help-38

[sharp vine]

ion have the cos graph memorised 💀

this is a certified skull moment

[hazy panther]

You can look it up

But definitely try to learn the general shape

These will become simple

[sharp vine]

yeah cuz they wont frickin give it to me in the exams 😭

[hazy panther]

Exactly

[sharp vine]

hmmm

wait does

@solid kiln

,help

[solid kiln]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[sharp vine]

,ask cosx

what does my boi wolfram have to say about it

oh.

this is the original graph

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[grave bane]

Any help would be appreciated 😊

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[rare cradle]

I'm not really sure how to find the limits for this and theta

[rare cradle]

The second one is my working so far

Rho*, not "this"

I got rho actually just theta

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[rare cradle]

Rho incorrect

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[wraith hinge]

alright , just to clarify, equating g(x) and f(x) and then integrating it will give me the shaded area?

[wraith hinge]

or do i need to do something else

[maiden dagger]

depends on what f(x) an g(x) is

[wraith hinge]

its written there but without the brackets

[maiden dagger]

f(x) = 4-x^2
g(x) = -2?

is that the bit ur talkin about

[wraith hinge]

yeah

nah sorry for the caps

[maiden dagger]

no worries

[wraith hinge]

yeah so equate them

i got +-sqrt6

and then intergrate with f(x)-g(x)

with the limits +-sqrt6

yeah i realised what the skull emoji was for now

wtf is this question!??!?!

[final sage]

what you've shaded is the area below between the x-axis and the graph of the function f(x) = 4 - x^2

you can find this by integrating f from -2 to 2

[wraith hinge]

i didnt shade the whole thing

[maiden dagger]

im not usre if u can combine f(x) and g(x) imma be honest

[wraith hinge]

😭

[maiden dagger]

not too sure abt that

i would jut find these areas separately and add them

[final sage]

ah, so you want to find the area of the shaded region between 4 - x^2 and -2

[maiden dagger]

idk if theres a faster way to do that

im assuming that is the total area u wanna find, correct me if im wrong

[final sage]

in general, whenever you have two functions f and g, if you want to find the area above g and below f, you can integrate f - g

[wraith hinge]

thats what im thinking

[final sage]

in this case, its even easier to see this because g(x) = -2, a constant. so the area between 4 - x^2 and -2 is the same as if you just shifted the parabola 4 - x^2 up by two units, to get 6 - x^2, and integrated it normally

you're essentially just moving the x-axis down which raises the parabola up

[maiden dagger]

yes

why - tho

why f(x) - g(x) anbt not +

[wraith hinge]

yeah i got (6 - x^2)

i intergrate to get

(6x-1/3x^3)

and then use the limits +-sqrt 6

[final sage]

well, you are integrating from -sqrt(6) to sqrt(6) since that's where the parabola intersects the line y = -2

[final sage]

if you think of the riemann sum, you want the small rectangles at each point (really, a small interval around the point) to have height |f(x) - g(x)|. it helps if you draw it. since we are assuming f(x) >= g(x), this is just f(x) - g(x)

[maiden dagger]

no clue what a riemann sum is

[final sage]

sadge. that's how they usually define integration the first time you see it, right?

[maiden dagger]

i mean

[final sage]

summing up tiny rectangles

[maiden dagger]

oh yeah

they didnt give a name to it

thats called the riemann sum?

[final sage]

yeah, named after the guy who invented them (and the riemann integral)

[wraith hinge]

so am i doing the right thing?

[final sage]

this is one small rectangle

[maiden dagger]

are we subtracting the functions cuz g(x) is -ve

[final sage]

as the rectangles get smaller, the height approaches the distance between the functions at some point in the rectangle

[final sage]

yeah

[maiden dagger]

bc if g(x) was like 10 wouldnt we do f(x) + g(x)

[wraith hinge]

so is this the answer

[final sage]

yep desmos agrees

[final sage]

you'd do g(x) - f(x)

[wraith hinge]

alright thanks bro

[final sage]

since the graph of g is now above the graph of f

[wraith hinge]

wow that is pretty cool , how do you do that?

[final sage]

so |f(x) - g(x)| = |g(x) - f(x)|

desmos.com/calculator

use this

it can help check your work and visualize stuff

try not to rely on it too much when you're learning, though...

[wraith hinge]

yeahhh

but how do you get the symbols

like the intergral sign and what not

oh wow

the functions button i didnt notice that ahhah

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[crimson rover]

if i were to write this in leibniz notation would it be

[crimson rover]

$\frac{d^2f(x)}{dx^2}=\frac{d^2}{dx^2}7x^3-\frac{d^2}{dx^2}6x^5$

[solid kiln]

Joshii

[stark bison]

Yeah sure, but it's more common to see $\dv[2]{f}{x}$ instead and I would put parenthesis around $7x^3$ and $6x^5$

[solid kiln]

A Lonely Bean

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[silver arch]

h

[silver arch]

can someone check my set proof

the question is prove that there is no sets a,b,c such that AnB = ∅ , AnC = ∅, and (AnB)-C =∅
Using contradiction i said, Assume there are such sets a,b,c, and take xe((AnB)-C), which means xeAnB and x∉C, but if xeAnB, then xe∅, which is not possible as the empty set has no elements. Am I correct?

[marble wharf]

!original

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If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[marble wharf]

such sets clearly exist

eg A=B=C=emptyset

[silver arch]

bruh

that doesn't make any sense, if A=B the intersection of A and B is the elements shared between a and b not the empty set

.close

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[pastel crypt]

.claim

a

[pastel crypt]

ok uh does anyone know how to do synthetic substitution

on a calculator

heres a question

Use synthetic substitution to find f(-3) and f(4) for each function.

