help-39

[midnight haven]

Ax=b doesnt always imply the 0 vector is an answer?

[tropic saddle]

A*0 will in general not be equal to b

[midnight haven]

makes sense

[tropic saddle]

any nonzero b is a counter example

[midnight haven]

makes sense

lemme try to wrap my head around why theorem 7 is relevant here

gimme a minute

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

Question: Suppose A is an nxn matrix with the property that the equation Ax=b has at least one solution for each b in R^n. Without using Theorems 5 or 8, explain why each equation Ax=b has exactly one solution.

What we know: A is an nxn matrix. Ax=b has at least one solution. Here, we are told each Ax=b has exactly one solution. Why do we know this?

Theorem 7 states that "an nxn matrix A is nonsingular (invertible) if and ONLY if A is row equivalent to In, and in this case, any sequence of elementary row operations that reduces A to In also transforms A into A^-1."

What does this mean? A and the identity matrix must have the same amount of rows. When this applies, we can use row operations to reduce A into the identity matrix. In doing so, we transform A into its inverse (A^-1).

Why is this relevant to question 2? Given that the identity matrix is In, we know it will also be an nxn matrix. We also know that it contains a pivot point in every column. As such, we can infer that the transformation taking place will only result in a unique solution.

I edited

[tropic saddle]

you havent used that Ax=b always has at least one solution

what would happen during row reduction if A is not invertible

[midnight haven]

lemme think on that

ya i got nothing

[midnight haven]

but why is this relevant if im told that Ax=b has exactly one solution

[midnight haven]

a lot of things...

we would have a free variable or no solution

[tropic saddle]

you arent told that. you are supposed to show that

[tropic saddle]

what would decide which of those happens

[midnight haven]

a free variable is obtained when a column has no pivot point. no solution is obtained when a system of equations becomes impossible (0=/=1)

free variable can also exist if one of the solutions is 0

meaning AT LEAST an entire row will be 0

[tropic saddle]

ok good

so we know we always have at least one solution

so we can never have a 0=1 situation

what does that say about the result of the row reduction on A

[midnight haven]

that it will either have a unique solution or many solutions

[tropic saddle]

how does the result look like

[midnight haven]

it depends

[tropic saddle]

can we have this?

[midnight haven]

yes

if we cant then idk why

[tropic saddle]

so we know that for every possible b we never get a 0=1 situation

[midnight haven]

yes

we will always have at least 1 solution

we will never have no solutions

[tropic saddle]

so can the matrix that we get after applying row reduction to A have a zero row?

[midnight haven]

yes

[tropic saddle]

no

[midnight haven]

why would the elimination of no solutions also eliminate 0 as a possible answer

[tropic saddle]

some b would then lead to a 0=1 situation

[midnight haven]

i dont understand that

what if its 0=0

[tropic saddle]

so, lets say we have applied some row transformations to A and now it has a zero row

[midnight haven]

is b not able to contain 0?

[midnight haven]

ok

[tropic saddle]

follow my thoughts for a second

[midnight haven]

okie

[tropic saddle]

ok

so we have our linear system

(matrix | vector)

and the matrix has at least one zero row

[midnight haven]

ok

[tropic saddle]

lets imagine the vector was (0,...,0,1)

so in the last row we have a 0=1 situation

ok?

[midnight haven]

no im still confused

what if the vector was 1,0, 0...0

[tropic saddle]

well humor me

[midnight haven]

i just dont understand why eliminating no solutions means u cant have 0 as a solution

[tropic saddle]

so we have our matrix that we got from applying row transformations to A. and we have the vector (0,...,0,1). what happens if we now apply all row transformations backwards

[midnight haven]

wdym by that

like undoing them?

[tropic saddle]

yes

[midnight haven]

IDK

ive never done that

[tropic saddle]

well we started with A and did row transformations to get that matrix

[midnight haven]

yes

[tropic saddle]

so if we undo the transformations what do we get

[midnight haven]

A

[tropic saddle]

good

and if we undo our transformations on the vector (0,...,0,1)?

we get something

[midnight haven]

b

[tropic saddle]

some vector

and this we now call b

so what can you now tell me about Ax=b for this b

[midnight haven]

b has at least 1 solution

[tropic saddle]

yes by the problem statement

but what happens when we solve Ax=b by row reduction?

[midnight haven]

we are left with pivot points

[tropic saddle]

not what I want to hear

[midnight haven]

am i overthinking this

[tropic saddle]

yes

we just talked about row transformations and so on

what do we get when we apply row reduction on (A | b)

[midnight haven]

we get the values of x

the x vector

[tropic saddle]

we started with (matrix | vector) and undid all row transformations and got (A | b)

so what do we get when we apply row reduction on (A | b)?

[midnight haven]

sorry had to take a step away

we get (matrix | vector)

[tropic saddle]

and what do we know about that?

[midnight haven]

it cant have 0=1

[tropic saddle]

but it has

so we have a contradiction

to the assumption that the matrix we get by applying row reduction to A can have a zero row

[midnight haven]

so wait

when it says b has at least 1 solution

does that mean b1, b2, b3, ..., bn, or does it mean [0,0,0,1, etc] can exist?

[tropic saddle]

what

[midnight haven]

for each variable within b

do they all have an answer

im so confused bruh

gonna crash out

jk

but this is the last problem on my homework

😭

[tropic saddle]

well you should take a break

and then read through this conversation again

[midnight haven]

okay

[tropic saddle]

fwiw, I dont particularly like this argument cause its pretty convoluted

but I have no clue what else they might be looking for

[midnight haven]

ok back

[midnight haven]

that is a good acronym

i reread a bit

[tropic saddle]

(for what its worth)

[midnight haven]

yeah i looked it up

gonna add that one to my collection

OK so quick question

does b eliminate the possibility of a 0 vector being the answer?

