help-49

[brittle cedar]

and i've already started with 5... is x =31 and y=64?

@wooden badger

[wooden badger]

Yes

[rugged coral]

are you certain that the two sides marked with “>” are parallel

[wooden badger]

Thats how parallel lines are generally marked

[rugged coral]

oh alr

[wooden badger]

Rnt they?

[rugged coral]

im not familiar with it but ok

then u know the value of S

[rugged coral]

and the >> are parallel also right

[brittle cedar]

@wooden badger lets proceed with 6

[wooden badger]

Kay

[brittle cedar]

do we just do 100 - x + 30+ 4x + 7x =180

[wooden badger]

Yes

Theres one more equation for y

[brittle cedar]

is x =5?

@wooden badger

[wooden badger]

Yes

[brittle cedar]

y=145

[wooden badger]

Yep👏

[hallow trout]

Guys can you help me on rational expressions

[wooden badger]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[brittle cedar]

@wooden badger how do u do 8

i feel that this is really easy but also hard

[wooden badger]

Can u make an equation using sum of angles in triangles?

[brittle cedar]

wdym

[wooden badger]

Sum of angles= 180

[brittle cedar]

ohh

the only one i could think of is 35+m = 180

but that's not correct because it's 145

[wooden badger]

There r three angles

2 are marked equal

So wouldn't it be m+m+35=180?

[brittle cedar]

hmm

2m+35=180

oh

145/2

[wooden badger]

Yes

[brittle cedar]

okay thanks

@wooden badger ty so so much

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[lone dock]

HOW??????????????

[outer saffron]

Because 3-2=1

And k-k=0

[lone dock]

what the...

[bold peak]

This is called partial fraction decomposition

Look it up

[lone dock]

ohh

ok

okok

[bold peak]

Essentially you can split a fraction into sum of fractions with simpler denominators

[keen imp]

what , i did not know that

[lone dock]

is it a part of calculus or just equation manipulation?

[bold peak]

It is expression manipulation

But the only use of it you'll see for a while is integration

[wooden badger]

U can write 1= (k+3)-(k+2)

Then split it up

[bold peak]

While I agree, that isn't a good way to understand what happened here

Partial fraction decomposition isn't always as simple as that

[wooden badger]

Isnt tht what happened here?

[lone dock]

makes perfect sense

[lone dock]

and i looked it up, makes sense

thanks a lot!

[lone dock]

gotit

[bold peak]

Cuz the OP asked about calculus, I imagined this was in the context of integration. So like I said, while you're correct, the idea is much more general

[tall crag]

its like when u have two functions

like

f(x)/g(x)

[wooden badger]

Well tht explanation was focused on that specific ques rather than anything general💁‍♀️

[tall crag]

u can write as $\frac{f(x)}{g(x)} = \frac{A}{x - r} + \frac{B}{x - s} + \cdots$

[grand pond]

Viͥђaͣnͫ

[lone dock]

oh dang...i see

[tall crag]

want example?

[lone dock]

yeah

[tall crag]

ex.
$\frac{3x+5}{(x-1)(x+2)}$

[grand pond]

Viͥђaͣnͫ

[bold peak]

Only if g is a polynomial

Let's not say this applies to all functions

[tall crag]

take this as an example

[lethal path]

the guess $\frac{1}{(k + 2)(k + 3)} = \frac{A}{k + 2} + \frac{B}{k + 3}$ for some $A, B$ makes sense

because after multiplying both sides by $(k + 2)(k + 3)$, you get $1 = A(k + 3) + B(k + 2)$

you need to choose $A, B$ such that the right-hand side has $0k$ and also has $1$: those are 2 simultaneous equations

[tall crag]

bro lemme help em

pfft

[grand pond]

south

[lone dock]

wow okay

[lethal path]

helping is not exclusive

[tall crag]

its inclusive?