1. f(x) = x^2 + 2x + 3

[dapper flicker]

remind me what synthetic substitution is

[fiery goblet]

just replace the x's with the given values

[dapper flicker]

I did it last year but I can remember exactly ;-;

[fiery goblet]

why do you want to do synthetic substitution on a calculator

[pastel crypt]

teacher told me to

so id rather do it the way she recommends, especially since the exam will be timeed

i believe im meant to do it on mode 5,3

[fiery goblet]

if you're using a calculator, doing synthetic substituion will be a very big waste of time

just replace the x's with the given values

synthetic substituion makes sense when you don't have a calculator

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[rugged spear]

I have a question

[rugged spear]

Regarding the 12^3

The problem requires to turn all into inches

But the right side equation is on Feet thus the 12

I don't know why it needs to be on ^3

https://mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-613-double-integration-method

Here's the link in case you need more references

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[lunar moat]

On part e, i have the following pattern:

N=1: 22
N=2: 322
N=3: 4 * 32

i dont know how to write it in form of an equation using n though

[sleek canopy]

$a_n=4-\sum_{k=1}^nk=4-\frac{(n+1)n}{2}$

[solid kiln]

everg

[sleek canopy]

for c)

[lunar moat]

i dont need C

i got c and d, i just dont know how to do e

[sleek canopy]

it seems like factorial

$a_n=2(n+1)!$

[solid kiln]

everg

[lunar moat]

thats what i thought too but idk if we are allowed to use factorial\

[sleek canopy]

this satisfy $a_0=2$ and the recurrence equation

[solid kiln]

everg

[sleek canopy]

there is no simpler writing for factorial ... if there was, we didn't need to define symbole "!"

so i think your are allowed

[lunar moat]

aight sounds good. thank you

.close

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[manic mirage]

can someone explain how to find the domain and range of this rational function

[manic mirage]

im abit confused on what was wrong

and what ymax and ymin mean

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@manic mirage Has your question been resolved?

[scarlet cape]

I need the channel cuh

close it NEOW!!\

[zinc ginkgo]

!occupied

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

[scarlet cape]

!Iknow

[zinc ginkgo]

Then wait

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[fallow ether]

are these formulas switched?

[fallow ether]

for spherical coordinates for triple integrals

nvm

theyre not

i thought it was the angle on the inside

between the line and the y axis in this case

but if its sin we get r/ distance away from origin on xy plane

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[wintry lily]

help

[low gazelle]

Pull out a negative 1 from the right side

[wintry lily]

ok

i was shitting mb

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[wintry lily]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[wintry lily]

<@&286206848099549185>

[wintry lily]

i got it never mind

<@&286206848099549185> do smth next time

/close

.close

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[void dew]

<@&286206848099549185>

is this right

[wraith hinge]

this is not math

[final sage]

This looks like something you could just google

Phospholipid bilayer

Also you said “is this right” but didnt say what your answer was

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[zinc ginkgo]

.close

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[trim kestrel]

how do i factor it

honestly i dont even know how i got the zeros

i just charted it on desmos and guessed what the fractions were

im clueless

[tulip violet]

A polynomial can be factored in the form of f(x) = (x - a)(x - b)...
When you find the zeros, you set each factor equal to 0, so (x - a) = 0, (x - b) = 0, then solve for x
So you would have x = a, x =b, etc
You have the zeros, you just need to work backwards

[trim kestrel]

but it should be the other way

around

the only way to get the zeros is by factoring it in quadratic form

[tulip violet]

Start with the x = a step then get them to look like (x - a) = 0, then combine the factors

[tulip violet]

You have the zeros from part b

[trim kestrel]

yea how do i answer part b

like how do i get the rational zeros

like i said, i just guessed it on desmos

i want to know how to do it

[tulip violet]

You just plug in each combination into f(x) and see when f(x) = 0

[trim kestrel]

i see

okay so 1/3 is a zero

now what

[tulip violet]

You do that and find all the rational roots

You have 3

[tulip violet]

Then for c, you do this

[trim kestrel]

so f(x)=(x+2/3)(x+3/2)(x-1/3)

[tulip violet]

Yes

[trim kestrel]

now what

[tulip violet]

That's it

[trim kestrel]

its not

[tulip violet]

That's f(x) in factored form

[trim kestrel]

this was wrong

[tulip violet]

What does it say in the blue, underneath that?

[trim kestrel]

do you know what to do? if not lmk

ill just ask someone else

[tulip violet]

Looks like before anything happened, you can factor out a gcf from the original equation

So you might be missing that gcf

[trim kestrel]

huhh

i didnt even know that was a thing

if i have the zeros why is there a gcf at all

[tulip violet]

Because if you facotr out a gcf, it will make things easier

[trim kestrel]

well what do i factor out

the -6?

[tulip violet]

A gcf is a number that is common with all the terms, 54, 99, 15 and 18

Is 6 a common factor between all those?

[trim kestrel]

nope

is it 3?

[tulip violet]

Yes

[trim kestrel]

so i

is it this?

[tulip violet]

Try it

[trim kestrel]

it also said to write it in ax-b

what does that mean

is a/b the numerator/denom?

[tulip violet]

It means that it doesn't want fractions because 2/3 is a fraction

[trim kestrel]

but it said to use fractions for any numbers in the expression

is it just lying?

