help-49

[midnight plank]

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[last slate]

what means A.P.

😔

[modern sapphire]

Arithmatic progression

it is a sequence a1, a2, a3, ... where a(n) - a(n-1) is constant for all n

[late palm]

<@&286206848099549185>

[late palm]

Someone please help

[fallen aurora]

ayo

[fallen aurora]

what's the matter?

[late palm]

The question

[late palm]

We have to proof

[fallen aurora]

what have you tried?

[fallen aurora]

anything?

[late palm]

Just used the sn formula

[late palm]

This question

[fallen aurora]

perfect

[fallen aurora]

in the first equation you made
divide both sides by p

divide by q in second one

and also in 3

you'll get the required things
a/p, b/q and c/r

then plug that information in the expression we want to prove
after that it is just algebra

[modern shard]

yo

[late palm]

Yo

[modern shard]

follow 77² approach and you will get to your ans

[late palm]

This is what I got

[fallen aurora]

yup

just plug those values in given expression

[late palm]

Value of a/p, b/q???

[fallen aurora]

yup

it will be a bit messy but will work

[late palm]

Ok

[midnight plank]

@late palm Has your question been resolved?

[modern shard]

everything will cancel out at last

[fallen aurora]

you literally tried it☠️

I just wrote a few terms and made a hypothesis

[modern shard]

na, its obv cuz we have prove it's 0

lmao

[late palm]

Sorry I'm not able to solve

[fallen aurora]

what happened

maybe share the progress

[late palm]

Ok sharing

[midnight plank]

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[late palm]

.reopen

[midnight plank]

✅

[late palm]

Idk what to do

[fallen aurora]

I've said that it will be messy

you have to expand all of it

then everything cancels out

[late palm]

Ohh

[modern shard]

BTW q-r will be multiplied in whole bracket and not only to (p-1)d

[late palm]

Ya

Doing...

[modern shard]

[fallen aurora]

hope you don't miss the 2 in denominator of 2nd term

[late palm]

Ya😅😅

Silly mistakes

[fallen aurora]

good thing we care those things here

[late palm]

That can ruin the whole ans

[fallen aurora]

fr

[modern shard]

that mistake would have costed 💀

[late palm]

Fr

[modern shard]

Sun Flour be like:
phew, close call

[late palm]

Lol

Everything got cancelled

Yayyy

The most fun part of a question when everything's start getting cancelled

[fallen aurora]

golden words

[late palm]

Thank you @fallen aurora @modern shard gaaizzz

[fallen aurora]

gaaizzz☠️

overpowered

[late palm]

Lol

[modern shard]

i was gonna tell a joke but i won't as it might get me banned 💀

[fallen aurora]

!done

[midnight plank]

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[pearl hull]

.close

[midnight plank]

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[pearl hull]

Tell plz

[modern shard]

nahh man

[midnight plank]

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[late palm]

Dm me bta do

[fallen aurora]

!done

[midnight plank]

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[fallen aurora]

.close

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[paper galleon]

Will someone help me with this integral? Integral of (10csc2x-9cotx-2x)/(xcotx-5)?

[paper galleon]

Anyone?

[midnight plank]

@paper galleon Has your question been resolved?

[paper galleon]

<@&286206848099549185>

[frigid nexus]

?

[gusty birch]

@paper galleon ?
$\int \frac{10csc(2x)-9cot(x)-2x}{xcot(x)-5} ,dx$

[grand pond]

woomy

[paper galleon]

Yes this one

[gusty birch]

try x = cos^-1(u)

[paper galleon]

Is it just a fluke?

[gusty birch]

wdym?

[paper galleon]

I mean will x=arccosx(u) work ?

[gusty birch]

cot(x) becomes 1/u, csc(2x) becomes 1/sin(2cos^-1(u)) use the double angle identity, and the fact that sin(cos^-1(x)) = sqrt(1-x^2)

[gusty birch]

not sure

i havnt tried it

with these types of integrals you just gotta throw stuff at the wall and see what sticks

try different substitutions

maybe by parts

[paper galleon]

Ok

[gusty birch]

have you made any progress so far?

!status

[midnight plank]

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

[paper galleon]

2

[gusty birch]

send the progress u made so far

[paper galleon]

Hold on

[surreal moon]

I would turn everything into either sinx or cosx

[paper galleon]

Ok and does that leads to anywhere?

[surreal moon]

it would probably help the most

just offerring suggestions

[paper galleon]

Ok I have found the answer with this weird trick

.close

[midnight plank]

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[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[void lily]

is 2x apart of the square root?

if so

would it be x^2x?

and if it isnt apart of square root what do i do

[feral sedge]

no

this expression is (2x) multiplied by sqrt(x)

[void lily]

so

2x multiplied by x^1/2?

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[midnight plank]

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[silent scaffold]

I don't understand where to start here

The base points

I don't understand it

I know that one and zero is set as base point, but Im not sure how to set up the log

Because if I try to convert it any way I get x = 5^x

Which doesn't give the right answers

[hearty rune]

if log_5(x)=y then x=5^y if thats what you mean?

