help-36

[fallow compass]

What do you mean by that?

[silver cargo]

exactly what I said: explicitly show the values of s and q too

from what I can see, unless I missed some blindingly obvious gotcha, you only found p and r = -1

[fallow compass]

Since q=-s
I don't think I need to find s and q

Oh wait

[desert mantle]

p+q+r+s=(p+q)+(r+s)=r+p

[fallow compass]

It should be -p = r+s

[desert mantle]

so finding s and q is not necessary

[fallow compass]

Wrong vietta formula

Dammit

[desert mantle]

oh true

[silver cargo]

ah, fair enough then

[desert mantle]

unlucky

[fallow compass]

Luckily the answer is still -2

[silver cargo]

I'm just making sure you avoid any instance of people asking "but what about the other two variables?"

but if you have a solid enough reasoning around that, that's completely fair, and my bad for not realizing it

[fallow compass]

So the answer is correct yes?

Then, thank you

.solved

[final saddle]

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[drowsy viper]

i need help

[drowsy viper]

1: random variable x has a probability density function of 3/8 (x^3-5x^2+6x) 0≤x≤2
A: sketch graph for this function
B: show that f(x) is a valid probability density function
C: to 3 significant figures find
i: its mean
ii: its mode
iii: P((1/2) ≤x≤ (3/2))

idk how to do part C at all, this was the question i was given

[barren hound]

mean should be $\int x f(x)\dd{x}$ right?

[soft zealot]

〈 kitten | teacup 〉

[drowsy viper]

i did i and iii, could i use the quadratic formula for ii? doing it rn seems like the right thing to do idk tho, dont have my textbook to check lol and ive gotta do this by tmrw

[barren hound]

yeah you just want to maximise f(x)

quadratic seems fine

[drowsy viper]

okay cool, thanks, you can like end the thingy now idk how to do it

[barren hound]

.close

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[dawn plover]

Sorry its not in english ill try to translate best i can

So basically a, b, c are any numbers that fit a + b + c = 1

and we have to prove the equation is correct

[tiny kraken]

any numbers? or like for positive numbers

[dawn plover]

sorry

positive number

[tiny kraken]

These problem seems like typical AM-GM type for positive number so I just want to check

[dawn plover]

yeah positive

[dawn plover]

you know how to solve it?

[tiny kraken]

nahh, this type of problem has pretty unique sol for each

But you can predict when the equality hold and try to AM-GM

[bronze grove]

can u put a = b = c?

[dawn plover]

well i assume you cant

since they are different letters

[bronze grove]

then u will get value of a that proves the equation a + b + c = 1 is correct

[bronze grove]

we are assuming

that a = b = c

and we will just a >= 1/3 after solving the equation

[dawn plover]

how do you know it so fast

[bronze grove]

?

[dawn plover]

wait so what do i do

a = b = c = 1/3?

[bronze grove]

put b = a

and c = a

[bronze grove]

no

[dawn plover]

so b = a and c = a

got it

[bronze grove]

yes

but im not 100% if u can do it that way but worth a try ig

[dawn plover]

ok lemme try

[tiny kraken]

that is not how you're supposed to do a inequality problemmmmmm

but ehhh

just try it ig

[dawn plover]

ive never done smth like this

ok

i have 3/(3a^2 + 2a) => 1/3a^2

[bronze grove]

solve it

[dawn plover]

i am

ok i got it

a => 0 and a => 1/3

ok well answer can be 1/3

but i strongly belive its incorrect for some reason

[bronze grove]

yea me too

i think we need to put c = 1-a-b

but the calculation is too long

[dawn plover]

well

if you want i got another problem

we can ditch this one

[bronze grove]

suree

[dawn plover]

this is last one i have unsolved

we have to find all combinations of a b c m n

and they are natural numbers (1, 2, 3, 4...)

[bronze grove]

substitute value of b in second equation?

[dawn plover]

what is substitute

[bronze grove]

putting value of something in another equation

so u get the value of b from first equy

and put it in second equi

[dawn plover]

im not sure if i can do that

probably can

ok but how does that help me

[deft terrace]

hmm ok

so first square the expression a+b+c = 1

and find the value of the sum of their squares

then put them greater or = 0

[deft terrace]

then get an inequality using a.m h.m inequvalence

[dawn plover]

im on different question tho

since i couldnt get the first one with help

[deft terrace]

[dawn plover]

so do we do 1st question or 2nd

i need to solve at least one today

[deft terrace]

i did 1st

[dawn plover]

ok

[deft terrace]

lemme do the second

[dawn plover]

ok

[dawn plover]

this one

[deft terrace]

did you get how i did the first one though

[dawn plover]

dont give me solutions right away tho

[deft terrace]

ohh mb

[dawn plover]

no not really but ill try it later

[deft terrace]

do you want me to send u a soln ??

[dawn plover]

noooo

i want to work on the 2nd one ill try 1st one later on

with the square you said

[bronze grove]

damn smart asf

didn't think of that

[deft terrace]

[dawn plover]

sum of squares would just be the same tho

= 1

im getting off tracked

[dawn plover]

can we try this

ok so what i think here

try getting value of b from first one

and from second one make c^m x c^n so we can also put in value of c^m from first one?

not sure how that helps in the long run

[bronze grove]

so when we get the second equation

what are u getting?

