help-43

[quartz yoke]

The only hard one to get is <PQA

you can notice that ||CQ bisects <PCM (you probably already know this) and MQ bisects <PMC (given in the statement). So what do these 2 things imply?||

[compact pewter]

@icy topaz Has your question been resolved?

[icy topaz]

That Q is the incentre of the triangle PMC

[quartz yoke]

yeah

this should allow you to angle chase and find <PQC

then you can do this

[icy topaz]

Got it.

Thanks.

.close

[compact pewter]

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[forest token]

adding x^3 and subtracting x^3 to that first integral

and then i thought x^3-sin^3x ill write as (x-sinx)(x^2+sin^2x+xsinx)

but i didnt know what to do from there

@native shard

[native shard]

should be + xsinx

[forest token]

yea thats what i meant

how do i proceed

or start

ykw just leave it

.close

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[twin idol]

Do you know the formula for summing the natural numbers from 1 to n?

[strange pendant]

n(n+1)/2

[twin idol]

Ye

Q 3 will need you to use it

Also use the fact that summation is linear

[strange pendant]

can you elaborate

[twin idol]

$\sum_{i = 1}^n ki = k\sum_{i = 1}^n i$

[boreal girder]

VulcanOne

[strange pendant]

what about it

how do I do i) for example

[twin idol]

You will need to use this fact to deconstruct the summation

And make it easier to use the formula

Also

$\sum_{k=1}^n \left(k + c\right) = \left(\sum_{k=1}^n k\right) + \left(c\sum_{k=1}^n 1\right)$

[boreal girder]

VulcanOne

[twin idol]

@strange pendant

Stop playing osu and focus here

[strange pendant]

dude can you help me

step by step please

[twin idol]

You will need to focus on the problem so I can help

[strange pendant]

I am focused, 100%

[twin idol]

On osu that is

[strange pendant]

im here dude

[twin idol]

I gave you the 2 most important facts for Q 3

[twin idol]

Fact 1

[twin idol]

Fact 2

[strange pendant]

well

how do I join the pieces together then?

[twin idol]

Well

Use fact 2

Then use fact 1

[strange pendant]

how ?

[twin idol]

I want you to reach the answer by yourself Renato

[twin idol]

There you go renato

[strange pendant]

@twin idol

[twin idol]

Now take n as a common factor

Can you do ii in the same way?

@strange pendant

[strange pendant]

how?

[twin idol]

ii has a bit of a trick up its sleeve

[strange pendant]

dude HELP

[twin idol]

[twin idol]

That's correct

You did well

Now are you able to do 2 in the same way?

[twin idol]

Yeah

Can you do it in the same way?

I'll need to give you another fact

Because of the trick

$\sum_{k=c}^n k = \left(\sum_{k=1}^n k \right) - \left(\sum_{k=1}^c k\right)$

[boreal girder]

VulcanOne

[twin idol]

@strange pendant yeah please exit osu for now

[strange pendant]

the problem is that the starting index is 6 not 1 dude

[twin idol]

Yeah that's what the fact is for

[twin idol]

Fact 3

[strange pendant]

now what?

[strange pendant]

@twin idol

are you in osu dude?

[twin idol]

Nop

I'm not on osu

4 hours and I'll go to my work week

[twin idol]

Alright nice

[twin idol]

That's great

[twin idol]

Exactly!

Now apply formulas

[strange pendant]

now what dude

I need a fact number 4

[twin idol]

Nah all the facts are given

No need for fact 4

Fact 4 is gonna be for Q4

[strange pendant]

?

[twin idol]

I won't have access to my laptop

[twin idol]

Exactly

Make sure to evaluate the sum from 1 to 6 of 5

[strange pendant]

how

[twin idol]

Well

[strange pendant]

please elaborate

[twin idol]

The same formula you used for $\sum_{k=1}^n 5$

[boreal girder]

VulcanOne

[twin idol]

But instead of n it will be 6

[strange pendant]

?

what about the 5

[twin idol]

$\sum_{k=1}^6 5$

[boreal girder]

VulcanOne

[twin idol]

Evaluate it

[strange pendant]

5*6

[twin idol]

Yeah

[strange pendant]

lmfao

[twin idol]

Lmfao

Now simplify

[strange pendant]

@twin idol @twin idol @twin idol @twin idol @twin idol

[twin idol]

Correct

[strange pendant]

NOW WHAT

@twin idol

[twin idol]

Now we can get into Q4

[strange ermine]

Stop asking this! Be patient!!

[strange ermine]

!noping

[compact pewter]

Please do not ping individual helpers unprompted.

[strange pendant]

what is the trick to q4?

[strange ermine]

And if you really want to ping, do it once

[strange ermine]

Because this sounds very very bad-manner, even if it's not your intention

[strange pendant]

you can always use discord from the phone

[cerulean bramble]

He's gonna be at work

[twin idol]

for a week straight

[strange pendant]

I didnt know

[twin idol]

Alright. Please tone it down with the pings

Alright now let's move on to Q4

$\sum_{k=0}^{n-1} p^k = \frac{1-p^n}{1-p}$

[boreal girder]

VulcanOne

[twin idol]

Fact 4

[strange pendant]

I dont think I can take this as true

[twin idol]

Why not?

[strange pendant]

I dont see how is this true

I dont think I follow

[twin idol]

This is the sum of a geometric series from k = 0 to n-1

[twin idol]

Check the indices

[strange pendant]

well

[twin idol]

I gave you the sum with different indices. It's a correct formula.

[twin idol]

This is also a correct formula

For the used indices

Anyways

[twin idol]

Use this as fact 4

[strange pendant]

how did you came up with this?