[silent scaffold]

Do they count the f(x) as y?

I just used f(x) as x since I was reffering to it as sm else

[hearty rune]

yeah im using y as a stand in for f(x)

[grand pond]

AℤØ

[silent scaffold]

ohh, I think I get it so then it'd be 5^y = x

and you'd fill in y for 1 and 0 so thats why the x's are like that

at least thats what I assume

Thank you so much!!

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[midnight plank]

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[rigid turret]

is x^2*y^2 in the log or only x^2

[hearty rune]

i would assume both are in

[grim vector]

Brackets missing, ambigious

Lets say both indeed

[hearty rune]

usually if its out theyd put it out front, but it is a dodgy presentation

[delicate wharf]

Yeah probably

[grim vector]

Ye you right

[rigid turret]

thanks guys

yeah its probably both

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[midnight plank]

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[merry peak]

it says the correct answer is -4/5(ln2)

[merry peak]

we’re supposed to express the answer as a constant tines ln2, which i did

[hearty rune]

,rccw

[merry peak]

they used u-sub, but i dont sre why we need that? unless i did somethibg wrong

[hearty rune]

you didnt integrate correctly

[merry peak]

is the integral of 1/(2-5x) not ln|2-5x|?

[hearty rune]

no, the integral of f'(x)/f(x) is ln|f(x)|

you didnt have that

if you can get it into that form by factoring something out, then you can do it

[merry peak]

heres the correction on my quiz, ignore that 1.2

,rccw

[merry peak]

all they did differently here, as far as i can see, is u-sub

[hearty rune]

yeah? What do you mean 'all they did differently'

[merry peak]

well why do we need u-sub?

[hearty rune]

you dont have to do a sub, but what you did doesnt work

[hearty rune]

you were trying to use this property

but it wasnt your integrand

[merry peak]

so what would the integral of that be?

if not ln

[hearty rune]

there is an ln

do the sub and see

[hearty rune]

or try to get it into a form where you can use this directly

[hearty rune]

dont forget to change your bounds

[merry peak]

ok i did u sub, so from here, i can't do u^-1

tes of course

but does it now become ln|u|?

[hearty rune]

it does, yeah

[merry peak]

what makes 1/(2-5x) different than 1/u?

[hearty rune]

the -1/5 is the difference

[merry peak]

ok lemme

go through this problem

[hearty rune]

$-\frac{1}{5} \int \frac{-5}{2-5x} dx$ is the whole picture

[grand pond]

AℤØ

[hearty rune]

this is in the form i spoke about

f'(x)/f(x)

[modern sapphire]

the 5x grows 5 times as fast wrt x, so you can think that 1/-5x decreases five times as fast. So you can intuitively see that the integral has to be scaled appropriately

[merry peak]

now using usub i get the right answer, and it makes sense

so we use usub to get it in that form?

[hearty rune]

didnt have to use the substitution if you knew what you were going for

[hearty rune]

could jump straight to this

but if unsure, the sub will often work

[merry peak]

ok no i see

if f(x) = 2-5x

then the top would have to equal -5 to be able to use that property

Ah

[merry peak]

tank u

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[bronze chasm]

can someone send me a video if i just entered geometry that will teach me alot pls

[midnight plank]

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[latent osprey]

inbox

[midnight plank]

@latent osprey Has your question been resolved?

[midnight plank]

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[granite smelt]

Say f is a function s.t. f : R —> R
f(a+b) = f(a)+f(b)-1
Can i conclude anything about the function? Hopefully f(x)=cx+1?

[granite smelt]

a and b are free variables btw

[viral dagger]

yo candies remember me

[pine wave]

it's quite close to the cauchy functional equation

you can probably transform it to the cauchy functional equation

then you can import all the theory from there

,w Cauchy functional equation

[grand pond]

Results provided by WolframAlpha

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Requested by element118

[granite smelt]

Yes

Actually

Generalize my question please

[pine wave]

what do you mean

[granite smelt]

F(a+b)=f(a)+f(b)-k

[pine wave]

yeah you can also solve for that too

[granite smelt]

Can i conclude f(x)=cx+k

[granite smelt]

How tho

[pine wave]

let g(x)=f(x)-k

[granite smelt]

Okay

[stoic sandal]

hello

what is x+3=439508

[granite smelt]

G(x)+k=f(x)
G(a+b)=g(a)+g(b)+2k

Uh oh

Where do i go from here

[stoic sandal]

uhhhhh candies you did wrong

[granite smelt]

How so

[stoic sandal]

wait

nvm

idk also

[viral dagger]

wtf 😭

[granite smelt]

???

What is bro on

[stoic sandal]

what grade r u

[granite smelt]

8

[stoic sandal]

im 7th

[viral dagger]

wh

[granite smelt]

Hi 7th im candies

[viral dagger]

arent you 14 yo like a year ago

[granite smelt]

I’m still 14 wdym

[viral dagger]

wrow

[stoic sandal]

hello 8 my name is gojo

[granite smelt]

Okay anyway

Can anyone help on my question

[rotund veldt]

you can get that the function is point symmetry

i think

[granite smelt]

Mind if you elaborate?

