help-28

[torn jolt]

Guys

[onyx glen]

@torn jolt you are interrupting a conversation in a busy channel, AND giving out answers, AND what you're giving out as answers is wrong

[torn jolt]

Shush u busy

Mad ??

Close ur eyes

And dont look at my text

[cobalt kiln]

please im just trying to get some help

[torn jolt]

Okay

[cobalt kiln]

youre just being disruptive

thank you

[torn jolt]

U can just tell me to stay quite

Im leaving

Bye

[cobalt kiln]

okay why do you take an infinite summation

nevermind it makes sense

okay

so you get this

this looks similar to the taylor series for e^x

r^x/x!

oh wait other way around

x^r/r!

but what about the lambda

maybe you take the lamba ^x on the side

oh wait its just e^lambda t

so its e^ -lambda * e^lambda t

then you can group the powers

e^ lambda -lambda t

e^ t (lambda - lambda)?

that cant be right

oh yeah its e^ lambda t - lambda

e^ lambda (t-1)

[onyx glen]

hold on, my internet is continually shitting itself

test

test

test

okay, what the hell

i can't send any fucking messages for some reason???

[cobalt kiln]

omg

haha they all came through at once

[onyx glen]

missing parentheses here

also zero is still o

[cobalt kiln]

oh yeah didnt even realise the 0

okay i think i got the derivation down

okay so Gx(t) = E(t^x)

G`x(t)=E(X)

so you differentiate this function

what do you differentiate in terms of?

<@&286206848099549185>

[torn jolt]

?

[cobalt kiln]

hey can i have some help

the person who i was working with seems to be having connection issues

heres my original question

[onyx glen]

okay, i'm no longer on foot for now

[cobalt kiln]

sounds good

is it d/dt?

[onyx glen]

your pgf has only one argument, t

so that's what you take the derivative with respect to

[cobalt kiln]

okay

i have a question though

[onyx glen]

yes?

[cobalt kiln]

why did we take the sum to infintity

infinity

[onyx glen]

didn't i already say

[cobalt kiln]

i thought i got it but after looking at how i derived the pgf for binomial i got confused again

[onyx glen]

the sum ranges over all values that our RV can take with positive probability

and for a poisson rv that's all natural numbers

[cobalt kiln]

okay let me write out what i got for the binomial pgf

and you can fact check it

[onyx glen]

yeah sure

[cobalt kiln]

then you can rewrite as

[onyx glen]

seems legit

[cobalt kiln]

(1-p+pt)^n

[onyx glen]

yup

[cobalt kiln]

so why is the bound here n

but with poisson its infinity

[onyx glen]

because a Bin(n,p) random variable takes values between 0 and n inclusive only

[cobalt kiln]

ahh

okaayy

makes sense now

okay so back to differentiating

so you jsut get lambda as a coeficcient

thats G`x(t)

so you sub t=1

so you get lambda

then for var

its also going to be lambda because i forgot that both mean and var for X~po(a) is a

but as an exercise id still like to derive it

so for var you differentiate again

or can you use the regular var formula

var(X)= E(X^2) - E^2(X)

we already know the latter half

lambda^2

so E(X^2)

is going to be

times something

<@&286206848099549185>

i give up

.close

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[hot herald]

depends on how much precision you want

if you wanted to round to the nearest million, it'd round to 0

ok, what's your issue with this specific problem
assuming you were able to round that one yourself

[hot herald]

that would be fine

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[tender thicket]

can get a pointer on how to start solving this? does it require actually sketching smthn

[hot herald]

required? no
recommended, highly

[tender thicket]

can u decide whether to integrate wrt y or x w/o sketching the region?

[hot herald]

depends on your visualisation skills, you're better off just sketching it regardless

[tender thicket]

alright looks like i should be integrating wrt y

[hot herald]

yes

[tender thicket]

so, should be intregrating this: x = 2 - y^2, no?

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[sterile halo]

Can anyone help with these 3 please?

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[simple totem]

x and y are not 24 and 18

they are 48 and 36

although now that i think about it that may not change much since everything is multiplied by 2

ok no it does

so your equality should be

2 * 48 * 24 + 2 * 36 * 18 = 2 * 60 * z'

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@devout idol Has your question been resolved?