[dawn plover]

what

what is your question

[bronze grove]

when u put b in second equation

whaf are u getting

[dawn plover]

you get rid of b's

yeah you do

[deft terrace]

i think i got

find the value of a and b

in the terms of c

then focus on what you get for a

[dawn plover]

why value of a

[deft terrace]

it will be of the form c^m(5-c^n)/24

[dawn plover]

isnt b just enough

i have no clue how you got this

[deft terrace]

now it is given that all of them are natural numbers right

[dawn plover]

im not looking for solution

just help

[deft terrace]

i am

[deft terrace]

ok then

step 1 : find a

[dawn plover]

from first one?

[deft terrace]

step 2 : make sure that a is positive and natural

done

[dawn plover]

ok well that didnt help 🙁

[deft terrace]

how would you find a or b in this system of equations

[deft terrace]

forget the right side for now

[dawn plover]

b = c^m - 5a

[deft terrace]

then

you use this value in the second equation

[dawn plover]

yeah

[deft terrace]

what do you get then

[dawn plover]

shit

well i got 5c^m - 24a = c^m x c^n

[deft terrace]

now find the value of a and tell me

[dawn plover]

i have no clue how to do that but wait

well i know if we can still use unknowns

[deft terrace]

take the terms with a on one side

and the terms with c on the other

[dawn plover]

yea

[deft terrace]

find it in terms of c

[dawn plover]

i get

a = (c^n x c^m - 5c^m) / -24

eh

[deft terrace]

yes

good

now in the bracket above multiply the -1 from the denominator

what do you get

[dawn plover]

what

my english is limited in terms of math

[deft terrace]

there's a -1 below right

[dawn plover]

we can just put it in front

you still get negative either way

[deft terrace]

yes good

now take that - inside the brackets

[dawn plover]

why

dont i get -c's then?

[deft terrace]

believe

[deft terrace]

thats what we want

[dawn plover]

ok 🙁

[bronze grove]

now make cases

[dawn plover]

a = 5c^m-(c^n x c^m)/24

[bronze grove]

lets start by putting a = 1

[deft terrace]

good

[deft terrace]

wait

no

[deft terrace]

now do you see 5c^m on both terms above

[dawn plover]

for some reason i assume it has to be interval as answer

[deft terrace]

take it common

[dawn plover]

i cant

its - and multiplication

i cant just do minus first

substraction

[deft terrace]

[dawn plover]

what

[deft terrace]

do you know 2X3+2X5 = 2X8

[dawn plover]

???

its not

[deft terrace]

[dawn plover]

2x^3 times 2x^5 would be 2x^8

[bronze grove]

hm what else?

[deft terrace]

we will take c^m common

it will make it c^m(5-c^n)/24

[dawn plover]

i still dont get what u want me to do

[deft terrace]

now we know a is +ve

[deft terrace]

wait a sec

[bronze grove]

and then apply the cases?

[deft terrace]

yes as a is positive so should the found expression

[dawn plover]

wait i think i understood

[deft terrace]

c^m is always +ve

so is 24

so 5-c^n >0

now we find all possible value

s

[dawn plover]

i got this yes

[deft terrace]

do you get what i did after that

[dawn plover]

um yea

but still 2 unknowns

[deft terrace]

now try to find combinations of values for c and n

[dawn plover]

waiiiiiittt

[deft terrace]

i am a pretty patient person

[dawn plover]

c^n < 5

[deft terrace]

yes indeed

now tell me all possible combinations of (c,n)

[dawn plover]

5^1

[deft terrace]

nope

[dawn plover]

oops

mb

MY BAD

wait does it have to be interval?

[deft terrace]

what other than that

[deft terrace]

nope just keep listing them

[dawn plover]

ok so

c = 2 n = 2

c = 1 n = any positive natural

well

[deft terrace]

and n=1 as well for 2

[dawn plover]

another issue

nvm no issue

[dawn plover]

ye

so c can be 1 and 2

n can be 1, 2 and any number

depends on scenario

[deft terrace]

well we get
c=1 n= any natural no.
c=2 n=1,2
c=3 n=1
c=4 n=1
right

[dawn plover]

why first one no

1^1808080 is still one

[deft terrace]

yes that's why i included it

[dawn plover]

you said no

[deft terrace]

umm where

[deft terrace]

thats the short form for number

[dawn plover]

oh ok

[deft terrace]

no.=number

[dawn plover]

ok

[deft terrace]

now second task is to make a and integer

[dawn plover]

ohhh

[deft terrace]

so no fractions

[dawn plover]

its starting to make sense slowly

c = 1, 2, 3, 4

[dawn plover]

what

[deft terrace]

now try to make A a natural number

[dawn plover]

input any?

[deft terrace]

by taking the case of c and n found above and using diffrent value's of m

[dawn plover]

so do i put a as 1

[deft terrace]

try it what all value's of m do you get

[deft terrace]

do that at the last

[dawn plover]

wha

what do you want me to do then

[deft terrace]

never mind

continue

[dawn plover]

a = 1?

