help-28

[sacred valve]

is there a specific working out of liek some sort of understanding behind this question?

[signal arrow]

To prove it, u need to know some calculus ig

[stiff musk]

what is your definition of e?

[signal arrow]

I can show the calculations if u want

[hardy charm]

Apply natural logarithm.

Then L' Hopital Rules.

[signal arrow]

Yeah

[sacred valve]

yes please

[signal arrow]

Do u know this stuff?

[hardy charm]

What do you mean?

[signal arrow]

I don’t think u will understand them if not

[signal arrow]

Sorry I was asking him not u

[hardy charm]

Oh mb.

Sorry.

[sacred valve]

no, it was a question in my lecture and i was just interested

[signal arrow]

Ic

[hardy charm]

Also, you really can't prove it.

[signal arrow]

Welp the simple answer is that is the definition of e

[hardy charm]

It's literally a definition.

[signal arrow]

True ig

Cuz the “proof” uses ln

[hardy charm]

It's like asking you to prove i² = -1.

[split hatch]

Well there is something to prove here, namely that the limit exists

[hardy charm]

When it's literally the definition.

[sacred valve]

okay then

ill jsut stick to it ig

[hardy charm]

Let that sink in.

[fathom saddle]

It's tougher to prove that definitions are equivalent. Did you have a definition of e^x you want to show is equal to this limit?

[sacred valve]

aye will there be questions where this definition will be used to proof something else?

[signal arrow]

It basically comes from compounding an infinitely small amount continuously

Compound interest

[signal arrow]

I doubt it

[fathom saddle]

We can't really know what your class will do. I'm going to guess probably not

Especially if you're just learning diff calc

[signal arrow]

When I was first shown this definition, I didn’t much about limits either so the teacher couldn’t rlly do anything mathematical with it

More of like a here is where it comes from thing

[sacred valve]

oh ic

alirhg tthank you guys

[signal arrow]

Np

[marsh trellis]

Yes

For example, evaluate $$\lim_{x \to 0} \left( 1+ 2\tan^{-1}{\sin x} \right)^{\frac{1}{e^x-1}}$$

[glossy valve]

robin.dabanc_

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[rare pine]

hey! i dont know if im being confused on random things right now, but for a sequence a{n}, what does lim n -> ∞ of a{n} actually mean? because the sequence is only defined on the natural numbers

[gritty rose]

Precise answer
https://math.stackexchange.com/a/3612043

[rare pine]

oh okay i see

.close

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[torpid gazelle]

hi how do i get total displacement vs total distance

[torpid gazelle]

i dont wanna do all that work of finding crit points

[hot herald]

can you show the original question
and the info you're given

[torpid gazelle]

uh

lemme make one

3t^2−12t+9

[hot herald]

what's that supposed to be for? displacement? velocity?

[fleet relic]

displacement function?

units would check out

[torpid gazelle]

i messed up equation

x-2

when you do intigrale

[brittle crystal]

What?

What is (x-2)?

[torpid gazelle]

\int_{2}^{5}\left(x-2\right)dx

is not the same as f(5)-f(2)

when its itigrated

[brittle crystal]

Assuming that f(x) = x-2, yes it isn't

[torpid gazelle]

bro i keep messing it up

[brittle crystal]

You use the limits on the antiderivative, by the FTC

Do you know to calculate the indefinite integral of x-2?

W.r.t. x?

[fleet relic]

$$\int_{2}^{5}\left(x-2\right)dx$$

[torpid gazelle]

yes its 1/2 x^2 -2x

[glossy valve]

Raod14

[torpid gazelle]

thats still a wrong equation im sorry i cant find what im trying to ask

[brittle crystal]

Right, so now call this F(x)

Then the answer to your definite integral is F(5)-F(2)

[torpid gazelle]

yes but sometimes it isn't and i have to make it into many definite intrgrals

im not really sure how to explain it

[brittle crystal]

Do you mean an integral like $$\int_{-3}^{5} |x-1| \dd x$$?

[glossy valve]

Tillman

[brittle crystal]

You need to split that into two definite integrals

Or if you're referring to your earlier total displacement predicament,

If you have the velocity-time graph... I mean, I still don't get your predicament fully

[fleet relic]

more context and specific one at that would be swag

[torpid gazelle]

yes

im sorry wifi got cut out

[brittle crystal]

okay so

is your question why we do that?

[torpid gazelle]

how do i get that into one intigral

[torpid gazelle]

yes

[brittle crystal]

well, you split into two integrals, calculate them separately

[torpid gazelle]

i know why but why cant it be one

because sometimes its like 3

[brittle crystal]

that's because $|x-1|$ is not integrable directly

[glossy valve]

Tillman

[torpid gazelle]

is there a way i can make it integrable directly

[brittle crystal]

not without splitting the integral into two (well, it depends on the limits)

are you aware of the definition of $|x-1|$?

[glossy valve]

Tillman

[torpid gazelle]

how its absolute value?

[brittle crystal]

So, $|x-1| = x-1$ when $x \geq 1$ and $|x-1| = 1-x$ when $x < 1$

[glossy valve]

Tillman

[torpid gazelle]

yeah i get that it just shifts it to the right by one

[brittle crystal]

yeah, so

[torpid gazelle]

because its in the same function as x

[brittle crystal]

we figure these two "portions" or "branches" can be nicely integrated

so we split

[torpid gazelle]

ok

[brittle crystal]

here, for $-3 \leq x \leq 1$, $|x-1| = 1-x$

[glossy valve]

Tillman

[brittle crystal]

and then for $1 \leq x \leq 5$, $|x-1|=x-1$

[glossy valve]

Tillman

[torpid gazelle]

yes

i get that

[fleet relic]

in terms of integrals?