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[torn jolt]

not sure where to start, i was able to foil it but thats all

[torn jolt]

please tag m ewhen responding

[plush egret]

something tells me usub will end up working

[torn jolt]

tan would be u right

[plush egret]

so we split into three integrals

$\int \qty( \sec ^2 x - 2\sec x \tan x + \tan ^2 x ) \dd x$

[glossy valve]

jan Niku

[plush egret]

first one is just tangent

[torn jolt]

yes

i see now

[plush egret]

middle one is a usub

[torn jolt]

i just used trig identities

[plush egret]

$\int \frac{\sin x \dd x }{ \cos ^2 x }$

[glossy valve]

jan Niku

[torn jolt]

am i always suppose to remember that tan^2x = sec^2x - 1?

[plush egret]

ye

[torn jolt]

the answer for this will be 2tanx - x -2secx + C

[plush egret]

you shouldnt just share answers

especially without asking

[torn jolt]

why -x?

i got 2 tanx - 2secx + x + c

[plush egret]

. this

[torn jolt]

oh i see

thank you

[torn jolt]

this should be memorized?

[plush egret]

itd simplify your life

[torn jolt]

gotcha ty

@plush egret

i have a quick question for this

so i can rewrite it as

then i can do 1/x - 1/sqrt(x) right

so integration of first part minus integration of second part which is lnx - (2 * sqrtx)

.clsoe

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[willow wasp]

Yo I got a question

[willow wasp]

When I tried to solve this the step I’m on errors on my calculator

Can anyone point out any errors that i made thanks

[wheat jewel]

have you multiplied both sides by (2x-1)^2?

[devout valley]

It looks like they did, but

[wheat jewel]

bcoz expansion of (a-b)^2 is wrong

[willow wasp]

this part?

[devout valley]

The red circled where you expanded the brackets

That -2x term

,w expand (2x-1)^2

[glossy valve]

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[devout valley]

Oh yeah and also the 2x^2 term too lol

[rough plaza]

or you can do it like this? (without multiplying two times (2x-1))

\begin{align*}
&\frac{x}{2x-1}<5 \
\implies & \left(x<5(2x-1) \text{ and }2x-1>0\right) \text{ or } \left( x>5(2x-1) \text{ and } 2x-1<0 \right)
\end{align*}

[glossy valve]

OldBiscuit

[willow wasp]

oooooh alright thanks for the help everynioe

got the answer

.close

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[tacit eagle]

where did i mess up

[tacit eagle]

,rotate

[rough plaza]

-(2+√2)+6 this part?

[tacit eagle]

what

i just plugged into x

-x^3+6x^2

[rough plaza]

yea

x²(-x+6)

[tacit eagle]

tes 2sqrt2

[rough plaza]

tes?

[tacit eagle]

yes

plugging that

2sqrt2 into x

[rough plaza]

yea, but you got -6 instead of +6

[tacit eagle]

i factored out

(2+sqrt2)^2

but not the negative

ah damn

i see it now

wait but

is everything else good

besides

that

like lets just say

it was meant to be negative

i wanna know if my math is passable

l0l

fixing it

OK

@rough plaza how about now?

,rotate

[tacit eagle]

my answer is still wrong apparently

[rough plaza]

(6+4√2)(4-√2)
=6+16√2-4•2-6√2

[onyx glen]

what's the original problem

[rough plaza]

missed one term, and √2 times √2 is 2, not 4

[tacit eagle]

my x values r good

already checked

[onyx glen]

,w -3x^2 + 12x = 6

[glossy valve]

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Requested by Ann#0413

[onyx glen]

ok that they are

so it is indeed only a matter of arithmetic screwups

caused, no doubt, by the high amount of messiness in your work

[tacit eagle]

true lol

[tacit eagle]

even after i corrected that problem, i still got an incorrect answer

[rough plaza]

show your work again please

[tacit eagle]

i need to buy erasers in bulk

[onyx glen]

may i suggest something that can maybe make our lives easier with the evaluation of this thing

[tacit eagle]

and youre right ann

i need a better way of writing things out

do you have any examples i can copy

yes

suggest away

[onyx glen]

instead of x^3 + 6x^2, evaluate x^2(x+6)

will be marginally easier this way

less chance to screw up

that or just use wolfram lmao

,w f(x) = x^3+6x^2, find f(2+sqrt(2))

[glossy valve]

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[tacit eagle]

,rotate

[tacit eagle]

,rotate

[onyx glen]

gonna need to reupload yours and rotate it prior to upload

[onyx glen]

or that

...

that's the same work you showed already

[tacit eagle]

both are incorrect, here are the answers:

he told me

to fix something

and i did

but yes ann

i can do ur method

but i worked so hard here

where did i mess up

[onyx glen]

ah. i missed the minus sign.