[deft terrace]

c=1 now what

nope

a was a = c^m(5-c^n)/24

right

[dawn plover]

yeah

i got it already now

[deft terrace]

so now if we use c as 1 we get a= 1/24

is that a natural number

[dawn plover]

not really no

[deft terrace]

so c cant be 1

now try c as 2

[dawn plover]

depends on n

[deft terrace]

use n=1 first

[dawn plover]

i am

i just said it does

[deft terrace]

check if a value of m is possible

[dawn plover]

wait this is gonna sound stupid

can i multiply 2^m x 3

[deft terrace]

yes

but it will change nothing

[dawn plover]

to get 6^m?

is that a thing?

[deft terrace]

uhh no

[dawn plover]

yeah ok so i think well

[deft terrace]

both should have the same power

[dawn plover]

2^m x 3 / 24

we dont got a

oh nvm i got it

mb

m can be 3

m can be 4

[deft terrace]

and anything greater than 3 right

[dawn plover]

yea

[deft terrace]

so we got m>3 for n=1 and c=2

now check for the other values

[dawn plover]

c = 3?

[deft terrace]

ye

[dawn plover]

not 3

unless its smth above 5

[deft terrace]

how about c=4

[dawn plover]

no 4

a = 6.75 then

oh wait i can finish this on my own now

well i think with enough time i could have finished this one ty for help

[deft terrace]

so not possible

[deft terrace]

wlcm

[dawn plover]

4 is possible too btw

[deft terrace]

use .close

[dawn plover]

should we try 1st problem real quick?

[deft terrace]

uhh not realy

c cant be 4

[dawn plover]

nvm

my bad

its 4 x 1

i forgot

[deft terrace]

if you have free time

[dawn plover]

i do

i mean i have 3 more problems but idk if i can do them all today

ok so

a + b + c = 1 would just be all squares = 1

[deft terrace]

square on both sides

[dawn plover]

1^2 = 1

[deft terrace]

and take 2ab + 2bc + 2ca the other side

[deft terrace]

(a+b+c)^2 = 1

expand this

[dawn plover]

can you even

you cant

[deft terrace]

you have a lot to learn

[deft terrace]

a^2+b^2+c^2+2(ab+bc+ca) =1

now solve from here

[dawn plover]

ok this is pointless for me then

ive never seen this expantion thing

ok anyways i dont think i could do this one so im done for today, its getting late so thanks for your time

.close

[final saddle]

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[final saddle]

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[radiant carbon]

how do i do this

[abstract bramble]

what is the small circle representing

[zinc geyser]

what do each of those circle represent?

[radiant carbon]

left is spanish, right is french

small circle is german

[stiff bear]

so not german so c or d

c?

[radiant carbon]

nah

its A

lol

idk why

[stiff bear]

but german isn’t shaded in

wait what

why would german be the small circle

spanish would be small circle

[radiant carbon]

.close

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[pastel karma]

How should i start this, i dont have much understanding

[radiant cedar]

do you know about the vertex form?

[pastel karma]

F(x) = ax^2 +bx + c

I just forgot how to implement it

Im reviewing for a test

[radiant cedar]

that's not the vertex form

$f(x) = a(x - h)^2 + k$

[soft zealot]

kizzyyy

[pastel karma]

Damn

[radiant cedar]

does this ring a bell?

[pastel karma]

Im washed

[pastel karma]

Yes it does

Looks familiar

[radiant cedar]

what does (h,k) represent?

[pastel karma]

I dont know

Something to do with a parabola

[pastel karma]

Oh

[radiant cedar]

anyway, you're given the vertex

so likke can you fill in

[radiant cedar]

the values of h and k here?

what would the new f(x) look like?

[pastel karma]

F(x) = a (x-5)^2+9?

[radiant cedar]

close

what is h?

[pastel karma]

X coordinate of vertex

[radiant cedar]

i mean what is it

in your case

[pastel karma]

-5

[radiant cedar]

,, f(x) = a(x - (-5))^2 + 9

[soft zealot]

kizzyyy

[radiant cedar]

that isn't this now

is it

[pastel karma]

Oh

No it isnt

[radiant cedar]

anyway so

since -(-5) = 5

it's f(x) = a(x + 5)^2 + 9

all u have to do now is find the value of a

you'd use this particular fact to aid you with that:

"quadratic function.... passes through the point (-7,-15)"

[spiral skiff]

Hi, I would like to know if question b requires a module, if so, how does this affect the result?

[radiant cedar]

uhh

you should ask your question in #help-11

someone's already using this channel

[pastel karma]

Im sorry but

[spiral skiff]

Okk sorry

[pastel karma]

How would i find this again

[pastel karma]

Dont worry about it

[radiant cedar]

y = a(x + 5)^2 + 9

this passes through (-7,-15)

which means u can plug it into your equation and solve for a

y = -15 and x = -7

[pastel karma]

OH

-15=a (-7+5)^2 +9

?