[marsh trellis]

you can directly find the area using the graph

1/2 base height, two triangles

[torpid gazelle]

yes

[timid relic]

Shit

Sorry

[brittle crystal]

oh, wait

that's it?

[timid relic]

Oh wait the picture wasnt sent?

[torpid gazelle]

yes when u intigrate that i understand that it has to be 2 seperate intigrales

[brittle crystal]

@torpid gazelle do you have a velocity function in terms of time $t$, or a graph of it

[glossy valve]

Tillman

[fleet relic]

total displacement is just integrating a v(t) iirc from a to b
and for distance iirc should be integrating |v(t)|
which iirc should work as splitting that integral into integrals to work around sign change or smth (and making sure everything you add at the end is positive)

[fleet relic]

i might be wrong

[torpid gazelle]

this is what i was working on

[brittle crystal]

okay, does this relate to your question about displacement?

or is that a different concern

[torpid gazelle]

yes because the intigrale is not the same as g(5)-g(0)

[brittle crystal]

okay i get your question

[brittle crystal]

the integral is the area the curve encloses (here, with the x-axis)

[torpid gazelle]

because the derivitive of the intigrale is negitive for some time, and displacement total distance traveled, positive and negitive

[brittle crystal]

area under the x-axis is considered negative

[torpid gazelle]

which makes me need to intigrate that seperate

is there a way to just get the absoulute value then intigrate that

so that the total distance is the same as total displacement

[fleet relic]

which function is velocity function in your graphs

[brittle crystal]

well, for the $f(x)$ you have,
$$\int_0^5 f(x) dx \neq \int_0^5 |f(x)| dx$$

[glossy valve]

Tillman

[torpid gazelle]

this isnt particle problem

[torpid gazelle]

YES

[brittle crystal]

if you want to make them equal you need to change the function

[torpid gazelle]

how do i change the function so that f(x)= |f(x)|

[brittle crystal]

yeah you need to define it like the absolute value function \
for $x \leq 2$ and $x \geq 4$, $f(x) = x^2 - 6x + 8$ \
for $2 \leq x \leq 4$, $f(x) = -x^2 + 6x - 8$

[glossy valve]

Tillman

[brittle crystal]

essentially splitting it up

[torpid gazelle]

so just flip all the signs

when its negitive to make it positive

[brittle crystal]

errr, yes

for $x$ where the function is positive, let it be as such

[glossy valve]

Tillman

[brittle crystal]

where it is negative, "flip the signs"

[torpid gazelle]

yes

so there is no way to make it one single formula

[brittle crystal]

nope

[torpid gazelle]

damn

thank you for helping me ima try to get better at asking shi

[brittle crystal]

you could write $f(x)=|x^2 - 6x + 8|$, but that would eventually be interpreted piecewise or 'split'

[glossy valve]

Tillman

[fleet relic]

sorry for being slow or not paying attention but is f(x) supposed to be a displacement function or smth

[brittle crystal]

i think @torpid gazelle is using a physical interpretation of integrating absolute value

so it's a calculus question but understood with physics

[torpid gazelle]

wait ima try taht

[fleet relic]

oh ok

is integrating absolute value just splitting the original integral into integrals and then adding the ones resulting in positive values to the absolute value of the ones that came out negative?

[brittle crystal]

essentially, yeah

[fleet relic]

swag

thanks for clearing that up for me

[torpid gazelle]

function is f\left(x\right)=x^{2}-6x+8
integral is g\left(x\right)=\frac{1}{3}x^{3}-3x^{2}+8x
so because i just need the derivitive to always be positive, so that it can only increase not decrease, it needs to be 2 seperate functions

how do i do that tool thing

[brittle crystal]

do you have a motive to make the function constantly increasing?

[fleet relic]

$f\left(x\right)=x^{2}-6x+8$
$g\left(x\right)=\frac{1}{3}x^{3}-3x^{2}+8x$

epic fail hold on

[torpid gazelle]

yes so that the integral will always be increasing

so i dont have to make many functions

[fleet relic]

function: $f\left(x\right)=x^{2}-6x+8$

[glossy valve]

Raod14

[torpid gazelle]

my calc teacher said there was a way but he told me he wouldnt teach me it yet

[fleet relic]

integral: $g\left(x\right)=\frac{1}{3}x^{3}-3x^{2}+8x$

[glossy valve]

Raod14

[brittle crystal]

hm.

[torpid gazelle]

but u see how if the "derivitive" of the intigrale was only positive, that would mean the intigrale is only increasing, which would mean that total displacement would be the same as total distance

[fleet relic]

oh bruh i was looking at the desmos screenshot wrong lol

[torpid gazelle]

it sucks having to get all the crit points, then to add them up,

i have to get the maximums

[prime pier]

what's another word for "derivative of the (indefinite) integral"?