[tulip violet]

So in each factor, you want to "remove" that denominator, by multiplying all the terms of that factor by the denominator
For example, (x - 6/5) = 5(x - 6/5) = (5x - 6), notice how I multplied all the terms by 5, which was the denominator

[trim kestrel]

so 3x+2?

for the first term

[tulip violet]

It might have been for something else

[tulip violet]

Yes

Do it for the rest

[trim kestrel]

do the signs change

[tulip violet]

No

[trim kestrel]

so 3(3x+2)(2x+3)(3x-1)?

[tulip violet]

Try it

[trim kestrel]

tysm

ive been struggling with this for a minute

i didnt realize all i had to do was take out the gcf

[tulip violet]

If you're done with the channel, you can close it using .close

[trim kestrel]

.close

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[worn sun]

https://i.imgur.com/0OhJAny.png

[trim jolt]

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

[worn sun]

I think you may need to use u sub and use arclength but I am not really sure if that's right

[broken pilot]

can you show your work?

[worn sun]

I havent started on it quite yet, but was wondering if that's the direction it wants me to go

[broken pilot]

Yes I think you’re supposed to use arc length, maybe after parametrizing

[worn sun]

This really is a mess of a problem

.close

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[scarlet cape]

Can someone help verify this for me?

[scarlet cape]

for B, one solution because a is bigger than b

For C, one solution because a and b are the same length

right?

SSA conditions

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@scarlet cape Has your question been resolved?

[broken pilot]

SSA is ambiguous

[scarlet cape]

yeah

i gotta find the number of solutions

using this

[broken pilot]

one sure way to know is to try constructing the triangle

[scarlet cape]

wait nvm

I cant do that for all 4 questions

[broken pilot]

by drawing a line of length 26 cm, which is c, and then constructing a ray starting from one end of l and having an angle of 76 degrees with it

[scarlet cape]

ill move on to the harder one

thnx

[broken pilot]

and then constructing a circle of radius 26 cm

[scarlet cape]

.close

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[left karma]

Twice the sum of a number and 6 is equal to 18. What is the correct equation to solve this problem

[left karma]

<@&286206848099549185>

[serene ferry]

2(x+6) = 18

[left karma]

how to convert celsius into farenheit

28 degrees celsius into farenheit

formula please

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[dry urchin]

If you just need a formula without explanation try to google your question. If you dont understand what is said there come back here and ask again.

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[mystic veldt]

how can i find min value of sin theta + cos theta with calculus

[mystic veldt]

i have found the max value, but cant find the min

i rechecked my work

it was correct

[ripe valley]

Show work

[mystic veldt]

yea ok

[ripe valley]

Its not loading...

Ah yes finally

[ripe valley]

You are solving for critical points, but u forgot to check for the second derivative for minima

[marble wharf]

well, what is f(5pi/4)

[mystic veldt]

0?

[marble wharf]

no

[mystic veldt]

no sorry that was coming out to be

neg sqrt 2 also

the same

[marble wharf]

what do you mean with "same"

f(pi/4)=+sqrt(2)

f(5pi/4)=-sqrt(2)

[ripe valley]

Wait u need to find a k such that f''(kpi + pi/4) is greater than 0

@mystic veldt just do the same thing for minima as u did for maxima, just find a solution using ur general solution such that the second derivative is greater than 0

[mystic veldt]

that is what i cant find

[mystic veldt]

wait let me see

[ripe valley]

Sub npi + pi/4 in the second derivative

[mystic veldt]

,w sin(5 pi/4)+cos(5 pi/4)

[solid kiln]

Results provided by WolframAlpha

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Requested by yajatjamal

[mystic veldt]

wait

okay got it

thankyou

.close

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[vagrant marsh]

Oh no not one of these

[pseudo parcel]

Can u guys help me?

[vagrant marsh]

Ill tey

Try

[pseudo parcel]

Ok. If u find please write

[sleek canopy]

garlicb is the best in this type of problem!

[vagrant marsh]

Thanks lol

[pseudo parcel]

Really?!

[vagrant marsh]

I try my best

[willow urchin]

19

[pseudo parcel]

How? Send me solution

[sleek canopy]

you can bet your house that he is goint to solve it

[vagrant marsh]

Ah yes, the everything is 19 rule

[willow urchin]

more formally its called the ann-liouville malicious compliance with underspecification theorem

[sour token]

First two look like sum of two numbers - 1 but then third is just weird

[pseudo parcel]

Yeah I saw that too.

[vagrant marsh]

Gimme some time ill brb

[pseudo parcel]

Ok. Np. I am waiting

[vagrant marsh]

I have to go to class, but ill keep thinking about it

[trim jolt]

@pseudo parcel Has your question been resolved?

[wraith hinge]

man you sure this has to have a unique answer there are some interesting things about that I could tell you if you want but I don't think there is a definitive answer

[vagrant marsh]

Probably doesnt

You could argue for any case

[wraith hinge]

theoretically every answer is right😂

but i dont see those creative ones here

[trim jolt]

@pseudo parcel Has your question been resolved?

[vagrant marsh]

Idk man, sorry

[rugged latch]

Even chatgpt doesn’t know, which means it can’t be solved

Originally I was thinking it was number 1 + number 2 - 1

But that broke at case 3

[sleek canopy]

you are tired ..maybe tomorrow you'll give one of your fantastic solution!

[vagrant marsh]

Perhaps

[rugged latch]

GarlicB #1 fan

[pseudo parcel]

Ok guys. Bye. I will find the answer soon. So I don't close it.