[lethal path]

yeah

[lone dock]

aight! thanks yall ill close this now i think i got it

.close

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[hallow meteor]

Why does the process of determinant calculation with cofactors and minors work

[tall crag]

hm

check the gilbert strang course on determinant

https://youtu.be/srxexLishgY?si=qDr9DypGEDYYeKXM

[hallow meteor]

Also why does moving the eigenvectors of a symmetric matrix to the basis, stretching them by the eigenvalues and then rotating them back to original basis work the same transformation as the symmetric matrix itself im having trouble making intuitive sense of why the composition of those 3 has the same effect

[tall crag]

yk the main things about the symmetric matrix

idk what its called in english, features?

idk

like

[hallow meteor]

Being that they have orthogonal eigenvectors etc?

[tall crag]

actually

orthonormal

[hallow meteor]

And S = S^T

[tall crag]

eigenvectors

which leads to what???

spectral theorem

yk spectral theorem?

[frank wolf]

ultimately i think it comes down to the uniqueness result for totally antisymmetric functions

[hallow meteor]

Eigenvectors are not unique, eigenvectors scaled by a scalar are also eigenvectors

[frank wolf]

by this i mean the following

[hallow meteor]

Yeah

[frank wolf]

suppose $V$ is a $n$-dimensional vector space. then up to scaling, there is a unique totally antisymmetric function $\underbrace{V \times V \times \dots \times V}_\text{n times} \to \mathbb{R}$

[grand pond]

Pseudo (Cat theory #1 Fan)

[tall crag]

yes but they are orthonormal cuz they have length one

lol

[tall crag]

what is it?

[hallow meteor]

[frank wolf]

you can prove the determinant cofactor expansion from knowing this result

[hallow meteor]

The magnitude is irrelevant for eigenvectors.

[tall crag]

bro wait let dont confuse her or him

[frank wolf]

i'm not a bro

[frank wolf]

do you know roughly how this result gets proved?

[hallow meteor]

No I'm trying to understand what you're saying actually

[frank wolf]

so we can specialise to 2D space

suppose we have a totally antisymmetric bilinear map $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$

[grand pond]

Pseudo (Cat theory #1 Fan)

[frank wolf]

then it has to be a scaled copy of the determinant

[hallow meteor]

Well

I'll look into the determinant theory

But can you explain the spectral decomposition intuition in the meanwhile

[hallow meteor]

This

I'll just repeat it here: Why does moving the eigenvectors of a symmetric matrix to the basis, stretching them by the eigenvalues and then rotating them back to original basis work the same transformation as the symmetric matrix itself im having trouble making intuitive sense of why the composition of those 3 has the same effect

[tall crag]

look

can u write the formuls for spectral theorem

?

[hallow meteor]

S = Q D Q^T

We first apply Q ^ T which takes the eigenvectors and makes them the basis

[tall crag]

ok so u know what are those matrices

[hallow meteor]

Then stretch them according to the eigenvalues in D

And then rotate the basis back

[tall crag]

For each eigenvector $v_i$ we have $A v_i = \lambda_i v_i.$

[grand pond]

Viͥђaͣnͫ

[tall crag]

yk u can decompose a vector x in terms of eigenvecs

$x = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n$

[grand pond]

Viͥђaͣnͫ

[tall crag]

now apply matrix A

it gives u $A x = c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2 + \cdots$

[grand pond]

Viͥђaͣnͫ

[hallow meteor]

I suppose the idea is that since the transformation is linear, in the eigenbasis, that linear transformation is gonna do the exact thing as stretching the basis vectors,

[tall crag]

which is stretching each coordinate in the eigenbasis.

[hallow meteor]

Since all vectors in that space are linear combination of the eigen vectors of the transformation

[hallow meteor]

Rught

Right

Yeah makes sense

Thanks a lot

[tall crag]

now why changing the cords does not distort distance or angles

wait bro

[hallow meteor]

[tall crag]

lemme

[hallow meteor]

Yeah?

