help-0

[warped topaz]

(-x-9)/1

[tardy stag]

look where the limit is going

[warped topaz]

oooooooooooooh

I forgot to check haha

Ok thanks again

❤️

.close

[lone heart]

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[warped topaz]

https://i.imgur.com/nKqvXqT.png

[warped topaz]

(x-5)(x+3)/(x-5)(x+8)

(x+3)/(x+8)

Wait

I made a miscalculation ahha

nvm sry

.close

[lone heart]

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[last ether]

Goofy

[lone heart]

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[full eagle]

help

<@&286206848099549185>

[full eagle]

okok

[primal lance]

!15m

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@full eagle Has your question been resolved?

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[lone heart]

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[torn isle]

Can anyone teach me how to express b in gamma function, the answer is provided

[torn isle]

,rotate

[torn isle]

I start with something like this

Or what value should I let u? Because initially I let you be ln x, let ignore the negative where I put

[winter bison]

Can anyone explain me what is Laplace Transform and what does it actually do?

[torn isle]

Another channel pls as I’m asking here

[winter bison]

I don't know what is meant by channel I am new to Discord can you explain me

[torn isle]

<@&286206848099549185>

[lone heart]

@torn isle Has your question been resolved?

[jolly isle]

$\int_{0}^{1} \sqrt[3]{\ln x} , dx = -\int_{\infty}^{0} (-v)^{\frac{4}{3}-1} , e^{-v} , (-dv) = \int_{0}^{\infty} v^{\frac{4}{3}-1} , e^{-v} , dv = -\Gamma\left(\frac{4}{3}\right)$

[ocean seal]

sotpau

[lone heart]

@torn isle Has your question been resolved?

[torn isle]

May I know what you have done here?

What do you let u to be?

[jolly isle]

is it not similar to what you have done?

[torn isle]

I got u^2/3

[jolly isle]

yes

[torn isle]

This is 1/3

Oh that one should be 3/2

[jolly isle]

yup I did it wrong in my head

[torn isle]

How you wake v becomes negative?

[jolly isle]

ok so what you did wrong is that you found that 2/3

you should have wrote 1/3

I did it correctly

$\sqrt[3]\ln x = (\ln x)^{1/3}$

[torn isle]

Oh yea

[jolly isle]

I cant type today sorry

[ocean seal]

sotpau

[torn isle]

I wonder how you change the e^v into -ve

And all others to negative

I think the problem now with me is e^u

I need to get something like e^-u then onky I can convert to gamma

[jolly isle]

look at what you wrote for x

and what you found for dx

you had a $e^{-u}$ and changed it

[ocean seal]

sotpau

[torn isle]

I let u=ln x

Then x=e^u

And got dx= e^u du

[jolly isle]

then how did you get those limits of integration?

[torn isle]

That one is the wrong part where I purposely simply put -ve

This is the one I’m working with

[jolly isle]

it should got to $-\infty$

[ocean seal]

sotpau

[jolly isle]

not $\infty$

[ocean seal]

sotpau

[jolly isle]

$\lim_{x\to 0} \ln{x} =-\infty$

[ocean seal]

sotpau

[jolly isle]

you did the limits of integration wrong

and in the wrong order

by letting $u=\ln x$ you get as $x=-0 \implies u\to -\infty$ and as $x=1 \implies u=0 $

[ocean seal]

sotpau

[torn isle]

You mean this?

[jolly isle]

yes and now let $v=-u$

[ocean seal]

sotpau

[torn isle]

Oh, I got know

Is it like this?

[jolly isle]

yes!

[torn isle]

Okay tyvm!!!

Finally I knew my mistakes and need to let u be -v

[jolly isle]

you're welcome!

[torn isle]

Thanks

.close

[lone heart]

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[marsh cape]

Could please explain what I need to do to draw the graph according to the question with examples. Question (All is just one question): (a) Draw an x(t) graph for a particle moving in 1 dimension of your choice.
(b) Choose t1 and t2 times on this graph. Draw a line that represents the average velocity between these two times.
(c) Show a time t3 between t1 and t2 where the instantaneous velocity is greater than the average velocity found in (b). Draw a line representing this instantaneous velocity.
(d) Show a time t4 between t1 and t2 where the instantaneous velocity is less than the average velocity found in (b). Draw a line representing this instantaneous velocity.
(e) Under what condition can't you answer (c) and (d)?

[nimble fern]

I would love to help, but this involves so many concepts behind. I don't know how well you know about movement of 1-D particles, have you tried something that can be shown?

[marsh cape]

I am actully a highschool 1 student and I am taking a course I know a bit as the teacher told me but I couldn't be sure how to do it.

Because of that I could ask some questiones but eventually I will understand

[nimble fern]

that's good

so, have you drawn a graph, like a draft one, as a start?

[marsh cape]

I did draw like 12 graphs but they did not accuire the enough information for the rest of the question

then I got mad and threw the unfortunately

[nimble fern]

ouch

[marsh cape]

I waited lke 15 minutes watching anime to calm

and now I am fine

I decided to take some help

[nimble fern]

glad you feel better

lemme draft one for you then

try start with this one

[marsh cape]

ok wait let me draw it fastly

[nimble fern]

sure

[marsh cape]

ok dd it

[marsh cape]

you fine?

