help-23

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[latent solar]

Can someone give me a hint as on where to start

[latent solar]

This can be solved with casework bash, but that’s 16 cases

I want to find a more “elegant” way to do it

[safe radish]

@latent solar Has your question been resolved?

[latent solar]

<@&286206848099549185>

[blissful tapir]

one thing that immediately comes to mind is you need to eliminate 1 from S, cuz a, b and c are not necessarily distinct, so 1 * 1 = 1 makes any set that has a 1 not product free

[latent solar]

Yes I know

[ionic spade]

@latent solar hi

Hello all of them

[blissful tapir]

what solution have you already done? do you have it written?

[latent solar]

Well I tried to use recursion

If a-n is the number of product free subsets, then a-n = 2a-n-1 +1 is n is prime

Other wise it’s 2(a-n - number of product free subsets in a-n-1 containing two numbers or a number when squared resulting in n) + 1 is n is composite

But that didn’t work

I think it’s flawed in some way

[blissful tapir]

what did you take as n?

[latent solar]

Sorry

A-n is the number of product free subsets of 1, 2, 3, … n

[blissful tapir]

okay

okay it might not work like that

[safe radish]

@latent solar Has your question been resolved?

[solar hazel]

i will try it

here are all the sets that can be made with 1,2,3,4,5:

2
3 23
4 34
5 25 35 235 45 345

for each of these, check if you can add 6,7,8,9,10 to "complete" them

e.g. for 2, you can add any of them

so that contributes 2^5 "completed" sets (5 yes/no options to add 6,7,8,9,10)

same thing with 3, except you can't add 9

for 23 you can't add 6 or 9

so 2^3 there

and so on

i'm not gonna call this elegant but it is fast

[solar hazel]

the key is that adding any one of 6,7,8,9,10 to a set does not change whether any other of them are allowable

,calc 2^5 + 2^4 + 2^3 + 2^5 + 2^4 + 2^5 + 2^4 + 2^4 + 2^2 + 2^5 + 2^4

[flat frigate]

Result:

220

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[solar hazel]

i didn't check super carefully btw

so that may or may not be right

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[mossy ridge]

suppose [f'(x)]² = 2f"(x) f(x) + [f'(0)]² if f(x) is a polynomial with highest degree of x as n, then whats the maximum value of n for which the equation holds

[solar hazel]

the left side will have degree 2(n-1) and the right side will have degree n - 2 + n (well, so long as deg(f) >= 2)

that should get you started

[mossy ridge]

wont it be 2(n-1)

and n-2+n

[solar hazel]

yes sorry

[mossy ridge]

that gives me n=n

[solar hazel]

well fuck

[mossy ridge]

LMAO

[safe radish]

@mossy ridge Has your question been resolved?

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[still grove]

Gotta find the second derivative. I don't understand what I'm doing wrong

[lament steeple]

sorry misapplied. you dropped the square on y' on the 6th step. also i dont see what x^2/y^2 is. thanks

[blissful tapir]

It's y'

She wrote at top

[lament steeple]

ok. dropped square still

[blissful tapir]

Yeah

[still grove]

ye mb I forgot to write the square but I did square it in the seventh step

just forgot to write it in the previous step

[blissful tapir]

Oh yes

[lament steeple]

oh. gotcha

didnt multiply y to the 2x term

7th to 8th step

(-2xy^4-2x^4y)/y^6

[still grove]

Why do I have to multiply it by 2x? I thought it was just by 2y

[lament steeple]

you multiplied y^4 to the denominator and part of the numerator

that is obviously not very correct

[still grove]

OMG right. That's such a stupid mistake lol

that makes sense

[lament steeple]

new thread please

[still grove]

so the final answer would just be y'' = (-2xy⁴ - 2x⁴y)/y⁶?

[lament steeple]

i'd cancel out one y on both numerator and denominator

otherwise good to go

[still grove]

alr alr

wait

the answer key says (-2xy⁴ - 2x⁴y)/y⁶ = (10x)/y⁵

Where did that come from then

[mossy lotus]

original equation

[lament steeple]

prolly took y back to the original func

yes

[blissful tapir]

Factored out xy

[lament steeple]

ok thats dumb. kills the whole purpose of implicit differentiation

[lament steeple]

no more than that

[blissful tapir]

Yeah then just put the value of x^3 + y^3

[lament steeple]

i mean, might as well operate on the original func and yield y=(5-x^3)^1/3

and differentiate that

dumb question

[blissful tapir]

🥀💔

[still grove]

right😭

imma just leave it then

Thanks guys!!

.close

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[ruby junco]

can someone explain why this is wrong 💔💔💔

[little finch]

where is u

[ruby junco]

u=arctanx

😭?

WAIT

omg

so itd be u times (tanu)^2 du?

[little finch]

yeah

[ruby junco]

OK

THANK YOU

.close

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[lean otter]

The origin is at the center of a regular polygon. What is the sum of the vectors from one fixed vertex to each of the remaining vertices? Give your reasoning.

[lean otter]

iunno what to do

[scarlet plank]

As a vector? Starting from the first and adding rest

[lean otter]

sum of vectors from one fixed vertex to the remaining vertices

[keen tulip]

Label the vertices of the regular n-gon by vectors r_1, r_2, ..., r_n, all from the origin. Then you know that their sum is 0.

Fix r_1.

Then just look at the sum.

[lean otter]

i dont understand 😭

[keen tulip]

Which part don't you understand?

[lean otter]

the part where we sum the vectors from one vertex to the rest

[keen tulip]

Suppose we fix r_1, the question is simply asking for

[
rb{
sum_{k=2}^n (veb r_k - veb r_1)
}
]

[flat frigate]

クーリー

[keen tulip]

@lean otter Does that make sense?

