help-23

[proud zealot]

now whats next?

[plucky elk]

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-3/v/sigma-notation-sum

[proud zealot]

or is the ...

right.

khan academy was never great at explaining

for me at least

thank you

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[teal yarrow]

I'm trying to study for my math test and I can't remember how to do this part of the test

Ik I need things like dr/dt

dr/dt = 3.4 in this case right

[dapper nacelle]

pi * r² = A
differentiate both sides wrt t

[teal yarrow]

wrt t?

[dapper nacelle]

thats how youll get dr/dt

2*pi*r*dr/dt = dA/dt

you have dr/dt = 3.4

[teal yarrow]

ohhh

i remember now

what about for b?

@dapper nacelle

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[coral dagger]

These functions seem mad useless to me, what uses do they have? Is it more so "if we don't know the sin but we know the csc, find da y coordinate" thing?

[unkempt blade]

and whats the question?

[plucky elk]

you'll probably never need cosecant outside math/science

[edgy isle]

I mean they all still have their geometric meanings

it's just sin, cos and tan are sufficient to solve any problems, pretty sure, sometimes cot may be useful

[coral dagger]

Word, and I think the rest of the functions are more so for just exams n shi like "if you know cot t" find " tan t" r sum idk

Thanks gus

guys

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[coral dagger]

Oooh

[lean otter]

Is this related to quadrant I ?

If the right angle triangle falls on the first quadrant , then sin,cos and tan will be +ve

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[plucky elk]

sir this is a math help channel

[lean otter]

?

[plucky elk]

not your mailbox for people

[lean otter]

be kind

What topic are you refering to?

[plucky elk]

if you're just gonna be an asshole and not ask a math question, go somewhere else

[lean otter]

i havent learnt that

[lean otter]

Im in highschool

I need help with this

@stiff halo

[warped roost]

i recommend starting your own help channel

#❓how-to-get-help

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[queen mango]

a, b in $Z^2$:
$$(E): 18a + 23b = 2001.$$

1. prove that for every couple $(a, b)$ solution to (E) a is a multiple of 23 and b is a multiple of 3

[flat frigate]

<rajel />

[queen mango]

i dont know how to start

[obsidian oracle]

it's mostly gauss's lemma

if you show that 23b is a multiple of 3 for example

then, knowing that 23 and 3 are coprime...

[queen mango]

23b isnt multiple of 3

e.g if b = 1

[obsidian oracle]

is that possible though?

if you have 18a + 23b = 2001

[queen mango]

idk , a would find a way ?

[obsidian oracle]

really?

[queen mango]

can u prove it

[obsidian oracle]

if you wanna find a counterexample, go through with it all the way

if you take b = 1

then knowing 18a + 23 * 1 = 2001

you have 18a = 1978

is 1978 a multiple of 3?

so I doubt it's gonna be 18 * something

[queen mango]

ok i see, but thats not enough i guess

how would i prove that 23b is a multiple of 3

[obsidian oracle]

well... what is 23b equal to

[queen mango]

2001 - 18 a

[obsidian oracle]

and isn't that a multiple of 3?

[queen mango]

fuck yeah

[obsidian oracle]

alr

so 23b is a multiple of 3

maybe write explicitly 2001 - 18a = 3 * (...)

[queen mango]

yep

23b = 3 (667 - 6a)

[obsidian oracle]

correct

so how do we go from 23b is a multiple of 3

to b is a multiple of 3

[queen mango]

i guess their is smth like if $a | b => a |bc$

[flat frigate]

<rajel />

[queen mango]

meaning $a =kb => a =kbc$ and c here is 23

[flat frigate]

<rajel />

[queen mango]

a|b: a divide b

@obsidian oracle what do u think ?

anyway give a hint if what i said isnt totaly correct

[obsidian oracle]

it's the other way around

you have 3 | 23*b

and you want to end up on 3|b

I can refer you again to gauss's lemma

maybe it has another name for you

you have to use that gcd(3,23) = 1

[queen mango]

Alr thx @obsidian oracle

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@quaint grove Has your question been resolved?

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[fervent fog]

can i get help with this task?

[fervent fog]

translation is find x ... so that ...

i understand almoast everything, its just at the end when some random numbers is intertwined

solution in norwegian

<@&286206848099549185>

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[main mural]

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[main mural]

this is chinese remainder theorem if i am not mistaken

[fervent fog]

yea, but i cant seem to figure it out

and yes it it that

but how does he uise euler to get the values "1066, 1353 and 780"?

use*

[gilded vault]

1066 is the number which is 0(mod 13,41) but 1(mod 3)

Similarly

1353 is 0 mod 3 and 41 but 1 mod 13

[primal parcel]

Quick question, what does f(x,y) mean ?

[fervent fog]

okay okay, but how do you calculate 0(mod 13,41)?

is it eulers method? so 41=3*13+2

and so forth?

[queen saddle]

you just multiply 13*41 by its inverse mod 3.

as 13 * 41 mod 3=2. It's inverse is easy to guess: 2 * 2 = 1 (mod 3)

[fervent fog]

then how do you get 1066?