[radiant cedar]

-15 but yes

[pastel karma]

Ok i got it right

Thanks

.close

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[rugged merlin]

Why does this hold [
\ceil{\4ab} = \floor{\4{a+b-1}b}
]

[bronze grove]

do u know what those square brackets mean

[rugged merlin]

I do, yes

[bronze grove]

can u simplify the right side?

by dividing it in three parts

[rugged merlin]

yes, [
\floor{\4ab+1-\41b}
]

[bronze grove]

do u know pascal's identity?

[rugged merlin]

i do not

[bronze grove]

it would be much easier with that ig

[rugged merlin]

Looking it up, it seems to be a combinatoric identity; how does it apply in our case?

[bronze grove]

we get the right equation directly by applying the first one

lemme think of easier way wait

it always holds for a = b

[rugged merlin]

I mean I managed to prove the case where a is a multiple of b

[robust mulch]

lex

[bronze grove]

whats ur approach

[drowsy epoch]

what about if there is a remainder?

[rugged merlin]

oh hi garlic! nice new pfp :D

ok yeah i think i figured it out

using the division algorithm, if $r = 0$, we have $a =qb$: [
\floor{\4{a+b-1}b} = \floor{q + \4{b-1}b}
]
and since $b\in\Z_{\ge1}$ that term will always be between 0 and 1, hence [
\floor{\4{a+b-1}b} = q
]

[rugged merlin]

if $r\ne 0$ then [
\floor{\4{a+b-1}b} = \floor{q+1+\4{r-1}b}
]
Since $1\le r< b$ the term $r-1$ satisifes $0\le r-1< b-1$. Hence [
\floor{\4{a+b-1}b} = q + 1
]

well at least i think thats right?

[bronze grove]

yess

[tiny kraken]

one question, how can you make sure r-1>0 are a and b integers or smth

[final saddle]

@rugged merlin Has your question been resolved?

[rugged merlin]

im using the division algorithm which only applies for integers a and b yeah

it says [
a= bq + r, \q 0 \le r < b
]

[rugged merlin]

r being bounded that because, conceptually, division is just partitioning a number a into b cells, and r is the leftover component. If it was greater than b then it would mean the divisor could have fit in at least one more time.

[tiny kraken]

But we're proving for real number no?

[rugged merlin]

for my case i needed this for integers a and b, so not in my case no

also i mean i think you can come up with a counterexample that disproves it for the real case

,w ceiling(9.5/3) = floor((9.5+3-1)/3)

[soft zealot]

Results provided by WolframAlpha

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

Requested by lexqa

[tiny kraken]

Ehhh I thought we're proving for real numbers

[rugged merlin]

well it doesnt hold for that anyways so

[tiny kraken]

Then I think your proof looks good. Or I'm not able to spot any flaw

[rugged merlin]

yeppie

have a good one

.close

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[harsh stratus]

hello requesting help with this problem

[harsh stratus]

the area of a sphere increases with 28 cm^2 per minute

with which speed does the volume of the cube increase when the radius is 6.5 cm

let me show my work

[tired walrus]

hol up

[drowsy epoch]

r is also a function of time

[tired walrus]

the area of a sphere increases with 28 cm^2 per minute
with which speed does the volume of the cube increase when the radius is 6.5 cm

volume of the cube?

[harsh stratus]

oh my apologies

the volume of the sphere

r(t)

[drowsy epoch]

What you computed was dA/dr not dA/dt

[terse crypt]

😔

[harsh stratus]

hmm

but I don't have a time

why the derivative with respect to time

[drowsy epoch]

it says the area of a sphere increases per minute

[harsh stratus]

oh okey fine

[drowsy epoch]

so you wanna know the rate of change in terms of time, not distance

[harsh stratus]

we're searching for

dv/dr

I assume

[drowsy epoch]

speed would more like be change in time

[harsh stratus]

one second

heading to the grocery market now

[final saddle]

@harsh stratus Has your question been resolved?

[signal cipher]

why tf are you she/they

you are an it

a fucking it

[severe canyon]

And so?

It's not relevant

[severe canyon]

!redir

[final saddle]

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

[signal cipher]

oh my bad dude i thought this was discussion channel because it was open by default. Sorry

[severe canyon]

Don't worry

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[pearl solar]

Hi im very confused on this proof

like

I understand that the cartesian product of A and B, order matters

I understand the idea of an ordered pair

but

Im unsure how the double power set comes in here

[worldly mesa]

x in P(P(A)) means that x is a set of sets which are all subsets of A

[pearl solar]

yh this is mind boggling

[worldly mesa]

so {{1}, {2}} is a set of sets each being a subset of {1,2}

[pearl solar]

ah yep, this makes sense

[worldly mesa]

this is a formal definitions

[pearl solar]

OHHHH

I SEE

[worldly mesa]

how you want to remember AxB is as the set of ordered pairs of elements (a,b) where a in A and b in B

and all this proof really does is demonstrate how from the axioms we can build this set

[pearl solar]

so what the first power set does is, say for (1,2) u get {1}, {1,2} sets, the seocnd power set {{1}, {1,2}} ?