[normal moss]

||Woah you have a lot of roles||

[keen tulip]

I am simply very qualified.

[normal moss]

Obviously 😂

[frigid spruce]

what's with the laughing emote?

are you doubting him

[lean otter]

if we set vertex as A
than the vector will be like AB, AC, AD, ....
AB is basically like B-A
so we do it vector style for all of them, then it becomes what u sent?

i think im wrong

[normal moss]

No, I meant it with genuine enthusiasm. I see how it can be interpreted differently though.

[frigid spruce]

:thumbsupsuperserious:

[keen tulip]

Hm. Okay. If we say the vertices are $A, B, C, D,$ etc. and let $\vec A, \vec B, \vec C$, etc. be their position vectors from the origin (so $\vec A = \vec{OA}$), then vectors from A to the others are $\vec{AB} = \vec B - \vec A,~ \vec{A C} = \vec C - \vec A,$ etc. 

We want the sum \[\vec{AB} + \vec{A C} + \vec{AD} + ... = (\vec B - \vec A) + (\vec C - \vec A) + (\vec D - \vec A) + ...\]

,tex ? AC

[flat frigate]

クーリー

[keen tulip]

lol

Anyways, does this make sense @lean otter

[flat frigate]

クーリー

[lean otter]

yes

[keen tulip]

Okay so

\bas
\vec{AB} + \vec{A C} + \vec{AD} + cdots &= (\vec B - \vec A) + (\vec C - \vec A) + (\vec D - \vec A) + cdots\\
&= (vec B + vec C + vec D + cdots) - (n umber~of~terms) cd vec A\\
&= (vec B + vec C + vec D + cdots) - (n-1) cd vec A
\eas

[flat frigate]

クーリー

[keen tulip]

Now use the regular-polygon fact that the sum of all vertex position vectors is zero (because they are symmetric around the origin)

Can you go from here?

[lean otter]

uh im not sure if i can

[lean otter]

i do understand this

[keen tulip]

Do you agree that the sum of all vertex position vectors is zero?

[lean otter]

yes

the previous part of this question was this

[keen tulip]

So, vec A + vec B + vec C + ... = 0

what is vec B + vec C + ...?

(obviously, it's a finite sum)

[lean otter]

iunno what im doing 😭

sorry

[keen tulip]

No worries

[keen tulip]

You agree with this?

[keen tulip]

From here, we can derive that vec B + vec C + vec D + ... = - vec A

[lean otter]

i dont quite get the (n-1) part

can anyone help? 😭

[feral linden]

They explained fully. a_1 a_k= O a_k - O a_1. Can’t be any clearer

[lean otter]

but what does it tell

[lean otter]

^ i do understand this

[feral linden]

Sum of O a_k is 0

[lean otter]

and?

[feral linden]

That leaves you -n Oa_1, -nv with the original one v given

[lean otter]

why the n

im so dumb 😭

[feral linden]

n many terms for a n-gon

[lean otter]

oh

wait actually

i dont get allat

sorry

😭

[feral linden]

Write it on paper then…

[lean otter]

i finna get it 😭

my brain wasnt working

thx

.close

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[trim vine]

what exactly am I doing wrong here?

[vagrant ice]

sin(pi/2) is not -1

[trim vine]

oh yeah

how do I calculate sin x = -1

[vagrant ice]

well it's going to be a multiple of pi/2

[lost jewel]

also x= pi/2+ 2npi for cos(x)=0 is wrong

[trim vine]

uh huh

an odd multiple of pi/2 no?

[blissful tapir]

npi should do

[trim vine]

(2n+1)pi/2

[trim vine]

okay

[vagrant ice]

you still haven't answered: sin(what?) = -1

just the value in [0, 2pi)

[trim vine]

sin(pi/2 + (2n-1)) = -1

[trim vine]

oh

[mossy lotus]

he has a pi/2 + (2n-1)pi tho, so that makes it -pi/2 + 2npi, which would be correct

[trim vine]

sin(3pi/2)

[vagrant ice]

oh fair enough

okay I see

okay yeah then hopefully you can put pi/2 + npi and pi/2 + (2n - 1)pi together

[trim vine]

wait okay so

[trim vine]

for cos(x) = 0, x = (pi/2 +npi) innit

[blissful tapir]

Yeah mate

[vagrant ice]

yep! the other case gets totally absorbed into pi/2 + npi

[trim vine]

and for sin(x) = -1, x = (pi/2 + npi)

[blissful tapir]

Nah mate

[vagrant ice]

no, that's still pi/2 + (2n - 1)pi

[trim vine]

yeah sorry i meant that

odd multiple of pi/2

mb

okay so for the y values i just plug in these values into x innit

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[terse atlas]

can anyone point out the mistake I made? this is over Zmod5 , please lmk if something is undecipherable

[terse atlas]

this is a solution for three different systems which only differ in the b column btw, that's why its augmented 3 times

I should have gotten a parameterized solution here but instead I got an actual solution. wasn't supposed to happen

<@&286206848099549185>

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[marble linden]

hi

[marble linden]

how do i find the surface area of a sphere

[burnt notch]

4πR²

[marble linden]

thanks

[burnt notch]

Where R is the radius

[marble linden]

thanks

[nimble vine]

!done

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[mossy urchin]

is AD also 12?

[vale vessel]

is $\angle{ADO}$ a right angle?

[flat frigate]

Mirror

[mossy urchin]

yes

[vale vessel]

also what are you aiming to find out in this problem?