[tall crag]

explain the whole thing

[hallow meteor]

Alright

Thanks

I'll wai5

Wait

[tall crag]

yk that Q is orthogonal

or orthonormal

or whatever

u just know that Q^TQ = I

which is a rigid rotation

[hallow meteor]

Yeah, it's an orthogonal matrix, => orthonormal column vectors, i.e., no stretching

Yeah

[tall crag]

mashallah

so u understood it all?

[hallow meteor]

I think so yeah

I'll now move on to SVD

[tall crag]

have any questions?

[hallow meteor]

I suppose in the eigenbasis, the eigenvectors being stretched by the eigenvals of the lintransform is the equivalent of the entire lin transform in the original basis

So we are just rotating into the eigenbasis, stretching the eigenvectors, and then rotating back

Cool

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[tall crag]

whats wrong bro

[twilit field]

Typing out my soln

[tall crag]

aight lemme see

[twilit field]

Let $v,u \in V$ be arbitrary. Let $\phi(ST(v)) = \lambda; \phi(ST(u))= \mu$.

[grand pond]

wai

[frank wolf]

i don't understand this notation

what is $ST(v)$

[grand pond]

Pseudo (Cat theory #1 Fan)

[bold peak]

I'm guessing S ∘ T?

[hard shard]

standard veviation

[twilit field]

yes , evaluvated at v

[frank wolf]

then this doesn't typecheck for me

$ST(v)$ is a vector

[grand pond]

Pseudo (Cat theory #1 Fan)

[frank wolf]

how are you applying $\varphi$ to a vector

[grand pond]

Pseudo (Cat theory #1 Fan)

[twilit field]

fair

I need to think of some other approach

[tall crag]

ye i think its not right

[tall crag]

yk let I be an operator on V

[twilit field]

I'm wondering if I need to work with a basis here

[bold peak]

You don't

[twilit field]

rank nullity feels like overkill here

[bold peak]

Let's do some typing

$\varphi: (Vec \to Vec) \to \mathbb{F}$

[grand pond]

Xavier 🌺

[tall crag]

say that $\Phi(I^2) = \Phi(I)^2.$

[grand pond]

Viͥђaͣnͫ

[tall crag]

yk why thats right?

@twilit field

[twilit field]

I did this , this gives me phi(I)=0/1

[tall crag]

no

it gives u 2 value

u should prove why its not 1

[twilit field]

sure, but that just tells me the identity goes to 0, not every LT

[twilit field]

okay

[bold peak]

If identity goes to zero, you can use composition

[twilit field]

ah fuck

[bold peak]

Every map composed with identity is itself

[twilit field]

yes

the identity is the zero vector of L(V)

so it must map to 0

[tall crag]

yes but if u have phi(I) = 0 then u can say phi(I)phi(V) = 0

[scarlet solstice]

hmm, the identity isn't the zero vector of L(V), is it? The constant map v |-> 0 is the zero vector

[twilit field]

hmm, the operation is function composition unless I'm forgetting something

[runic hamlet]

L(V) is not a group under composition

do not mix up concepts

[bold peak]

That isn't the important part here though

It's a semigroup

(is that the one without inverse?)

[runic hamlet]

monoid

[scarlet solstice]

yeah, but L(V) is a vector space under pointwise addition, not under composition. If you include composition as an operation L(V) turns into an algebra, ie. a ring plus a vector space

[bold peak]

Monoid yes

[twilit field]

can I have a hint pls

Oh, might have it

phi(OI)= phi(O)Phi(I)=Phi(O)phi(O)

the problem here ofcourse is division by zero

[runic hamlet]

well you havent used linearity yet

so do that

[twilit field]

So chose two functions such that their sum is 0

and work with that

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

how

[runic hamlet]

honestly I did proceed by picking a basis

[twilit field]

this is probably wrong

but

[runic hamlet]

well I didnt need the full basis

just a lin independent set

[twilit field]

okay, so fin dim is important here

[runic hamlet]

dim > 1 is important

[twilit field]

Maybe I'm just not in the right frame of mind, but I don't see what to do here

should I just maybe come back to this in a bit, when I'm in a better mental state

thanks !