[tropic saddle]

we are given that Ax=b always has at least one solution for every possible b

so our strategy in the proof will be to find a good b

that we can use somehow

we could set b=0 but that wouldnt give us anything

[midnight haven]

so then its in our best interest for b to not = 0

[tropic saddle]

yes

[midnight haven]

so then that is why we cant have a 0 row

because a 0 row becomes impossible if b is anything other than 0

[tropic saddle]

well it could still happen by accident

for some b's it will happen

we have to find some b so that it doesnt

which we did by undoing all row transformation starting with the vector (0,...,0,1)

cause then by construction we know that if we apply our row transformations again, we definitely get a 0=1 situation

because we literally made up the example that way

[midnight haven]

so then are we assuming this b will not be = 0 and doesnt accidentally have a 0 row?

[tropic saddle]

not assuming

we literally constructed it that way

we started with the end result of 0=1

and then undid our steps

to go back to the start

so that when we now start again we already know what we are going to end up with

[midnight haven]

so we are back at square one

but with helpful information

We know Ax=b has to have at least 1 solution. We know it can't have no solution.

You asked at which point can we see there is at least 1 solution for Ax=b.

I think that becomes a given once we know there can't be no solution.

We know the difference between obtaining a free variable and a unique solution is pivot columns and having more columns than rows

Since we know A is nxn, columns and rows are aligned. Since we know we won't have a 0 row, we can say that every column will have a pivot position.

It becomes impossible for a free variable to be obtained, which is why we know each Ax=b will only have a unique solution

[tropic saddle]

yes

[midnight haven]

im so goated

So that's my answer then?

Am I missing anything?

[tropic saddle]

yes

[midnight haven]

Swag

Thanks Denascite. I appreciate your patience.

[tropic saddle]

yw

[midnight haven]

.close

[pearl pond]

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[pulsar escarp]

how do I apply the Direct Comparison Test to these problems? I was looking at the other examples in my textbook, and they seem to suggest that you should remove everything but the "most dominant term" in the denominator to create the second series.

So, for A, should I remove +2, and leave my second series with (2sin^2(n^2+2))/n^2?

[summer imp]

Indeed. The idea is that $\frac{1}{n^2+2}<\frac{1}{n^2}$ for any $n$, so you may compare the terms.

[jolly parrot]

Azyrashacorki

[summer imp]

The other simplifying step would be to consider how big that sin^2 can be

Then you’ll be able to compare directly to a known convergent series

[pulsar escarp]

oh, I need to simplify it further for the second series?

[summer imp]

Well the end goal is to compare with a series you know converges, like the series with terms 1/n^2

[pulsar escarp]

This is what I have so far

Yeah, either the p series test or the geometric series test, right?

[summer imp]

Yeah

[summer imp]

Here to compare to a p-series, you need to get an inequality that gets rid of that sin^2 factor

How big can that be?

[pulsar escarp]

I don't know what that really means of how big it can be. I guess it goes around a circle and becomes pi or something.

[summer imp]

No but like the actual number sin^2(n^2+2)

Can you bound it for all n?

How large can sin^2 of anything get?

[pulsar escarp]

I don't know, I know sin(n) diverges but not I don't know if sin^2 does

[summer imp]

What’s the range of, say, sin(x)?

[pulsar escarp]

negative infinity and infinity

[summer imp]

Really? What number could you plug into sin to get, say, 35?

[pulsar escarp]

I'm trying not to say I don't know, but I really don't know

[summer imp]

I guess what I’m trying to get you to say is that sin(x) always spits out a number between -1 and 1

So then what about sin^2(x)

[pulsar escarp]

-2 and 2?

[summer imp]

Not quite. We’re squaring sin(x), which lies between -1 and 1

The result will be >= 0, and how big can it get?

[pulsar escarp]

I don't know!

[summer imp]

It has to be <= 1

Any number in [-1,1] you square ends up in [0,1]

[pulsar escarp]

Well, I never knew that. So what does that do for the original problem?

[pulsar escarp]

And was this still the right call for the second series?

[summer imp]

So now, knowing that sin^2(anything)<=1, what can you say about sin^2(n^2+2)?

[pulsar escarp]

It's all just going to become 1?

[summer imp]

It works, you just need another step to compare to 1/n^2

[summer imp]

Yes, you may bound it above by 1

What are you left with?

[pulsar escarp]

2*1/n^2?

[summer imp]

Yep

Should look familiar now, since that’s just twice a p-series

[pulsar escarp]

converges?

So this is a good answer?

Sorry, phone auto aligned

[summer imp]

It’s ok. I would further add that the sum ends up less than the series 2/n^2 explicitly after your first inequality

But that’s the idea

[pulsar escarp]

I don't understand what you said

sounds important

Like, put a symbol of 2/n^2 being greater than 1/n^2?

Anyway, moving on to problem B. I guess it's just the same thing?

[pearl pond]

@pulsar escarp Has your question been resolved?

[pulsar escarp]

@summer imp this is what I have so far for part b

I guess 2 comes out, it becomes n/n^2, and diverges for it?

[pulsar escarp]

Can I ping helpers if someone has already attempted to help me?

Is this an acceptable answer?

[pulsar escarp]

Whatever, I looked through the textbook, this probably works.

and I need to move on.

.closed

.closed

.close

[pearl pond]

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[honest spear]

for $x(1+x)y''+2y'+3xy=0$ I need to say if x=0 is a ordinary or singular point and whether it is regular or irregular.

I have that $A=\frac{2}{x(1+x)}$ and that $B=\frac{3}{1+x}$

i can see that B is convergent cuz its just a geometric series so around 0 it converges to 1 but A should blow up to infinity no? and so the function A is not analytic at x=0 and x=0 must be an irregular singular point.

But my memo says its regular :/

[jolly parrot]

Nyxzore

[honest spear]

[rough forge]

ops didnt see

[honest spear]

so am i tweaking or ...

[rough forge]

Do we actually check it like this, I havent seen this before

[honest spear]

oh i mean how else?

[rough forge]

oh ok i looked it up

[honest spear]

an analytic function has a valid convergent power series representation at a point x=a

[rough forge]

frobenius method never seen that before

[honest spear]

thats what ive been struggling with for ODEs and u think i dont understand any ODE

[rough forge]

it seems to me you forgot the (x-x0) parts?