[queen saddle]

so, it is 13 * 41 * 2=1066

[fervent fog]

so its

1341 the inverse

13

*41

• the inverse

[queen saddle]

(13 * 41)^(-1)=2^(-1)=2 (mod 3)

[fervent fog]

and then we have 13 * 41 * the leftovers of the inverse of the two previous mentioned numbers 13 and 41?

[queen saddle]

yes. this number is the 0 mod 13 and 0 mod 41 as it is divisible by them. And it is 1 mod 3

[fervent fog]

if it were just two then?

so if we for example drop the 7x=10(mod41)

how would you then solve this?

[queen saddle]

first you make it of the type x=a (mod 41).

So you multiply by inverse to 7 mod 41. which is 6.

6 * 7x=6 * 10 (mod 41)
42x=60 (mod 41)
x=19 (mod 41)

[fervent fog]

so the answer to that is 19?

[queen saddle]

to 7x=10(mod41) yes

[fervent fog]

and if you now add on 2x=2(mod3)?

[queen saddle]

just divide by 2. Or multiply by 2

its the same in this case

[fervent fog]

okay okay i think i get it

thank you very much

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[snow stag]

no idea where to start

[rustic goblet]

just differentiate term by term

[snow stag]

wdym

like the right handside

[rustic goblet]

mhm!

the series is just an infinite sum

we know how to take the derivative of a sum

[snow stag]

how?

[rustic goblet]

,, (f + g)' = f' + g'

[flat frigate]

higher!

[snow stag]

like this

[median vigil]

note that only x is a variable here

[snow stag]

so on questions like these I only have to differeniate the x

not the with the exponent of n

[median vigil]

yes, because in each term n is a constant

[snow stag]

ohhhh ok

that makes sense thanks

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[frail beacon]

Hey y’all. Why does the answer show x^3 and x^4 cancel out just to leave an x?

[lucid stream]

x^4 doesn't cancel lut

you take 3 factors of x out to cancel with the x^3

that leaves 1 x remaining

[frail beacon]

Ohh oka tysm

I should have written out the xs then

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[lyric blaze]

so like why is there a 1dy there

oh i put it in terms of x

so i have to do 1 - tan(y)

and then integrate it from 0 to pi/4

nvm then

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[round maple]

I don't understand how to set up the integral equation. All I have so far are the intersection points at pi/3 and 5pi/3.

[tame raft]

you want to integrate over the difference in r wrt. theta to get the area basically

[round maple]

Sorry, what is r wrt?

[tame raft]

wrt. means 'with regards to'

so if you have integral x^2 dx the integral is wrt. x

integral sqrt(t) dt, that's wrt. t

just an abbreviation

so here you want to integrate wrt. theta

[round maple]

Oh sorry, I was unfamiliar with the abbreviation.

Well what I don't understand is what I put in the integral.

Would I be integrating from pi/3 to pi/2 for:
1/2 * ((1 + cos(theta))^2 - (3 cos(theta))^2))?

[tame raft]

yyyes i think so

[round maple]

I'm unsure how to choose the bounds and the order of the curves in the integral (Which curve - which curve)

[tame raft]

well you want the big one minus the small one

so it's not negative

i mean you could do small - big and then just absolute value

it doesn't super matter

[round maple]

I can't tell which one is which 😅

Is it the longer one?

[tame raft]

well in the diagram you can see that one of them is further out than the other, in the region that you care about

so that one will have a bigger r^2

[round maple]

Okay, that makes sense.

How do I choose the bounds of the integral?

[tame raft]

how did you get pi/3 to pi/2 before

talk me through it

[round maple]

So the intersection starts at pi/3. And then I follow the curve, and it seems like it's a quarter of the limacon, so it should end at pi/2?

[tame raft]

so yeah i think you want to do one integral from pi/3 to pi/2

and then another integral from pi/2 to pi

[round maple]

Okay, got it thanks so much!

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[quiet plume]

So here I'm kind of stuck at part b).

Given a model $\mathcal{A} = (A, I) $ of $\Gamma$, and given that $A$ is infinite, I've invoked Loweheim-Skolem to get a countable model $\mathcal{A}'$ for $\textbf{Th}(A) = {\phi \text{ closed} : \mathcal{A} \vDash \phi}$.

It's pretty clear that for $\phi$ closed, if $\mathcal{A} \vDash \phi$, then $\mathcal{A}' \vDash \phi$, but I can't seem to show the converse.

What am I missing?

[flat frigate]

Azyrashacorki

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@quiet plume Has your question been resolved?

[quiet plume]

Wait. Is Th(A) necessarily complete?

[quiet plume]

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[quiet plume]

I'm back working on this.

I'm still stuck. I have no idea how to show the backwards implication.

Sorry my <@&286206848099549185> friends

[rustic goblet]

I don't know any model theory, but wouldn't it be easier to find help in #foundations?

[quiet plume]

You are probably right! I seldom ask for help here so I'll try my luck there and see if someone has an idea.

Thanks!