[nimble fern]

back, wife gotta sleep, i was preparing

[marsh cape]

oh, ok fine np. Though life is going on

[nimble fern]

take t1=0, t2=4

[marsh cape]

1 sec

[nimble fern]

so what is the average velocity

[marsh cape]

I lost there. ||sorry||

Just dont know the functions in my memory

I wll look it from the internet fastly

[nimble fern]

ok

btw x is the displacement

[marsh cape]

yeah I know delta x as another name

[marsh cape]

owh ok ok I caugth on I didnt write t1,t2 in the graph because of that I was confused

V = x/t

then avarage velocity for it is 2/4?

[nimble fern]

nah, the average would be
(ending displacement - starting displacement)/(total time)

which is just
(0-0)/4=0

[marsh cape]

owh I thought the avarage for t1 and t2

[nimble fern]

nah, it's between t1 and t2

[marsh cape]

isnt ending displacement 2 and starting 0?

[nimble fern]

both are zero

[marsh cape]

sorry for many questiones I am really trying hard to learn it but why

[nimble fern]

brb wife's crying

[marsh cape]

owh, ok

[lone heart]

@marsh cape Has your question been resolved?

[nimble fern]

sorry for the wait

[marsh cape]

No problem she is your wife. She is more precious for you then everything

[marsh cape]

Lastly asked this

[nimble fern]

x representing displacement

[marsh cape]

Yes

Owh ok I now get it

Ok ok

Then what do I need to do

[nimble fern]

lemme direct myself back to the question

(a), (b) done

now (c)

now it's about instantaneous velocity

so, can you tell from the graph which times are of velocity non-zero?

[marsh cape]

2-3 I guess

[nimble fern]

hmmm, for t=2 to 3,
the partical is always 2 units from the original place

so technically it didn't move

so the velocity is 0

[marsh cape]

Not very sure but isnt t1 and t2 these

Owh

[nimble fern]

yea

[marsh cape]

Sorry I mis understood your tezxt

[nimble fern]

no worries

[marsh cape]

Wait isnt it 0 everytime

Because we start from 0 and end with 0 x

[nimble fern]

lemme try and draw it out

from t=0 to 1, the partical moved from 0 to 2, that's why the velocity during that time is non-zero

same as when t=3 to 4, also non-zero

[marsh cape]

Owh it is very right sorry didnt see it

Easy but I sometimes cant see it

[nimble fern]

it's okay

sometimes, i have to draw it out to figure what's happening

okay, so (c)

[marsh cape]

Same as my old teacher

[nimble fern]

which t would you choose which is greater than the average velocity?

[marsh cape]

Avarage velıcity of all yes?

[nimble fern]

i would choose t=0.5

as the particle is moving from 0 to 1 uniformly at a speed of 2units per second

[marsh cape]

Cause 1 is the best thing that we can divide something and take the biggest result am I right

[nimble fern]

which is larger than 0

[marsh cape]

İt is saying between t1 and t2 in the question though

I guess

Maybe I am misunderstanding

[nimble fern]

find a t such that it is between t1 and t2 meaning
find a time t which is greater or equal to t1 and smaller or equal to t2

which is any number between 0 to 4

[marsh cape]

Owh ok

As I said I misunderstood

[nimble fern]

it's ok👍

so, is it okay for (c)?

[marsh cape]

Yes

[nimble fern]

now (d) is very similar

[marsh cape]

Now we find less then avarage

[nimble fern]

yes

which means we need the velocity below 0

[marsh cape]

Then we gotta find a negative number

[nimble fern]

yes

[marsh cape]

But can V get below 0

[nimble fern]

yes

negative v just mean it's the opposite direction

[marsh cape]

I guess we can choose a negative direction

[nimble fern]

yes

[marsh cape]

Then we can choose -t as the previous one?

[nimble fern]

nah

[nimble fern]

try reading this again to see which times it's moving backwards

[marsh cape]

Between T3 T4

[nimble fern]

the times will be t=3 to t=4

yes

so which number will you pick

[marsh cape]

3.5

[nimble fern]

nice

[marsh cape]

Because othervise it will be 0

[nimble fern]

and the velocity is ?

[marsh cape]

And it is nonzerp

[nimble fern]

yep

the velocity this time is -2units/s

[marsh cape]

About to say that

I dropped my notebook

Found the page and you wrote

:D

[nimble fern]

oh lol

😆

[marsh cape]

Then we found d as well

[nimble fern]

now (e)

there are many conditions, to be honest, that we can "make" (c) and (d) fail

but the easiest for me would be changing t1 and t2

[marsh cape]

Hmmm I guess the most pbvious one is to form a new value to t1 and t2

[nimble fern]

high five

[marsh cape]

(Trying to find gif

[nimble fern]

😆

[marsh cape]

https://tenor.com/view/high-five-apir-news-report-gif-24480072

Yeah found the mood

@nimble fern I am about to finish another question when I am finished with it can I send it to you for you to say me either it is wrong or true

By opening a new help request

[nimble fern]

but imma gonna sleep soon

maybe next time

[marsh cape]

Ok had a great time

Would you mind adding friends I liked your mood very much

[nimble fern]

sure, I won't mind

[marsh cape]

Send it

[nimble fern]

accepted

[marsh cape]

Have a nice night see you another time

[nimble fern]

cya!

[marsh cape]

.close

[lone heart]

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[sand pawn]

help

[sand pawn]

how to solve this?

Guys I have no idea

[minor needle]

you mean proving it's true

[sand pawn]

yeah

[minor needle]

some trig identities required, I'd start from tan(2A)

[sand pawn]

ok

[minor needle]

well, you can only solve this starting from the one side and getting the another?

or you can manipulate both

[sand pawn]

manipulate both sides till both sides are equal to each other'

[minor needle]

okay

so first step should be taking tan( . ) of both sides

and using fact that tan(arctan(x)) = x

[sand pawn]

ok, then?

