help-17

[dry orchid]

greek letter guy's channel

[hidden kelp]

Yeah!

[short pilot]

Fr

[bitter pilot]

ah

[short pilot]

We making it to Adonis mind with this one fr 🔥🗣️

[hidden kelp]

Ok so Oppai here's a cool one

[short pilot]

Imma scared bro

[hidden kelp]

,, \int_{0}^{1}\frac{x-1}{(1+x^2)\ln x}\dd x

[twin meteor]

𝔧𝔪𝓣𝛾𝜑𝜽

[short pilot]

Oh

Oh

Okok

Let's do it

[hidden kelp]

Let's see if Adonis' huge brain can do this

[short pilot]

Lol bro

He so pro

He gonna make me regret in this one 🗣️🙊 🔥😭

[bitter pilot]

@drowsy igloo

[torn timber]

I feel like u should use lnx as t

[short pilot]

I feel like we should parameterise it

[bitter pilot]

I feel nothing

[short pilot]

No I feel like we should do

[hidden kelp]

This is the play

Yall ever heard of the best integration technique of all time? || Feynman's trick||

Anyhow, I'll be pleased to answer questions about this integrals in dms, but I'll close the help channel in order to let someone else have it

.close

[vocal sleet]

Channel closed

Closed by @hidden kelp

Use .reopen if this was a mistake.

[bitter pilot]

oh

that could work actually witht that

[hidden kelp]

hell yeah it does. I mean, lnx in the denominator? Come on

[short pilot]

Wait I reached some where here

Let me calculate now

Yes ik

We can do x^s

And eliminate lnx from denominator

@hidden kelp

Class duis moment

Using that I have reached here

,w calculate digamma(s+1/2)-digamma(s/2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by oppailol.

[short pilot]

Noo

Aaah

@hidden kelp

, w calc digamma((s+1)/2)-digamma(s/2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by oppailol.

[hidden kelp]

Getting close 👀

[short pilot]

Yes

Now I have to integrate

Yes we will get gamma

Yay

Fr

Aah

Give me 1 min

Oof

Imma getting heated

@hidden kelp

Bro

@hidden kelp

Help me find c bro

I got I(s)=log(gamma(s+1)/2)/gamma(s/2))-2log(gamma(s+1)/4)/gamma(s/4)

@hidden kelp

+C

I need the C

@hidden kelp

.reopen

[vocal sleet]

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.

[untold arrow]

Hints?

[vocal sleet]

@untold arrow Has your question been resolved?

[knotty lynx]

Im not sure how to actually solve it but I have something

Try the most obvious u-sub

[surreal basin]

then using the same $u$-sub, find $\int_0^1(1-x)e^{1-x}f(xe^{1-x})\dd{x}$

[twin meteor]

mtt

[untold arrow]

the most obvious u sub which means...?

[surreal basin]

as another hint, you would usually choose the u to be something which would make this easier to read

right now it has an f(xe^(1-x))

if it was an f(u), it would be easier to read

[surreal basin]

so looking at this, what would be the "most obvious" u-sub noticing that its f(xe^(1-x))?

[knotty lynx]

I suppose that was a bit pretentious 😅

[surreal basin]

you can reread those words as "the u-sub that you would be the least surprised about"

[untold arrow]

how can it easy to read

darn im dumb

[knotty lynx]

They meant the u-sub would make it easier to read

[knotty lynx]

look at what's in bold in their message

[vocal sleet]

@untold arrow Has your question been resolved?

[vocal sleet]

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.

[vast shale]

I need help with my problem

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

[vast shale]

<@&286206848099549185>

[vocal sleet]

@vast shale Has your question been resolved?

[wise pilot]

Solve the 2 equations using elimination and then use (x*,y*) as an arbitrary point on the line and use 2 point form of an equation of a line

For the 1st part

[vocal sleet]

@vast shale Has your question been resolved?

Channel closed

Closed by @glass basalt

Use .reopen if this was a mistake.

[vocal sleet]

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.

[wary mantle]

If we have a permutation $\pi = \mr{3 & 1 & 2}$ in one-line notation, then [\mr{3 & 1 & 2} \overset{\tau_{12}}{\to} \mr{1 & 3 & 2} \overset{\tau_{23}}{\to} \mr{1 & 2 & 3}] and so [\pi = \tau_{23} \circ \tau_{12} \circ \operatorname{id}.] But wouldn't it be more intuitive if it was instead $\pi = \tau_{12} \circ \tau_{23} \circ \operatorname{id}$? We first apply $\tau_{23}$ on $\operatorname{id}$, then $\tau_{12}$ on that to go the reverse direction.

[wary mantle]

So how come it's still the former?

[oak magnet]

id is the first one ?

[wary mantle]

?

[oak magnet]

id stands for [3 1 2] ?

[wary mantle]

No

As I wrote, pi stands for that, id is [1 2 3]

[oak magnet]

Thats why i was confused

[wraith venture]

You've proven that t23(t12(pi(i))) = i
Hence inverting all the way
pi(i) = t12(t23(i)) because transpositions are involutive

[hard atlas]

well no

cause t12(t23(1))=2 is not 3

the first line makes no sense to me

I suppose yes it does

but like

ugh I just hate the notation

the arrows should be in the other direction or something

[wraith venture]

Shall we return to xfg notation rather than g(f(x))?

[hard atlas]

cause you arent switching the symbols 1 and 2, you are switching the first and second position

no I mean one line notation instead of cycle notation

[wary mantle]

oh

[wary mantle]

Why pi(i)?

Just t23(t12(pi)) = id, no?

[wraith venture]

Yes but that's notationally weird as that's not really usual function composition notation

[wary mantle]

Is $\tau_{23} \circ \tau_{12} \circ \pi = \operatorname{id}$ fine?