[rotund veldt]

a = x and b = -x then f(x) + f(-x) = 2

[granite smelt]

f(0)+k (aka a constant) = f(a)+f(-a)

I see

What does that contribute

[rotund veldt]

so cx+1 would work because its also symmetrical to (0,1)

not sure if that concludes it tho

[granite smelt]

Hm

[granite smelt]

Also a problem is cauchy’s is over Q so

[rotund veldt]

or maybe u can try getting the derivative

f'(2x) = f'(x)

[pine wave]

there are the pathological solutions

[rotund veldt]

or maybe just f(x) = g(x) -k, g(a+b)=g(a)+g(b) would work

[pine wave]

said that earlier lol

[rotund veldt]

yea

[pine wave]

https://en.wikipedia.org/wiki/Cauchy's_functional_equation#Existence_of_nonlinear_solutions_over_the_real_numbers

basically you need to choose a Hamel basis

[granite smelt]

Wait setting f(x)=g(x)-k gives that?

[rotund veldt]

f(a+b) - k = f(a) - k + f(b) - k

[midnight plank]

@granite smelt Has your question been resolved?

[midnight plank]

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[bright shoal]

Where to submit solution for challenge problems

[bright shoal]

Closed

[fallen aurora]

where are challenge problems?

[bright shoal]

It closed in 2021 for real

[modern shard]

[bright shoal]

Yes I feel this bro

It closed

.close

[midnight plank]

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[manic bison]

Let a,b,c,d be positive integers such that GCD(a,b) = 24 and GCD(b,c)= 36 GCD(c,d)= 54 and GCD(a,d) lies between (70,100) which must be a divisor of a
A)5
B)7
C)11
D)13

[manic bison]

I have factorised 24,36 and 54

Dunno what to do next

Nvm got it

It's 13

.close

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[dusky cradle]

$a_n=(a+bi)^n$ such that $\sqrt{a^2+b^2}<1$

[grand pond]

Raviti

[dusky cradle]

how to find general term of it?

[dreamy lichen]

Try polar form

[surreal moon]

Do you mean to find the real and imaginary parts of the final values?

[dusky cradle]

Yeah

Just general term

[dusky cradle]

i only know that e^(i theta)= cos theta+i sin theta

[surreal moon]

Polar form is good advice imo. Also you can do binomial expansion if you want to ignore the trig work.

[surreal moon]

Lore dump: $(e^{i\theta})^n=e^{i(n\theta)}=\cos(n\theta)+i\sin(n\theta)$

[grand pond]

SWR

[gleaming spear]

fax

[dreamy lichen]

The point is that if you convert it to a polar form, that is r*e^(i theta), then you can apply what SWR showed above and immediately get the real and imaginary part

r is the magnitude, and theta is the argument / angle

[dusky cradle]

But it is (a+ib)^n

Not (cos theta+i sin theta)^n

[dreamy lichen]

right, for now just focus on the a + bi part

[dusky cradle]

a,b are unknowns

[dreamy lichen]

so we have certain number a + bi in the complex plane

[dusky cradle]

I am agree with it

Only b

[dreamy lichen]

Yeah, right, sorry

[dusky cradle]

B on the imz

a,b are real

[dreamy lichen]

We can also define that number through the angle θ and magnitude r

the way we could write it is r * e^(iθ)

e^(iθ) would give us the point on unit circle with argument (angle) θ, and the r then rescales it to magnitude r

So now, if we can express r and theta in terms of a and b, we can next do the following:

(a+bi)^n = (r * e^(i theta))^n = r^n * e^(i theta n) = r^n * (cos(θn) + i*sin(θn)) = r^n*cos(θn) + i*r^n*sin(θn)

Which gives us the real and imaginary parts

[dusky cradle]

Actually i am trying to understand epsilon definition of sequence

[dreamy lichen]

epsilon definition of limit of sequence?

[dusky cradle]

Yes

[dreamy lichen]

and you are supposed to find the limit of the sequence above?

and prove it with the epsilon definition

[dusky cradle]

https://www.desmos.com/calculator/fackubmjh5

[dusky cradle]

Someone gave me this to feel it

So that i can understand better

[dreamy lichen]

The can help you see that the limit is 0

[dusky cradle]

Limit is 0 no doubt

yeah now try to find M such that for all N>M |a_n-0| < epsilon

I am not understanding N,M amd a_m

[dreamy lichen]

you need to find some term a_m, such that every term (a_n) that comes after a_m is at most epsilon apart from the limit (0)

[dusky cradle]

Is at what

[dreamy lichen]

ill try to explain it on an example

[dusky cradle]

Sure sir let's do it

I can give an example

And picture too

[dreamy lichen]

ah okay, that's nice

let's take the second pic

epsilon is 0.4 there

[dusky cradle]

Yeah

[dreamy lichen]

we need to find some term M, such that every term that comes after a_m is closer to 0 than epsilon

the colored range is the range of all values that are within the distance of 0.4 (epsilon) from 0

