help-43

[eternal pulsar]

We were here.

[vivid raven]

schools only teach you to use the first two decimal digits: 1 and 4 to calucate stuff related to circles and the first 2 decimal digits of 22/7 is the same as Pi

[gaunt wave]

I take half of that because the area of one half of a petal is 1/4th of the circle - a half of 1/4 of the square

that triangle

[next fiber]

ah i see mb

[eternal pulsar]

The area for a triangle is 1/2 times base times height.

[gaunt wave]

Yeah

[next fiber]

yeah i didn't realize they were doing it by the triangles mb

[gaunt wave]

That makes sense because you get a half of the petal

and since there are 4 petals you multiply it by 8

[signal mist]

Wait…

So we can confer that the curve is 1/4 the perimeter of a circle, right?

[gaunt wave]

yeah

this is what they are intending you to think off

[signal mist]

So if we were to take the area of the petal, and one of the 2 unshaded areas of the quadrant, we would get 1/4 the area of a circle

[gaunt wave]

correct

[signal mist]

Moreover, if we found the area of the square, we could subtract it with the unshaded areas to get the shaded area

And times it by 4 to find the area of ALL petals

So Area_Sq - Area_US = Area_S

[gaunt wave]

how did you get the unshaded parts?

i think i am being dumb rn

[signal mist]

No, you’re fine

It’s a more intuitive process

[signal mist]

Using this, we can prove that Area_S + 2Area_USHalf = Area_Quad

Or Area_US Half = Area_Quad - (Area_C)/4

[gaunt wave]

yeah but how do i know what's the are of the unshaded parts?

[signal mist]

Therefore, find the area of the entire quadrant, then find 1/4 of the area of the imaginary circle

[signal mist]

…using this statement

@gaunt wave What is the area of a square equal to?

[gaunt wave]

one side is 14cm

so 196

[eternal pulsar]

Which area are you trying to find?

[gaunt wave]

the area of 4 petals

[eternal pulsar]

The shaded here?

[gaunt wave]

the inside of the petals

[eternal pulsar]

Okay.

[signal mist]

It’s okay, I’ve got it from here

Anyways

[eternal pulsar]

Then how will this work?

[signal mist]

@eternal pulsar Start here

[signal mist]

That’s for the entire square, divided into 4 quadrants.

196/4 =49

@gaunt wave Next, how do you find the area of a circle (Hint: Give me the formula and we’ll work it from there)

[gaunt wave]

since one side of the square is 14 cm

then the radius is 7

[signal mist]

Correct

@gaunt wave What next?

[gaunt wave]

144

[signal mist]

That’s radius squared 👍

[gaunt wave]

not really

[signal mist]

Oops

[gaunt wave]

that's the area of the entire circle

[signal mist]

Nope

Close, though

[gaunt wave]

doesen't matter

[signal mist]

Why do you say that?

[gaunt wave]

they ask you to use 22/7 so i just reduce 7/7 to 1

that's their implication

they want you to do that in this test

[signal mist]

Hold up…

[gaunt wave]

the are of the circle is 22/7 multiplied by 49

[signal mist]

Give me the formula of the Area of a Circle

Ok

What’s 22 * 7?

[gaunt wave]

OH I AM DUMB

154

[signal mist]

Correct

That’s the area of the circle

But to find the unshaded area, we need to use 1/4 of that value

[gaunt wave]

but how to we derive that conclusion?

[signal mist]

By using this

Wait

I updated the equation

[gaunt wave]

i don't quite get how do we get the unshaded half by removing 1/4 circle from the entire square

[signal mist]

@gaunt wave Go back to here

1/4 * Area_Circle = Area_Shaded + 1/2 * Area_Unshaded

[gaunt wave]

that statement is true

OH

i think i am getting what you mean

[signal mist]

Ok

So, Area_USPart = Area_Quad + (Area_Circle)/4

Let Area_USPart = X

X = 49 - 154/4

Once you find X, then 49 - 2x = The area of the shaded part per quadrant

[gaunt wave]

HELL YEAH

THAT MATCHES MY ANSWER

I'VE DONE THE CALCULATIONS

I get 112 either way

i did the system of equations

Unshaded + Shaded = 1/4 circle= 38,5

[signal mist]

That works

[gaunt wave]

2 Unshaded + Shaded = 49

Unshaded = 10,5

Shaded equals 28

There are 4 petals

112

[signal mist]

YES!!!

[gaunt wave]

Thanks man

[signal mist]

Good job!

Yw

To close the chat, type .close

[gaunt wave]

that's what i was trying to do. i just wanted to check if i was correct or not

.close

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[deft tangle]

Teacher says G function is not bounded in set S

Why?

[light hedge]

No help gang

Bro where are the numbers

Im not joking

[pine osprey]

cuz abs(g) isn't upper bounded

[light hedge]

I like maths but WHERE ARE THE NUMBERS

[stark marsh]

there won't be 🙂

[light hedge]

I'm still in grade 9 doing grade 10 add maths🙏

[pine osprey]

guys please #discussion

[deft tangle]

Bro stop spamming

🙈

[copper sierra]

they showed $|g(z)|=e^{-y}\app[y][-\infty]\infty$

[boreal girder]

ロケットジャンプ

[deft tangle]

e^iz so it will be e^ix,e^-y

[pine osprey]

yep

[deft tangle]

Yes

But when S is domain like -π<=x<=e^π

Can it be not bounded?