[mossy urchin]

i js wanna know if AD will be 12

12 is the radius here

[vale vessel]

pythagoras would like to have a word then

[mossy urchin]

oh

ohhhhh

pythagoras

ah yes

long time no see

thanks

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[mossy urchin]

how is AO 12

[mossy urchin]

fine, they said 12 is the radius so OC is 12

[autumn sun]

A is on the circumference of the circle isn't it?

so why wouldn't OA be 12m as well?

[mossy urchin]

then why is AD not 12m?

[autumn sun]

is D the center?

[blissful tapir]

AD is not connected with center

[mossy urchin]

Ohh

so a radius is a radius when it is connected to centre

[autumn sun]

that's the very definition of a radius, yes.

[mossy urchin]

👍

[blissful tapir]

yeah, and the other point should be on the circumference of the circle

[mossy urchin]

thanks!

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[mossy urchin]

how to find the red angle now

[autumn sun]

is ABCD a rectangle? what other context do you have?

[mossy urchin]

yes abcd is a rectangle

APCOA is a sector of a circle centre O with radius 12m

[autumn sun]

then angle ADC is one of the four corners of the rectangle, isn't it?

[feral linden]

cos rule

[autumn sun]

you can directly get angle ADC from that fact

no cosine rule or anything required unless you wanna check or something

[mossy urchin]

wait what

i was gonna do cosine

[autumn sun]

that's why I asked you if ABCD is a rectangle

and you said yes

[mossy urchin]

so?

[autumn sun]

do you not know the interior angles of a rectangle...?

[primal bone]

The angles of a rectangle are...?

[opaque fern]

A rectangular has 90 degrees angles

[primal bone]

no spoil

[opaque fern]

Oh welp

[mossy urchin]

ohh

[rigid blaze]

triganametry?

[primal bone]

bruh

[opaque fern]

Soz

[primal bone]

We got there at last tho lol

[autumn sun]

trig to find the right angle?

[rigid blaze]

no

[autumn sun]

but that's the angle OP is asking for!

[primal bone]

I mean I've had students doing that lol

[rigid blaze]

Wait wha

[opaque fern]

But yeah cos rule is a nuke here

[rigid blaze]

that’s clearly 90

[mossy urchin]

right thats 90

[autumn sun]

we're beating a fly with a flammenwerfer here if we use cos rule

aight, anything else?

[rigid blaze]

that’s just overkill

I thought the question was to find another angle

[mossy urchin]

nope, ty!!!!!

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[unkempt delta]

Let m_0 and n_0 be district positive integers. For every positive integer k, define m_k and n_k to be the relatively prime integers such that m_k/n_k = (2m_{k-1}+1)/(2n_{k-1}+1). prove that 2m_k +1 and 2n_k+1 are relatively prime integers for all but finitely many positive integers k.

[unkempt delta]

I’m not sure how to start

[vale vessel]

it sounds like induction

but i'm not 100 on it (i can maybe get you started but have a final to write so will need to tag out)

[unkempt delta]

Oh, please just focus on your final!

I’m sure someone else can help

[vale vessel]

i still think induction will be good, especially if you can write $2m_{k+1}$ and $2n_{k+1}$ in terms of $m_k$ and $n_k$

[flat frigate]

Mirror

[unkempt delta]

I’ll give it a try

[keen tulip]

Here's a (probably) shorter way to solve this problem. Consider the difference |m_k - n_k|. When you move from k-1 to k the difference is multiplied by 2 and also divided by gcd( 2m_{k - 1} + 1, 2n_{ k - 1} + 1 ) which is odd if > 1.

[unkempt delta]

Sorry I don’t really get it

[safe radish]

@unkempt delta Has your question been resolved?

[feral linden]

a_j=m_j-n_j
b_j=gcd(m_j, n_j)
a_(j+1)=a_j / b_j
a_0 = b_0 b_1 … b_j a_(j+1)
Any odd prime p, if p divides one b_j, then p divides a_0. So finitely many p such that p divides one b_j. If one p divides infinitely many b_j, then again a_0 = b_0 b_1 … b_j a_(j+1), a_0 can have as arbitrary large many p in its factorization as we want, contradiction. Thus finitely many p, and each p divides finitely many b_j. Done

[safe radish]

@unkempt delta Has your question been resolved?

[safe radish]

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[astral glacier]

x^7-7x+1 = 0 find all real solutions

[honest perch]

,w x^7-7x+1 = 0 find all real solutions

[astral glacier]

wel how to solve

[flat frigate]

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Requested by mayervietorus

[twilit spindle]

you pray, or use Newton-Raphson

I doubt the roots can be written using roots/operations

[quasi timber]

You can either graph it, or Newton-Raphson

[quasi timber]

sniped 😔

[astral glacier]

Lol what is the Newton -Raphson method

[twilit spindle]

(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)})

[flat frigate]

ΠαϳαμαΜαμαΛλαμα

[blissful tapir]

Bro will get disappointed

[honest perch]

a way to approximate teh roots

[astral glacier]

bruh

i am sed

:(

[twilit spindle]

,w gcd(x^7+x+1,x^4-x)

[flat frigate]

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Requested by xxmrfancyu2xx

[covert yoke]

So the thing about this equation, and most equations of degree 5 or higher, there is no way to solve the equation using "elementary" operations, such as multiplication, addition, nth roots, etc.

[quasi bison]

where did you get this question

did you think it up yourself or

[astral glacier]

No

one of my friends sent

me

[blissful tapir]

Did he think it up himself

[quasi bison]

and your friend is known to be a reliable source of known-solvable questions?

where did he get it from?

[astral glacier]

Well idk

he is decently good at maths

And he recently qualified a decently tough exam

so I thought maybe it's correc

[blissful tapir]

That doesn't imply he isn't trolling

[astral glacier]

tthank you for ur guidance

[covert yoke]

well, he might not know any better regardless.