[wary mantle]

Ok so now we have this

Then we "left-multiply" by t_23

No?

$\underbrace{\tau_{23} \circ\tau_{23}}{\operatorname{id}} \circ \tau{12} \circ \pi = \tau_{23} \circ \operatorname{id}$

[hard atlas]

in what direction are you reading composition

[wary mantle]

Ah

[wary mantle]

From right to left

[hard atlas]

but then t23 circ t12 circ pi = id is just wrong

[wary mantle]

what

[hard atlas]

t23(t12(pi(1))=t23(t12(3))=t23(3)=2

[wary mantle]

t23(t12(pi)) = id

tf

Ok start from here again

[\mr{3 & 1 & 2} \overset{\tau_{12}}{\to} \mr{1 & 3 & 2} \overset{\tau_{23}}{\to} \mr{1 & 2 & 3}]

[wary mantle]

So: $\tau_{23}(\tau_{12}(\pi)) = \id$

How come no

We literally follow this?!?

[wraith venture]

No
Positions!= Values

[wary mantle]

my tau stands for positions..

tau_12 means positions 1 and 2 switched

Not values

[wraith venture]

You monster

[wary mantle]

this is how I learned it lol

And how my prof does it

[wraith venture]

What's t12 as a permutation?

[hard atlas]

this is legitimately disgusting

[wraith venture]

Makes no sense

[hard atlas]

I have no clue how to deal with that

[wary mantle]

It switches the positions 1 and 2. So id = [1 2 3] becomes [2 1 3]

[wraith venture]

That's ok

But in (3 1 2) you swap values 1 and 2, not positions

Otherwise you're doing left composition rather than right composition

[wary mantle]

In the first step?

Here?

[hard atlas]

how are you computing the composition [213] circ [132]

[wary mantle]

I look at [1 3 2]. First position gets mapped to 1, then the first position in [2 1 3] is 2. Now second psoition gets mapped to 3, that's 3 in [2 1 3]. Third position gets mapped to 2, that's 1 in [2 1 3]

So [2 3 1]

[hard atlas]

so and now we have a huge problem

cause thats not the same as t12 circ [132]

but t12 circ [132] = t12 circ id circ [132] = [213] circ [132]

[wary mantle]

Oh

Ok so this is all about this left and right composition stuff

Thanks

.close

[vocal sleet]

Channel closed

Closed by @wary mantle

Use .reopen if this was a mistake.

[hard atlas]

thats not the takeaway from this

the takeaway is that this notation sucks

[wary mantle]

$\frac{\Xi}{\overline{\Xi}}$

[twin meteor]

Kepe

[loud folio]

what is the integration of root(cos(x^2-1) * sin(x^2-1))

[wraith venture]

#❓how-to-get-help

[vocal sleet]

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.

[robust badger]

wolfram gives me an approx and im not how to find exact value of $\sum_{n=1}^{\infty} (-\cos(\frac{1}{n})+1)$

[twin meteor]

alaska v.2

[vocal sleet]

@robust badger Has your question been resolved?

Channel closed

Closed by @robust badger

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.

[ivory violet]

does this equal this?
im aware of the b^log(base b)(whatever) = whatever
but if theres smth multiplied in the power does it just stays in the power

[ivory violet]

if so anyways, is there any other rule regarding logs to the power i should just know

[outer warren]

this is just being combined with normal exponent laws

a^(bc) = (a^b)^c

$e^{\frac13 \ln|x^3+8|} = \br{e^{\ln|x^3+8|}}^{\frac13}$

[twin meteor]

ℝαμΩℕωⅤ

[vocal sleet]

@ivory violet Has your question been resolved?

[vocal sleet]

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.

[warm horizon]

find the exact value of:

$\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+\binom{n}{2}\binom{n}{3}+...+\binom{n}{n-1}\binom{n}{n}$

[twin meteor]

BOHO

[warm horizon]

this looks like a binomial identity where u equate coefficients but idk what coefficients r being equated

[vocal sleet]

@warm horizon Has your question been resolved?

[warm horizon]

<@&286206848099549185>

ok i got the answer lol but now idk how to explain it

so the general form of equating coeffs is

$(1+x)^n(1+x)^n=(1+x)^{2n}$

[twin meteor]

BOHO

[warm horizon]

and then i just equated coeff of $x^{n+1}$ on both sides

[twin meteor]

BOHO

[warm horizon]

and then i used symmetry identity to get the LHS look like the q

but now idk how to know that u need to equate x^n+1 coeff

.close

[vocal sleet]

Channel closed

Closed by @warm horizon

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.

[hushed cosmos]

cant u technically do this by doing 1-p(x=50)-p(x=49)

[vocal sleet]

Please don't occupy multiple help channels.

[bitter wren]

i have so many answers in my head that probably wouldn't solve the quetion

but i can try ig

[bitter pilot]

should work as well

[hushed cosmos]

wait would it be 1-p(x=50)-p(x=49) or p(x=50)+p(x=49)

[vocal sleet]

@hushed cosmos Has your question been resolved?

[ivory violet]

perchance is this qcaa ia3

[ivory violet]

i think scnd

• 48

cuz >= 8

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[hushed cosmos]

yuh howd u know

[vocal sleet]

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.

[lethal heart]

$f(x)= 1$ if $x \geq 0 , x+1$ if $x<0$$

[lethal heart]

I need to see if the function is limited above

[twin meteor]

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

[hard atlas]

$f(x) = \begin{cases} 1 & x\geq 0 \ x+1 & x<0 \end{cases}$

[twin meteor]

Denascite

[lethal heart]

Thanks

I need to see if the function is limited above

[hard atlas]

what have you tried

[vocal sleet]

@lethal heart Has your question been resolved?

[lethal heart]

I solved it graphically

[vocal sleet]

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.

