help-10

[warm shale]

stabulo

[rigid pine]

LaTeX on phone. 😢

[grave halo]

yes thank you thats what i was thinking it makes so much sense thank so much

[rigid pine]

Good luck in your computations. 🙂

[grave halo]

hahaha im sorry to make you go through that thanks for all the help 🙂

[rigid pine]

Glad to help. 🙂

[grave halo]

.close

[obtuse pebble]

Channel closed

Closed by @grave halo

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

Im so confused

[wooden cipher]

Is this the graph the derivative of a function?

Can you show us the directions too otherwise we have no idea

[obtuse pebble]

@timid silo Has your question been resolved?

[wooden cipher]

Ok this is the graph of the derivative

So when the graph of a function has a critical point, what is the value of the derivative?

[timid silo]

0

[wooden cipher]

So when is the graph of the derivative 0?

[timid silo]

At those numbers

-1, 4.5

and -5

[wooden cipher]

The graph is the derivative

When does the function in the picture hit 0?

[timid silo]

Ah i see

didnt see the the '

im partially blind

.close

[obtuse pebble]

Channel closed

Closed by @quartz orchid

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sudden patrol]

i need little help

[limber umbra]

yeah

[sudden patrol]

here it is

[limber umbra]

okay

so

[sudden patrol]

i need to arrange it

[limber umbra]

1- (cos(x))^2 = ….

[sudden patrol]

but i dont know how

[limber umbra]

.

[sudden patrol]

what equals that

sin^2x

?

[limber umbra]

yes

cos(2x) = …

[sudden patrol]

sin2x

[limber umbra]

no

cos(x+x) =

[teal turret]

luka are u familiar with cosine double angle identities?

[hollow kite]

cos(2x)= 2cos²(x)-1 or 1-2sin²(x)

also 1-cos²(x)=sin²(x)

[sudden patrol]

yeah i see

2sin^2x

[limber umbra]

1 - sin^2(x)

[sudden patrol]

no

[limber umbra]

1 - cos(2x) = ?

@sudden patrol

[sudden patrol]

2sin^2x

[limber umbra]

yes

[sudden patrol]

i wrote it up

hahah

[limber umbra]

now you can simplify the sin

[limber umbra]

ohhh 😂😂

then a primitive of 1/2 ?

[sudden patrol]

yeah

bcs i have 2sin

on the bottom

so ill have 1/2 before intehral

integral

[limber umbra]

yes

[sudden patrol]

so ill

get

this

[limber umbra]

okay

[sudden patrol]

then

can i this

sin^2x/sin^2x

wrote like its 1

[limber umbra]

yeah

primitive of 1 is ?

[sudden patrol]

c

[limber umbra]

x

[sudden patrol]

x+c

i meant

[limber umbra]

yeah

[sudden patrol]

but i wrote that on photo math

[limber umbra]

easy

[sudden patrol]

and answer is different

see?

[limber umbra]

HAHAHAHAHAHAHAHA

[sudden patrol]

what

[limber umbra]

💀

[warm shale]

Stephen

[limber umbra]

no he wrote it right

but the program just went nuts i think

[sudden patrol]

yeah

okay

answer is 1/2x+c

thanks guys

[limber umbra]

you get 1/2( x+c)

[sudden patrol]

yeah

[limber umbra]

yes

[sudden patrol]

i did it on photo math

just to be shure

thanks

.close

[obtuse pebble]

Channel closed

Closed by @sudden patrol

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[balmy atlas]

Hi can you help me with this problem please ?

[sudden patrol]

let us see

[balmy atlas]

I have no idea of how to start

[timid silo]

are M_{n,m}(K) the matrices of size n-by-m, and rg is rank?

[obtuse pebble]

@balmy atlas Has your question been resolved?

[kind hawk]

if you have rank 1, what can you say about the relationship between the rows

what about the columns

in a product CL where C is a column and L is a row, what can you say there

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[long hornet]

Help

[long hornet]

Does anyone know where I can learn to solve this problem I have never seen it before and I don't see similar example on yt

[tardy epoch]

are you doing 4?

just solve it case by case

IF X >= 2

and then
IF X < 2

[long hornet]

Okay I have a few questions about that but what does
|X| do here?

[tardy epoch]

what answer are you looking for?

|x| is a function of x

[long hornet]

To add the two solution

[tardy epoch]

,w plot |x|

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[tardy epoch]

,w plot x

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[tardy epoch]

what is "add" here?

math add?

[long hornet]

So I find what the solution is
IF X >=
lets say its 2

And IF < 2

lets say its 1

And then I have to add them

So solution is 2+1 =3

[tardy epoch]

you do not add solutions

you're only looking for solutions to your equation

they do not add like that

[long hornet]

Well they told me to find

x for both cases and add them for some reason

[tardy epoch]

show the question

[long hornet]

Okay one sec let me translate it

But it just says
Find the
the sum of the solutions of the equations

Let me write the solution on paper

[tardy epoch]

then this is correct

[long hornet]

Okay let me see if |X| is needed here

And I to find a way to ask this question in a way that makes sense

[tardy epoch]

looks good

[long hornet]

Okay here , so I don't see |X| interacting with anything in this problem so I don't really see the point of its existence

