help-36

[inland flame]

1

[proud roost]

now solve for x and substitute it on lhs

[inland flame]

By elimination right?

[proud roost]

in terms of alpha

[proud roost]

yep

[pliant elk]

no

[inland flame]

Oh?

[proud roost]

oh wait this is not simultaneous equations

[pliant elk]

by rearranging terms

[proud roost]

like x^3+3=alpha right from this equation get x in terms of alpha

[inland flame]

Uhm

How do i do that

By using sin?

[proud roost]

no

x^3+3=alpha
x^3=alpha-3
x=cube root of (alpha -3)

[inland flame]

Oh change lanes

[proud roost]

now keep this equation in 2nd line

[inland flame]

?

Oops its sideways

[proud roost]

,rcw

[proud roost]

yes

[inland flame]

2nd line as in the right side?

[proud roost]

yes

[inland flame]

Oh!

[proud roost]

in the left side try to guess what keep inside f inverse

[inland flame]

Hmm

X?

[proud roost]

im talking in the 2nd line , inside f inverse what do we keep

its x^3+3 already

what would make it simpler

[proud roost]

are you there

;-;

[inland flame]

Wait im still thinking

[proud roost]

okay

[inland flame]

Is this the second lane?

[proud roost]

think about our assumption maybe

[proud roost]

bro it was x^3+3 not 3x+3 😭

[inland flame]

X³?

[inland flame]

Oops

[proud roost]

yes

[inland flame]

When you move something into a root the +/- changes right?

[proud roost]

nope

[inland flame]

Oh

[proud roost]

x^3+3=alpha

x^3=lpha-3

now cube root on both sides to get x in terms of alpha

[proud roost]

why did you ask that btw?

[inland flame]

Bc i heard something about changing the -/+ when you turn the number into something but i forgot, iirc it was root?

[proud roost]

mb ****

[proud roost]

idk what the exact context is so i cannot confirm

[inland flame]

Oh ok

[proud roost]

cube root of x^3

not x

[proud roost]

we are not taking square root we are taking cube root to elimiate powers of x

on both sides btw

[proud roost]

yes so x=cube root of (alpha-3) right

now substitute that in left hand side of 2nd line equation

[proud roost]

because, cube root of x^3 would just be x

[inland flame]

Ouh

Wait

[inland flame]

Of so you mean

[proud roost]

i think my method is confusing you a little bit too much you should just continute with ren's method

[inland flame]

Is it wrong 🥲?

#help-36 message

I don't understand ren's method either 😭

[proud roost]

how about this

[inland flame]

Is it like this?

[inland flame]

Hm?

[proud roost]

you need to find y interms of x

[inland flame]

Oh uh, so x^2 becomes 2x & so on?

This si so hard 😭😭

[proud roost]

if the given eq is
y=sin(x^2) to find inverse function

1. Switch x,y it gives x=sin(y^2)
   2.find y in the newly made equation so first begin with taking sin inverse
   giving sin inverse of x=y^2 now take the square root which makes it
   sqrt root of (sin inverse of x) = y so now this would be inverse function

try doing the same thing for your problem

[inland flame]

Like this?
Does the sin just go away? Without purpose?

Im gonna go to bed now my parents are telling me to

Just leave your input I'll read it tomorrow 😅😅 tysmm for helping me!!

[heady forum]

I believe what the person above was saying, is that the left side should become the Inverse of Sin, also known as Arcsin or Sin^-1, of x

[final saddle]

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[polar spruce]

ight man

the issue is?

[ebon agate]

?

[polar spruce]

dude this is for math

[vital crag]

<@&268886789983436800> spam

[polar spruce]

moderators can you not delete these

[cyan kayak]

@lean rain do you have a math question?

[polar spruce]

this is funny as shit

[hushed sequoia]

@lean rain please don't use these channels for spam

.close

[final saddle]

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[hushed sequoia]

(yeah I muted they obviously didn't have a question!)

[old quarry]

bro its SIUU not SUII come on

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[heavy plaza]

im so clueless on this

considering just writing a python script to find it

heres some definitions we're given

[lofty sinew]

Writing a python script to find them is honestly not a bad idea, I might've done something similar when I took my first number theory class

[heavy plaza]

im just terrified something like this will show up in an exam 💀

[lofty sinew]

Well hopefully on an exam they'd pick a smaller number like 7 lol

[heavy plaza]

if i cant find a reasonable solution then ill go with a progam ig

[heavy plaza]

right?? 😭

[steep marsh]

thats funny

[desert mantle]

one of 2 or 3 will very likely be a solution

[drowsy epoch]

-# use python in the exam 💀

[steep marsh]

i used fusion 360 to cad out half of my geometry things lol

[heavy plaza]

TRUE ill get a calculator running python 13

[steep marsh]

ti nspire cx ii

[desert mantle]

but I'm not sure how you are supposed to do this without at least the knowledge that you dont have to check every power

[heavy plaza]

glad im not missing something obvious then

[desert mantle]

rip neither of them

[heavy plaza]

just so weird to me that an exercise in the notes wants a programmed solution

[desert mantle]

well you can do this by hand

it just takes a bit

[heavy plaza]

im sure u can

[desert mantle]

not even that much

20 mins max

[lofty sinew]

You could likely do a sieve method, where if 2^(2n) = 1 then 4^n = 1, so you don't need to bother checking 4

[desert mantle]

you only need to calculate up to ^11

[heavy plaza]

good point, just check primes

ahh ill give it a shot then

[desert mantle]

so there really isnt much to do

[lofty sinew]

Yeah just calculate the 11th power of everything lol

[desert mantle]

only need to check a^2 and a^11 arent 1 mod 23

thats really not that much

[lofty sinew]

Maybe remember that a = -(23-a)

[woven ledge]

also checking a^11 may be simplified cus thats checking if its a qr

[desert mantle]

oh I suppose yeah you could see that 2 is a qr

[heavy plaza]

oo facts

[desert mantle]

3 is harder to see

[heavy plaza]

could you elaborate ?