For A a factor x and B a factor x²

[honest spear]

fck

im actually silly

its just strange where those parts come from

supposely they are a "correction factor"

and so if the point is well behaved given the correction factor

frobenious' method will succeed

[rough forge]

might take a look at the proof

\color[HTML]{181A5E}
Seems this comes from some power series approach in the form of
[ y=(x-x_0)^\alpha\sum_{n\ge0}y_n(x-x_0)^n ] where they are trying to find the unknown coefficients.

[honest spear]

and the values of alpha

i know how to do it

just no idea why it works

[rough forge]

it is related to some theorem of Fuchs

[honest spear]

and then there's these strange cases that arise for the values of alpha

[honest spear]

[rough forge]

https://en.wikipedia.org/wiki/Frobenius_method

it's translated from German lmao english site sucks

[honest spear]

neither of these makes sense ngl

[rough forge]

yeah i think that will help
https://grokipedia.com/page/Frobenius_method

[honest spear]

if $x P(x)$ and $x^2 Q(x)$ are both analytic at $x = 0$. Equivalently, in the multiplied form above, $p(x)$ and $q(x)$ must possess Taylor series expansions around $x = 0$

but why

[jolly parrot]

Nyxzore

[honest spear]

saying that the singularility is mild enough is so vague

like for y' term multiplying by x ensured "mildness" and for y multilpying by x^2 "ensures mildness"

probably wouldnt be crazy to assume it follows this trend and the y'' is actually being multiplied by x^0

[rough forge]

ts reminds me kinda of complex analysis, encountered laurent series, maybe there is a complex background, but as said i am not familiar with that method anyway, might ask your prof

[honest spear]

probably best

tysm!

.close

[pearl pond]

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[rough forge]

i think there is a mathematical reason why but i cant figure it now

@honest spear pretty sure this relates to some theorem of Fuchs

it also related order of poles to solutions of that form of differential equation

https://www.mat.univie.ac.at/~gerald/ftp/book-ode/ode.pdf

found something fr this time 😄

section 4.2

[honest spear]

TS looks complicated asf

[pearl pond]

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[manic goblet]

chat how do i do this question

i tried doing z = qudratic formula

[thorn shale]

where did you get stuck on

[manic goblet]

solving the formula

wait no

i need help with i

[humble root]

What's the definition of |z|?

[manic goblet]

the distance on the argand diagram

sqrt(a^2 + b^2)

[humble root]

Uhuh, which is ..? (Algebraically)?

[humble root]

Nice.

In particular, where z = a + bi.

So, using this fact,

How might you rewrite the given equation?

[manic goblet]

USING RE^ITHETA

[humble root]

Mmmm nope :)

[manic goblet]

oh

[humble root]

After all, you did already tell me this.

Perhaps it might be helpful to use this definition.

[manic goblet]

okie dookie

idk u just sub the values in

[humble root]

Wait

The equation in the question is,
|z| = |z - 4 - 3i|
Correct?

[manic goblet]

yes

[humble root]

And we also know that if z = a + bi, then |z| = sqrt(a^2 + b^2), right?

[brisk steeple]

of course

[manic goblet]

yes

[brisk steeple]

z - 4 - 3i = a + bi - 4 - 3i = (a-4) + i(b-3)

[humble root]

Nice. So, if we use this exact definition to rewrite our equation, what would we get?

Like
... =
What would go in the ...?

[manic goblet]

ohhhhhhh

abs(z - 4 - 3i) = sqrt(a^2 + b^2)

is this right bro

[brisk steeple]

no

[humble root]

Good, but how would you rewrite |z - 4 - 3i|?

[manic goblet]

with a plus minus

[humble root]

Hint: Use what Meolve mentioned earlier

[brisk steeple]

The complex number z - 4 - 3i is equal to (a-4) + i(b-3), and so |z-4-3i| = root((a-4)^2 + (b-3)^2)

[manic goblet]

oh i see what you need to do

but why put it in the form (a-4) + i(b-3)

[manic goblet]

oh i see

[humble root]

As you said, the function |z| is asking for the distance of the complex number z, right?

[manic goblet]

ya

[humble root]

That means you actually need to find the complex number in the require format

Like we need to put it in component form:
A + Bi

[manic goblet]

yeah i get it

[humble root]

Because these are the components used for calculating distance - the real and imaginary part.

Nice

[manic goblet]

ok thank you guys @humble root @brisk steeple i think i can do the rest

.close

[pearl pond]

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[pearl pond]

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[lapis dove]

How does matrix multiplication work? I have a test in 2 days so I'm not bothering to read any of the content im just jumping in and doing it as I go along

[sterile python]

If AxB then column A =row B

Or, if A is mxq and B =qxn then AB =mxn

[lapis dove]

uh

what

[thorn shale]

gl for your test ❤️‍🩹

[lapis dove]

I'll NEED it 🙏🙏🙏 Ty 😭

[sterile python]

Size

[ivory basin]

That's a bad idea

Read the content

[lapis dove]

hhhhhh

You're prolly right

[ivory basin]

Nothing anyone says here will be meaningfully different from what's in the content

[lapis dove]

This is the explanation it gave

but how does columns equate to rows

[ivory basin]

What's the it here

[lapis dove]

it ?

Igg

Ohh*

Mathsspace

I don't get it at all

[ivory basin]

Do you not have a book or notes

[lapis dove]

I was away every lesson they taught this

[ivory basin]

Book?

[sweet junco]

condolences

[ivory basin]

https://youtu.be/2spTnAiQg4M

[lapis dove]

My book is at school

[ivory basin]

This is a good youtube channel

[ivory basin]

Sister you can't do this

[lapis dove]

I usually just use like an online whiteboard or something

[ivory basin]

Well go through the video

[lapis dove]

I can't think rn im just gonna wing it and go to sleep LOL (absolutely failing)

.close

[pearl pond]

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[ivory basin]

Good luck lol

It's okay, take a look when you're awake

[pearl pond]

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[desert cliff]

why did he added the numerator

[desert cliff]

that doesnt follow the projection formula

[smoky gull]

where did he add?

[desert cliff]

w_1

3(2) + 5(4)

[calm wing]

that's the scalar product

[smoky gull]

Thats how you find dot product right?