So do I just say oh in this problem I found |X-2| and then go , okay I was given some rules for x-2 let me look at them .
And since there is no |X| in
|X-2|+X×2:3-4=0 I literally just ignore it since it has no reason to exist

[tardy epoch]

|x| is just compact notation

that's like saying $\pi$ doesn't have a point for existing because you can just say "number equal to the ratio of a circle's circumference to its diameter"

[warm shale]

riemann

[long hornet]

Okay so will there be a problem that is being changed by |X| practically,I will read more on it soon but for now do I have to ever look at |X| while solving problems like this one

[tardy epoch]

what do you mean "do i have to ever look at |X|" ?

are you asking if you have to read the equations you're being asked?

then the answer is yes

[long hornet]

Really I just feel if I ripped the |X| part from the paper and did this:
Find solution IF X >=

Find solution IF < 2

Add the solutions
My solution will still be the same

[tardy epoch]

will still be the same as what method?

[long hornet]

As the method where I did not tore the |X| out of the paper in this problem

[tardy epoch]

well yes they're mathematically equivalent because that's just the definition of |X|

[long hornet]

That is interesting Eve if I did something like this?

Or is it more something like pi where you cannot change it

[tardy epoch]

i have no idea what you're even doing here.

[tardy epoch]

no. you cannot change the definition of |x|

[long hornet]

Oh okay so it is literally saying
Pi = 3,14 at the start of every problem involving pi

[tardy epoch]

i don't know what you're saying anymore

just do a new problem

[long hornet]

Thanks I got it will read more on WHY X is like that but for now just knowing that it is like that is enough

[tardy epoch]

a good start
https://en.wikipedia.org/wiki/Absolute_value

[long hornet]

Yeah that looks like it will explain it thanks a lot! just have to take a break my head hurts a little I should sleep more than 4h

[obtuse pebble]

@long hornet Has your question been resolved?

Channel closed

Closed by @long hornet

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[rustic heath]

any idea on how to start this problem?

[rustic heath]

i tried to parametrize it but idt it's the right approach

anybody?

i just need a little hint and i think i can solve it on my own

[obtuse pebble]

@rustic heath Has your question been resolved?

[obtuse pebble]

@rustic heath Has your question been resolved?

[obtuse pebble]

@rustic heath Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

hey, for any people good with trigonometry, i had confusion in regards to two different answers
the two images have an answer that is 60 degrees (pi/3 rad) vs another answer that has 30 degrees pi/6 rad

[timid silo]

my first question in the first image had a regular triangle that i drew as its supposed to be in the answer

in the second question however, i made the same similar shaped triangle and assumed it would be an equilateral triangle (which was right, but the execution was wrong?) because the triangle was the other orientation and was facing upwards
why is it suddenly 60 / 2 = 30deg? it was just a bit jarring to me even after reading the answer explanation

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[bold nest]

Is there a nice limit for this sequence of functions? Can I use l'hopitals

[bold nest]

I have brain rot

b) for full q

i have dominating g I just am too ??? to get f_n -> f

nvm

i have brain rot

.close

[obtuse pebble]

Channel closed

Closed by @bold nest

Use .reopen if this was a mistake.

[bold nest]

IM SO DUMB

WTF

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[gleaming ridge]

I am having a small confusion about the notation Rudin uses. In chapter 2 he mentions that (a, b) is open in R but not in R2. First one is immediate but I don't understand the second one.

[gleaming ridge]

Is (a, b) like an open rectangle in R2?

He generalises closed intervals in higher dim spaces as k-cell. So, a 1-cell is an interval, 2-cell is a rectangle and so on...

Someone on math.stackexchange said that we have to identify (a, b) with (a, b) x {0} to view it as subset of R2

But Rudin never mentioned anything even close why is why I am confused

[civic zealot]

what is Rudin's definition of open?

open ball around every point?

[gleaming ridge]

That, p is an interior point of E if there exists some real number r>0 such that Nr(p) is contained in E. If every point is an interior point then the set is open.

[gleaming ridge]

Yeaah

[civic zealot]

ah. So Rudin is basically avoiding using some clunky notation.
The interval (a,b) is open in R because you can put a ball in it (other open intervals)
The interval (a,b) is meaningless in RxR unless you think of it as a line from a to b. or (a,b)x{0}

if you want it to be rectangle in RxR then you need a second interval for the 'height'
So (a,b) doesn't have a height, it's just a line.

[gleaming ridge]

I thought that a and b are 2d coordinates specifying the width and height but it seems I was wrong

Like a=(a1, a2) and so on..

[civic zealot]

Ah, nope. He's literally saying 'the one dimensional interval (a,b) in RxR"

[gleaming ridge]

Fkin Rudin

So, for higher dimensions I'd think of (a,b) as (a, b) x 0^k

[civic zealot]

basically, yeah

[gleaming ridge]

Alright, thanks ❤️

[civic zealot]

the intuition is that the line (a,b) in R is the same one dimensional line in R^n

[gleaming ridge]

@unreal musk thanks for the reacts, very helpful

[unreal musk]

Hey, Rudin is amazing, yea

[thick fog]

What edition are you using? I don't see this notation in my version

[gleaming ridge]

3rd edition

[thick fog]

Oh okay, I have a 2nd edition

[gleaming ridge]

I heard 2nd has worse notation lol

[gleaming ridge]

Plus for cheap SEA version 3rd edition was the only one I could get

[unreal musk]

Mind you, they do say "if we regard it as a subset of R^2" (and the way C and R^2 seem to be identified there!)