[desert mantle]

by general group theory, the order of the element is a divisor of the group order

which is 22 here

so the only possible orders are 1,2,11 and 22

so you only need to exclude 1,2 and 11

1 is obvious

[heavy plaza]

ohh right

and 22 follows from flt

icu

thanks for the simplifications makes this much more managable

[heavy plaza]

waitt is it not 23??

are we not referring the group Z/23Z?

[desert mantle]

the multiplicative group has order 22

the additive group is boring in comparison

[heavy plaza]

oh right excluding zero

because 23 is prime

yeah yeah

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[wraith glen]

i don't get it

[wraith glen]

no one ever said you can use negative x values

[blissful meadow]

The area isn't "bounded by x=0" if you add that part

[timber leaf]

Thats the grey part?

[wraith glen]

sorry

[wraith glen]

what?

i had a feeling they wouldn't ask us to do a trapezoid sorta shape though

that's just a bad question

[blissful meadow]

The region should have one of its sides x=0

Otherwise it's not bounded by that curve

[wraith glen]

ah i see

so either side woulda worked

[blissful meadow]

If anything you could argue about using just the red side

[wraith glen]

i gotchu bro thanks

.solved

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[wraith glen]

hopefuly i don't bomb this exam hahahaha

[timber leaf]

Good luck

[round anchor]

hola

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[chrome minnow]

Is this correctly calculates

[chrome minnow]

The question is to calculate wektor of AB if a = (1,4) and B = (3,2).

[proud roost]

y2-y1 would be 2-4 or -2

[chrome minnow]

Apologies it’s [2,-2]

Yes

And so what does this tell me? @proud roost

Wector AB = [2, -2].

[proud roost]

AB vector = 2i-2j

or you can also understand like if A was origin, B would be 2,-2 ig

basically like B with respect to A

[chrome minnow]

I don’t understand…

It’s the length of the wector???

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[chrome minnow]

Diora

Idiots

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[hollow gust]

What is pentation

[hollow gust]

Nvm

.close

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[warm python]

I'm unsure of what T looks like here

it surely isn't just p^4

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[long cave]

Hi guys
So this particular question is kinda throwing me off
Im having trust issues with the less/greater than or equal too sign
Is my shaded graph right or wrong,

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@long cave Has your question been resolved?

[stable helm]

It’s correct

[long cave]

Really, is it ok if you can elaborate

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[abstract nymph]

I gen just don’t know how to do this question i feel dumb

[south salmon]

can u find the slope of g

[abstract nymph]

I forgot how

My brain is empty

[vapid hound]

do you know what slope is? let’s start from there

[chilly cipher]

yk x/a +y/b= 1?

where a and b are intercepts

[abstract nymph]

I learned it in like 8th grade i think but forgot everything

[chilly cipher]

you can get the eq of line g from there

[vapid hound]

slope is rise over run. for a straight line that doesn’t curve or change, you can calculate it by finding how much the line rises (through y) then divide that by how much it runs (through x).

[abstract nymph]

Oh yeah

[tranquil pine]

\frac{+1}{+2}

[vapid hound]

yeah my latex skills are a bit off lately

[vapid hound]

for example, if a line goes up 1, then right 2, you’d have $\frac {+1}{+2}$

[soft zealot]

foxxeil

[vapid hound]

as the slope**

with this concept, can you try to find the slope of g?

[abstract nymph]

I can try

[vapid hound]

go for it

[abstract nymph]

Sorry for taking long im writing

[vapid hound]

no worries, take as long as you need

ur not in a rush at all

the goal here is to have you understand the concept, not to have you finish a problem as fast as possible

[abstract nymph]

3/3?

[vapid hound]

yes! this is also just 1, anything divided by itself is 1

don’t answer the problem for op

[stark coral]

i am new to this server

[stark coral]

means?

[vapid hound]

means don’t answer the problem for op

the point isn’t to give them the answer

[abstract nymph]

I have a dba later i have to actually learn the material

[vapid hound]

it’s to help them learn

[vapid hound]

don’t mind it, keep going

[stark coral]

ah ok sorry mate

[vapid hound]

so now you have the slope as 3/3

which is equal to 1, as anything divided by itself (except 0) is 1

so now, do you know how to find the slope of a line perpendicular to another line?

[abstract nymph]

I need a refresher

[vapid hound]

all good

[vapid hound]

perpendicular just means that the lines intercept creating a 90° angle, and the slope of a line (let’s call this slope A) perpendicular to another line is the negative reciprocal of the original slope (let’s call this slope B). so, if line A is perpendicular to line B, then slope A will be equal to $\frac {-1}{B}$

[soft zealot]

foxxeil

[vapid hound]

here, do you know if you have slope A or B?