[twin idol]

https://www.cuemath.com/geometric-sum-formula/

[twin idol]

Use this as fact 4

[strange pendant]

[twin idol]

Check the proof after scrolling down

[twin idol]

Buddy I'll have to leave now. I gotta do some stuff before going to work

[strange pendant]

$\sum_{k=0}^{n-1} p^k = \frac{1-p^n}{1-p}$

[boreal girder]

Renato

[twin idol]

Let j = k+1

$\sum_{j=1}^n p^j = \frac{1-p^n}{1-p}$

[boreal girder]

VulcanOne

[twin idol]

Same thing

Anyways gotta leave now

Good luck. You can do this on your own I know that

[strange pendant]

[strange pendant]

have fun at work dude, ping me if you want to osu

[compact pewter]

@strange pendant Has your question been resolved?

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[frank root]

Did I set this up right?

[vivid breach]

do Win + Shift + S to take screenshots

also, go show your answers for (a) through (d) before we consider (e)

[frank root]

A. 15,957 B.797.85 C.”It measures the rate of change of people” D.9092.15

[vivid breach]

(c) is too vague, theres multiple ways to measure that rate of change

you need to be specific that its the average rate of change from 1980 to 2000

(d) btw is exactly the same as (b)

[frank root]

Alright so we’re back on question e

[vivid breach]

you know why (d) = (b), right?

[frank root]

The average rate of change divided by 20 right?

[vivid breach]

the average rate of change already includes a /20, as you did in (b), so no

you dont divide the average rate of change by 20 a second time

[frank root]

Can u help me on e now?

[vivid breach]

do you know what average rate of change is

[frank root]

797.85

[vivid breach]

a number isnt going to convince me you know why its supposed to be like that

where did you even get 9092.15 from

[frank root]

181843/20

is where i got 9092.15 from

[vivid breach]

oh alr, thats wrong but I see you wont be doing that anymore

now for this

unless youre a fan of writing down numbers with decimals on paper all the time, we gotta get the calculator to do this for us

[frank root]

i got desmos pulled up

[vivid breach]

now the best-case scenario is telling the calculator a and b and it tells us back the average rate of change

to do this, we have to make something that can calculate this

sort of plan the calculation out

lets say youre figuring out the average rate of change on [5, 10]

what would you calculate to get this answer?

[frank root]

797.85-2000/10-5?

[vivid breach]

lets look at the numerator again

also dont forget ()s

for average rate of change on [5, 10], you need (something) / (10 - 5)

797.85 - 2000 is not the correct idea

we're nto considering average rate of change of average rates of change

first,

do you understand this part of the problem

[frank root]

That's the formula I should be using

[vivid breach]

surely theres more to P(t) than "oooh its a formula"

maybe that it means something about the population?

[frank root]

p(t)=population

[vivid breach]

P(5) represents the population in what year?

[frank root]

1985?

[vivid breach]

yea

[vivid breach]

try redoing this using P(t)

[frank root]

(1985-1.04) / (10 - 5)?

[vivid breach]

???

why 1985?

are you just coming up with random things to put in the numerator?

[frank root]

I thought thats why u asked me this

[vivid breach]

Im asking you that because, no, you dont just put 1985 - 1.04 for the numerator

lets try something else

earlier you said P(5) means the population in 1985

so we have a way to go between 5 and the year 1985

based on that, [5, 10] represents what 5-year span?

[frank root]

1980

[vivid breach]

1980 is not a "5-year span"

1980 is just a single year

lets restart

10, what year does this represent?

(you already know 5 represents the year 1985)

[frank root]

1990

[vivid breach]

good

[5, 10] is a closed interval from 5 to 10

what does that mean?

[frank root]

1985-1990

[vivid breach]

so where did this come from?

[frank root]

I went backwards instead of forward

[vivid breach]

??????????????

thats a bold move

anyways [5, 10] represents 1985 to 1990

so since P(t) represents the population in 1980 + t,

P(5) represents the 1985 population
P(10) represents the 1990 population

[frank root]

(1990-1985) / (10 - 5)?

[vivid breach]

those are just the years

thats 1

thats saying 1 year happens every year, on average

think about what youre doing

[frank root]

mb

[vivid breach]

let me continue

so P(5) tells us the population at the beginning of [5, 10]

and P(10) tells us the population at the end of [5, 10]

we know the years that these represent, but we dont need to use that information for now

I was just giving context for what this (otherwise vague) thing means

[vivid breach]

you understand these, right

[frank root]

P(5) is the population at the beginning of the interval [5,10]

[vivid breach]

right

we can calculate P(5) to be 185444
and P(10) to be 189117

so whats the average rate of change for the population from 1985 to 1990?

[frank root]

where did u get these numbers from 185444 189117

[vivid breach]

ok you dont understand this then

I want you to actually know what this image means

read this closely, what does this mean?

theres not like a hidden message or anything, I just want you to know what P(t) even is supposed to be about

[frank root]

you just multiplied 181843(1.04) and got 189117

[vivid breach]

kaleb you do see the big formula in the middle of the picture right

P(10)

[frank root]

so you just replaced t with 10 and did the same with 5?

[vivid breach]

yes, thats how you use a function

moving on,
185444 is the population in 1985
189117 is the population in 1990
so whats the average rate of change in population from 1985 to 1990?

going to remind you btw that you didnt have a problem with this problem when it was phrased like (b), so this time you have no excuse

look back on the calculation you did for (b) if youre stuck on finding this rate of change

[frank root]

734.6

[vivid breach]

there you go

now lets consider another problem with the way we're doing things

remember that we were trying to plan out a calculation, right?