[thick fog]

Well... its from the 60s and does have some pain associated with reading any math book from that era

[gleaming ridge]

Hmm... I thought it would be something like generalisation of segment in higher dimensions

[wild swallow]

segment

[unreal musk]

Ah, like the k-cells but open instead right?

[wild swallow]

then it would be a rectangle

[gleaming ridge]

Yep

[gleaming ridge]

Yes I was thinking of open rectangles as it's more natural

[wild swallow]

[gleaming ridge]

I thought maybe there's some problem at the corners but now that I think about it open rectangles would be open

[wild swallow]

it would be a bad name otherwise

[gleaming ridge]

Lol yeah

Anyway, thanks for the input everyone

.close

[obtuse pebble]

Channel closed

Closed by @gleaming ridge

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[idle spade]

ok so

[idle spade]

shit it didnt send

one sec

im basically just unsure if i did what they asked me to

or if i'm missing something

or if i did it right in general

[fierce lagoon]

Seems justified to me

,w Limit[Sum[(6/n * 1/(6 k/n + 1)), {k, 1, n}], {n, Infinity}]

It broke

Seems fine to me

[obtuse pebble]

@idle spade Has your question been resolved?

[idle spade]

haha thank you

much appreciated

.close

[obtuse pebble]

Channel closed

Closed by @idle spade

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[north forum]

question b

[fierce lagoon]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[north forum]

3

i got 7.5

[fierce lagoon]

Show your work

[north forum]

it’s 7.8

[fierce lagoon]

_ _

[royal basin]

@north forum "show your work" does not mean "blurt out another answer"

[north forum]

yeah i know, i am trying to do it again because i did it on my calculator

and my working out went

i did

this

261.8/2

which is

130.9

130.9= 1/3 pi h r^2

(130.9 times 3)/pi= h times r^2

[royal basin]

where did 261.8 come from...?

[north forum]

the volume of the cone

full

[royal basin]

,calc pi/3 * 5^2 * 10

[warm shale]

Result:

261.79938779915

[royal basin]

okay, checks out.

[royal basin]

asterisk for multiplication

[north forum]

oh alright thank you

[royal basin]

what did you do from here?

[north forum]

i divided it by two

and then

tried solving for h

the ratio for h to r is

2:1

[royal basin]

do you have your work written out on paper?

[north forum]

yes but it’s messy

[royal basin]

it's hard to decipher what you are doing from this fragmented retelling

[fierce lagoon]

Rewrite the work then

I think I know what they're getting at

[north forum]

i tried doing the square root

but that wouldn’t work would it?

[royal basin]

this seems like it is not your full work

[north forum]

i’ll send it all to you

[fierce lagoon]

I'm assuming what they did was they got $V = \frac{250\pi}3$ for the full volume. And then since $\frac r h = \frac{1}{2} \implies r = 0.5 h$, $V = \frac{h (0.5h)^2 \pi}{3}$.

[warm shale]

Umbraleviathan

[fierce lagoon]

But then you need to solve for h when it's V/2 instead of V

[north forum]

hold on

[fierce lagoon]

You did hr instead of hr^2

And even then, you'd need to solve for r in terms of h

that's why r/h = 5/10 implies r = 0.5h

[north forum]

yeah

so r would equal 3.7

[fierce lagoon]

Since the volume of the full up is 250pi/3

You don't even need to solve for the radius

Just replace r with 0.5h

[fierce lagoon]

The half volume would be 250pi/6

[north forum]

what about the square root?

[fierce lagoon]

There should'nt even be a square root

If you replace r with 0.5h and the volume with half-full volume:

$$\frac{250\pi}6 = \frac{h(0.5h)^2 \pi}{3}$$

[warm shale]

Umbraleviathan

[obtuse pebble]

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.

[fierce lagoon]

Did you seriously nosols me lol

[north forum]

no

it was someone else

[fierce lagoon]

Nah EndTimes just popped in

[north forum]

i swear to god

[fierce lagoon]

Goofy moemnt

Anyways you understand why I replaced r with 0.5h right

[north forum]

yeah

[fierce lagoon]

So I don't understand where the square root comes from

[fierce lagoon]

I want you to simplify the numerator of the right hand side

[north forum]

would it be 0.25h^3 pi?

[fierce lagoon]

Yeah

So now it should be easy to solve for h

[north forum]

h would be 5.5

[fierce lagoon]

That's not what I got

I got around what your answer key says

[north forum]

what did you solve for?

you solved for h using what?

[fierce lagoon]

Algebra

[north forum]

yeah so you did

[fierce lagoon]

$\frac{250\pi}6 = \frac{0.25h^3\pi}{3}$

[warm shale]

Umbraleviathan

[north forum]

i didn’t get that

i did

what you did

but didn’t divide by 3

because i forgot the original question

[cinder bane]

<@&286206848099549185>

[north forum]

what?

[frigid prism]

@cinder banebroo

first this channel is occupied

[north forum]

i got 7.9

[frigid prism]

second dont ping helpers until you waited 15 min

[fierce lagoon]

man.