[abstract nymph]

Uhh

[vapid hound]

well actually i’m not sure it matters, that’s on me😭the rule will stay the same

ignore that

from here, can you use the slope you were given and plug it into the equation $\frac {-1}{slope A}$ to find slope B?

[soft zealot]

foxxeil

[abstract nymph]

Would i be plugging in -1/1 or

Like 1 into the slope a thing

[vapid hound]

yes, since your slope you found was 1

now you have -1 / 1 , which is slope B. this represents the slope of the line perpendicular to your equation. you’re looking to write the equation of this line in slope-intercept form, y = mx + b. from here do you know what to do?

[abstract nymph]

Lemme see

[vapid hound]

[abstract nymph]

Idk what to do

[vapid hound]

alright, now that you have your slope, do you know which variable in y = mx + b that represents?

[abstract nymph]

B i think?

[vapid hound]

the slope is m, no worries! now you plug that in, and have $y = -1 *x+ b$

[soft zealot]

foxxeil

[abstract nymph]

Ohhh

[vapid hound]

for slope intercept form, you now have just the b to solve for! the b is where your line crosses the y axis

what information are you given in the problem that could be useful for this?

[abstract nymph]

Idrk

I’m trying to use the graph, but I’m confused

[vapid hound]

the line on the graph is your original line, not the one you’re solving for

don’t get mixed up with those !

[abstract nymph]

Oh

😭

[vapid hound]

you need the perpendicular line’s equation, so far you’ve:

• found the slope of your original line

• found the slope of the perpendicular line

and you need the perpendicular line’s equation in y= mx + b, and we are solving for b, where the perpendicular line crosses the y axis. we are given that the perpendicular line passes through the point (1, -2).

[abstract nymph]

Do i plug that one in too

[vapid hound]

hold on i just got a charley horse

myb

okay so

[vapid hound]

now your equation is y = -1x + b , and you have the point (1, -2). can you use these to find b?

think about what the numbers in (1, -2) represent

[abstract nymph]

Wait

so like

X and y? Or

[vapid hound]

yes exactly, your 1 is your x, and -2 is your y, you can plug these in to find b now!

[abstract nymph]

This is what I have down

[vapid hound]

yes, going great so far

[vapid hound]

be sure to remember to include how you got -1/1 so you don’t forget, to find a perpendicular line you always do $\frac {-1}{slope}$

[soft zealot]

foxxeil

[vapid hound]

and then now that you have b you can write the equation of the perpendicular line

[abstract nymph]

Wait what was b

Ih wait nvm

[vapid hound]

so now can you write the equation of the line in y = mx + b form?

[abstract nymph]

I js have that one

Uh

Without the b

[vapid hound]

so you have your m (slope) and your b, right?

you can plug those in to make y = mx + b

[final saddle]

@abstract nymph Has your question been resolved?

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[bronze grove]

hello

[shell condor]

hey, what's your problem?

[bronze grove]

i lost it wait a min

[drowsy epoch]

c'est la vie

[bronze grove]

helllooo

i thought there was just a single method to solve singluar solutions

but they are using like 3 diff types in examples

[vital crag]

which line do you get lost first

did you see 1(b) of Art. 4.5?

[bronze grove]

well from the method i learned that

[bronze grove]

this is example problem type

[bronze grove]

we need to put p =c and then

diff wrt c by forming it in y = sth

[bronze grove]

this book is very badly written man but lemme see

Art 4.5 is not in my syllabus

like not from the section we are doing cant find it

and i think i saw art 4.1 before this so idk how it reached 4.5

[vital crag]

it's not surprising that it takes more than one course to cover a whole book

[bronze grove]

the art before this is 4.17

idk where 4.5 is coming from

[vital crag]

well you should read it to understand the solution in Ex 10 or show it here if you don't understand

[bronze grove]

there is no art 4.5 😭

what should i read

im here to understand the solution

[severe canyon]

Ask the teacher, because we can't do anything with that

[vital crag]

what about Art 4.4 or Art 4.6

[bronze grove]

yall can explain the solution?

[severe canyon]

Or look at the previous pages and find something similar to this

[bronze grove]

like help with this ig?

[bronze grove]

the only art before this is 4.17

[vital crag]

there are 16 positive integers before 17, you're saying none of them exist?

4.1, 4.2, ... ,4.16 aren't in the book?

[bronze grove]

our syllabus starts from 4.17

damm idk

[vital crag]

.

[bronze grove]

can i just get help with this?

@vital crag

[vital crag]

maybe if you don't keep ignoring me

[bronze grove]

?

[vital crag]

.

[bronze grove]

its like 2 am bro

[vital crag]

your solution mentions this example which gives context to understand

[bronze grove]

this was my response?

[vital crag]

.

[bronze grove]

IDGAF ABOUT SOLUTION NOW

[vital crag]

oof

[shell condor]

can you please share 4.5 irrespective of whether it's in the syllabus or not, for context

[bronze grove]

idk where it is bro

does no one know how to solve it from equation?

[vital crag]

you suddenly deleted the solution you're asking about

[bronze grove]

bro

its 2am for me man

[vital crag]

wants help. doesn't want to provide relevant context to help.

¯_(ツ)_/¯

[vapid hound]

is this assuming $p = \frac {\dd y}{\dd x}$?