[copper sierra]

x is restricted not y

[pine osprey]

y in R

[deft tangle]

Opps

Thanks

🤭

[copper sierra]

np

[deft tangle]

Can I send you guys a request?

Just to be in friend list i will not DM

🤭

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[cedar stump]

When differentiating (taking a derivative) and in the question there is a variable called a/b/h/whatever and it is a constant do I treat it like a constant when taking the derivative

[kind viper]

if it is a constant then you treat it as a constant yes

show us the original question just to be sure though

[cedar stump]

Even if it's called X or Y

[kind viper]

!xy

[compact pewter]

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.

[kind viper]

let's just make sure there's no funny business there

[cedar stump]

I don't have a question a want to solve

Just couldn't find the answer on google

[kind viper]

ok then we cannot possibly answer

you need to actually look at the equation and its variables in context and what is asked of you

[rigid perch]

a constant is a constant no matter what it's called. so you have to use your judgement on whether a given letter represents a constant in the context of a given problem

[tribal iris]

if you're taking the derivative of a function anything inside that function that does not depend on the variable in respect of which you're taking the derivative is a constant

so unless they say something like a = sin(x) then you should just treat it as a constant

[cedar stump]

So let's say I'm taking the derivative of a×X³

Where a is a number

[tribal iris]

then a is a constant

you answered by yourself

[cedar stump]

The derivative is 3aX²

No funny bussiness

[tribal iris]

yep

[strange ermine]

Why not writing aX³ though? 😅

[cedar stump]

Habit

That's all of the question

Thanks for all ur help

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[runic nimbus]

How would we find OQ:QM? A friend sent me this

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@runic nimbus Has your question been resolved?

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@runic nimbus Has your question been resolved?

[humble mortar]

Imagine lots of lines from O to AB, and what the ratio the line AP splits into

https://www.desmos.com/geometry/c1apmoahfj

Can you see how everything moves proportionally

If M = B, then OQ:QM = OP:PB = 2:1. And if M = A, then OQ:QM = 1:0. So, when M is between A and B, the ratio OQ:QM is proportionally between OP:PB = 2:1 and 1:0

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[mighty gate]

given this expression, why does it simplify to this:

and not (-cosn+ xn) / (sinxn +1)

[stoic dove]

well you could just multiply the negative sign to the denominator see it yourself

[mighty gate]

but the numerator and denominator must be multiplied by -1

[stoic dove]

no thats not how it works, your saying -(-2/3)=(2/-3), see where your wrong?

[mighty gate]

yeah

[stoic dove]

so a negative sign should only be multiplied to either the numerator or the denominator

[mighty gate]

-(a/b) = a/-b

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[signal mist]

Pretty much

e^-inf approaches 0

And 2/-inf approaches 0

So…

[mortal elbow]

when do i get rid of the limit notation

cuz my teacher is strict abt it

[keen granite]

note that:

[L=\lim_{x\to -\infty}\frac{\frac{-4e^x}{x}+\frac{2}{x}}{\frac{3e^x}{x}+1}]

[boreal girder]

PajamaMamaLlama

[mortal elbow]

why we multiplying by 1/x

i thought it should be multiplied by 1/e^x

@keen granite

[compact pewter]

@mortal elbow Has your question been resolved?

[weary stone]

because then ur dividing by 0 if u plug -infinity in

[keen granite]

if we did we'd have:

[L=\lim_{x\to -\infty}\frac{-4+\frac{2}{e^x}}{3+\frac{x}{e^x}}]

Note that as (x\to -\infty) we have (\frac{2}{e^x}\to\infty) and (\frac{x}{e^x}\to-\infty) which in total gives us:

[L=\frac{-4+\infty}{3-\infty}]

or that too lol

[boreal girder]

PajamaMamaLlama

[weary stone]

yeah

u get infinities

diviidng by x gives a much easier ansdwer

[mortal elbow]

ok ty

[weary stone]

do u know what the definition of continuity is

[mortal elbow]

lowkey no

is it just

continues

[weary stone]

theres a more formal definition

f is continuous at a if $f(a) = \lim_{x \to a} f(x)$

[boreal girder]

shiru

[mortal elbow]

okay

[weary stone]

try using this to find the answer

since f is defined separately left and right of 0

to find the limit as x tends to 0, u need to find the left and right limits separately

if they are the same then you have a limit

then see if that limit is the same as just regular old f(0)

[mortal elbow]

so use 1-2sinx

how do i use e^-4x

[weary stone]

u have to use that to find the right hand limit

[mortal elbow]

so find the limit of both and if they are equal then the function is continuous at that point?

[weary stone]

no

find the limit of both

if they are equal then the general limit of 0 exists

and is equal to that number that is the limit of both the left and right functions

if that limit then equals the actual function value at 0, then it is continuous at 0

[weary stone]

this last step is needed cuz imagine this function

the left and right limit is at the empty dot but the actual function value is the solid dot way above

this function is therefore discontinuous

for this function, the left limit is at the solid dot but the right limit is at the empty dot

the general limit then doesnt even exist since the left and right limits dont match
this is also discontuinious

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[mortal elbow]

.reopen

[compact pewter]

✅

[mortal elbow]

r u sure its not continuous

[weary stone]

for this?

yes

i am sure its not contiunuos

[mortal elbow]

no for my function

the one i posted

[weary stone]

u have to figure that out

on ur own

i am here to teach u how to solve this question not to do it for u

[mortal elbow]

ok so the limt as x approaches zero of 1-2sinx is 1

[weary stone]

ok

and the limit of the other function?