[fierce lagoon]

Yeah that's close enough

They probably did intermediate rounding

Or they curtailed the number

Yeah if you do intermediate rounding, it's 7.8 something

7.9 is more accurate

[north forum]

i thank you so much and i am so sorry for the hassle 😞 you’re so helpful i really appreciate it

[fierce lagoon]

Np

[north forum]

can i close this channel?

.close

[obtuse pebble]

Channel closed

Closed by @north forum

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

Where do I start?

[flat roost]

By solving the differential equation first ??

[fathom flicker]

do you know how to do seperable ODEs?

[timid silo]

no

[fathom flicker]

Did your class cover them?

[ruby path]

Group all the ys with the dy and all the xs with the dx, and integrate both sides without limits

That's the general method

[fathom flicker]

For ODEs that are seperable

[ruby path]

For seperable ODEs i.e.

[timid silo]

alright

[flat roost]

What knowledge do u have of differentential equations ?

[ruby path]

Don't forget to add the constant of integration btw

[fathom flicker]

and realize that you can group constants at the end

[timid silo]

ln|y| = -xcos(x) + sin(x) + C

[ruby path]

ln y?

$\int y \dd{y} \neq \ln |y|$

[warm shale]

NEONPerseus

[timid silo]

y^2/2

[ruby path]

,w int xsin x

[warm shale]

Results provided by WolframAlpha

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

Requested by NEONPerseus#2128

[ruby path]

yup with that correction you're good

now put in the values for x and y to solve for the constant

[timid silo]

C= 81/2

[ruby path]

,w solve dy/dx = (xsin x)/y if y(0) = -9

[warm shale]

Results provided by WolframAlpha

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

Requested by NEONPerseus#2128

[ruby path]

thanks wolfram

but yeah it make sense you got it right I think

[timid silo]

So the answer is just the value of C?

[ruby path]

Nope

that whole equation

$y^2 = 2\sin x -2x\cos x + 81$

[warm shale]

NEONPerseus

[timid silo]

this?

[ruby path]

this

theres no ln y

[timid silo]

oh yea

[ruby path]

Replace c with the constant calue

[jolly ginkgo]

Hi obchusegoose

[timid silo]

hello

[jolly ginkgo]

When are tou starting to help others

[timid silo]

how come c is 81?

[timid silo]

not in a while, i suck at math

[jolly ginkgo]

Us

[jolly ginkgo]

Use initial condition

y(0)=-9

[timid silo]

oh nvm we multiplied by 2

thanks for the help

[jolly ginkgo]

No problem

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @quiet folio

Use .reopen if this was a mistake.

[ruby path]

I just multiplied both sides with 2

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[proper narwhal]

does someone mind explaining how f(x) is = y. I can't really comprehend why

[royal basin]

yeah cause it isnt a Rule™️ or anything

it is simply common sometimes when talking about a function to write the letter x for its input and y for its output, but there is really nothing special about those letters from a formal pov

[proper narwhal]

Oh really?

I thought there was some theory behind it

And explanation on why because my teacher said something about it and couldn’t really understand

[obtuse pebble]

@proper narwhal Has your question been resolved?

[royal basin]

there is none

[frigid prism]

y is just a function of x

f(x) means a function of x

so y can be the same as f(x)

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[worldly trench]

Is there any value in using matrix algebra as opposed to normal elimination/substitution techniques in solving 2 variable linear equations? Or is it always much more efficient in using normal algebra for 2x2?

[kind hawk]

2x2 it doesn't really matter I would say

[worldly trench]

I know that with more variables, matrix algebra becomes more computationally fast. But what about just 2x2?

[kind hawk]

you are doing the same steps, just slightly less writing down

cause you dont have to write down the variables

[worldly trench]

If you have to repeat the same techniques with changing constants and coefficients, then what it become simpler with matrix?

or still the same?

still a 2x2

[kind hawk]

if you are just talking about row reducing, the matrix variant is really just a different way to write it down

imo a better way cause you only write down the coefficients and there is less chance to mess something up with the different variables

but in the end its just a different way to write it down

and of course with matrices you get all the theory from linear algebra

[worldly trench]

Oh ok, so it's the same efficiency in all 2x2 cases?

[kind hawk]

well you are doing the same steps

whether you subtract the first row in a matrix from the second or the first equation from the second doesnt make a difference

same result

latter is just more stuff to write down cause of the variables

[worldly trench]

ah ok, so it would only start to make a difference with more variables?

[kind hawk]

well more variables means more extra stuff to write down

already in the 2x2 case you have to write down more stuff

but really thats the only difference

[obtuse pebble]

@worldly trench Has your question been resolved?

[fossil stump]

I need to find the fraction

Decimal = 0.2777
Fraction = ???
Perfect = 27.77%

I need to find the fraction

Decimal = 0.2777
Fraction = ???
Percent = 27.77%

[timid silo]

@fossil stump #❓how-to-get-help

This is someone else's help channel, please grab an available one

[obtuse pebble]

Channel closed

Closed by @worldly trench

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[half silo]

When does the equality hold true for integral form of triangle inequality? I'm dealing with real valued functions

[half silo]

:

[wild swallow]

do you know when equality holds for the usual triangle inequality?

[half silo]

yes

[wild swallow]

which is when?

[half silo]

when all the 3 points lie on the same line

[wild swallow]

3 and -2 certainly lie on the same real line

[half silo]

im dealing with distances no?

