help-28

[delicate torrent]

hint: use the identity $\sin^2 \theta + \cos^2 \theta = 1$

[glossy valve]

1 divided by 0 equals Infinity

[craggy tapir]

there's nothing to prove

[lime trellis]

do you mean simplify?

[torn jolt]

Ye Ig

[delicate torrent]

@torn jolt

solves all problem

[torn jolt]

Alr lemme try

Yall im good i was afraid to try

Thanks

[glacial pendant]

Hello

Guys

[delicate torrent]

!done

[torn jolt]

.close

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[delicate torrent]

!redir

[full forum]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

Channel closed

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[delicate torrent]

so you did know the commands lol

[craggy tapir]

are you trying to ask a question? if so you should use a channel in the available category

[glacial pendant]

Is it tan^4A

[craggy tapir]

!nosols

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glacial pendant]

Oh sry

[delicate torrent]

!nosols

[dusk nebula]

no lol

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glacial pendant]

How to check then?

[torn jolt]

U stole my help channel

[dusk nebula]

idk

[glacial pendant]

What's the solution? I'm still in high school

10th standard

[torn jolt]

I don’t remember I’m on another problem now and am too lazy to turn back

[torn jolt]

I’m 11th

[glacial pendant]

I just solved this with identities I knew

[glacial pendant]

Indian?

[torn jolt]

There’s only like 20 different identities

[torn jolt]

No

[glacial pendant]

Oh ic then syllabus is diff

[glacial pendant]

In India they taught only 3

[torn jolt]

Ig that’s for 10th

[glacial pendant]

Ye

In 11th it is more diverse

[torn jolt]

I promise there gonna be 20 more next term

[torn jolt]

XD

[glacial pendant]

Yes

[torn jolt]

I’m actually gonna cry over trig so bye

[sour shore]

Way more than 20

[glacial pendant]

I just know sin^2 + cos ^2 = 1

[torn jolt]

So fr

[glacial pendant]

cosec^2-cot2= 1

sec^2 - tan^2 = 1

[torn jolt]

😭

[sour shore]

And they uglyy too

[glacial pendant]

Nope

Trigo is fine till 10th

Idk about 11th

[torn jolt]

It gets more cooked when u get flashbacks in trig sub

U substitution ftw

[glacial pendant]

Can u solve this one? @sour shore

[sour shore]

[glacial pendant]

I was right who the hell said no @torn jolt

[sour shore]

Yh..

[glacial pendant]

Just confirming tan^4A?

[sour shore]

(TanA)^4

[glacial pendant]

Yes ofc

I started doubting my self

[torn jolt]

U were wrong

It’s 4tanA

[glacial pendant]

What

No

It's in exponent

[glacial pendant]

.

[torn jolt]

Nuh uh my key says otherwise

[glacial pendant]

Crazy key

AI also says the same

[delicate torrent]

what does your key said then

[glacial pendant]

4tanA

How could she get 4😭

[delicate torrent]

😭

that's the damn problem

[delicate torrent]

yeah, this is the key to this problem

[glacial pendant]

Wtf

Ya I mean ya

[full forum]

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[harsh cove]

how can i solve log3 11 by hand, i know how to work out logs when the base and n can be represented as a power of a common factor ( like 4 and 8 can be represented as 2^2 and 2^3), but 3 and 11 are both prime numbers so im not sure how i can work it out

[onyx glen]

by hand? in exact form? log_3(11) is already as simplified as it possibly can be

are you required to compute it to 5 decimals or something (BY HAND) or what

[harsh cove]

well the whole question is:
3^2x = 11, workout x

so what iv done so far is use log on both sides to get to
2x = log_3 11
so if i work out what log_3 11 is then half it i can get the answer, the decimals isnt really specified

[delicate torrent]

$3^{2x} = 11$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

find $x$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

well $\log_3 11$ is irrational

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

you can keep it as it is

if you tried to find the decimal representation then it would only be an approxiate => wrong answer

@harsh cove

[harsh cove]

so should i just answer with x = 1/2(log_3 11)

[delicate torrent]

yes

$x = \frac{\log_3 11}{2}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

this is valid already

or if you want to convert to $\ln$, you can

[glossy valve]

1 divided by 0 equals Infinity

[harsh cove]

im gona check the answer sheet to see if thats what i can put down

[delicate torrent]

unless if your question asks to approximate

[harsh cove]

it says x = 1.09,

[delicate torrent]

then you would need to work on the decimal representation depend on what the question asks you

!original

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[delicate torrent]

give me the whole context

[harsh cove]

it just says

[delicate torrent]

keep it as it is

,w calc log_3(11)/2

[glossy valve]

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Requested by iaminfinityiq

[delicate torrent]

this is just a decimal representation

.

,w calc 3^(2*1.09)

[glossy valve]

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Requested by iaminfinityiq

[delicate torrent]

this is what happens if x = 1.09

[harsh cove]

tbh the question doesnt specify whether its calculator or not so i think that it is calculator given that giant decimal

[delicate torrent]

keeping it as this is enough

or safer you can do $\frac{\ln 11}{2\ln 3}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

if your teacher said x = 1.09 then show your teacher this

[harsh cove]

okk thank you

.close

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[round gust]

what does it mean to rotate this figure around the x axis

i need to rotate it around the x axis

so that TBC is in the xy plane

and S has a positive y- coordinate

do i just

move it towards the right?

like

towards the positive y direction

[spiral vigil]

if i understand it correctly it's like this

[round gust]

ok so

i know the points

T(0,0,0)

B(5,5,12)

C(-5,5,12)

S(0,0,24)

now i rotate

i need to find the new coordinate of S

my reasoning is:

the distance from T to BC is 13

and then i go an extra 12 (distance from abcd to S)

so the new y coordinate for S would be 13 + 12 = 25?

because i am thinking that d(T, BC) = d(T, B'C')

=13

[round gust]

this is correct

[round gust]

but this isnt

[spiral vigil]

the distance is the same but what's the angle?

may be helpful to look at the yz projectin

[round gust]

,rccw

[round gust]

I thought that

I can calculate the angle at which TB rotates

So the angle between TB and the xy plane

Which is 59.49°

So then i though

That

TS rotates at the same angle as TB

So TS should also rotate at 59.49 degrees

While keeping its x coordinate unchanged

So i made a triangle

[covert remnant]

answer this

i can t solve it broo

[round gust]

try u sub, u=tan(x)

[round gust]

do you think my method is correct?

or im i approaching it wrong

[turbid badge]

!occupied

[full forum]

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

[covert remnant]

no bro i still no get it

[full forum]

@round gust Has your question been resolved?