[torpid gazelle]

function

they cancel out

[prime pier]

so, if the function is positive, then f(x) = |f(x)|, which is what you said earlier

[brittle crystal]

yes, so you want to keep your function itself positive, which is what our absolute value discussion yielded

then the indefinite integral in terms of x is increasing

[torpid gazelle]

yes but i want to be able to calculate it myself without having to add every maximum and minimum

[fleet relic]

are they asking if they can tweak f(x) to be just like |f(x)| without having it piecewise and therefore not have to split integral
or nah

[torpid gazelle]

yes

[torpid gazelle]

YES

[brittle crystal]

yeah that was the question

[fleet relic]

i arrive to the question 30 years later lol

[prime pier]

however, you should try to integrate something like |x^2 - 6x + 8| over [1,5], because that actually requires splitting the integral

[brittle crystal]

and i said, no it needs to be piecewise, but

@torpid gazelle to get the absolute value, you have to split at the zeros of the function

not the critical points

if that's what your confusion now is

[brittle crystal]

notice for your prev example, $x^2 - 6x + 8$ becomes zero (switches from +ve to -ve or -ve to +ve in the graph) at $x = 2$ and $x = 4$

[glossy valve]

Tillman

[fleet relic]

are they talking about critical points because they are looking at g(x) instead of f(x) in their desmos?

[torpid gazelle]

how do i intigrate a absolute value

[brittle crystal]

we're regressing

[torpid gazelle]

f(x) is og function, g(x) is the intigrated

[brittle crystal]

go back here

[torpid gazelle]

yes

thats the 0s

so i HAVE to make that a seperate function there is nothing else i can do

[brittle crystal]

so you split piecewise there, yea?

[torpid gazelle]

yes

[brittle crystal]

well you're not "making" it a separate function, you're expressing the absolute value in terms of two separate branches

[torpid gazelle]

the way i do it is i find maximum on intigrale, lets call it m1 and minimum n1, then i add |m1 - n1| to get total displacement

[torpid gazelle]

true

Distance ≠ Displacement because:

The function changes direction

Distance counts ALL movement

Displacement only looks at start and end

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[dire spear]

can someone help me wiht this

[dire spear]

i tried taking u as sqrt(x) and then doinf partial fractions

wiht 1/(u^2(u-1))

[signal arrow]

this is my first thought when looking at this

du would be 1/(2sqrtx)dx

and u could replace that sqrt(x) with u

so 1/2u dx

[dire spear]

mmm

i think i messed up there

[signal arrow]

ic

it simplifies to a partial fraction integration

[dire spear]

yeah

but my answers off

[signal arrow]

can i see ur work

[dire spear]

alr just a sec

[scenic timber]

you can take x = t² it should work

[signal arrow]

yeah thats what we were talking about

[scenic timber]

cool

[signal arrow]

welp u = sqrt(x)

same idea

[dire spear]

yeah

mistake was

[scenic timber]

its not exactly same because denom is x^3/2 - x so you want to take substitution so that you have integral powers in denominator

[dire spear]

not replacing dx wiht du

[signal arrow]

huh where did u get that

[signal arrow]

ic

[scenic timber]

it just makes it easier the end expression is gonna be same obviously

[signal arrow]

it reduces to a simple partial fraction expansion with whole number exponents in terms of u

anyways ping me if u still need help ig

[dire spear]

Seem right?

[signal arrow]

looks right to me

[dire spear]

Alr thanks man

[signal arrow]

no prob

[scenic timber]

After the 2/u(u-1) step u can also do without partial fraction

like this

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[thick hedge]

For $X_n$ , I was thinking as $T_n$ goes to $a$, $T_n>a/3$ eventually, and thus $X_n$ converges to $1$.

[glossy valve]

wai

[thick hedge]

$Y_n$ goes to $0$ for the same reason

[glossy valve]

wai

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[raven mango]

Hallo

[raven mango]

Best way to fit a ermm stock curve?

Such that the curve is smooth (😭)

Considered polynomial regression but was too worried about overfitting. Any thoughts would work, I’m just boiling in thoughts

[spiral vigil]

could try a stack of sine waves

youre right that polynomial has issues (it always goes to +oo or -oo)

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[thin lark]

why is this the equation of tangent at a point for t=to in a parametric curve

[split hatch]

what is that εφαπτ

[thin lark]

i wrote its tangent

[split hatch]

is that actual greek language

[minor crater]

greek lol

[thin lark]

yea

its short for εφαπτομένη

[strange flicker]

thats just the function in terms of variables

oh what nvm

who uses greek 😭

[split hatch]

the greeks?

[thin lark]

greeks?

[strange flicker]

wait ur greek?

mb

[thin lark]

yea

[split hatch]

also this sentence makes no sense? but anyway

[woeful girder]

Clearly the sigmas among us do

[strange flicker]

i thought you just used for the thrill of it

[thin lark]

can someonee explain it

[split hatch]

Do you know what the r'(t_0) represents?

[strange flicker]

yeah sorry mb

[thin lark]

also r is r(t) = (r1(t), ..., rn(t))

[thin lark]

yea

(r1'(to), ... , rn'(to))

[split hatch]

Ok but geometrically what does it mean

[thin lark]

the vetor that is tangent to the curve

oh yea i see it

this is the span of that vvector

[split hatch]

yes

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[raven sail]

Hello,
this is about Logic and truth tables. Specifically about contraditciton.

[raven sail]

this is the truth table according to the solution

I assumption, was that the Truth table shouldve shown, that everything is wrong.