[wild swallow]

-2 is not a distance

the usual triangle inequality for real numbers goes like
[
\abs {a + b} \le \abs a + \abs b
]

[half silo]

oh, i just relate it with the usual "triangle

And consider |a+b| as one distance

and |a| , |b| as the other two

sides

[wild swallow]

sure

although for real numbers your triangle will always be degenerate

[half silo]

yes so for this

does the equlity hold when |a| |b| = ab?

[wild swallow]

so when a and b are the same sign you mean

[half silo]

alright, so do we have the same continuation for the integral as well?

[wild swallow]

basically yes

you want f to have the same sign almost everywhere

[half silo]

int f(x) . int(g(x)) should have the same sign, both going from a to b?

do i consider int f(x) * int(g(x)) >=0

[wild swallow]

no the integral itself is already a kind of sum

so you want f(t) to be the same sign for almost every t

[half silo]

we dont consider the limits?

[frosty spoke]

almost every t within the limits lol

the limits are just a fancy way of saying you're integrating over the whole real line the function 1_[a, b] times your function

or is it the other way around? hmm idk but whatever lol

[wild swallow]

ye

[frosty spoke]

yeah okay nevermind that thing doesn't work so nicely for Riemann integrals

[wild swallow]

when i said almost every t obviously i meant the ones that mattered

[frosty spoke]

but yes almost every t within the limits

[half silo]

wait can i get help over this example so i understand it fully of how i should use this equality?

[frosty spoke]

isn't that jensen's inequality or something

[half silo]

consider this:

$$ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} { |(\sin{x})-a|} \geq |\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}{ (\sin{x})-a}|$$

[warm shale]

saltyballs666

[half silo]

equality holds whenever $$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} {(\sin{x})-a}>=0$$

[warm shale]

saltyballs666

[half silo]

or is it

(sin(x) -a) >=0

for x in ( pi/4 to 5pi/4)

[frosty spoke]

well certainly the function being nonnegative a.e. is sufficient to ensure they're equal

[half silo]

ok thank you i have got it!!

[frosty spoke]

but hmm I'm trying to come up with an argument for why it's necessary but I'm sleepy

[half silo]

i think the normal discrete analog makes sense

like i can't really derive triangle inequality for integrals, i just know it works since its the continuous analog for discrete stuff, we arent taught Riemann sums yet, we are just told to know that since it works for discrete, it will work for continious too lol

[frosty spoke]

oh yeah it's because the integrals are equal on the positive parts

[half silo]

signed area?

[frosty spoke]

so if it's negative for some non-negligible set

then you're never making up that difference

[half silo]

makes sense

[obtuse pebble]

@half silo Has your question been resolved?

Channel closed

Closed by @half silo

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[rough parrot]

i used mkdir to create new directory y9 and then i want to create a new file called myfile inside y9 how to do it

[frosty spoke]

this is a math server lmfao

[obtuse pebble]

@rough parrot Has your question been resolved?

[opaque galleon]

bruhh

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[glossy yew]

help

[sour jacinth]

whats the question

[glossy yew]

wait

i will show you

[royal basin]

image on its way?

[glossy yew]

yes

this i need help with

exponential and logrithmic moddling

[royal basin]

is the 14 answer to part (a) written in by you?

[glossy yew]

is wriiten by me

because

thats the start

do you need a clearer photo

clear picture

me waiting for help

[royal basin]

sorry, was called away for a moment.

At the time of death, a particular plant has 100 units of carbon-14 in its tissues.

the "14" in carbon-14 is, for the purposes of this problem, simply a name. it is not a piece of numerical data for you to work with.

[glossy yew]

oh okwy

[royal basin]

have you started on part b or are you paralyzed by it

[glossy yew]

paralyzed by it for sure

[royal basin]

ok do you know what a halflife is

[glossy yew]

geussing 50% of the plants life

[royal basin]

incorrect

and you do not need to guess

if you don't know then say you don't know and i'll point out where in the text it says what the word "halflife" means

[glossy yew]

okey

whats halflife

[royal basin]

the time taken for half of the original amount to decay

or to put it another way, the halflife of a radioactive substance is the time during which said substance shrinks to half of its amount

[glossy yew]

okey i understand

[royal basin]

can i ask you now to write something down in symbols

[glossy yew]

yes i can

[royal basin]

\textit{read these instructions to the very end before proceeding.}

referring to the formula $$Q(t) = Q_0 e^{-kt},$$ and knowing that the halflife of carbon-fourteen is $5568$ years,

write down an equation (but \textbf{DO NOT} manipulate, rewrite or simplify it in any way) that states:

\textit{``The amount of carbon-fourteen remaining after 5568 years is half of the initial amount.''}

use the letter $Q_0$ for the initial amount; \textbf{DO NOT} replace it with any numerical value.

if you have read and understood the above instructions, send the message ``Okay, I will work on the equation now.'' before proceeding.

[warm shale]

Ann

[glossy yew]

okey lets look at this

okey

Okay, I will work on the equation now.

was the initial amount 100

@royal basin

[royal basin]

use the letter Q_0 for the initial amount; DO NOT replace it with any numerical value.