[soft zealot]

foxxeil

[shell condor]

for me too, but you have to co-operate

[bronze grove]

cooperate what 😭

[vapid hound]

if it’s PDE i may be able to help, sleep brain of mine can’t tell tho

[bronze grove]

idk where 4.5 is man i just have a section of the book in my syllabus

[bronze grove]

nope

[vapid hound]

shucks

[bronze grove]

i think so

[vapid hound]

wait then wouldn’t it be pde

[drowsy epoch]

co operate or get arrested

[bronze grove]

i told u that i dont know where 4.5 is and u still keep asking man- what do u want me to do?

[bronze grove]

it shouldnt be

we start pde after midsems

[vapid hound]

hmm

[vapid hound]

ahh

well i mean it could be solved like pde but that’s how id do it, unsure if you haven’t covered that yet

[lunar plaza]

it's an ode right

[vapid hound]

hm maybe

[bronze grove]

can u guys not solve it without soltuion?

i will just ask chatgot bruh

[leaden spire]

!noai

[bronze grove]

oh hell nah

no ones helping me here just ragebaiting me

[leaden spire]

let me think about the problem

can i ask something?

?

[vapid hound]

first step would be to solve for x, then put p=c, then rearrange to quadratic and solve for y, i believe

i wouldn’t use chatgpt for this, it seldom solves problems correctly that are higher than like middle school math

[drowsy epoch]

omg what

[rocky tusk]

i mean this isn't true

[leaden spire]

it's a cubic equation so can't you just factor it out?

[rocky tusk]

may have been true like 3 years ago

[vapid hound]

my thought process would go here first but i’m unsure

[bronze grove]

why was no ones just saying this instead of bugging me about art 4.5

[vapid hound]

eh not the place for this discussion

[bronze grove]

😭 see the chat omg

[vapid hound]

no worries on that, just focus on the problem

[leaden spire]

unsure like?

[vapid hound]

of when you’d factor. at the start?

[drowsy epoch]

too many helpers ig, i am sorry

[leaden spire]

but if you factor then what would be the problem you are thinking you will face?

[bronze grove]

solve for x like in write the whole term in form of x?

[bold turtle]

And I've pointed this out before - at a stage of learning, it is much harder to check for when/if it does hallucinate, because you won't know better

[bronze grove]

x = ...

[bold turtle]

Also, this is an ODE, isn't it?

[vapid hound]

yes

[vapid hound]

i believe so (?)

[bold turtle]

Which I don't think other people have noticed

[bronze grove]

more like ragebaiters man 😭 where is 4.5 after i told them idk 5 times

[bold turtle]

i.e. an ordinary differential equation - p has been stated earlier to be the dy/dx

[bronze grove]

i have c3 + 8y2 / 4yc = x
now what

[vapid hound]

exponents?

sorry

wait

i have $\frac{p^2}{4y} + \frac{2y}{p}$

as x

woah

texit

$\frac{p^2}{4y} + \frac{2y}{p}$

[soft zealot]

foxxeil

[vital crag]

you're just too lazy to look

[vapid hound]

no need imo

[bronze grove]

yes same

[vapid hound]

just work

[bronze grove]

I DONT HAVE THE BOOK BRO

[vapid hound]

don’t focus on him,

[bronze grove]

its like 2 am for me and im suffering from stomach infection man

[vapid hound]

we can work with the problem at hand

[vital crag]

uh huh

[vapid hound]

no need

[bronze grove]

let me just undertabf how to solve it

[vital crag]

you posted the solution, then deleted it

making it take longer

if you really help faster, you wouldn't be so uncooperative

[vapid hound]

it’s being worked on rn, you’re not helping

arguing will take this nowhere

you as well @bronze grove , don’t fuel into it

[bronze grove]

so now what

[vapid hound]

take your focus onto the math

okay

so

[bronze grove]

we have equation

[vapid hound]

rearrange it into a quadratic

[bronze grove]

in terms of?

y?

[vapid hound]

y

yes

[bronze grove]

isnt rhat just our initial equation?

[vapid hound]

bro

[soft zealot]

foxxeil

[vapid hound]

i think i did that wrong

wait

[bronze grove]

how is this quadractic

[vapid hound]

wait

wait.

ignore

wait no

i think that’s right

then $4xyc = c^3 + 8y^2$

[lunar plaza]

i can help you out with the singular solution (as it is easier to find)

[jovial grotto]

Wait. We have x interms of p and y. We know p = dy/dx so then we can just differentiate both sides. dx/dy = 1/p and the rest will be a DE in just p and y

[vapid hound]

texit hurry up

[lunar plaza]

treat the function as F

and differentiate it wrt p and set it to 0

[vapid hound]

that’s the next step i believe

yes

and diff in respect to c?

or are we on different wavelengths rn

[jovial grotto]

Basically. But I don't know why you renamed p to c

[vapid hound]

easier for me

😭

i learned it as c, i do the same

i also rewrite things to x and y

[lunar plaza]

@bronze grove

(then u will get value of xy which u can put it into the original eq )

[vapid hound]

-# i’m gonna nap now, if it’s been quite a bit ping me if you still need help

[leaden spire]

can somebody also explain me the solution of this problem?

actually nevermind, first help her

[jovial grotto]

@bronze grove Are you still there?