[full forum]

@round gust Has your question been resolved?

[full forum]

@round gust Has your question been resolved?

[round gust]

omg i did it

after 4 hours

yay

.close

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Channel closed

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[toxic sage]

can someone explain the alternating subgroup to me? i don't get it

[cedar cloud]

i can try

did you understand the permutation group?

S_n

[toxic sage]

Not really

[fathom saddle]

This channel is closing

Will lock soon. You'll need to grab your own. #help-2 is available

Well, then again may not lock soon. It's been open for an hour and a half lol

But will lock randomly

[cedar cloud]

we can use #groups-rings-fields ?

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[molten torrent]

Hello! I need a bit of help with verifying trigonometric identities! I've been stuck on my homework for the last hour T-T

[regal thicket]

What exactly are you struggling with?

[molten torrent]

Ive never been too good with fractions, so far I've made it here

[stone matrix]

try and simplify a bit now in the denominator of the denominator

[molten torrent]

I don't know how, I forgot 😅

[digital void]

can you explain the step you took from line 1 to 2? what did you multiply with exactly?

oh nvm sorry

you expanded the expression

hmm perhaps you shouldnt expand quite yet, to avoid confusion in the next steps

do you know how to cross multiply? id vouch for that in this case

[prime pier]

what about $\frac{\sin^2(x)}{\cos^2(x)} + \frac{1}{\cos(x)},$ can you simplify just that?

[glossy valve]

ηασιβ ♥

[delicate torrent]

fractions inside fractions are ugly, so try to get rid of that quickly

and the 1 on the numerator will flip the fraction you will have in the denominator

[molten torrent]

I can but I don't know how I'd do that one it's addition

[molten torrent]

I don't remember how to simply that

[delicate torrent]

do you know what a common denominator is?

[molten torrent]

It's cosx right?

[delicate torrent]

no

but you're close

[molten torrent]

Cos2x ?

[delicate torrent]

yes

$\cos^2 x$

[glossy valve]

1 divided by 0 equals Infinity

[molten torrent]

So what do I do with the cosx ?

[delicate torrent]

then make it the same denom

[molten torrent]

I can just make cosx into cos2x ?

[delicate torrent]

ya?

[molten torrent]

Ok

[delicate torrent]

$\frac{1}{\cos x}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

multiply both numerator and denominator by $\cos x$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

what you got @molten torrent

[delicate torrent]

also, don't just remember how to do things in math, you need to understand why it works too

that helps a lot when you're struggling in math

[delicate torrent]

yep

[molten torrent]

I think I'm getting it a bit more but I don't know how to get the fraction out of the fraction

Should I start over and separate the tan2x and secx ?

[formal snow]

the fraction below the 1?

[molten torrent]

Yeah I think it's stuck

[formal snow]

ok so take that below fraction, pull it out and flip it

kiểu

1/(a/b) = b/a

$\frac{1}{\frac{a}{b} = \frac{b}{a}$

[glossy valve]

^the shark
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[formal snow]

$\frac{1}{\frac{a}{b}} = \frac{b}{a}$

[glossy valve]

^the shark

[molten torrent]

Like this right?

[formal snow]

yeah !!

[molten torrent]

Thank you all! I did it!

[formal snow]

aight now type .close

[molten torrent]

.close

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[formal isle]

Can anyone help me with question 9. I dont know why my answer is different from the answer key ( ignore the writing its wrong )

[sleek torrent]

<@&286206848099549185>

[ocean sluice]

,rccw

[outer lotus]

Can you show your work?

What's your answer

[ocean sluice]

at the top

well above the problem

i think

[formal isle]

That’s the formula for part a

[ocean sluice]

why would it be (1+0.045)^5

[formal isle]

Cause like 0.045 is monthly and it’s compounded interest so it to the power of 5 months

[ocean sluice]

ohh oops 😂

[outer lotus]

That's not wrong

[ocean sluice]

yeah i saw that it was compound interest

[outer lotus]

Via, what is your answer for a)

[formal isle]

For A I got 4185.09 and for B I got 3825

[outer lotus]

What are the right answers

[formal isle]

It’s 3738.55 for a and 675 for b

[ocean sluice]

,w (17000(1+0.045)^5) - 17000

[glossy valve]

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[ocean sluice]

hmm

[outer lotus]

It's because you first need to calculate what the interest would be without down payment

And then find the difference

[formal isle]

Ohhh bruh

😔

[ocean sluice]

ah

[formal isle]

Tysm

[outer lotus]

No problem 😎

[formal isle]

Wait.

How does that get the answer for a tho?

[formal isle]

OHHHH

Tysm !

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[old edge]

wait is there anyone that knows how to factor x^4+1 using something similar to completing the square

[rare dock]

would you know how to factor x^2 + 1?

[old edge]

over imaginary numbers but my math teacher told me i can factor x^4+1 without imaginary?

[rare dock]

oh well yes sure

[rare dock]

this could still help you though

[old edge]

am i slow idk how to do it

[whole lily]

Refer to Sophie Germain Factorization

\begin{aligned}x^{4}+4y^{4}&={\bigl (}(x+y)^{2}+y^{2}{\bigr )}\cdot {\bigl (}(x-y)^{2}+y^{2}{\bigr )}\&=(x^{2}+2xy+2y^{2})\cdot (x^{2}-2xy+2y^{2})\end{aligned}

[glossy valve]

Roy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[old edge]

wait ur a genius

TYSM

[whole lily]

No worries. It's used a lot in integration

\begin{align*}
x^{4}+1 &= x^{4}+1+2x^{2}-2x^{2} \
&= \left(x^{2}+1\right)^{2}-2x^{2} \
&= \left(x^{2}+1-\sqrt{2}x\right)\left(x^{2}+1+\sqrt{2}x\right)
\end{align*}

[glossy valve]

Roy

[whole lily]

And yes, it's the same idea as completing the square

[old edge]

where do the 2x^2 - 2x^2 come from

[whole lily]

2x^2 - 2x^2 = 0

We basically add those terms to force the "completing of the square"

[old edge]

oh but how do you come up with them

[whole lily]

Because $\left(x^{2}+1\right)^{2}=x^{4}+1+2x^{2}$

[glossy valve]

Roy

[whole lily]

Replace each x^2 term with x and it's basically the same as completing the square

[old edge]

ok got it ty

[whole lily]

what is this problem for for may I ask

[full forum]

@old edge Has your question been resolved?