[glossy yew]

okwy

so this is what you mean

@royal basin \

[royal basin]

no, this is only the general radioactive decay equation

you have not yet written down what i asked you to write down

[grizzled shore]

The only variable in your equation should be t

Well I suppose initial amount is also a variable…

[grizzled shore]

No we want t to stay

We don’t want k

(And don’t just replace it with a number)

[glossy yew]

oh okey

[royal basin]

the "half of the initial amount" is missing.

@grizzled shore you might be misleading here a bit

[royal basin]

no

no

no

no

bad

[grizzled shore]

Ok I’m just gonna hush I feel like I’m not helping at all

[royal basin]

your feelings are true

@glossy yew Q(5568) = [half of the initial amount]

that's what i want you to write down, partially paraphrased

can you put it entirely in symbols

[glossy yew]

what do you mean by symbols

@royal basin

[royal basin]

what symbol did i tell you to use for the initial amount?

[glossy yew]

q

so dont use q

[royal basin]

wrong on both counts.

[glossy yew]

okey

[royal basin]

do you have problems with reading comprehension?

[glossy yew]

kinda

[royal basin]

eurhg

then it is going to be very hard for me to instruct you

[glossy yew]

yes it will

[royal basin]

like ok

what i was looking for is $Q(5568) = \frac{1}{2} Q_0$

[warm shale]

Ann

[glossy yew]

oh

[royal basin]

and THEN i would have you write down $Q_0 e^{-k \cdot 5568} = \frac{1}{2}Q_0$ and walk you through solving this equation for $k$

[warm shale]

Ann

[glossy yew]

yes\

do we use elimanation for this

[royal basin]

i do not know what you mean when you say "elimination"

[glossy yew]

so

@royal basin whats the next process after this

lol this looks like log

so

[royal basin]

the next step would be to divide both sides by Q_0 unless you want more headaches

[glossy yew]

that would elimate it

then you get

e^-k5568 = 1/2

[arctic meteor]

What the actual

[glossy yew]

yah like thats my expression

[royal basin]

e^-k5568 = 1/2
e^(-5568k) = 1/2, yes

[glossy yew]

do i flip it and turn it into to a log?

[royal basin]

not at all the words i would use

but i guess yes

[glossy yew]

so log e ^1/2 = k5568

[royal basin]

no

[glossy yew]

what

[obtuse pebble]

@glossy yew Has your question been resolved?

[glossy yew]

instead of the log

@royal basin do i solve it manually

[royal basin]

you tried to take the logarithm of both sides (i assume) but you fucked it up

or maybe you did not write it correctly

you should have ended up with $-5568k = \ln(1/2)$ perhaps

[warm shale]

Ann

[glossy yew]

is seee

and we get k = in(1/2)/-5568

@royal basin

[royal basin]

it's ln (lowercase L, lowercase N) not "in".

[glossy yew]

okey my bad

so now that we got k

whats next

@royal basin

bro we already solved it

bro im a monekey

that hasn't envo

[obtuse pebble]

@glossy yew Has your question been resolved?

Channel closed

Closed by @glossy yew

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

How do i find what z is, in my class we have only done trig sub where z does not have a coefficient

i see how the solution works i just want to know how to come to the conclusion ig

[sage geode]

You cannot solve for anything given no statements (An expression alone provided zero information)

[timid silo]

this is a trig sub intergral

[sage geode]

Ah, so you are integrating?

[timid silo]

yes I am sorry

[sage geode]

If the sub is unclear, consider factoring 4 out

You should get 2sqrt(1 - 9z^2/4)

You may also notice this is the same as 2sqrt(1 - (3z/2)^2)

Meaning you can do 3z/2 = sin(t) for example

[timid silo]

i think the idea is to sorta play around with it so that you are able to factor it out into something that has a perfect square root and then the trig identity right?

i was doing another trig intergral where it was like sqrt(13 + 25x^2)

[sage geode]

Yeah, after some practice it will become obvious that the sub is z = 2sin/3

[timid silo]

and the solution subbed in sqrt(13)/5

[sage geode]

Yeah

[timid silo]

gotcha

im also unclear about the theta

like finding its actual value

[sage geode]

You mean changing the bounds of integration during a sub?

[timid silo]

I mean like putting the final answer in terms of x and not theta

[sage geode]

Ah

E.g. z = 2sin(theta)/3 means that theta = arcsin(3z/2)

Just use inverse trig functions

[timid silo]

when does the pythagorean come in

like for instance if you had cos theta right

you would have to solve from theta = arcsin to get cos

[sage geode]

Like how would you get cos(theta) in this case?

Or how to simplify cos(arcsin(something))

[timid silo]

oh sorry instead of using inverse you could just san that 3z/2 = sin theta and then use the triangle and pythagorean theorem instead of directlky subing in the inverse right?

[sage geode]

Yeah, if you want to evaluate some other trig functions of theta

[timid silo]

gotcha thank you so much

.close

[obtuse pebble]

Channel closed

Closed by @devout tapir

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[worldly narwhal]

why is being multiplied by 8x

cant we do this with just a 4

[supple granite]

LCM is 4x

[worldly narwhal]

what about 4x

[supple granite]

4x works too

[worldly narwhal]

why did you tag on the x

[supple granite]

Doesn't really matter

[worldly narwhal]

or how do we know when to include the variable

in that lcm

[supple granite]

If you just multiply by 4, you can't get rid of the x in the denominator

Remember, 4 is not divisible by 2x, whereas 4x is divisible by 2x

[rich plume]

both sides have been multiplied by a common (here 8x) so the value of x remains same
It is mostly done to cancel the denominators and make the calculation easier

[obtuse pebble]

@worldly narwhal Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[neon gulch]

,rotate

[neon gulch]

Hey guys, wanna ask for help. Maybe you know a yt vid relating to this topic

[knotty thorn]

is 't' not defined anywhere else?

like is all you know about t that it is parallel to the line y?