Now this term here can serve as our a_m

every single term that comes after it falls within the range

the green term is around 0.5, so it's not within the epsilon range yet. But the next term looks to be at somewhere around -0.35, which is less than epsilon apart from 0

so it falls within the range

and the next term is somewhere around 0.3 maybe, that one is less than 0.4 apart from 0 as well

so it again falls within the range

and all the following terms fall within the range as well

and that's what we needed

And if we can proof that we can find such term a_m for every value of epsilon (>0), then we have proved that the limit is 0

[dusky cradle]

Awesome

Thank you thank youuuuuu

So much

[dusky cradle]

Now let's back to this one

here limit is 0

[dreamy lichen]

Right

Firstly, it would be great to determine a formula for magnitude of a_n

you can assume that magnitude of a_1 is some r < 1

we know that the r = sqrt(a^2 + b^2), which is given to be less than 1

so if a_n = (a+bi)^n, and |a+bi| = r, do you know what will |a_n| be?

[dusky cradle]

r^n

[dreamy lichen]

correct

[dusky cradle]

And so?

[dreamy lichen]

so in fact, we just need to prove that the sequence
r, r^2, r^3, r^4, r^5... converges to 0

so now we need to do the epsilon part

[dusky cradle]

And why we are looking magnitude only?

[dreamy lichen]

yeah now try to find M such that for all N>M |a_n-0| < epsilon

it's in the definition

we only care about the magnitude

about |a_n - 0| in this case

so we need to find some m, such that for n > m, r^n < epsilon

[dusky cradle]

So we must know the limit 0 before it?

As we are putting into it

[dreamy lichen]

Yes, ideally

epsilon delta is mainly used to prove the limit, once we get a good guess

[dusky cradle]

Suppose If I imagine a fake limit?

Like limit is 1

[dreamy lichen]

then you won't be able to do the epsilon proof, because you wont be able to find the M

[dusky cradle]

|an-1|<epsilon for all n>m

[dusky cradle]

i want to see it why

[dreamy lichen]

because there will be no such m.

here is your diagram, we can see that the terms are getting further and further away from (1, 0), which is your fake limit

so if we chose let's say epsilon = 0.01, then there would actually be no term that's close-enough to (1, 0)

because the closest point is at around (0.8, 0.1)

[dreamy lichen]

Much less would there be a term a_m, such that all the terms that follow it would be close-enough to (1,0)

[dusky cradle]

For all n>m or just one?

I guess all n right

[dreamy lichen]

For the real limit, we will need to prove it for all n > m. Yes

[dusky cradle]

Epsilon describes the distance right?

[dreamy lichen]

Yes

[dusky cradle]

Around that point

A_m

[dreamy lichen]

not quite, ill add it to the desmos graph, wait a moment..

here i chose epsilon = 0.2

the blue circle shows all points whose distance from L, not from a_m is smaller than epsilon

If I put it really informally, what we are proving is that for any epsilon, there is always an infinite tail of the sequence in the circle

https://www.desmos.com/calculator/vkuzx0neps

here is the graph btw

I can also demonstrate the problem of choosing a "fake" limit on this

Here I chose a fake limit -0.2

we can see that it's not anymore true that for any epsilon, there is an infinite tail in the circle. Now there is only 5 points inside that circle.

And actually, if i chose a smaller epsilon, there would be no points at all

[dusky cradle]

Why you said l not a_m?

You meant we are looking at points inside the circle not that bright red dot right sir?

[dreamy lichen]

the red point is a1. It's the first term of the sequence
the purple points are hen the other terms of the sequence

Then we have the orange point L. It's the limit to which the sequence converges

and it's also the center of blue circle with radius epsilon

Now to prove that the sequence converges to L, we need to prove that there is some am, such that every single term an that comes after it falls inside the blue circle

as shown on the pic

[dreamy lichen]

and we need to prove this for every possible epsilon

[dusky cradle]

Ahhha

Awesome

You made my soul happy

With this concept

[dusky cradle]

Here i can say any point inside radius circle is a_m

[dreamy lichen]

Yes, any point inside that radius would serve as a valid a_m

https://www.desmos.com/calculator/t3jmz3pcpe
Final version of the graph btw. Here is the epsilon sequence convergence restated:

The sequence converges to limit L, iff for all values of epsilon, there is some term after which all the following terms are orange

[midnight plank]

@dusky cradle Has your question been resolved?

[midnight plank]

@dusky cradle Has your question been resolved?

[dusky cradle]

Wonderful

One more doubt arises about the sequence

When I set a_n equals to 1

.close

[midnight plank]

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[midnight plank]

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[lapis forge]

I would simply like an explanation on how this step was computed for solving the integral of xsqrt(2x+1)

[lapis forge]

I don't get where the 1/4 is coming from, nor why there is now subtraction in the integral

[lavish venture]

ok so if u=2x+1

what does x=

[lapis forge]

(u - 1) / 2

I think, yeah?