[tacit eagle]

ill do ur method after

[rough plaza]

yea, and you still missed a term, which is -6√2

[onyx glen]

and i see where you messed up now

[tacit eagle]

where at

[onyx glen]

you missed the parentheses in (4+4sqrt(2)+2)[4-sqrt(2)]

[rough plaza]

6+16√2-4•2 "- 6√2"

[rough plaza]

oh, and it should be 6 times 4 instead of 6 for the first term here

my bad

lemme try and write it out

[tacit eagle]

no paren needed? foiled it out

the x^2 part

then i distributed the 4sqrt2 by the inside

4, and -sqrt2

[rough plaza]

,rotate

[tacit eagle]

where is ur x^3'd at

[rough plaza]

although my handwriting is bad, you can kinda follow that the rough work sticks to a side, and the steps are aligned with = sign

[tacit eagle]

(x)^3 + 6 (x)^2

is that

a factor out

ah yes it is nvm

why does 3rd to last row have parentheses

you foiled it out already

[onyx glen]

the point is that no, you didn't foil it out properly. the 6 was left intact.

[tacit eagle]

what 6

i addded 6 and -2

which made 4

[tacit eagle]

i did the same thing he did on 3rd row from bottom

[tacit eagle]

how do u get the last row

yes @onyx glen i understand what you mean

but why does there have to be parentheses

after you foil out (2sqrt2)^2 ?

[rough plaza]

(2+√2)² times (4-√2)

[tacit eagle]

ah

[rough plaza]

so, even after you expanded (2+√2)², the whole thing would still have to be multiplied by (4-√2)

[tacit eagle]

i see it

ty

had i known about ann's method

i would of been chillin

jk would still be screwed

woooooooooooo

https://tenor.com/view/michael-scott-happy-throw-confetti-the-office-celebrate-gif-14368600

we did it

thank god

ok

now time for anns method

lets see how fast i can do this

https://tenor.com/view/forblad-forbladgaming-maximum-concentration-concentration-maximum-gif-18285466

you guys are so smart

tyvm ❤️

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[last scaffold]

can someone help me with this

[last scaffold]

i have no idea what i should do

[marsh surge]

what's the question?

[marsh surge]

what have you tried?

[last scaffold]

i tried splitting the top

like this

[keen bolt]

you're trying to find the height right? Have you tried trig on it?

[last scaffold]

do you mean something like sin?

[keen bolt]

yeh, but for this one use tan, as you have the O side and trying to find the A side.

[full forum]

@last scaffold Has your question been resolved?

[last scaffold]

i did use tan for this one

it was wrong

i got 10.7

[torn jolt]

hey : )

do you still need help for this question?

[full forum]

@last scaffold Has your question been resolved?

[last scaffold]

yes please

[torn jolt]

well

let's start with part a

we have the triangle, let's call it abc

and it's an isosceles triangle

can you tell me the property of isosceles triangles?

@last scaffold do tag me when you're online again if you still need help 😊

[full forum]

@last scaffold Has your question been resolved?

[marsh surge]

correct, now using trigonometry, you can find the height (left vertical side) of the triangle

add that to the height of the rectangle and you will get your answer

[last scaffold]

so do i use sin for this

?

[marsh surge]

tan

since you don't have hypotenuse

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[bronze spire]

can i isolate dy/dx to the right side of the equation?

[plucky thorn]

you can develop the parentesis and isolate dy/dx to the right, but not sure i answer your question

[bronze spire]

so i cant directly transfer this to the right side?

[hallow walrus]

its multiplied isn't it?

part of a product

[bronze spire]

$18\left(sin^3\left(2x\right)+y\right)^2\left(sin\left(2x\right)+y\right)cos\left(2x\right)-4x^3y^2=\:\frac{dy}{dx}\left(-1+2x^4y-2\right)$

how do i write this

[glossy valve]

Kuugang

[bronze spire]

is this simplified version correct?

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[glass jetty]

is this 0.5?

[hallow walrus]

how did you get 0.5?

[glass jetty]

idk

its just appear in my head

sr

;-:

[hallow walrus]

no worries

think of it in this way

[glass jetty]

i suck at probability - conditional

[hallow walrus]

the total probability always has to be 100%

[glass jetty]

yes

[hallow walrus]

and you're given the probability of choosing Apple and Banana and the fact that the probability of Orange and Pear is the same

so

if i take the probability of Orange = probability of Pear = x%

[glass jetty]

0.3?