[calm wing]

$u\cdot v = 3(2) + 5(4)$

[jolly parrot]

artemetra

[desert cliff]

ohhh righght

[calm wing]

he underlines it here too with blue and green

[desert cliff]

this works with 3d vectors too right

[calm wing]

yes

it works in any number of dimensions

[desert cliff]

yeah for some reason its wrong

under the blue line

[smoky gull]

Wait you didnt just add AB and AC directly did you?

no you didnt ig

Well you dot product isnt correct

Try again

[desert cliff]

isnt it -2 * 0 + -3 * -4 + 4 * -4

[smoky gull]

Yeah

And that is?

[desert cliff]

0 + 12 -16 = -4

[smoky gull]

Yea

And you wrote -6

[desert cliff]

oops

now we need to find the length of AB

which i already found

[smoky gull]

yeah so try again with -4

[desert cliff]

I got it wrong again

[empty goblet]

What's that

[desert cliff]

Projection

[empty goblet]

What was the question

[smoky gull]

its pinned

[empty goblet]

Oh I see it

[smoky gull]

wait you applied it worng

The denominator wont be |AB|^2

It would be |AC| ^2 check again

[desert cliff]

But I calculated length of AB heree using AB dimensions

[smoky gull]

No no thats not right!

Apply the formula correctly!

[pearl pond]

@desert cliff Has your question been resolved?

[desert cliff]

Yeah I got it correct now, I was solving for part b

But it’s wrong

,rcw

,rcw

[pearl pond]

@desert cliff Has your question been resolved?

[pearl pond]

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[amber hull]

v_\text{cart} = \text{speed of the axle}, \quad \omega = \frac{v_\text{cart}}{R}

[jolly parrot]

professor paradox

[pearl pond]

@amber hull Has your question been resolved?

[smoky gull]

whats the Question?

[pearl pond]

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[left kite]

anyone can help?

[honest spear]

Anyone can indeed help

Might anyone help is the question

!status

[pearl pond]

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

[left kite]

hey

[proper nova]

Agreed

[left kite]

im on 2

[honest spear]

Show where you got stuck

[proper nova]

!show

[pearl pond]

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

[honest spear]

[tiny cloak]

yooooooo wht the flip my brother's name is ashton

yoooooo w ashtonnnn

[left kite]

hahah yo bro

[tiny cloak]

yoooooooo

im way younger then u i wanna learn more math

[left kite]

u got it bro

clearly ur interested being in this server 🔥

[tiny cloak]

how old r u

[left kite]

im 18 this year

[honest spear]

Why do people not learn the $\sqrt{u^2} \neq u$

[jolly parrot]

Nyxzore

[proper nova]

It's $\abs{u}$

[left kite]

wait how come?

[jolly parrot]

1 divided by 0 equals Infinity

[honest spear]

Plug in -3

[left kite]

ahh

[tiny cloak]

im 15

[honest spear]

Don't solve for u

Just do the differentiation

[tiny cloak]

no1 math in my class

😉

[left kite]

i dont really get the working out tbh

[left kite]

awesome dude

[tiny cloak]

im trying to do it the whole grade 9

i did it just need to be more consistent

[left kite]

normally u have the equation in terms of somethign but in this case its just completely all over the place

[honest spear]

Just take the derivative of both sides of x=u^2+1

[left kite]

did i do anything wrong

[honest spear]

Well you have it in terms of u

[left kite]

yeah thats the aim right when u do u sub

[honest spear]

x = F(u)

[proper nova]

Yooooo whatup

[proper nova]

!redir

[pearl pond]

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

[tiny cloak]

heyyyyyyyyyyyy 1 divided by...

nice pfp

thts from indonesia

[proper nova]

!redir heyyyy

[pearl pond]

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

[proper nova]

Please, this channel is for your brother solving something

[tiny cloak]

yeaaaaaaaa brother sorry

[left kite]

did i get the dx right?

[proper nova]

@honest spear im sorry to disturb but please continue

[honest spear]

Ty sis

[thorn shale]

<@&268886789983436800>

[steep saddle]

no u dont want it with x in it you want the integral in terms of u

if you sub in x = u^2 - 1 what do you get for dx

[left kite]

id get (u^2 -1)u^-1 ?

respect of du / 2sqrt(x+1)

since earlier i made it where dx becomes the terms of du

[left kite]

i substitute dx to that to the original equation

[steep saddle]

ok yes but you still have x after you sub dx

[left kite]

ahh do i put in the u again?

[steep saddle]

you want the integral in terms of u

[left kite]

since i know that u = sqrt(x+1) ?

it would be dx = du/2u

[honest spear]

[left kite]

lemme try it out

[tiny cloak]

smarties

[left kite]

appreciate it thanks

[pearl pond]

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

[left kite]

.close

[pearl pond]

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[pearl pond]

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[fair tide]

how to integrate trigonometric stuff? like f (x) = 2 sin (x)

[green aurora]

Remembering what the derivative of sin(x) comes from

[fair tide]

How to derivate trigonometric stuff? Like f (x) = 2 sin (x)?

[green aurora]

Yes

[green aurora]

.

[fair tide]

Yea how..

What is the derivative of sin (x)?

[honest spear]

By definition of the derivative

[fair tide]

And where it comes from?

[honest spear]

The limit definition

Or you could expand its Taylor series and take that derivative

[fair tide]

what is the limit definition

[plush bramble]

and read
https://tutorial.math.lamar.edu/Classes/CalcI/DiffTrigFcns.aspx

[fair tide]

omg wow I just truly understood what Taylor's series is

[pearl pond]

@fair tide Has your question been resolved?

[pearl pond]

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[pearl pond]

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[pearl pond]

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

[woven falcon]

1

[rough forge]

\color[HTML]{181A5E}
Maybe try to take the logarithm on both sides
[ L\coloneqq \lim_{n\to+\oo} \f{1}{n^{\f{3}{2}}}\Big(\prod_{k=1}^{n} (n+k)^{(n+k)}\Big)^{\f{1}{n^2}} ]

[pearl pond]

@woven falcon Has your question been resolved?

[woven falcon]

How does taking log work on limits??

[plush bramble]

limit log(stuff) = log(limit stuff)

[woven falcon]

if i do break it in logs

then convert the product into sum

im quite not sure how to proceed further with tht

[rough forge]

Show your work

[pearl pond]

@woven falcon Has your question been resolved?