But it showed everything was true. Therefore, this only showed its a tautalogy? or not

[minor crater]

why

[minor crater]

Specifically about contraditciton
there is no contradiction involved here

[raven sail]

I mean the initial statement assumed the whole statement is true

therefore, I shouldve shown its not true,

[minor crater]

you are trying to do a proof by contradiction, but that's wrong

you aren't supposed to use contradiction here

it says "using a truth table" (which is a direct proof), it means you make the table and you see that all values match up

[raven sail]

so what does "Prove the Law of Contraposition"

oh

[minor crater]

contraposition and contradiction are different things

[raven sail]

there was never a "contradicition"

[minor crater]

this property is called contraposition

and you are proving it

[minor crater]

indeed

[raven sail]

hahaha

this is funny

Ive read it ateast 4 times

But whats the Law of Contraposition anyway

[minor crater]

an intuitive example is like this
let A => B be "if it rains, i will take my umbrella with me"
then ~B => ~A is "if i didn't take my umbrella, it's not raining"

think about the fact that these are equivalent statements

[raven sail]

and ~ does mean opposite right?

[minor crater]

yup

[raven sail]

always get hung up, by the opposite. Its always easier to understand for me:
Leave the Music Box on, instead dont turn the music box off

is that normal?

Ive once read childrens have a hard time understanding "not", . Kinda feel like my brain hasnt evolved over this

[minor crater]

this sounds like you are struggling with double negatives

[raven sail]

yes

[minor crater]

but here there are none

[raven sail]

like,

-- 3

or 1 - - 3

is 1?

or 4

[minor crater]

4

[minor crater]

yes it's normal, you get used to it with time and practice

[raven sail]

so, its : I have 1 coin , someone steals me 3 coins, but also then he reverts his action

thats how I read --

[minor crater]

pretty much

[minor crater]

“Young man, in mathematics you don't understand things. You just get used to them.”
― John von Neumann

[raven sail]

so then I never got 3 or lost 3 coins

[minor crater]

well okay not like that

[raven sail]

:>

[minor crater]

i think thinking in terms of distances is easier. say going +1 meter means going to the right 1 meter, and -1 meter means going to the left 1 meter

subtractions means you go the opposite way

[raven sail]

so I have a negative coin

I remember

I once saw a drawing of someone going meter to the right = +1

then from position 0 he went 1 to the left = -1

then he changed his direction of looking / direction of facing which direction. then he went 1 to the right but was looking in anoter direction or smth

I have 1 coin someone steals me a negative coin therefore Ill recieve a positive coin since $"1 \neq" -1$

[glossy valve]

PainAndLight

[raven sail]

like ${x | x \neq -x}$

[glossy valve]

PainAndLight

[raven sail]

ill leave this open, and let it close itself. Maybe someone will leavea comment until then. I think its 25 min until now

Ive got "1 Coin and theres infinte numbers of negative coins, If subtract a negative coin, I have 1 more coin", therfore 1 - (-1) = 2

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[lethal sable]

hey, can you help me get this last one

[sturdy valve]

why did u put sqrt(2)/2

from the others u know x=cos t y= sin t

and u got cos pi/3 correctly

sqrt(2)/2 is for pi/4 not -pi/3

[lethal sable]

nvm i got it

1/2

[sturdy valve]

yeah

👍

u can close this if ur done

[sturdy valve]

the 2nd one isnt correct..ur getting a negative angle when pi/2 is bigger than theta

[lethal sable]

could you explain how to start this problem

[sturdy valve]

theta is the angle made by the line (OP) with positive horizontal axis

now when we do pi/2 + theta, we are rotating that line by (90+theta) degrees

thats why in the first one P is already rotated by theta degrees so when we rotate it by 90 degrees more we get F

[lethal sable]

so is the second one C

[sturdy valve]

think of it like this:
start from the horizontal line and rotate 90 degrees anticlockwise and theta degrees clockwise from there

which point will u get

[lethal sable]

ohh

so it’s A

for the third one I got E

[sturdy valve]

yes 👍

let me check

yep

its E

[lethal sable]

is the last one B

[sturdy valve]

yeah

[lethal sable]

ok im done with hw. Thanks for the help!!

.close

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[twin venture]

An urn contains 3 red and 2 blue balls.
Two balls are drawn without replacement.

Calculate the probability that
a) at least one ball is blue

[twin venture]

is it 70%?

[late epoch]

hi

[twin venture]

i just wanna double check

[twin venture]

hi

[rare dock]

,calc 1 - 3/5 * 2/4

[glossy valve]

Result:

0.7

[rare dock]

yes

[twin venture]

aight thanks

[twin venture]

is that just the probability against 2times red?

[rare dock]

not (at least one ball is blue) is equivalent to 0 balls are blue

i.e. all balls are red

[twin venture]

that makes sense

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[twin venture]

how do i find x ?

[delicate torrent]

hm

can you factor on this thing?

try substituting each factor of 8 (rational root theorem) into each $x$ and see if some of them is equal to 0

[glossy valve]

1 divided by 0 equals Infinity

[twin venture]

huh

[delicate torrent]

we need to guess a root here

so that we can factor

[gritty rose]

,w roots x^3 - 6x^2 + 8

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[delicate torrent]

💀

[woeful girder]

wow

[gritty rose]

learn numerical techniques or use the solution for the general cubic

[delicate torrent]

what is this cubic

[woeful girder]

dont think we're sliding with RRT here king

[proven plover]

i see there's literally no chill here

[delicate torrent]

I THOUGHT EACH CUBIC MUST HAVE AT LEAST ONE REAL ROOT

[queen crater]

That smells of LLM

[proven plover]

my bad guys I'm gonna go

[atomic hare]

did you copy the problem right

[delicate torrent]

each cubic must have at least one real root

that's a guarantee

[woeful girder]

real

[queen crater]

Yes, they do

[woeful girder]

no pun intended

[queen crater]

In fact all 3 of those solutions are real

[delicate torrent]

and why are there complexes in both 3 roots in here

[delicate hearth]

yes

[woeful girder]

eh

[twin venture]

i asked chatgpt to generate questions for Curve discussion, didnt know that he asks taht hard questions

[delicate torrent]

elaborate how

[delicate torrent]

!noai

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[queen crater]

Calculate them

[delicate torrent]

AI is not recommended

[twin venture]

i dont think i will face an such hard f(x) for tomorrows exam

[woeful girder]

you wont

[queen crater]

It just spat out some random cubic

[delicate torrent]

how the hell it's real

[woeful girder]

whats the syllabus for the exam?