[lavish venture]

mhm

so then that’s the extra 1/2

because du=2dx

dx=du/2

[lapis forge]

I saw that one from the du = 2dx

[lavish venture]

that multiplied by^

gives the 1/4

that makes sense?

do you need to walk you through the rest?

for the subtraction

[lapis forge]

just don't I end up with

[lavish venture]

there should be another 1/2

[lapis forge]

that's the inside of the integral I mean so the 1/2 shown there would get pulled out

[lavish venture]

but notice the numerator is precisely u^3/2 -sqrt(u)

[lavish venture]

mhm

[lapis forge]

is usqrt(u) the same as u^3/2

[lavish venture]

well use your exponent rules

[lapis forge]

that was the disconnect I had then

[lavish venture]

recall that (x^n)(x^m)=x^(n+m)

[lapis forge]

I didn't realize it was the same thing, so yeah you can consider this solved then

[lavish venture]

in this case you have u^1 and u^1/2

[lapis forge]

I understand then

Thanks a bunch

[lavish venture]

you’re welcome a bunch

[lapis forge]

.close

[midnight plank]

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[midnight plank]

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[craggy owl]

Could someone check my work please? Og equation is $-34(x^6)+9(x^(68))y+(y^8)=-24$

[grand pond]

nothin

[small jasper]

,w tangent to -34x^6+9x^(68)y+y^8=-24 at (1,1)

[grand pond]

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[sinful trout]

...

[quiet hinge]

Checks out

[craggy owl]

Thanks

Wtf

[quiet hinge]

💀

[rain wasp]

<@&268886789983436800>

[crude summit]

yo what is this

[sinful trout]

UH

tf was that

[rain wasp]

frin

[craggy owl]

Crazy

[crude summit]

https://tenor.com/view/new-boot-goofin-goofing-gif-18618992

[sinful trout]

*shudders*

[craggy owl]

. solved

[sinful trout]

it's .close

[craggy owl]

.solved

[midnight plank]

Channel closed

Closed by @craggy owl

Use .reopen if this was a mistake.

[sinful trout]

ok

[olive matrix]

they both work

[crude summit]

works anyway xd

[sinful trout]

yea i didnt know lol

[midnight plank]

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[small jasper]

,w \sum^{\infty}_{n=1} \frac{(-1)^{n-1}x^{(3n-1)/2}}{n}

[grand pond]

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[small jasper]

Yes

Show me

You’re right until here

You solved the inequality incorrectly tho

Hint: Recall that |x| represents the distance of x from 0

[wispy fulcrum]

orz

[small jasper]

In light of this, what does |x|<1 represent?

Yup

So what values of x work?

Don’t forget ||negative numbers||

✅

Oh you might be using the convention that fractional exponents of negative numbers are undefined

If so, then it’s (0,1), then you need to check endpoints

See what I said above

[midnight plank]

@grave haven Has your question been resolved?

[small jasper]

Omit this for brevity, the rest is fine

[midnight plank]

@grave haven Has your question been resolved?

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[midnight plank]

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[tepid atlas]

how can i find the max while there is 2 variables

how can i eliminate a

and what does it mean by upper boundary

nvm

one of the solutions of derivative = 0 is an integer and thats the answer

.close

[midnight plank]

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[bright shoal]

💀

[midnight plank]

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[tepid atlas]

im backkkkkkkkkkkk

for c) i get 10

i plugged in 1,40 to original function

[bright shoal]

Yoooo

[tepid atlas]

to solve for a

[bright shoal]

See

[tepid atlas]

for part b the answer is 1

[bright shoal]

Find the x first for which

U get maxima

[tepid atlas]

yeah 1

so i thought we got (1,40)

[bright shoal]

Good

[tepid atlas]

ez plug in and solve

but nahhhh

[bright shoal]

Now put it

💀

[tepid atlas]

i got 10

answer is 0.1

[bright shoal]

See

[tepid atlas]

is it a mistake

See what

[bright shoal]

dy/dx is 0

If x=3 or 1

[tepid atlas]

technically x is a(x-3) or 1

[bright shoal]

Imma show u

[tepid atlas]

wait maybe calculator tripping

[bright shoal]

, w compute 3a(x-3)(x-1)=0

[bright shoal]

What

What

,w evaluate 3a(x-3)(x-1)=0

[tepid atlas]

nevermind

[bright shoal]

Why

[tepid atlas]

i missed a multiply sign

yeah x is 1 or 3

[bright shoal]

See

Ok

[tepid atlas]

the max is at 1 tho

[bright shoal]

See the domain is

0 to 3

[tepid atlas]

yer

[bright shoal]

Did u check for both 1 and 3

[tepid atlas]

i did the table thing

[bright shoal]

U sure?

[tepid atlas]

lemme try again

well the answer to x value for max is 1

plus if i sub in 3 to the original function u get 0

[bright shoal]

What

Okok

[tepid atlas]

the x value for the height to be a maximum is 1

[bright shoal]

So x=1 is max

[tepid atlas]

yea

[bright shoal]

,w 40=a(1-3)^2

[grand pond]

Results provided by WolframAlpha

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Requested by oppailol.