[viscid bridge]

hi guys

[past anchor]

[viscid bridge]

what does that xer mean

[mild flower]

the E symbol means "element of" and the R symbol means "the real numbers"

[viscid bridge]

okk thank youu so much

.close

[vocal sleet]

Channel closed

Closed by @viscid bridge

Use .reopen if this was a mistake.

[viscid bridge]

.reopen

[vocal sleet]

✅

[viscid bridge]

guys so

[past anchor]

what

[viscid bridge]

the third property, what does it mean

[past anchor]

so you know that |x| = x also |-x| = x?

[viscid bridge]

yess

because - and -x is + right

[past anchor]

so we can say

$|x| = |-x| = x$

yes?

[twin meteor]

Newt

[viscid bridge]

oohh

yesyes

ooooo

THANKYOUUUU

[viscid bridge]

for the second property,
|x| can be x or -x right

[past anchor]

no

[viscid bridge]

what does it has to do with square root and square?

[past anchor]

|x| = x for all positive number and |x| = -x for all negative number

because square root only have positive answer

[viscid bridge]

this is unimportant but i'm curious why is it written "for all positive number" and "for all negative number"

doesn't -x always mean negative?

[past anchor]

no

imagine ur putting value of x = -4 for equation y = -x

y = - (-4) = 4

[viscid bridge]

ooooohhhh

guys sorry to ask again i know this is so dumb

[past anchor]

what just ask

[viscid bridge]

how is this related with the first principle

[queen root]

homie had enough fr

[viscid bridge]

[past anchor]

by its definiton |x| is for making every number became positive

[viscid bridge]

OOOHHHHH

so they say
|x| = x, if x is a positive number
|x| = -x, if x is a negative number

this means if x is a positive, it'll still be positive.

and if x is a negative, then it'll be positive because there's a negative infront of the (negative x)?

[queen root]

yes

[queen root]

u made it sound so complicated lmao

[viscid bridge]

i think this is more complicated

[queen root]

Thats so easy

[viscid bridge]

guys is it rlly important to understand what's written here bcs i have a hard time understanding concepts written this way

[queen root]

It works for any value that can replace x

[queen root]

yeah, theyre important for algebra

[viscid bridge]

okaayyy

thank you so much @queen root and @past anchor

.close

[vocal sleet]

Channel closed

Closed by @viscid bridge

Use .reopen if this was a mistake.

[vocal sleet]

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.

[verbal haven]

in this step, where do the dx's come from?

[verbal haven]

this is a bernoulli ODE via IF

previous to this step, I have this

this is pre multiplying everything by the IF, however the IF in this case was just x^2 with no dx

ok I have figured out I just forgot to multiply by dx with the IF to put the equation in differential form, however im now wondering what the jump from x^2dv+2xvdx=6x^2dx to d[vx^2]=6x^2 dx is

it doesnt fit grouping form so im not sure how he did that simplification between steps

this

[vocal sleet]

@verbal haven Has your question been resolved?

[verbal haven]

<@&286206848099549185>

.close

[vocal sleet]

Channel closed

Closed by @verbal haven

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.

[vital wave]

$y = cot^2(sin(x))$ how do i derive using power and chain rules?

[twin meteor]

wakamole

[twin horizon]

Rewrite as ( cot(sinx))²

Sub u = cot (sinx

[vital wave]

can u show me steps .plz

[twin horizon]

Ok if I am still around

I hope I did this properly

I'm error prone but this is basically what you should do

,w d/dx((cot(sin(x)))²

[twin meteor]

Results provided by WolframAlpha

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

Requested by ishigamisenku_

[vocal sleet]

@vital wave Has your question been resolved?

[vocal sleet]

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.

[ancient vapor]

someone please help or pm me

[pallid zenith]

what have you tried

[pallid zenith]

.

[pallid zenith]

https://en.wikipedia.org/wiki/Convolution_of_probability_distributions this might be helpful

maybe

yes, here you go

look at the end there

so assuming you are able to determine the constant combo indicator function which defines f_X and f_Y, you are set

[floral pike]

Is that discrete though? Looks continuous

[pallid zenith]

theyre all continuous

only the statement at the top is about discrete

it continue

i included that portion since it defines the variable Z

but, it follows more or less directly here

the desired pdf is $f_Z (z)$

[twin meteor]

jan Niku

[pallid zenith]

so assuming that @ancient vapor can describe a uniform pdf as a function

the answer is just there

[vocal sleet]

@ancient vapor Has your question been resolved?

[ancient vapor]

idk how i only know the uniform dist of X and Y individually

[pallid zenith]

you need to click ❌

[ancient vapor]

also i dont think i even learned convolution

[pallid zenith]

or the channel is going to close

idk you need to learn it necessarily

its described in that paragraph there

[ancient vapor]

can you use the cdf method?

[pallid zenith]

idk what that is

[ancient vapor]

when you get the cdf of Z = X+Y and differentiate it

to get the pdf

if you dont know could you just walk me through the convolution then

never learned it before

[pallid zenith]

all youll need is the pdf's of the uniform distributions

its probably helpful to write them in the form of like

so define some set $S = [a,b]$, then write the indicator function as $\qty(\frac{1}{b-a}) \mathbb I _S (x)$

[twin meteor]

jan Niku

[pallid zenith]

so what this means is that I_S (x) is 0 if we arent inside [a,b]

the 1/(b-a) ensures that its a pdf, so that like

if its really wide, the prob is scaled down

does that make sense?

this is just one way to write or use uniform distributions

i just think its helpful to have these indicators

[ancient vapor]

absolutely not idk what I_S(x) is

i just defined them as

f_x(x) = 1/2 for 0 < x < 2
f_y(y) = 1 for 3 < y < 4

[pallid zenith]

sure

okay

now you cite the convolution theorem

[ancient vapor]

f_x(x) * f_y(z-x)?