But, can u solve it?

[sleek canopy]

for the non-sniper people

[trim jolt]

@pseudo parcel Has your question been resolved?

[pseudo parcel]

Can someone look this AMO question please ?!

[trim star]

pretty sure its a but should i try help with the reasoning

[trim jolt]

@pseudo parcel Has your question been resolved?

[main mesa]

do you know how to rationalize?

you would start off by multipling the whole equation by the roots that are present in the denominator,

for example of the propety ^

[trim jolt]

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[pearl field]

I'm not sure if these are correct

[pearl field]

<@&286206848099549185>

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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.

[pearl field]

I'm not sure if my answers are correct, can someone help?

[edgy ermine]

What are the checks that need to be done before we can determine differentiability?

[pearl field]

If f(x) is continuous right?

[edgy ermine]

There's another rule

Both must be true

[pearl field]

Wdym?

Like equal to each other?

[edgy ermine]

The limit as x approaches a for f'(x) must also exist right?

[pearl field]

Yeah

[pearl field]

That's the 3 step continuity rule right?

[edgy ermine]

Yeah

[pearl field]

What's the next step?

[edgy ermine]

Tbh when I was taught this I was told that with absolute values, the place to check for kinks and discontinuity and vertical tangents was where you switch the two different equations (x = 0) in this case

Since we know cos(3x) is always differentiable, same with -4

[pearl field]

But -3sin(3×0)=0

Ohh wait but d/dx(-4) is 0 as well

[calm quarry]

Check the continuity at 0. I think that'll be enough.

[pearl field]

It's continuous at 0 over the interval except (-1,1) right?

[calm quarry]

It's discontinuous at 0. RHL = cos(3*0) = 1
LHL = -4

[pearl field]

Oh ok ok I got it

But why do the questions have (-1,1) if that's not the interval?

[calm quarry]

Maybe to confuse as cos has range [-1, 1]

[pearl field]

But cos3×0=1, so it would be no and it would not be if the interval was (-1,1), since cos(3x) is 1

[calm quarry]

Yes, correct

[pearl field]

Ok thank you so much

What about the other one? Ik that 3(8)-9=15, so that should be continuous, but 3(9)-9=18

[calm quarry]

Which que? Your both pics are same

[pearl field]

Oops sorry

[calm quarry]

For this you even need not to check continuity or differentiability.
First find the domain of the function.

Domain sry

Hint : the function is considered non-differentiable over values outside it's domain

[pearl field]

I used a graph calculator and the domain is (−∞,8)U(9,∞)

[calm quarry]

Now can you get the answer?

[pearl field]

D?

The answer is just the domain?

[calm quarry]

Yupp

No I mean, you need not to check the differentiability here.

For this question

[pearl field]

How do you spot that?

How do I know when I need to check it or not?

[calm quarry]

Because in rest other answers the interval which is given has some values over which the function is not defined only

If function is not defined at a point, straight it's discontinuous there

[pearl field]

So you only need to find the differentiability when there's an interval?

[calm quarry]

Find the differentiability when it's required. Sometimes questions are bit simpler.

[pearl field]

Ok thank you so much, you're a great help

[trim jolt]

@pearl field Has your question been resolved?

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[remote pulsar]

How do I apply the transformation $T(x, y) = (x - 2y, x + 2y)$ onto the circle $x^2 + y^2 = 1$? What is the image of the circle under $T$?

[solid kiln]

o.O

[remote pulsar]

I suspected I need to reparametrize it as $x = \cos\theta$ and $y = \sin\theta$ with $0 \le \theta \le 2\pi$

[solid kiln]

o.O

[remote pulsar]

since now x and y can be expressed in terms of the same variable

but like, I am confused how to actually apply the transformation onto the circle with this

[trim jolt]

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[maiden nova]

How to work out the perc multiplier? Please ping if can help ty

[wraith hinge]

add 100% to multiply if increase, subtract from 100% if decrease
example: increase $5 by 15% = 5x(100+15)%= 5x115% on calc or 5 x 1.15
decrease $20 by 45% = 20x(100-45)%= 20 x 55% or 20x .55

[maiden nova]

ohk

So for the first one since its increase by 11% I would hve to do 10 * 1.11, right?

[wraith hinge]

yep

[maiden nova]

ok perfect so thats 11

secound would be 40 * 1.04 so 41.6

third is 500 * 1.11 = 555

Is the fourth one 34 * .40 @wraith hinge

ok thanks

fifth is 60 * .98

😭

what I do wrong

[wraith hinge]

you missed out on decimals

[maiden nova]

oh

[wraith hinge]

should be 11.1 on 1

[maiden nova]

oh ofc

for the second one its 40 * 1.04 right, i got 41.6?

[wraith hinge]

2 should be correct

[maiden nova]

hm

I'll skip no.2 for now cos it might be an issue in sys maybe

whats wrong w 4

I did 34 * 0.4

.close

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[trim jolt]

@rancid aspen Has your question been resolved?

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[primal sage]

help me please

[wraith hinge]

[
a^m \cdot a^n = a^{n+m} \tss{and} a(b\cdot c) = (a\cdot b)c
]

[wraith hinge]

the above is all you need to solve your particular problem

[primal sage]

i am not sure

[wraith hinge]

of?