[worldly mesa]

yeah

saying x in P(A) mean x is a subset of A

so x in P(P(A)) means x is a subset of P(A)

[pearl solar]

i feel like these without example would kill me

ahh i understand it now

tysm

[pearl solar]

yea its kind of like builind on the sets

alr ty

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[rugged merlin]

\def\blah{\xrightleftharpoons{\hspace{0.667cm}}} [ \int_{-\infty}^\infty g_1(t)g_2^(t-\tau)\dd t \overset{\mathcal F}\blah G_1(f)G_2^(f) ]

[latent dragon]

what does the left right harpoons mean here?

[rugged merlin]

this is relating the time domain representation of the left function with the right which is the frequency Fourier domain representation

[latent dragon]

i’ve never seen this notation before.

*in math atleast

[proper raptor]

physicists ahh notation

[rugged merlin]

Anyways, I am trying to make sense of what this is meant to be telling me

[sick roost]

Well, its specifying a change in domain, = would be incorrect here

[proper raptor]

what is G, g ?

[sick roost]

G is the fourier transform of g here

[proper raptor]

ahh ok and * is the complex conjugate ?

[rugged merlin]

indeed

[sick roost]

This is a cross correlation between functions i believe, whats your question exactly

[rugged merlin]

understanding what "cross correlation" is, i suppose

[sick roost]

https://en.wikipedia.org/wiki/Cross-correlation

Its very similar to a convolution in function

[rugged merlin]

It seems highly reminisicent of what convolution is to be honest

yeah

exactly

so there are three things: convolution, cross correlation and autocorrelation

[sick roost]

Autocorrelation is like a convolution with itself

I mean the name literally means ‘correlation with itself’

Actually, the cross-correlation is precisely a convolution but it takes one functions conjugate

[rugged merlin]

so cross correlation and convolution resolve to the same thing with real valued functions?

[sick roost]

Well, the time domain is slightly different

[rugged merlin]

this is highly confusing

i cant conceptualise the difference between the two

[sick roost]

I believe the time domain is negative over real functions

Thats how theyre related

I might be insane rn, lemme check something rq

Right, because the convolution is a function of t, but the cross correlation is of tau

So the time domain is negative for reals

They are otherwise the same here

We know this directly via a simple u sub

u=t-tau

[sick roost]

We should take the convolution d\tau

Not dt

Its just symbol changing here

[rugged merlin]

well yes

then it should be dt

as i wrote

[sick roost]

But its important we keep tau and t consistent across the domains

This is probably where the confusion is

One sec

[rugged merlin]

the thing on the right will have $\exp(j2\pi f\tau)$ as its kernel

[rugged merlin]

uhh

[sick roost]

the convolution should be put
$$(f*g)(t)=\int_\mathbb{R}f(\tau)g(t-\tau)d\tau$$

[soft zealot]

Cycadellic

[sick roost]

is what i meant

compare with the cross-correlation
$$(f\star g)(\tau)=\int_\mathbb{R}f^*(t-\tau)g(t)dt$$

[soft zealot]

Cycadellic

[rugged merlin]

yeah im trying to derive this formulation starting from convolution

[sick roost]

so, take f's conjugate, convert to negative time domain, then they are related

[sick roost]

start here, take the u-sub u=t-tau

[rugged merlin]

aight

[sick roost]

choose f^*, g for convolution

as opposed to f, g

apply convolution theorem, the result should follow immediately

[rugged merlin]

,, \int_\R f(\tau)g(t-\tau)\dd t \overset{u= t-\tau}= -\int_\R f(t-u)g(u)\dd u

i think i did that correctly

[sick roost]

now we want f^* instead of f

[rugged merlin]

well do we conjugate twice?

[sick roost]

no, just take a new f=h^* or something like that

or just plug in f^* directly

[rugged merlin]

right

you end up with [
-\int_\R f^*(t-u)g(u)\dd u
]

[rugged merlin]

what this is like

[sick roost]

wait

lose the -

u = t-\tau \implies du=dt

[rugged merlin]

that should be with respect to tau

not t

i typo'd

[sick roost]

we want u here

[rugged merlin]

yeah im saying u = t- \tau implies du = -dtau

[sick roost]

oh i see

right

[sick roost]

then this one needs fixed

should be taken d\tau

[sick roost]

then do convolution theorem here

[rugged merlin]

well i mean it seems like [
(f\star g)(t) = f^*(-t)*g(t)
]

urgh

[sick roost]

yeah

and thus $$=(f^**g)(-\tau)$$

[soft zealot]

Cycadellic

[sick roost]

and so it is a modified convolution

[rugged merlin]

oh i see

so applying the convolution theorem

wait how would that -tau affect our result from the convolution theorem

or would it

[sick roost]

well, continue writing it out

what does the convolution theorem say about $(f^**g)(-\tau)$, now?

[soft zealot]

Cycadellic

[rugged merlin]

wait but before that

im a bit unconvinced still with that formulation: [
(f^**g)(-\tau) = \int_\R f^(t)g(-\tau - t)\dd t \0{red}\ne \int_\R f^(t)g(t-\tau)\dd t
]

[rugged merlin]

the latter being the actual correlation

[sick roost]

right, i might have said something stupid

ill be more precise

[final saddle]

@rugged merlin Has your question been resolved?