[pallid zenith]

and you say our desired pdf is $f_Z (z) = \int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x$

[twin meteor]

jan Niku

and you say our desired pdf is $f_Z (z) = \int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x$

[ancient vapor]

ok

what are the bounds for the integral then

and what do i do with the z inside the indegral

[pallid zenith]

this is where the indicators help

nothing, its along for the ride

the idea is like this

lets start with ... f_X

[pallid zenith]

you say f_X (x) is only nonzero for 0 < x < 2

yea?

i mean, its 1/2 there

[ancient vapor]

yeah

[pallid zenith]

but everywhere else its 0

[ancient vapor]

yeah

[pallid zenith]

$\int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x = \int _{- \infty} ^{0} f_X (x) f _Y( z-x) \dd x + \int _{0} ^{2} f_X (x) f _Y( z-x) \dd x + \int _{2} ^{\infty} f_X (x) f _Y( z-x) \dd x$

hope this isnt too wide

[twin meteor]

jan Niku

$\int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x = \int _{- \infty} ^{0} f_X (x) f _Y( z-x) \dd x + \int _{0} ^{2} f_X (x) f _Y( z-x) \dd x + \int _{2} ^{\infty} f_X (x) f _Y( z-x) \dd x$

[pallid zenith]

sure lets work with it

you see what i did?

i split up the bounds

hopefully its suggestive

can you simplify this?

look at the interaction between f_X and the bounds

[ancient vapor]

yeah

but dont the improper integrals diverge

[pallid zenith]

nope

you get lucky

were integrating over x right

so lets say that first integral, x is swept over all values from -inf to 0

but, thats not a concern

f_X (x) is 0 over that entire region

[ancient vapor]

so the improper integrals are 0?

[pallid zenith]

yea

they vanish

[ancient vapor]

oh yeah i see because theyre out of bounds

[pallid zenith]

yup

$\int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x = \int _{0} ^{2} f_X (x) f _Y( z-x) \dd x$

[twin meteor]

jan Niku

$\int _{- \infty} ^{\infty} f_X (x) f _Y( z-x) \dd x = \int _{0} ^{2} f_X (x) f _Y( z-x) \dd x$

[pallid zenith]

now what

[ancient vapor]

is f_x(x) * f_y(z-x) = 1/2?

[pallid zenith]

no, but you have the right idea i think

f_X (x) is 1/2

because 0 < x < 2

[pallid zenith]

see, here

[ancient vapor]

ok

then what is f_y(z-x)?

[pallid zenith]

$f_Z (z) = \frac 12\int _{0} ^{2}f _Y( z-x) \dd x$

[twin meteor]

jan Niku

[ancient vapor]

wait does this turn f_y into a cdf or something?

[pallid zenith]

lets say $f_T (t) = 1$ for $3<t<4$

[twin meteor]

jan Niku

[pallid zenith]

no, its a little weird

convolution is harder to explain in this context than in others

but you can think of it as like, combining two things by pulling out values at each point in some domain

its sweeping the two pdfs over each other

[ancient vapor]

so 3 < z-x<4?

for f_Y(z-x)?

there's gotta be something you can do with that interval

[pallid zenith]

sorry i thought i had this but i need a sec

[ancient vapor]

no problem

[pallid zenith]

man this gets more complicated than i thought

i really want to use an indicator function

[ancient vapor]

do it with that instead

[pallid zenith]

i think i have the rough answer but theres case work to do

[ancient vapor]

is it for each interval?

[pallid zenith]

it honestly makes me wonder if you could just figure out the answer by casework

well okay

you understand what indicator means if i write it

call $S = [3,4]$

[twin meteor]

jan Niku

[pallid zenith]

so we want $f_Z (z) = \frac 12 \int _0 ^2 \mathbb I _S (z-x) \dd x$

im not writing a 1, here

$f_Y (y) = (1) \mathbb I _S (y)$

[twin meteor]

jan Niku

jan Niku

[pallid zenith]

now it gets kind of screwy

we know I_S has to be nonzero to have any value

[ancient vapor]

when you say do it by cases, do you mean for each sub interval of z in [3,6]?

[pallid zenith]

then it must be that $3 \leq z-x \leq 4$ so that $z-4 \leq x \leq z-3$

[twin meteor]

jan Niku

[pallid zenith]

cases just appear

which you maybe anticipate

[pallid zenith]

i dont know if we can start to sub in values for z, here

it makes our life easier

[ancient vapor]

is the problem describing the region here?

[pallid zenith]

like i believe that we can say $-1 \leq z-4 \leq x \leq z-3 \leq 3$?

[twin meteor]

jan Niku

[pallid zenith]

yea

im really starting to think that this is the wrong way to do it

and youre supposed to just suss it out by casework from the get-go

[ancient vapor]

i have no idea how to do it but graphically it's really easy

[pallid zenith]

yea

[ancient vapor]

red and green triangles are 1/2, and the blue thing is just 2 of those, so it's 1

[rugged orchid]

hi jan

[pallid zenith]

hi frosst

[ancient vapor]

do you know how to get the area of the blue shape in the middle using double integrals? I forgot all of multivariable calc

[pallid zenith]

u come to save me again

I suggested convolution

but im beginning to think this is overkill

i imagine we can just casework it out

[rugged orchid]

so we got X ~ U(0, 2), Y ~ U(3, 4)

[pallid zenith]

but its not becoming obvious to me

[rugged orchid]

and we want the pdf of X + Y

[pallid zenith]

ye

[ancient vapor]

cdf method?

[rugged orchid]

Let $W = X + Y$.
[f_W(w) ={f_X*f_Y}(w) = \int_{-\infty}^{\infty}f_X(x)f_Y(w-x),dx]

[ancient vapor]

okay hold on how did you get those bounds

is it by definition?