[neon gulch]

Yes

[neon gulch]

Also its f, sorry for my handwriting

[knotty thorn]

nw

to be clear

is it f that is parallel to y

or is it the tangent line to f

[neon gulch]

Yes

Wait

[knotty thorn]

this question is oddly minimal in the information it gives

presuming f isn't a straight line, where a tangent would be redundant, the phrase 'the tangent line to the graph of f' is an odd statement

you cant give a tangent line of a graph

you give a tangent line at a point on a graph

[neon gulch]

Our teacher says that just from that, we could tell that m=1/2

[knotty thorn]

i'll presume the question is trying to just teach you about parallel lines, then

if something is parallel to another straight line, it shares the same gradient

[knotty thorn]

this is how they arrived at this

[neon gulch]

btw, our lesson is about Intuitive Notion of the Derivative

[neon gulch]

How?

[knotty thorn]

if two straight lines share a gradient, they are parallel

you are told the tangent is parallel to y = x/2 + 5

the gradient of y = x/2 + 5 is 1/2

so the gradient of the tangent must also be 1/2

[neon gulch]

Ohhh, but how do we use this to find the equation of the tnagent line?

[knotty thorn]

a tangent is a straight line

so it is of the form y=mx+c

you know m = 1/2

you cannot determine c without any further information

=> y = x/2 + c

[neon gulch]

So Is it unsolvable?

[knotty thorn]

if there is literally no other information about graph f than 'it exists'

yeah

[neon gulch]

My teacher solved a bit of the equation, maybe this could help

[knotty thorn]

.rotate

[neon gulch]

,rotate

[neon gulch]

,rotate

[knotty thorn]

wheeeere are you getting f(a) = sqrt(x-2) from

if that is what f is you should've just said lol

you're also getting

'find the tangent at point x=0'

out of nowhere

[neon gulch]

Ohhh

I seem to have not understood the question thoroughly

[knotty thorn]

i don't think you've written it out, correctly

your question should be akin to
(Using the limit definition of the derivative/gradient), Find the equation of the tangent to the graph f(x) at x=X)

[neon gulch]

I did write the question properly, my teacher mentioned to refer to the given formula where lim is ∆x->0

[knotty thorn]

this is from this equation

the limit definition of the gradient at point a

in your question, you did not provide f(x)

yet it seems to be defined enough from the limit written above

[neon gulch]

Ohhhh

[knotty thorn]

if you did not have a f(x) writing sqrt(x-2) and sqrt(x+Δ x-7) would be impossible

so you do, the question is just missing information

[neon gulch]

Ohhh, thank you

So the missing info is just x=0? Which is already there but isn't mentioned?

.close

[obtuse pebble]

Channel closed

Closed by @neon gulch

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[mental fiber]

help

[jolly ginkgo]

Sum of what?

[mental fiber]

p^2q/r + p^2 r/q + q^2 p/r + ...

[jolly ginkgo]

Ok so it's different permutations

[mental fiber]

yes

[jolly ginkgo]

So show your work

[mental fiber]

this is what i got after expanding the summation

p^2q/r + p^2 r/q + q^2 p/r + ...

also the info i have is

p+q+r = 2/3

[jolly ginkgo]

You must've the roots

[mental fiber]

pq + qr + pr = 0

pqr = -1/3

[mental fiber]

im sure we need to use these to find the answer

[jolly ginkgo]

Alr

So (p+q+r)(pq + qr + pr) = p²q + pq² + pqr + .....

But after this gotta think

Ok anyway find this expansion and put -1/3 for every pqr and equate it to 0

[pine owl]

3 equation 3 unknown u can solve them

[mental fiber]

yeah i also tried making the numerator same

ig i'll just have to put in the values of the roots

[jolly ginkgo]

Then we can go ahed

[jolly ginkgo]

Did you try this

[mental fiber]

ok one second

[jolly ginkgo]

Tell me what you get

[mental fiber]

ok i got this this

$$p^2q + pq^2 + q^2 r + qr ^2 + p^2r + pr^2 = 1$$

[jolly ginkgo]

pq² twice?

??

[mental fiber]

yeah i made a mistake

lemme fix that

[warm shale]

wise

fäf

[mental fiber]

yeah

the denominators kinda makes this hella confusing

[jolly ginkgo]

Yeah a bit

We must be close tho

[mental fiber]

nah this is the hard part

😭

[jolly ginkgo]

Yes

[mental fiber]

so did u find anything

[timid silo]

hi

[mental fiber]

hi

[timid silo]

need any help?

[mental fiber]

yeah

[timid silo]

ok

which sum?

i'll try 😂

[mental fiber]

this

[jolly ginkgo]

Nu

[mental fiber]

that sucks

[jolly ginkgo]

Yeah

Well I stopped thinking about it for a whil

[timid silo]

wait wdym summation of p2q/r?