[gritty rose]

do real problems your teacher gives. or other ones from a book. not random ones on the internet

[queen crater]

Well you can get random ones online, just don't use LLMs

[twin venture]

Exam: Probability and Curve Sketching. There is a section where no aids are allowed and another where we are permitted to use a calculator, and there will be two application problems in the latter section.

[twin venture]

sounds better

[delicate torrent]

you can use a calculator to find the roots of a cubic tho

so dw

[twin venture]

aight

thanks yall

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[woeful girder]

you're welcome

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[proven lynx]

hello can someone help im frustrated

[proven lynx]

the task is to find the locus of inflection points

im stuck how the hell do i solve that

[gritty rose]

set u = e^t and use quadratic formula or factor

[proven lynx]

wtf

can i not solve it the way im doing it

cuz ive never done quadratic formula with e functions

[narrow ermine]

Yeah you can just divide through by e^t if you want.

Since e^t != 0

[proven lynx]

cant i just pull the ln?

[proven lynx]

wdym t?

[narrow ermine]

If you take the log now, you'll have ln(a+b) and that doesn't look great

[narrow ermine]

I mean you can divide the equation by e^t

Or factor e^t out if you want

[atomic hare]

what is the question asking

[narrow ermine]

They're setting f''(x) = 0 to solve for inflection points

[proven lynx]

why e^t

its e^x

[atomic hare]

same thing

[narrow ermine]

Yes

I just read your x's as t's that's all

[proven lynx]

sorry😭

okay

so

how do i factorize

so

hold on

they got differet exponents tho

[narrow ermine]

Yes, but $e^{2x} = (e^x)^2$

[proven lynx]

oh god

[glossy valve]

Azyrashacorki

[proven lynx]

wait but (e^x)^2 is

2e^x^2

or no

[narrow ermine]

You have $4e^{2x} - ce^x = 4(e^x)^2 - ce^x$.

[glossy valve]

Azyrashacorki

[narrow ermine]

Both terms have e^x in common yeah?

[proven lynx]

yes

[narrow ermine]

So you can factor e^x out

[proven lynx]

okay so

yes

but what do we do with the ^2

[narrow ermine]

The same thing you would do if it wasnt e^x under there.
If you have 4a^2 - ca how would you factor a out?

[proven lynx]

take the a out

and 4^2?

[narrow ermine]

You factor like this
4a^2 - ca = a * (4a - c)

[proven lynx]

oh

so

okay let me write

like this then

[narrow ermine]

So now you have a product = 0

[proven lynx]

but is that better than doing ln

[narrow ermine]

You couldn't have done ln straight away

It wouldn't have helped

[proven lynx]

oh you just cant do that?

[narrow ermine]

Well you'll get ln(a+b) and that doesn't have nice rules

[proven lynx]

hmm okay

whats the rule for

ln(a+b)

[narrow ermine]

There isn't one

That's why

[proven lynx]

no i mean

how did u know ln(a+b) comes out

[narrow ermine]

a and b are just to show that you'll end up with the logarithm of a sum

[proven lynx]

oh

because

[narrow ermine]

Wait I was thinking of something else actually

[proven lynx]

yeah makes sense

[narrow ermine]

You could do ln in fact

If you start from $4e^{2x} = ce^x$

[glossy valve]

Azyrashacorki

[narrow ermine]

Sorry I wasn't paying attention

Maybe that's easier for you then

You can ln both sides here and use log rules

[narrow ermine]

Essentially from here, you would conclude that e^x = 0 or 4e^x - c = 0 and solve those separately to find solutions

[proven lynx]

because isnt

[proven lynx]

yes

[narrow ermine]

But from this you can do ln yes

I was mistaken

[proven lynx]

hmmm

wait

when you have

e^x = e^2x

and do ln

does it cancel out both e

[narrow ermine]

No you would have $ln(4e^{2x}) = ln(ce^x)$

[glossy valve]

Azyrashacorki

[narrow ermine]

You would have to use log properties to get x alone

[proven lynx]

damn what

[narrow ermine]

So like ln(ab) = ln(a) + ln(b)

[proven lynx]

okay i think ur methodnis better

😭

[narrow ermine]

Okok

They're both equivalent in the end

[proven lynx]

still gotta take a look at logarithmic rules tho

[narrow ermine]

My method makes it so you don't have to use them really.

But it depends on what approach you want

[proven lynx]

btw how do i solve this

[narrow ermine]

e^ln(something) = something

Just gotta be careful with that first term

Write it like $(e^x)^2$ first

[glossy valve]

Azyrashacorki

[proven lynx]

hmm

okay

but im confused

generally speaking isnt (e^x)^2 = 2e^x^2

[narrow ermine]

No

$(e^x)^2 = e^{2x}$

[glossy valve]

Azyrashacorki

[proven lynx]

okay

yeah

yes

[proven lynx]

wait so then we just have ln(1/4c)^2?

because the ln cancels the e

[narrow ermine]

Wait no

[proven lynx]

no we just have 1/4c^2?