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[empty gyro]

model theory scares me

[rustic goblet]

me too

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[tall lance]

trig identities

[tall lance]

not sure what to do for the last couple of steps i dont think i should expand

[fathom kraken]

First off, cos^4 + sin^4 does not equal (cos^2+sin^2)^2 lol

(a+b)^2 = a^2 + 2ab + b^2, not a^2 + b^2

[tall lance]

oh shii

my bad

oh yea

[lime dust]

Just factor the original

[tall lance]

from that do i get 2 or do i do something else

wdym

[lime dust]

It is one step exercise

[fathom kraken]

Hint: this

[tall lance]

do you mean like a^2 ab b^2

[fathom kraken]

||treat sin^2 as a, cos^2 as b||

[tall lance]

that factoring

[fathom kraken]

yeah

[tall lance]

ok thanks

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[civic raven]

does anyone know how to do this

I need to “determine the most general type” of this

$\lambda x . \lambda y . (x y 0)$

[flat frigate]

m.

[civic raven]

$\lambda x . \lambda y . (x y 0)$

[flat frigate]

m.

[civic raven]

basically x(y(0))

[versed wave]

you might have better chance asking in #foundations

[civic raven]

cool

thanks

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[fluid tiger]

the answer is 441/16. But idk how to get that answer

[fluid tiger]

i put x=a/b, y=b/c, y=c/a to make it easier to solve

[desert pasture]

I think you can use linear algebra to solve this

think about an appropriate vector space equiped with an inner product

[fluid tiger]

idk my teacher hasn't taught us this

[desert pasture]

hmm

what have you learnt

[fluid tiger]

some inequalities like am-gm, bunyakovsky

smth like that

[fluid tiger]

my teacher gave me this solution

idk why P = 3+ (a/b)^6…

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[unique crypt]

Let S be p in P1 such that [p]B1 = [p]B2. Prove that S is a subspace of P1 and find a basis for S

if it matters this is the change of basis matrix

B1 and B2 are in the image

[median vigil]

can you show how you got that?

[unique crypt]

tha change of basis matrix?

it's correct

[unique crypt]

i was going down this path

i think it could work?

wait no

i don't think that's right

not the change of basis matrix that is correct just the solution i was attempting

no yeah i think that's right

because you're asking for a polynomials which have the same coordinates in B1 and B2

alpha, beta being the coordinates

[median vigil]

you have c + d = -9 and neither of them are negative?

[unique crypt]

oh

i forgot the minus

alright so i guess it wasn't correct then, lol

[median vigil]

well if you already have the change of basis matrix you can use that

[unique crypt]

oh

well

too late lol

i guess at this rate i can say it's 4beta(-1+x)

and since beta is any R we can remove the 4

[unique crypt]

yeah this was wrong

i mean

5 beta = alpha is correct

it's just that setting them equal leads me to 6beta = 6beta

[median vigil]

maybe it will become more clear if you consider what it represents in matrix form

[unique crypt]

oh

well

[median vigil]

but it's not necessarily unexpected to end up with a free variable here

[unique crypt]

i've arrived at the correct solution but how can i look at this in matrix form

i tend to do things very um...

i try and see what sticks

if that makes any sense

sometimes i forget what it is i'm actually doing and how i could do it in other ways

[median vigil]

well we know that the change of basis matrix will change from one basis to the other. so what would it mean if we multiplied by the matrix and it ended up with the same basis representation?

[unique crypt]

so like

C(B1,B2) v1 = v1

?

[median vigil]

yes

[unique crypt]

hm

i see

and this is the same as 5 beta = alpha

just a more straightforward solution

i can apply this vector to either basis

and obtain the basis for S

[unique crypt]

same as i did here

[median vigil]

but what kind of vector is this with respect to that matrix?

[unique crypt]

these are the coordinates of a vector in B1

which when multiplied by C(B1,B2)

return the coordinates of the vector in B2

which are the same as in B1

which satisfies our subspace

[median vigil]

consider that any vector multiplied by 1 is itself

[unique crypt]

sure

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[gleaming laurel]

I want to turn a polar form of a complex number to a trigonometric form, specifically I want to understand how to find an argument, something related to arctan(b/a)

[gleaming laurel]

As far as I understand, depending on which quadrant the complext number is, it will have different ways to find an argument, I am confused about that part

nvm, I found the answer

.close

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[trail otter]

How many pairwise non-isomorphic trees with seven vertices exist if exactly one vertex has a degree of four? Determine the possible degree sequences of such trees and, for each degree sequence, specify the number of non-isomorphic trees with that sequence.

[trail otter]

ive only found two possible degree sequences

(4,2,2,1,1,1,1) and (4,3,1,1,1,1,1)

am i right?

[safe radish]

@trail otter Has your question been resolved?

[trail otter]

<@&286206848099549185>

pls help

[devout scroll]

yes

u r right

[trail otter]

nice and now how should i find all non-isomorphis trees with eah score?

[devout scroll]

try constructing them

[trail otter]

i can only thik of these two for the first degree sequence

[devout scroll]

yes

[trail otter]

there arent any more?

[devout scroll]

no

[trail otter]

alright and for second sequence is there only one?

[devout scroll]

yes

[trail otter]

alright thanks

[devout scroll]

welcome

[trail otter]

.CLOSE

.close

[safe radish]

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[safe radish]

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[old inlet]

help

[old inlet]

proof of the marginal density function for respective bivariate continuous rv X and Y

Anyone with notes or pdf or anything

@helper

<@&286206848099549185>

[young nexus]

!show

[safe radish]

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

[old inlet]

I do not know the proof of it and need help

[old inlet]

I am reading for next year statistics and the notes miss 1 page

which has the proof

<@&286206848099549185>

[safe radish]

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.