[frank root]

yes

[vivid breach]

when I ask, all youre giving are the ending results

you didnt show what you did

we cant plan just on the final result

[frank root]

how do i sc again

[vivid breach]

Win + Shift + S

theres an option in the windows settings to remap this to the PrntScreen button, if you want

not sure if itll do the same thing though
last time I checked, PrntScrn copies your entire screen to your clipboard

[vivid breach]

yep

ideally you dont want to reveal your entire screen

[frank root]

mb

[vivid breach]

select an option at the top of the Win + Shift + S menu to change which kind of selecting you can

default option is whole screen, but you can also chosoe to screenshot a window or screenshot a rectangle of your choice

Im screenshotting rectangles with Win + Shift + S to get my pictures

[frank root]

ok thanks

[vivid breach]

anyways for here, small change we gotta do for convenience later

we change the denominator to 10 - 5 instead

that way, its closer to the t we used

10 - 5 and 1990 - 1985 both calculate a 5-year difference

since the units are in years

now for the final change

189117 and 185444 werent just any numbers

we had to calculate them, from a formula

what function gave us those two numbers?

you dont have to write out the whole formula or anything, you can just mention the function

[frank root]

P(10)=181843(1.04)^t/10

mb

[vivid breach]

dw

youre being precise

youre forgetting ()s though, ^(t/10) instead

and that t should be a 10 as well

[frank root]

wouldnt that just be 1 for 10?

[vivid breach]

yep

dont simplify it though

[frank root]

alr

[vivid breach]

we're going to leave that up to the calculator

really, this is our formula

keep in mind what we based this all on: (b)

(b) had you do the same kind of calculation for the average rate of change

now we see the same formula again, but for P

(P(b) - P(a)) / (b - a) finds the average rate of change of the function P from a to b

on [a, b]

[vivid breach]

this is an example for [5, 10]

from here, Ill show you two ways you can get the calculator to calculate this

[frank root]

ok

[vivid breach]

first, by default desmos doesnt know what P is

define the function P in a separate line

desmos then knows that its a function, so P(5) and P(10) can be calculated

for desmos in particular, use x for the input variable

using another variable (like t) might mess things up if you use t for something else

to further shorten how much we need to do, replace 5 and 10 with a and b

that way instead of replacing both 10s with 9s, you can just replace the =10 with =9

[vivid breach]

yep

theres a closely-related second way to this

take a look at the guess you started with

I dont understand how you cooked any of this to the way it is

but does look like you want to reduce it all into as few lines as possible

all in one big formula

[frank root]

Trial and Tribulations

[vivid breach]

Trials and Tribulations

consider instead if we had something like this

[frank root]

little too advanced for me

[vivid breach]

compared to the previous one, theres a lot more going on here, but its the same calculation

first

this is a function that takes in two inputs

it is written f(x, y)

f(5, 10) will have the input x=5 and y=10

yes?

[frank root]

yes

[vivid breach]

second

this is just P(y) - P(x), isnt it

you dont seem to recognize it given your immediate reaction

[frank root]

yeah but i probably wouldnt come up with this on my own

[vivid breach]

my brother in christ, you copy-paste P into there

you can come up with copy-pasting

thats all I did, nothing more

youre seeing what Im seeing right? I didnt do any further steps than just writing down P(t) into there

[frank root]

yes i can see that

[vivid breach]

wdym "come up"?

[frank root]

like for me I see all the steps i did to come up with answer to actually understand it

[vivid breach]

well we didnt reach the final step yet

[vivid breach]

so you mean this for the whole formula then

[frank root]

wym

[vivid breach]

sure

final thing, instead of a and b we use x and y

no big reason other than for desmos to like it better

[frank root]

Oh alright

I put in the other numbers thats it right?

for the problem

[vivid breach]

yep

we made the function so that you can do that

f(5, 10) for [5, 10]

you can also see the function so you can be sure that its calculating the right thing

its doing (P(y) - P(x)) / (y - x) for the x and y you put in

[vivid breach]

(P(a) - P(b)) / (a - b) also works

you can multiply by -1 / -1 to swap between this and the other one (P(b) - (a)) / (b - a)

[frank root]

But what I did is correct right?

[vivid breach]

yea

[frank root]

Alright thank you so much for your help idk how u put up with me I was trying but some of this just didn't make sense to me

and im just using my calculations i used for part E on part F right?

[vivid breach]

(f) has you do something different

consider this

(f) wants you to consider how the numbers change as you go from [5, 10] down to [5, 6]

then use that to guess what the "instant" rate of change at 1985 is

now f(5, 5) wont work, since thats a rate of change with / (5 - 5) which divides by 0

so they want you to roughly estimate it from these

[frank root]

Its getting closer to 727

[vivid breach]

you can also just do f(5, number really close to 5)

it gets close to 727.596

Im not too sure what they mean by the second question

[vivid breach]

yep

[vivid breach]

second question, it should just be P(6) - P(5) but not sure on why theyre just asking you to do that

seems too easy

but it should be correct

[frank root]

Alright thank you

[vivid breach]

np

[frank root]

Don't go to far ill be back tomorrow your one of the best tutors I've ever had

[compact pewter]

@frank root Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
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[hushed tapir]

I need help with part 2 of question 1.5. This is real analysis 1

[hushed tapir]

<@&286206848099549185>

[old mica]

have you tried anything yet?

[hushed tapir]

Yes

I’ll send it right now

I think it’s wrong though

Would you like to see my professor’s hints?

[old mica]

hm, you're picking z to be an element of X, but that isn't what we want to prove

[upper bane]

you are claiming that sup(X) = 1/3, right? why not find the difference between 1/3 and f(x) in terms of x?

[old mica]

we'd like to show that if s > z, then z isn't an upper bound of X - that's not the same as proving that no element of X is an upper bound for X

[hushed tapir]

We haven’t learned functions yet @upper bane

[upper bane]

replace f(x) with that expression

[hushed tapir]

Huh?