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[keen harbor]

What does the equal sign with a tilde means?

[dawn dagger]

usually isomorphic

[lyric charm]

isomorphic, yeah

[dawn dagger]

Ann is more correct than me

[clever sedge]

Isomorphism

[clever sedge]

Am I the most correct

[midnight plank]

@keen harbor Has your question been resolved?

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[keen harbor]

thanks

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[toxic prism]

absolute

[paper prism]

generally, conventionally, area bounded by a region should be taken with the sign

that is how we defined the definite integral

[toxic prism]

you're right, but the context matters

when area between curves is taken we usually consider it to be absolute

integrals are where we take signed

[paper prism]

fair enough, i guess

[visual tiger]

y' = (F(t^4))', not F'(t^4)

chain rule

[midnight plank]

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[novel cobalt]

Hi, I need to describe what this transformation matrix does. It is in the homogeneous coordinates. I know that the transformation scales an object to 0 and then translate to (2,1). But in order is this happening? Scale then translate or translate and then scale?

[novel cobalt]

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[glass panther]

Hey guys, i am 16 years old and i want to start self studying math. I just really dont know where to start. It has so many different fields and stuff and i am completly lost...

[hybrid widget]

depends on how much you've already learn at school

maybe number theory is one of the easiest ones, atleast for me

[glass panther]

I am now learning logaritms at school, i dont know if that answers your question?

[glass panther]

What do you mean with that?

[hybrid widget]

number theory is uh the study of properties of natural numbers?

and it doesn't really need too much of pre knowledge kindof

[hybrid widget]

geometry and number theory I guess

[glass panther]

So you think without any basics like knowing like those really wierd signs and stuff i should already get into fields?

[hybrid widget]

for book recommendations go to #book-recommendations

[hybrid widget]

you may need to know some terminology and new syntax

but if you're talking about doing the mathematical 'field' and algebraic structures, that may take a while

[glass panther]

Ah ok

thanks!

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[last slate]

Twenty balls are placed in an urn. Five are red, five green, five yellow and five
blue. Three balls are drawn from the urn at random without replacement. Write down
expressions for the probabilities of the following events.
(ii) The three balls drawn have different colours.
Hello! I'm not too sure if I have done this right. I thought it could be (20/20)(15/19)(10/18), as the first ball could be anything, the second has to be not the first one, and the third has to not be the second or first? Would there be a better approach ?

[forest iron]

i think its good

[last slate]

Ok thank you !

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[ember sage]

How do you start

[lethal path]

can you convert tan^2 x into sec^2 x + something?

[main current]

I personally suggest writing everything in terms of sin/cos

[ember sage]

I don't know

[visual elk]

Or maybe everything in terms of sec x? You'll get a cubic equation in sec x though

[lethal path]

they don't know this identity (which they really should learn)

[visual elk]

Yeah this works, the cubic you get is really easy to solve

[midnight plank]

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[rapid inlet]

where did the ln come from here?

[pearl mist]

result of integrating cosec theta

[fallow scarab]

you can differentiate log(csc(t) - cot(t)) to verify

[rapid inlet]

ah so this is going to be something i have to memorize ?

[pearl mist]

probably

[rapid inlet]

okay thank you !

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[hallow meteor]

What if we have a matrix with non orthogonal eigenvectors

[hallow meteor]

And we move those eigenvectors to the basis

Do the stretching etc

And move back

i.e., doing a spectral decomposition-like activity with non symmetric matrices

Why does this not work

Or does it work

[runic hamlet]

the moving back part is harder now

you cant just transpose anymore

and you might not have enough eigenvectors for a basis

[hallow meteor]

Right, well have ti find inverse the hard way

[runic hamlet]

but other than that thats just called diagonalisation

[hallow meteor]

Oh

[hallow meteor]

We will always have no. Of eigenvectors≤ no. Of basis vectors?