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[whole lily]

I was gonna say, in case if it was for integrating $\int \frac{1}{x^{4}+1}dx$ using partial fractions, there are more efficient methods

[glossy valve]

Roy

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[still marlin]

help

i forgot how to solve this typa stuff

[keen vector]

!status

[full forum]

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

[keen vector]

@still marlin

[still marlin]

1

[still marlin]

.

[keen vector]

really you don't know where to start?

how about simplifying the denominators

[still marlin]

i mean are u supposed to take x2=t y2=p and then solve 4 equations?

[keen vector]

yes basically

it's a system of linear equations

[still marlin]

dang i thought there was a better solution which wasnt click in my head

[keen vector]

so you did know where to start

[still marlin]

yeh ig so lol

[undone radish]

just set them equal

[still marlin]

wdym

[keen vector]

they're all equal to 1

[still marlin]

can someone find the telescoping series here? ik i can solve this with gamma functions but

x+n-x-1 gives n-1 but how do i make it a telescoping sum

[full forum]

@still marlin Has your question been resolved?

[jovial junco]

I think you can express the denominator as a product of gamma values

[round sparrow]

You can do it without gamma function

Let’s call each term u_n

What is u_n+1 in function of u_n (and of n) ?

[still marlin]

yeh as i said ik i can solve this with gamma functions but

still wanna do it with telescoping

i gtg now ill come back to this later

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[summer creek]

8c2?

Or 8c2/2=14

[queen flame]

This

[summer creek]

Why?

A handshake to B

B handshake to A is same

[queen flame]

1 hand shake =>1 pair of people so we find how many pairs of two can be made

[queen flame]

Yes and while choosing we consider them same

So 8C2

[onyx glen]

already accounted for by the fact that you're doing C rather than P

[summer creek]

8P2/2=8C2?

@onyx glen

[stone matrix]

Yes. nPr/r! = nCr

[summer creek]

I see

Thanks

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[grave elm]

If we have a rhombus with ratio of side to height 5 : 3, what's the ratio of diagonals?

[grave elm]

what i did is u1^2 + u2^2 = 100 and u1*u2 = 30, divide it and get r + 1/r = 10/3. But is there anything more elegant?

r is the ratio u1 / u2

[torn jolt]

10?

shouldnt that side be 5

[grave elm]

i chose it to be 10 : 6 instead, same thing

[torn jolt]

oh ok i thought the height was 3

[torn jolt]

take the altitude so that it starts from the vertex

[grave elm]

the 3 is half-height

[torn jolt]

and you get a 3-4-5 right triangle

[grave elm]

oh shit

wait no

[torn jolt]

oh yeah nvm

[grave elm]

oh u do

idk what next though

[torn jolt]

oh ok yeah

[torn jolt]

you can explicitly calculate the diagonals with pytagoras

[grave elm]

oh thats nice

[torn jolt]

with this

[bright bronze]

you can see the diagonals as hypotenuses of two right angled triangles of sides 3&9 and 3&1

[grave elm]

That's so cool

and the ratio is obviously 3

[bright bronze]

yes

[grave elm]

beautiful

thanks

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[torn jolt]

Hi

[torn jolt]

So
Consider 4 boxes , where each box contains 3 red balls and 2 blue balls . Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

[onyx glen]

!status

[full forum]

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

[torn jolt]

3 because i am over counting but i dont get how

Each box has 3 red ball and 2 blue ball so firstly i picked 8 red balls to ensure 1 red ball and 1 blue ball is taken from each box and then we can find the ways to find the other two balls from the remaining. So 3⁴.2⁴ would be the no. Of ways to choose . In this case for 8 balls

[hoary ember]

what did you do with the 2 other balls that are to be chosen?

[torn jolt]

I made 2 cases

Ist one where 2 balls can be choosen from 1 box and 2nd case where balls are choosen from 2 boxes

Ofc subcases of colours will be formed

Total 66 ways

[hoary ember]

did you consider the fact that a blue ball (call this one A) chosen from the first 8 balls then another blue ball chosen from the 2 balls (call this one B) is counted twice with B first then A?

[torn jolt]

No its not counted twice

While forming the cases it was considered

<@&286206848099549185>

[vapid shell]

Hi

[torn jolt]

Hi bro i hope you can help me

[torn jolt]

This one

[vapid shell]

Tell I try

[torn jolt]

What?

[vapid shell]

Problem

[torn jolt]

Maybe show ur work

[torn jolt]

@vapid shell this one

[torn jolt]

Read the chat i already did

[onyx glen]

ok so yeah you are overcounting in a way that's kinda hard to account for

bc like, you're essentially marking the first red and first blue ball with a sticker

and sometimes this leads to overcounting by 2, sometimes by 4, sometimes not at all

here's a better way to do it

there are only two distributions of ball counts:
4-2-2-2 and 3-3-2-2 (up to box order)

[vapid shell]

1296 is ans

[onyx glen]

!nosols

[torn jolt]

No

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[torn jolt]

Mhmm

[vapid shell]

How than what is ans

[torn jolt]

21816

[onyx glen]

so for 4-2-2-2: first pick which of the 4 boxes you take 4 balls from, then find the number of ways to pull from each box

with the 2s it'll only be RB, with the 4 it will be RBBB or RRBB

[torn jolt]

4C1

[onyx glen]

that's the first factor yes

[vapid shell]

How it can be ans is 1296

[torn jolt]

No its not

[onyx glen]

maybe your logic is also wrong

[torn jolt]

Pls let be understand

[onyx glen]

but "yes it is" "no it's not" is unproductive so leave it now

unless you wish to share your own logic to see where you must have gone wrong

[vapid shell]

Let me try again

[onyx glen]

!nosols, as a reminder

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[torn jolt]

I already showed you my logic

Why is baseline method wrong

[vapid shell]

I checked its wrong

[torn jolt]

Whats wrong

[vapid shell]

21816

[torn jolt]

The answer is 21816 bro

[vapid shell]

How

[torn jolt]

Idk where u checked from

Even ai overcount here

[vapid shell]

I solved

[torn jolt]

Show your work

[vapid shell]

What ai

[torn jolt]

Chat gpt etc

I mean its pretty obvious the answer wouldn't be a 4 digit

[vapid shell]

What is chat gpt sayed i don't what is it

[torn jolt]

He also overcounted at first

His answer was the same as mine which is 85536

<@&286206848099549185>

[vapid shell]

Bro your qualification

[torn jolt]

Why do u ask

[vapid shell]

I agery my answer can be wrong but more than 10000 is never be ans of it

[torn jolt]

Nvm bro you did well you can leave

[queen flame]

-# RRRB instead of RBBB but yeah going this way is easier

[onyx glen]

why 10000

[onyx glen]

oh hold on i mixed up which color was 3 balls and which was 2

[onyx glen]

!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).