[tepid atlas]

see u get 10

[bright shoal]

Oof

I see it

It's given 40cm

Brother

40 cm

And in the question x and y are in m

[tepid atlas]

oh shize

oh shit

oh fuck

oh crap

[bright shoal]

Fr

[tepid atlas]

oh dinglididy dong

[bright shoal]

Fr what the shit moment

[tepid atlas]

im selling in life

[bright shoal]

[tepid atlas]

its over for me

ggwp

[bright shoal]

By your pfp

[tepid atlas]

lay down my weapons kinda momeny

[bright shoal]

I can confirm

U sell souls too

[tepid atlas]

i dont listen to travis

[bright shoal]

Jk jk

[tepid atlas]

years ago i had nitro and it was cool and animated

back when i thought i was sigma

[bright shoal]

Okok

Fr

Everyone is a sigma

[tepid atlas]

but i grew up

no point changing pfp

[bright shoal]

Till one becomes ligma

💀

[tepid atlas]

wow

anyway thank you

[bright shoal]

Ok 👌

.close

[tepid atlas]

farewell oppailol

.close

[midnight plank]

Channel closed

Closed by @tepid atlas

Use .reopen if this was a mistake.

[bright shoal]

farewell lebob

[wary thorn]

lebob

[viral dagger]

lebob

[verbal solar]

lebob

[wary thorn]

@bright shoal what happen

use lebob channel

[bright shoal]

Lol

see

I use to ping people when I wrote something which would help in their question when they were off or to ask consent for closing

[wary thorn]

.reopen

[midnight plank]

✅

[bright shoal]

But at a point the mods saw that it was getting too much

So I got timed out

💀

[wary thorn]

brooo why

did you apologize

[bright shoal]

;-;

[wary thorn]

they ban you for being too helpful

smh

[bright shoal]

To whom I should apologise

Idk who did

[wary thorn]

@shadow scaffold

We can DM this one

[pearl hull]

ig there are some hidden secret

[wary thorn]

probably @normal valley

[viral dagger]

honestly the only nitpick about you is you just randonly come in to say hello

[wary thorn]

oh big math is definitely hiding many things from us

[bright shoal]

Na i did not get banned for that

I got timed out for pinging

[pearl hull]

lol

[wary thorn]

@bright shoal

ping me any time brother

⚔️

[bright shoal]

But today I learned

@pearl hull this guy is definitely not bad at maths and English

.close

[wary thorn]

I will do it sir

One day

[viral dagger]

.clowse

[wary thorn]

we will avenge your helpful rank

Trust me

⚔️

.close

[midnight plank]

Channel closed

Closed by @wary thorn

Use .reopen if this was a mistake.

[bright shoal]

no prob

We here to help

Not for roles 👑

[wary thorn]

🔠

Exactly why you should have it

[bright shoal]

The only problem is that I can't access helpers lounge

💀

[viral dagger]

i sometimes help so that i can pass time while waiting for someone to help me lmao

[wary thorn]

nooo

[bright shoal]

I have seen it

[viral dagger]

skigma

[bright shoal]

Skill issue often comes

When I am there

@wary thorn let's close this for someone

.close

[wary thorn]

Sir yes sir

.close

I already ran the command

it will come close soon

[bright shoal]

Ok ok sir

[midnight plank]

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[fallen ridge]

Shouldn't here be n?

is this an error on their part or am I missing something?

[grim vector]

(-x)^n = (-1*x)^n = (-1)^n * x^n

Well yes it should be (-1)^n

With parentheses, not only the n on the 1

[midnight plank]

@fallen ridge Has your question been resolved?

[fallen ridge]

Ok thanjs

[midnight plank]

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[tired wind]

Hey y'all, can anyone help me in identifying what law of logical equivalence I can use to prove this without a truth table? I've looked for bloody ages and can't find one

[runic hamlet]

well that boils down to what your previous definitions, results etc are

if you for example know that a->b is the same as (not a) or b then your claim quickly follows

[tired wind]

I mean, I know that. But I haven't written that down or proved it, its just prior knowledge from lecture notes. But the task is to prove it from only logical equivalences. Does this mean I've got to go back to proving a->b ≡ ~a v b, and then use that to prove this?

[tired wind]

How can I prove that with logical equivalencies rather than truth tables? I'm given this, but I can't see anything there to use to convert if statements

[runic hamlet]

well how is a->b defined in the first place

if its defined using a truth table then you have to use truth tables (at least initially)

[tired wind]

Its not defined I don't think. I've got a whole big conditional statement that has multiple premises given in the form a->b, b->c ~c->a, etc. Then a, ->r. I have to prove that this statement is a tautology. Its easy enough once its in the form ~a v b, and i can use these laws of logical equivalencies from there. The issue I have is I have no justification to get it in that form at the moment. I hope thats what you were asking?