[bronze grove]

i think i understand the basic structure

so im gonna leave it there and sleep

[bronze grove]

nicee

@vapid hound thanks mann !!

.close

[final saddle]

@bronze grove Has your question been resolved?

[vapid hound]

ofc, lmk if u need anything else

[final saddle]

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[jagged flare]

im trying this problem and this was one of the hints

[jagged flare]

hint 57

why is it phi(n)/2 and not phi(n)?

for any coprime number a to n, there is b=n-a that is coprime to a cause gcd(a,b)=gcd(a,n-a)=gcd(a,n)=1, so shouldnt there be phi(n) options for a, and phi(n) pairs of (a,b)

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

[bold turtle]

I'm guessing (it's past 1am here) but at a glance, (a, b) is the same pair as (b, a)

You'd be overcoumting

So in fact you could cut that upper bound in half, i. e. you'd have phi(n) /2

[final saddle]

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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[elder bridge]

i’m dumb and i can’t figure out wtf to do for 1&2 (i used ai for 1) and i have a test tomorrow and i needa lock in on math bc i got a 44%

and i don’t wanna fail gr10

so someone please help me tysm

[last anvil]

do u know what y equals to when the graph crosses the x axis

[elder bridge]

no

maybe like 0

[last anvil]

yes

[elder bridge]

how do we do num 2

[last anvil]

so for 1a look for one that makes sense when y = 0 and x = 5

[elder bridge]

okay

well i already have the answers but i need help for 2

[last anvil]

both intercepts are at the origin

but you need two distinct points to draw the graph

[timber leaf]

No need AI lol

So which questions are you concerning at

We will help 1 by 1

[elder bridge]

well

how do i find x and y through an equation

[timber leaf]

For example, y=x+1?

[elder bridge]

yeah

[swift cove]

I’m genuinely failing geometry

[timber leaf]

Then, let x=0 and y=0

[elder bridge]

like 2x-5y= 10

wait acc thats bad exmaple

[timber leaf]

Let x=0, then y=0

[elder bridge]

its 5,2 i think but

what abt

2x+y= -5

bc y is by itself

[timber leaf]

Then it will intercept at y and x axis right?

Since its a linear function that have positive gradient

So, for the case that it intercept the y axis, the x value must be 0

And for x axis, the y value is 0

[elder bridge]

but then its js 2x = -5

[timber leaf]

Then find x

What is x if 2x=-5?

[elder bridge]

um idk

-7

[timber leaf]

-5/2

[elder bridge]

so it will js become a fraction

[timber leaf]

Yes, then the point is (-5/2,0)

[elder bridge]

oh

[timber leaf]

And do the same for x axis, this case, make x =0 and find y

[elder bridge]

y = -5

[timber leaf]

Then, the coordinate will be (0,-5)

[elder bridge]

ohh

okay

[timber leaf]

Thats the coordinate

[elder bridge]

this oage isnt rlly relevant but

i need help with soemthing rlse

[timber leaf]

Okay, please tell?

[elder bridge]

my teacher told me

this is rlly close to the quiz

like probably same things but different numbers

and graphs

[timber leaf]

Which question are you specifically concerning about?

[elder bridge]

how do we do this

[timber leaf]

I understood.

[elder bridge]

how do we find out if it’s a one to one function and a function

[timber leaf]

But you need to show which one specifically so we can work it out.

[elder bridge]

maybe for

iii

i already did it

but i was wondering if i got it rught

and also for vi i know the range is already y > 0 but i don’t know how to get that andwer because it’s literally infinite

[timber leaf]

A one to one function is when you have an output y, it come from exactly from a x value

And a function is an input x have exactly 1 output y

[elder bridge]

ohh oksy

so

ii isn’t a function

[timber leaf]

Question iii is correct.

[elder bridge]

because it has 2, -3 x

[timber leaf]

But, it is a one to one function.

[elder bridge]

how is it a one to one function

bc its inf?

[timber leaf]

iii is a one to one function because when you give it a value x, it give exactly 1 output y and vice versa

[elder bridge]

ohh okay

[timber leaf]

Or, each output y have one unique input x

[elder bridge]

oh

how do we iv

[timber leaf]

Have you tried to graph ii?

[elder bridge]

because it’s a circle

and i don’t know how fo circle

[elder bridge]

no

[timber leaf]

4 is NOT a function.

4 is a relation.

[elder bridge]

omg😭😭😭

but did i atleast get the domain and range right

[timber leaf]

Yes

But c, it is not

How will you manage to survive tests with this.

[elder bridge]

i don’t…

[elder bridge]

for iii?

[timber leaf]

Yes.

[elder bridge]

how is it not right

[timber leaf]

Fix iiic

[elder bridge]

its infinity

[timber leaf]

Which one?

[elder bridge]

iii the one u said

[timber leaf]

Ah, i mean iv

iv c is wrong

[elder bridge]

oh

i fixed it

to no

this is answer key

how did they get y > 0 on the last question

because it’s infinity

and why tf is it a one to one function

[timber leaf]

Vi?

Range of it start from 0

Because it is exponential

And it is a 1-1

Because for any y value you have 1 unique x value

[elder bridge]

ih

wait so one-to-ones are when

the things dont cross right

[timber leaf]

Yes

[elder bridge]

ik ii isnt a function bc it has 2 of the same donain’s

[elder bridge]

so

that automatically makes it an

y > 0

[timber leaf]

Yes

[elder bridge]

what about this

[timber leaf]

Which one?