[torn jolt]

Can you explain why the baseline method overcounted

[warped frost]

please do not ping helpers when some are already in your channel.

[torn jolt]

Where i manually counted as 3⁴.2⁴

[onyx glen]

actually sometimes also by 3

[torn jolt]

At the time i did nobody was here

[queen flame]

^

[queen flame]

^

[torn jolt]

Well if you do it by your way you are still marking stickers on balls

How does that differ

[onyx glen]

i am not, actually

maybe i can illustrate with a simpler example if you are ok with that?

[torn jolt]

I think maybe if we choose first 8 balls and then the other two balls , it could have been possible that these extra two balls would have been choosen in initial 8 balls and rest in extra , so it forms the same final result which results to overcount. Thats what i can understand as of now

[onyx glen]

yes, that's basically the logic

[torn jolt]

Yeah i get it

[onyx glen]

let's say we have a standard deck and we want to know the number of ways to draw 3 cards from it such that there is at least one spade

a mistaken way to do it would be 13 * 51C2, with the reasoning of "draw a spade first then two other whatever cards"

but then like, if you look at the hand {Q♤, 6♤, 2♤} how can you tell which one was the first drawn spade

you can't

[torn jolt]

Now can you please explain the correct method to solve this

[onyx glen]

i did already, no?

[torn jolt]

You didnt solve fully ig

Only 1 case

[onyx glen]

of course i did not solve it fully

[torn jolt]

Oh i see you want me to do the rest

[queen flame]

-# basically no free labor

[onyx glen]

yes that's always the idea on this discord

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[sour torrent]

$\int^{+\infty}_{-\infty} \cos(t) \delta(t) \mathrm dt$

[glossy valve]

alee

[sour torrent]

Can I say that since the delta is different from 0 in a very small neighborhood of 0, then the cosine assumes a constant value?

i.e cos(0) ?

in the sense that in that small neighborhood the cosine function almost always assumes the value 0?

so

$\cos(0) \int^{+\infty}_{-\infty} \delta(t) \mathrm dt$ ?

[glossy valve]

alee

[sour torrent]

and now I can think of the delta as a rectangle with base $\epsilon$ and height $1/\epsilon$ then area = 1 ?

[glossy valve]

alee

[sour torrent]

so the result would be cos(0) = 1 ?

[rigid crescent]

Yes that works. In general the delta function obeys this property given f is continuous at t=t_0

[sour torrent]

Yes, I'm trying to see how these results come out using some approximations (like the normalized Gaussian).

[rigid crescent]

the reasoning is because the delta function concentrates most of the weight near t=0

[rapid rain]

If you really need to understand dirac delta as a function, yes... though it is a bit wrong when we don't refer to it as a distribution

[sour torrent]

i was thinking about this graph so when epsilon -> 0 the hight increase

[rapid rain]

In the meantime you can think about it like a limit of functions like that where a -> 0

[sour torrent]

and the base get smaller

so area = epislon * 1/epsilon = 1

[sour torrent]

yes

that is, how should I justify those results that follow from the properties above?

i was thinking about that

ie in an small interval espilon i can say that che function always assume the value 0

i mean $\approx x(0)$

[glossy valve]

alee

[rapid rain]

welllllllll depending on the definition of delta, you don't require f to be continuous

[sour torrent]

instead if i have like $\delta(t-1)$ then

[glossy valve]

alee

[sour torrent]

$\approx x(1)$

[glossy valve]

alee

[sour torrent]

can I imagine an interval epsilon that contains the entire bell in point 1 and in that interval the function is approximated to a constant?

[rapid rain]

yes for now you can think of $\delta (\cdot - t_0)$ as a limit of "plateau" functions, rectangles of width $\varepsilon$ around $t_0$, and of height $\frac 1\varepsilon$

[glossy valve]

Rafilouyear2026

[sour torrent]

instead

how can i think this as ?

[rapid rain]

if a > 0, this is just a variable change right

if a < 0, you get $\int_{-\infty}^{\infty} x(t)\delta(at) dt = \frac 1a\int_{\infty}^{-\infty} x(\frac ta)\delta(t) dt$

[glossy valve]

Rafilouyear2026

[rapid rain]

the bounds are swapped because "u = at" runs through the real line from right to left

[rapid rain]

I guess to understand this formula, think that delta is once again the limit of those plateau functions

if you now look at plateau(at) instead of plateau(t), you change the pace at which you run through the plateau

so going faster/slower at a rate of "a" means the area is changed by a factor of 1/|a|

it's the same as if you changed the height of the plateau by 1/|a|

[sour torrent]

mm ok

[full forum]

@sour torrent Has your question been resolved?

[sour torrent]

ok so basically now the base is epsilon/|a|

so $\int^{\infty}{\infty} x(t) \delta(at) \mathrm dt$ = $x(0) \int^{\infty}{-\infty} \delta(at) \mathrm dt$

[glossy valve]

alee

[sour torrent]

and $\int^{\infty}_{-\infty} \delta(at) \mathrm dt = 1/|a|$

[glossy valve]

alee

[sour torrent]

since Area = epsilon/|a| * 1/epsilon = 1/|a|

right ?

so i can also this as

$\int^{\infty}_{-\infty} x(t) (1/|a| \delta(t)) \mathrm dt$

[glossy valve]

alee

[sour torrent]

so $\delta(at) = \frac1{|a|} \delta(t)$

[glossy valve]

alee

[rapid rain]

exactly

[sour torrent]

thanks

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[restive nimbus]

could someone help me solve this

[onyx glen]

$\cos(z) = \frac12 (e^{iz} + e^{-iz})$

[glossy valve]

Ann

[onyx glen]

though one is inevitably led to wonder why we are asked to find "the" value of z satisfying this equation when in fact there's infinitely many of them (as cos does not lose its 2pi-periodicity when transitioning from R to C)

[restive nimbus]

so theres infinite amount of answers..?

wait that makes sense

but like how do i get one of the solutions

i did rewrite it as cos(x+iy) and the used some formulas to get it to cosxcoshy -jsinxsinhy

[onyx glen]

did your i become j at some point

anyway like. ok. sure.

you want sin(x) sinh(y) = 0 and cos(x) cosh(y) = 3 and x, y both real.