[runic hamlet]

if you dont have a proper definition of what the notation a->b even means then you can do literally nothing

[tired wind]

Isn't -> just an if? So, p->q means if p, then q

[runic hamlet]

well thats the intuitive interpretation. but its not a definition

[tired wind]

When dealing in discrete math I'm fairly certain its part of mathematical convention and is defined as an implication

I thought I'd be allowed to just assume that

[runic hamlet]

well its usually either defined by its truth table or as (not a) or b

but it needs some sort of definition

[midnight plank]

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[cobalt dragon]

answer is 1/4

conditional probability. but I'm having difficulty defining events

[bold beacon]

So how many random variables are there?

[cobalt dragon]

only 1?

[bold beacon]

For this question try list all the possible outcomes

[cobalt dragon]

TTT

[bold beacon]

on the right track but no

There are three random variables X1, X2, X3

This is because there are three coin tosses

[bold beacon]

Now you need to figure out the coin tosses that consist of at least two tails

[cobalt dragon]

binomial?

3C2(1/2)²(1/2)¹

[bold beacon]

Almost

[cobalt dragon]

3/8

am I going right

[bold beacon]

No need to use binomial distribution to calculate probability we will take another approach

List the outcomes where the coin tosses that consist of at least two tails

[cobalt dragon]

TTH
TTT

[bold beacon]

keep going

[cobalt dragon]

that's it

[bold beacon]

Okay you are assuming the first and second coin tosses are tails. But we don't know that.

[cobalt dragon]

oh

×3!/2!

[bold beacon]

So the other possible outcomes are THT and HTT

no need for combinatorics

[cobalt dragon]

i mean number of arrangements

[bold beacon]

Notice that P(TTT) = P(TTH) = P(THT) = P(HTT)

[cobalt dragon]

TTH THT HTT are 3 outcomes

[bold beacon]

yep

[cobalt dragon]

are you sure there would also be TTT

[bold beacon]

So we are going use of the cardinality operator i.e. ||. This counts the number of elements in a set. A set is just collection of objects

[bold beacon]

At least two tails would mean TTT is included

num_outcomes_at_least_two_tails = |{TTT, TTH, THT, HTT}| = 4

Because there four elements in the set

|{TTT}| = 1

So we get 1/4

[cobalt dragon]

💀

suppose even A and B
A= getting 3 tails
B= at least 2 tails

what event will be A intersection B @bold beacon

can we use this

[bold beacon]

The formula highlight in blue box is the definition of conditional probability

In this particular case A intersection B is just TTT

Since we know the probability of each event is equal we can just count total number of events.

[cobalt dragon]

so how will we find probability of it? P(TTT}

[bold beacon]

So we end up with num_events /total_num_events

[cobalt dragon]

will it be 1/8

[bold beacon]

You can find the P(TTT) but there is no need to in this question

Just use the binomial distribution.

[cobalt dragon]

it'll be numerator in conditional probability formula

[bold beacon]

Yep but approach I took above was different

[cobalt dragon]

if 3 tosses, 8 total cases.

each case will have equal probability

[bold beacon]

I think you need to take time to go through my explanation above

[bold beacon]

correct

[cobalt dragon]

you reduced sample space

[bold beacon]

correct

[cobalt dragon]

my teacher didn't teach reduced sample space. he used different approach

[bold beacon]

different approaches can be used

[cobalt dragon]

see I'm getting what you explained

but I can't have a different explanation for all questions under this same topic. i need to have a general concept which will solve questions

I won't remember it

[bold beacon]

It is good to be able to think about topics in multiple ways but it is up to you

[misty carbon]

if you want to do it as your pic says

you would do it like this

$P(A|B) = P(A \cap B)/P(B)$

[grand pond]

Heartuary

[misty carbon]

calculate $P(A \cap B)$ first

[grand pond]

Heartuary

[misty carbon]

in this case $A$ is 3 tails

[grand pond]

Heartuary

[misty carbon]

b is 2 tails

[cobalt dragon]

i understand that if question asked 3 coin tossed, and probability of 3 tail will be 1/8.
but if already occurred event is at least 2 tails, then we take 'at least 2 tails cases' out of sample space with 8 cases. then we'll get 4 outcomes sample space and then we choose one from it so 1/4. i

@misty carbon I'm seeing what u wrote

[bold beacon]

correct

[misty carbon]

then basically since the event A is a subset of B

[cobalt dragon]

please help

[misty carbon]

you only need to calculate $P(A)$

[grand pond]

Heartuary

[misty carbon]

P(A) is just prob of 3 tails

you said its a fair coin

so thats just 1/2^3

[cobalt dragon]

1/8

[misty carbon]

or 1 way in a sample size of 8

yes

now you calculate the denom

which is P(B)

i.e 2 tails

[cobalt dragon]

no, help me with P(A intersection B)

[misty carbon]

yeah i am

you need to calculate 2 parts

as in your pic

u see on the rhs

there is a numerator and denom

u just calc each part

[cobalt dragon]

which method will I use for P(B)

binomial?