[astral glacier]

.close

[safe radish]

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[blissful tapir]

It is correct, but not solvable with elementary stuff

[safe radish]

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[old nebula]

I want to do 16 using shell method

[old nebula]

so I integrate wrt y and I do integration of -pi(y)(lny)

from 0 to 1

but the integral is incorrect somehow

please explain

[blazing swallow]

right now you are saying that $x = -\ln(y)$ is the same function as $y = e^{-2x}$

[flat frigate]

ηασιβ ♥

[blazing swallow]

right away from looking at that, you can see that the 2 has disappeared, which is a problem

[old nebula]

I actually cancelled the 2

the 2 is in the shell method formula

it cancels out with the -ln(y)/2

[blazing swallow]

oh right, sorry i thought it was pi for a second

[old nebula]

but what about my integration though

it is incorrect, im getting the wrong answer

my calculator is not even processing it, it throws a math error

[blazing swallow]

here's the region in question

[old nebula]

yes

[blazing swallow]

your shells are sideways, so when they reach approx. y=0.13, the height function isn't the orange curve, it's "capped" at x=1

so if you wanted to integrate with shell method, you'd have to split it into two integrals, one for each piece of the piecewise height function

trying to figure out where that math error is coming from, though, and im not having much luck

[old nebula]

maybe cuz my integral involves ln and ln0 DNE

[blazing swallow]

well, that too

are you dead set on using the shell method? fwiw disc is more suited to this problem

[old nebula]

just for learning purposes yk

[old nebula]

ok so one from 0 to 1/e^2 and the second from 1//e^2 to 1?

[blazing swallow]

yes

and this is why i always recommend a sketch of the region, even if its quick and messy

it may help you to find intersections and bounds you didnt think of before

[old nebula]

even if i get 1/sqrt(1-x^2) ? 💀

its not always feasible

altho i agree w you, sketching helps

i cantt figure out the two integrals

can you help 😭

[blazing swallow]

sometimes you can cut corners, though not so easily in this case. but if you calculate a few values and connect the dots, it at least gives you an idea of where the intersections could be

[old nebula]

i mean the second one (from 1/e^2 to 1) is gonna be -pi ylny right

wat abt the other one

i should do it using disk method

its just a rectangle that i gotta revolve

[blazing swallow]

you could, but you can also do it using shell method

[old nebula]

wait really

how

[blazing swallow]

remember, y is the radius of the cross-section, and f(y) is the height

thinking of "sideways height", how far up does our region go in this area?

[old nebula]

so pi/e^2 or wat

the height is 1/e^2

oh well the sideways height is 1

[blazing swallow]

it's 1, yes

[old nebula]

why do i have to take the sideways height?

[blazing swallow]

1/e^2 is our bound, it is going perpendicular to the height we care about

[old nebula]

oh

[blazing swallow]

because we're integrating dy

[old nebula]

gotcha

ok so integral of 2 pi y from 0 to 1/e^2 ?

[blazing swallow]

that should do it

[old nebula]

got it

ty

appreciate it!

.close

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[safe radish]

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[tranquil geyser]

Ok im really struggling here haha

is A, B supposed to be 2x2 matricies?

[blazing swallow]

yes, they are 2x2 matrices and so are e^A and e^B

did your class happen to cover the Jordan normal form of a matrix and what the exponential of that is?

[quasi bison]

it says in the text that yes they are

[tranquil geyser]

yes

[tranquil geyser]

k

We will take $A = \operatorname{diag}(2\pi i,0)$. Then $e^A = \operatorname{diag}(e^{2\pi i }, e^{0}) = I$.

[flat frigate]

toast

[quasi bison]

ok, that works for A

but then you need to find sth that doesn't commute with this yet still gives identity when exp'd

[tranquil geyser]

hmm

@quasi bison Lmaooo i love gpt

[quasi bison]

!noai

[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).

[tranquil geyser]

yeah ik i was doing it as a joke but anyways i was curious to see

[safe radish]

@tranquil geyser Has your question been resolved?

[tranquil geyser]

looks good?

[safe radish]

@tranquil geyser Has your question been resolved?

[tardy mango]

looks good

[safe radish]

@tranquil geyser Has your question been resolved?

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[proud vector]

hello. if there are 21 available concepts, 7 will be on the final exam, and i need to answer 4, how many do i need to study?

[cedar widget]

all of them

[proud vector]

ok i get that education is important but mathematically how many

[cedar widget]

21 - (7 - 4) = 18

[rustic goblet]

18, because even if 3 out of the 7 are the ones you don’t know, you’ll know the other 4

[proud vector]

ok seen thanks guys

happy holidays

[safe radish]

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[pulsar pecan]

what are you stuck on?

[safe radish]

@old halo Has your question been resolved?

[safe radish]

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[safe radish]

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[swift surge]

Hey there! Having some trouble with these division problems

[rustic goblet]

your channel is closed already :<

you deleted your original message

I suggest you open a new channel since this one will lock soon

[swift surge]

Augh yeah sorry, I got my leg in there lol😭

Preciate you man!

[safe radish]

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[vestal ether]

@safe radish show me how this would look on a graph

[pulsar pecan]

That’s not what the bot is for

[pulsar pecan]

you can try plotting it yourself if you have graph paper

[safe radish]

@vestal ether Has your question been resolved?

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[trail portal]

(d)

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[trail portal]

2

[open wedge]

induction

do you know how that works

[trail portal]

Yes wait

I was gonna latex it

But I’m on iPad

I’ll just show my working

[open wedge]

aight

you got stuck midway?