[rugged orchid]

i think it's w-x right?

[ancient vapor]

yeah

[pallid zenith]

yea but the bounds look sus

[rugged orchid]

it's either w-x or x-w

oh

it should be inf

[twin meteor]

frosst

[rugged orchid]

my bad

this is by definition of the convolution

are we following so far

[pallid zenith]

we worked through this part, if its helpful

[ancient vapor]

yes

[rugged orchid]

ok

[pallid zenith]

got to here

[rugged orchid]

let's look at f_X(x)

[pallid zenith]

but i dont want to stop you

[rugged orchid]

i'd rather do it from scratch cos there's a lot to read and this isn't that long

[f_X(x) = \begin{cases} 0.5 & 0 < x < 2 \ 0 &\text{otherwise}\end{cases}]

[twin meteor]

frosst

[rugged orchid]

[f_W(w) = \int_0^2f_X(x)f_Y(w-x),dx]

[twin meteor]

frosst

[rugged orchid]

because when x is outside (0, 2), the first function returns 0 so the integrand is 0

so the integral is 0

[ancient vapor]

yup got that

[rugged orchid]

okay let's look at the 2nd function

[f_Y(w-x) = \begin{cases} 1 & 3 < w-x < 4 \ 0 &\text{otherwise}\end{cases}]

[twin meteor]

frosst

[ancient vapor]

yeah this is where we stopped pretty much

[rugged orchid]

so here this is the same as saying 3 + x < w < 4 + x

oh no that's bad

[3 < w - x < 4\iff -4 < x - w < -3 \iff -4 + w < x < -3 + w]

[twin meteor]

frosst

[rugged orchid]

so when x is inside these bounds, f_Y(w - x) = 1, when it's outside we get 0

does that make sense

[ancient vapor]

yeah but what do we do with these intervals

[rugged orchid]

yeah so

[twin meteor]

frosst

frosst

[rugged orchid]

so in order for our integrand f_X(x)f_Y(w-x) to be a non-zero function, we need both to not be 0 at the same time

inside the bounds where at least 1 of the pdfs are 0, the integral for that part is just 0

[rugged orchid]

this is the reasoning for

outside 0 to 2, the integrand is 0 because of the first pdf

what this means is we only care about when [x \in (0, 2)\cap (-4 + w, -3 + w)]

[twin meteor]

frosst

[rugged orchid]

or, we can say [\max(0, -4 + w) < x < \min(2, -3+w)]

[twin meteor]

frosst

[rugged orchid]

hence we get
[\int_{\max(0, -4+w)}^{\min(2, -3+w)}f_X(x)f_Y(w-x),dx]

[pallid zenith]

this is where i stopped lol

[twin meteor]

frosst

[rugged orchid]

now, inside these bounds, the pdfs are both non-zero

and since they are only piecewise 1 way, ie, they are 0 or just 1 function

we can just replace it with the non-zero part

so,
[f_W(w) = \int_{\max(0, -4+w)}^{\min(2, -3+w)}1\cdot 0.5,dx]

[twin meteor]

frosst

[rugged orchid]

[f_W(w) = 0.5(\min(2, -3+w) - \max(0, -4+w))]

[twin meteor]

frosst

[rugged orchid]

now this looks ugly asf

do you follow up until here tho?

[ancient vapor]

yeah i see but why do we say max(0,-4+w) instead of min(0,-4+w)

[pallid zenith]

you want to be sure that you dont go below 0

[rugged orchid]

so you were lost here?

[ancient vapor]

oh got it

[rugged orchid]

going from here

[rugged orchid]

to here

is the hard part

[rugged orchid]

at this point you just need to test the boundary values and see what the piecewise looks like

for example, when -3 + w > 2, the min always returns 2

this happens when w > 5

[ancient vapor]

wait before that, is W in [3,6]?

[rugged orchid]

now when w > 5, max(0, -4 + w) returns -4 + w because 1 > 0

[rugged orchid]

dont worry about that

it'll pop out later

oh no

[rugged orchid]

you're right

you do need to care about htis

so you also have 3 < w < 6

so let's build the pdf

[f_W(w) = \begin{cases}
0.5(2 - (-4+w)) & 5 < w < 6 \
0.5(\min(2, -3+w) - \max(0, -4+w) & 3 < w < 5 \
0 & \text{otherwise}
\end{cases}]

[twin meteor]

frosst

[rugged orchid]

[f_W(w) = \begin{cases}
3 - 0.5w & 5 < w < 6 \
0.5(\min(2, -3+w) - \max(0, -4+w) & 3 < w < 5 \
0 & \text{otherwise}
\end{cases}]

[twin meteor]

frosst

[rugged orchid]

now when w < 5, the min will return -3 + w instead

when w > 4, max(0, -4 + w) still always returns -4 + w

[f_W(w) = \begin{cases}
3 - 0.5w & 5 < w < 6 \
0.5((-3+w) - (-4 + w)) & 4 < w < 5 \
0.5(\min(2, -3+w) - \max(0, -4+w) & 3 < w < 4 \
0 & \text{otherwise}
\end{cases}]

[twin meteor]

frosst

[rugged orchid]

[f_W(w) = \begin{cases}
3 - 0.5w & 5 < w < 6 \
0.5 & 4 < w < 5 \
0.5(\min(2, -3+w) - \max(0, -4+w) & 3 < w < 4 \
0 & \text{otherwise}
\end{cases}]

[twin meteor]

frosst

[rugged orchid]

now between 3 and 4, what happens to the min and maxes

@ancient vapor do you think you can work through that?

[ancient vapor]

it's 0.5?

wait

0.5(-3+w)?