[primal sage]

the equation

@wraith hinge

[wraith hinge]

have you thought about this

[primal sage]

yeah i dont know what it is

[merry owl]

its an exponent rule

since you already have 10^m and youre multiplying by 10^n, your final answer would just be 10^m+n

[primal sage]

How would you find that

What are the steps of solving this question

@merry owl @wraith hinge

@wraith hinge

.close

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[wraith hinge]

how to dertermine if |x| =<1 and |x| =<1 is clsoed or open or neither

[wraith hinge]

in (x,y) belongs to R^2

[nova spire]

you wrote the same thing twice, did you mean $B(0,1) = {(x,y)\in \bR^2, |(x,y)|<1}$ on one hand and $\overline{B(0,1)} = {(x,y)\in \bR^2, |(x,y)|\leq 1}$ on the other?

[solid kiln]

rafilou2003

[wraith hinge]

oh my bad

how to dertermine if |x| =<1 and |y| =<1 is clsoed or open or neither

[nova spire]

ok

So let $A= {(x,y)\in \bR^2, |x|\leq 1$ and $|y| \leq 1}$

[wraith hinge]

yes

[nova spire]

Do you have an opinion as if it is closed or open (or both or neither)?

[wraith hinge]

not is =<

[nova spire]

oops sorry yes

[solid kiln]

rafilou2003

[wraith hinge]

i would guess

closed

[nova spire]

that's a good intuition

So let's try to prove it's closed, what tools do we have to prove it?

[wraith hinge]

mhh

the set from (1.inft) is open

[nova spire]

if there are no obvious tools, we can go back to the definition

[nova spire]

this is a possible approach, what do you have in mind?

[wraith hinge]

complement

so |x| > 1

and |y| > 1

[nova spire]

and?

[wraith hinge]

well this is open, right

[nova spire]

no, I was talking about this "and"

[wraith hinge]

this how the question is written

with an and in the middle

[nova spire]

this is how A is written

is this how the complement will be written?

(what's the opposite of "X and Y"?)

[wraith hinge]

i'd say there are a plethora of answers to this

but i assume you're hitting for or?

[nova spire]

well yes

the opposite of "X and Y" is "notX OR notY"

so, $A^c= {(x,y)\in \bR^2, |x|> 1$ or $|y|> 1}$

[solid kiln]

rafilou2003

[nova spire]

so, what did you have in mind with this?

[wraith hinge]

will (-inft-0) and (1-inft) will not be in that set

in A

and those sets are open and are the complement to A, so a is clsoed

i think that is how it goes

[nova spire]

I think it's lacking the explanation of the tool you're using

[wraith hinge]

might be

i dont really have any

[nova spire]

If $f$ is a continuous function and $O$ is an open set, then can we say something about $f^{-1}(O)$?

[solid kiln]

rafilou2003

[wraith hinge]

closed then?

[nova spire]

the preimage of an open set by a continuous function isn't closed, no

[wraith hinge]

oh

i thought i was just reverse

[nova spire]

so?

[wraith hinge]

what is preimage

F^-1(O)?

[nova spire]

If you don't know then you probably can't use that tool

you have to go back to the definition of an open set

[wraith hinge]

i have the definition

and mine doesnt say anything about preimage

[nova spire]

yes i was trying to make you use a certain tool

but you most likely haven't seen it

[wraith hinge]

what would be the easiest way to do it then

[nova spire]

If you don't know the following theorem : "If $f:X\to Y$ is a continuous function and $O\subseteq Y$ is an open set, then $f^{-1}(O)$ is an open set. Similarly, if $F\subseteq Y$ is a closed set, then $f^{-1}(F)$ is a closed set." then we have to use the definition of an open set

[solid kiln]

rafilou2003

[wraith hinge]

i just know of episolon neighborhood

[nova spire]

And we will have to prove that this set is open by the definition of openness

yep

So show that every element of A^c has an epsilon neighborhood

[wraith hinge]

im not sure i can do it like imagine when it two variables

[nova spire]

ok, here's a hint

[wraith hinge]

since it says that both x and y is between (0,1]

[nova spire]

Show that $B= {(x,y)\in \bR^2, |x|> 1}$ is open

[solid kiln]

rafilou2003

[wraith hinge]

i dont know to show it

but like obv it is

[nova spire]

Well how to show it :

but with B

Prove the statement "for all (x,y) in B, there exists an epsilon neighborhood B((x,y),epsilon)"

[wraith hinge]

idk how you prove it tho

i can just tell you can add or subtract epsilon and it'll still be within

[nova spire]

So in order to do this, the following may apply :
"Let (x,y) in B.
We introduce epsilon = ...
Let (x',y') in B((x,y), epsilon).
...
Thus (x',y') is in B.
Thus (x,y) has an epsilon neighborhood.
Thus B is open"

[wraith hinge]

how is this a proof

"i will let this satifisfy the already known condition for it to be an open set, hereby this is an open set"

[nova spire]

?

Where the "..." are you need to fill in

this is the sketch of the proof, the value of epsilon chosen will depend on what B is, as well as how you prove that (x',y') is in B

recall that epsilon can depend on either x or y or both

[nova spire]

in any case, work your way backwards, suppose you've already chosen your epsilon, and you want to show that (x',y') is in B. So you want to show that (...) ?

[wraith hinge]

is there no algebraic way of this

where i can just modify it into something @nova spire

also what does it even when they say "and"

@nova spire i can solve for x+2y=1 or something

but what does "and" mean in this context

[nova spire]

? means and

[wraith hinge]

yeah but can i seperate them?

[nova spire]

sure if you have any tools that would benefit you from separating them

how would you plan that?

