersection to consist of a single point, for example you could take I_n = [0,1] for all n

[stoic imp]

care to elaborate?

[pastel umbra]

We'll have that an is monotonic increasing and similarly bn is monotonic decreasing.
Further, we must have that all the a's are smaller than all the b's.
They must each have limits (say, A and B).
If they were not the same, then (it should be obvious that) the Interval-Limit would be [A, B] which... is still an interval. (why not A > B?)
If they are the same, then say A = B = x, then the infinite intersection of intervals would be...?

[pastel umbra]

Bungo's intervals don't satisfy the second requirement - it should be obvious what the infinite intersection is

[pearl pond]

@stoic imp Has your question been resolved?

[last flower]

No

[stoic imp]

i don't follow

[pastel umbra]

Care to elaborate?

[last flower]

No play minecraft instead

[pastel umbra]

Why don't you follow? What's tripping you here?

[stoic imp]

why do you say lim an converges to A

[flint basalt]

don't troll in help channels

[pastel umbra]

Because it's a monotone increasing sequence and it's bounded above

[last flower]

Einstein protests this rule

[pastel umbra]

You should recall something from Analysis about this

[pastel umbra]

The <@&268886789983436800> are less likely to
-# Suggestion: either a warning or a mute, at your discretion

[flint basalt]

if you'd like general conversations you can bring that over to #discussion . please don't do that while someone is trying to learn

[rigid sphinx]

can you stop bothering people who are trying to learn?

[radiant terrace]

Einstein should think about why trolling in the help channels gets in the way of people helping/receiving help

[rigid sphinx]

also, I get you're trying to help but backseat modding is also against the guidelines

[pastel umbra]

oh shit is it? mb

However we ask not to be rude about it and to avoid “backseat moderation”: making authoritative decisions about whether someone violated the rules and what punishment awaits them.
oh it is in the rules, apologies

[pastel umbra]

But yh @stoic imp this implies that an has a limit

[stoic imp]

it converges to sup

right?

what theorem is this or def?

[stoic imp]

.

[pastel umbra]

It's the Monotone Convergence Theorem

(This should be in your analysis notes)

(and it's a pretty easy one to remember)

[stoic imp]

its in section 4

but we are in 3.2

[stoic imp]

are you sure we need monotone convergence theorem for this? strange

@pastel umbra

[pastel umbra]

You can do it with MCT pretty easily

That said, do your notes there not have a proof of 3.1?

[pastel umbra]

https://es.wikipedia.org/wiki/Principio_de_los_intervalos_encajados Heck, even the Spanish Wikipedia article of this uses MCT to prove this pretty fast

-# tf is it with maths articles and embeds

[pastel umbra]

By notes I mean the PDF file that has your lecture notes on them

The screenshots you keep sending

[stoic imp]

the proof uses this proposition that if an is monotonically bounded then the limit exists or equivalently an converges

[pastel umbra]

But even so, there you go there's already a visual representation of the intervals

[pastel umbra]

...Yh, that's MCT

[stoic imp]

alr

[stoic imp]

right

why not A > B?

[stoic imp]

if A = B we are talking about a singleton

i guess

[pastel umbra]

Yes; that is what the theorem ultimately states

[pastel umbra]

the proof is left as an exercise to the reader

[stoic imp]

i am the reader and I need help doe

[pastel umbra]

Genuinely you can read the Wiki article I sent, it's there

The paragraph beginning

La prueba de este teorema es una aplicación del teorema de las sucesiones monótonas

[stoic imp]

i dont see no proof

[pastel umbra]

AHEM

[stoic imp]

i understand that if lim an = l1 and lim bn = l2
since by hypothesis we have lim an - bn = 0 then l1 = l2

but shit gets fuzzy after that

[pastel umbra]

Find the finite intersection of these intervals up to some K'th interval

That finite intersection will just be that K'th interval

Take limits

[stoic imp]

wat?

@pastel umbra

[pastel umbra]

Look, take these four intervals, I1 down to I4

Can you see that the intersection of these intervals - i.e. a finite intersection - is just I4, the yellow one?