[mental fiber]

@timid silo

[jolly ginkgo]

Thought you'd figure it out

[mental fiber]

nope

cant find anything to factor

[timid silo]

ohkk

[jolly ginkgo]

Feeling sleepy rn

[timid silo]

wait

did u find the roots?

[mental fiber]

ur supposed to use this @timid silo

[timid silo]

ok

so

in cubic

sum of roots = what

[mental fiber]

-b/a

[timid silo]

ye

so

oh wait

nvm

bruh

[mental fiber]

i wrote all of that stuff above, im just having trouble factoring this thing

[timid silo]

ok imma find the roots

just take t = x^2

no that doesnt work

so 3x^3 = 3t^y

whats y

[mental fiber]

what are u trying to find

[timid silo]

relation btw t and x

t^2 sorry not y

[mental fiber]

why are u using t^2

[timid silo]

it would be easy to solve as quadratic

wait 1 sec lemme solve

[mental fiber]

it wont be a quadratic

[timid silo]

1 sec bro

so t = x^(3/2)

[mental fiber]

yeah now u cant write the x term in terms of t

[timid silo]

ye sed

[mental fiber]

also i can solve the cubic

but thats not the issue

we not to use the sum of roots and stuff..

thats the problem

[timid silo]

ok

do u have any idea of what to use?

any special features?

[mental fiber]

nope

faf and i tried to think of ways to factor

but we got nothing useful

[timid silo]

oh

hmm

i found 1 root

wait no

[mental fiber]

,w solve 3x^3 - 2x^2 + 1 = 0

[warm shale]

Results provided by WolframAlpha

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

Requested by wise#9945

[mental fiber]

yeah i dont think we are supposed to find the roots lol

[timid silo]

lol

yep

and i thought i was good at math 😂

[mental fiber]

nah the problem is wack

not your fault

atleast u tried

[timid silo]

nah bro i feel like smth sus

did ur teacher tell anything?

/professor

[mental fiber]

nope

[timid silo]

wait we'll use the p+q+r and pqr

so pqr = -1/3

[mental fiber]

aight

[timid silo]

ok

so im taking common

brujh

still not getting anything soo useful

wait a sec

ok denominator cancelled

so finally im getting this

3(p^2q^2r - 2p^2q^2/3 + p^2r^2q - 2p^2r^2/3 + q^2r^2p - 2q^2r^2/3)

[mental fiber]

u multiplied by pqr then put it = -1/3?

[timid silo]

not exactly but

somewhat substitution

bruh

its coming like infinite loop

i've simplified to max

is the answer 3/4?

[mental fiber]

idk what the answer is

[timid silo]

lol

this is so dragging man

do u mind asking ur teacher or smth?

cuz idk after this what to do

[mental fiber]

aight but i doubt they'll respond rn

[timid silo]

😦

u in skl?

im sorry bro i tried my best

[mental fiber]

nah man why u saying sorry

thanks for helping

[timid silo]

🙂

ill get u ans tmrw bro

[warm shale]

yenn

[sudden lagoon]

now we have

[warm shale]

yenn

[sudden lagoon]

so we just need to be able to compute $\sum{1/x}$ and $\sum{1/x^2}$ to get what we want

[warm shale]

yenn

yenn

yenn

[sudden lagoon]

if i havent made any mistakes which you can veryfy, now all the variable algebra is done and you're left with just numbers

[sudden lagoon]

here I merged x^3y^2 + y^2x^2 as x^2y^2(x+y) for all pairs (x,y), (y,z), (z,x)

[warm shale]

yenn

[sudden lagoon]

@mental fiber you can analyse it if you want

[obtuse pebble]

@mental fiber Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[fierce lagoon]

Alright so I just wanna make sure I'm on track with understanding Big-O.

If I have something like f(x) = x + O(x^3), this means that f is asymptotic to x^3 or something like that?

[fierce lagoon]

Because I came across this a month ago and decided to read about it on google

And then it came across me again and now I'm like "O"

[drifting cypress]

'o'

[fierce lagoon]

And asymptotic, I mean that the remaining terms remain beneath x^3

So like

Something like that

[fierce lagoon]

And so does this mean that f(x) < x^3?

[timid silo]

from wiki:

which intuitively means that f(x) grows at most as fast as g(x)

[fierce lagoon]

Oh I see

[timid silo]

< M x^3 for some constant M

[fierce lagoon]

Hm

[timid silo]

so like 2x^3 = O(x^3)

[fierce lagoon]

So like

I could say 2x = O(x^3)?

[timid silo]

yes

[fierce lagoon]

But not x^5 = O(x^3)

[timid silo]

it's also O(x^n) for any n≥1

[timid silo]

yup

[fierce lagoon]

Alright thanks

I'll play around with this more

.close

[obtuse pebble]

Channel closed

Closed by @fierce lagoon

Use .reopen if this was a mistake.

[fierce lagoon]

@timid silo there's more

That's scary

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[static fulcrum]

How can I solve (1-x)^2 by u substitution?

[static fulcrum]

Would it be u^3/3 ?

[mental fiber]

yes

[static fulcrum]

I got -du, looks weird

[mental fiber]

its not weird

[mental fiber]

oh yeah it will be -u^3 / 3

[static fulcrum]

ah

[trim portal]

then you have to substitute back

[mental fiber]

^^