[wraith hinge]

idk

this is the first time im ever doing this

and the book has like 2 pages about it, so i came here

[nova spire]

ok, do you know about the cartesian product of closed sets?

[wraith hinge]

no

[nova spire]

so it won't help

[wraith hinge]

only thing that is being said that open said are with epsilon neighborhood

and closed are when all "outer-points" are within

[nova spire]

ok, so start writing this sketch of proof, and try to see how to fill in the blanks

[wraith hinge]

well if (x'y') is in b(x,y,), epsilon) then because epsilon is in it will also be in

but that circles back to the fact i already know its open

so i know there must be a epsilon +/-

[nova spire]

epsilon is in what?

you have to find the epsilon that works

that you can express in terms of x and y

(maybe just in terms of x)

[wraith hinge]

x+epison

is in

[nova spire]

just checking in, do you know the sequence caracterisation of closed sets or nah?

[wraith hinge]

nah

[nova spire]

if a sequence in a closed set converges then the limit is in the closed set?

[wraith hinge]

that is in the next chapter

so idk

[nova spire]

ok

[nova spire]

What does it mean for (x',y') to be in B?

[wraith hinge]

waht?

that they r in B?

[nova spire]

What's the definition of B?

[wraith hinge]

?

idk

[nova spire]

right there btw

[wraith hinge]

is that the definitionm

what it equals

[nova spire]

we defined it this way

yes

so what does it mean for (x',y') to be in B?

[wraith hinge]

i dont know what you want me to say

it means that it is in B?

to me, this is like asking "what does it mean that 2+2= 4?"

[nova spire]

No, this is is more similar to asking "What does it mean that $x\in {y, y$ is even$}$"

[solid kiln]

rafilou2003

[nova spire]

it means that x is even right?

so

[wraith hinge]

if x belongs to a group called y

and that group is always even

then yes x is even

[nova spire]

The group is not called y, it's saying that it regroups all elements y that are even

you're perhaps not fully familiar with set notation and what it means

[wraith hinge]

i am

but not in english

so its half translate

[nova spire]

what's your main language?

[wraith hinge]

danish

[nova spire]

alright can't help then (in danish)

[nova spire]

So back to this question

[wraith hinge]

is this any different than (x,y)?

[nova spire]

We're just using different letters to distinguish

[wraith hinge]

so its the same?

[nova spire]

nothing else

[nova spire]

serves the same purpose

[wraith hinge]

is this your question

so the set B consist of all points in a plane where is greater than 1

[nova spire]

yes

so if (x',y') is in B, what does that imply about x' or y'?

[wraith hinge]

the exact same as (x,y)

just that they are also in B

[nova spire]

which is, if you use x' and y'?

[nova spire]

literally re use this sentence with x' and y' instead of x and y

[wraith hinge]

how does this hlep me solve the question

[nova spire]

it will

[wraith hinge]

so the set B consist of all points in a plane where is greater than 1

ok i said this now but for x' y'

[nova spire]

where what is greater than 1?

[wraith hinge]

x'

[nova spire]

no, not x'

[wraith hinge]

then what

[nova spire]

|x'|

[wraith hinge]

omg

[nova spire]

are we good on this?

So if you want to prove that (x',y') is in B, you need to prove that |x'| > 1

[wraith hinge]

why would it not

[nova spire]

?

[nova spire]

if you're replying to this, ok then

so, we want to prove |x'| > 1, what info about x and x' can we use?

[wraith hinge]

how can i prove this

i have 0 idea

maybe you give give one example instead of asking me 100 questions?

[nova spire]

ok

for example, the point (2,100) is in B

because |2| > 1

and if I let epsilon = 0.5 for example

well for any (x,y) in B(2,0.5), we have |x| > 1.5 > 1

so (2,100) has an open neighborhood

[wraith hinge]

yes ofc

that is trivial

this cant be my proof

because i can find one number

[nova spire]

well the point is, how can you generalize finding this epsilon that works to ANY (x,y) ?

you can make a graph for example

[wraith hinge]

ok

but surely you can do that without making a graph

i cant understand how there isnt a straight way forward step by step

[nova spire]

well you can by working your way backwards in the proof

[wraith hinge]

like everything else in math that you follow and and reach at the goal

[nova spire]

There are sometimes when you will have a ridiculously complicated set, not even part of R^2, to prove is open or closed

and you won't be able to graph it

[wraith hinge]

ok

can you solve this one for me?

i have modfied the question since i assume you arent keen on solving my homework

[nova spire]

sure

any intuition as to what this set will be?

[wraith hinge]

open

[nova spire]

yes

[nova spire]

So Idk, name this set C

We want to prove it's open, so we will use the definition with epsilon neighborhoods

And so we can copy paste the proof sketch we had earlier :
"Let (x,y) in C.
We introduce epsilon = ...
Let (x',y') in B((x,y), epsilon).
...
Thus (x',y') is in C.
Thus (x,y) has an epsilon neighborhood.
Thus C is open"

So again, finding epsilon either results of going backwards in the proof to make sure that epsilon works, or graphing C

[wraith hinge]

i think you forgot the middle part

[nova spire]

again, "..." are blanks to be filled

[wraith hinge]

can you fill them

that was the idea of the example

[nova spire]

ok, this once because it's not the original question

[wraith hinge]

yh, that was my idea

[nova spire]

Let's start with the second blank

what do we need to prove in order for (x',y') to be in C?

[wraith hinge]

that B = {(x',y'})?

[nova spire]

?