[sick roost]

oh wait im literally going insane

LOL

theyre the same theorem here

but with G_2^* instead of G_2

no u-sub necessary

just swap t and tau, and take G_2^* instead of G_2

[rugged merlin]

Oh I see

Right ok

Yeah well that was easier tahn we made it out to be

[sick roost]

the u-sub with the negative sign made me lose my mind lol

[rugged merlin]

😅

Still a bit confused by what convolution and correlation are really saying

I mean i get convolution but not this

[sick roost]

this is very similar to the statistical cross-correlation with some extra assumptions

stats takes the expectation of it, is the real difference

i think this image on the wiki does a good job explaining exactly whats going on

[rugged merlin]

But didnt you see autocorrelation is the same signal

So why are f and g diggerent

[sick roost]

well, hang on, the autocorrelation is on f\star f, and then g\star g

[rugged merlin]

I see

Well visually

It seems like correlation is convolution but without the flipping that occurs bcuz of the g_2(t-tau)

[sick roost]

specifically the correlation is a convolution, but with a sign flip and conjugate on the first function

[rugged merlin]

Seems like ghe conjugation is irrelevant here

Considering these seem to be real

[sick roost]

over R, they really come out very similarly, except the convolution is commutative, which the cross correlation flips the domain on commutation

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[rugged merlin]

.reopen

[sick roost]

.reopen

[final saddle]

✅ Original question: #help-36 message

✅ Original question: #help-36 message

[sick roost]

you forgot to react to the bot

[rugged merlin]

Yeah soz

[rugged merlin]

What sign flip though

Ok wait

[sick roost]

in the time domain

[rugged merlin]

I think we need to establish something

[sick roost]

we showed that $$(f\star g)=(f^*(-t)*g)$$

[soft zealot]

Cycadellic

[rugged merlin]

Are these horizontal axes t in both?

[sick roost]

well, hold on

convolution is in t-domain, correlation is in tau-domain

[heavy mural]

Hi

Is there someone here to help

[sick roost]

#❓how-to-get-help

[rugged merlin]

Is this -t reallt correct though in hindsight

It just doesn't check out with the convolution theorem

But im eyeballing it rn

[sick roost]

i dont really think this distinction is particularly significant in application tbh

[heavy mural]

I got the answer D for this is that right

[sick roost]

use your help channel, this one is occupied

[heavy mural]

Oh ok

[rugged merlin]

Well i mean im asking bcuz im trying to figure out what those axes in thr images are supposed to be ig

[sick roost]

they are both time domains, we should probably think of the axes as the same there

[rugged merlin]

Correlation seems to be just 1. Flip t and tau 2. Chug a * from the conv formula

[rugged merlin]

So i dont know where the -t here is coming from

[sick roost]

$$(fg)(t)=\int_\mathbb{R}f(\tau)g(t-\tau)d\tau$$
let $u=t-\tau\implies du=-d\tau$
$$(f^(-t)g)=-\int_\infty^{-\infty}f^(u-t)g(u)du$$

and so, $$f^*(-t)*g=f\star g$$

[soft zealot]

Cycadellic

Cycadellic

[sick roost]

because u = infinity implies tau = -infinity
so thats how we deal with the minus sign here

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[solar crest]

hi im a bit confused, if i have a matrix m x n why is it that:
if m > n, there is no way that it's surjective?

[solar crest]

and why is it that if m < n, there;s no way it's injective?

i wanna understand the logic behind i t

[static plinth]

i would say it’s easier to understand that logic if you know that any matrix represents a linear transformation

are you familiar with that idea?

[ivory sandal]

If m>n, the matrix maps from a smaller space to a larger one. There aren’t enough independent directions in the input to reach every point in the higher-dimensional output. So it can’t cover the whole output space, meaning it’s not surjective.

[static plinth]

matrices/linear maps cannot grow dimensions

it can only keep its current dimension or reduce it

[solar crest]

yeah

[solar crest]

i still dont really understand this idea..

[solar crest]

o so

if i have r3 --> r2

then it's surjective

but if it have r3 --> r3 it cannot be?

[static plinth]

well it can still be surjective

because we have an input space of dimension 3 and an output space of dimension 3

there would exist some linear map that would retain its "size"

[ivory sandal]

Only if invertible

[static plinth]

but take T: R^3 -> R^4

theres just no possible linear map you can make such that the codomain is all of R^4

thats the idea of how linear maps cannot grow dimensions

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@solar crest Has your question been resolved?

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[main mirage]

Hi. I'm not sure how to use the hint in this question, can anyone help?

The function $\delta(x)$ refers to the dirac delta function. That is:
[
\int_a^b \delta(x)dx = \begin{cases} 1 \quad \text{if: } 0 \in (a, b) \
0 \quad \text{otherwise.}
\end{cases}
]

[soft zealot]

Good Luck HSC Students!

[rocky tusk]

did you look at integrating delta(ax)?

[main mirage]

I did take the integral and set x = au, where do I go from there though?

Or did you literally mean integrating delta(ax) straight from the start?

[rocky tusk]

well with the test function

[main mirage]

I'm not sure how you would do that sorry.