[trail portal]

idk how to manipulate the bottom part

lmk if u want me to latex it

[open wedge]

wait a minute

you want to prove that it is true for $k + 1$

[flat frigate]

1 divided by 0 equals Infinity

[trail portal]

yes

[open wedge]

and if it's true for $k + 1$, then it must have a factor of $k + 1$

[flat frigate]

1 divided by 0 equals Infinity

[trail portal]

if true for k+1 then its true for all natural numbers

[open wedge]

so try manipulating $k(2k - 1) + 3$ so that there is $k + 1$

[trail portal]

how tho

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

distribute the brackets

and simplify into a quadratic

im at school rn lmao

[open wedge]

what do you get after that?

[snow fjord]

nothing

[trail portal]

,,k(2k-1) +3 =2k^2 -k +3

[flat frigate]

calvin

[trail portal]

Hey im also 14

[open wedge]

good

now since we know we need a $k + 1$ factor

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

you need to find a way to seperate your terms such that there is $k + 1$ common

[trail portal]

add 2k and subtract 2?

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

yep, add 2k and subtract 3k

wait a moment

this part is wrong btw

[trail portal]

,,\begin{align} k(2k-1)+3 &= 2k^2 -k+3 \ &= 2k^2 -k+3 +2k-2 -2k+2\end{align}

[open wedge]

you forgot a $2k + 1$ on a factor of 3

[flat frigate]

calvin

1 divided by 0 equals Infinity

[trail portal]

wdym

Did i screw my algebra up

[open wedge]

yea lol

[trail portal]

☠️

i knew it looked odd

so its 3(2k+1)

[open wedge]

if you turn into a quadratic

and do what i say

[trail portal]

instead of just 3?

[open wedge]

yep

[trail portal]

okay thanks

[open wedge]

do your algebra correctly next time

and you should have what you prove

[trail portal]

woops

[open wedge]

i gtg

bye\

[trail portal]

bai

.close

ty

[safe radish]

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[safe radish]

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[latent solar]

can anyone give me a hint on how to start here, idk where to start

[tardy mango]

Consider the behavior of $\frac{(k+1)^3}{k^3}$.

[flat frigate]

Civil Service Pigeon

[tardy mango]

The fact that ||this decreases|| suggests that ||Q(m) is eventually just 1||

[safe radish]

@latent solar Has your question been resolved?

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[safe radish]

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[patent laurel]

Hi guys, Can i ask what the difference of three examples in linear order differential equals? I'm confused of various way to slove it. In this d, we can assume the general solution y = k after replace into the formula and find k. But the next two problem is different....

[safe radish]

@patent laurel Has your question been resolved?

[safe radish]

@patent laurel Has your question been resolved?

[safe radish]

@patent laurel Has your question been resolved?

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[nocturne birch]

So i calculated zero points and i have a square root and idk how to mark that on the x axis

[safe radish]

Please don't occupy multiple help channels.

[nocturne birch]

.close

[safe radish]

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[safe radish]

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[upbeat drum]

.close

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[glossy orchid]

(√7+√5)/(√7-√5)+(√7-√5)/(√7+√5)=?

[kind seal]

!status @glossy orchid

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[glossy orchid]

2

[open wedge]

Common denominator

Please

[glossy orchid]

Ok it doesn't work for me

I have tired i got monster numbers

[crimson python]

!show

[safe radish]

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

[open wedge]

Specifically multiply with the conjugate

[near atlas]

is this your expression: \ \
$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$

[flat frigate]

T&C

[glossy orchid]

(√7+√5)/(√7-√5)•(√7+√5)/(√7√5)+(√7-√5)/(√7+√5)•(√7-√5)/(√7-√5) gets me stuck

[worldly plover]

It's gonna he some full Square and square identity

[glossy orchid]

Yes

[near atlas]

you could either take the LCM or multiply each term with the conjugate of the denominator of each term

[worldly plover]

First multiply the dominators by each other

And the whole fractions basically

[glossy orchid]

Ok just give me the answer✌️

I will figure out the solution to similar problems later

[worldly plover]

Lemme calculate

[near atlas]

i dont think thats allowed because many kids copy off the answer directly from here when they have assignments which do not require to show the work

[open wedge]

!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.

[glossy orchid]

Ok

[near atlas]

anyways if you take the lcm, u get a difference of squares

[glossy orchid]

Well thanks for help

[worldly plover]

It's gonna be 24/2

[glossy orchid]

.close

[safe radish]

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[open wedge]

!nosols

[safe radish]

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.

[near atlas]

and in the numerator you have something like (a+b)² +(a-b)² = 2(a²+b²)

[worldly plover]

Alright lemme send the whole solution

[open wedge]

!nosols

[safe radish]

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.

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[worldly plover]

Ima just send a pic of what to do to get to the answer lol

[worldly plover]

Ah shit

.close

[safe radish]

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[open wedge]

Same thing

[open wedge]

You're bascially giving the answer by doing that

[worldly plover]

Shit I didn't think about that

Eitherway, you are gonna have few identities which will get you to the result if he still wants help

[worldly plover]

Why does 1 deviced by 0 equals infinity btw

I wondered in the past few days but didn't research about it

[near atlas]

it does not equal infinity. it is simply undefined

[worldly plover]

I heard it is undefined for pre-university students. But I think someone told me it's actually defined somewhere

[near atlas]

look at the graph of 1/x near x=0 from both directions

[worldly plover]

I don't know what graphs are yet

[near atlas]

do you know how to use a calculator?

try dividing 1 by numbers like 0.01, 0.0001, etc

and then by the negatives of these

[worldly plover]

Oh wait

I just asked chat gpt. Basically when u devide 1 by a number almost as close as 0 you get a huge number

So that's why some call it infinity ig. But deviding it by 0 isn't possible appreantly

[near atlas]

it isn't possible precisely because when we try to divide 1 by numbers closer and closer to 0, we get drastically different values depending on which direction we choose to approach 0

for any other number, that is not the case

[worldly plover]

Ah yea

[near atlas]

again id say you try this experiment

[worldly plover]

We can say in some cases it can approximately be infinity, but deviding by 0 itself isn't possible

[worldly plover]

So devide 1 by -0.1?