[wraith hinge]

i really have no idea what you mean proof

i can set x'y' to be in C

why do i need to prove it is in C if that is done by my will

[nova spire]

The problem is we set (x',y') in B((x,y), epsilon).

and we want to show that (x',y') is in C

so no "done by my will" is not an argument

[nova spire]

what do we need to prove for this (x',y') to be in C?

[wraith hinge]

because B(x,y,epsilon) in the set

[nova spire]

and why?

[wraith hinge]

because its open

so you can always find an episolon less or greater

[nova spire]

But we want to PROVE that it's open

so we don't know that it's open yet

[wraith hinge]

yh but idk that part

[nova spire]

you're running in circular arguments

[wraith hinge]

yes, that i what i said 100 msg ago was the problem

[nova spire]

So we DON'T know that B(x,y,epsilon) in the set

We want to PROVE it

[wraith hinge]

i know we do

i am just telling you i dont quite know how to

[nova spire]

in order to do so, we show that every element of B(x, y,epsilon) is in C

so we picked a random element (x',y') in B(x, y,epsilon)

that's where we stand for now, is that alright so far?

[wraith hinge]

yh

[nova spire]

ok

So what do we need to prove, to show that (x',y') is in C?

[nova spire]

again, use the definition

[wraith hinge]

B(x, y,epsilon)

that this is in the set

[nova spire]

No, we already said this was a circular argument

[wraith hinge]

ik

but dont we wanna prove that this is in first

[nova spire]

Let me take a smaller example

[wraith hinge]

then the other is also in

[nova spire]

A = {bananas}

B = {fruits}

to show that A is a subset of B :

"Let x be an element of A.
Thus x is a banana.
Thus x is a fruit.
Thus x is in B"

So every element of A is an element of B. So $A\subseteq B$

[solid kiln]

rafilou2003

[wraith hinge]

yes...

[nova spire]

Is this proof clear?

[wraith hinge]

yes, but here

you've returned the language of the common

so one can express themselves with ease

[nova spire]

yes, so I'm using this example to address more complicated examples later

[wraith hinge]

in math, it isnt as simple, if you dont know the correct terms

[nova spire]

If I'm asking you what it means for x to be in B, what would you respond?

[wraith hinge]

that x is a fruit

[nova spire]

yes

Let's take another example

D = {fruits bigger than 5 inches}

What does it mean for x to be in D?

[wraith hinge]

bit of a mockery

but

that its larger than 5 inches

[nova spire]

yes

So if E = {x such that |x| > 1}, what does it mean for x to be in E?

[wraith hinge]

that it adheres to the condition that it must be larger than 1

[nova spire]

that |x| > 1 yes

and reciprocately, if |x| > 1, then x is in E

So now

[nova spire]

what does it mean for (x',y') to be in C?

[wraith hinge]

and C is which set again

[nova spire]

this set

[wraith hinge]

just means it adheres to set conditions which such set provides

[nova spire]

and the conditions being?

[wraith hinge]

which ever conditions the set had given

x+y < 5

x-y^2 > 2

[nova spire]

In this case, the conditions are?

It's always important to restate what the conditions are, otherwise we fail to progress

[wraith hinge]

that x must be positive

but below 1

so inbetween (0,1)

[nova spire]

Are you sure x has to be positive?

-0.5 works

[wraith hinge]

what is |-0.5| equal to?

you think its -0.5?

[nova spire]

|-0.5| = 0.5 which is < 1

[wraith hinge]

i assumed that step was one step

any interger which gets inserted into || will become positve

[nova spire]

first of, non-negative*

and it's the output that is non-negative, not the input itself

[wraith hinge]

alright

so the values own existence precedes the computation which is done upon it

[nova spire]

yes

[wraith hinge]

ok, so -0.5 is also an option

[nova spire]

So, just say that |x| < 1

and |y| < 1

"(x',y') in C" means that "|x'| < 1 and |y'| < 1" and vice versa

is this alright?

[wraith hinge]

yes

[nova spire]

In general, when we say "what does it mean for x to be in such set", usually it's best to write out the condition(s)

without modification

and then build from that

and the other way around :

[wraith hinge]

now that we have concluded that, yes indeed, x' and y' are also contained within a set

now what

[nova spire]

If we prove that "|x'| < 1 and |y'| < 1", then we will have proved that (x',y') in C

[wraith hinge]

yes i set x' = 0.5 and y' = 0.5

ok they're in the set C

[nova spire]

ok, now we need the generalization

[wraith hinge]

yes

[nova spire]

so let's go back to this

Since you're struggling, i will add more steps in between so you see what needs to be done

"Let (x,y) in C.
We introduce epsilon = ...
Let (x',y') in B((x,y), epsilon).
This means, by definition of B((x,y), epsilon), that...
...
Thus |x'| < 1 and |y'| < 1.
Thus (x',y') is in C.
Thus (x,y) has an epsilon neighborhood.
Thus C is open"

Is this becoming clearer as to what to do?

[wraith hinge]

no

i dont see why any of this is necessary

[nova spire]

because in order to prove "(x',y') is in C" (what we want), we will need to compute |x'| and |y'| or at least reach some inequalities on them

[wraith hinge]

i dont see how episolon isnt the set

why does that need to be proved

ofc it is in the set

[nova spire]

which part?

[wraith hinge]

why do we introduce x' and y'

which could just be represented by our x and y

x' and y' is nothing but notation

[nova spire]

Well no, (x,y) are already in use

It's like I said "Let x be an even number.
Let x be an odd number.
We have odd number = even number"

[wraith hinge]

yes that means x will be an even number

why do we have to write "if it is an odd number, then it becomes even"

we have two options

and you can only pick one

guess which one you will pick?