[pearl pond]

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[sand kelp]

I require assistance with the following:

f(x,y,z) = (x/z)^y

$\frac{\partial f}{\partial x} 😒

[sand kelp]

$\frac{\partial f}{\partial x}$

[jolly parrot]

sans.

[snow moss]

You have to differentiale with respect to x and think of y and z as constants

[sand kelp]

Okay. I just need help because my teacher only taught us partial derivation with only 2 inputs in the function...

[summer imp]

It's the exact same thing.

[snow moss]

its the same for all number of inputs

[sand kelp]

Okay. Thanks.

[snow moss]

Couldn't agree more

[dapper kraken]

dudr what

[pearl pond]

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[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[dense jasper]

<@&268886789983436800>

[pearl pond]

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[empty junco]

are we able to evaluate directional limits like this because the only way to have a limit not exist on a continuous function be if the left and right approach negative infinity at x = a?

[empty junco]

also, can you use less precise numbers here? im assuming you're able to eyeball what number you can plug in based on how complex the function is

[rough forge]

because the only way to have a limit not exist on a continuous function be if the left and right approach negative infinity at x = a?
That's not the only way, and to answer your question, yes you can do that

[rough forge]

But I see you plugged in numbers that's wrong notation

You rather make a table of values then write it like this, cause as it reads it's plainly wrong, taking the limit of a constant is that constant again

[empty junco]

true, true, but not my work

took this screenshot from a brian mclogan youtube video

[empty junco]

what other way would there be?

[rough forge]

If two limits dont meet for example

[empty junco]

jump discontinuities would only occur on a piecewise function, and i would be able to recognize if i was given one in a problem

[empty junco]

wouldnt that only be because they both go different directions

[rough forge]

f(x)=|x|/x

[ashen ivy]

the limits could agree, but there's a hole; the function isn't defined at that point, or it's defined as a different value

[empty junco]

true

[empty junco]

ahh

so i would just need to recognize what kind of functions cause what discontinuities

and then solve accordingly

obviously if theres a hole, the directional limits agree

so solving analytically on a test for example, i could just put the same answer for both

[pearl pond]

@empty junco Has your question been resolved?

[pearl pond]

@empty junco Has your question been resolved?

[ashen ivy]

this is probably not the easiest/best way.
my method would be, look for points where f(x) may not exist (VAs, holes, denominator is 0, square root is negative, etc.). those points cannot be continuous
if you have a 'dubious point' (like the joining of two piecewise functions), examine the limits there
otherwise, the addition/subtraction/multiplication/composition of continuous functions will be continuous

this probably doesn't cover every single case, but you are looking for points where:

• left-sided limit doesn't exist
• right-sided limit doesn't exist
• left limit doesn't equal right limit
• f(x) is not defined
• f(x) is not equal to the limit approaching f(x)

[empty junco]

ok, ty

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[fresh solar]

<@&268886789983436800> multiple help channels are spammed with this

[pearl pond]

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[rough forge]

You should open a new channel

[sharp musk]

oh i didn't read occupied

mb sorry

[pearl pond]

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[trail linden]

i dont know if im doing it correctly hahaha

[pearl pond]

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[sharp smelt]

First an answer: The polynomial ring $\left(1,x_1\right)$ is not princpal

[jolly parrot]

Wai

[sharp smelt]

For my question : How is $R[x]$ principal.

[jolly parrot]

Wai

[sharp smelt]

the entire ring that is

.close

[pearl pond]

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[pseudo oxide]

hi wai
bye wai

[pearl pond]

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[wooden flare]

when doing substitution for integration, I let u = ..., and i sub it in and i have stuff i can cancel it with, and i end up with some x which is just a rearrangement of my u, can i sub that in after or does it all have to be in the one step?

[wooden flare]

$\int \frac{x^3}{\sqrt{x^2 + 1}} dx$

[jolly parrot]

KB

[wooden flare]

if i let u = x^2 + 1, i end up with:

$\int \frac{x^3}{\sqrt{u}} \frac{du}{2x}$

[jolly parrot]

KB

[wooden flare]

am i allowed to cancel the x's and then make x^2 = u-1?

[dapper kraken]

yes

[wooden flare]

ok cool ty, was just wondering if i could do that

.close

[pearl pond]

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[wooden flare]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[wooden flare]

is it possible to get a hint for this problem: $\int \frac{1-\sin(x)}{\cos(x)} dx$

[jolly parrot]

KB

[wooden flare]

oh wait nvm

ok this one idk what to do: $\int \sec^3 (4x) dx$

[jolly parrot]

KB

[wooden flare]

like, after u=4x, idk what to do with sec^3 (u)

[thorn shale]

try splitting it into two parts

[wooden flare]

ibp i assume is after that right?

one with secu, and sec^2 (u)

[thorn shale]

[wooden flare]

ty

how does this one work? Im trying /figuring out trig sub:$\int \sqrt{x^2 - 16} dx$

[jolly parrot]

KB

[wooden flare]

and i think this is right but i've ended up with: $\int 16\sec(\theta)\tan^2(\theta)d\theta$

[jolly parrot]

KB

[buoyant cypress]

seems good

[wooden flare]

but how do i integrate it...

[thorn shale]

what is tan^2

[wooden flare]

1-sec^2 theta right?

oh

wait is it

[thorn shale]

other way around

[wooden flare]

sec^2 theta -1

[thorn shale]

[wooden flare]

oh no, not sec^3 again

and uh, once i integrate it... how do i recover my x

bc i end up with (...) with thetas

[thorn shale]

!occupied

[pearl pond]

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).