[mortal elbow]

and the limit as x approaches zero of e^-4x is e^0 so also 1

[weary stone]

ok so the limit of the whole function as x approaches 0 is 1 since the left and right limits are both 1

now whats the actual function value at x = 0

[mortal elbow]

we use the 1-2sinx function since that part of the piecewise contains zero

so its also one

[signal mist]

Hi

Nvm y’all are good

[weary stone]

great

so we now have

left limit = right limit = function value

that means that the function is continuous

[weary stone]

by this definition of continuity

[mortal elbow]

ty

i got another question

[weary stone]

ok

[mortal elbow]

do i just factor the bottom

simplify

and then find the limit?

[weary stone]

did u try the first thing to try when u have a limit problem?

[mortal elbow]

substitute?

[weary stone]

yeah did u try that

u only have to do extra math if simple substitution doesnt work

[mortal elbow]

ya i got 1/6

oh ok

if i get like

0

does that mean something else

[weary stone]

if u get stuff like

0/0

infinity/infinity

0^infinity

u have to do extra math to try and resolve these problems

these are called indeterminate forms

0 is not an indeterminate form

its just 0

these are problematic

and they require u to try and resolve them

simply getting 0 or infinity is not problematic

thats just teh answer

anyway i gtg now sorry

[mortal elbow]

ok

ty

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[mortal elbow]

.close

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[buoyant knoll]

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✅

[buoyant knoll]

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[kind crane]

You could find y = f(x) using polar to Cartesian transformation

But I suspect there's a better way I can't think of right now

[buoyant knoll]

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[kind crane]

tan(theta) = y/x and y = rsin(theta), etc

[buoyant knoll]

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[kind crane]

dy/dx = dy/dtheta / (dx/dtheta) might be easier

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[buoyant knoll]

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[native isle]

woas

[buoyant knoll]

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[native isle]

jakaj

I like

[buoyant knoll]

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[native isle]

Better

[boreal girder]

Kenzo

[native isle]

woas

Again

[buoyant knoll]

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[native isle]

I don't understand

[next fiber]

on the right track

also just a reminder than it's r^2 when integrating

[buoyant knoll]

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[next fiber]

yep

1/2 integral r^2 dtheta

[buoyant knoll]

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[boreal girder]

Kenzo

[buoyant knoll]

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[next fiber]

yeah that should give you the intersection point

[buoyant knoll]

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[next fiber]

alternatively (assuming you're allowed to have practice doing it by hand), you could use a graphing calculator

to find the interseciton

[buoyant knoll]

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[next fiber]

though by hand is always good

[buoyant knoll]

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[next fiber]

what's wrong about it?

[buoyant knoll]

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[next fiber]

you do

[buoyant knoll]

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[next fiber]

theta is just a weird decimal in this case

[buoyant knoll]

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[next fiber]

alr so that's our point of intersection

what's the next step

[buoyant knoll]

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[native isle]

yeah

[buoyant knoll]

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[native isle]

Yes

ig

[buoyant knoll]

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[native isle]

well

[next fiber]

ok well we're looking for the common area

so let's try setting up an integral

[buoyant knoll]

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[next fiber]

missing the 1/2 and the squares

and almost

not quite

also for reference u don't add πn

you just use the decimal

polar functions have domains

[boreal girder]

Kenzo

[next fiber]

1st integral also has 1/2

[buoyant knoll]

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[next fiber]

yeah

[buoyant knoll]

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[next fiber]

ah

mb i missed that

[buoyant knoll]

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[next fiber]

ok you're almost there

[buoyant knoll]

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[next fiber]

not quite

dyk how a cos(theta) graph traces

or have abt this

im gonna give you theta values for cos

you tell me what they equal

[buoyant knoll]

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[next fiber]

cos(0)

[buoyant knoll]

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[next fiber]

cos(π/3)

[buoyant knoll]

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[next fiber]

cos(π/2)

[buoyant knoll]

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[next fiber]

cos(4π/3)

this is the last one

[buoyant knoll]

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[next fiber]

yeah

so basically i just made you plot out some points from r=cos(theta)

[buoyant knoll]

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[compact pewter]

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[slim osprey]

I’m having a slight confusion. The problem goes like this: we choose 3 points with uniform probability independently on a circle. What is the probability that after they form a triangle that one of their angles is greater than 90 degrees? I get that we use Thales theorem for this, and I understand part of the solution where we rotate the circle such that we can extend one of the points to form a diameter with both other points lying on one side, but here is where I’m confused. Why is the probability of the angle at one of the points not formed with the diameter 1/4, and why is the same as the probability of the point being formed with the diameter?

[slim osprey]

the problem is basically out of all triangles inscribed in a circle, what is the probability that the triangle is obtuse

Oh wait I think I get it now

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[upbeat bramble]

Hello! I’ve missed a lot of probability, and I was wondering how I should go about starting this question? Thank youuu!

[upbeat bramble]

I was wondering if Bernoulli comes in here at all? For part (a). I’m pretty sure expected value is what I need to use for part b.