I’m just confused where to start

And where to go

[old mica]

you assumed that z has the form (x^2 - 1)/(3x^2 - 1), when this doesn't have to be the case

as for what to do, Hanako's suggestion is useful

[upper bane]

the idea is, you want to prove that sup(X) = 1/3, so you have to prove that there is no z < 1/3 that is an upper bound

[hushed tapir]

Here’s what my professor wants me to do

[upper bane]

from here, find the difference between 1/3 and the expression.
then, let z < 1/3 and set $\epsilon$ = 1/3 - z > 0

[boreal girder]

Hanako

[upper bane]

from there, the difference between 1/3 - z is exactly the difference between 1/3 and the expression found earlier

and we want this difference to be smaller than epsilon

[old mica]

here's a more explicit plan of action:

first, prove that the sequence (n^2 - 1)/(3n^2 - 1) converges to 1/3. spoilers for the rest: ||then if z < 1/3, the distance from 1/3 to z, given by the difference |1/3 - z|, is equal to some positive number epsilon. but since that sequence converges to 1/3, the difference between (n^2 - 1)/(3n^2 - 1) and 1/3 goes to 0 as n goes to infinity, so for a sufficiently large value of N, you can pick an element of X that's larger than z (distance to 1/3 is less than epsilon), hence z isn't an upper bound.||

[hushed tapir]

I can’t use sequences

[old mica]

then what can you use?

[hushed tapir]

It’s like very limiting with what I can use

The definition of LUB

And ordered sets

The hints doc I sent pics of lays out kinda what I should do

Also, Archimedean property could be useful

[old mica]

I must apologize for misleading you then

I don't think I can figure out a way to do this problem with what you have

[hushed tapir]

Then what should I do

I’ve been stuck on this one

I can show a problem where I proved GLB, which might help

<@&286206848099549185>

[quiet minnow]

?

[hushed tapir]

I need help with this analysis problem, and other people can’t figure out how to do it with my current level of knowledge

[compact pewter]

@hushed tapir Has your question been resolved?

[hushed tapir]

No

<@&286206848099549185>

[twin anchor]

can you repost the question and your progress so far

im not sure how far youve gotten

[hushed tapir]

I have basically no progress

[upper bane]

i have an attack plan, but i don't know if i'm correct

sorry

[hushed tapir]

It deals with smth like this

[twin anchor]

it seems like you have a lot written

thats good

[hushed tapir]

I only proved that it’s a upper bound, not that it’s a LUB

[twin anchor]

okay, cool

there's 2 ways to show that an upper bound s is a LUB

way 1: show that all other upper bounds are >= s

way 2: show that any number < s is not an upper bound

I think way 2 is easier here

[hushed tapir]

That’s what the professor said

[twin anchor]

let's try that then

[hushed tapir]

Did you read the hints?

[twin anchor]

let's take some number less than s. call it s - ε for some ε > 0

no, do you want me to? i have a fairly certain idea of how to go about this but if theres a specific way you want this done we can look at that

[hushed tapir]

Can you? It’ll give kinda an idea of how he wants it. I only just started analysis so we have learned: ordered sets, LUB GLB, BFA, and archemdian property

[twin anchor]

that seems like enough

where are the hints

is that the red

[hushed tapir]

Yeah, it’s typed

[twin anchor]

everything they said seems pretty reasonable and normal

i dont think theres a special way you have to do this

[hushed tapir]

So what should I start doing

[twin anchor]

.

[hushed tapir]

I have that already, I called it z

[twin anchor]

we want to show that for some n in N, (n^2 - 1)/(3n^2 - 1) > 1/3 - ε

i think it might be easier to work with 1/3 - ε instead of z

because then you isolate a positive part and the limiting part

[hushed tapir]

Why would that prove this

[twin anchor]

are you asking why we want to show that for some n, (n^2 - 1)/(3n^2 - 1) > 1/3 - ε ?

or why i decomposed it

into 1/3 and ε

[hushed tapir]

So you’re saying trying to find an x s.t. It’s inbetween the LUB and some variable ε

[twin anchor]

no, let's just use your z variable

[upper bane]

(sorry to bother, but when you're done, snowflake, could you pls look in bots and see if my version works? thank you, and sorry for imposing. it's ok if you can't though!)

[twin anchor]

i want to show that for any z < 1/3,

(n^2 - 1)/(3n^2 - 1) > z

for some n. I just want to show that z isn't an upper bound, and this is the definition of that

[hushed tapir]

Yes

I understand

[twin anchor]

okay great. it's just equivalent that if z < 1/3, then I can write z as 1/3 - ε for some ε > 0, and then the inequality just becomes finding n such that

(n^2 - 1)/(3n^2 - 1) > 1/3 - ε

which is equivalent to finding n such that

ε > 1/3 - (n^2 - 1)/(3n^2 - 1)

and it's generally easier to aim for something to become arbitrarily small/close to 0 than to work around an annoying variable like z that has implicit stuff floating around it

that's just the motivation I had for decomposing it, it makes things a little more straightforward this way

for all ε > 0, we just need to find some n such that 1/3 - (n^2 - 1)/(3n^2 - 1) < ε

do you still follow

[hushed tapir]

Yes

[twin anchor]

great

[hushed tapir]

Let me write down you’re thinking real quick

Why are we allowed to make this equality

We don’t know if such n exists yet

[twin anchor]

this is the goal, we don't know this works yet

the goal is to obtain an n that makes this true

if it does, then we've found an n that disproves 1/3 - ε being an upper bound

if we can do this for arbitrary ε, then 1/3 must be the LUB

a common method (the common method) in analysis is to "work backwards" from inequalities that you want to be true, to solve for appropriate conditions that make them true

since for each ε, we only need to find a single n that satisfies the inequality, we have a good amount of flexibility to solve for valid n

does that makes sense

[hushed tapir]

Yeah, so we use the inequality we want to be true to solve for something? That seems circular