[minor needle]

RHS is obvious then (just the fraction under arctan)

LHS requires double tan identity

$\DoubleTan$

[ocean seal]

modus7591

[proper vine]

Can anyone help me in gradee 11 functions?

[minor needle]

#❓how-to-get-help

[sand pawn]

use another channel

[sand pawn]

ok dude, what's next?

[minor needle]

it leaves you with

[ocean seal]

modus7591

[sand pawn]

yeah

I got that too

[sand pawn]

then?

[minor needle]

now some algebra stuff, apply tan = sin/cos

and try to do something

also tan(A-B) could be useful

[sand pawn]

did that, but it went to look horrible quick

@minor needle what to do actually?

<@&286206848099549185>

[tacit arch]

Show it

[last mural]

this smells like a weierstrass sub

[sand pawn]

will u even understand my handwriting?

anyways, I am typing it

[last mural]

yep it works and is probably cleaner than what youre doing

sub x=tan(alpha/2) and y=tan(pi/4-beta/2)

and solve for sin, cos

[sand pawn]

wdym?

can you show me?

[last mural]

for example, if x=tan(alpha/2) then sin(alpha)=2x/(1+x^2)

try to also find formulas for sin(beta) and the cosines

then just fill it in and it becomes an identity of rational functions which is much easier

[last mural]

after also doing this

[sand pawn]

after this, I should do a substitution?

@last mural

[last mural]

yes

[last mural]

this one

[sand pawn]

how do I solve for sin and cos?

it's one equation only

I need to prove left hand side is equal to right hand side

[last mural]

https://mathworld.wolfram.com/WeierstrassSubstitution.html

here they give half already

I'm not going to type it out

[sand pawn]

it's very hard for me still

I still am not able to understand how to correlate with that

[last mural]

What do you mean?

Here, it's also on wikipedia with some more steps apparently: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

[lone heart]

@sand pawn Has your question been resolved?

[sand pawn]

still I don't get it

how to prove that problem?

<@&286206848099549185>

oh I get it

so anyways

.close

[lone heart]

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[fervent ferry]

i know i can use the reverse chain rule for 2 functions, but how would it work if i wanted to reverse it for multiple functions? all the examples i've encountered so far only deals with reversing the chain rule once.

[fervent ferry]

is it even a thing?

[young finch]

give an example

you mean integration by parts right

[alpine sable]

Integration by parts isn't reverse chain rule, though, is it? It's u-substitution.

[fervent ferry]

i mean u substitution yes

[alpine sable]

Yes, can you send a concrete example?

[fervent ferry]

mmm not quite concrete but

this is applying the chain rule multiple times

[alpine sable]

Oh, I see what you mean

[fervent ferry]

and in u subst exercises i only do it once

i've never seen a situation where i have to reverse it more than 1

[alpine sable]

You have $f(x) = e^x, g(x) = \sin(x), and h(x) = x^2$. And you are looking at integrating functions of the form $f(g(h(x))$?

[ocean seal]

kenfps

[fervent ferry]

yea yea

[alpine sable]

Well first off, your example is not integrable using elementary functions, so I think we should choose a better one.

[mortal trellis]

these integrals turn impossible very quickly

[alpine sable]

Yeah.

[fervent ferry]

if i were to recognize the pattern i could reverse though , no?

because i guess this is what u subst is teaching: recognize the pattern

[alpine sable]

But to be honest, you can decompose some functions into a multiple chain of functions. For example, let $f(x) = x^2, g(x) = \sin(x),$ and $h(x) = 2x$.

[ocean seal]

kenfps

[alpine sable]

Then you have $\sin^2(2x)$.

[ocean seal]

kenfps

[alpine sable]

Which is really just f(g(h(x)))

[fervent ferry]

mmm not following

this should be the function i want to back integrate

so i dont know the 3 original functions

i mean, yes i can see the functions as decomposed like that as explained on wiki

if i only had
cos(x^2)2x i could do usubst

but in this case i have another level basically.

are those patterns simply too complex for the human brain to recognize?

[lone heart]

@fervent ferry Has your question been resolved?

[placid zinc]

Do you have much experience with u-sub? Try starting by setting u = x²

This is a pretty easy example of u-sub.

[lone heart]

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[junior kraken]

can someone help me out with this, I asked my teacher for help and all she did was repeat the questions to me

[junior kraken]

If anyone could help It would be greatly appreciated

[lone heart]

@junior kraken Has your question been resolved?

[lone heart]

@junior kraken Has your question been resolved?

[lone heart]

@junior kraken Has your question been resolved?

[lone heart]

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• 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.

[novel oriole]

My math question is for a game that I'm making, and I need a way to calculate the probabilities of some things, maybe in an excel spreadsheet if necessary, so that I can model the problem and determine the probabilities of some things when I add modifiers to it

[novel oriole]

I will show the problem along with the details

Here we have 8 elements, and they are aligned at north, south, east, west, and the in-between diagonals. The lines represent interactions between these elements

[fallen verge]

Oh jeez

[novel oriole]

I have reduced those to this set of dominos

Each tile is a combination that represents mutual participation in an interaction

then we have "pure" and a 9th element if necessary

I have already found the abstracted forms here

1.2
1.3
1.4
1.5
1.6
1.7
1.8
2.3
2.4
2.5
2.6
2.7
2.8
3.4
3.5
3.6
3.7

each number represents a tile, where each color has been replaced arbitrarily with a number

here is the question

given a random draw from this set of tiles, if a player draws 3 tiles from another set comprised of an arbitrary selection of these tiles (their build), what is the chance that they draw a tile that shares a color/number with the tile shown

for example, if Re/Con was shown, then Re/Fa would be an acceptable play, as would Re/For, Con/Fa, Con/Dri, and so on, like Dominos