[hallow walrus]

can I say $5% + x% + x% + 35% = 100%?$

[glossy valve]

kheerii

[hallow walrus]

yes

[glass jetty]

oh damn

am smart

asf

tyy

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[frail cradle]

So I have no idea how to do this

[fleet briar]

so when you put x as -2

where do you think it goes

[frail cradle]

In x

[fleet briar]

@frail cradle is this the first time youve seen

a function like this

[frail cradle]

Yes

[fleet briar]

its called a piecewise function

basically, you see the inequalities on the right @frail cradle

the 4 of them

[frail cradle]

Ya

[fleet briar]

what you do is that, whichever x value the inequality fits for

you use that equation that is right next to it

ill give you another example

[frail cradle]

ok thank for understanding

[fleet briar]

say we had this piecewise function@frail cradle

what happens is that, if i want f(5)

since 5 is >= 1

we use

f(x) = x^2

and so, f(5) = 5^2 = 25

but if i wanted f(0.5)

since 0.5 is < 1

we use

f(x) = x

so f(0.5) = 0.5

[frail cradle]

so all x turn to 5?

[fleet briar]

the first step

is finding the inequality

that the x = value is for

so x = 5 is in the x >= 1 inequality

then when we find the inequality

we say f(x) = the equation next to the inequality

which is x^2

in this case

since thats >= 1

and then, we solve for f(5) knowing that f(x) = x^2

[frail cradle]

ok ight thank you

[fleet briar]

this is what it looks like on a graph

after x >= 1, the function turns into x^2

but before that

the function turns into x

[frail cradle]

thank you for the help i think i can get it my self will our converstion still be here after we close cuz i want to take nothes

notes*

[fleet briar]

yh, you can just search for the messages you have done

and find this chat from it

[frail cradle]

do i have to cloes this chat or you?

.close

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[torn jolt]

can someone expalin to me why the 1st part of fundamental theorem of calculus says that if f(x) is continuous on [a, b] F(x) is continuous on [a, b] and differentiable on (a,b)

[torn jolt]

like why these intervals?

[fleet briar]

f(x) might not be continous on other intervals that arent listed

like 1/x isnt continous on x = 0

so its not differentiable at that point

or integrable at that point

[torn jolt]

i know but why F(x) continuous and differentiable

[fleet briar]

do you disagree?

[torn jolt]

yeah like what about if f(x) has vertical tangent lines on [a, b]

[fleet briar]

then its not continous at that reigion

[torn jolt]

o

rly i thought vertical tangent lines still make it continuous

[fleet briar]

pretty sure the entire point of vertical tangent lines is that its not continous

[atomic blade]

It can still be continuous

x^(1/3) is continuous

Not all continuous functions are differentiable for all values of x

[fleet briar]

wait yh youre right

[atomic blade]

A vertical tangent will make f'(x) discontinuous where the vertical tangent will be

But f(x) would still be continuous

x^(1/3) is a case where f(x) is continuous but f'(x) is not (assuming a domain of R)

[torn jolt]

nvm i think intuitively iti doesnt matter if there is a veritcal tangent

but still why does F(x) is continuous on [a, b] and differentiable on (a,b) have to be true

[atomic blade]

Endpoints are not differentiable

Think of a line segment

Or a curved segment

[torn jolt]

ooh

[atomic blade]

And assume concavity to be the same

(To avoid confusion of what a tangent line really is)

Then the endpoints are not differentiable

You can have an infinite number of tangent lines

[torn jolt]

just to be clear a single point is not differentiable

?

[atomic blade]

Well if it's literally by itself then yeah it's not differentiable

[torn jolt]

its cuz its not continuous right

[atomic blade]

Supposedly yeah

Not continuous

[torn jolt]

so thats why endpoints are not differentiable?

[atomic blade]

I guess you could make that argument

[torn jolt]

cuz one side of it is not continuous?

[atomic blade]

I like to say that if you have a segment, then there's an infinite number of tangent lines at the ends

[torn jolt]

yeah taht makes sense

[torn jolt]

yeah i knew what you meant

[atomic blade]

Perhaps it's something to do with continuity

[torn jolt]

so that means F(x) is not necessarily differentiable on a or b because it would just be an endpoint?