[twin anchor]

no

we are not using the inequality at all yet

we are going to use basic facts to prove the inequality for some n first

then, we can use the inequality to prove our claim that 1/3 is the LUB

the trick here though is that often times, operations on inequalities and equations are "invertible"

we can obtain equivalent inequalities by adding the same thing to both sides for instance, or multiplying by a nonzero number

the idea is if we can do some "allowable" manipulations to this inequality, constantly "thinking backwards" about what these manipulations imply, then we can eventually find some n that will directly imply the original inequality

you did this all the time in grade school

when i want to solve 3x > 18 for x, I first look at 3x > 18, and then divide by 3 to get x > 6, right?

maybe that seems circular. I started with 3x > 18 to find x values that satisfy 3x > 18, how does that work?

well, because dividing by 3 is an invertible operation on inequalities, I knew I could always just work backwards from wherever I end up to get the inequality I wanted

[hushed tapir]

Alright

So how does it apply here

[twin anchor]

we havent even gotten to the solving step yet

[twin anchor]

are you okay with this

[hushed tapir]

Yeah

[twin anchor]

this is the important logical understanding

okay

[hushed tapir]

It’s the goal?

[twin anchor]

yes

but we're gonna try manipulating it directly first

in ways that we can easily reverse

or at least reason about

so that we can find valid n that we can apply the steps in reverse for to get the desired goal

so let's start manipulating it

[hushed tapir]

Yes!

[twin anchor]

ideally, we get the n by itself

a polynomial / a polynomial is pretty annoying

do you have any ideas on how to make it simpler

[hushed tapir]

We should create a simpler rational fraction

[twin anchor]

yup

[hushed tapir]

n^2-1/3n^2

It’s gonna be smaller

[twin anchor]

that isn't something we can legally do though

we need valid operations

you could potentially make an argument to do that, what would your argument be

[hushed tapir]

Idk

That’s what the hints said to do

[twin anchor]

right, youre close already

[twin anchor]

just finish this thought

[hushed tapir]

n^2-1/3n^2-1 > n^2-1/3n^2

[twin anchor]

if that was true, that wouldnt be helpful for us

if I only know that c < a, I can't use c in any way to prove that a < b

if c > a, then i could just prove c < b to show a < b

[twin anchor]

the issue is that this isnt true

[hushed tapir]

Why?

[twin anchor]

oh wait your denominator includes -1

do you know latex?

[hushed tapir]

I didn’t include parenthesis, want me to rewrite

Yeah but my computer is far and my roommate is sleeping

[twin anchor]

parentheses would be rlly helpful yeah

$\frac{n^2 - 1}{3n^2-1} > \frac{n^2-1}{3n^2}$

[boreal girder]

snowflake

[twin anchor]

this is what you wrote right

[hushed tapir]

Yes

[twin anchor]

okay perfect

yes this is true

now how can we manipulate this to relate to our goal inequality

[hushed tapir]

Idk how to do that

We gotta introduce z

I think

[twin anchor]

reminder that this is our goal inequality

no we dont

just look at the LHS

our goal was to simplify it right

[hushed tapir]

Yes

[twin anchor]

and we said that (n^2 -1)/(3n^2) is smaller than (n^2 - 1)/(3n^2 - 1)

[hushed tapir]

Yes

[twin anchor]

if i subtract a smaller thing from the same number, whats the effect on the overall difference

[hushed tapir]

Huh?

[twin anchor]

you can figure this out

if $a < b$, then what's the relationship between $x-a$ and $x-b$

[boreal girder]

snowflake

[hushed tapir]

x-a bigger

[twin anchor]

yes

and we can prove that from simple inequality rules, multiply by -1 on both sides and add x

so with that in mind, what does the inequality look like with 1/3 and our rational expression

[hushed tapir]

1/3 greater than our expression?

[twin anchor]

$\frac{1}{3} - \frac{n^2 - 1}{3n^2-1} < \frac{1}{3} - \frac{n^2-1}{3n^2}$

[boreal girder]

snowflake

[hushed tapir]

Yes!

[twin anchor]

do you see why we care about doing this

[hushed tapir]

Not really

[twin anchor]

we wanted to show that the LHS was < ε right

[hushed tapir]

Yeah

[twin anchor]

what can we instead show now

[hushed tapir]

Um?

Something with our new fraction

[twin anchor]

this is still our goal

instead of showing this directly

we can do something else that will prove this indirectly

[twin anchor]

look at this

think about transitivity of inequalities

[hushed tapir]

What’s transitivity of inequalities

[twin anchor]

if a < b and b < c then a < c

[hushed tapir]

But our new inequality is bigger, so how would that help

[twin anchor]

we know a < b, and we want to show that a < c

what do we have to show for that

[twin anchor]

based on this

[hushed tapir]

I feel so stupid, I’m really confused. I get the idea of using another inequity but idk how that applies here

[twin anchor]

okay, so we had our original quantity a (a is our 1/3 - fraction).

we want to show that it's smaller than this other quantity c (c is epsilon).

we found a new quantity b that's bigger than a (b is 1/3 - the new fraction).

if we can show that b is smaller than c, and we know that b is bigger than a, then a must also be smaller than c

right now, we know b is bigger than a, so all we need to do is show that b is smaller than c. if we can do that, then we've indirectly proven that a is smaller than c

and a < c is our goal

[hushed tapir]

Yes!

[twin anchor]

so what do we want to prove right now

[hushed tapir]

1/3 - new fraction is less than ε

[twin anchor]

perfect

in other words, we need to find n such that $\frac{1}{3} - \frac{n^2 - 1}{3n^2} < \epsilon$

[boreal girder]

snowflake

[hushed tapir]

Yes

[twin anchor]

what are your ideas to do next

[hushed tapir]

Solve for n in terms of epsilon

[twin anchor]

right yes

how should we do that

[hushed tapir]

Make a new inequality?