[surreal meadow]

so they have 1 tile from that list

and you get to see 3 more

and what is the chance one of those 3 shares a color with your tile

[novel oriole]

i'll give you the greater picture so it makes sense

the DM is a fortune teller

they are presenting a challenge with a tile that is the issue at hand, representationally

the player needs to ~draw upon themselves~

they are able to draw 3 tiles from their build, which is the collection of tiles they chose to represent their personal leanings

they are seeking to draw a tile that is appropriate for the task

i can tell you that for the corner colors, they're represented 7 times, and N/S/E/W are represented 5 times each

i'm trying to make something such that I can view the mathematical probability of getting a success, given a challenge and build

I can see that the number of tiles drawn matters, not only because it increases chances, but also because you can't draw the same tile twice

additionally, i want this to be a game with multiple players, and you can choose to not participate in a challenge, so it would be nice to be able to consider the chances with multiple decks available

i don't really know how to consider the math on this, the most i can do is painstakingly create lookup tables or something

[surreal meadow]

so they have 1 tile from the list

and they get to see three more

and you want to find the chance that at least one shares a color

[novel oriole]

the dm has a separate set. they can show any tile

but the player has a collection from that full set, that they chose

they draw 3 from their own set

then what is the chance that at least one shares a color

if possible, what is the chance you have multiple matches

[surreal meadow]

so there is 1 tile present, and they get to draw 3
what is the chance you have a color match

[novel oriole]

yes but it is important to note that although you can have the same tile as the one that is shown originally, you cannot duplicate within the set you draw 3 from

that's the distinction i was trying to make

[surreal meadow]

i am aware

[novel oriole]

ok cool

[surreal meadow]

you are drawing without replacement

[novel oriole]

then yes

[surreal meadow]

then you have one tile with colors x and y

how many times are in the deck we draw from

[novel oriole]

Undetermined at this point, but for an arbitrary count, I will say 10.

[surreal meadow]

let’s keep it variable

call it D for deck

or d

[novel oriole]

right on

[surreal meadow]

do we know ahead of time how many tiles have the color x or y

in the deck that is

[novel oriole]

the player's deck is made by the player, as a way of making them more likely to be successful at one thing over another

[surreal meadow]

ok

so yes

[novel oriole]

in this way, it seems to me that when thinking about it more, the question is "how many tiles are relevant vs how many tiles are not"

[surreal meadow]

it’s the opposite

we want to find how many are not relevant

[novel oriole]

aha

[surreal meadow]

then do 1 - the chance of pulling 3 of those

[novel oriole]

ah so, it is the chance of pulling for one color, but also the chance of pulling for another color

each challenge is two challenges

[surreal meadow]

let’s say you know for sure that in your deck

n of the tiles contain the color x or y

then we have that the chance of pulling at least 1 match is

[novel oriole]

( 1 - (n(x) + 1 - (n(y) ) / 2 ?

[surreal meadow]

$\frac{n(n-1)(n-2)}{d(d-1)(d-2)}$

oh it’s flipped sorry

[ocean seal]

maximofs

[novel oriole]

yo is that a in-discord latex parser

[surreal meadow]

oh let me change the variable meaning

n is the number of tiles that don’t contain the colors x or y

[novel oriole]

omg

[surreal meadow]

then that fraction is the probability of drawing something without the colors

to find the probability that you do draw the colors

do 1 - the fraction there

this is also given by $\frac{{n\choose 3}}{{d\choose 3}}$

[ocean seal]

maximofs

[surreal meadow]

and then 1 - that

[novel oriole]

hrmmmm

[surreal meadow]

this reads:
(how many ways we can pick out 3 tiles that have no color matches)/(how many ways to pick out 3 tiles)

[novel oriole]

ok, so the number of tiles not containing an accompanying color, in the whole set, is 19 for each NESW direction, and 17 for each diagonal

but among d it'll be known for you how much you have of each thing

[surreal meadow]

say you have an 18 tile deck

and you are shown a tile with colors x and y

and the deck has 12 tiles that do show those colors

then n = 18 - 12 = 6

then the probability is

,w 1-(6 choose 3)/(18 choose 3)

[ocean seal]

Results provided by WolframAlpha

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Requested by maximofs#0

[surreal meadow]

so we have a 97% chance to see our color

[novel oriole]

damn

i did not know about this choose 3 function but I bet it's in excel or something, and that I can even find a way to just use it within an array

thank you, you've been so helpful

[lone heart]

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[eager shell]

How are these equations the same?

[wary stream]

Distribute (h + 1)^2 to each term inside the parentheses

[eager shell]

How?

[naive crystal]

Its like a(a+b) => a²+ab
Consider (h+1)^2 as "a"

[ocean seal]

mellowdramallama

[sour dove]

beat me to it 🙂

[fathom grove]

nah you got the full explanation

[eager shell]

Thank u ❤

.close

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[carmine bronze]

hola. Some discrete math help por favor

[carmine bronze]

help on 5

I have

p->(qvr)
q

∴ ¬q->r

[young finch]

i like your name

[carmine bronze]

sorry but it's not for sale

[young finch]

keep it then

[carmine bronze]

IT crowd reference

[tardy stag]

that's correct

i mean, that's the right translation

[carmine bronze]

Well now I gotta prove it using rules of inference and laws of logic

[tardy stag]

yep either prove or disprove it

[carmine bronze]

I was gonna go with domination law or something but idk that may not work

I know I gotta use q somehow

was gonna do modus tollens

wait a minute...double negation?