[atomic blade]

Mmhm

It could be an endpoint

[torn jolt]

it just feels weird that F(x) has to be continuous too because i know there are some integrals of non continuous functions

like piecewise functions with jump discontinuities still have integrals

actually

nvm those integrals are continuous

what the hell

[atomic blade]

Eh not always

Can't think of any rn but I did encounter one

Although you can force them to be continuous lol because of constants

[torn jolt]

not sure what you mean

[atomic blade]

Basically you can adjust the pieces

In terms of their vertical translation

So that they're continuous

But that's just a funny moment

I wouldn't worry too much

[torn jolt]

huh

what do you mean

you're changing the function itself?

[atomic blade]

You're familiar with integrals right

$\int f(x) \dd x = F(x) + C$

[glossy valve]

Umbraleviathan

[atomic blade]

Each "piece" of the piecewise integral (if the antiderivative is another piecewise function) will be associated with their own unique F(x), whose unique C can be adjusted such that each piece "connects"

[torn jolt]

i meant that a function like this has a continuous integral right

[atomic blade]

It could

[torn jolt]

ohh crazy crazy

[torn jolt]

ok thank you i think i sort of get that F(x) has to be continuous on [a,b] and differentiable on (a,b) now

das still crazy tho

that means literally any continuous function has a antiderivative

.close

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[atomic blade]

Some go into complex planes but

Don't worry about that

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[fleet epoch]

help, so i am doing a mathematical report comparing correlations of two subjects, and I did pearsons product for one of them and the correlation is 0.118

[fleet epoch]

should i do spearsman, because i visualized the data and it looked like this:

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@fleet epoch Has your question been resolved?

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@fleet epoch Has your question been resolved?

[plush egret]

this looks like it needs a transform

at the very least

idk spearsman, sorry

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[versed cape]

What would be the answer for this

[full forum]

@versed cape Has your question been resolved?

What step are you on?
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2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
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5. I have a question about someone else's worked solution
6. None of the above

[versed cape]

<@&286206848099549185> any help would be appreciated

I dont need the full answer i just need an idea on how to factor it out so that it wont equal 0

[gritty rose]

Try x=t^3 substitution and factor the denominator

[versed cape]

alright

[versed cape]

Stuck on this step

[versed cape]

Original question <@&286206848099549185>

[void widget]

Just put the limit in

[severe topaz]

is the cube root only on x or on x-8 ?

[versed cape]

just x

[severe topaz]

then its just 0

[versed cape]

Cant equal zero bro

[void widget]

Just put the value of x in

[severe topaz]

cube root of 8 is 2 so the denominstor is 2-8 = -6

[versed cape]

My guy

[void widget]

Why it can't be 0?

[severe topaz]

and the numerator is 0 , so the limit is 0

[void widget]

,w factor x²-11x+24

[glossy valve]

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[void widget]

What it can't be 0?

[versed cape]

?

[void widget]

It can be 0

[severe topaz]

yeah

[versed cape]

You have to factor it out so that it wont equal zero

[severe topaz]

the limit is 0

[void widget]

It is not in in determined form

0/-6 is something that you can evaluate

[severe topaz]

why? that only is the case when the denominator vanishes

[severe topaz]

yeah

[void widget]

,w plot (x²-11x+24)/(x^(1/3)-8)

[glossy valve]

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[void widget]

See the graph

[severe topaz]

,w limit (x²-11x+24)/(x^(1/3)-8) at x -> 8

[glossy valve]

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[severe topaz]

the limit is 0

[void widget]

It is 0 in 8

[severe topaz]

yeah

[void widget]

,ti

[glossy valve]

The current time for 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛 is 12:47 AM (IST) on Tue, 21/02/2023.

[void widget]

I'm sleeping bye

[versed cape]

What about this one

<@&286206848099549185>

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@versed cape Has your question been resolved?

[gritty rose]

!show

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

[versed cape]

.close

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[steep falcon]

can someone pls explain this

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@steep falcon Has your question been resolved?