[twin anchor]

(do you see how to solve from here)

no

we can solve from this one

what is the literal first step we can take right now

why was it helpful to get rid of that -1 in the denom

[hushed tapir]

Distribute

[twin anchor]

exactly

after doing that and simplifying, what do we get as our new equivalent condition for n

[hushed tapir]

But we are using something we are trying to prove

[twin anchor]

yes, distributing is a reversible operation

[hushed tapir]

No, I mean with the epsilon and stuff

[twin anchor]

so i know i can undo my steps once I can get to n

we haven't specified anything on epsilon yet

its fully arbitrary still

and our n is allowed to depend on epsilon

we're trying to prove that for all epsilon, there exists some n such that the inequality holds

so that n can intrinsically depend on the specific epsilon

our goal was to show that inequality right? if we can apply a series of reversible steps to reach a direct constraint on n based on the given epsilon, then for that epsilon, we can find a specific value of n and work backwards to show that the inequality holds

it might make more sense once we finish

[hushed tapir]

The inequality should be switched in the last step

[twin anchor]

n > right

that seems good to me

so after applying our reversible steps, we got that n should be > sqrt(1/(3ε))

because all of our steps were reversible, this means that if n > sqrt(1/(3ε)), then (n^2 - 1)/(3n^2) > 1/3 - ε, meaning that (n^2 - 1)/(3n^2 - 1) > 1/3 - ε, proving that the RHS isn't an upper bound

you can easily verify that this is true by proving this starting from n > sqrt(1/(3ε)). we just don't have to because all of our steps were reversible

[hushed tapir]

Huh

So we found an inequality satisfying n, so we using the transitive properties of inequalities to show that since n exists, then the inequality exists

[twin anchor]

we found an inequality describing the necessary conditions on n for our goal inequality to be true

and then yes, we showed the goal inequality through transitivity using our modified inequality

[hushed tapir]

One sec

Let me formulate this

So where from here

[twin anchor]

pick some integer n such that n > sqrt(1/(3ε))

then it follows that 1/3 - ε is less than that corresponding element of the set

that's basically it

[hushed tapir]

So what did that prove

In that since we found a condition for n, then 1/3 must be the sup

[twin anchor]

we proved that every number less than 1/3 isnt an upper bound

since you already showed 1/3 is an upper bound, that makes 1/3 the least upper bound

[hushed tapir]

Awesome

Thank you so much

[compact pewter]

@hushed tapir Has your question been resolved?

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[wheat seal]

Help needed

[wheat seal]

What do you mean by max value of x + y + z
When they are complex

[compact pewter]

@wheat seal Has your question been resolved?

[rotund sphinx]

1. Real numbers are complex
2. The imaginary parts can cancel

[compact pewter]

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[carmine garden]

wrong problem

my bad

[mild sky]

Well every element of H has odd order

[carmine garden]

well, how

[mild sky]

Lagrange's theorem

[kind viper]

order of element divides order of group

[carmine garden]

ah right

Lemme prove that

right

follows from the order of a group, divides the order of it's subgroups

[mild sky]

The other way

[carmine garden]

oops yea

[mild sky]

Order of subgroup divides order of group

Brb

[carmine garden]

I'm not sure if it's 0 or infty here

[mild sky]

Infty

[kind viper]

order of an element is never 0

[carmine garden]

right, I got confused

sorry

[compact pewter]

@carmine garden Has your question been resolved?

[carmine garden]

so like if rotating by 90deg preserves symmetry we're done

[kind viper]

not every regular polygon survives a 90° rotation

it doesn't work for n=6

[carmine garden]

yeah I just realised what to do

[kind viper]

note that having a subgroup of order 4 isnt the same as having an element of order 4 -- the latter property is stronger!

[carmine garden]

We have each interior angle is $\frac{180 (n-2)}{n}$

[boreal girder]

wai

[carmine garden]

I'm trying to come up with a rotation of order 3

[kind viper]

why 3

[kind viper]

correct but useless

[carmine garden]

I have to have the 0 rotation

[kind viper]

a rotation of order 3 cannot possibly belong to a subgroup of order 4

[carmine garden]

now I can have the 180 deg rotation

that gives me a group of order 2

[kind viper]

a 180° rotation does generate a subgroup of order 2 indeed

[carmine garden]

I could then add a vertical and horizontal reflection

would that work

[short ferry]

have you tried it

[carmine garden]

yes

seems to work

[short ferry]

btw you are currently talking about a group generated by vertical refleciton, horizontal reflection and 180° rotation?

[carmine garden]

No

I have a n-gon

I'm looking at a dihedral group of order n where n is an even integer

And it's subgroups of order 4

[short ferry]

Yeah, and by adding a vertical and horizontal reflection, what did you mean?

you had a group generated by 180° rotation (this one was of order 2)

and you wanna add vertical and horizontal reflection now?

That could add way more elements than you think

unless you intend to add only vertical and horizontal reflection, in which case you should recheck the definition of a group

[carmine garden]

I mean a vertical reflection done two times brings us back to where we started

[short ferry]

yeah, its an element of order 2

[carmine garden]

Same for a horizontal reflection

Ah I see what you mean

I see what you mean

[short ferry]

What do I mean then?

Can you tell what's wrong with your idea now?

[carmine garden]

Well a horizontal reflection , followed by a vertical reflection won't bring me back to any of the elements in the group

[short ferry]

indeed

doesnt satisfy the closure axiom if that's how you call it

seems like you'll have to do a small tweak

[mild sky]

for any two differents elements of order 2: a and b, ab is not the identity

[short ferry]

Btw in general, if you want to add some elements to your group / subgroup, then the smallest subgroup closed under the group operation is the one generated by the elements you added (+ elements or generators of the og subgroup). That's why i thought you meant generating

[carmine garden]

I see what you mean

[carmine garden]

Oh

Got it

.close

[compact pewter]

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[wheat seal]

How to do this

[wheat seal]

I tried transferring 5 and 13 to other side

Then make a polynomial q(x)

= p(5x) + p(x) - 18

That didn't give any ideas

Can anyone give more strategies?