[stoic imp]

but I want to use logic and math to prove it dawg

[pastel umbra]

Bruh this is basic logic

The yellow bit is inside ALL the other intervals

So it MUST be definitionally inside the intersection

Similarly if you pick a point not inside the yellow interval, it's not in the intersection; so the intersection must be inside the yellow interval

But if A \subseteq B and B \subseteq A, then double inclusion implies A = B

End of proof

[stoic imp]

yeah i need to add this proof to my notes

[pastel umbra]

y finito

[pastel umbra]

Okay, you're making progress - genuinely I want you to give proof writing like this more of a shot

yk the status thing here

!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

[pastel umbra]

This thing

A lot of the time you've been on step 1, and you wanna make sure you can get at least onto step 2 more frequently

[stoic imp]

say lim an = l1 and lim bn = l2
then by the hypothesis of the theorem we know
lim bn - an = 0 and so l1 = l2 = x for some x
then since mct tell us that lim an = sup(an) if an is bounded above and increasing sequence, we get that x = sup(an) = inf(bn)
so x >= an, and x <= bn forall n in N
so x in [an, bn] so x in In forall n in N

so x in the big intersection of In forall n in N

wtf does this proof though?

that x is in the intersection of all the intervals In

then we proof is the only one

[stoic imp]

okay I get it completely now

[stoic imp]

shit is tough tho, ngl

i need to fix my notes tho

ty for the help dawg

[pastel umbra]

Ultimately, this is a skill you need to learn, because that's what differentiates high school from university

You're not gonna get far without it

-# and without all the high school knowledge too, you still need that as well

[stoic imp]

.close

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[pearl pond]

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[wooden flare]

I am finding the Cartesian relation for r(t). But idk why my answer is wrong and also, how is the correct answer x= 1/4 (5-y^2) i don’t see how we get that as an answer

[dense jasper]

$(5-y)^2$ is not $5-y^2$

[jolly parrot]

Civil Service Pigeon

[wooden flare]

Yea ik

But that’s the answer

Idk how they get that

[dense jasper]

People make mistakes.

[wooden flare]

Ah was it a typo in the solutions

Ok yea that makes sense then, but why is my approach wrong

[dense jasper]

And what makes you say that?

[wooden flare]

Wdym?

My original approach with the square root

Y= 5-2sqrt(x)

[dense jasper]

$$y=5-2\sqrt{x} \implies 2\sqrt{x}=5-y \implies \sqrt{x}=\frac{5-y}{2} \implies x=\left(\frac{5-y}{2}\right)^2=\frac{1}{4}(5-y)^2$$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

I'm not following

[wooden flare]

Oh so it is fine like this then

Well, a pm would make it a bit better

But it’s fine to either be in y= (…) or x = (…) then

[dense jasper]

well yes ofc

[dense jasper]

a valid equation is a valid equation

[wooden flare]

Ok

Ty for the help

.close

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[dense jasper]

[wooden flare]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[wooden flare]

Got another Q gimme 2 secs to type it out

Why do both of these give me the correct “shape” but wrong domain

And how do I find the correct domain for these

.close

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[pearl pond]

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[rain vessel]

I'm completely stumped on the differential equation $(x-y)y'=x+y$

[jolly parrot]

greaterthan.333

[runic shoal]

this is a homogeneous equation

let y=ux (you can check x = 0 case later on)

[rain vessel]

I don't really see how I can get it into f(y/x) form

[runic shoal]

divide both sides by x

[rain vessel]

thanks

I think I got it

.close

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[vital hollow]

yllll

[pearl pond]

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[rich portal]

n = (tan a + tan x) / (tan a - tan x) . (1 + tan a tan b) / (1 - tan a tan b)

Anyone may use LeXit to make it easier to understand

[modest jackal]

Well

Tan(a+b)/tan(a-b)=n

And convert tan into sin/cos

Now try it

@rich portal

[rich portal]

ok

https://giphy.com/gifs/oooh-WuGSL4LFUMQU

@modest jackal tysm

[modest jackal]

Did u do it

[rich portal]

yeah man it was very easy

how did I not see that coming oof

thanks alot

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[pearl pond]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[warped violet]

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[thin root]

how can I start this...

I put A=B=C=π/6 it satisfies

[fathom oasis]

I would try simplifying sin(A+B+C)

[thin root]

Sin(A+B+C)=1

Sin((A+B)C)

Sin(A+B)cosC+cos(a+b)sinC

sinAcosBcosC+cosAsinBcosC+cosA.cosBsinC-sinAsinBsinC

[fathom oasis]

looks good!