[narrow ermine]

Yeah whole thing squared

(c/4)^2

[proven lynx]

thats just 1/16c^2 then

i hope

[narrow ermine]

Yes

[proven lynx]

wait sorry can you explain this im kinda confused

why isnt it 2e

[narrow ermine]

In general, $(a^x)^y = a^{xy}$

[glossy valve]

Azyrashacorki

[narrow ermine]

That's a law of exponents

Just like $a^x \cdot a^y = a^{x+y}$

[glossy valve]

Azyrashacorki

[proven lynx]

oh shit

oh my god

yes yeah

ty

okay so

yeah got it

okay ty man, really helped me

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[twin venture]

i am clueless on both of these tasks

[sweet onyx]

how can you translate growth rate in a function ?

[twin venture]

idk wym

derivate

[sweet onyx]

what is the "growth rate" of a function between 2 instants ?

yes exactly

[sweet onyx]

so the growth rate is the derivative

[twin venture]

between 2 points its that

[sweet onyx]

i wanted to lead you to derivative

[twin venture]

oh

i just need the extrem points of the derivate for d right?

[sweet onyx]

yes exactly

[twin venture]

oh

that makes sense

[sweet onyx]

now for e

v is a function that returns a growth rate relative to a time

how can you understand it, acknowledging how we interpreted d)

[twin venture]

uh

[sweet onyx]

since the growth rate in the question d is the derivative of a function

[twin venture]

its an derivative

[sweet onyx]

yes

and you want to find the total height

so, in d) you made the transformation "height (h) -> growth rate (h')"

now you want to go from a growth rate to a height

[twin venture]

so i need an Antiderivative

[sweet onyx]

exactly

yes

[twin venture]

is it that ?

or do i just input an random number for c

[sweet onyx]

yes, but now you have something in the exercise that says that g(2) = 120

so just find c with that result

[twin venture]

oh okay

last question about d

after i get the extrem point what then?

[sweet onyx]

then if that extremum is more than 27, he's right

else he's lying

[twin venture]

so the y value which i get by doing the x value of that extrem point in the h(x) right?

[sweet onyx]

what ?

the value you have to check is h'(x)

so you have to find if h'(x) < 27 or if h'(x) >= 27

with x the extremum

[twin venture]

i thought ill have to h'(x) = 0 and then input the x into h(x) to find y of that extrempoint

wait

[sweet onyx]

no

[twin venture]

i get it now ill need solve the equation from h'(x) < 27

lemme try it

[sweet onyx]

uhh you just have to find the maximum value of h'.. or you can find the solutions for "h'(x) < 27"..

but finding the maximum value is way easier i believe

[twin venture]

how

[sweet onyx]

how do you normally find the extremum of a function ?

[twin venture]

i think there is confusion of what extremum is

1sec

[sweet onyx]

the maximum(s) if you want

extremum means maximums + minimums

[twin venture]

im german and that often confuses the words for me because i cant speak it in math terms very well

[sweet onyx]

dont worry I understand

[twin venture]

so u dont mean that with extremum

the points inthere

[sweet onyx]

yes

[twin venture]

h'(x) = 0 and then input the x into h(x) to find the height

right?

[sweet onyx]

but you want to find the extremum of h'(x), not h(x) !!

so treat h' as a function

[twin venture]

oh wait

but why

[sweet onyx]

and find x s.t. h''(x) = 0

[twin venture]

why do i need the derivate of the derivate for the extremum

wouldnt taht be the inflection point

[sweet onyx]

the function you want to study is the growth rate

you want to find the maximum value of the growth rate

meaning that you want to find the maximum of h'

[twin venture]

oh

[sweet onyx]

is that clearer ?

[twin venture]

ye

thanks

[sweet onyx]

no problem

[twin venture]

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[spice grail]

can someone check if this is correct so far

Also I don't know what to do after this stage

[uneven cargo]

If I'm not mistaken $x''=-16 \lambda sin(4x) - 16 \mu cos(4x)$

[glossy valve]

Emil

[spice grail]

oh yeah

[spice grail]

yep

and after I've simplified this

[uneven cargo]

I'm reading over it one second

is this -16 x lambda x sin(4x)

[spice grail]

yes

no

-16lambasin(4x)

[uneven cargo]

oh yeah i meant multiplication

[flint hull]

<@&268886789983436800>

[spice grail]

so I get thus

[uneven cargo]

The algebra is right

looks good

[spice grail]

I don't get what to do after assigning the coefficents

[uneven cargo]

I think you just need to solve the linear system

[spice grail]

like why do I equate the cosine coeffients to1

[uneven cargo]

solve for mu and lambda

[uneven cargo]

no no wait

[spice grail]

^

[uneven cargo]

-13mu + 16lambda = 0

because there is no cosine term in the differential equation

oh wait mb im high

sorry

[uneven cargo]

since right hand side is sin(4x)+0*cos(4x)

we want to match the coefficients

[spice grail]

I don't get it

[uneven cargo]

okay so you are plugging in $x=\lambda sin(4x)+\mu cos(4x)$ into the original differential equation right

[glossy valve]

Emil

[uneven cargo]

And after plugging this into the differential equation x''+4x'+3=sin(4x)

you get $(-13 \mu + 16 \lambda) * cos(4x) + (-13 \lambda -16 \mu) * sin(4x) = sin(4x)$

[glossy valve]

Emil

[uneven cargo]