[placid zinc]

Note the question is a true/false

[tardy stag]

idk what laws you have to work with but I normally just translate all the -> into conjunctions

[carmine bronze]

well I think it's true, tryna prove it now

righto. sooo alright I got ¬p v(qvr)

conditional identity

[placid zinc]

Note that v is associative, so you can drop those last parenthesis. Writing this is valid:
~p v q v r

This means at least one of the following is true:

• I don't study hard
• I get A's
• I get rich

Why does the second line kind of "eliminate" the first?

[lone heart]

@carmine bronze Has your question been resolved?

[carmine bronze]

can't q happen even if p is false?

if p is false, q can be either T or F, doesn't matter

[lone heart]

@carmine bronze Has your question been resolved?

[carmine bronze]

anybody?

[carmine bronze]

hmm double negation path didn't work out

[tardy stag]

here's a foolproof way to do it, if a bit annoying
you have {[p -> (q + r)] & q} -> [~p -> r]
so you can prove that by again just converting all of the implications into their logical equivalents and seeing if everything falls away or if you're left with something

[carmine bronze]

wth is this

I've never seen this

what is this?

[tardy stag]

oh i'm using + and & instead of V and /\

because I hate those symbols

but I took your entire setup, with both premises and the conclusion, and translated it into logic

[carmine bronze]

so far I just have

1. p->(q&r) Hypothesis
2. ~p+(q+r) conditional identity, 1

[tardy stag]

what do you get if you combine that with your other premise?

[carmine bronze]

I can just chuck q in there? wouldn't I need to follow some laws or something?

I was looking at how I can do it with disjunctive syllogism but it doesn't seem like it would work

wait a minute

I can do addition? so q+r

nah that wouldn't do anything

[carmine bronze]

well slap my ass and fuck me sideways it was false all along

ended up doing a truth table

is false

thanks for the help guys

.close

[lone heart]

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[iron oasis]

I have a question. I'm not sure if it belongs here, but I'll try. I have two curves. One is the circular arc curve y1 = sqrt(1 - (x - 1)^2) and the other is the identity function y2 = x. I would like a set of curves that "interpolates" between these. You can imagine the circular arc curve slowly flattening to meet the identity function. What's the best way to achieve this? I could use Cartesian coords or polar, but in the end, I would like to have something relatively simple I can program. Please help if you can.

[waxen flame]

Off the top of my head, treat the semicircle like an ellipse that flattens onto the x-axis. As it flattens, rotate that equation 45 degrees.

[iron oasis]

I'm not sure how to translate that into a function, but I'll think about it. Thanks for your help.

[sly mantle]

we can define d=y2-y1 and plot y1+cd for values of c between 0 and 1

[iron oasis]

I believe that makes the curve quite flat near x = 1 though correct? I was hoping for a circular shape that sort of "deflates" (uniformly?) towards the identity function.

I was thinking a polar function might be good? Not sure how to define a polar function for a line though. It's been a while since I studied that stuff.

Here's the plot of the function. It's not exactly circular but it's close.

[tardy stag]

a polar function can't make a line that goes through the origin

[iron oasis]

Good to know thank you

If we consider (1, 0) in the graph I posted as the origin, is it possible to get the in-between curves as a polar function?

[tardy stag]

how much do you care about where it is located?

it's not possible like that since the same radius would have multiple points

[iron oasis]

I don't care at all, I just need the shape

[tardy stag]

ok hang on let me try something

[iron oasis]

Please and thank you!

[tardy stag]

going to move it to quadrant 2

[iron oasis]

Any rigid transformation would be lovely

[sly mantle]

im not sure what you mean

[iron oasis]

Imagine a line of symmetry that runs through y3 = 1 - x. I want the curve to flatten so it stays symmetric

[tardy stag]

this can probably be simplified but https://www.desmos.com/calculator/91tu3uqczh

[iron oasis]

WOAH DUDE! YOU'RE THE BEST

This is for an image processing app. Would you like a demo when it's done or nah?

[tardy stag]

yes ofc!

i'm very curious how the points move on that curve so an image would be helpful

[iron oasis]

Can I friend you to DM you an image or short video? Or can you just DM anybody out here?

[tardy stag]

right now i have my DMs turned on but I might turn them off so just ping me in #discussion or whatever

[iron oasis]

What channel is that? It says No Access for me

[tardy stag]

oh it's # discussion

[iron oasis]

Gotcha. I'll do that. Thank you so much. That would've taken me hours surely

[tardy stag]

why don't you have access whatever

[iron oasis]

I disabled social channels when I joined. I can probably just turn that back on

[tardy stag]

smart

[iron oasis]

thank you and goodbye for now

.close

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[nocturne robin]

.close

[gilded magnet]

Nice name

[nocturne robin]

Thanks

[gilded magnet]

It's Felix ryt

Without the yn

[nocturne robin]

what does ryt mean?

[gilded magnet]

Right

[nocturne robin]

Yes, without yn, it's Felix

[lone heart]

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[sterile storm]

Can anyone recommend a website to take a complete course in python?