[misty carbon]

if you like

if you know it

then sure go for it

[cobalt dragon]

3C2 .(1/2)².(1/2)¹

2 tails out of 3

[misty carbon]

nope

look at your event B

B is at least 2 tails

so 3 out 3 is also ok

[cobalt dragon]

oh

[misty carbon]

here

you said it yourself

so do binomial for 2 out of 3

and add it to 3 out of 3

to get at least 2 tails

[cobalt dragon]

so 3C2 .(1/2)².(1/2)¹ + 3C3 .(1/2)³.(1/2)⁰

[misty carbon]

yes

that gives 1/2 right

[cobalt dragon]

this correct?

[misty carbon]

ye its right

[cobalt dragon]

3/8 + 1/8

4/8= 1/2

[misty carbon]

ok so you have both

of the pieces on your right hand side

of your pic

[cobalt dragon]

wait what both

[misty carbon]

P(B) = 1/2 you got

correct?

[cobalt dragon]

um ok

[misty carbon]

you just did that right

3/8 + 1/8

like you said

[cobalt dragon]

and P(AÛB)

Û intersection

[misty carbon]

it should be upside down

$P(A \cap B)$

[grand pond]

Heartuary

[cobalt dragon]

I CAN'T WRITE IT OKAY IGNORE IT LMAO

[misty carbon]

just think about what this means

what is event A

no

go back to what you wrote

[cobalt dragon]

3 tails

[misty carbon]

yes

what is event b

[cobalt dragon]

at least 2 tails

[misty carbon]

ye

then so what is A and B

[cobalt dragon]

P(at least 2 tails) = 1/2

[misty carbon]

but you want to compute

the intersection

for the numerator right

like you said

[cobalt dragon]

but we haven't calculated AÛB

[misty carbon]

ye we are doing that now

you know what event A and B are

what is the event for which both A and B are satisfied

[cobalt dragon]

how will you define it

[cobalt dragon]

that's what I don't know

[misty carbon]

A n B means A and B

right?

[cobalt dragon]

yes

both condition

[misty carbon]

so how do u satisfy both

conditions

[cobalt dragon]

all 3 tails + at least 2 tail= all 3 tail

[misty carbon]

yes

exactly

[cobalt dragon]

yay

wooho

[misty carbon]

so can you calc that

[cobalt dragon]

but that'll be out of what

[misty carbon]

P(all 3 tails)

you can use your binomial

[cobalt dragon]

TTT out of 8 sample space?

[misty carbon]

like before

ye

[cobalt dragon]

1/8 | 1/2 ?

[misty carbon]

ye

[cobalt dragon]

is it done

[misty carbon]

ye

[cobalt dragon]

thanks heartuary 💋

I was missing out on at least 2 tail. instead i took 2 tails

and since event A is 3 tail, and that is only 1 case, and if question is solvable, then it means intersection will also be 1 case only? @misty carbon

[misty carbon]

wdym question is solvable

[cobalt dragon]

that means intersection won't be null, so if it asks intersection then we can just look at the event with minimum number of cases and take that as intersection. i mean not considering common cases, but if event has only 1 case

[misty carbon]

it can be null

then the probability is 0

if you like sets you can think of it like this

[cobalt dragon]

that's what I mean

[misty carbon]

A = {TTT}, B = {TTH, THT, HTT}

[cobalt dragon]

if question is solvable then it won't be null

[misty carbon]

so A n B = {TTT}

it can be null

for example if i ask you

what is the probability of 3 heads

given I flipped 2 tails

it is just 0

[cobalt dragon]

also.
at least 2 tails= TTH and TTT
we can go like
(1/2×1/2×1/2)×3 + 1/2×1/2×1/2
we can also calculate like this. apart from binomial we uses

×3 because of arrangements

[misty carbon]

sure you can do that

using binomial just gives you a general framework

supposed the coin is not fair

and it flips head 1/4 and tails 3/4

[cobalt dragon]

yeah but all ends open clears it a lot

[misty carbon]

binomial is easy replacement

[cobalt dragon]

the concept

[misty carbon]

ok

[cobalt dragon]

what would this indicate at @misty carbon , if it's coin toss

10C2.(1/2)².(1/2)⁸
does it mean exactly 2 head

[misty carbon]

2heads 8 tails

or 2 tails 8 heads

yes

[cobalt dragon]

if question asks only first 2 tosses are heads?

how would only first 2 heads differ from exactly 2 heads

[misty carbon]

what do you think

[cobalt dragon]

its just order

[misty carbon]

right

[cobalt dragon]

i still don't understand

[misty carbon]

if this is the case then what must be the sequence of flips

[cobalt dragon]

2H8T

[misty carbon]

ok

thats right

what is prob of that sequence

[cobalt dragon]

1/2¹⁰

[misty carbon]

ye

compare that to exactly 2 heads

see the diff

[misty carbon]

here

[midnight plank]

@cobalt dragon Has your question been resolved?

[midnight plank]

Channel closed

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Use .reopen if this was a mistake.

[midnight plank]

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[last slate]

[verbal solar]

calculator

[last slate]

sorry

[visual tiger]

well you do digit by digit

start with units

9-9

[verbal solar]

Just do it.

[visual tiger]

then tens digits