[rocky tusk]

$\int_a^b \delta(ax) \phi(x) \dd{x}$

[soft zealot]

knief

[rocky tusk]

u = ax

[final saddle]

@main mirage Has your question been resolved?

[main mirage]

Hmm. Still a little unsure, but this is what I am thinking of:
[ \int_{-\infty}^{\infty}\phi\left(x\right)\delta(x)dx=\phi\left(0\right) ] [ \int_{-\infty}^{\infty}\phi\left(au\right)\delta(au)adu=\phi\left(0\right) ] [ \int_{-\infty}^{\infty}\phi\left(x\right)\delta(x)dx=a\int_{-\infty}^{\infty}\phi\left(au\right)\delta(au)du ] If we let $\phi(x) = 1$, then: [ \int_{-\infty}^{\infty}\delta(x)dx=a\int_{-\infty}^{\infty}\delta(au)du ] Is this on the right track?

[soft zealot]

Good Luck HSC Students!

[main mirage]

I'm guessing there should really be |a| at the front of the integral, because when we make the substitution x = au, we don't want the bounds of the integral to flip?

However, does this mean the integrals cancel out? What is the justification for that?

[final saddle]

@main mirage Has your question been resolved?

[final saddle]

@main mirage Has your question been resolved?

[pliant shore]

"good luck HSC students" is a great message

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[main mirage]

It's that time of year again :)

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[jagged flare]

lemme send the hints

[fiery bluff]

First thing i think of is euler line, draw centroid

[jagged flare]

599=||This is just a statement about distances to line OH; ignore the areas.||
152=|| You can replace line OH with any line through the centroid G.||
598=|| Without loss of generality, B, C lie on the same side of the line. Let D be the midpoint of BC.||
545=||Use the fact that AG = 2GD.||

yeah im stuck on how i would prove AX=BY+CZ

i havent used the last hint yet so its prolly related to it

[fiery bluff]

pigeonhole principle - two sides of the line, 3 points so two points are on the same side

[jagged flare]

yeah i just realized that

[fiery bluff]

You have to use point d somewhat

Consider ||distance from d to the line||

[jagged flare]

huh wait ok hold on this is interesting

DK is half of AX

since BY//DK//CZ, and D is the midpoint of BC, by similarity DK=(BY+CZ)/2 <=> AX=BY+CZ, QED?

ok yeah i think this is it

thank you so much!

.solved

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[dusky isle]

Over here, is the purpose of finding a general equation?

I dont understand what we are exactly doing here, the idea is to use differential equations:

Sorry if the handwriting is bad

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@dusky isle Has your question been resolved?

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@dusky isle Has your question been resolved?

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[timber moth]

Hey, extremly rough draft as i just learned little o notation 5m ago but is any of this correct

[timber moth]

Im mostly asking for the potentially botched use of some sort of triangle inequality with h1, and h2

forgot to say by using squeeze theorem we can bind g(h) between 0 and 3h at the end there **

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[ivory lynx]

hi guys, dumb question

[ivory lynx]

its mathematical induction

how did we get (k +1)(k + 2) in the thrid step?

[opal plinth]

This, but with n = k+1

[ivory lynx]

OH

i need

i see**

k + 1 = k+ 1(k + 1 + 1)

right right

thank you

🙏

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[onyx peak]

<@&268886789983436800>

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[latent dragon]

what happened. 👀

spam?

[onyx peak]

yeah, some rando spammed some scam ss's

[latent dragon]

lotsa those these days.

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[warm python]

$\int_{0}^{1} \int_{0}^{1} 2x^2y+Cy^5=1$
\
Solving this will give me the value of C

[tiny gorge]

no

why should that integrate to 1?

[warm python]

it's the joint PDF \

oops

[tiny gorge]

the joint PDF doesn't involve integrals..

[warm python]

over (0,1)*(0,1)

[tiny gorge]

yes

integrate over the full domain to get 1

[soft zealot]

wai

[tiny gorge]

yep

[warm python]

lemme solve it rq

C=2

[tiny gorge]

i get a different answer

,w integrate (2x^2 y + 2y^5) for x = 0 to 1 and y = 0 to 1

[soft zealot]

Results provided by WolframAlpha

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

Requested by bungo.the.original

[warm python]

Okay, lemme send my steps

you've got to be kidding me

[warm python]

I integrated 2x^3+Cxy^5

[tiny gorge]

haha numbers are hard

[warm python]

😭

[warm python]

I did a worse mistake in my probablituy midsem

$\int x \cdot 2x^2 dx = x^2$

[soft zealot]

wai

[tiny gorge]

😭

[warm python]

This messed up eveyrthing

and cost me 5 marks I think

[tiny gorge]

yea stupid calculus mistakes are the source of many problems in higher level courses, alas

[warm python]

Sorry, more than help, I need to rant

[tiny gorge]

nw, it happens to all of us

[warm python]

gets better, while calculating the product of 2 cycles I read 2 as 1

and while calculating something I got 30 instead of 35 ( cost me 5 marks)

[tiny gorge]

harsh grading!