[hollow aurora]

The result is: n(n+1)/2

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[eager anchor]

How can I reduce this further

I got here by trying to find a second solution to 9y'' -12y' + 4y = 0

Since it's characteristic function has a single root of multiplicity 2

but I got stuck trying to reduce for a second solution v(t)e^(2t/3)

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[torn jolt]

so is it D=RT

distance = rate time

[steep falcon]

No

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[tall elbow]

what would be the easiest way to factor 6x^2 - 5x - 21 = 0

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[jade radish]

hmm

complete the square

[glossy valve]

Jukelyn

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[ornate compass]

help

[ornate compass]

please help

[torn jolt]

when does the graph go blelow x axis

there f(x)<0

wherever the graph is above x-axis f(x)>0

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[dull mauve]

If I want to apply segond grade equation and have this 1=2x^2+4x, when I pass the 1 in the other side It turns to negative or not? 0=2x^2+4x-1

[rocky vale]

What did you do to both sides of the equation?

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[tacit eagle]

(2x^7) /(x^6) cant be simplified to 2x?

[torn jolt]

it can

assuming x \neq 0

[tacit eagle]

what about (3x^2)/(4x^2)?

can it be simplified to 3/4

i looked it up and it can be simplified to that

[sly frigate]

yes

[tacit eagle]

so why cant you simplify this? the x^2

at the top

3x^2 and 4x^2

[sly frigate]

thats.... not how fractions work

[tacit eagle]

but u just did this

a second ago

[sly frigate]

(a+b)/(c+d) is not equal to a/c + b/d

[tacit eagle]

so only

if its isolated

u can manipulate it?

[sly frigate]

yes

I'm not so good at explaining fractions

[onyx glen]

@tacit eagle you can only cancel it common factors in the num and denom of a fraction if both are written as products.

[tacit eagle]

only exception

is if theyre products

so (a x b)/(c x d) is equal to a/c - b/d?

[full forum]

@tacit eagle Has your question been resolved?

[jade radish]

nope, (ab)/(cd) = (a/c)(b/d)

i'm not sure where u got that from but the canceling is just when u have something in the top and bottom of a fration as a factor so (ab)/(cb) = a/c

and a, b, c can be anything

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[fast peak]

dont troll in help channels

.close

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[normal tree]

If f(t) is the inverse Fourier transform of g(omega), then is F(t) = an antiderivative of f(t) the inverse Fourier transform of g(omega)/(i * omega)?

[normal tree]

my mind is basically dead

I remember the result that the Fourier transform of f'(t) is g(omega) * i * omega

[torn jolt]

oh hello again lol

hmm

[normal tree]

yeah basically same problem as before but I decided I wanted to figure out the cdf

using some sort of fast fourier transform algorithm

[torn jolt]

was the integral not too nice?

[normal tree]

integral is okay numerically, but naively computing it is quadratic in the granularity of it

like it only gives you F(a) for a specific a

[torn jolt]

i see

[normal tree]

ideally I'd like F(a) for a whole bunch of them

and that's basically where fast fourier transform comes in

There's the result that the inverse Fourier transform of a function is just the Fourier transform of the reversed function

[torn jolt]

wait i have to go eat but i will see what you mean in a bit if no one else does

[normal tree]

and obviously you can put all of this together to conclude something

also very unfortunate that the classes I took involving the Fourier transform used the convention 1/2pi and e^(-iwt) as opposed to the crap I learned in probability theory which is just E[e^itX]

[stiff musk]

2pi factors are the bane of your existence when doing fourier stuff and translating between different authors' conventions

[full forum]

@normal tree Has your question been resolved?

[normal tree]

hmm I guess the distribution I'm dealing with is actually discrete

so I can probably survive off a characteristic function to pmf formula, which is just some inverse fft

but still would be nice to know

also I think any fourier method is going to yield slightly funny-looking cdfs for probability distributions with atoms, so I'll need to correct for those

[stiff musk]

certainly within a scale factor, it's gonna look like g(omega) / (i * omega)

(referring to the question about the antiderivative)

[full forum]

@normal tree Has your question been resolved?

[normal tree]

@torn jolt so I did an FFT to just compute the probability mass function of the thing

and it looks absolutely ridiculous

like im honestly not sure if it's supposed to look like that

[torn jolt]

oh sorry thanks for reminding me about this

i just came back

[torn jolt]

i dont think this is exactly true though

[torn jolt]

yuck

[normal tree]

does the shape make sense though

like why is it so spiky

actually I think it kinda does

the big drop is probably around 1169

.close

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[muted quiver]

can someone explain to me what they actually want me to find

for number 1

like do they want me to break it into like integration by parts

box style method

[rocky vale]

Not integration by parts, you're meant to do a trig substitution

[muted quiver]

right

the only thing i can see

by looking at this problem

is that 1/sqrt(1+x^2) = tan^-1(x)