[restive nimbus]

okay so i got x=2kpi and y cant be 0 i think

[restive nimbus]

so just cosh(y)=3?

also yeah we use j lol cause i gets confused with current or something like that

[onyx glen]

yes

[restive nimbus]

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[glossy valve]

Couldn't find an attached image in the last 10 messages.

Jan 7, 2026, 7:25 PM

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[signal wasp]

Huh?

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[pliant hornet]

i get no result and I don't know what i did wrong, do anyone know?

[gritty rose]

solve for b here and plug it into the line below

then you can try to solve for lambda

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[timber vapor]

hello how would i solve this if there’s a hole on coordinates that have the x value of 0?

[steep elbow]

if x is close to 3, what values does f(x) take?

[timber vapor]

0 i think?

[steep elbow]

well not exactly 0, right?

[timber vapor]

right

[steep elbow]

but you are correct in the sense that, if x is close to 3, then f(x) is close to 0

now let's set u = f(x)

if u is close to 0, what is f(u) close to?

[timber vapor]

im not sure if i understand correctly

because i thought then you’d find the y value when x is 0?

[full delta]

Think of the limit in terms of left and right hand limits

lim (x->3)f(x) approaches 0 but only from one direction

[steep elbow]

there's a difference between "at" 0 and "near" 0

[timber vapor]

i see

[steep elbow]

for example, if x is near 4, then f(x) is near -3

even though f isn't defined "at" x = 4

[timber vapor]

got it

how would i write an answer tho cuz i tried another method and it said it was wrong

[full delta]

You know what left and right hand limits are right

[timber vapor]

yes

but this one doesn’t specify

[full delta]

Can you tell me if we should be concerned with the right hand limit of f(f(x)) in this case

[timber vapor]

yes i think

[full delta]

Why do you think so

[timber vapor]

because it approaches 3

[full delta]

x approaches 3

what does f(x) approach

[timber vapor]

f(x) is the y value right?

[full delta]

Uh yeah

[timber vapor]

there’s two points that aren’t filled on the y axis

so one of those?

[full delta]

as x approaches 3, f(x) approaches only one value

[timber vapor]

this?

[full delta]

Yeah

It approaches 0 but only from the negative side

That's why we need not be concerned about the right hand limit of f(f(x))

Think of f(x) as u

u is approaching 0 only from the negative direction

[timber vapor]

i see

[full delta]

What is the blue box in the image

Is that the answer you gave

[timber vapor]

no thats an option if the answer is does not exist

[full delta]

Yeah so what do you now think the answer is

[timber vapor]

it doesn’t exist?

[full delta]

Why do you think that

Since we are being asked the left hand limit only

What could the reason be for it to not exist

That can only happen if it blows up to infinity or has a gap in the domain

Around* that point

In its neighborhood

Both of which isn't teue

Theres only a point discontinuity

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[cold sail]

how do u obtain the pressure scale without using the stokes eqs i dont get it

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[nimble ridge]

My answer was marked incorrect, but I'm not sure where I went wrong

[nimble ridge]

,rccw

[nimble ridge]

I believe the Riemann sum is supposed to be uneven and based off the values in the table in this case.

Will be AFK until updated.

[queen crater]

Holy handwriting

,calc 42724+63108+33697+74283

[glossy valve]

Result:

70616

[queen crater]

I don't see the problem with 9.a

[nimble ridge]

Hmm. Could just be my teacher, he always overlooks errors when making online quizzes on which I submit these answers.

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[gloomy rampart]

@nimble ridge is this composite functions

[nimble ridge]

Riemann sums

[gloomy rampart]

type of precalc?

i mean

algebra

[nimble ridge]

Calc AB, why?

[gloomy rampart]

i js wanted to know, it looked very similar to functions

i mean

it is a function

im stupid

nvm

dw

wait is it..?

it is

right

@nimble ridge

[nimble ridge]

what the hell are you doing here still

[gloomy rampart]

im sorry 😭 im tryna learn calc

[drowsy niche]

It has do to with functions, yeah

[drowsy niche]

Try not to be so rude, please, the guy simply asked you a question...

[nimble ridge]

Understandable but what was he trying to do?

[drowsy niche]

That, I do not know tbh

[gloomy rampart]

I just wanted to see what kind of math, and what level it was so I could know what I was going to be studying soon

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[nimble ridge]

Oh got it

[gloomy rampart]

sorryfor bothering you

[nimble ridge]

No you're fine partially my fault

[hoary current]

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[timber vapor]

i tried -3 and 2 and it marked it as wrong?

[devout valley]

Why -3?

(+ do you think there may be more?)

[timber vapor]

im looking at examples and some of them mark it correct when there’s a hole and like a filled dot on top of it

[timber vapor]

im not too sure 😅

[devout valley]

Like if it looked like this, but without the dot on top and the hole filled in, you mean?

[timber vapor]

yes

[devout valley]

Fair, but here, it's a bit different do you understand what limits basically are?

[timber vapor]

yes i do!

[devout valley]

Could you explain limits a little bit to me then?

[timber vapor]

so like for this example it’d be the values of y approaching 1 on both sides

[devout valley]

Yep, and from both sides, what do those y values approach?

[timber vapor]

wait sorry where do i look on the graph to find that out

[devout valley]

Basically, if you take your finger, and put it on the graph, and then move towards the point, doing that from each sides

[timber vapor]

so one of them would be 2 right

[devout valley]

Which one do you mean?

[timber vapor]

there’s a point at (2,1)

[devout valley]

Yep, that would be a valid answer for our question, and we're also lucky that there are no holes there

[timber vapor]

[devout valley]

I wanted us to also work through that point -3 though, to see if we were right or not

[timber vapor]

okay got it

would the point at (-3, 1) override the hole

[devout valley]

It wouldn't, no basically for limits, you don't really care what actually happens at the point you're going to, so for this example, we don't care that the function value at -3 is actually 1, we only care about what it looks like it's going to be when we approach from either side

If that makes sense?

[timber vapor]

ooh yeah okay

so -3 wouldn’t be an answer

and same with -5 and -7

[devout valley]

Are you sure?

[timber vapor]

the points don’t overlap the hole tho right

[devout valley]

They don't, but we don't care too much about what those points are, or if they don't overlap the hole or not

All we care about is if we try to approach the point from both sides, what does it look like we're going to get to (and if those happen to be the same thing!)

[timber vapor]

ooooh okay

so -7 would be an answer then

[devout valley]

It would

[timber vapor]

okok got it

[devout valley]

Any others too?

[timber vapor]

hmmm

6?