So we know that $-13 \lambda - 16 \mu = 1$

[glossy valve]

Emil

[uneven cargo]

Cuz sin(4x) and cos(4x) are linearly independent

and $(-13 \mu + 16 \lambda) = 0$

[glossy valve]

Emil

[uneven cargo]

If you look at your original differential equation the RHS is sin(4x) btw just for clarity

[spice grail]

ah

I see now

thanks

[uneven cargo]

no problem

[uneven cargo]

when you do the common denominator it should be $-169 \mu - 256 \mu = 16$

[glossy valve]

Emil

[spice grail]

oh

[uneven cargo]

ur multiplying both sides by 16

so $\mu = \frac{-16}{425}$

[glossy valve]

Emil

[uneven cargo]

I believe

[spice grail]

yeah it does

thanks

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[uneven cargo]

no problem

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[mellow yacht]

how do I solve this DE?

[robust slate]

It's linear

the integrating factor integral seems gnarly, but they put it together to work out nicely

[mellow yacht]

yea Ik

[robust slate]

so

[mellow yacht]

I meant ik it's linear

[robust slate]

So can you show what you've done so far with that?

ex. have you evaluated the integral for the integrating factor?

have you set up the integrating factor?

etc

[mellow yacht]

the integrating factor is a mess. I am not able to integrate it

I tried to force it into an exact DE but it won't work either

[robust slate]

In the future, please give additional context like this - it gives us more to work with and saves time from explaining things unnecessarily. Anyway, what was your integral for the integrating factor?

Perhaps you're off by a sign or something

[mellow yacht]

ohh I'll take a note of it for the next time.

yea

just a sec

$\mu(x) = e^{\int \frac{e^x (x-2)}{x(x^2 + e^x)}}$

[glossy valve]

Prathmesh

[mellow yacht]

this

[robust slate]

Try dividing the numerator and denominator by $x^3$.

[glossy valve]

Civil Service Pigeon

[mellow yacht]

then substitute $\frac{e^x}{x^2} = t$ right?

[glossy valve]

Prathmesh

[robust slate]

I suppose that would work

Or just do $\displaystyle t=1+\frac{e^x}{x^2}$ all at once

[glossy valve]

Civil Service Pigeon

[robust slate]

not that there's much a difference in the end

both integrals are equally trivial

[mellow yacht]

suree

you made it look so easy

[robust slate]

Observe the 2

[mellow yacht]

thanks!

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[mellow yacht]

ahhhh

i got it

thanks

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[atomic venture]

just wondering what it means by 'order' in 4

[atomic venture]

i've established previously that R/I is the quotient ring with representatives of the quadratic polynomials in F_3

[spiral vigil]

order of an element is the order of the subgroup generated by that element

[fast peak]

how often do you need to multiply X+I with itself to get the identity

(the things we said are the same)

[atomic venture]

oh right

uh lemme try both of those see if i see wym

[atomic venture]

using this, $(X+I)(R/I)={(X+I)(Y+I):Y\in R}={XY+I:Y\in R}$, and $XY \sim 0$ so $XY+I=0+I\quad\forall Y\in R$. so the subgroup generated by $X+I$ is the trivial subgroup?

[glossy valve]

lyric

[fast peak]

why should XY~ 0 ?

[atomic venture]

because $I=\langle X \rangle$ so $R/I$ is the set of equivalence classes of $\sim$, where $P\sim Q$ if $P-Q\in I$ which is the same as $P-Q$ having a factor of $X$

[glossy valve]

lyric

[atomic venture]

so $XY\sim 0$ because $XY-0=XY$ has a factor of X

[glossy valve]

lyric

[atomic venture]

oh i misinterpreted i think

i was thinking of the ideal generated by it

not the subgroup generated by it

[fast peak]

I is generated by the poly of degree 3 you chose earlier

[atomic venture]

oh yeah

but thats different to the subgroup generated by it

$I=._R\langle X \rangle={PX:P\in R}$ is how we've defined a ring generated by $X$

[glossy valve]

lyric

[atomic venture]

but $\langle X \rangle={X^n:n\in\mathbb{N}}$ is the subgroup generated by it

[glossy valve]

lyric

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@atomic venture Has your question been resolved?

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[mild hatch]

i dont understand this question... I sat in class the other day and went through the proof, but now im really confused looking back at it

[rapid rain]

You're confused about the statement or the proof? If it's the latter, you should also share it if you have it

[mild hatch]

can you explain the statement first...

i dont even know what it means

something about vertex connectivity, edge connnectivity, and the minimal degree of connected graph G

[rapid rain]

Well, "[quantity] can be made arbitrarily large" means no matter which size N you pick, you will always be able to find a graph G such that [quantity] >= N

Meaning for example if I pick N = 10

I can find a graph where delta(G) - lambda(G) is bigger than 10

But I could also pick another graph G where that same quantity is bigger than 100

Or 1000

Etc...

There is always some graph G that makes that quantity as big as we want

[mild hatch]

ohhhhh i was like what does that mean lol

uhmmm let me post the proof wait

[rapid rain]

Oof for the handwriting 😭

[mild hatch]

i hate every teacher who does this-

[rapid rain]

So wait that's the proof of the inequality chain

Did you have a problem with understanding this proof?

[mild hatch]

yeah...

i actually dont under what lamda G means..