I also have a solution but I failed to understand
Can anyone explain that too?

[rotund sphinx]

Let $r$ be the integer root. Then, $$p(x)=(x-r)q(x)$$ for some polynomial $q$ with integer coefficients.

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

Further hint (this gives away a lot): It follows that ||10-r divides 5|| and ||2-r divides 13||

[rotund sphinx]

Я не вижу такого решения

[wheat seal]

Ok sir
On it

[wheat seal]

Interesting

[wheat seal]

Ok got it sir

[wheat seal]

Can I share the solution?

[rotund sphinx]

well I kinda need to see the solution to explain it

and you also need to be specific about what part you don't understand

[wheat seal]

Ok

[wheat seal]

Why write p(r) as that?
What is the intuition?
How will one come up with it

[rotund sphinx]

this is translating "integer coefficients with at least one integral root" into math

and they give you p when x=2 and x=10

so it's natural to put in those two values

and at this point you see it and go

oh

question done sans number crunching

yay

[wheat seal]

Huh?

[rotund sphinx]

they're using a somewhat similar idea here

[wheat seal]

Oh

[rotund sphinx]

except they're trying to make a polynomial that's equal to zero at x=2 rather than 13

because 0 is generally nice to work with

not that it really matters

[wheat seal]

Wait a minute
So they are trying to make a polynomial just as I did?
What is the difference in their and my formation?

[rotund sphinx]

I explained their formation using the logic I used since they're basically identical

idk what your intuition was tbh

[wheat seal]

I mentioned it

[wheat seal]

Oh

Just make a polynomial that is 0 at 2

[rotund sphinx]

Ig you're trying to combine the conditions into one?

but a lot of the time you lose information that way

ex. Let's say you're given a=7 and b=5

then you know a+b=12

but if you now only work with a+b=12, you have now lost information

because you can no longer deduce the values of a, b with just a+b=12

[wheat seal]

Ok
A bit too complex

Nice then

How did they reduce r to +- 1 etc.

[rotund sphinx]

(r-2)q(r) = -13

q(r) is an integer because q has integer coefficients

aka you have two integers multiplying to -13

[wheat seal]

It's totally unimaginable for me
Why would you only use the p2 = 13

Yes

[rotund sphinx]

13 is more restrictive since it's prime

fewer possibilities for factors

so it's easier to narrow down with that first

before turning to p(10)

[wheat seal]

Oh i see now

My vision is clearer

You idea was way smarter

It would make r restricted to only a multiple of something

Sorry not r
X

[rotund sphinx]

yeah this is a classic case of "do the easy stuff first because if it doesn't work, you didn't waste that long. but if it turns out to be the main idea and you glossed over it, well then you feel like shit"

the easy stuff being rewriting the condition as p(x)=(x-r)q(x) and plugging in the values of x corresponding to the given values of p

[wheat seal]

Right

Thanks

Bye

[rotund sphinx]

[wheat seal]

.close

[compact pewter]

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[foggy sandal]

can somebody see if this proof is correct , if it has errors , where and how can i fix it please? Appreciated much.

[chilly basalt]

hello, we'd prefer screenshot lest the link is malicious.

[foggy sandal]

sorry will do

[weak cobalt]

!nopdf

[chilly basalt]

nah, I don't like to use factoids.

[weak cobalt]

Why f,g continuous on I?

I know it's not the proof part but I don't think it's necessary, is it?

[foggy sandal]

I suppose its not neccesary as it will follow from them being differentiable.

[weak cobalt]

No, it does not at a

[foggy sandal]

True, i suppose it needs to be continuous at a for some reason, but im not sure what

[weak cobalt]

I don't think it needs to; anyway, it's probably irrelevant

[foggy sandal]

Yeah its just what it said in my book for the theorem

Oh actually i think i know why

I think its because to use the generalised mean value theorem in the proof, the function need be continous on the entire interval

[weak cobalt]

You sure? It says "f,g [...] are continuous on the open interval. So we can use the generalised mean value theorem..."

The open interval doesn't include a

Tbh I don't quite understand the proof

[foggy sandal]

Really which part

Pretty much what i said was

<@&286206848099549185>

[kind crane]

if you're just checking your proof, there are plenty of proofs online to check against

https://www.macmillanlearning.com/studentresources/highschool/mathematics/rogawskiap2e/additionalproofs/proofoflhopitalsrule.pdf

https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/04%3A_Applications_of_Derivatives/4.08%3A_LHopitals_Rule

[weak cobalt]

I don't know if V is usual notation for what I assume to be a neighborhood; you might want to define it

Also the formula before the proof is missing an implies symbol

[foggy sandal]

Yeah theres not much standard notation for neighbourhoods, my book uses V but ive seen N and B before

[weak cobalt]

If I understand your proof correctly, it looks ok; but I'm not the best to ask

I'd follow riemann's advice

[foggy sandal]

Okay

I always have troubke verifying my proofs; im yet to find somewhere suitable or somebody to help

Do you have any advice

[weak cobalt]

No... I think verifying things is boring to most people, never mind proofs

And again I'm not the best to ask, sorry

It could be the case that some helpers would be willing and confident to review your proofs but they haven't stumbled upon your channels, idk

This is not a paid service after all

You might have more luck asking a professor in your university (or any nearby one if you aren't in uni)

[compact pewter]

@foggy sandal Has your question been resolved?