[warm python]

haha , see hlounge for harsher grading

anyways, just to be sure

[tiny gorge]

this kind of crap is why i always stayed until the end of every exam even if i finished early
double and triple check everything, i nearly always found some stupid arithmetic mistake

[warm python]

here once I get C, I integrate wrt y from 0 to 1 fo get F_X(x) and similarly for F_Y(y)

[tiny gorge]

well not from 0 to 1

you want to find $F_{X,Y}(0.8, 0.6)$

[soft zealot]

Bungo

[tiny gorge]

so you would integrate from x = 0 to 0.8, and y = 0 to 0.6

[warm python]

I also need F_X(x) and F_Y(y)

[tiny gorge]

oh ok i overlooked that, lemme go reread

[tiny gorge]

yea assuming you meant f and not F in both cases

(this gives you the marginal pdf's, not the marginal cdf's)

[warm python]

t

tq

[tiny gorge]

yw

[warm python]

.close

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[stone cradle]

its on german so dont wonder. My question to this is what does it means when s = -1.5? like a = -1.5c, but what does that means ?

[terse crypt]

${\vec{a} = -1.5\vec{c}}$

[soft zealot]

k

[terse crypt]

it means that vector a is in the opposite direction of vector c

and its magnitude is 1.5 times that of c

[stone cradle]

oh, so if c is 10m long a is 15m long ?

wait

no

[terse crypt]

yes

and is pointing in opposite direction

[stone cradle]

and what would it be if its 1.5c

same direction?

[terse crypt]

yup

[stone cradle]

oh got it thanks

.close

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[stone cradle]

used an translator and idk if its the right topic

lines in space,
the vector parametric equation

anyways, my question is how do i use that with an line in an 3d coordiante system?

[stone cradle]

like this one

with x y z

[thin thicket]

die vector gleichung von der linie die durch A und B geht? sorry mein deutsch ist nicht so gut

[stone cradle]

beim ersten bild wird das ganze ja mit einem 2d koordinatensystem gemacht also mit x und y.
ich vestehe aber nicht wie ich den vector AB bei x y z herausfinde, da man ja bei nur x y ablesen konnte

ich verstehe nicht wie die das mit 7, -1, 2 herausgefunden haben

[thin thicket]

der vector (7 | -1 | 2) ist B-A ( 4 - (-3) | 0 - (1) | 2 - 0)

vector

[stone cradle]

oh stimmt, das hatte ich sogar mal gemacht

voll vergessen

danke dir

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[toxic tree]

I dont get how they solved this one

I understand that without the restriction its 54^5

[thin thicket]

yes

[toxic tree]

But idk how they got the 26^5

to do 54^5 - 26^5

[thin thicket]

we need the restriction and the 26^5 gets rid of the number of passwords with only lowercase

cuz 54^5 includes the only lowercase and the mixed lower-upper

[toxic tree]

Wait

I do get it

Cause its just the same again

n = 26 and k = 5

[thin thicket]

yes

[toxic tree]

Gives all the only lower case ones

Okokokok

[thin thicket]

yep

[toxic tree]

Got it

Thanks!

❤️

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[thin thicket]

np

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[warm crater]

Write the pseudocode for an algorithm that determines if a function is injective, assuming that you can iterate through all the elements in the domain and the co-domain in a finite number of steps.

[warm crater]

Im in a Discrete math course following Kenneth Rosens Textbook for discrete math

and im just wondering how should i refer to the co domain in pseudo code

should i just reference it, like b1,...bm: corresponding elements of co-domain?

[glossy zephyr]

Thats just up to your liking tbh given its pseudocode

If we assume f: A -> B
A = {a1,a2...an}
B = {b1,b2...bm}
is intuitive enough

[warm crater]

ok

so could i technically also just refer to the output using f(a1)?

in my algorithm

[glossy zephyr]

Yes, but remember that would be the image of A instead

Not the Codomain

[warm crater]

ahh i see

ok thank you

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[glossy zephyr]

If you remember the general idea of Surjection and Injection this one is pretty easy

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[manic helm]

ive formed DEs $\frac{dx}{dz} = \frac{-(2x + z)}{x}, \frac{dy}{dz} = \frac{2y}{x}$ but stuck form there

[soft zealot]

ashyboi

[glossy zephyr]

They should be d/dt

anyways, this is more like solving 1 diff. equation and 1 diff.system

[manic helm]

how come btw

i kinda just assumed the flow was steady state

and there was no t dependence

[glossy zephyr]

We basically want a function for the position, right?

[manic helm]

yes

[glossy zephyr]

Velocity is the derivative of position through time

[glossy zephyr]

So here we got this:

$\u = \frac{dR}{dt}=(4x+2z,-4y,-2x)$

[soft zealot]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[glossy zephyr]

where R is the position

[manic helm]

yes

lemme give it a try

[glossy zephyr]

Or, in another words

[manic helm]

wait but#

oh ye

[glossy zephyr]

Here you can see that x and z depend on eachother

and y depends on itself

Soooo, they are basically
1 system
1 equation

From solving those two separately, you can combine them back into vector notation

[manic helm]

ye i see it

[glossy zephyr]

tbh, a lot of physics problem are usually written down through dt

[manic helm]

how come

in here they dont do this

[glossy zephyr]

They did