[lament dawn]

??? why are you saying this here

get tf out

[fathom oasis]

I actually don't know if that was the right path to take or how to proceed

[thin root]

Ohh

[lament dawn]

leverage the idea that sin(pi/2 - x) = cos x and cos(pi/2 - x) = sin x

[thin root]

Yeah but couldn't see any path

[lament dawn]

okay, let's start over, i dont think using sin(A+B+C) is going to be helpful here

lets work from the original expression, recall that sin^2(A) can be written in terms of cos(2A)

right now, my motivation is to convert the first two terms sin^2(A) and sin^2(B) into forms of cos(2A) and cos(2B) respectively

i will leave sin^2(C) untouched because i want to only use the sum to product indetity purely on cos(2A) and cos(2B)

in doing so, i will get an expression somewhat resembling the form of (A+B) and what not, which i can use the idea that A+B = pi/2 - C

hope this provides more help

[fathom oasis]

awww

[opal vapor]

Why is this guy awing

[fathom oasis]

because my idea sucked

[timber cape]

happens to the best of us.

[lament dawn]

nono happens to the best of us

[timber cape]

!topic

[pearl pond]

Please read the channel description before posting, and stay on topic.

[lament dawn]

woah

jinx

[timber cape]

-# sniped

[fathom oasis]

Another idea would be to take the cosine of A+B+C and then maybe that could help eliminate something in here

[lament dawn]

differing degrees

at least intuitively it wouldnt work

[fathom oasis]

well there is the sinAsinBsinC term

which is present

and then after the possible elimination maybe you can take a square

but I kind of need to take a nap so I might just check in later if you all got it solved

[blazing mesa]

idea:

[thin root]

I guess same solution available over gemini

[blazing mesa]

the expression is symmetric so you could differentiate with respect to A and show that it equals zero

[blazing mesa]

then use this

oh yeah after setting C = pi/2 - A - B and keeping A and B free of course

feels like at that point you would just need to bash with sum formulas

because everything is the same degree i think

[thin root]

Thanks everyone

.close

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[timber cape]

a

wait

:|

[pearl pond]

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[lofty storm]

hello, i am going and i made certain equations but i cant proceed further

it is like i use, x+y= ab and i finally get x-y=8 which is already given.

[honest oyster]

ab?

[lofty storm]

it was an example, i prefer u provide the path

<@&286206848099549185>

[honest oyster]

Can you see how the side of length x can be divided into 2 line segments

[lofty storm]

after that?

[honest oyster]

Call the length of the longer line segment a, so the other one is x-a

[lofty storm]

okay then

[honest oyster]

We see the triangle circumscribing the circle has sides a, x-a and hypotenuse y

[lofty storm]

yes

[lofty storm]

now?

<@&286206848099549185>

[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.

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.

[proper nova]

@lofty storm

[lofty storm]

im here

[proper nova]

so all of this is a square

[lofty storm]

yes

[proper nova]

prob you didn't use a + b = x

[lofty storm]

i did a lot of stuff but, it just throws me back to x-y=8

[lofty storm]

i did

and hence i get, x+y= ab/4

nobody helped... i figured it out myself, thats helps even more fr.

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[pearl pond]

Please don't occupy multiple help channels.

[hoary relic]

<@&268886789983436800>

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[midnight haven]

how do I proceed

All you have left to do is sub in the x value they gave you

You're pretty much done

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[icy hornet]

i tried that approach

but that gives me y's to deal with

and thats not the form the equation is supposed to be

[pearl pond]

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[craggy pebble]

@icy hornet have you tried substituting the equation of the line

into the distance formula

to get rid of y

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[icy hornet]

can i then substitute p as x for the equation

so

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[fallen marten]

I'm wondering if I answered this question algebraically correct. What the questions wants me to find out is the "center angle" of a circle sector.

• The radius is 7 cm
• the area is 33cm^2
• the formula I used is at the very top of the calculation

[edgy stone]

no, as from the second step to the third you should multiply both sides by 360

so you would get 33 * 360 = v * pi * 49

and then from the third step to the fourth step similarly you should divide both sides by 49pi

[fallen marten]

oh shoot

thanks! I just found out what I did algebraically wrong xd

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[gritty citrus]

how do i solve this?

[light helm]

why do have a -sign multiplied to your 1/2

[gritty citrus]

im not rlly sure did this a long time ago haha

just reviewing now and have no clue

[light helm]

look up ftc

in simple cases like these
where the upper bound is x
and the lower bound is a constant
simply replace the t(s) in your integrand with x

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[foggy mason]

if their traces are the same

A and B are similar if B = P^{-1} AP
tr(B) = tr(P^{-1} AP) = tr(P^{-1} A) + tr(P) = tr(AP^{-1}) + tr(P) = tr(AP^{-1}P) = tr(A)

[mental hull]

ok

i might be able to do this

do you still need help

[foggy mason]

did you read my proof ?

if you want you can reduce the matrices, but they are already diagonal

[rough stream]

They are similar. They are both diagonalizations of some other matrix, where we chose a different order for the eigenvalues.