[old inlet]

ok

I am not real beluga

[tawdry portal]

Sry

[old inlet]

Yes

[tawdry portal]

I can’t help u

[old inlet]

👌

[tawdry portal]

I’m under qualified for that

[safe radish]

@old inlet Has your question been resolved?

[safe radish]

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[hot portal]

Am I doing this correctly

[safe radish]

@hot portal Has your question been resolved?

[old inlet]

I forgot and by the way you stole my channel

i want it back

.close

[opaque lintel]

the channel automatically closed itself

[old inlet]

.reopen

[old inlet]

Nahhh i want my channel back

[opaque lintel]

it's a bit late now

i feel like it'd be easier if you just got a new channel

[old inlet]

You must help me then

[opaque lintel]

my math isn't good enough for that sorry 😅

i haven't passed high school yet

[hot portal]

https://tenor.com/view/young-thug-mall-dap-up-hubie-hubert-gif-8502918938711374979

[old inlet]

Once a thieve always a thieve🤣

[hot portal]

https://tenor.com/view/flightreacts-tongue-laugh-flightreacts-flight-tongue-tongue-laugh-gif-739589491343629007

[safe radish]

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[hushed mirage]

Hello, I'm currently in my first year of uni, and the level of math is a bit above what I'm used to. We're studying sequences and series, and I can't wrap my head around them. The end goal , from what I understood, is to be able to then learn, understand, and use the taylor series with lagrangian remainder. I would appreciate if someone could point me in the right direction to get started with learning about sequences and series, alongside with some exercises. My math knowledge isn't bad but I've not studied these in highschool.

[fresh wasp]

I would start with prof Leonardo on yt, his lectures are long but they will leave you confident in your abilities ❤️ ( just watch it on 2x speed)

[hushed mirage]

These are basically on what they're based, but I'm pretty sure Tao is waaaay above my level

[hushed mirage]

Is PatrickJMT also good?

[fresh wasp]

I dont know who that is

[hushed mirage]

Math yt channel, name came up a few times while searching for an answer to this question online too

[fresh wasp]

Doesn’t hurt to try

[hushed mirage]

I'll look into prof leonardo for now, I'll gladly take any more help too

Thank you

[fresh wasp]

You’re welcome ❤️

[safe radish]

@hushed mirage Has your question been resolved?

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@hushed mirage Has your question been resolved?

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[full lotus]

$\int _0^3\int _0^{\sqrt{9-y^2}}6e^{3\left(x^2:+y^2\right)}:dx:dy:=:\int _0^3:\int _0^{\frac{\pi :}{2}}:\int _0^{6e^{3r^2}}r:dz:dt:dr:$

[flat frigate]

Chuti | Argentina

[full lotus]

This is what I got

$\frac{\pi }{2}\left(\frac{1}{3}\cdot :e^{27}-\frac{1}{3}\right)$

[flat frigate]

Chuti | Argentina

[fathom jewel]

,w Integrate[6e^(3x^2+3y^2),{y,0,3},{x,0,sqrt(9-y^2)}]

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[rotund gull]

,calc pi/2(e^(27)-1)

[flat frigate]

Result:

2.9523569611303e-12

[fathom jewel]

,w Integrate[r(6e^(3r^2)),{r,0,3},{theta,0,pi/2}]

[flat frigate]

Results provided by WolframAlpha

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

Requested by adoniszgks

[rotund gull]

Huh

[full lotus]

I'm clearly in the wrong

but I cannot see my error

[rotund gull]

Where is the factor of 1/3 from

[full lotus]

I can send my work

[rotund gull]

When you sub into polar you get this integral

[fathom jewel]

you prob did a mistake during the calculations

cause the set up is right

[full lotus]

okay

hmmm

[rotund gull]

Also integrate wrt r first

[full lotus]

why?

isnt it the same?

[rotund gull]

Well yeah but it's just easier I'd say, since there's no pi/2 factor to account for while doing arithmetic

[fathom jewel]

You did u = 3r² but you have 3u

[rotund gull]

The first integral gives you e^3r^2

[fathom jewel]

So the extra factor 1/3 is not meant to be there

[full lotus]

puch

hahahaha

[rotund gull]

$\int_0^3 6re^{3r^2}dr = e^{3r^2}$ at the upper and lower bounds

[flat frigate]

ari

[rotund gull]

Since you can easily see taking the derivative of e^3r^2 is giving the 6r factor

[fathom jewel]

the rest should be fine other than that

[full lotus]

I'm trying again

apart from that, there is no other way to calculate it right?

[fathom jewel]

i wouldnt say that

[full lotus]

cause idk the integral of e^x² + y²

[fathom jewel]

but polar is ideal because of the power

[full lotus]

I found it a bit difficult to get these integral limits

[rotund gull]

The bounds?

[full lotus]

yes, bounds

[rotund gull]

Well like you drew, it's a quarter circle so the bounds for r and theta are easy, and for the actual integrand when you make the substitution you just have to remember dxdy = rdrdtheta

[full lotus]

ok, got the right results now

[rotund gull]

$\int_0^{\pi/2} \int_0^3 6re^{3r^2}drd\theta$

[full lotus]

nice

[flat frigate]

ari

[full lotus]

so it looks like I took the right decision?