[devout valley]

Not 6, because we don't get close to the same thing if we go from left to right

[timber vapor]

ah i see

[devout valley]

I'll give you a hint, you said the other one already somewhere

[timber vapor]

-5?

would that only be if i approach it from the right side

[devout valley]

Yep that one and -5 works from both sides, if e.g. you moved from left to right, or right to left, you would be getting closer and closer to 1 as you get closer to -5

[timber vapor]

got it thank u sm

so far i have 2, -7 and -5

[devout valley]

Yep, do you think there are any more, or do you think those are all?

[timber vapor]

it can’t be -2 cuz both points there r holes right

[devout valley]

It isn't -2, no, but more because while moving from left to right we do get closer to 1, when we move from right to left, we don't get close to 1, but instead to -1

So the [overall] limit doesn't even exist there

[timber vapor]

got it

so those 3 should be it i think

[devout valley]

(it's a similar story for 6, moving right to left we approach 1, but moving left to right we actually approach 3 instead, so no limit there either )

[devout valley]

Yep I would agree

[timber vapor]

got it thank u so much!! have a good one

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[timber vapor]

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[onyx glen]

arithmetic bullshit, no doubt

[void jay]

for 1) i did: 0·2 + 1·4 + 2·4 + 3·x + 4·2 + 5·2 + 6·4 = total wins (=60) , so x =2 ..

is there a faster way ?

[onyx glen]

there's 20 teams total

2+4+4+2+2+4+x = 20

[void jay]

is there some kind of cue here? i was a bit skeptical to use the method you wrote because of the freq = 2 for 0 wins

i mean when would i count 2 and when wouldn't i count it if the other column shows a numerical value of 0

not sure if my question makes sense

[onyx glen]

those loser teams are still teams

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[blissful girder]

Hello! I'm curious, what does the bx term in the quadratic equation tells us about the graph?

If the leading term of the quadratic equation, ax, tells us about the shape of the parabola, while the coefficient tells us about the elevation of the graph in the y-axis... then what does bx tells us about the graph? What is the role of bx? This is my first question.

For the second question, please refer to the image attached. In this case, I am using x^2 - 2x + 1 as the equation. I deposited each term one by one on the Desmos app, and then I noticed that when I finally put -2x, the vertex shifts to -1. But the question is, why did it not shift to -2 if that's what I explicitly indicated? Why is it sitting there instead of -2? Does the roots of the graph influence the position of the vertex?

Thank you in advance!

[cursive pier]

ax not necessarily means what the shape is rather, the sign of a dictates if the parabola opens upwards or downward

in a very theoretical way, bx tells you the distance between two roots as well, larger the value the farther the two roots are

[nimble ridge]

If the whole parabola was to be shifted c units downward or -c units upward to provide more detail

[blissful girder]

Tell me more! I like this! They don't explain these stuff in school

They just tell you to apply the quadratic formulas to get the roots and then you move on with the next topic without telling you how did even those terms or equations come to life

[nimble ridge]

Factoring any quadratic expression w/ c=0 will reveal the two roots of the parabola and the x-coordinate of the vertex will be their average

[blissful girder]

Or what role do they even play lol

[blissful girder]

Sorry, but what does this mean? Are you referring to the c term?

Why are we even equating everything to 0 in the first place

Like we find the roots by equating it to 0, but why?

[nimble ridge]

Wish I could tell you more at the moment but I can't

[blissful girder]

Yeah, I'm sorry if I'm too demanding TvT

[nimble ridge]

They might explain stuff like this in greater detail in some college math classes when/if you get to that point but idk

[cursive pier]

hmm okay

• a -> gives you as you said the shape
• sign of a -> gives you direction of opening
• b -> the size of gap between two roots
• c -> as you said the y axis intercept

---

now you can just rearrange ax^2+bx+c = 0 into a(x-h)^2 = y - k

then you will see, (h,k) is just the position of the parabola head and r is just the shape.

---

and as for the needs, the simplest example of a parabola is maybe the glass inside a car's headlight and you need to understand where to put the bulb so that even with less power bulb you can get the most luminence and yada yada

[blissful girder]

Oh is that why the vertex sat on -1 because the average of the two roots are its x coordinate position

[blissful girder]

Where did they derive the h and k though

[nimble ridge]

You can expand that expression back out to get h and k from a,b,c

[cursive pier]

lol wait i shouldnt have skipped steps

ax^2 + bx + c = y
a(x^2-bx/a) + c = y
a(x^2 - 2*b/2a*x + b^2/4a^2) + c - b^2/4a = y
a(x-b/2a)^2 = y - c + b^2/4a

LET b/2a = h, b^2/4a -c = k
a(x-h)^2 = y - k

AND please put it in desmos to feel it move in real time instead of believing me by words

[blissful girder]

Thank you very much for taking the time to expand the equation... If I am not mistaken, is this factorization and perfect square?

[cursive pier]

it wasnt necessarily but i did algebraic manipulation to turn it into a perfect square

[blissful girder]

Thank you very much for demystifying these concepts. I really appreciate the effort

Thank you to @nimble ridge too!

[blissful girder]

Bold of you to assume that I know how to operate Desmos

[cursive pier]

Oh wow lol wait

[scarlet pike]

https://www.desmos.com/calculator/zu16y1xicz

just slide these sliders to change the values of a,h,k and see how it changes the graph

[blissful girder]

I thought you were supposed to put all the equations mentioned 👽

Anyways, yes, this is very illuminating

[cursive pier]

https://www.desmos.com/calculator/4stet8dfus

[stiff summit]

I have some intuition for what b does here #help-8 message

I’ll just copy it here

[stiff summit]

if you think about a parabola defined by the equation ax^2 + bx + c, and then change b as if on a slider, then you

might have noticed that the parabola will go through the same point on the y axis for any value of b

(if a and c are fixed)

this is because if you substitute x = 0, then y = ax^2 + bx + c = c, so the place where the parabola crosses the y-axis doesn't depend on b

next, since a doesn't change, the shape of the parabola (how wide it opens) doesn't change

so it's like you're sliding a rigid (you can imagine it as being glass), infinitely large, parabola up and down through that point on the y-axis, and keeping its direction the same

[sonic stratus]

consider that ax^2 + bx + c is also a construction based on
ax^2, bx and c
ax^2 is the quadratic term, and it essentially defines the general shape at big scale.
c is the constant term, and it defines the elevation at x = 0
i wont bore you with the why of it but;
bx is the lineal term, and it defines the inclination of the curve at x = 0.

between the three you can define any parabola.

parabolas of the form
ax^2 + c have inclination 0 at x = 0.
So they are all centered at the y-axis.