[tacit arch]

Don't use help channels for non math things

.close

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[fading ivy]

determine the formula of function f(x) indicated by the blue line

[tacit arch]

Pretty similar to
https://en.m.wikipedia.org/wiki/Sawtooth_wave

Modify sawtooth to fit your function

[fading ivy]

sorry i dont get how it works

[tacit arch]

What part

You just need the function definition

https://youtu.be/OYOXMyFKotc

https://youtu.be/Uzw9tsGq2Pw

Watch those two

[lone heart]

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[fading ivy]

thanks

[lone heart]

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[fallen verge]

.close

[lone heart]

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[fallen verge]

No trolling

[lone heart]

✅

[fallen verge]

Im too tired for this

.close

[lone heart]

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[fallen verge]

Lol thanks

[lone heart]

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[distant meteor]

Chemistry question: how do I transform cubical decimeters in liters? And how do I know if and when I have to do it? Also after finding something, how do I know the measuring unit?

[wind cloak]

,w cubic decimeters to litres

[ocean seal]

Results provided by WolframAlpha

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Requested by neonperseus#0

[wind cloak]

depends on the units you used to find it

If I say $m_1V_1 = m_2V_2$

[ocean seal]

neonperseus

[distant meteor]

Also, does it change between different fluids?

[wind cloak]

And I plug in mass in kg, and volume in cubic lightyears, my answer would be in cubic lightyears and kg

[wind cloak]

nope

[distant meteor]

😯

[wind cloak]

The density may change but 1 l of liquid is always 1 l of liquid

[distant meteor]

So, 3,45 dm^3 of nitrogen are 3,45 L?

[wind cloak]

yes

[distant meteor]

Thanks

Oh another thing

How do I know which measuring unit do I have to use to have a correct result

For example, do I have to use liters or m^3? pascals, ectopascals, bars or atmospheres

[wind cloak]

it's a matter of convience

chemistry uses CGS a lot since it makes sense

also a bar and an atm is basically the same thing

[distant meteor]

So, the numeric result doesn’t change using different measuring units

Right?

[wind cloak]

well the quantity doesn't change

But there will be a difference with different units

1 m != 1 cm

[distant meteor]

I suck at maths

😭

Basically I have this

And I need to find the final temperature

M remains stable, and density too (I think)

Also it’s nitrogen

Temperature and pressure change

And seems like V changes too, but I know only the final one

What can I do to find the final temperature

[lone heart]

@distant meteor Has your question been resolved?

[distant meteor]

<@&286206848099549185>

Pls

[lone heart]

@distant meteor Has your question been resolved?

[tardy stag]

if you don't know the initial volume I don't think this is solvable

oh you can use m and d to get the volume maybe?

and then $PV = nRT$ for the start to calculate $n$ and then use $PV = nRT$ again to calculate $T_2$

[ocean seal]

Hayley

[distant meteor]

Thanks

[lone heart]

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[prisma tulip]

Yo, anyone help me;
using mathmatica to calculate these partial derivatives but they seem wrong?
https://gyazo.com/d17c20c02107cdf16c8f4d3ec81a0af5

[prisma tulip]

https://gyazo.com/070a282ab8ac9c341fa8a44f8b7c8ce6

But I get this on derivative calculator?

[tardy stag]

not quite sure how that even gave you anything
the syntax is D[x^x + y^y - 1000, x]

[prisma tulip]

no way

Thank you @tardy stag

.close

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[prisma tulip]

.reopen

[lone heart]

✅

[prisma tulip]

What's wrong in this picture

:L

Is there a like mathmatica guide to formatting

[tardy stag]

surprised you're not using Mathematica's formatted input

that was my favorite thing about it

i think you might need to say b*z - b*c*t in that mode?? not sure

or it might not recognize sech? probably capitalized Sech

[prisma tulip]

Formatted input?

:L

[lone heart]

@prisma tulip Has your question been resolved?

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[placid swallow]

hi

[cinder thicket]

Hello Thomas

[placid swallow]

Can anybody help me figure out whats the time complexity of the following algorithm? G is a graph, w : V -> R is a function that maps every node to a weight, pop is theta(1).

this the work ive done so far

[placid swallow]

Hello

[cinder thicket]

Damn can't help with whatever that is

[wary stream]

You might be better off asking in a computer science server, check #old-network

[torn elk]

the while loop should be O(V^2) i think

[lone heart]

@placid swallow Has your question been resolved?

[tardy stag]

specifically O(|E|)

which is probably O(V^2)

[lone heart]

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[brave junco]

I'm having some trouble with this one

[brave junco]

I've gotten that C_n+1 = (C_n-1)/(n+1), where plugging in for n=1 solves for C_2

so n=3 -> C_4 = C_2/4 = C_0/8

but I'm not sure how so solve for even values of n other than filling in the blanks based on C_2,4, and 6 since n=2 results in C_3 = C_1/3

and I have no idea what C_1 is

[upbeat hornet]

this equation is separable, you could solve it the normal way to check your answers

[alpine sable]

Yeah, but I think the question is asking for a series' solution

In that case, we have $\sum_{n=1}^\infty n a_n x^{n-1} =\sum_{n=0}^\infty a_n x^{n+1}$, I think?

[tardy stag]

a_n but yes

[alpine sable]

oops!

[ocean seal]

breathe88rising

[alpine sable]

So, then we can do $\sum_{n=1}^\infty \left(n a_n x^{n-1} - a_n x^{n+1}\right) - a_0 x = 0.$

[ocean seal]

breathe88rising

[tardy stag]

yes... that doesn't seem to help us much though

[alpine sable]

The tricky part is rearranging this sum so that we can obtain a recursive sequence

[tardy stag]

oh i see it

[alpine sable]

You do? Go on then

[tardy stag]

you have two seeds, the linear term and the constant term

[brave junco]

Fellas, I'm sorry but what you guys are saying is math jargon to me right now

[alpine sable]

Have you done series' solutions to differential equations before?