Nvm I got it T_T

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[safe silo]

how to solve this

[main pendant]

^ Answered in #help-31

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[whole vigil]

what is the difference between a satisfiable set of wffs and a consistent set of wffs

[whole vigil]

@mortal scroll

[compact pewter]

@whole vigil Has your question been resolved?

[compact pewter]

@whole vigil Has your question been resolved?

[compact pewter]

@whole vigil Has your question been resolved?

[mortal scroll]

I assume you're talking about propositional logic?

Or are you talking about first-order logic

[whole vigil]

this is just propositional logic along with forall right

[whole vigil]

does the meaning change with it

[mortal scroll]

the basic meaning is the same, the details change

which one are you learning about right now

[whole vigil]

i asked chatgpt for a proof of compactness theorem for wffs and it was using the completeness theorem

but in the book the theorem was in the propositional logic section

[mortal scroll]

okay
satisfiability = you can assign each variable a value (true or false) such that all the wffs become true
consistent = there is no proof, using the rules of propositional logic, that shows that the wffs imply a contradiction

the soundness and completeness theorems tell you that these two conditions are equivalent for propositional logic

which is perhaps very surprising

[whole vigil]

by variable you mean the sentence symbols right

[mortal scroll]

any true implication, you can prove

[mortal scroll]

the propositional variable symbols

I'm not sure what you mean by sentence symbols

[whole vigil]

i think they are the same
like the symbols by applying the functions to which you can construct all wffs?

[whole vigil]

okaayy
so satisfiability is you start from the propositional symbols and try to make wffs true
and consistency is you start from the wffs, assume them to be true, and hope, with the rules of propositional logic, to not arrive at a contradiction?

[mortal scroll]

yeah

[mortal scroll]

yeah, satisfiability is about semantics and consistency is about proofs (syntax)

[whole vigil]

can you explain what you mean by semantics here? doesnt it not matter what the propositional symbols mean as long as they are assigned values such that the set is true?

[mortal scroll]

the true/false is the meaning

with consistency you don't ever talk about propositional symbols being true/false

you talk about syntactically manipulating them in various ways instead

[whole vigil]

ahh

[whole vigil]

like proofs

understood ty @mortal scroll

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[worn cape]

Can someone give me a brief rundown of how fields work? My teacher is not very good and he refuses to help outside of class.

[mortal scroll]

what class are you in right now?

[dull raft]

You might be better served just opening an algebra book

[worn cape]

Modern Algebra

I take abstract next semester

[mortal scroll]

ah

[worn cape]

I can't read very well so textbooks have never been that helpful

[mortal scroll]

the prototypical examples of fields are the rational numbers, real numbers, and complex numbers

[worn cape]

Are fields just sets then?

[mortal scroll]

basically sets where you're allowed to add, subtract, multiply, and divide objects

[worn cape]

So sets with closure?

[kind viper]

(except for division by 0)

[mortal scroll]

yes they are sets with operations with them

[kind viper]

"closure" is too unspecific

[mortal scroll]

and the operations have to satisfy certain axioms

the axioms make sure that the operations capture all the essential properties of addition, multiplication, etc.

[worn cape]

The commutative associative transitive thing?

[mortal scroll]

yes, although idk what you mean by transitive

so a field is a set F along with two binary opeeations on F, + and *

and there are a number of axioms, like for example a * (b + c) = a*b + a*c for all a,b,c in F

[kind viper]

transitive
that one doesn't belong

also it must be said

the field axioms essentially encode the basic rules of arithmetic that you are/should be already familiar with

[worn cape]

Yeah these look familiar from analysis

[woeful schooner]

https://en.wikipedia.org/wiki/Field_(mathematics)
I think this would be helpful

[worn cape]

I think most of our fields are regarding matrices for this class

[kind viper]

[worn cape]

Apparently we're doing Gilbert groups today

Whatever that means.

[mortal scroll]

do you have any questions about the field axioms?

take a look at the full axiom list on wikipedia

[worn cape]

Wow Google does not like that search. Started giving me real estate agents

[mortal scroll]

the link is right above in chat

[worn cape]

For the most part it seems understandable

I'm trying to study for this quiz on Gilbert groups in 20 minutes. I've been extremely sick for the past 2 weeks so I'm pretty behind

I'm still sick, but I can't be missing quizzes

[mortal scroll]

😔 that sucks

[worn cape]

Not when the grading is so wonky

[mortal scroll]

the prof won't let you move the quiz?

[worn cape]

He's not a very understanding person. Every move requires documentation, and I don't have any since I do not have health insurance.

[mortal scroll]

I hope you feel better soon

[young kestrel]

Yeah wtf is a gilbert group

Did the prof just name something after themselves ?

[worn cape]

What am I even looking at?

[worn cape]

I still haven't found out. My classmates don't understand it enough to give a description.

Bruh I'm dumb. Gilbert is the author of the textbook. It's just groups.

[mortal scroll]

Oh I see you're not just talking about fields, also about groups and algebras

[worn cape]

Additive inverse is they add to 0 and multiplicative means they produce to 1?

[mortal scroll]

yup

that allows you to do subtraction and division

[worn cape]

This looks like the big idea for this quiz.

[mortal scroll]

yeah that theorem is pretty important

it seems like your quiz is on groups then, not fields? maybe you should focus on groups

[worn cape]

Yeah not fields. Fields is next week apparently

How does this work? Order is like how many times you need to operate on an element to get to 1?