[nova spire]

Are you even understanding what I'm trying to point out?

If you want to take a random element of C first, and then pick another random element in B((x,y), epsilon), I can't use the same notation can I?

[wraith hinge]

our set is |x| < 1 and |y| < 1

why cant this be done simply by the number line

[nova spire]

If you can find a way to "do it simply by the number line", be my guest

[wraith hinge]

i am asking why

[nova spire]

because of the very nature of B((x,y), epsilon)

it can't be expressed as something on the number line

[wraith hinge]

we have a plane that never reaches one?

or is that incorrect

but gets closer and closer

[nova spire]

In what sense does a plane of pairs of real numbers reach 1?

[wraith hinge]

what does of couples mean

[nova spire]

tuple, pair

[wraith hinge]

i mean their value

[nova spire]

sure, you can technically represent this set, C, as two number lines, one for x and one for y. But then how would you represent B((x,y), epsilon)?

It can't be done using number lines, we need a plane

[wraith hinge]

episolon is just added onto x and y

[nova spire]

well definitely not added

[wraith hinge]

it is supposed to change the value of x and y?

yes?

[nova spire]

yes, and not by "adding" and not independently either

[wraith hinge]

maybe you can go against your will and finish this here

i think you drastically underestimate the power of actual examples

not certain why you refuse to try to give one

[nova spire]

I will show you the full example here

"Let (x,y) in C.
We introduce epsilon = min(1-|x|,1-|y|) > 0.
Let (x',y') in B((x,y), epsilon).
This means, by definition of B((x,y), epsilon), that sqrt[(x-x')^2 + (y-y')^2] < epsilon.
This implies that |x-x'| < epsilon and |y-y'| < epsilon.
By triangular inequality, |x'| = |(x'-x)+x| <= |x'-x| + |x| < epsilon + |x| < 1.
Thus |x'| < 1. Similarly, |y'| < 1.
Thus (x',y') is in C.
Thus (x,y) has an epsilon neighborhood.
Thus C is open"

[wraith hinge]

how was i supposed to this

[nova spire]

as I said, working backwards

[wraith hinge]

i've never seen anything like this

[nova spire]

it's your first time so I can understand why you were lost a bit

But I now have to show you why this example works otherwise you won't get anything from it

[wraith hinge]

do you have any vidoes to watch on this

i could not find anything on this with two variables in R^2 to solve for open,closed or neither

[nova spire]

The method for "two variables" is the same as with one variable

because "two variables" = "a vector with two coordinates" = "one variable, the vector"

How do you prove, for example, that (1,3) U (5,6) is open?

[wraith hinge]

because this here

is an interval on a numberline

[nova spire]

(without using the result that open intervals are open and closed intervals are closed)

[wraith hinge]

[-1,1] is closed

(-2,2) is open

(-2,2] is neither

then i have no idea

[nova spire]

Well then it's back to the same problem

so I recommend you to this video maybe: https://www.youtube.com/watch?v=-QjTtHe27Bc

Or this https://www.youtube.com/watch?v=pnWgj8jjs3w

[wraith hinge]

A C_ R is closed if (in R) if A^C = R\A is open

[nova spire]

Because clearly the problem is not that you can't prove it for 2D, it's that you've never proved it using the epsilon neighborhood definition

[wraith hinge]

for 1d you have a number line

so its very intuitive which are open or closed

[nova spire]

ok*

Is R\Z an open set?

[wraith hinge]

what is Z in this case

[nova spire]

integers

[wraith hinge]

R is real numbers?

[nova spire]

yes

[wraith hinge]

All real numbers that arent intergers?

[nova spire]

yes

[wraith hinge]

what is the definition on an interger in english

[nova spire]

0, 1, -1, 2, -2, etc...

[wraith hinge]

i'd say its open

[nova spire]

ok, can you prove it?

[wraith hinge]

no

divine intution

[nova spire]

well then, i thought open sets in R were obvious

[nova spire]

back to this problem

[wraith hinge]

i cnat prove any sets

as we've just come to

[nova spire]

yes, well you can guess what your teacher will reply to "it's very intuitive"

[wraith hinge]

i can

[nova spire]

yes

[wraith hinge]

sadly

[nova spire]

So I would recommend this video to find examples of proofs of open sets

[wraith hinge]

would a proof be that between each interger is infintel many numbers?

[wraith hinge]

i've already seen this video

[nova spire]

well no, "infinitely many numbers" doesn't guarantee "every non-integer has an epsilon neighborhood"

[nova spire]

ok, so you must have seen that in the part "examples of open sets" for each x, we find an epsilon that gives us B(x,epsilon) an epsilon neighborhood

[wraith hinge]

i have seen examples of open sets

[nova spire]

I'm not talking about the open sets, I'm talking about the examples of proofs

[wraith hinge]

but he is just making a number line and saying

[nova spire]

and saying delta = ...

[wraith hinge]

"wow look guys i can find an epislon in this set"

so its open

[nova spire]

No, he EXPLICITELY gives epsilon

"take delta = 2"
"take delta = min(x-a,b-x)"

[wraith hinge]

that is no different from what i am doing

[nova spire]

it is, you're not explicitely giving epsilon = ...

[wraith hinge]

yes

but the idea isnt that only two works

5 should also work in the first one

so should 800 million

or -300