[rugged orchid]

where'd you get this

[rugged orchid]

im missing a close bracket on line 2

it's 0.5(min - max)

[ancient vapor]

min(2,-3+w) is gonna output -3+w right?

[rugged orchid]

yes

[ancient vapor]

and max(0,-4+w) is gonna output 0

[rugged orchid]

uh huh

[ancient vapor]

so it's 0.5(-3+w)...

[rugged orchid]

oh that's your answer

yes

well idk i didn't do it i was waiting for you to reason through it

lol

[rugged orchid]

i can't tell if this is right without reasoning through it

but yes it's right

[ancient vapor]

my bad

okay buts we have the answer now

[rugged orchid]

so -1.5 + 0.5w

[ancient vapor]

however the thing is i didnt learn convolution or any of these methods in my class

[rugged orchid]

so what did you do in class then

[ancient vapor]

If we take F(W) = P(W < w) = P( X+Y < w), then differentiate F(W) to get f(W)

[rugged orchid]

ok

that's the other way

we did it via convolutions of pdfs

[ancient vapor]

yeah

[rugged orchid]

this is the way via cdfs

[ancient vapor]

do you have time to do it this way

my trouble is just figuring out the bounds

or really just setting up the integrals and intervals

[rugged orchid]

im trying to remember how you do this

cos i've only done cdf method for transformations of 1 RV

not functions of multiple RVs

[ancient vapor]

i can do it, but only if X and Y are the same disribution

but theyre not so idk

[rugged orchid]

[P(X+Y<w) = \int_{-\infty}^\infty P(X < w - Y| Y = y) P(Y = y),dy]

is it something like this

[twin meteor]

frosst

[ancient vapor]

oh wait i think you can do it like that, but isn't this just the convolution?

[rugged orchid]

that doesn't even work P(Y = y) would just be 0

[rugged orchid]

well but how does your course teach you to do it then

cos idk how else

[ancient vapor]

The reason im having trouble is because the notes are insufficient to do the problem

[rugged orchid]

can you show the notes

[ancient vapor]

yeah hold on

here they are

[rugged orchid]

okay

that's not too bad

so, since X and Y are independent

we colour in the box [0, 2] x [3, 4]

this would be the [0, 1] x [0, 1] box

we dont want this one

[ancient vapor]

like this

[rugged orchid]

yes that rectangle

ok so then, since we want Y < z - X, we want below the diagonal lines

[rugged orchid]

so the area woule be, from z ∈ [5, 6], we have the red + blue + varying green corner

from z ∈ [4, 5] it's the red + varying blue middle section

from z ∈ [3, 4] it's the varying red corner

[ancient vapor]

when starting the problems how am I supposed to even know theres intervals i have to solve the area for

[rugged orchid]

i've never seen this before but it's pretty clever

[rugged orchid]

well

the equation X + Y < z

you can rearrange it into Y < something of X

so it's like a half-plane on the X-Y plane

yeah?

[ancient vapor]

yeah

[rugged orchid]

well

as you vary the z

the line moves

if i give you some arbitrary z value

so the line is just somewhere

how do you find the area of the shaded region?

[ancient vapor]

oh i see it doesnt work

[rugged orchid]

you do this to it

the red part

you already know

the blue part

that depends on the z value

[ancient vapor]

okay i guess the only way to do this problem is graphically i guess, i dont have enough intuition to do it without it

[rugged orchid]

from z = 4 to z = 5, you need to find the area of the blue part

for your particular z value

[ancient vapor]

alright since it's been a while since ive taken multivariable calc, how do i find the blue part using a double intergral?

[rugged orchid]

which is given by a strange geometric parallelogram shape

i mean it's just a shape

[ancient vapor]

yeah but just in case

[rugged orchid]

i would advise against using multivariable calc

[rugged orchid]

since you literally end up here

well if you really wanted to know

i'd integrate this particular blue region part against the dx direction first

[ancient vapor]

yeah but isnt it just $\int_{3}^{4} \int_{4-x}^{5-x} \frac{1}{2} dx$

[twin meteor]

gaydan jew

[rugged orchid]

where did the half come from

[ancient vapor]

1/2 being f(x,y)

f(x,y) = f_x(x) f_y(y)

by independence

[rugged orchid]

yeah but then you're looking at the volume under the joint pdf

which isn't wrong

but you said area

if you want to find the area you'd just put 1 for the integrand

and dont forget the dy at the end as well

[ancient vapor]

ohhhh

this doesnt work for the triangles though since they are both 1/2 in area

wait nvm

[rugged orchid]

it will work for the triangles, but the bounds of dy will change with z

[ancient vapor]

okay so i get how to do the areas and all that

but then the answer isnt gonna be the same as when we did convolution

[rugged orchid]

it should be

[ancient vapor]

the areas are 1/2 for each triangle and 1 for the shape in the middle

[ancient vapor]

this is completely different

unless theres a step im missing

[vocal sleet]

@ancient vapor Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vocal sleet]

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.

[earnest pendant]

is this a valid notation?

[earnest pendant]

this is supposed to be the chain rule

while i understand the chain rule, i dont understand this notation

[surreal basin]

plugging in a number into a function is shown as function(number)

so plugging in h(x) into a function is shown as function(h(x))

the function is dg/dh

so dg/dh(h(x)) means to calculate dg/dh then plug h(x) into it

the video intends for you to pick up on that

[earnest pendant]

is the same for dh/dx?

[surreal basin]

yes

[earnest pendant]

however how do you knows it meant to be pluged as a function instead of multiplied? is there a difference?