[mortal scroll]

yes, an example is the rotations of an equilateral triangle

if you rotate it three times, you go back to where you started

so the order of a 120-degree rotation within that group is 3

[worn cape]

That's a lot of letters.

[mortal scroll]

you should be able to draw those tables just from the descriptions given in that picture

[worn cape]

I know a cyclic group is just cutting the unit circle up into p sectors.

That's how it was described, anyways

[mortal scroll]

yeah, g could be a rotation of angle 360/p degrees

and then g^2 is applying g twice

and so on

g^p is then equal to the identity, so the order of g is p

[worn cape]

Heard. I should be OK for today then. Fields is our last unit then the final

I just need to figure out what an isomorphism is

[mortal scroll]

an isomorphism is basically a way of saying two groups have the same structure

do you know what a homomorphism is?

[worn cape]

Is I the identity matrix and J a rotational matrix here?

[mortal scroll]

yes

[worn cape]

Not yet

[mortal scroll]

J is a rotation by 90 degrees

you should think of aI + bJ as representing the same thing as the complex number a + bi

because multiplication by i is also a 90 degree rotation

[worn cape]

Oh so it's a hyperbolic number?

[mortal scroll]

complex, not hyperbolic

this is an example of an isomorphism, the 2x2 matrices of this form and the complex numbers behave the same way with respect to multiplication

[worn cape]

I don't really understand the purpose of this problem

There isn't really a problem stated at the beginning. He just does math.

Oh well. We starting now

[mortal scroll]

good luck!

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[maiden estuary]

How do I make a rational approximation for irrational numbers?

The easiest way is probably using decimals, for example.$\$

$\sqrt{2}=\frac{1414}{1000}\\$
However, I was wondering if there's a better and more accurate way to approximate irrational numbers. I'm curious because I came across the approximation of $R=\frac{25}{3}$ and calculations are a lot easier with fractions like this.

[boreal girder]

Gilsfen

[keen granite]

the infinite fraction of sqrt(2)

[maiden estuary]

I don't think I'm familiar with that.

[keen granite]

(\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\cdots}})

[boreal girder]

PajamaMamaLlama

[carmine garden]

You could use an infinite seqeunce that converges to \sqrt{2}

[keen granite]

if you cut it off at some interation then simplify you'll get a good fractional approximation

[maiden estuary]

But isn't this recursive?

[keen granite]

fun fact:

if you repeat this exercise for pi, you can recover the famous 22/7 and 355/113 approximations :)

[carmine garden]

Maybe the seqeunces $\sqrt{\frac{2n^2}{n^2+1}}$

[boreal girder]

What a wonderful world !

[carmine garden]

chose an n of your choice to appxomate $\sqrt{2}$

[boreal girder]

What a wonderful world !

[carmine garden]

but yeah, this is the best

[maiden estuary]

But that's specific to just √2. What about the other irrational numbers?

[keen granite]

,w pi continued fraction

[boreal girder]

Results provided by WolframAlpha

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Requested by xxmrfancyu2xx

[keen granite]

,w phi continued fraction

[boreal girder]

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[keen granite]

oh well golden ratio broke

but every number has a continued fraction :)

[maiden estuary]

I think I saw a similar series for pi. However the series followed an unusual pattern, not so clean like √2.

[keen granite]

indeed, they're not necessarily easy to remember, I just know sqrt(2) and golden ratio

bcz golden ratio is really nice:

[\phi=1+\frac{1}{1+\frac{1}{1+\cdots}}]

[boreal girder]

PajamaMamaLlama

[keen granite]

notice that if we cut it off at some random point:

[\phi_1 = 1+\frac{1}{1+\frac{1}{1}} = 1+\frac{1}{2} = \frac{3}{2}]

or a different one:

[\phi_2 = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}} = 1+\frac{1}{1+\frac{1}{2}} = 1+\frac{1}{\frac{3}{2}} = 1+\frac{2}{3}=\frac{5}{3}]

notice what if we continue we get the ratio of the Fibonacci terms, which is expected

[boreal girder]

PajamaMamaLlama

[maiden estuary]

Oh, I was thinking about
[
\phi=1+\frac{1}{1+\frac{1}{\phi}}
]

[boreal girder]

Gilsfen

[maiden estuary]

But this is a lot easier.

Anyway, I guess I'll just memorize this for commonly used irrationals.
Thank you very much!

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[urban glen]

for part c am i getting 1/2 from estimating based on 1/4 from the graph?

[molten coral]

its a point on f

s

so you can just calculate it

[urban glen]

wait i thoughts its like x=1/4 and its estiamting it might be 1/2 for y

i can plug and find it right?

[molten coral]

i dont think its an estimation

[urban glen]

oh ok

[molten coral]

1/2 comes from solving f'(x) = 1

this gives some x, which can just be plugged into f(x)

[urban glen]

if x=1/4 f(x) = y = 1/2

do i need to check g too?

[molten coral]

you could check that the solution is actually a solution, if you wanted

i dont think you need to

youd need to back out the equation for the tangent line to f at P

[urban glen]

i graphed it but it gives two different y value

[molten coral]

then make sure its actually perpendicular to g

[urban glen]

wait the point P is on f

[molten coral]

it is

[urban glen]

so i only need to conisder f right

[molten coral]

yea

[urban glen]

besides from getting x value

i get it

[molten coral]

that the line intersecting P and tangent to f will be perpendicular to g is by design

[urban glen]

yea

[molten coral]

its the first step in the problem

[molten coral]

we dont care about the intersections location, yea

[urban glen]

yea finding x that makes f'=1/g' and using that x to find the x,y of point P

i get it

thanks 🙂

.close

[molten coral]

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[shell pilot]

hiiii

can we ask q of 8th grade math??