[runic hamlet]

yes

well, number of linearly independent eigenvectors

[hallow meteor]

Y3ah

Whys there never more

Ah right

They can't be independent

Makes sense

Also I wanted to ask about mxn matrices where m>n

What do they do exactly

R^n -> R^m

[runic hamlet]

what kind of answer are you expecting

[hallow meteor]

Sorry I don't have a clear idea gimme q sec

So am I correct in my understanding that if a matrix collapses dimensions down, then there is a matrix which does the exact inverse but we just have no way to calculate it out?

Or does that inverse just does not exist

[runic hamlet]

doesnt exist

we lose information

we cant get it back

imagine projecting from R^3 down to the xy plane

[hallow meteor]

But aren't there matrices which are like 4x3

[runic hamlet]

you just simply lose information

you cant undo that

[hallow meteor]

So like a matrix which is

3x2

What does it do

Doesn't it map R2-> R3

[runic hamlet]

yes

but the image is just a plane

(or a subset of a plane)

never all of R^3

[hallow meteor]

Wait whazd the image

What's*

[prime hornet]

the set of outputs of a function

[hallow meteor]

Oh

Oh

So there is nothing that can add a dimension

?

Yeahh makes sense

Theres just not enough poinnts in R2 to somehow map to all of R3

[prime hornet]

there's no linear map from R^2 that fills all of R^3, if that's what you mean

there are nonlinear maps that work

[hallow meteor]

non linear maps...

[runic hamlet]

linearity is a very strong condition

[hallow meteor]

Yeah

[runic hamlet]

if you have a 3x2 matrix then every vector you can get is a linear combination of those two columns

so the set that you get is <=2 dimensional

[hallow meteor]

Can u give example of non linear map which adds a dimension?

[runic hamlet]

google hilbert curve

[prime hornet]

https://en.wikipedia.org/wiki/Space-filling_curve

[hallow meteor]

This seejs like something which requires understanding of some other topics

[prime hornet]

R and R^n are bijective for any n in fact, so there always exists a bijection between the two sets

[hallow meteor]

I don't think I'm familiar with them

[prime hornet]

of course, the bijection won't be linear

[hallow meteor]

Makes sense!

[runic hamlet]

there are some examples of functions that give a bijection [0,1]->[0,1]x[0,1]

[hallow meteor]

Yeah ... So ofc there exist such mappings but with looser and looser conditions

[runic hamlet]

which are essentially based on decimals

[prime hornet]

linear maps are really important though

despite the condition being really strong, they end up showing themselves all over the place

[hallow meteor]

yeah

Not NxN

P(N)

Anyway thanks yall

❤️

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[bright plinth]

Just wondering, on integration questions which involve two equations and finding the area, in which cases can you take the integral of the Simultaneous Equation?

[lofty knoll]

Never

[bright plinth]

For example, In the question below tried just taking the integral of the simultaneous equation, and i got the wrong answer. But im sure ive done this before and got the right answer. So im confused

[lyric charm]

can you show your working for this one

"integral of the simultaneous equation" does not really make sense

[bright plinth]

Sorry, im horrible at explaining things

[lofty knoll]

I think they mean when they set the two equations equal to each other to find the intersect point

And then they want to integrate the function they get from that to find the area, which obviously doesnt work

[lyric charm]

ok so the thing is you integrate the difference of your two functions

[bright plinth]

yeah, i saw that in the markscheme. However im sure ive done it like this before and got the right answer? maybe i fluked it?

[lyric charm]

there's a complication in that your region is not bounded by only l and C

[bright plinth]

so if its bounded by the x axis you cant do?

[lyric charm]

less "can't", more "requires more work"

let me desmos this to explain:

• what your calculation actually gave you;
• how to proceed properly

[bright plinth]

which is? because i was originally planning to do it the right way, But i thought if i just took the integral of the S.E it wouldve been quicker than finding two seperate integrals and doing it like that.