[rotund gull]

Yeah

[full lotus]

cool

[rotund gull]

Just algebra mistake evaluating the inner integral

[full lotus]

yes

perfect

alright

cool

thanks 🩵

[lean otter]

!done

[safe radish]

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[acoustic surge]

f(x)= X^2-4/X+2
(I need to show my work on all steps)
1.Determine the domain of f(x).

2.Determine the range of f(x)

3. Determine any intervals of increasing or decreasing.

4. Determine the equation of the vertical asymptotes, or state that there are not any.

5. Determine the equation of the horizontal asymptotes, or state that there are not any.

6. Calculate the y-intercept.

7. Calculate the x-intercepts.

8. Calculate at least 4 appropriate points.

9. Sketch a graph.

[safe radish]

@acoustic surge Has your question been resolved?

[real arch]

What are you confused on

[acoustic surge]

Everything

We have a test coming up I was told I can have this problem on the notecard as a reference for the test and use on the test

[real arch]

What can you do and what can't you do

I can help you, I won't do it for you.

Try making a table and drawing the function out on a graph.

[acoustic surge]

Nothing… we’re not too far in, and I missed most of the days because I was sick

[real arch]

A function assigns an element (x) to an element in another set (f(x))

Domain is the set of numbers represented by x

Range is the set of numbers that x outputs.

An interval is the range of x-values

A vertical asymptote is a vertical line indicating where the function does not have a value that exists in the domain.

A horizontal asymptote is a horizontal line indicating where the function does not have a value that exists in the range

The y-intercept is all values where the function crosses the y-axis

The x-intercept is all values where the function crosses the x-axis

I think you need to read your textbook and learn the concepts.

[acoustic surge]

We don’t have textbooks

[real arch]

What class is this

https://openstax.org/books/college-algebra-2e/pages/2-introduction-to-equations-and-inequalities

I think reading chapter 2 and 3 will help you identify how to solve the problems you have.

[acoustic surge]

Math 1314 College Algebra 2

I never took algebra before tho

[real arch]

read the material i sent and practice the examples in the text.

[safe radish]

@acoustic surge Has your question been resolved?

[acoustic surge]

I don’t get it😭

[acoustic surge]

<@&286206848099549185>

[young nexus]

you should explain what exactly you dont get.

[acoustic surge]

Can someone solve it😭🙏

[young nexus]

!noans

[safe radish]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[real arch]

@acoustic surge The responsibility of your own education lies within yourself and it is up to you to decide if you want to put in the work to be successful. You are doing yourself a disservice. Nobody can help you if you don't help yourself. You need to learn the concepts by studying the material I gave you.

[acoustic surge]

But there’s 1 week left until school finishes 😭

[real arch]

Then you better start now.

[real arch]

.close

[safe radish]

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[magic vale]

Just looking for some simple explaining. I can do part a perfectly but for part B I don't understand why I can't just minus the two values to get the correct answer. Thanks in advance.

[teal thicket]

which two values are you talking about?

Are you talking about writing the answer zero for the part b?

[safe radish]

@magic vale Has your question been resolved?

[safe radish]

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[charred cedar]

Hi

[charred cedar]

Why gamma here is accepted as less than zero not less than 4 ?

[raven heart]

you want gamma < 0 and gamma < 4 at the same time

take gamma = 2 as an example, does it satisfy these two inequalities ?

@charred cedar

[charred cedar]

nope

😮

gamme less than zero satisfy both terms to be negative

[raven heart]

the thing is gamma < 0 is more restrictive than gamma < 4

yeah

[charred cedar]

silly me..

Thanks alot

imma close this

.close

[safe radish]

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[sacred shard]

is logv10(x) used as ln(x) cause my professor said so and i am a freshman so i dont know anything and the internet says there's no such thing

[light shoal]

i assume you mean $\log_{10}$

[flat frigate]

Bungo

[sacred shard]

yeah

[light shoal]

it's not the same as ln

ln is log base e

[sacred shard]

sorry i dont know how to type it...

[light shoal]

nw

[sacred shard]

yeah thats why i thought so but

idk why he said that

[severe pond]

you mean writing $\log x$

[flat frigate]

knief

[light shoal]

he might have meant "log" means "ln"

which in math it usually does

(log without any subscript)

[sacred shard]

no he said log e base is not ln and just logx

like the other way

[severe pond]

😦😦

[light shoal]

that's just false

[sacred shard]

in so confused bro ...

[light shoal]

log base e and ln (natural log) are exactly the same thing

[sacred shard]

yeah bro

i need to change my school i guess

[edgy isle]

ln is always with base e, no exceptions, log is mostly with base e, at least in pure maths, though you might see it with base 10 in some contexts

[light shoal]

haha

[severe pond]

yep

[light shoal]

just ask your prof to clarify

they make mistakes sometimes, it's not the end of the world

or there was a misinterpretation

[sacred shard]

i will🙏

[sacred shard]

i hope so cause no way he said that

[light shoal]

yea

[sacred shard]

thanks yall

✊🏻

[safe radish]

@sacred shard Has your question been resolved?

[safe radish]

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[river wigeon]

if im being asked to decompose sin (13pi/12)

[river wigeon]

into two remarkable angles (a+b)

how do i go about it?

do i just write down every remkarable angle?