[brave junco]

Just now learning it, I think

[tardy stag]

start with the series on both sides (either use sigma notation or write out the first like 4 terms) and ask "how does the constant term get balanced" then "how does the x term get balanced" then "how does the x^2 term get balanced" etc

[brave junco]

if it helps, this is the method the textbook is applying

that's kinda what this is doing, right? ^

[alpine sable]

I see it too. The left sum becomes $\sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n$.

[ocean seal]

breathe88rising

[alpine sable]

By shifting the index of the sum

[brave junco]

oh I forgot to add 1 to the n in the first series

but how can I use that so solve c_1 or c_3 in terms of something that can be plugged in back for c_0?

[tardy stag]

well... it doesn't have to depend on c0 it just can't depend on anything else

[alpine sable]

$\sum_{n=0}^\infty (n+1) a_{n+1} x^n = \sum_{n=0}^\infty a_n x^{n+1}$

[ocean seal]

breathe88rising

[alpine sable]

Now you need to balance the powers of $x$

[ocean seal]

breathe88rising

[alpine sable]

I fear I've done too much and you won't learn if I do the rest for you

[brave junco]

I already combined the series I think, that's how I got

right?

[alpine sable]

I don't know. I haven't done the work necessary to come to that conclusion. But if you think that's right, then all you need to do is just plug in numbers

[brave junco]

uhh one sec I'll upload my current work

[alpine sable]

go on

Please don't send it as a file

I won't download one

[brave junco]

trying to rotate it one sec

took out the n=1 for the first series (c_1)

shifted them to have the same starting value for the series and same x^n

then combining them I got the c_1 (series bracket)[C_n+1(n+1)-c_n-1)x^n]=0

so dividing x^n and c_1 over I was left with

C_n+1(n+1) - c_n-1 = 0

into

but I'm still not sure how I'm supposed to solve even values of n in terms of c_0

since plugging in for n=2 -> c_3 = c_1/3

and I don't have values to plug into c_1 to get back to terms of c_0

[tardy stag]

what are the first few terms of y'?
what are the first few terms of xy?

[brave junco]

y' starting at 1-> C1, C2 2x, C3 3x^2, C4 4x^3

xy starting at 0-> C0 X,

oh

[tardy stag]

ye

[brave junco]

okidoki, I think I got it, thanks

[tardy stag]

[brave junco]

.close

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[pastel prism]

what is the formula i need to find c, i dont remember

[lone heart]

@pastel prism Has your question been resolved?

[fathom grove]

i believe it’s the formula for the barycenter which i do not know and recommend you google it

[pastel prism]

okay
i did it making an ecuation with the barycenter formula

.close

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[somber nimbus]

Can anyone please help me understand how they get the eigenvectors for lambda = -2 in this example

[alpine sable]

Do you know how to find eigenvectors?

[somber nimbus]

I thought by just solving for x1 x2 and x3 and putting them into parametric form

[alpine sable]

Yes, you do solve for x_1, x_2, and x_3

[somber nimbus]

Then x_1 = -x_2 -x_3 etc, right

[alpine sable]

You are given $A = \begin{bmatrix} 1 & 3 & 3 \ -3 & -5 & -3 \ 3 & 3 & 1 \end{bmatrix}$. To find the $\emph{eigenvalues}$, you find the determinant of the matrix $A - \lambda I$ and solve the characteristic polynomial. Then, once you've the eigenvalues you subtract them from A using the identity matrix and then solve $A\mathbf{x} = 0$. Does that make sense?

[ocean seal]

breathe88rising

[somber nimbus]

It does yeah

[alpine sable]

So, you basically are stuck with the final step? (Solving Ax = 0)

I shouldn't call it A... It's $(A - \lambda I)\mathbf{x} = 0$.

[somber nimbus]

Yeah, I don't get how they end up with the v2's

[ocean seal]

breathe88rising

[alpine sable]

For the lambdas you've found

[somber nimbus]

(-1, 1, 0) and (-1, 0, 1)

[alpine sable]

Yes

So, can you show your work?

[somber nimbus]

v1 I get

Sure 1 sec

[tardy stag]

,rccw

[lone heart]

@somber nimbus Has your question been resolved?

[somber nimbus]

@alpine sable

[worn fox]

none of your vectors satisfy x1 + x2 + x3 = 0

[somber nimbus]

What should I do?

[worn fox]

well for example you let x3 = 0 and see what that results in

[somber nimbus]

Then x1 is just -x2?

[worn fox]

yeah

[worn fox]

note thats the first option here

then you're looking for another independent solution, setting a different variable = 0 will do that for you

[somber nimbus]

I don't really get why the top entries of the v2 vectors are -1 though

[worn fox]

well if you need x1 = -x2 then just let x1 = -1, and x2 = 1

its not a unique choice

[somber nimbus]

I got it!

Thank you

Dunno why I couldn't wrap head around it for so long

[worn fox]

yeah you can always times everything by a scalar and it still be an evector

[somber nimbus]

Alr, ty 😉

.close

[lone heart]

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[warped topaz]

https://i.imgur.com/SalqO8s.png

[warped topaz]

What does function rule even mean

I know the derivative is x/2

[long axle]

It just means the function

k(x) is the function tangent to the curve at (2,7)

Just that dotted line

[warped topaz]

Ok, so wouldnt it just be x/2?

Cause its the derivative

[indigo lotus]

they are asking for the equation of that dotted line

[warped topaz]

And the function tangent is just the derivative

[long axle]

That’s the function that represents the slope of any point on f(x)

[somber nimbus]

y = ax + b

[warped topaz]

Oh so I just plug in the values to find the y intercept?