[compact pewter]

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[thorny ridge]

guys

[thorny ridge]

I have a few questions

first

angles subtended by the same arc are equal

So MLP is 58 deg

yet because it is a cyclic quad its 180 - 58

Whats up with that

For this one I have some working

[kind viper]

angles MNP and MLP dont subtend the same arc

also uh

!1q

[compact pewter]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[thorny ridge]

Oh thank god

ok ok

can you explain one question at a time I have 3 questions

[kind viper]

angles MNP and MLP subtend not the same arc but arcs which add up to 360°

(ie a full circle)

[thorny ridge]

what does that mean

All I see is that the both come from points M and P

so that must mean they are from the same arc

and they both form an angle at the circumfrence

[kind viper]

here let me scribble on the diagram...

[thorny ridge]

fearful for my exams

everyday i discover new things

[kind viper]

MNP subtends the red arc and MLP subtends the green arc

red + green = full circle = 360°

[thorny ridge]

WHT

[kind viper]

that make it clearer?

[thorny ridge]

WHAT

I feel like I just broke the fourth wall

How am I supposed to know if they came from opposite arcs

Like the 360 deg things you mentioned

[kind viper]

...

by LOOKING?

i actually dont know how else to answer that lmfao

[thorny ridge]

Looks can be deceiving

[kind viper]

you have to be able to imagine like

standing at the vertex of one of these angles

looking out of it

and seeing what parts of the circle you have line-of-sight to

[thorny ridge]

Ok so its just a guessing game

You can probably deduce it

Thanks

[kind viper]

it is not a guessing game

[thorny ridge]

Can I ask my second question

[kind viper]

it is a "look at the diagram" game

also i am gonna say

[thorny ridge]

i cant wait to be like you

youre a gamer

and youre smart

im getting the second part real fast but

[kind viper]

when you have N ≥ 2 unrelated questions, it's better to close your previous channel and open a new one for each next question

so if we're done w this one i will say you should .close and take a new channel

[thorny ridge]

Ok see you or a helper in help 30

.close

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[craggy vigil]

this is so hard

[native shard]

lmao

which one

[worldly ledge]

sequence A (b) is in AP lol..common difference d = 4

[native shard]

you’re brilliant

[eternal pulsar]

Don’t do the work for them.

[eternal pulsar]

Which parts?

[chrome cedar]

Woah I just got part b

It's tricky lol

[compact pewter]

@craggy vigil Has your question been resolved?

[craggy vigil]

its easy but i forgot

what is the common ratio

inv

in b

[chrome cedar]

Look at 3 (a) for hint

It's something like that

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@craggy vigil Has your question been resolved?

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[young raft]

i am trying to understand what's going on here: https://groups.google.com/g/sci.math/c/ofkao7iugNg?pli=1

[young raft]

The family $\mathcal{ F }$ of open subsets of $[0,1]$ with measure < 1 has cardinality $\mathfrak{ c }$.
So it can be well-ordered in such a way that every member of $\mathcal{F}$ has fewer than $\mathfrak{c}$ predecessors. [why? no clue]
Since the complement of each member of $\mathcal{F}$ has cardinality $\mathfrak{c}$, there is a function $f: \mathcal{F} \to [0,1]$ such that for each $A \in \mathcal{F}$:
$$
f(A) \not\in A
$$
and
$$
A \not\subset { f(B)+ q: B \text{ is a predecessor of } A, q \in \mathbb{Q} }.
$$
Thus, the set ${ f(A) : A \in \mathcal{F} }$ contains at most one representative of each coset. Complete $V$ by putting in representatives of all equivalence classes not already represented. Then the outer measure of this $V$ is 1.

[boreal girder]

artemetra

[young raft]

question is – how?!

what makes the outer measure of V equal to 1?

i think i am lacking a visualization for what kind of ordering or whatever we are working with

[compact pewter]

@young raft Has your question been resolved?

[compact pewter]

@young raft Has your question been resolved?

[compact pewter]

@young raft Has your question been resolved?

[compact pewter]

@young raft Has your question been resolved?

[young raft]

I AM SO CLOSE

[woeful basalt]

check DMs

[young raft]

please feel free to scrutinize this

[woeful basalt]

Theoretically what do you need help with.

[young raft]

see if this proof is fine or not

[woeful basalt]

Proof on what

[young raft]

that you can construct a Vitali set of outer measure exactly 1

[woeful basalt]

ok

let me read it

[copper sierra]

this should go in #real-complex-analysis

[compact pewter]

@young raft Has your question been resolved?

[red bloom]

what is it called when there is an exponet that is in front of the number

[frank ibex]

A left superscript is quite uncommon

Do u have the context

[weary stone]

a tetration

or mass number

[compact pewter]

@young raft Has your question been resolved?

[earnest dew]

this?