[merry swift]

I assume remarkable angles are just those in (0, pi/6, pi/4 pi/3, pi/2, pi) and multiples thereof?

[river wigeon]

yes

[visual bone]

13/12 = 1 + 1/4 -1/6

[river wigeon]

do i just memorize

[visual bone]

No

[river wigeon]

all of the remarkable angles

and put them over 12

[visual bone]

use your skills

[river wigeon]

i cant have 3 terms

it has to be two

[visual bone]

only to would be pi + pi/12

[river wigeon]

pi/12 isnt remarkable

[visual bone]

and recall sin(pi+x) = -sinx

then do 2 remarkables with pi/12

[river wigeon]

im breaking it down so i can use sin (a+b)

[visual bone]

-sin(pi/12)

break that one

as pi/4-pi/6

[river wigeon]

i can do that?

[visual bone]

1/4-1/6=1/12

[river wigeon]

i guess i can

removing one pi and then

so what is - sin

[visual bone]

yes

[river wigeon]

if i have -sin pi/6

does it mean that - (sin pi/6) = -(0.5)

is the minus just like a * - to my final answer

[visual bone]

u dont have that

[river wigeon]

to flip the sign

but therotically

[visual bone]

u have -sin(pi/4-pi/6)

[river wigeon]

yes

but im confused about something

[visual bone]

that is -sin(pi/4)cos(pi/6)+cos(pi/4)sin(pi/6)

[river wigeon]

so completley unrelated

if i have -sin(pi/6)

is it -(0.5)

what does - siginfy when attached to sin

is it just a *-1

[visual bone]

yes

[river wigeon]

gotcha

i can do (sin 10pi./12 + 3 pi/12)

instead of doing pi-x

[visual bone]

yes

3 and 12 have 3 factor

so thats pi/4

and 10 and 12 have 2 factor

thats 5pi/6

[river wigeon]

YES

i have a trick for it now

thanks

.close

[safe radish]

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[charred cedar]

Hi

I got this question

and it was solved by UC method and final ans was:

I tried to solve this by variation of paramters but I got different result

I'll appreciate if anyone take a look and lemme know what I did wrong

[safe radish]

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[solar briar]

need help with orienting my graph

[solar briar]

i am given sin(pix/6) and need the integral graph

so i integrate it and get -(6/pi) * cos(pix/6) + C

so i plug in (0,3.67) and get C=5.58

at x=0 the integral graph indeed starts at 3.67, but it curves upwards? i need it to curve downwards

and it should end at y=-0.146 at x=6, but this is not the case

what part of my math went wrong?

[fresh wasp]

If you need it to curve downwards the make its positive

Just cos (pi x/6) (pi/6)

[solar briar]

oh and then plug in the initial conditions

[fresh wasp]

Or you can add on a phase shift but i dont know if that would help

and it it ends at a negative value than i think your c is calculated wrong

[solar briar]

thanks it worked

[fresh wasp]

Not that you need to change the function

[solar briar]

6/pi * cos(pix/6)-1.764

[fresh wasp]

That worked?

[solar briar]

after solving the integral i took out tghe negative sign

and solved for +c with the conditions x=0 y=3.67

[fresh wasp]

Did they give you those conditions?

[solar briar]

but i didnt know i had to remove the negative

[fresh wasp]

i havent taken stats to it would probably be better if someone who has helped you

[solar briar]

uh statics not statistics

[fresh wasp]

i dont know what that is sorry

ill ping

[solar briar]

well no need

i already got it

[fresh wasp]

huh

um ok?

i think it should be checked tho i dont understand the problem

ok

[solar briar]

thanks again

[safe radish]

@solar briar Has your question been resolved?

[fresh wasp]

youre welcome ❤️

[safe radish]

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[lean otter]

is it true that ref is always an upper triangular matrix

[median vigil]

by definition, yes

[safe radish]

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[neon summit]

Hello, are we allowed to use limit laws when the limit does not exist?

[neon summit]

Like in the second line where I immediately split up lim (sinx^2 / x^2 + 2cos....) into -lim (sinx^2 / x^2) + 2lim( cos.....

[icy lance]

yeah

[safe radish]

@neon summit Has your question been resolved?

[neon summit]

Thank you

[safe radish]

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[rapid tinsel]

I need help

[safe radish]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[rapid tinsel]

The 1/2 answers are wrong

How do I fix this

[thin bridge]

check the outcomes that satisfy the event to count

[rapid tinsel]

I did but it was wrong

I unchecked everything

Idek if the probability right

[thin bridge]

do it again, show what you checked

[rapid tinsel]

Only event A is wrong

[icy lance]

whats the probability of each combination from three tosses?

[rapid tinsel]

Those were correct but the ones I put for A were wrong

[thin bridge]

show what you checked for A

[icy lance]

you sure you read the event right?