[lyric charm]

don't call it "integral of the SE" please, nobody will ever understand you

[bright plinth]

sorryy what do i call it?

[lyric charm]

integrate the difference of the functions

so

[bright plinth]

okay okay!

[lyric charm]

you see this picture?

green = what you need to find, the region R that the question talks about

what your integration did was find the area of green + purple, as you integrated the difference between the red and blue curves from 1.5 to 6

[bright plinth]

Ohh okay, so it doesnt bind it by the x axis?

[lyric charm]

there are two routes here that you can take:

1. work out the purple area as an appropriate integral of just the quadratic function, then subtract from the area you found already
   or
2. work out green + orange as just a triangle, and then work out orange as an integral of just the quadratic, then subtract

[lyric charm]

there's no x-axis boundary there, indeed

[lyric charm]

my personal suggestion is route #2 btw

[bright plinth]

This is the markscheme, which is route 1?

[lyric charm]

their suggestion is actually route 2, by the looks of it

they don't suggest 1 in the ms at all

though if you pull it off and do it right, it can earn you full credit

[bright plinth]

Okay thanks, but how would you find orange ?

[bright plinth]

Wait, after a second look, i think i actually understand. Thanks alot for the help. Much appreciated!

[lyric charm]

integral of just the quadratic from 5 to 6

[bright plinth]

okay i get it now! Again, Thanks alot much appreciated!

.close

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[hallow meteor]

Why are singular values of left and right symmetric matrices equal

[hallow meteor]

And the leftover eigenvalues are 0

Why

[ancient ridge]

If you have eigenvector v of A^TA so that A^TAv = lambda v, then AA^TAv = lambda Av so that Av is an eigenvector of AA^T with the same eigenvalue.

[midnight plank]

@hallow meteor Has your question been resolved?

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[bold peak]

Think of it this way

When you go "towards" the angle it's cos

When you go "away from" the angle it's sin

[leaden seal]

let me show you

[bold peak]

No

But why does that matter

[leaden seal]

becus im trying to understand why Fs=sintheta

[bold peak]

Draw a rectangle with F as the diagonal

[leaden seal]

i got it ty

[crude adder]

Is this question from an intro applied aero physics or math class? I noticed another problem with Cl and Cd which are related related varaibles just I was just curious

[leaden seal]

no

sorry

yh

[midnight plank]

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[last slate]

I need help

[placid spoke]

!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

[radiant roost]

I think "at least one club" can be calculated as 1 - no clubs

[last slate]

wait a sec bro i posted the wrong question heres where im stuck on

im stuck on this

[last slate]

1

[radiant roost]

The "1 -" trick should work well here too

1 - 0 heads - 1 head

Wait...

I'm dumb

[last slate]

Can you tell me what i did wrong?

I did C(n-1, 2) * 2

why is it wrong

we have n spots

and n-1 need to be heads

[radiant roost]

It should be the probability of n-1 head, + probability of n heads

[last slate]

so P(n-1, 2) /2! Gives us the possibilities for the 4 spots to have heads

[radiant roost]

Right?

[last slate]

yeah apparently but idk why man

im so frustrated

[radiant roost]

Because those are the two possible ways to get at least n-1 heads

They're the only two possible ways

And they're disjoint, so you can add them

[last slate]

whats wrong with what i did

[radiant roost]

You said n spots and n-1 need to be heads

[last slate]

Yes

[radiant roost]

Wouldn't that be C (n, n-1)?

Idk how you got that expression

C (n-1, 2) is choosing two from n-1

That's totally different

[last slate]

so yeah this id wrong

[last slate]

But can you please explain to me why its this

[radiant roost]

C(n, n-1) is the number of ways to choose n-1 heads from n flips

[last slate]

whats the intuition

[radiant roost]

That's pretty much the definition of C

[last slate]

im so frustrated

bro please

[last slate]

why bro😭

im about to give up

{Heads}^n-1 @radiant roost

1^n-1

BROROROOOOOO

[slow thorn]

for part a?