Then 1 by -0.01?

[near atlas]

yeah

[worldly plover]

Alright lemme see

[near atlas]

and then with the positives of these

[worldly plover]

The number keeps decreasing

[safe radish]

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[grim plover]

me

[worldly plover]

Ah we can't tell which way it goes

Now I get it

[grim plover]

sorry for the interruption, i just didnt want to eat more pings >_<

[worldly plover]

Cuz - infinite 0 then a 1 is as close to 0 as + infinite 0s +1 at the end

[worldly plover]

If we didn't count the -s we could somewhat maybe say its infinity but yea now we can't tell

[open wedge]

Someone's using the channel here

!redir

[safe radish]

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

[worldly plover]

I was mid type when he joined

[grim plover]

the first is technically a direct antiderivative but i want a more proper method to solve this

[worldly plover]

Oki I'm done rn

[vagrant ice]

it's not your channel anymore

you were typing in a channel that you closed yourself

[worldly plover]

I'd like to allegedly apologies to all of our dear mathematicians.

Anyway Thanks alot for the help.

[grim plover]

ORIGINAL PROBLEM

[open wedge]

I don't really know about integrals

But one note: ||don't forget to add c||

[grim plover]

well then u really should go learn them first

[near atlas]

im trying to work the 1st one out pls wait

have you started with anything?

[grim plover]

yes the first one had options which revealed me to it was a direct antiderivative

but i want to know how to solve it without knowing that

ie cases like the generalised one

[near atlas]

i see. well with problems like these, i know that we start by taking the radical under the square root as u, but ive never worked with one where theres a quadratic expression in the numerator. ill have to see

[grim plover]

i did try that

ended up at this ugly piece of crap

[near atlas]

ohh

[peak estuary]

well in principle solvable by partial fractions

[grim plover]

how exactly?

[peak estuary]

factorize denominator, do partial fractions, profit

[grim plover]

but the roots will be ah

not even real lol

[grim plover]

surely that isnt the only approach here?

[vagrant ice]

,w (3t^4 - 2t^2 - 2) / (t^4 - 2t^2 + 2)^2 pfd

[flat frigate]

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Requested by sooouth.

[peak estuary]

the roots are like +-sqrt(1+-i)

could be much worse

[grim plover]

[vagrant ice]

still pretty ugly

but doable

[grim plover]

no hopes for the closed form of a generalisation?

[peak estuary]

I dont see a reason to believe that there should be a much nicer way

at least in general

if there was a nicer way to integrate rational functions I surely would know about it

[grim plover]

well if ur saying no i assume theres no nicer way

thank yu

.close

[safe radish]

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[fierce osprey]

.reopen

[lean otter]

Can someone tell me if this is supposed to be the distance between a plane and a line or what

[lean otter]

Idk what I’m supposed to do here

[last wren]

the length of the vector is the distance between the plane and the point, and the direction of the vector is orthogonal to the plane

do you know the formula for finding an orthogonal projection?

[lean otter]

Yes but not ona span

Just on another vector

<@&286206848099549185>

[tired spindle]

damn this channel still open?

[lean otter]

I reopened it a minute ago

[tired spindle]

🙏 mb bro

[lean otter]

It’s okay

I’m just tryna figure this out cuz test is coming up

I need to just know what to do with this problem and that’s it

[last wren]

alright, well if you have a set of orthogonal vectors (or an orthogonal basis), you can orthogonally project onto their span. if you want to project u onto the span of v1 and v2, then the orthogonal protection is

u - [(u • v1)/(v1 • v1)] v1 - [(u • v2)/(v2 • v2)] v2

[lean otter]

Wait so that’s just the gram process or what

[last wren]

yes

[lean otter]

Duuuude

Really?

Alright thank you very much

[last wren]

yep, np!

[lean otter]

The only confusing this is that they’re saying span

Like a projection on the other two’s span

Which I didn’t hear before lol

[last wren]

oh yeah, first you just find a vector that's orthogonal to another. then they try to generalize it usually

in the formula i gave you, the first two terms make u orthogonal to v1, and the last term takes that and makes it orthogonal to v2, so the whole formula makes u orthogonal to both. but note that this only works bc v1 and v2 are orthogonal to each other

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[weary ravine]

l need help

[weary ravine]

|GF|?

[brave wolf]

try finding the height of A

the one u drew

u only need this part for that

do u know how to do that or do u want a hint?

[safe radish]

@weary ravine Has your question been resolved?

[safe radish]

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[north orbit]

hey

for b

i basically said

suppose AB is diameter
AB = 2 * sqrt(26)

radius = 0.5 * AB = sqrt(26)

midpoint of the circle will be half way from A to B
M = (2, -5)

so, the equation of the circle will be (x-2)^2 + (y+5)^2 = 26
then i subbed in all of the points to show that the equation worked for them
this proves that ab is the diameter

does this work as a proof? this is not the way they did it on the mark scheme

[tardy mango]

looks good

[north orbit]

ok ty

.close

[safe radish]

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[tardy mango]

[round osprey]

how do i answer bottom

[safe radish]

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[verbal kite]

.(sec(c))^2 - (tan(c))^2 = 1 for all c

.The most direct path is to convert everything into sin() and cos() and let it unfold itself.