[rapid tinsel]

I’m using ChatGPT ngl

[icy lance]

this dude

'alternating'

[rapid tinsel]

Due by 11:59

[iron leaf]

lmaooooo

[rapid tinsel]

Chill on me

I got a bunch of these random problems due by 11:59

[odd quartz]

alternating

[rapid tinsel]

I ain’t know it was gonna take this long

[odd quartz]

what does that mean

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[rapid tinsel]

Bot im sorry but please hook me up with a tutor that can provide me the correct answers then

So none of you know how to do this

[odd quartz]

what does alternating mean

[rapid tinsel]

Idk

[odd quartz]

search up

and you probably can find the answer

[rapid tinsel]

That’s why I used you know what

But even you know what is giving me wrong answers

[odd quartz]

i know the answer

but im tryna find a way to help u

[rapid tinsel]

We can review after, this as many other problems are due before 11:59

[odd quartz]

Is htt alternating?

[rapid tinsel]

So I can’t really do thing s the “right” way right now

[rapid tinsel]

Maybe

[odd quartz]

Is hth alternating

[rapid tinsel]

Yes?

[odd quartz]

correct

because it switches one to an another

Htt is not alternating cuz after second alter it dosent switch

[rapid tinsel]

What’s wrong here

[odd quartz]

Is OEE an even number first?

[rapid tinsel]

Odd and two evens

Bro I’ll pay you if you just give me the answers

Kinda on a tight timing rn to be trying to learn and answer at the same time

[odd quartz]

i think reading the boxes u selected and the question will help

[rapid tinsel]

I have over 124 questions left

[icy lance]

your timing isnt anyones fault here, so youre not just gonna get solutions

[rapid tinsel]

Due before 11:59, is anyone willing to just give me answers or is that against rhe rules here

Oh ok

[icy lance]

indeed it is

[rapid tinsel]

Quick question, why?

Are y’all real math teachers or something?

Why care about morality with a discord, just wondering

[odd quartz]

Am i allowed to give answer?

[iron leaf]

💀

[safe radish]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

@rapid tinsel Has your question been resolved?

[safe radish]

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[open mountain]

hi guys

[open mountain]

i’m currently stuck on this question. I know this is supposed to be solved using difference of cubes, but what is the next step from here?

[median vigil]

at that point you can quadratic formula

[open mountain]

thank you!

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[frosty mist]

how did the 8 get squared

[frosty mist]

i got 8sin(8x^2)(8)

[plucky elk]

(8x)^2 = 8^2 * x^2

,tex .exp rules

[flat frigate]

riemann

[frosty mist]

uh is it the power of power? 😭

someone just showed me this too and i had to guess which rule it was

[timid ridge]

distribution

[frosty mist]

oh oki

thank u

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• Show your work, and if possible, explain where you are stuck.
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[tepid osprey]

Can anyone explain how to do those

[desert pasture]

,rotate

[desert pasture]

which problem!

[tepid osprey]

The first one

<@&286206848099549185>

[tepid osprey]

Say my name

[runic spruce]

Heisenberg

[lean ruin]

The question is find x yea?

[tepid osprey]

Yep

[lean ruin]

And you wanna do part a yea?

[tepid osprey]

Is this some quadratic equations or smth

[tepid osprey]

Yeah

[lean ruin]

What have you tried

[tepid osprey]

Havent tried anything yet

If the 10x would disappear its gonna be ez and how can i find the x with this

Sorry for asking ez question

[safe radish]

@tepid osprey Has your question been resolved?

[tepid osprey]

No

[quick crater]

let the top part be equal to zero, if you got the zeroes of the top part check for the denominator if it becomes zero, as you cant divide by 0

[tepid osprey]

Oh

As what ive been thinking

Like this?

<@&286206848099549185>

[queen ingot]

yeah

• or -5

[tepid osprey]

Ty

.close

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[faint seal]

so I've proven (i) and (ii)

for part (iii) I see that if we take the upper sign identity from part ii we have $$\left(\dv{}{x}+x\right)\left(-\dv{}{x}+x\right)y=-\frac{\dd ^2 y}{\dd{x}^2}+x^2 y+y$$ or $$\left(\dv{}{x}+x\right)\left(-\dv{y}{x}+xy\right)=2y$$

[flat frigate]

kheerii

[faint seal]

so with w it becomes dw/dx + xw = 2y

[restive niche]

Dont you want it to act om xy

[faint seal]

what?

[faint seal]

I just replaced -dy/dx + xy with w here

[restive niche]

I mean why arent you replacing w by -dy/dx + xy in the equation they want you to check it satisfies

Does that not lead anywhere?

[faint seal]

it seemed tedious

and I thought I was supposed to use part (ii)

[faint seal]

wait this does help me though doesn't it?

[restive niche]

Oh yeah sure, so multiply this by x i guess?

[faint seal]

we have $\left(\dv{}{x}+x\right)w=2y$ so $\left(-\dv{}{x}+x\right)\left(\dv{}{x}+x\right)w=-2\dv{y}{x}+2xy$

[flat frigate]

kheerii

[faint seal]

or -w'' + x^2 w - w = -2 dy/dx + 2xy

which is just 2w

but like what is the point of this

it seems like we got very similar DEs for y and w

[restive niche]

Not sure, given they even said "hence or otherwise verify..."

[faint seal]

yeah

[restive niche]

Yeah, the only change was in the constant multiplying w

[faint seal]

yeah

the y differential equation also has much nicer solutions

so

idk

[restive niche]

I think the solution changes by a constant factor in the exponent

[faint seal]

,w -y'' + x^2 y = y

[flat frigate]

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[faint seal]

,w -w'' + x^2 w = 3w

[flat frigate]

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[faint seal]

D_n is the "parabolic cylinder function" apparently

[restive niche]

I thought it was useful because the first equation had nice(comparatively) solutions, so you could solve that and convert your solution so that it satisfies the third one, but you cant be sure that it gives you all solutions to the third one

[faint seal]

how would you convert your solution though?