[wooden flare]

^ ^

so i end up with something like:

$16\left[\frac{1}{2}\sec\left(\theta\right)\tan\left(\theta\right)+\frac{1}{2}\ln\left|\sec\left(\theta\right)+\tan\left(\theta\right)\right|-\ln\left|\tan\left(\theta\right)+\sec\left(\theta\right)\right|\right]+c$

[jolly parrot]

KB

[wooden flare]

i think thats correct idk

but what do i do with the thetas, bc i need my x back don't i? for indefinite

[thorn shale]

(you should combine the second and third term first)

usually when you have thetas like this you should draw a reference triangle

[wooden flare]

yea i did that at the start, but idk how to "recover" x

[wooden flare]

it should be -1/2 ln(...) shouldn't it

[thorn shale]

yeah

[thorn shale]

since you had x=4sec(θ), how does the triangle look like

[wooden flare]

sqrt(x^2-4^2) as opposite, 4 as adj and x as hypot

[thorn shale]

tan(θ) should be trivial to find then

[wooden flare]

ah, 1/4 (sqrt(x^2 + 4^2)

and sec(theta) is just x/4 ?

[thorn shale]

yep

[thorn shale]

-*

[wooden flare]

this is so much work for one integral 😭

[thorn shale]

it is

and the end result is pretty disgusting

[wooden flare]

yea...

[wooden flare]

-- ?

where did the - come from?

[thorn shale]

i mean

you wrote a + there

[wooden flare]

oh

mb, didn't notice

so we get...

$16\left[\frac{1}{2}\left(\frac{x}{4}\cdot\frac{1}{4}\sqrt{x^{2}-16}\right)-\frac{1}{2}\ln\left|\frac{x}{4}\ +\frac{1}{4}\sqrt{x^{2}-16}\right|\right]$

and just do a bit of simplifying

this is horrible

[thorn shale]

should be a +

[jolly parrot]

KB

[thorn shale]

but more or less that should be the final answer

😭😭😭

[wooden flare]

no wonder they say bounded integrals are easier

bc this is just a semicircle

tysm!

[thorn shale]

yw !!

[pearl pond]

@wooden flare Has your question been resolved?

[pearl pond]

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[tardy path]

can someone tell me how the sign of the expression (a1)(a2) + (b1)(b2) tells us which equation gives us acute angke bisector and which equation gives us obtuse angle bisector and why do i need to first make the constants (c1 and c2) positive in the equation of the two straight lines ?

[cyan heron]

Hang on the derivation should be in my notes

As of why C1 and C2 should be positive, its because we use the concept of bisector containing origin for this

[tardy path]

can u share it here?

[cyan heron]

Gimme a sef

[tardy path]

and what is the concept?

[cyan heron]

I have an idea but its Best to let someone else respond on this
Geometry isnt my strong suit

[tardy path]

ok np

[tardy path]

thanks for sharing

what form is this?

@cyan heron

[cyan heron]

Normal form of the line

[tardy path]

okay

[buoyant cypress]

Hi

I'm back

So based on what csp said,
Do you know about direction cosines?

[buoyant cypress]

.

[tardy path]

lemme check idk the english terms of things very well

[tardy path]

yeah i do know

[buoyant cypress]

ok

[tardy path]

why does the normal form look like this,
what i know is that the normal form of a straight line is
xcos(theta) + ysin(theta) = r

where r is the perpendicular distance from origin to the line and theta is the angle the perpendicular line segment r makes with the X axis in the positive direction

[cyan heron]

It is precisely that

Try adding the squares of coefficient of x and y

Theyll result to 1

Sin²x + cos²x = 1

This is how you convert the general form to normal form

[buoyant cypress]

so if you see
take a line l1 and l2
l1 : a1x +b1y +c1 = 0
l2 : a2x +b2y +c2 = 0

you can express them as a dot product of a vector normal to the line and the the vector (x,y)

denote the vector normal to l1 as n1 and l2 as n2
l1 : n1.(x,y) = (a1,b2).(x,y) = -c1
l2 : n2.(x/y) = (a2,b2).(x,y) = -c2

if you divide that by \sqrt(a^2+b^2) you get the unit vector of n

now to find the angle between the lines you take the dot product of their normal vectors
n1.n2 = |n1||n2|costheta
cos theta = n1.n2/|n1||n2|

from here you can bring the a1a2 +b1b2 expression

altho i donot understand why do they consider c1,c2>0 you can deduce w/o ig

[tardy path]

u mean the two normal vectors are like this right?

[buoyant cypress]

yes

[tardy path]

okay

[tardy path]

but then this means that this expression tells us or the sign of it tells us about the angle between the two normal vectors, how does it help us identify which bisector bisects which angle?

[buoyant cypress]

so because the magnitude is positive, you consider the numerator for theta, if n1.n2 is negative the angle is obtuse , if it +ve the angle is acute
now if angle is acute

for acute case

if you consider the equations given itsjust perpendicular distance of points , you need to consider the sign

simplify the equations a bit for positive and negative case,

a1x +b1y +c1 /\sqrt(a1^2+b1^2) = +- a2x +b2y +c2/ \sqrt(a2^2+b2^2)

=> x(a1 / \sqrt((a1)^2+(b1)^2 - a2 / \sqrt((a2)^2+(b2)^2)
+y (b1/ \sqrt((a1)^2+(b1)^2 - b2 / \sqrt((a2)^2+(b2)^2)

• c1 / \sqrt((a2)^2+(b2)^2) -- c2 / \sqrt((a2)^2+(b2)^2)

if you find the normal vector to this bisector

it is n1\cap - n2\cap
similarly for the other case it would n1\cap + n2\cap

now since the angle is acute the the normal vector for bisector should lie between the normals of two lines hence you go with n1\cap - n2\cap for acute angle obtuse for other

might better latex ts

[tardy path]

ya i should

[jolly parrot]

professor paradox

professor paradox

professor paradox

professor paradox

[pearl pond]

@tardy path Has your question been resolved?

[tardy path]

i think i am starting to grasp what's going on here . i prolly need to go through all the explanation a few more times and i'll figure it out

i really appreciate your time and effort to help me, it means a lot. thank you very much @buoyant cypress @cyan heron

.close

[pearl pond]

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[pearl pond]

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[lost zodiac]

I ended up solving this but every country had a different answer key lol also I was looking for a faster way to solve this. Any ideas?

[pearl pond]

@lost zodiac Has your question been resolved?

[clear shore]

I don't quite get what the question mean, like take any 2 sides then the sum of vertices on both side are equal?

[spring pine]

same

[lost zodiac]

It means the sum of every number on points on each hexagon in this giant hexagon are equal.

So take any small hexagon within this bigger shape and add the numbers on each of its 6 points and they would be the same.