[surreal basin]

the difference is parentheses

think of how you tell the difference between $fx$ and $f(x)$

[twin meteor]

mtt

[surreal basin]

that is the difference

the same difference is being used here

[earnest pendant]

i get it now

thank you

took me 2days to understand it 🤦

a bit shocking its simpler than i thought

[surreal basin]

function notation is not intuitive and not designed well

the dthis/dthat notation youre looking at was specifically worked on for years to be intuitive
functions were not

youll expect it next time

[earnest pendant]

unfortunately

.close

[vocal sleet]

Channel closed

Closed by @earnest pendant

Use .reopen if this was a mistake.

[vocal sleet]

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.

[past yacht]

Hi, could any one help me to solve this problem? - Given set S = {1,2,..., 2023}. Consider subset T of S. If T does not contain any number a minus any b can get the result in set E
, then how many elements are there at most, with:

a) E = {3; 6; 9}

b) E = {4; 7}

[vocal sleet]

@past yacht Has your question been resolved?

[vocal sleet]

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.

[rain shale]

A square with a side of 2 m has a circle inscribed in it and in turn this circle has a square inscribed in it. If this square also has a circle inscribed in it, what is the area between the last square and the last circle.

[rain shale]

How do I do this question?

[thick warren]

Did you start with it? Atleast the diagram?

[tired gate]

^

Hints:

1. ||Find the area of the circle inside and square inside of that circle, then subtract them||

2. ||To find the area of the inner square, you know the lengths of the diagonals||

[rain shale]

No, it's ok I got it

.close

[vocal sleet]

Channel closed

Closed by @rain shale

Use .reopen if this was a mistake.

[vocal sleet]

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.

[vocal sleet]

@crimson willow Has your question been resolved?

[vocal sleet]

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.

[vagrant ibex]

What’s the DBM/OVS method when writing a report for comparison in statistics

[vagrant ibex]

UM WHAT DO I DO IM SCARED

[vocal sleet]

@vagrant ibex Has your question been resolved?

[vocal sleet]

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.

[tough onyx]

How can I represent this transition matrix?

I don't get the third dot point

[vocal sleet]

@tough onyx Has your question been resolved?

[pallid forge]

the third point just says that Tasmania and Western Australia's populations have zero migration to each others states

[vocal sleet]

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.

[worldly skiff]

hello

[worldly skiff]

<@&286206848099549185>

would the ans be 70 cm squared

[civic otter]

It's 60

The square has area 16 cm², the rectangle 44

[worldly skiff]

is the thinking correct?

<@&286206848099549185>

[gentle thicket]

It is correct but there's a simpler method

[short pilot]

Yo

[gentle thicket]

here you can calculate the area of rectangle ABCD and then add the area of square CEFG

[vocal sleet]

@worldly skiff Has your question been resolved?

[vocal sleet]

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.

[floral willow]

Normally, a very simple topic, two variables, and two equations

But I ain't getting this

[steel marsh]

for visualisationg u can try

setting p+q = a; pq = b

it will be a lot easier for u to get the solution

[floral willow]

Hmm, I will give this a try rn

Hey, this is working

Thanks you so much

[steel marsh]

ikr 💀

[floral willow]

🫶

[steel marsh]

np ^-^

[floral willow]

.close

[vocal sleet]

Channel closed

Closed by @floral willow

Use .reopen if this was a mistake.

[floral willow]

.reopen

[vocal sleet]

✅

[floral willow]

Hey, it was looking all good

But then it turned into a cubic equation of b

How much complicated can a two-variables and two-equations problem get?

Ik, that we can find the roots of b, but just imagine how lengthy it gonna be

Is there any simple solution at all?

Here's the problem again:

[steel marsh]

huh

;-;

i dont get what ur asking 😭

[floral willow]

Uhm

I solved with your idea

[steel marsh]

ye

[floral willow]

That let a = p+q and b=pq

And I ended up with a cubic equation of b

I put pictures above

[lament linden]

I did it in head and got a quadratic instead

Can you write your steps more clearly? Its subtle to understand current ones

[floral willow]

Uhmm, ok

You clear with question?

[steel marsh]

my idea is just for visualising bro ToT

[lament linden]

Multiply all sides with pq directly

Yeah i am

[steel marsh]

if u write pq and p+q with the same way u get the same prob

so u have to use a different way

[floral willow]

So, you have a way to find answer?

[steel marsh]

oh i havent try it yet

1 sec

[floral willow]

Yeah, please

[thick warren]

The last step can be b^2(b-4)-1(b-4) = (b^2-1)(b-4)

[floral willow]

?

[floral willow]

Oh yeah, how I did not got it

Lemme me try now

[lament linden]

$$ pq + \frac{2q + 2p}{pq} = 4$$ $\implies$
$$(pq)^2 + ( 2q +2p) = 4pq = a^2 + 2b = 4a$$

[floral willow]

What you gonna do with this now?

[twin meteor]

Cyrenux

[floral willow]

Like, how you proceed after this?

[lament linden]

You hold on until you do the same for the second equation

Then subtract or add them

[floral willow]

Oh, lemme try then

[steel marsh]

hmm i just draw a graph

and b has 3 possible roots and all 3 are not rational

like rlly bad number

[floral willow]

Oh

So it might brainfuck to solve these equations

[thick shadow]

in moments like this we let Photomath carry

Jokes aside, it can be useful for finding mistakes

[steel marsh]

i agree..

[steel marsh]

how to get to a new line

[floral willow]

What's this now

[thick shadow]

It's an app that solves problems

[floral willow]

Oh, cool

[lament linden]

Double \

Or auto new line when you use $$

[steel marsh]

oh thanks

[floral willow]

Umm I want to tell y'all how I got this problem/question actually

Because we been really trying for a while

[steel marsh]

🤔

[floral willow]

So,

[steel marsh]

[floral willow]

I was trying to find roots of a cubic equation

[steel marsh]

umhm 🤔 ?