[ornate wyvern]

!nosol

dont send witten solutions to problems asked by OP

[stone spruce]

@round osprey just try to remember the main 3-4 trigonometric properties

[ornate wyvern]

explain what to do how to do to OP

[round osprey]

alright i will thanks again

[safe radish]

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.

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[verbal kite]

.reopen

.force reopenness

[vestal carbon]

This is not your help channel.

[stone spruce]

sorry i didnt know ill remembr now

tell me what to do

[vestal carbon]

Only the op and people will helpful roles can reopen and close channels

Type .reopen

[stone spruce]

okkk

.reopen

[vestal carbon]

Wait actually you cant do anything

[ornate wyvern]

oh u in rjnt too?

nice

[vestal carbon]

Nah you gotta make another channel

This will lock soon

[stone spruce]

yup

okk

[safe radish]

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[full sigil]

hey, i want someone to help me with some things in math

Calculus in R
Linear and Quadratic Problems
Polynomial
Arithmetic
Vector Calculus
Centroid (or Barycenter/Center of Mass)

I have an exam and am a total imbecile in math, no need for explaining all, just one of them so i can get at least a point or smth in my exam, sorry for the big request

[abstract pendant]

Send your questions, dont ask to ask if you can get any help :) Peopel will help you dont worry

[full sigil]

i just did, i don't have anything else, i just want somebody to at least tell me how to do any of these so i can pass the exam

[opaque fern]

Arithmetic -> vector calculus is quite the, uh, jump if i may add

[full sigil]

i translated that from french to english

Calcul dans R
Porblem 1ere degre second degre
Polynom
Arithmetiq
Calcul vectoriel
Barycentre

that's the french version

i will do anything to get at least 5 out of 20 in the exam

i am literally beyond fried that i had to join a discord server about math to get help

😔

<@&286206848099549185>

[plucky elk]

Then start doing problems or asking math questions

[plucky elk]

Read

[full sigil]

i don't have problems, i really just need some simple explanations and then i find problems to practice what that explanation explained

[plucky elk]

Then read your book or notes

[full sigil]

my book only has problems, the teacher is the one explaining which i didn't understand the explaination that he gave

can you please just at least help me with one of these just to do smth in the exam?

[plucky elk]

There's nothing to help because you haven't asked any math questions

[plucky elk]

Read your notes and ask questions about them

[somber cape]

You dont have any notes to review and see what subtopics you are specifically struggling with within those topics you listed? thats a very general list you gave

it would take days going over all of that, you should find what problems you arent able to solve and ask help on those problems

[full sigil]

i literally did, i asked if someone can help me with these, i just want some explainations and then ill get some problems and my book and solve them

[full sigil]

at least one of them will get me at least a below average grade

i don't want to ask ai or smth

[somber cape]

I dont think anyone here will have 3 hours or more to spare to help you understand those topics, your best bet is to try looking over notes or looking up youtube videos on the topics, then trying problems and coming back with questions about a specific problem

[burnt notch]

And that's an awesome thing

[full sigil]

fair

[burnt notch]

But what topic of this infiniiiiite list do you need help with specifically?

[full sigil]

any one, beggers can't be choosers

[somber cape]

If you had to choose which to study for your final then these 3 are pretty related
Linear and Quadratic Problems
Polynomial
Arithmetic

so try looking into those and see if you professor gave any information on what specifically he will test you on within these topics and if so look up some videos on them, theres a lot of good youtube videos out there

[full sigil]

tho i can get a calculator with me

could that help?

[somber cape]

depends on the questions, but if you know how to graph equations in your calculator it could most likely be of help

or your instructor could design the questions where a calculator wouldnt be helpful

[full sigil]

she said the exam will be extremely easy

like it's my first year as a computer science student, we are still new to this

but i really don't know what to do

[somber cape]

Im also a CS student and the math just takes a lot of practice. The big thing is not getting behind, so making sure you attend all lectures and keeping up with the homework and going to office hours when you are struggling

[full sigil]

i live in a third world country we just go in for an hour and leave

[somber cape]

The best thing you could do for yourself then as a cs students is rely heavily on books starting back at pre-algebra and reading up through Multivariable Calculus (Calc 3). During the summer after my freshman year I reread an pre-algebra and algebra book to refresh on my knowledge since I had never learned it properly. Even if your not prepared for this upcoming exam and you cant prepare in time you are still early in your life to start preparing for you future. I dont know how accessible books are for you, but I would do some research on good math books to read for each subject. Take a look at this video for book reccomendations for the early subjects
https://www.youtube.com/watch?v=ZWH-4GJaErM&t=13s

These books can be found online for free most of the time by various means

[full sigil]

its not my final

this is my second exam in the first trimester

[somber cape]

Oh ok, well good luck on your studying I have to go finish my work now

[full sigil]

good luck

.close

[safe radish]

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[safe radish]

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[safe radish]

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[harsh crow]

hello, i just need some quick help with something that idk how to do in probability

[harsh crow]

this is in french so let me translate

rq

it's more for the 2. part the 1st one i did it

[harsh crow]

we throw n times a coin, and we set X to how many flips we had to do to get the first tail

the probability of doing tail is p

and that's all

we have to find P(X=k)

then express it as G_X(x)

then find its average E(X)

[harsh crow]

for that i just changed the P(X=k) into p^k

in the formula they give

[left gyro]

you sure?

[harsh crow]

and changed removed the k=0 case

[left gyro]

whats P(X = k)?

[harsh crow]

p^n if k=0 or p^k if k>0

[left gyro]

thats not correct

[harsh crow]

(1-p)^n

mb

[left gyro]

thats not correct either

[harsh crow]

why??