[elder bridge]

2 a)

then 2b

then 2c

the top one

[timber leaf]

Use rise/run formula

It stated that it is a line

So, the formula is y=ax+b

To find a, which is the gradient, do (y2-y1)/(x2-x1)

Then for the rest just subs the points coordinates in

[elder bridge]

kk

okay i got

-7/-8

[timber leaf]

👍

Wrong

-1-(-7) is not -8

[elder bridge]

what

oh

[timber leaf]

X2-x1, calculate again

[elder bridge]

-6

so its -7+1

[timber leaf]

Correct

[elder bridge]

bc the 2 - in thr middle becomes a positive righttttt

[timber leaf]

Yes

👍

[elder bridge]

so its -7/-6

okay im gonna do b nkw

i think its

0/-5

wait

is it -7+2

or -7-2

[timber leaf]

...

It is 7/6

Then the equation is y=7x/6 +b

substitute the point coordinates in the formula

To find b

[elder bridge]

what

where is b then

[elder bridge]

no its -7/6

[timber leaf]

Huh

-2-5=-7

And -7/-6 =7/6

[final saddle]

@elder bridge Has your question been resolved?

[final saddle]

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[runic phoenix]

How to prove x < y iff x^r < y^r? x, y > 0 and r is in Q.

[sick roost]

this isnt true as stated

consider r=2 and x=0, y=-1

[runic phoenix]

Does not y - x > 0 imply that y, x > 0?

I think

[sick roost]

so, you want to show that for x, y>0 and r in Q, then
x<y iff x^r<y^r?

[runic phoenix]

yes

[unborn forge]

No, take y=-10 and x=-15 for example

[sick roost]

well, first note that multiplying a positive preserves <

[runic phoenix]

okay

[sick roost]

the general method is to do a direct proof
suppose in general than x, y>0

start with x<y, prove x^r<y^r
then, start with x^r<y^r and prove x<y

[runic phoenix]

how

[runic phoenix]

You just said everything here

[sick roost]

lets do the first part
let x, y>0 and x<y

[sick roost]

well, we need to do more work than just invoking this

[runic phoenix]

okay

x < y => x^r < y^r

[sick roost]

so, how do you think we should approach this?

[runic phoenix]

r is a rational

We are assuming x < y, so we are obviously not gonna prove that, right?

[sick roost]

right

[runic phoenix]

I will just work from the definition of r

[sick roost]

heres a hint: ||start with integer r||

[runic phoenix]

That will help me with preserving <

[sick roost]

because theres a problem when you get into rational exponents naively

$$a^\frac{1}{b}=c$$
means by definition that
$$a=c^b$$

[soft zealot]

Cycadellic

[runic phoenix]

(x^1/n)^m < (y^1/n)^m

[sick roost]

so i think the best course of action here is to prove this first for integers, then it fallows very simply from this for all rationals

[runic phoenix]

Fine

Let us do m/n

and k/j

such that m/n = k/j

[sick roost]

just do m/n for both

[runic phoenix]

okay

[sick roost]

but first we should consider that m and n are both integers

[runic phoenix]

yes

[sick roost]

it would really be a lot easier if we first showed that x^m < y^m for integer m

then we will use that to prove the more general rational case

[runic phoenix]

x^m < y^m

(x^m)^1/n < (y^m)^1/n

and so x^m/n < y^m/n => x^r < y^r as desired

good start! Assuming x^m < y^m

We did that, right?

@sick roost

[sick roost]

well, we need to be a lot more careful than this

[runic phoenix]

Fine

m/n <= r

so x^m/n <= x^r

by density of Q over R

[sick roost]

thats circular

[runic phoenix]

there is an n in N such that nx < m < ny

[sick roost]

we should prove it by induction

[runic phoenix]

ah

[sick roost]

of course, the base case is the assumption that x=x^1<y^1=y

our hypothesis is x^n<y^n, can you show that x^(n+1)<y^(n+1)?

[runic phoenix]

yeah

[sick roost]

prove it

[runic phoenix]

the natural numbers are unbounded above by R

[sick roost]

hold on

[runic phoenix]

for x, y > 0 and y > x, there...

holds on

[sick roost]

this is a nonsequitur, we dont need this

[runic phoenix]

Good

I don't know what that means, but I agree

[sick roost]

it means it doesnt have significance in this proof

[runic phoenix]

I got that from context

but I ain't remembering all that

[sick roost]

given: x,y>0, x<y, x^n<y^n, and multiplying a positive preserves <
prove: x^(n+1)<y^(n+1)

[runic phoenix]

ah, strong induction

I see

[sick roost]

how do you prove it?

[runic phoenix]

Using algebra

[sick roost]

what is the actual chain of reasoning

[runic phoenix]

x

x * x

x * x * x

[sick roost]

no

[runic phoenix]

...

then

[sick roost]

we have the hypotheses here
x,y>0, x<y, x^n<y^n, and multiplying a positive preserves <

we need to invoke these hypotheses to concretely produce x^(n+1)<y^(n+1)

[runic phoenix]

How

I mean

[sick roost]

start with x^n < y^n

[runic phoenix]

n + 1 is in N

that is all I know

[sick roost]

heres a hint for you then

x>0, we know that multiplying a positive preserves the <, how do i get a statement about x^(n+1)

[runic phoenix]

Rewrite n+1 as m in N

yes?