[young raft]

@karmic ravine alright you asked me to ping you so this is the construction of a vitali set of outer measure 1

and then the argument you provided is almost the same word by word

Apply Carathéodory's criterion: Let $T=[0,1]$.
Then we have LHS:
$$
m^(T\cap V)+m^(T\cap V^c) = m^(V)+m^([0,1]\cap V^c)
$$
Notice that by translation invariance, as well as the fact that $V\cap (V+q) = \emptyset$, we get that for any $q \in \mathbb{Q}$:
$$
(V+q) \text{ mod 1} \subset [0,1] \cap V^c
$$
Thus:
$$
m^(V)+\underbrace{ m^([0,1]\cap V^c) }_{ \geq1 } \geq 1 + 1 = 2
$$
while
$$
m^*(T) = 1
$$
Contradiction.

[boreal girder]

artemetra

[karmic ravine]

Neat! Does splitting V + q at 1 and moving the right half backward mess with the outer measure? I know translation invariance makes the backward shift ok, but I haven’t thought about problems with splitting

Oh hold on, use Catheodory’s criterion again with T = V + q and the other set = [-oo, 1)?

[fast nacelle]

wtf is this

[earnest dew]

its called tetration

instead of doing 2 * 2 * 2 as you would with standard exponentiation

you do 2 ^ 2 ^ 2

[fast nacelle]

wow

never heard that before

ty

👍

[earnest dew]

yea it’s rarely used because it very quickly creates unreasonably large numbers

[young raft]

it doesn't: if you think of the interval [0, 1) as a circle, translating things is the same thing as rotating them

so i don't think splitting it should be a problem

[young raft]

wdym

T is an arbitrary 'testing' set (not necessarily measurable), V is Vitali that we assume and test for measurability

[karmic ravine]

Put V + q as the testing set and [-oo, 1) as the set we want to try for measurability. Then Catheodory’s criterion says splitting V + q at 1 gives two parts whose outer measure adds to 1. Now shift the right part backward to get that (V + q) mod 1 has outer measure 1

[young raft]

$m^(V+q) = m^((V+q)\cap[-\infty,1)) + m^((V+q)\cap[-\infty,1)^c) = m^((V+q)\cap[-\infty,1)) + m^*((V+q)\cap[1,+\infty))$

[boreal girder]

artemetra

[young raft]

huh that's.. interesting

not sure if this is needed tho

[karmic ravine]

¯_(ツ)_/¯

But we have a proof of what we wanted, so we’re good!

[young raft]

alrighty then

[young raft]

thank you so much!

can finally close this lol

.close

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[devout mountain]

i don't understand this how a derivative is equal to a func with two parameters

[eternal pulsar]

Chain rule...?

!xy

[compact pewter]

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.

[devout mountain]

euler method for 1 order differencial equations

[south sage]

what is I

[devout mountain]

interval in R

f is a function in I x R

[rigid perch]

that seems to just be the differential equation to be solved?

e.g. du/dx = x + u^3 is a differential equation which you could solve with the euler method, using f(x,u) = x + u^3

[white igloo]

send the full question dont just take snippets

[devout mountain]

https://fr.wikipedia.org/wiki/Méthode_d'Euler

[white igloo]

what?

[devout mountain]

that's where i got this

[white igloo]

where exactly?

aah ok ic

[devout mountain]

so basically i just need to solve a diff equation of the first order using euler method

but i don't know why du/dx = f(x, u(x))

[white igloo]

wym why?

f is just a function that makes a linear combination

[devout mountain]

why does it take two parameters

[devout latch]

u’(x) = f(x,u(x)) is the relationship u(x) must obey for each x in I for it to be a solution

[white igloo]

give me an example of a first order diff equ

[devout latch]

u is your candidate solution you plug in the ode

[devout mountain]

let's say : y' = t^2

[white igloo]

y'(t) = t² , f(t, y(t)) = t² + 0*y(t)

thats a candidate f

[devout mountain]

i still don't get it you can just write F(t) = y^3/3

[white igloo]

whats F?

i dont see F in the context you stated

[devout mountain]

the y

integrate

[white igloo]

huh?

use more words in ur answer

[devout mountain]

t it's not y my bad wrote it wrong

[devout latch]

the ode could also depends on y(t) so for instance y’(t) = t y(t) is also a valid ode f(t,y(t)) = t y(t)

or even simpler y’(t) = y(t)

you plug a function y(t) in this equation and you see if it satisfy the relationship

for y’(t) = y(t) the solution is C e^t

and sure enough if you differentiate y’(t) = C e^t = y(t)

[devout mountain]

okay okay now i get it thank youu

[compact pewter]

@devout mountain Has your question been resolved?

[compact pewter]

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[low pulsar]

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[compact pewter]

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[low pulsar]

im stuck on -3 = a(x-6)^2

idk how to solve a or x

is this an inequation?

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this is an equation

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alr but i have no clue on what to do

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you want to solve for x or a first

?

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i guess a

like that i can find x

cause i need x for my problem rn

the actual equatio nwas this

5 = a(x-6)^2 + 8

-8 both sides

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is a a given number like 3 or is it fixed constant (you don’t know what it is)

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then -3= a(x-6)^2

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idk what it is

i could tell u the problem though

so a company that fabricates golf sticks implies that their newest product, the stick propulse PLUS, contributes to boost performances of golfers when they encounter an obstacle. In their newest publicity the company affirmed making a experimentation in which a golfer had to hit a ball by surpassing a tree of 5m of height.