P =
0 0 1
0 1 0
1 0 0

@peak iris

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[mint scarab]

How else can I further set up this problem?

I initially wanted to say that the amplitude of the current is the amplitude of E(t) = 120sin(12t), but this does not yield the right answer. Why is this the case?

Alternatively, do I just try replacing q' with E(t) because current is dQ/dt (as if it's a given), find q, take its derivative to get current, and find the amplitude of q'?

Or do I just set q' = E(t), and solve for the amplitude of q(t)?

[mint scarab]

(Ordinary Differential Equations)

The current differential equation comes from Kirchhoff's Junction Law (the sum of the voltages around a circuit is 0).

.close

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[stable badge]

So, my thoughts...

Probability = No. of successful outcomes / No. of all outcomes

= Ways to choose 3 cards out of (52 cards - 1 ace of spades - 1 king of spades) / Ways to choose 5 cards out of 52 cards
= 50C3/52C5

Is this the correct method?

Follow up question:

[rough stream]

Yes that looks good

What assumptions does the binomial distribution make?

This should be in your textbook, but is definitely on Wikipedia

[stable badge]

My thoughts:

• Binomial distribution requires that each trial has the same probability
  • I suppose if you draw one card, then another, the probability changes

[rough stream]

One of the assumptions is that the trials (that would be, a draw) are independent

And yeah, each the same probability

They're not independent so yeah, binomial isn't useful here

(Drawing a favorable card reduces the probability of another favorable card)

[stable badge]

But if you group each drawing of 5 cards into one trial, each trial has the probability 50C3/52C5, right?

Then, all the trials become independent.

[rough stream]

Well, yes, and yes this would be binomial

But a binomial with only one trial isn't exactly what they mean lol

[stable badge]

I'm just wondering what they meant by "how could you change the scenario to fit a binomial distribution"

With replacement, is that what they want? Hmm

[rough stream]

That's exactly it. Once you throw replacement in, this becomes binomial

Worth looking up is the hypergeometric distribution, which is useful for things like drawing from a pile

[stable badge]

That's great, thank you!
Just out of curiosity, why is it unorthodox to lump everything into 1 trial, and then use binomial distribution? It would fit the criteria, right?
(Or you just use another distribution)

[rough stream]

The problem is that you'd just get p chance of success and 1-p chance of failure

You don't need to use a binomial model to get that info, you knew that already

Could you? Yes.
Will you ever? No.

[stable badge]

Understood. Thank you very much!

[rough stream]

Np! Feel free to ask if you have anything else

[stable badge]

All right! Appreciate it, will close now. 😃

.close

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[unborn lotus]

Hi! Could someone please help me with this math problem please? "Betty goes to the store to get flour and sugar. The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar, and is no more than twice the amount of sugar. Find the least number of pounds of sugar that Betty could buy." I have tried writing an inequality to be able to solve that, but any of the inequalities I tried to write made sense lol.

[pearl pond]

@unborn lotus Has your question been resolved?

[unborn lotus]

<@&286206848099549185>

[knotty thistle]

What inequalities have you come up with?

[unborn lotus]

s= sugar f= flour 2s<f6>s/2

[knotty thistle]

not quite. So 1. at least 6 more pounds of flour so $s/2 +6 \leq f$ and no more that twice the sugar so $f \leq 2s$

[jolly parrot]

glittersparkles

[knotty thistle]

so $s/2 + 6 \leq f \leq 2s$

[jolly parrot]

glittersparkles

[unborn lotus]

OHHHHHHHHHHHHHH OMG RIGHT

THANK YOU SO MUCH! Have an amazing day!

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[short grove]

I'm trying to prove: Suppose M is a module with a(n) composition series; 0<N_1<N_2....N_{n-1}<N_n=M. Show that M has ACC and DCC (IE is Noetherian and Artinian) - I tried looking this up and the proofs all include some reference to Jordan-Holder, which was not covered in my class and thus not allowed. My attempts so far have been induction. The base case is obvious - (N is simple) and the inductive step looks promising, but I cannot garuentee any member of my chain is even related to any of the N_i's - so I'm stuck.