[sick roost]

no

[runic phoenix]

But

fine

[sick roost]

right now we just want to show that x<y implies x^n<y^n for n in N

[runic phoenix]

okay

[sick roost]

to do so, we just do induction, strong or weak, theyre equivalent

[runic phoenix]

omg

We just

proved

[sick roost]

x^1<y^1 follows from x<y
now x^n<y^n
use the fact that x>0 and multiply x to both sides

[runic phoenix]

Bro

[sick roost]

x>0, so it preserves the relation

[runic phoenix]

That is what I did

I multiplied

by x

or by z

or by d

all positive reals

[sick roost]

i have no idea what youre saying about z, d, etc

[runic phoenix]

so, now what? we have just proved the obvious?

[sick roost]

no, what do you get after multiplying both sides by x in x^n<y^n?

using the fact that x>0 and multiplication by positives preserves <

[runic phoenix]

x^n+1 < xy^n

[sick roost]

yes

now show that xy^n < y^{n+1}, how?

[runic phoenix]

x^n+1/xy < xy^n+1/xy

[sick roost]

hold on

start with our assumption that x<y

what to do to this?

[runic phoenix]

x^1 := x

[sick roost]

we want to say this

[runic phoenix]

x < y

then multiply both sides by y^n

[sick roost]

good, how do you know y^n>0?

[runic phoenix]

because you say so

[sick roost]

its hiding in the assumptions actually

[runic phoenix]

and because it is basic algebra

assumed math

[sick roost]

we cant say that or else youre using circular logic

[runic phoenix]

sorry--autocorrect

[runic phoenix]

Uhm, so it is like an axiom someone very smart axiomed based on very historical and phylosophical yap that was shaped through centuries

[sick roost]

oh wait are x,y in R?

[runic phoenix]

probably

it is real analysis

I think it is assumed

[teal coral]

it's already given in the question rt?

[runic phoenix]

no

we are proving just that

[sick roost]

really we can leave y>0 implies y^n>0 for later, i think its safe to assume here

[runic phoenix]

yes

[sick roost]

so whats the actual chain of inequalities?

[runic phoenix]

algebra is good

[runic phoenix]

for all x > 1, x<x^2<...r<x^n<...

similarly for y

[sick roost]

no for the induction proof

[runic phoenix]

1,2,3 all the way to n

[sick roost]

also this isnt true for all positive R (consider 0<x<1)

we have x^n<y^n
thus, x^{n+1}=x^n*x<xy^n<y^{n+1}

that completes the induction

now, we know that x<y implies x^m<y^m, for all integers m

do you follow so far?

[runic phoenix]

sure

[sick roost]

sure?

are you unclear about something?

this is the important part, the part about exponent Q follows pretty immediately from this

[runic phoenix]

oh

cool

[sick roost]

so, do we need to revisit the induction? or do you want to move to Q?

[runic phoenix]

Q

[sick roost]

consider integer n now
let a=x^{1/n} and b=y^{1/n}

let x<y

show a<b

so, by definition of fractional exponents, we have a^n = x, and b^n = y

n is a natural number here

[runic phoenix]

okay

x < y

a^n < b^n

we just proved that inequality is preserved

[sick roost]

OH i see, we need to do the iff by induction actually

right so, you see immediately how this follows from the natural case

lets do that one now

consider natural m again
let x,y>0 and all the global assumptions etc
let x^m<y^m

how do we show x<y now?

this is the only if case

hint: ||use contrapositive of what we showed for natural numbers||

[runic phoenix]

x^n

(x^n)^1/n

by definition

it is x

and it preserves inequality if done the same on y

[sick roost]

to be clear, i did jump to Q too soon, right now we should only consider our induction from before
so, let x,y>0
we know that for all natural n, x<y implies x^n<y^n

we need to show the converse here now

let n be an arbitrary natural number and suppose x^n<y^n

prove x<y

[runic phoenix]

idk

[sick roost]

we do it by contradiction

suppose BWOC that x\ge y

case: x=y, then trivially x^n=y^n

what does this contradict?

[runic phoenix]

x%n<y^n

[sick roost]

(we supposed x^n<y^n)

yes

[runic phoenix]

ik

[sick roost]

case 2: x>y

[runic phoenix]

i corrected it before

skesn

d

[runic phoenix]

uh

sleep

please

[sick roost]

(were so close)

[runic phoenix]

ok

[sick roost]

recall that we showed for all natural n, a<b implies a^n<b^n

we are supposing x^n<y^n
in case of x>y what does this contradict

[runic phoenix]

We

uh

induction

[sick roost]

changed this to a,b so the clashing doesnt confuse you, its a result that holds for all a, b>0 here

[runic phoenix]

that is being contradicted

x is less than y iff x^n < y^n

[sick roost]

hold on

we only know x<y implies x^n < y^n

for natural n and for x.y>0

we are currently trying to show the other direction

i think using x,y here is whats confusing you

so think of it like a<b implies a^n<b^n

we already know that x,y >0 by assumption so, x^n, y^n >0 too

now we let x^n < y^n
we are going to show BWOC that x<y

suppose instead that x\ge y
case 1: x=y implies x^n = y^n, a contradiction

case 2: x>y

this is what we have so far

[sick roost]

how can you use this to get a contradiction

[runic phoenix]

x<y

[sick roost]

start with x>y, that is y<x

what does this tell you about y^n and x^n

[sick roost]