[long axle]

But u have to plug into the derivative x/2 to find the slope at (2,7)

[warped topaz]

Ok I will try 1 moment

[indigo lotus]

could do that without too as we know the two points the line passes through

[long axle]

Yep

[warped topaz]

https://i.imgur.com/DVLLLrI.png

Why is my slope not correct

I subbed in 7=(2)(x/2)+c to get c=6

So then the equation should be x/2+6, no?

[warped topaz]

I did this but apparently its wrong? Or did I make a mistake?

I got it

.close

[lone heart]

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[alpine sable]

How do I do this question

[alpine sable]

Wait

I messed up the perfect square formula

It’s still wrong tho

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

.close

[lone heart]

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[tender anchor]

Did I do this right?

[tender anchor]

LHS meaning left hand slope

[crimson totem]

i think 4th point is diffrentiable

[tender anchor]

o

why would that be?

[crimson totem]

i might be wrong though, but i highly believe 4th point is diffrentiable

[crimson totem]

you would get a almost perpendicular slope

[tender anchor]

oh shoot yeah

is 1 not differentiable?

[wet nest]

Yeah

Sharp corner

[tender anchor]

ah

ty guys

.close

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[round wind]

hi

[round wind]

the q

[lavish sedge]

hi

[round wind]

rationalise

[gilded magnet]

Were not ur workers

[lavish sedge]

multiply each fraction by its conjugate and simplify

[crimson totem]

brother chill

[round wind]

who asked u to help baka

[gilded magnet]

Ok

[gilded magnet]

!status

[lone heart]

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

[round wind]

how

[gilded magnet]

!show

[lone heart]

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

[round wind]

lol

[gilded magnet]

Mult den and nu by √7-√2

And √3+√2

[crimson totem]

brother chill

[gilded magnet]

And √5+√3

[lavish sedge]

like this

do this for every fraction

[round wind]

thanks

[lavish sedge]

should simplify pretty nicely

[lavish sedge]

np

[gilded magnet]

This server ain't meant for spoonfeeding

[lavish sedge]

can you shut up

[crimson totem]

True. +1

[round wind]

damn

[gilded magnet]

Bruh it's what all mods point out u should shut up

[crimson totem]

i forgot the command...

!helper

[lone heart]

Please read #❓how-to-get-help

[crimson totem]

hm

[lavish sedge]

🤓 👆

[crimson totem]

there was a command for this

[gilded magnet]

You can Google that but I'll shut up now

[ebon sparrow]

nosolos?

zamn

[crimson totem]

!nosolos

!nosols

[lone heart]

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.

[crimson totem]

yep

thnx

[ebon sparrow]

yeah i think i miss spelled lol

np

[round wind]

@lavish sedge thank u

[lone heart]

@round wind Has your question been resolved?

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[hollow zephyr]

how to prove that the directional cosine matrix is orthonormal?

[hollow zephyr]

thank you

.close

[lone heart]

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[tender anchor]

Derivative of f(x) = 3 / 5x^3

[tender anchor]

How do i start it?

[gray isle]

do you mean
$$\frac35 x^3$$
or
$$\frac{3}{5x^3}$$

[ocean seal]

ℝamonov

[surreal meadow]

!nosols

[lone heart]

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.

[tender anchor]

could u explain how you got that please

[surreal meadow]

@tender anchor answer this

[tender anchor]

o second one

[surreal meadow]

what differentiation rules do you know

[tacit arch]

can you not copy paste from ai engines like chatgpt

[round wind]

Which class is this from

[tacit arch]

not buying it

<@&268886789983436800> and their history

[gray isle]

nor should you be giving out solutions

regardless

[round wind]

Chill who cares

[tender anchor]

only the power rule

[tacit arch]

read #rules

mad because you won't do your own thinking

[ebon sparrow]

?

[surreal meadow]

do you know another way to write x^-3

[tender anchor]

-3x^-2?

[safe tartan]

Lol caught red handed

[surreal meadow]

well that’s the derivative

[sly mantle]

@stray hearth please dont post full solutions. instead talk with them to guide them to an answer

[tender anchor]

oh

no

[surreal meadow]

i’m asking more like

[tender anchor]

uh

wait

[sly mantle]

@ebon sparrow gpt doesnt actually answer whether they wrote a response

[ebon sparrow]

I see

[surreal meadow]

do you know what a negative exponent represents

can you guys take the gpt convo elsewhere

[round wind]

What country

[tender anchor]

wait so it would be

1/x?

^3

[sly mantle]

ok so now thats out of the way, lets move on from here

[tender anchor]

like if its x^-2 its 1/x^2?

[round wind]

Trust her

[surreal meadow]

yeah that’s the idea

[sly mantle]

only block and those wanting to help should talk here

[surreal meadow]

so try writing your original function in that form

[tender anchor]

so like do the inverse of that kinda

[surreal meadow]

yep

can you try that

also don’t worry too much about the 3/5, if you recall we can pull out the constants before we take the derivative

[tender anchor]

so it turns to x^-3?

[surreal meadow]

right

(with the constant in front, but yes)

[tender anchor]

yeah

[surreal meadow]

so now we have a new form

where we can just apply the power rule

what would the derivative of (3/5) * x^-3 be

[tender anchor]

-3(3/5)x^-2

[surreal meadow]

yep

[tender anchor]

wait

no its -4 right

[surreal meadow]

oh

yes

[tender anchor]

since u -1 but its a negative

[surreal meadow]

good catch

and then if you need to you can turn it back into a fraction