[short grove]

https://en.wikipedia.org/wiki/Composition_series

[pearl pond]

@short grove Has your question been resolved?

[short grove]

no

[short grove]

I'm stepping out for a moment to walk my god.

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[pearl pond]

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[empty canopy]

so im stuck on a quadratic equation. i currently have -9 +- Squareroot of 9^2-4(2)(8) divided by 2(2) I got 17 under the square root but i dont know where to go from there

[open mist]

keep it at sqrt17

[empty canopy]

gotcha but im confused as to what to do next

i am left with -9 + Sqrt17 Divided by 4 and -9 - sqrt17 divided by 4

i dont think i can add or subtract 9 to 17 since 17 is under the square root symbol

[open mist]

correct, you cannot

it doesn't need to be simplified further from there

unless you want to put it into a calculator and get a decimal approximation

[empty canopy]

that would be nice but the answer sheet says -5/2 , -2 so there is a method but im just super confused

[open mist]

i'm not sure the entire question

[empty canopy]

(2s+7)(s+1)=-3

[open mist]

your numbers are wrong

[empty canopy]

probably lol

[open mist]

your c value to be specific

[empty canopy]

omg i added and did not multiply. God its been a long night haha thank you

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[open mist]

don't forget you have an equation equal to -3 and you need it equal to zero in order to use the quadratic formula

[empty canopy]

yea thanks mate

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[stuck junco]

hello

[stuck junco]

can anyone help to solve this

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[idle lintel]

can someone help me

[pearl pond]

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<@&286206848099549185>

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[languid phoenix]

i have to calculate the eigenvectors and eigenvalues of a matrix

[languid phoenix]

this is the matrix

[pearl pond]

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[languid phoenix]

<@&286206848099549185>

[languid phoenix]

ok 😦

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[clear moon]

.reopen

yoo i'm too late lol

@languid phoenix let's do thisss

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[languid phoenix]

yea?

[languid phoenix]

.reopen

@clear moon

[clear moon]

this is gonna suck but let's compute the eigenstuff of a 3x3

it's just tedious that's why noone answered prob

do you know how to do it?

i mean where to start with computing eigenvalues?

[languid phoenix]

well, i have to find a det

of the matrix - lambda * I

[clear moon]

yep!

[languid phoenix]

i did something, but im not sure thats alright

because i got stuck

[clear moon]

ok so far we do det of this:

| a - l | b   | b   |
| b     | a-l | b   |
| b     | b   | a-l |

[languid phoenix]

yeah

i calculated the det as well

[clear moon]

ok good lemme catch up then haha

[languid phoenix]

its (a-l)^3-3b^2(a-l)+2b^2

but, after that, im stuck

first year of college here in romania, we are going crazy with maths

and im kinda behind since we didnt really do this stuff in highschool

[clear moon]

cool, no worries!

ok i have the det

(a-l)^3 - 3b^2(a-l) + 2b^3

i think you didn't get the b^3

[languid phoenix]

i did

[clear moon]

oh you had it earler but then wrote it wrong in the end

[languid phoenix]

yeah

[clear moon]

well anyway from here you just solve for lambda

kinda sucks we have a cubic

[languid phoenix]

how

do we solve it

[clear moon]

i assume we have to solve this by hand?

prob rely on this being a depressed cubic

https://en.wikipedia.org/wiki/Cubic_equation#Depressed_cubic

or maybe we try to factor, but i don't see anything obvious

[languid phoenix]

tried to factor

[clear moon]

yeah same lol

currently trying to use this
http://www.sosmath.com/algebra/factor/fac11/fac11.html

[languid phoenix]

nope, doesnt work

[clear moon]

what do you mean?

[languid phoenix]

cant factor

[clear moon]

oh yeah for sure

[languid phoenix]

no program will

xd

thats sad

[clear moon]

i haven't pulled out sage, but that's a last resort

[languid phoenix]

and i have to send it at :30...

[clear moon]

almost done tho! this is the hardest part

[languid phoenix]

ive waited 2 hours for an answer but nobody would help

[clear moon]

sorry fam 😦

so we kind of have u^3 - 3b^2 u + 2b^3, with u = a - l

[languid phoenix]

yeah

[clear moon]

we want an s and t where
3st = -3b^2 and t^3 - s^3 = 2b^3

one solution will be s - t

the "real" solution

the other two are complex

i think s = -b and t = b work

[languid phoenix]

yeah

[clear moon]

so the real solution u = s - t is a - l = -2b so lambda is a + 2b 😄

[languid phoenix]

and u have another solution l1=l2=a-b?