[rocky vale]

well

the reason they did the substitution x=tan(theta)

[muted quiver]

mhm

[rocky vale]

is because that gives 1+tan^2(theta) in the denominator

which is very convenient with the pythagorean identities

[muted quiver]

wait

isnt pythagorean idenitity sin^2 theta + cos^2 theta = 1

[rocky vale]

Yeah

[muted quiver]

nvm i just looked it up

[rocky vale]

But if you divide both sides by cos^2(theta)

[muted quiver]

ur right

so the whole bottom

gets replaced

by

sec^2x

or theta

right

[rocky vale]

yeah

but also it's in a radical

[muted quiver]

right

now u have

integral ((tan(x)^3)/sqrt(sec^2(x))

[rocky vale]

theta, not x

and what about dx?

[muted quiver]

oops

oh right

i forgot that too

also

tan(theta)^3

is the same as

tan^3 theta right

[rocky vale]

$\tan^3(\theta) = (\tan(\theta))^3$

[glossy valve]

tatpoj

[muted quiver]

ok cool

so now i just have to integrate this

i can break tan^3 into tan^2 theta and tan theta right

[rocky vale]

off the top of my head I'm not sure

[muted quiver]

the integral of tan ^2 (theta)

[rocky vale]

But that's as far as the problem you posted asked you to go

[muted quiver]

oh it is ?

wait

then how did they get

[rocky vale]

They just wanted you to identify the integral

[muted quiver]

right but in 1b

how did they get

sec^2(theta) on the numerator

[rocky vale]

Make the substitutions
x = tan(theta)
dx = ?

[muted quiver]

sec^2(theta)

wait

is this just a u sub problem...

[rocky vale]

sec^2(theta) d(theta)

It's trig sub, not quite the same as u-sub

I mean I guess it's kinda the same

[muted quiver]

right so

i get what u did here but

im still confused how they turned the integral of

tan^3(x)/sqrt(sec^2(x) into the answer they got

cuz i found the sub value for x

which is tan(theta)

[rocky vale]

You're leavning off the d(theta)

[muted quiver]

right my fault

the dtheta would be sec^2(theta)

dx

right

[rocky vale]

that's how it ends up in the numerator

yeah

[muted quiver]

but what about the sqrt

does it just go away?

[rocky vale]

$\sqrt{\sec^2(x)}$

[glossy valve]

tatpoj

[rocky vale]

I think we can simplify this?

[muted quiver]

sec x

[rocky vale]

yea

[muted quiver]

but we said

du = sec^2(theta)

dx

to get this to work

we would need this whole du to be 1/sec du = sec(x) dx

right

?

[rocky vale]

$$\int \frac{x^3}{\sqrt{1+x^2}} dx$$

Let $x = \tan\theta$, so $dx = \sec^2\theta$

[glossy valve]

tatpoj

[muted quiver]

right

so if you plug in the values

we get integral ( (tan^3(x))/(sqrt(sec(x))dx right

[rocky vale]

Then when you make the subs, you should get

$$\int \frac{\tan^3(\theta)}{\sqrt{1+\tan^2(\theta)}} \cdot \sec^2(\theta) d\theta$$

[glossy valve]

tatpoj

[muted quiver]

wait where did u get multiply sec^2(theta) from

again?

[rocky vale]

dx = sec^2(theta) d(theta)

[muted quiver]

oh my god

its not u sub

we sub in dc

dx with dx

i get it now LOL

im sorry it took so long

[rocky vale]

Lol

All good lmao

[muted quiver]

thx for ur help

[rocky vale]

It's kinda the same, just from the other side

so to speak

[muted quiver]

i really appreciate it man

[rocky vale]

No problem 👍

[muted quiver]

.close

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[glossy fern]

wow so many help channels open today

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@glossy fern Has your question been resolved?

[glossy fern]

<@&286206848099549185>

[torn jolt]

hello : D

what have you done till now?

[glossy fern]

nvm i got the sol

but thanks anyway!

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[eager falcon]

Suppose (a b) is some arbitrary permutation in Sym(4). How do we simply (a b)(1)(a b)^-1?

[torn jolt]

write out the inverse of the transposition

[eager falcon]

@torn jolt Like (a b)(1)(b a)?

[torn jolt]

yes

[eager falcon]

Then, what do you do next?

[torn jolt]

do the multiplication

[eager falcon]

Honestly, not sure how.