The inclination can move around the vertex.

[blissful girder]

Sorry if I'm replying too long, I'm readine ach reply very carefully

Wait I replied to the wrong person lol

[stiff summit]

on why you find the roots by setting y = 0: this is because y = 0 is the equation of the x-axis. So when you require both y = 0 and y = ax^2 + bx + c, in other words setting 0 = ax^2 + bx + c, you find the position of the intersection of the x-axis and the parabola, which is exactly where the two roots are

[cursive pier]

also do not take so much stress, the more time passes and you come by new ideas and math these will automatically get more cleared, none of us cleared it overnight

[stiff summit]

feel free to ask if you need to clarify

[full forum]

@blissful girder Has your question been resolved?

[blissful girder]

I feel like my intuition is telling me to go higher... so I can stand on the summit and view everything else from the vantage point in order to get the big picture. There's a brick missing, but I can't pinpoint what. I am seeing something and nothing at the same time.

I think most of my inquiries are stemmed from a broader epistemic question as to why algebra is the way it is. Why is mathematics the way it is? Like, what's the point of doing all these things? (not alluding to something existential by the way) Why are we doing all these steps? Why are we factoring? Why are we equating to zero? Why do we have to add and subtract the two sides of the equation at the same time? Why are we doing this and that?

There is just something so utterly symmetrical about mathematics that I can't simply pinpoint as to where it is coming from...

And most importantly, what framework exactly is mathematics operating on? If philosophy uses reasoning, logic, dialectic, syllogisms... then what does mathematics utilizes?

I believe my epistemic inquiry has something to do with the mathematical tools besides abstraction. I'm only familiar with abstraction, I don't know yet about the others. I feel like my questions are eyeing on the study of mathematical structure. I'm not looking for formalisms, I'm looking for the lense as to how I can perceive mathematics.

It's really weird when you're standing at the negative space of things where both questions and answers simply don't exist yet, only a strange mix of intuition and vertigo. I'm sitting at the pre-formal edge of a framework that is quite elusive

This feels like knowing how to move the pieces in chess but you don't know what to do because you don't have the tools, the strategy, the philosophy, and the way of properly thinking about it yet

[stiff summit]

uhh, first, tldr, changing b is sliding the glass parabola through (0, c), since you were asking that in the first place, so just making sure you have that for now

[stiff summit]

the goal in chess is checkmate, right? and then you can learn how to capture pieces

and then you learn some basic opening moves, and then you have a general direction of how a game goes

[stiff summit]

maths uses logic, and can be based on set theory or type theory. You might want to look at the Natural Number Game for an example of building properties of the natural numbers up from axioms https://adam.math.hhu.de/

It is possible to build absolutely everything in maths up from undefined primitives and logic and axioms and definitions about various objects. It’s just that, beyond the undergraduate level, things get unwieldy, and you learn to fill in the details of the proof steps yourself, so you can make larger leaps and go farther

but when you’re learning about quadratic equations, the focus is more on calculation and manipulation, because those tend to be the things that get applied most easily and clearly

when you’re going shopping and have to calculate with discounts, you need to know how to calculate with percentages. If you’re calculating the path of a cannonball, or rates of reaction with two or more chemical species when they’re reaching equilibrium, or anything to do with parabolas, then it’s very useful to know how to translate a quadratic equation to and from general form, and vertex form, and find its roots, and so forth

[full forum]

Channel closed

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[stiff summit]

that’s one of the goals of middle/high school maths, which is to give you tools you may not see the use of yet, but which you might be able to apply to something unexpected in future. So if you didn’t learn this stuff, then you would not know what you don’t know

.reopen

[full forum]

✅ Original question: #help-28 message

[blissful girder]

Yeah, I think I should learn how to think about mathematics first, or how to think mathematically. This feels like I'm using a chopping board with the technique of a welder to cut vegetables, lol. I'm not the type of learner who will simply jump head-on into an abyss without knowing what happens on the other side, or someone who uses formulas blindly

I'm actually trying to teach myself calculus and I'm going back to algebra to treat some blind-spots. But right now, I had a realization. I realized that I don't want pre-requisites anymore, I want to know how to think about mathematics and how to think mathematically. I want to understand the rules of the game and not just follow it

[stiff summit]

Interesting

[blissful girder]

Why do we even have rules, how do these rules even come to exist . I want to understand the rules so I can twist and break it eventually

Those types of questions itch my brain

[stiff summit]

You might like to try out the Natural Number Game, or maybe Reintroduction to Proofs

Or you could have a look at abstract algebra, say through here https://judsonbooks.org/aata-files/aata-html/aata.html

[stiff summit]

I had a hard think about why you must do the same thing to both sides of an equation while I was working through the early chapters of this book

(I decided that the answer was because, if you do the same thing to two equal things, then you end up with two equal things, and this is some kind of fundamental property of equality, maybe an axiom, but I didn’t think about it further until right now. For now I’d say that it can be taken as an axiom)

[blissful girder]

Oh yeah, my intuition was sort of right about abstract algebra. I was circling around this territory and I felt like I just needed to study it

I just bought a book

Let me share the author

Wait

A Book of Abstract Algebra by Charles C. Pinter

Is that good?

[stiff summit]

Yes, some people recommend it

[blissful girder]

Thank you very much

[stiff summit]

I didn’t really read it because I had read another abstract algebra book by then

But it looks fine from what I can tell

[stiff summit]

this one is free and you could consult it if you don’t like any part of Pinter

it’s useful to have several books on one topic that you can reference

you might be able to find a pdf of Herstein’s book online

[stiff summit]

you can look in the pinned messages in #book-recommendations for more book recommendations, or have a look at #books-old or #books probably

[stiff summit]

Although I read somewhere that some people, upon first learning abstract algebra, think that it seems like a giant chess game, with no clear purpose except to provide a puzzle

[blissful girder]

Okay, I'm writing down your book recs

By the way, can I ask you a question?

[stiff summit]

If you feel like that, maybe you could try real analysis, because that eventually leads to a justification of calculus and rigorous proofs for all the differentiation things, and, say, the Fundamental Theorem of Calculus

[stiff summit]

Sure

[blissful girder]

So I sent this thing to GPT and I just want to confirm if his reply is adjacent to what a mathematician will typically say about this matter

Is it okay you confirm its veracity?