[clear moon]

so we'd have to divide our determinant expression to compute the other two but i'm pretty sure they would be complex

we can look though

[languid phoenix]

ive got half of the answers xd

[clear moon]

do you know synthetic division?

[languid phoenix]

thanks a lot for helping, man

[clear moon]

hey of course

[languid phoenix]

yeah, i did the other part

[clear moon]

oh nice!

[languid phoenix]

but had problems getting that

and i wasnt sure about it

[clear moon]

it's kind of rough haha

gl with classes!

[languid phoenix]

thanks a lot man

😉

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[languid phoenix]

.reopen

[pearl pond]

✅

[languid phoenix]

@clear moon

also, could you help me

one more time?

[clear moon]

oh haha sure

i'll do my best

[languid phoenix]

what does this mean

[clear moon]

just the notation?

[languid phoenix]

yeah

[clear moon]

i think it's a shorthand of writing a big matrix

[languid phoenix]

ok

and how

at 4.2.6

and how do you reach the final form

it is already solved

im just asking

i have to explain it in front of the class

so, thats a shorthand of writing a big matrix

but how do the transformations work

u have to show

b a ```

when a and b are 2 matrix

are equal to the multiplication of the polynom of (A+B) and (A-B)

[clear moon]

ok lemme label these haha

[languid phoenix]

ill wait, thanks

[clear moon]

1. given det ...
2. shorthand (B is never on the diagonal so no need to subtract lambdas)
3. using gaussian elimination / row reduction, substitute row1 with row1 - row2
   https://en.wikipedia.org/wiki/Gaussian_elimination
4. not sure yet lol
5. definition of determinant
6. using some characteristic equation function

kinda stuck on 4, but i'll post this so you can take a look
for 4, pretty sure its another gaussian move but kinda hidden since we're abstracted away with matrices of matrices

actually looking at step 3 again, i think it's not quite so simple as subtraction, unless there's a typo with double negative on the identity matrix part

[pearl pond]

@languid phoenix Has your question been resolved?

[languid phoenix]

no, not yet bot

[clear moon]

is C in M_n(C) here referring to the complex numbers?

[languid phoenix]

yes

what is at 3

and at 4

?

[clear moon]

yeah steps 3 and 4 i'm pretty stuck

very non-obvious...

my first thought was row-reduction but it doesn't seem like that's the case

there's no relations between A and B, right?

[languid phoenix]

nope

nothing

[clear moon]

looking at this atm: https://en.wikipedia.org/wiki/Determinant#Block_matrices
also thinking there might be some diagonalization involved?

hmm actually no diag 🤦

i think i've run out of progress on this problem...
i'd definitely talk to your professor to see how that step is done
if you do or you figure it out you should lemme know haha
sorry i couldn't be more help 😅

[pearl pond]

@languid phoenix Has your question been resolved?

[languid phoenix]

I found the thing

it's -row2+row1

and c1+c2

thanks a lot

sorry

i was presenting the homework

thanks a lot man

[pearl pond]

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[wheat trail]

(I asked this earlier, but I'm still a little confused)
how does one integrate $$\qty(arctan(x-9.5) + 3.70)^2$$

$$ = \int arctan^2(x-9.5) + 7.40 \cdot \int arctan(x-9.5) + \int 3.70^2 dx$$

[jolly parrot]

amy.ames.aims

[wheat trail]

So to integrate the integral with the arctan^2 it would be an integration by parts

does this mean u would be arctan^2(x-9.5) or arctan^2(x)?

[pearl pond]

@wheat trail Has your question been resolved?

[pearl pond]

@wheat trail Has your question been resolved?

[plush bramble]

The first one. Why would it be the second one?

[wheat trail]

would u mind showing me how the arctan^2 integration is done; I'm entirely unsure as to how i should approach this

$$\text{le u = } arctan^2(x - 9.5)$$
$$du = \frac{2arctan^2(x-9.5)}{(x-9.5)^2 + 1}$$
$$v = x$$
$$dv = dx$$
$$\int^{4.20}_0 arctan^2(x-9.5) dx = u \cdot v - du \cdot v dx$$

[jolly parrot]

amy.ames.aims

[wheat trail]

$$= x \cdot arctan^2(x-9.5) - \int^{4.20}_0 \frac{2x \cdot arctan^2(x-9.5)}{(x-9.5)^2 + 1} dx$$