According to the calculations of the company, this experimentation showed that with the new propulse plus stick, the golf ball got a height of 5m way quicker than the other sticks
By calculating the height obtained by the ball (in m) in function of the time spent (in seconds), heres some details in which the company used

Attempt 1 First competiteur of golf stick: The ball is hit. The ball got a max height of 8m after 6 seconds
Attempt 2 Second competiteur of golf stick: The ball is hit. The ball is in flight for 12 seconds before getting grounded. It's max height was 10m
Attempt 3 The propulse PLUS stick: The ball is hit. It obtains a height of 6m after 4 seconds. But it hits the ground 14 seconds after being hit.

IS The company that makes the stick propulse PLUS right to say that their product lets the golf ball get a height of 5m more quicker than it's competitors?

this is all the info

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give me a sec

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alr

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if i parsed this correctly i think you have to find the equation of the three parabola one different for each attempts then once you have their equations compare which is the one which gets to y = 5 m the quickest

[low pulsar]

yes sounds about right

we would have to find x for each equation

which would give us the time

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that’s once you have the parameters of each parabola imo

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probably yeah

the thing is im hard stuck rn cuz idk a

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yeah you need the parameters of the parabolla first for each attempts, what attempt is this

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attempt 1

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give me a sec i’ll try it out on my own then i’ll comeback

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alr

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you are on the right track

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yep and rn im doing the attempt 3 while ur doing the 1

so i can be useful

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but idk why you put the 5 on the left here

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since its the

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.

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ill explain

the 5m represents y

seconds are x

unless u wanna put it in a

but idk how will that work

i would need to understand

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I think you are doing 2 step at once and you should not, first we need a so that a(x-6)^2+8 = 0 when x = 0

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wait ur right

but if x = then y = 0

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cause x at 0 seconds means the ball hasn't departed

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basically we assume the competitors launch the ball from (0,0) and that the field is flat

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yea

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but the key to find a is to force the parabola to go through (0,0)

then we will have the parabolla attain y=8 at x=6 and the trajectory will start at (0,0)

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ye

do we replace x and y with = 0?

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in y=a(x-6)^2+8

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alr

it gives me a = 2/9

-/29

-2/9

which makes sense because

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the parabola is downwards

right?

yup

alr

now we should probably find x when y = 5

or not

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you can

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ill try t

what is -2/9 x 6

it gives me a decimal number

-1.33333333

what should i do

its -3 = -2/9 (x-6)^2

i did -2/9(x) and -2/9(-6)

= -2/9x -1.33333333

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first get to a point where you only have (x-6)^2 on one side and a number on the other

[low pulsar]

-3 + 2/9 = (x-6)^2

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i just got the 2 point where y=5 and it’s not pretty

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my math teacher said since we're in advanced maths its ok

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is this good tho?

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no you need to divide by -2/9 (or multiply both side by -9/2) to get rid of the -2/9 in front

[low pulsar]

oh ur right

13.5 = (x-6)^2

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now how could you get rid of the ^2

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uhh

that stuff is where im weak at

i think its the rationalisation

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x^2 has an inverse function

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wait btw i have 13.5 = x^2 + 36

are we on the same page rn

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oh you expanded

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yeah

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i took square roots on this it’s a bit quicker imo

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I think you are missing a term when you folded

the middle term

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where

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(x-6)^2=(x-6)(x-6)=x^2-6x-6x+36

[low pulsar]

oh i thought it wouldbe

ok ok

ur right

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do you know square roots

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so igot 13.5 = x^2 -12x +36

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yes

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to get x on this you would need to use the quadratic formula

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whats the quadratic formula again

ax^2 + bx +c?

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but using square roots on this both sides then you get x-6 = sqrt(13.5) and 6-x=sqrt(13.5)

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wouldnt that lead to inequations

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to find x1 and x2?

wait

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it gives you x when you have ax^2+bx+c=0

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thats to find x

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but this is longer than taking square roots each side

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lets do square roots

scrap whaty i said

ok so im back to 13.5 = (x-6)^2

now we put square roots

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the square root of a is by definition the number b such that b^2= a

so for example sqrt(4) =2 because 2^2 =4

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yeah

but we dont need b tho right

theres no b in the canonique formula

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yeah here a b and be where placeholders

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they are not related to this

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oh

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or the problem

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where did u do the sqrt tho

in the equation

was it at 13.5 = (x-6) ^2

alr

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I’m trying to explain that taking square roots will get rid of the ^2 around x-6

[low pulsar]

oh

so i gotta do sqrt of 13.5 aswell then

9.67 = x

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that’s one of the 2 solutions

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alright so when y = 5 for attempt 1 the time it takes to reach 5m is 9.67 seconds

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waot whats the 2.32577

and the 5

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the parabola reaches y=5 for two different x

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did u do another attempt?

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one way to see it is that when you take square roots on each sides you get two equations

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i got only 1 tho

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in a simpler setting if i were to tell you to get the solution to the equation x^2=1

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i would do

sqrt1

and sqrt x^2

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1 is a solution

meaning 1^2=1

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ahh

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but also (-1)^2=1

[low pulsar]

so actually the balls real distance when it hit 5 m was at 2.32 seconds

but im havin trouble understanding the math behind it

i do understand

what ur saying

im just like not understanding HOW to do it

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maybe ladder up to the og question

(x-1)^2=1

what are the two x satisfying this equation