[floral willow]

Like, I then assumed the roots to be p, q, r

[steel marsh]

nice number

[floral willow]

So there will be 3 equations on the basis of relationship between roots and coefficients

[steel marsh]

is it a 3 or a five

[thick shadow]

4

[floral willow]

4

[steel marsh]

wth\

ok

[thick shadow]

wait let me try something

[floral willow]

So, first equation will be p+q+r

Then pq+QR+rp

Then pqr

[thick shadow]

2 is a solution

we can now solve easily

[floral willow]

Ik ik but that is by trial and error

[thick shadow]

not exactly

give me a second

[floral willow]

I wanted to find using relationship between roots and coefficients

[lament linden]

Rational root theorem

[floral willow]

Explain please

Never heard about this, mb

I was bored of trial and error and Cardan's method of solving cubic equations

[thick shadow]

Wait

[lament linden]

If there exists a rational root for the cubic equation ax³ + bx² + cx + d = 0, then its in the form x = +- u/v
Where u is a factor of d and v is a factor of a

Sign doesnt matter

[floral willow]

Cubic?

[thick shadow]

Aren't the solutions 2, 1/2 and -2

At least that's what I get

[floral willow]

Yeah, they are

All corrct

Uhm, how you got em?

Trial and error?

[thick shadow]

Not exactly

[lament linden]

Trial and error with an educated guess

[thick shadow]

Since we can easily see that 2 is a solution

We can therefor

[lament linden]

Try the guesses with the way i said

Factors of constant, over factors of x³ s coefficient

[thick shadow]

Yeah I am searching for the exact name of the method since I forgot

[floral willow]

This?

[lament linden]

Yes

You can google it for better explanation too

[floral willow]

how it's cubic if it is ax^2

Ok it may be a typo by you

[steel marsh]

or u can use a calculator

it gives u the root

[floral willow]

Hah always

[steel marsh]

from that one root u can find

the other 2

wdym always 😭

[thick shadow]

Here it is

I used Synthetic division

[floral willow]

I meant that calculator option is always available

[floral willow]

Ik this

But you gotta know one root for this

[thick shadow]

Yeah

I mean cannot this be accepted as a solution?

Have I been living a lie

[lament linden]

Yeah typo, let me fix

[floral willow]

This seems quiet simple and useful

ig, the issue is kinda sorted

[lament linden]

Fixed

[floral willow]

Thanks y'all for your time

[floral willow]

This one took way more people I think

[lament linden]

Wth white text became grey

💀

[floral willow]

LoL

[lament linden]

Oh its bc background is transparent

[floral willow]

Downloaded the image, will always remember this one

tysm

[lament linden]

Is downloaded version fixed?

[floral willow]

?

Fixed what?

[lament linden]

Or it got gray background still

[floral willow]

Oh, lemme see in photos app

Nah, it's still grey

But it's ok

again, tysm all

.close

[vocal sleet]

Channel closed

Closed by @floral willow

Use .reopen if this was a mistake.

[steel marsh]

.

hmm y do my text turns red

like just right before

[floral willow]

Nah, it's normal

[lament linden]

https://en.m.wikipedia.org/wiki/Rational_root_theorem

[steel marsh]

what happened ;-; !#?

[floral willow]

?

[lament linden]

Wikipedia one is easier to read

[steel marsh]

like y do my texts turn red

[floral willow]

They ain't

At least, it's white for me

[steel marsh]

here

[lament linden]

He thinks i used latex to change color while i sent an image

[floral willow]

It kinda more detailed

[lament linden]

Oh thats a failed text

[steel marsh]

elier from the back but ye

[lament linden]

That wasnt sent

[steel marsh]

💀

is it bc internet connection ?

[lament linden]

Anyway, reminder that -2 is a factor of 4 too, so dont only think of positive numbers while considering factors of a number

[floral willow]

yh

And this is like for making guesses right?

[steel marsh]

ye it is called possible roots

[lament linden]

Yes

[floral willow]

Ok, got it

[lament linden]

Try for x² = 2

It wont work, guess why

(Look at name of the theorem)

[floral willow]

Got it

This won't help with equations with irrational roots

[lament linden]

It wont work if there isnt a rational root

👍

[floral willow]

🫂

[steel marsh]

[lament linden]

It could have a rational root still

Like the polynomial (x-2)(x²+x+4)

x=2 is the only rational root

[floral willow]

Yeah, I meant for the ones which have only and only irrational roots

And one small doubt

I did closed this help-17 or waeva

How are we still chatting?

nvm joined discord yesterday only

New to things here

[lament linden]

It takes a while for help channels to reset

[floral willow]

Oh, ok

[vocal sleet]

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.

[halcyon oracle]

https://www.desmos.com/calculator/cik091pzjx

[halcyon oracle]

hello is there any way of simplifying this equation so that I can easily linearize it?

e.g reducing it to something like this and graphing D^3 against h

.close

[vocal sleet]

Channel closed

Closed by @halcyon oracle

Use .reopen if this was a mistake.

[vocal sleet]

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.

[solar topaz]

is it just me, or there shouldnt be a minus sign in the [-ln(K-P)] and instead shouldnt it just be [ln(K-P)]

[willow pike]

using chain rule, differentiating K - P gives -1, so you need the extra minus

[solar topaz]

oh right i just redid that on my notebook, thanks

[vocal sleet]

@solar topaz Has your question been resolved?

Channel closed

Closed by @solar topaz

Use .reopen if this was a mistake.

[vocal sleet]

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.

[finite hatch]

.reopen

excuse me can I do it like this or is it illlegal in maths

[finite hatch]

.close

[vocal sleet]

Channel closed

Closed by @finite hatch

Use .reopen if this was a mistake.

[vocal sleet]

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.

[lone igloo]

(5-6d)^2

[lone igloo]

Expand equation

Is answer 5^2 - 60d - 36d^2