So it gave me 2 numbers opposite to each other on this honeycomb shape and told me to figure out x given this property

[clear shore]

that's weird, you have solutions right? can I take a look at one of them

[lost zodiac]

It says solution is A

I have answer key not worked solutions though

[clear shore]

really

I really believe what they mean is every sum of two vertices on a side are the same

Cuz that would lead to the ans to be 1

[lost zodiac]

Yes that's what I meant

Unless I made you misunderstand

[clear shore]

what you said basically implies the sum of all vertices of all the hexagons are the same

[lost zodiac]

Oh

Seems I misread it

But then the answer A makes sense

Each side is 5

This question is too easy it feels wrong

[clear shore]

yeah lol

[lost zodiac]

Anyways did you do it manually?

I just traced the numbers up to point x

[clear shore]

maybe ask your prof or smth, I don't think sum of hexagon is solvable even

[clear shore]

Not sure, but I won't bother finding one tho, gl

[lost zodiac]

Finding what

I meant the answer

[clear shore]

oh wait

[lost zodiac]

lol

[clear shore]

I thought you meant do I know a way to do it manually

I just jump from 1 to x lol, idk if that count manually

like this

[lost zodiac]

Yes that's what I did as well I just meant if there is another way other than manually assigning vals.

[clear shore]

idk, gl finding one

[frozen lantern]

i mean

1. what class is this for

youre prolly supposed to just rawdog it.

[lost zodiac]

It's just a problem I found online but it was extremely easy when I solved it but I thought there might be other ways of solving.

[frozen lantern]

2. even if there's a thing somewhere in the annals of math its prolly not smth youre supposed to use?

and 3) well graph theory is huge theres prolly smth

[lost zodiac]

I'm sure there is

I just wanted to see if anyone had anything different but I guess it's too straightforward...?

Anyways thanks.

.close

[pearl pond]

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[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[plush bramble]

<@&268886789983436800>

[pearl pond]

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[plush bramble]

<@&268886789983436800>

[pearl pond]

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[buoyant cypress]

ts seems to be the fav of scammers

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[royal galleon]

Why can't we have a negative c?

[rough forge]

The domain of log is x>0

[honest spear]

[royal galleon]

But we haven't specified that yet. All we said is we can't have x equal to 0

[rough forge]

Still, c<0 only if x<0 cause else f is undefined

[plush bramble]

you need to show the definition from 6.4

[royal galleon]

Where it says off it should be of f

[royal galleon]

Why would it be undefined say we have c=3 and x= -2 we would have $f(-2)-f(3)=\int_3^{-2} \frac{dt}{t}$

[jolly parrot]

BigBen

[royal galleon]

What would be the issue with this?

[plush bramble]

the function 1/t is not continuous on [-2, 3]

[rough forge]

1/t is not integrable on that interval

[royal galleon]

Oh I see the 0 gets in the way. Thx

.solved

[pearl pond]

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[warm elbow]

\subsection{Problem 1.} \textbf{Prove that multiplication on $\mathbb{N}, \cdot_\mathbb{N} :\mathbb{N}^2 \to \mathbb{N}$ is commutative. [You may freely use
that $+\mathbb{N}$ is commutative and associative.]}
\\
\textit{Proof:} We want to show that for all $(n, m) \in \mathbb{N}$, $g_n(m) = g_m(n)$. Let $()_n$ represent $(\forall m \in \mathbb{N})g_n(m) = g_m(n)$. We will show $\forall n \in \mathbb{N}, ()n$ by induction on $n$.
\\
$(*)0$. Pick any $m \in \omega$. Then $g_m(0) = 0$ by definition of $g_m$. We want to claim that $(\forall m \in \omega, g_0(m) = 0)^{[1]}$\\
Proof of $[1]$. If $m = 0$, clearly by the definition of $g$, $g_0(0) = 0$. Assume $g_0(m) = 0$. We want to show that $g_0(S(m)) = 0$. By definition of $g_n$, $g_0(S(m)) = g_0(m) +\mathbb{N} 0$. By the inductive hypothesis, this is equivalent to $0 +\mathbb{N} 0 = f_0(0) = 0$. $\square$
\\
So for all $m \in \omega$, $g_0(m) = 0 = g_0(m)$, hence $()_0$.
\\
Assume $()n$. We want to show $(*){S(n)}$. Select $m \in \omega$ arbitrarily. We want to claim $g{S(n)}(m) = g_m(S(n))$. By definition of $g$, $g_m(S(n)) = g_m(n) +\mathbb{N} m = g_n(m) +\mathbb{N} m $ by inductive hypothesis.

[warm elbow]

I am a little stuck as to what to do next

[jolly parrot]

toast

[warm elbow]

Evidently i want to show that $g_{S(n)}(m) = g_n(m) + m$ since $(n+1)m = nm + m = g_n(m) + m$

[jolly parrot]

toast

[warm elbow]

but i feel like im missing a claim

or maybe i'm dumb

[jolly parrot]

toast

[warm elbow]

wait

hmm

Proof of $[2]$. If $m = 0$, then by the definition of $g$, $g_{S(n)}(0) = 0$. Assume $g_{S(n)}(m) = g_n(m) +\mathbb{N} m$. Now, look at $g{S(S(n))}(m) = $.

[jolly parrot]

toast

[warm elbow]

fuck i am dumb

Proof of $[2]$. If $m = 0$, then by the definition of $g$, $g_{S(n)}(0) = 0$. Assume $g_{S(n)}(m) = g_n(m) +\mathbb{N} m$. Observe that $g{S(n)}(S(m)) = g_{S(n)}(m) +\mathbb{N}S(n)$ by definition of $g$. By the inductive hypothesis $g{S(n)}(m) +\mathbb{N}S(n) = g_n(m) +\mathbb{N}m +\mathbb{N}S(n)$. By associativity of $+\mathbb{N}$, we get $g_n(m) +\mathbb{N}m +\mathbb{N}S(n) =g_n(m) +\mathbb{N} (m +\mathbb{N}S(n)) = g_n(m) +\mathbb{N}f_m(S(n)) =g_n(m) +\mathbb{N} S(f_m(n))$

[jolly parrot]

toast

[warm elbow]

okay i think i actually did it right this time

but i feel like this might be the wrong? direction?