[left gyro]

try not to guess quickly

whats the chance you get HHHT?

[harsh crow]

oh yeah it's binomial im stupid

[left gyro]

no

[harsh crow]

i mean no it's not

but it's simmilar

[left gyro]

no its not

it might be close to negative binomial, but youre looking for "geometric"

[harsh crow]

it's (1-p)^k-1 * p

or something like that

[left gyro]

whats the chance for H?

[harsh crow]

1-p

[left gyro]

then yea P(X = k) = p (1 - p)^(k - 1)

go try on the problems again

[harsh crow]

okay

yeah and now i have something in the form of a derivative

with k(1-p)^k-1

i mean p*sum_1<k<n ( k(1-p)^k-1 )

if i do G'_X(1)

no wait it's not the thing

[left gyro]

you can simplify this

[harsh crow]

i mean im sure it is but i dont get why it would be since 1-p is a constant

[harsh crow]

i dont get how tbh

[harsh crow]

(so im not sure)

[left gyro]

just to be sure, did you simplify G_X(x)?

[harsh crow]

i have p*sum_1<k<n ( (x^k)(1-p)^k-1 )

[left gyro]

you should simplify G_X(x)

[harsh crow]

idk how lmao sorry

like nothing is related

[left gyro]

it may not look like that at first, but think about where all the ks are at

[harsh crow]

i have only one x and only the powers are simmilar kindof

[left gyro]

the only k in the thing is as an exponent, right

[left gyro]

also your sum starts at 0, not 1

[harsh crow]

should i multiply and divide by 1-p to get them together under the same power of k??

[left gyro]

sure

[harsh crow]

0 is another case isnt it???? i mean P(X=0) = (1-p)^n

[obsidian oracle]

it looks weird

[harsh crow]

it's just a number it doesnt depends of k

it's a constant

[left gyro]

in general 0^0 should act as 1

Im not worried about it

Ill go check though

[obsidian oracle]

What did you find for P(X=k)?

[obsidian oracle]

this right

[harsh crow]

(1-p)^k-1 * p

[obsidian oracle]

that's for k > 0

what about k = 0

[harsh crow]

(1-p)^n

[left gyro]

oh whoops

[obsidian oracle]

there we go

and so the sum starts at k = 1

and there's an extra (1-p)^n on the side

[harsh crow]

yea

h

[left gyro]

P(X = 0) is p instead

[obsidian oracle]

?

[left gyro]

shouldnt P(X = 0) be p instead of (1-p)^n?

[obsidian oracle]

X = 0 if there were no heads

[harsh crow]

no why would it be this would just mean winning

[obsidian oracle]

so, each of the n flips were tails

at probability 1-p each

$G_X(x) = (1-p)^n + p\sum_{k=1}^n(1-p)^{k-1}x^k$

[flat frigate]

Raphaelisius Maximus MMIII

[left gyro]

X is how many flips you had to do for a tail though

not the number of flips that were tails

[harsh crow]

yeah so to get X=0 you have to fail each times

oooh yeah

i get u

wait

[left gyro]

thatd be > n flips for a tail, not "0 flips"

0 flips means you somehow already got a tail before even flipping the coin

[obsidian oracle]

it is specifically written

[harsh crow]

in the thing they say it explicitely

[obsidian oracle]

X = 0 if you never got tails

[harsh crow]

sorry

forgot to mention

but yeah 0 if no heads are gotten

no tails*

[obsidian oracle]

urrrrgh translating heads and/or tails in french pile/face shouldn't be that difficult

[harsh crow]

lmao

no but i should have said it

[obsidian oracle]

no that's not the problem, in my head idk if heads means pile or face

[harsh crow]

oh

face is on your head

your face

that's the way to remember it

[obsidian oracle]

that's what I thought first, except I'm told different things

[harsh crow]

nah

face is heads and tails is pile

[obsidian oracle]

yeah

[harsh crow]

and tbh it doesnt change much as we would understand eachother

if we believe the same thing

[obsidian oracle]

it's weird how we say pile first but in english we say heads first

anyways

$G_X(x) = (1-p)^n + \frac{p}{1-p}\sum_{k=1}^n(1-p)^{k}x^k$

[flat frigate]

Raphaelisius Maximus MMIII

[harsh crow]

yeah i have that

[obsidian oracle]

do you recognize anything

in the sum

[harsh crow]

newton binomial??

nah

lmao

[obsidian oracle]

nah

should I rewrite it more explicitly

[harsh crow]

derivative thingy but idkkkkk

[left gyro]

I think you should give a hint

[obsidian oracle]

$G_X(x) = (1-p)^n + \frac{p}{1-p}\sum_{k=1}^n[x(1-p)]^k$

[flat frigate]

Raphaelisius Maximus MMIII

[harsh crow]

i just know there are some tricks with derivatives with that type of things

[left gyro]

thats a big hint

[obsidian oracle]

is it clearer now what the sum is

[harsh crow]

yeah i have that too tbh

[obsidian oracle]

and even with that you don't know what the sum is???

[harsh crow]

oh it's just a 1-q

thing

i feel a bit stupid

[obsidian oracle]

geometric sum

[harsh crow]

yeah

but then i dont really need G_X i need to know G'

[obsidian oracle]

but... if you simplify G

[harsh crow]

oh yeah

sorry

[obsidian oracle]

then you can compute G'... using the simplified version of G

[harsh crow]

i can differentiate G after simplifying it

you're right

sorry

tho it's a mess to diferentiate

that's annoying

like it's U/V

[obsidian oracle]

at least the expression will be simplified, and not a sum

[harsh crow]

i feel like im doing something wrong seeing how long it is