[stiff summit]

what did it say?

seems a bit strange

[blissful girder]

Sorry for the long string of images, lol

[split hatch]

category theory (the true summit)

[stiff summit]

Mathematical thinkers don’t compute —
they switch perspectives until the answer becomes unavoidable.

this is not true of several researchers. I’m not a researcher, but when I was thinking how to solve a certain problem which I didn’t know where to read about online (so I couldn’t really ask for help), I didn’t “switch perspectives until the answer became unavoidable”. I wrote down stuff I noticed until somehow I got some results and disproved my original guess. I’m guessing this is true of researchers, but with way more wrong turns and dead ends

[blissful girder]

GPT proposed group theory, is that adjacent to my trajectory?

[stiff summit]

group theory is a part of abstract algebra

it sounds like a crank

[blissful girder]

I must have missed that on the table of contents, lol

[blissful girder]

I thought that was the standard for every GPT model

I guess the way it talks says a lot about me

[brittle sun]

What the fuck

[stiff summit]

also, generally I mistrust LLMs, because anything they say might have a subtle error, because they’re language models, not truth models

[wild sleet]

Smiguel

[maiden vapor]

what the blud

[blissful girder]

Yeah, that's why I asked for you to confirm just in case

[rare dock]

what the blud

[blissful girder]

What's up with the influx of people so suddenly

[wild sleet]

his retribution, swift

[stiff summit]

oh yeah there’s a bot command for this

!nogpt

[brittle sun]

Everywhere I go, the wind follows

[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).

[wild sleet]

be the moon, and be the sun

[brittle sun]

Either way, lemme read through the channel and engage with you in good faith

[rare dock]

this channel was linked in a secret chat for green names

[blissful girder]

Anyways, thank you for keeping up with me! I really appreciate it! This community is so amazing. I'll take up your recommendations and I will definitely dive deep into Number Game, Abstract Algebra, and Proofs

[stiff summit]

no problem, good luck!

[stiff summit]

you can check the pins in #book-recommendations for more

[blissful girder]

[rare dock]

it’s an honor to have your channel shared there

[maiden vapor]

honour*

[rare dock]

pure you know i love you but you’re wrong on this one

[blissful girder]

Thank you, lol. I was a bit worried for a second

.close

[full forum]

Channel closed

Closed by @blissful girder

Use .reopen if this was a mistake.

[full forum]

Available help channel!

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• Type the command .close to free the channel when you're done.
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[patent crow]

For 12b how would I be able to graph this?? Graph each transformation?

[slate violet]

yes, you need to know what kind of transformation f(x) -> f(kx) is

[rigid crescent]

graph log(x) and log(10nx) with some n

[patent crow]

Okay so I set an arbitrary amount for n?

[slate violet]

or for logarithms you can use log rules to simplify that to log(10n) + log(x)

[patent crow]

Like n = 1?

[slate violet]

if that helps you, sure

[patent crow]

Yeah I did \

$logx + log10 + logn$

[glossy valve]

Sep

[slate violet]

yeah, you should know what adding some constant to a function does to the graph

[patent crow]

Okay thanks guys! Appreciate it

[slate violet]

no worries!

[patent crow]

Alr I'll graph that then

.close

[full forum]

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Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[waxen raven]

take $D = {(x,y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1}$
\\
$$f: D \rightarrow R$$
$$\ (x,y) \mapsto x^2 + xy + y^2$$
\\
Find all extremum points of $f$.

[waxen raven]

could someone please help me out with this?

[glossy valve]

Moonful

[spiral vigil]

find all relative extrema on the interior and then on the boundary

[waxen raven]

sorry i couldnt do curly brackets

[waxen raven]

okay yes
\\
so (1) $E \cap D^{\circ}$
$$\nabla f(x,y) = (2x + y, 2y+x) = (0,0)$$
$$\implies (x,y) = (0,0) \ \text{ so } E \cap D^{\circ} \subseteq {(0,0)}$$

now how do i know if it's a maxima or minima

[spiral vigil]

\cap for intersection

[waxen raven]

ah

sorry how do i do curly brackets 😭 😭

[spiral vigil]

backslash i think? \{

[glossy valve]

Moonful

[waxen raven]

ah thank you!

ok so that

now how do i know if it's a maxima orminima

sorry E is the set of extremum points of f

[spiral vigil]

two options -- use the second derivative with the like matrix determinant thingo

or just pick a point very very near it and check xD

second derivative is much more reliable

[waxen raven]

i believe the hessian matrix yields a 0 eigenvalue

i can try 1 sc

sec

oh never mind!

it got positive eigenvalues so minima

the plan was for it to get a 0 eigenvalue so then i'd be like "how do i know if its max or min" 😭

ok lets say it did get a 0 eigenvalue, what would i do?

[spiral vigil]

it would be neither a max nor a min

think of f(x) = x^3

[waxen raven]

its a saddle point if one eigenvalue is positive and the other negative

but the test is inconclusive if an eigenvalue is 0

so what do i do when theres a 0 eigenvalue 😭 😭

[spiral vigil]

apparently that means it's flat

(or sloped)

[waxen raven]

sloped?

/flat?

[spiral vigil]

like a ramp

but if you're only looking at spots where f ' = 0 then it'll be flat

[waxen raven]

so like z = x^2?

[spiral vigil]

no... more like z = 0

[spiral vigil]

or z = x + y

oh

right two dim

yes

[waxen raven]

ah let me see

ohh

[spiral vigil]

actually yeah go ahead and check

for z = x^2 i'm curious

[waxen raven]

well z=x^2 when you set gradient to 0 you get x=0 and y E R

[spiral vigil]

no yeah it'll have a 0 eigenvector which makes sense since the surface is not curved

[waxen raven]

hmmmm i see

so does this not count as a max or min?

also now looking at boundary

(2) $E \cap \partial D$
$$\text{Define } \varphi : [0, 2\pi ) \rightarrow \mathbb{R}^2$$
$$ t \mapsto (\cos t, \sin t)$$

We know that on $\partial D$, $(x,y) = \varphi (t)$. We want $f(x,y)$.

Hence,
$$f(x,y) = f(\varphi (t)) = f(\cos t, \sin t) = 1 + \frac{1}{2} \sin 2t$$

Now define
$$g: [0, 2\pi ) \rightarrow \mathbb{R}$$
$$ t \mapsto 1 + \frac{1}{2} \sin 2t$$

what do i do from here?

[glossy valve]

Moonful

[waxen raven]

actually, i know that we have to solve g'(t) = 0

but im confused how we just transformed f (which was a 2d problem) into a one dimensional problem

yes i know we used polar corrdinates by letting x=cos(t) and y=sin(t)