[rapid rain]

Ok, the first part is easier

[mild hatch]

understand*

like the edge connectivity

[rapid rain]

So a graph is made up of vertices (like nails) and edges (like ropes)

[mild hatch]

yeas

[rapid rain]

Lambda(G) is the minimum number of ropes (edges) you need to cut (remove) to disconnect the graph

[mild hatch]

ohhhh okayy

[rapid rain]

So

For example, take the vertex with the smallest amount of ropes (edges) that go to it

And cut those ropes

[mild hatch]

a single vertex?

oh

K_2

[rapid rain]

? Sorry I'm not understanding what you mean

Is this a specific graph?

[mild hatch]

can i use any connected graph as an example?

[rapid rain]

Well if you want to understand how the proof works, the example should be a little more complex

At the very least K_4 right

[mild hatch]

yeah thats okay

[rapid rain]

Well, take one of the (here many) vertices with the least amount of outgoing edges

And cut those edges

Well, we've successfully disconnected the graph

[mild hatch]

yeah we will be left with a isolated vertex

yes..

[rapid rain]

Exactly

[mild hatch]

what does that mean xD

then-

[rapid rain]

So it must be worse (or the same) as the most efficient way to disconnect the graph by cutting edges

[mild hatch]

oh so it took the least amount of edge removal to disconnect G?

[rapid rain]

To be more precise, what we did required us to cut at least the minimum amount of edges to disconnect G

So we cut the same amount, or more than lambda(G)

[mild hatch]

in this case, we cut the same amount

[rapid rain]

Well in our example maybe, but not in every case will it be like that

Imagine, as a counterexample

You take two graphs K_4

[mild hatch]

yes

[rapid rain]

You take exactly one vertex from each, and you connect them by one edge

You get a new graph, call it G

[mild hatch]

yes

[rapid rain]

Now, you could try to disconnect G by finding one vertex and cutting every outgoing edge from it

[rapid rain]

Or... you could just cut that one edge

One of which is much more efficient

[mild hatch]

so wait so lambda is then 1

[rapid rain]

Yes

While delta(G) is 3

[mild hatch]

ohhhhhh

that made a lot of senseee wowww

whats the second part?

i just know that S is just a set of vertices in G that disconnects it and its like cut set or something..?

or uhmm edges..

[rapid rain]

Ok so, remember vertices are like poles and edges are ropes that connect them
Lambda(G) is basically "what's the most efficient way of cutting roped to disconnect the graph"
And kappa(G) is "what's the most efficient way of removing poles to disconnect the graph"

Respectively counting the number of ropes and poles needed to remove to disconnect

So for the proof

Take the most efficient way of cutting ropes

And we're gonna create a way to remove poles that's as efficient, if not more efficient, that the ropes cutting method

[mild hatch]

the uhmmm handshaking lemma or is that not it

nvm

i need to reread your text xD

so removing one vertex could disconnect a lot of edges

from the two K_4 joint by a bridge

[rapid rain]

For context, we consider that a graph made of a single point is disconnected

So in K_4

Instead of cutting the 3 ropes that go outward some vertex

We remove all the poles that are connected to the vertex

And it has the same effect

[mild hatch]

in K_4's case its K(G)=L(G)

right...?

[rapid rain]

Yes

Now to find a counterexample

You once once again take 2 copies of K_4

And you introduce one new vertex in the middle

And connect everyone to it

That new graph, call it G

You could try to remove edges in the most optimal way

So you take one of the vertices of K_4 and remove every outgoing edge once again

That would be 4 edges in total

Or... you could just rip out the vertex in the middle

[mild hatch]

ohhh so K(G) is 1 and L(G) is 4 in this case

sorry i had to read why we were introducing a new vertex im dum but its G

omg that makes so much sense

so now i have to find family of graphs for delta(G) - L(G) and a family of graphs for L(G) - K(G)

i just realize that he gave us a hint

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[bitter tree]

im not sure how they simplified it like this

in the blue box

[pearl thicket]

through the definition of x and y in polar coordinates

i.e. x = r cos t, y = r sin t

then plug those in

[olive olive]

yea and then they simplified r² with r

[bitter tree]

This is where I’m at..

i simplified top to r^2costhetasintheta

[narrow ermine]

Right. You can factor r^2 from inside the root in the denominator and bring that outside

[bitter tree]

yer

[narrow ermine]

In general though you wouldn't have had to plug x=rcos(theta) and y=rsin(theta) into the denominator as r is defined as r = sqrt(x^2 + y^2)

[bitter tree]

oh

[narrow ermine]

It's not wrong

You'd get the same thing

[narrow ermine]

Just a tiny bit more work with this

[bitter tree]

ohh

pythag

rule

and then i just get r

ohh

[analog furnace]

Then sinx^2 + cosx^2 = 1

[bitter tree]

ok thx

.close

[full forum]

Channel closed

Closed by @bitter tree

Use .reopen if this was a mistake.

[bitter tree]

.reopen

[full forum]

✅ Original question: #help-28 message

[bitter tree]

how come we take the absolute?

[bitter tree]

here

[narrow ermine]

It's a quick way to show that the quantity in the limit is being squeezed from above and below

Essentially $\abs{\frac{xy}{\sqrt{x^2+y^2}}}\le r$ is equivalent to $$-r \le \frac{xy}{\sqrt{x^2+y^2}} \le r$$

[glossy valve]

Azyrashacorki

[narrow ermine]

So the limit is being squeezed above by $\lim_{r\to 0} r =0$ and below by $\lim_{r\to 0} -r = 0$

[pearl thicket]

Which implies r going to 0 is a looser condition

and so you can use r to 0 instead

[glossy valve]

Azyrashacorki