(we already proved this)

[runic phoenix]

idk

thanks

[sick roost]

(just plug and use it directly)

[runic phoenix]

y^n < x^n

[sick roost]

yes

why is that a contradiction

[runic phoenix]

cuz

y^n < x^n

and because

[sick roost]

what contradictory assumption did we make here

[runic phoenix]

no

we just proved that

y^n

is

less than

x^n

[sick roost]

yes that follows by the proof
now we want to show x^n<y^n implies x<y
we do this by letting x^n<y^n and showing x\ge y is a contradiction, from which it follows that x<y

we already know x=y implies x^n=y^n is a contradiction to x^n<y^n

in case x>y: we now know that since x>y we have y^n < x^n

if you can show a contradiction then youve shown x\not\ge y

so x<y

but we need to know exactly what is being contradicted here

[runic phoenix]

i nane ro gio

bye

[sick roost]

gl

[runic phoenix]

thanks

'close

[sick roost]

its .close btw

[runic phoenix]

.close

[final saddle]

Channel closed

Closed by @runic phoenix

Use .reopen if this was a mistake.

[final saddle]

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

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

[jagged flare]

,rccw

[jagged flare]

why is 9 wrong?

by mass point geometry, the mass of X=9, Z=12, Y=8 as 12/9=4/3 and 8/12=2/3
then the mass of R is 9+12=21 and the mass of Q is 12+8=20. then XP/PQ=20/9, and YP/PR=21/8. sincd the questiom is asking about the product, them we would get 35/6, but thats wrong?

[timber leaf]

Your textbook using half english half chinese?

It should be 35/9 i believe

We want massY.2 = massz.3 and massx.4=massz.3

[jagged flare]

OHH ok ok i see

[timber leaf]

Then assume massz= any positive int, let massz=4

[jagged flare]

the image was innacurate so i swapped the ratios of Y and Z

ok ok ty

[timber leaf]

Nw

[jagged flare]

.solved tyyy

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[rugged merlin]

why is [
\sum_{n=-\infty}^{\infty} e^{-j2\pi fnT_s} = \frac{1}{T_s} \sum_{k=-\infty}^{\infty} \delta(f - kf_s)
]

[rugged merlin]

why is [
\sum_{n=-\infty}^{\infty} e^{-j2\pi fnT_s} = \frac{1}{T_s} \sum_{k=-\infty}^{\infty} \delta(f - kf_s)
]

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@rugged merlin Has your question been resolved?

[opaque ember]

the question is rather loaded bc both sides of the equation are distributions instead of normal functions. might i suggest this SE post? it seems correct at a glance https://math.stackexchange.com/questions/827598/prove-that-sum-k-infty-infty-e-j2-pi-f-k-t-sum-k-infty-infty-del

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@rugged merlin Has your question been resolved?

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[left trail]

I have a couple of questions regarding the second image. He says that first we assume that f is continuous in some neighborhood around the point x. Does a neighborhood need have values greater and less than x or can it just be one. Second question is I don't see where our the theorem he says that f can only be continuous for one point. Is it where he says that A'(x) = f(x) only when f is continuous at x?

[vital crag]

For now you can think A'(x) = f(x) only when f is continuous yes

[left trail]

wait for which question?

[left trail]

so then the issue he is pointing at is is that he only concludes that A'(x) = f(x) when f is continuous at the point x and we don't know anything else about f. So assuming that f is continuous on [x,x+h] is greatly limiting the scope of this theorem

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[left trail]

so we know that $|f(t)-f(x)| < \frac{\epsilon}{2}$ but doesn't that mean that $\int_x^{x+h} |f(t)-f(x)|dt < \int_x^{x+h} \frac{\epsilon}{2}dt$ not $\leq$?

[soft zealot]

BigBen

[rugged merlin]

Yeah but I guess its not like it matters much

Something being < implies it being <= too

[left trail]

but than the use of $\frac{\epsilon}{2}$ is useless

[soft zealot]

BigBen

[left trail]

so we know the standard defintion says $<\epsilon$ we can say $<\frac{\epsilon}{2}$ since it is still positive so we can view that as our new $\epsilon$

[soft zealot]

BigBen

[left trail]

but why do all of those steps?

[pliant elk]

thats just analysis for you

[rugged merlin]

It does seem a bit gimmicky yeah

[left trail]

ok so I can just call it pointless and move on then

.solved

.close

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[left trail]

So we said that P is a primitive if $P'(x) = f(x)$. which means that $P(x) = \int_c^xf(t)dt$ . Now for the second fundemental theorem of calculus we have $P(x) = P(c) + \int_c^x f(t)dt$ but that just seems to be $P(x)$ then we would have two conclusions. That $P(c)=0$ which seems correct since $\int_c^c f(t)dt = 0$ but then we would also have $P(x)-P(c) = P(x)$ which cannot be true since we have show that the difference between two primitives is a constant not one of the primitive functions. I also see that he says that $A(x) = \int_c^x f(t)dt$ but P is supposed to represent any primitive so shouldn't $P(x) =A(x)$.