How do you multiply a variable permutation by a constant permutation?

[torn jolt]

(1) is the identity

[eager falcon]

Yeah.

[torn jolt]

so it doesnt change anything

[eager falcon]

Right.

[torn jolt]

so what is (a b)(b a)?

[eager falcon]

Just the identity element?

(1)?

[torn jolt]

yup

[eager falcon]

Awesome, ty. What about (a b)(1 2)(3 4)(a b)^-1?

@torn jolt

[torn jolt]

I think you have to do cases here

[eager falcon]

Some solution I found says to do this

Not sure if this is right though

They did (i j)(1 2)(3 4)(i j)^-1

Not sure why they didn't simplify it to (i j)(1 2)(3 4)(j i)

[torn jolt]

they did

[eager falcon]

Oh yeah

(i j) = (j i)

right?

[torn jolt]

yup

[eager falcon]

Anyway, does this look correct?

Like (i j)(1, 2)(i j) = (ij 1, ij 2)?

How?

[torn jolt]

I dont see how this works

(i j) doesnt have to be disjoint from (3 4)

[eager falcon]

Could you use the fact that permutations are associative and simply write (i j)(i j)(1, 2)(3, 4)?

[torn jolt]

idk that notation, sorry

[eager falcon]

which is (1, 2)(3, 4)?

[torn jolt]

I think you are getting associativity and commutativity mixed up

[eager falcon]

Oh yeah lol you're right

Anyway, how would you do this then

[torn jolt]

what's the original question

[eager falcon]

I'm trying to show that for all elements in A and G, that xjx^-1 is in A.

where x is in G and A is in j

To show that A is a normal subgroup of G

so, I'm going through every element of A

[torn jolt]

you would have to do cases, and show that if (a b) = (1 2) or (1 3), then its still in A

[eager falcon]

Why is a and b restricted to be 1 2 or 1 3?

is it because of sym(4)?

[torn jolt]

wlog

all other cases look the same

eg (a b) = (3 4) looks the same as (1 2)

[eager falcon]

how does (3 4) look the same as (1 2)

?

[torn jolt]

if you do the calculation, it would be very similar

[eager falcon]

Yeah, I see now

Ok, thank you for your help!

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[azure grove]

Hi people! I wanted to ask this — In questions like this (18), am I supposed to list all elements of $D_4$, and then figure which ones satisfy the condition? Or is there some other "trick" or approach? How would you go about solving it?

[glossy valve]

n11

[rose rain]

well

D_4 has 8 elements

it's not too bad

for K just look at what it means

[azure grove]

True true XD
But for bigger ones?

[rose rain]

honestly figuring out what K is is much faster

well there's a reason they aren't asking you to do it for bigger ones

[azure grove]

Or I supoose there just won't be too big groups?

Right!

[rose rain]

I mean you very well know we have infinite groups

so....

[azure grove]

Hmm

Alright, thankyou!

[rose rain]

np

[azure grove]

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[rose rain]

btw there should be a

theorem

about square elements

I can't recall it off the top of my head but it might be useful

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[tough sandal]

is
a + b x c

[tough sandal]

the smae as

ac + b

ame*

same*

sorry its late...

[onyx glen]

no, a + bc is not the same as ac + b.

[tough sandal]

dang

[onyx glen]

what led you to ask this?

[tough sandal]

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[tough sandal]

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✅

[tough sandal]

@onyx glen im getting missed messages here from learning

like this for example

which version is true

[fleet briar]

they are both the same thing

[vast fossil]

They are equivalent

Same story, different names

d = delta x
vf = v
v0 = vi

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[tribal gust]

What kind of algebraic structure is a sigma-algebra as used in probability theory? I'd think its an algebra or a field but maybe its a structure itself.

[torn jolt]

https://en.wikipedia.org/wiki/Σ-algebra#Definition_and_properties

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@tribal gust Has your question been resolved?

[tribal gust]

I've read the article, the reason i ask here is because im still confused.

Im leaning more toward it being a structure itself after reading it tho

[torn jolt]

it's it own thing I think

its not related to algebra, unfortunate naming ig

(it is related to an algebra in measure theory tho)

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[worldly jacinth]

Help

[worldly jacinth]

Is there any easy way to do this

[torn jolt]

row reduce

[stoic portal]

subtract the 2nd row from the third row

[worldly jacinth]

Yep i tried that

It gives 1, 1, 1

[torn jolt]

then either keep row reducing, or use cofactor