[compact pewter]

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[pale tiger]

i'm really confused on how to solve this problem. i'd assume they're asking for a 95% confidence interval, so i tried plugging it in to the formula, but how am i supposed to calculate a confidence interval w/ out the sample size (n)?

[static timber]

We don't really have context, but I guess the problem wants you to assume that the maintenance costs follow an approximately normal distribution. In this case, recall that, if $X \sim N(\mu, \sigma^2)$, then $\mathbb{P}(\mu - 2\sigma \leq X \leq \mu + 2\sigma) \approx 0.95$.

[boreal girder]

ucheo

[pseudo willow]

One message removed from a suspended account.

[pale tiger]

soooo

do i plug in the values of my mean and standard dev. into the second part of that

to end up with an inequality?

would that inequality be the interval?

[pseudo willow]

One message removed from a suspended account.

[pale tiger]

sry, im not actually taking stats rn, i have no idea what this means 😭

[static timber]

Yes! If a random variable follows a normal distribution, then the probability of its value being within 2 std. deviations of the mean is approximately 95%

[kind crane]

maybe you know it as empirical rule

https://en.wikipedia.org/wiki/68–95–99.7_rule

[pale tiger]

ohhh

tysm!!

so then i'd end up w/

2708-3692?

and that's that?

[kind crane]

show all your steps

[pale tiger]

for the left side

3200-(2*246) =2708

and for the right

3200+(2*246)= 3692

?

[kind crane]

,calc 3200-2*246

[boreal girder]

Result:

2708

[kind crane]

,calc 3200+2*246

[boreal girder]

Result:

3692

[kind crane]

sounds right

[pale tiger]

k, thanks a lot!! have a nice day :))

.close

[compact pewter]

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[compact pewter]

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

[vivid wedge]

Around 2

[compact pewter]

Show your work, and if possible, explain where you are stuck.

[vivid wedge]

Idk where to go from here in calculating the tension

[twin idol]

Ooo a statics question

[twin idol]

Complete the angles in the triangle

[signal willow]

Wait why are you multiplying 200 by 9.8

[cold widget]

it weights 200N you are already given the force

notice the unit

[vivid wedge]

ohhh

I see

[cold widget]

so can you find out T3

[twin idol]

Do you know how to relate the angles and the sides?

[vivid wedge]

Is it just the same?

[cold widget]

show why is it the same if they are same

[vivid wedge]

Not really lol

[twin idol]

Hmm

[timid dune]

Start with T3, what do you think T3 equals?

[vivid wedge]

Because the direction is the same ?

I'm kinda clueless

[vivid wedge]

Is it 200 N ?

[timid dune]

Yep

Now let’s consider the forces acting on the intersection between the 3 lines

What can you say about the forces acting in this point?

[vivid wedge]

Is the net force 0 because it's still?

[twin idol]

Yeah

[timid dune]

But we need to get clear what exactly is experiencing 0 net force

[vivid wedge]

The system as a whole?

[twin idol]

Sure

But we mean something between the 3 tension forces

[vivid wedge]

I'm not completely sure tbh

[twin idol]

Hmm...what connects T1, T2, and T3?

[vivid wedge]

Cables ? 😭

[twin idol]

Well they are cables

But what location makes T1 and T2 and T3 intersect?

In the diagram

[vivid wedge]

the beginning of T3 ?

[twin idol]

Yea...

Can you show where that is on the diagram?

[twin idol]

Exactly

Now we know the system as a whole doesn't move

But this point is important because it allows us to find the forces in T2 and T1

[vivid wedge]

so that means T1 + T2 = -T3 right

[twin idol]

Not quite

Now do you remember the 2 equations of statics for stationary points?

Hint: it's related to the x and y directions

[vivid wedge]

fnetx = 0 and fnety = 0 ?

[twin idol]

Yep yep

Now we need forces in these directions to make use of these 2 equations

But it appears that T2 and T1 are not in full x or y directions

[vivid wedge]

do we take the sin and cos of each of them and find the direction with the pythagorean theorem

of each one individually I mean

[twin idol]

Yea

sin and cos will come into play here

[vivid wedge]

and since T1 is in the fourth quadrant it would be sin319 and cos319 ?

[twin idol]

Wait

No no it's not in the 4th quadrant when you inspect it from the point you highlighted

[vivid wedge]

Oh it would be the second then

[twin idol]

Ye

[vivid wedge]

Okay but how exactly do I find the tension in the cable from that

[twin idol]

By using these equations

You found the x and y components for T1 and T2?

[vivid wedge]

is it just the sin and cos of the angles relative to the point ?

[twin idol]

Ye

[vivid wedge]

is that right

[twin idol]

Yeah but don't forget that they're not only the sin and cos.
The sin and cos must be multiplied by a magnitude

Namely T1 and T2 themselves

T1x = T1 cos(131)

Wait

139

T1x = T1 cos(139)
T1y = T1 sin(139)

[vivid wedge]

why 139

[twin idol]

180-41

[vivid wedge]

I just did 90 + 41 to get the reference angle in the second quadrant

is that wrong?

[twin idol]

Sadly that is not correct

[vivid wedge]

thank you for your patience btw

[twin idol]

We're happy to help :)

[vivid wedge]

Okay and how do I connect this to the tension in each of the cables

[twin idol]

Draw a free body diagram

Then use the statics equations

And you'll find T1 and T2

[vivid wedge]

Alright I think I can figure it out from here

Thank you so much for your patience

[twin idol]

[vivid wedge]

I'll let you know if I need any more help (assuming you're not busy)

[twin idol]

If you need any help, the help channels are open

[vivid wedge]

Okay thank you again

.close

[compact pewter]

Channel closed

Closed by @vivid wedge

Use .reopen if this was a mistake.

[compact pewter]

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[smoky heart]

wondering if anyone can help