OMG

I THINK I GOT IT 🥰

[muted orbit]

How do you type like this omg😭

[warm elbow]

hi

<@&286206848099549185> can someone review my proof :-)

[left steppe]

#latex-help

[warm elbow]

i probs need to clean up quite a bit

[muted orbit]

Oh what's the command

[left steppe]

,tex help

[jolly parrot]

Please use ,help tex for command help.

Mar 31, 2026, 8:34 PM

[left steppe]

or that

i forgot 😂

[warm elbow]

\textbf{Definition - $g_n$}: Define $g_n: \mathbb{N} \to \mathbb{N}$ \
$g_n(0) = 0$ \
$g_n(S(m)) = g_n(m) +_{\mathbb{N}} n$

Btw definitions

[jolly parrot]

toast

[muted orbit]

Ohh

[warm elbow]

Fix $n \in \mathbb{N}$. \\
Define $f_n: \omega \to \omega$,
$\begin{cases}
f_n(0) = n. \f_n(S(m)) = S(f_n(m))
\end{cases}$

[jolly parrot]

toast

[muted orbit]

,tex help

[jolly parrot]

Please use ,help tex for command help.

Mar 31, 2026, 8:38 PM

[muted orbit]

,help

[jolly parrot]

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

[warm elbow]

proof for reference again

[pearl pond]

@warm elbow Has your question been resolved?

[pearl pond]

@warm elbow Has your question been resolved?

[warm elbow]

also how would you compute $\operatorname{rank}(\cdot_\mathbb{N})$

[jolly parrot]

toast

[warm elbow]

testy

[pearl pond]

@warm elbow Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[warm elbow]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[calm kindle]

Is this one occupied?

[pearl pond]

@warm elbow Has your question been resolved?

[thorn shale]

<@&268886789983436800> hi ray

[dim linden]

hello

[pearl pond]

@warm elbow Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

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[pearl pond]

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[spare rivet]

im not sure where to even start to get this, is it some sort of am gm or cauchy schwarz, im very new to these type of inequalities

[spare rivet]

not sure what to manipulate here

[torn flicker]

interesting one let me try

[pearl pond]

@spare rivet Has your question been resolved?

[pearl pond]

@spare rivet Has your question been resolved?

[torn flicker]

guess what

I don't even know how to start 😂

[spare rivet]

literally me

[spare rivet]

definitely feels like amgm with the 25 but i just cant think of how to get it

not sure about what to do with a+b=c+d=2 also

[smoky gull]

The first condition given, are a b c d multiplied or the question just means that all are non-negative?

[spare rivet]

all non-negative

[smoky gull]

I see

Hold on i think i got it

[spare rivet]

opening again tmrw

i need sleep

.close

[pearl pond]

Channel closed

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[pearl pond]

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[modest tartan]

what does

xRy is and only if y=max{x,1}

mean

[modest tartan]

A={-2,-1,0,1,2,3}

we need to find nomber of elements in R

dont hv phone on me rn so cant take photo of question

[rough forge]

well for example if x=0 then max{0,1}=1 but if y is not 1 then xRy is false

so xRy if they satisfy y=max{x,1}

[modest tartan]

so we have elements

[rough forge]

So if you consider A²

you want to find all valid (x,y) pairs

[modest tartan]

like (-1,1) and also -1,-1)

[rough forge]

ye gng

[modest tartan]

for all the elements in A

ok last thing

[rough forge]

wait (-1,-1)?!

[modest tartan]

not there?

x=-1

[rough forge]

but then max(-1,1)=1

[modest tartan]

oh

im dumb

meh

you learn new things everyday

[rough forge]

whats the last thing

[modest tartan]

also

omg star

[rough forge]

i prefer discover

[modest tartan]

how u been

[rough forge]

i been good hbu star

[proper nova]

learn by understanding

not memorizing

[modest tartan]

it was to find number of elements needed to make R symetric

[modest tartan]

W

[modest tartan]

it my first time seeing such an incantation

my apoligies for not understanding

i though y would be either x or 1

not either but both

[rough forge]

recall the def of symmetry: aRb => bRa so what that means is if we have y=max(x,1) then that implies x=max(y,1)

[modest tartan]

alr got that part

just didnt understand what the max part meant right away

[rough forge]

take the max between some number x and 1

[modest tartan]

thank a lot m8

[rough forge]

i think symmetry fails if x < 1

[modest tartan]

yep

got it now

[rough forge]

try some example

[modest tartan]

aa

no its asking how many element we gotta add

to make R' ymetric

symetric

[rough forge]

A={-2,-1,0,1,2,3}

[modest tartan]

yeah

[rough forge]

only add?

[modest tartan]

yeah

[rough forge]

huh

[modest tartan]

we can only add elements

like

R={(1,2),(3,1)}

[rough forge]

i see

[modest tartan]

we have to add (2,1)

and (1,3)

[rough forge]

are you sure these work?

[modest tartan]

um not

that just an example

in actualy question the number of element u ned to add to make symetric is 3

[rough forge]

\bl[\max(x,1)=\begin{cases}x & x\ge1 \ 1 & x<1\end{cases}] so for $x\ge1$ you get $y=x$ already

[modest tartan]

yep

[rough forge]

yeah but it's wrong cause x=1 never maps to y=2

[modest tartan]

i just took that R as an example

wasnt related to question at all

[rough forge]

ok

[modest tartan]

should i type out the R accrd to question

[rough forge]

yes that'd be good to avoid misunderstanding

[modest tartan]

kk wait like 5 mins

[rough forge]

but i think if you write it like this it should be obv which pairs would make R symmetric

[modest tartan]

R={(-2,1),(-1,1),(0,1)(1,1)(2,2)(3,3)}

ya needa add {(1,-2)(1,-1)(0,1)}

to make symetric

[rough forge]

no

you need to remove them

😮

[modest tartan]

we cant remove :c

but ans still same

caus it asked the sum on the numbers

[rough forge]

i think what the task means is what to add into R from A² to make R sym

[modest tartan]

yeah

was that not what i said...

just started relations

like 5 days ago