[restive niche]

if y solves the first one, -dy/dx + xy solves the third one

[faint seal]

oh wait

yeah

lol mb

[restive niche]

But you cant be sure it gives you the whole solution set

[faint seal]

hmm there should still be two linearly independent solutions

like surely we can find two y such that their -dy/dx + xy are lin ind

[restive niche]

Well we can check that since we know what y looks like

,w -d(e^(-x^2/2)*erfi(x))/dx+ x(e^(-x^2/2)*erfi(x))

[flat frigate]

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[restive niche]

Oh wait @faint seal they arent linearl independent because -dy/dx + xy for y = e^(-x^2/2) is 0

[faint seal]

ahhh

[restive niche]

You just need to check the bases, which are e^(-x^2/2)erfi(x) and e^(-x^2/2)

[faint seal]

wait what?

[restive niche]

The first one gives you this function

[faint seal]

it isn't

[restive niche]

Wait

[faint seal]

it's 2x e^(-x^2/2)

[restive niche]

,w -d(e^(-x^2/2))/dx + x(e^(-x^2/2))

Oh wait lmao sign mistake

[restive niche]

Wait so we have two linearly independent solutions. This one and 2xe^(-x^2/2)

[faint seal]

Yep

Maybe wolfram is just high

[restive niche]

Im not sure what the cylindrical functions were about

[faint seal]

Or maybe d1 and d_-2 have representations in terms of erf

Still, pretty weird question

[restive niche]

Someone in #odes-and-pdes will know

[restive niche]

I mean we get the full solution set for the third one by solving the first one, which is marginally nicer

[faint seal]

Hmm

Yeah I guess that tracks

The 3 seems arbitrary, maybe we could find slightly different operators so that that could be any number

Or actually wait

What if the original DE in y was -y''+x^2 y=0

[faint seal]

Then from here we'd get -w''+x^2 w = 2w

[faint seal]

So we can write the solutions of this in terms of solutions of this

,w y''=x^2 y

[flat frigate]

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[restive niche]

[faint seal]

Whatthe fuck

Nah I know this one for sure

Oh no wait

Nvm

[restive niche]

The index for the parabolic functioms changed???

I guess there is some relation between the indices

[faint seal]

Perhaps

,w -y''+x^2 y = cy

[flat frigate]

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[faint seal]

Fascinating

[safe radish]

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[oblique jay]

Calculate the required number of rounds.

Imagine a game in which the paricipant must play multiple rounds to finish a play. However many rounds needed are the score.

The player first receives a random slot number {1,2,3,4}. Then he has 12.5% chance to get a point once for that number.

1. How many rounds does it aproximately take to win two slots?

[oblique jay]

So first the player has a 100% chance to hit a slot he did not hit before.
Then he has a 12.5% chance to win the slot. So winning one slot takes aprox 8 rounds.

_ _
Then the player has a 75% chance to hit a slot he did not hit before.
He has a 12.5% chance to win the slot again. And now I don't know how to reason further.

[thorn wagon]

do you know the answer?cause i’ll explain if my ans matches

[oblique jay]

No I don't, sorry

[thorn wagon]

aah ok

[oblique jay]

The thing that makes it difficult for me is that each round it can hit any slot again.

So for example you throw 8 balls and you win slot 2.

[thorn wagon]

i would approach it like this basically

[oblique jay]

Now you start throwing 8 balls again, but this time if you win the slot it only counts if you actually win on a new slot.

It confuses me sooo much lol.

[thorn wagon]

cause they said approximate take it as a infinite progression like first case he gets both rounds as the same number second case he gets 2 of 3 as the same 3rd case 3 of 4 and so on and you get the total cases right?

so that would be the probability of winning 2 slots ,and using that you can say the approx turns for 2 wins is what im thinking

[oblique jay]

So basically if you would play the game you first have a random chance to get any slot.

In this case there are only 4 slots, so you get a random number of the slot. Slot 1, slot 2, slot 3 or slot 4.

Then you have a chance of winning the slot which is 12.5%.
So you throw ball after ball, every time it might be on a different slot and every time you only win if the slot is not already won with a 12.5% chance...

[thorn wagon]

is the answer 4

that’s what im getting from my approach

[oblique jay]

That should not be possible.
Look you have a 12.5% chance. If you throw 8 balls you would on average win one slot with that...

[thorn wagon]

hmmm maybe that approach is wrong then

[oblique jay]

Cuz if slots would not matter then it would just be:
1/0.125 = 8

[thorn wagon]

wait is it 16?

i wish you had the answer

i forgot to consider cases before

[oblique jay]

but since slots matter the chance is way lower:

T = total number of slots
K = number of winnable slots
1/(0.125*( (1/T)*K ) )

This is not the correct answer I assume...

[thorn wagon]

wait so the slots part implies that the chosen slot has less than 1/4 prob of appearing?aaah

so its not fair probability

[oblique jay]

No no, I think you got it wrong!