[last slate]

yes

[slow thorn]

im not sure why its C(n-1, 2)

[last slate]

its not that was wrong

But what is it

whats the logic

whats the cartesian product

for part a

[slow thorn]

ok the sample space is $2^n$

[slow thorn]

right?

each flip has two choices

[last slate]

yes

I agree

sample space is 2ⁿ

[slow thorn]

and the cases where theres at least n-1 heads

so there could be n-1 heads or n heads

[last slate]

Yes

[slow thorn]

for an example of n = 3

say we find the cases where there are n-1 = 2 heads

[last slate]

yes

[slow thorn]

the cases would be something like
HHT, HTH, and THH

you can see that only one tail is in all these cases

[last slate]

yes

Order doesnt matter so like {heads}^2 x {tails}

?

[slow thorn]

okayy so

you're saying for each case

the probability is $(P(heads))^2 \times P(tails)$

[last slate]

No no im doing cartesian product

[slow thorn]

hmm did you do this cartesian product or see it in some other problem?

[last slate]

im just tryna figure it out

i came up with this one particularly

1^n-1????????

[slow thorn]

yeah but you can't just invent something like this without seeing anything related

like whats {heads} supposed to mean

[last slate]

Can you vc?

i cant type anymore

i cant express what im tryna say

if not its ok

[slow thorn]

not possible rn

[last slate]

but like you’re trying to get it in the order of { HHT }

wait order does matter

what’s happening bruh

how

help please man

[slow thorn]

cartesian product in the sens

${H, T} \times {H, T} \times {H, T}$?

[last slate]

yes

but I agree that its C(n, n-1)

i figured out why

[slow thorn]

ok cool

and for n heads its 1

HHHH...

[last slate]

its 1?

theres 1 way you can get n heads

[radiant roost]

Yeah

[last slate]

Thanks guys

@slow thorn thank you

but wait

how do i express C(n, n-1)

[radiant roost]

Write n-1 instead of k

If you do it right, it will simplify to n

[last slate]

thanks bro

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[twilit field]

Just wanna make sure I'm doing this right

the confusing thing for me is setting up the convolution here

like f(t)=f(z-t)=1/4

rightt

[ancient ridge]

I don't really like that notation but yeah

[twilit field]

and then the convolution gives me -2<t<2; z-2<t<z+2

but this isn't enough to find the bounds, is it

[ancient ridge]

I should be able to follow what you are saying without recasting it into notation that I am more comfortable with, I'm not.

[twilit field]

sure

[ancient ridge]

Start by defining $X, Y \sim U(-2,2)$ so that $Z = X + Y$ Then you have $f_Z(z) = \int_{-\inf}^{\inf} f_{Y}(z-x)f_{X}(x) dx = \frac{1}{4}\int_{-2}^{2} f_{Y}(z-x) dx$ since $f_{X}$ is zero outside of its support.

From there I think the Calculus is a little less confusing

[grand pond]

JessicaK

[twilit field]

mhm

yea, it is

[ancient ridge]

I just find not explicitly defining your random variables, and using a dummy variable t to make it a lot harder to follow

[twilit field]

it's just a matter of z-x=t now , is it not

[ancient ridge]

The integrand is zero unless -2 <= z-x <= 2

[twilit field]

yes

[ancient ridge]

yeah

Anyway you should be able to take it from here

[twilit field]

one minute

don't I have to change the bounds of integration

so it become $\int_{z-2}^{z+2} \frac{1}{16} dt$

[grand pond]

wai

[ancient ridge]

I need to write down the details because that was just definition grinding but you will need to do something like that yeah

[twilit field]

cool

thanks

another question

here I surely don't have to write down the table

nvm got i

it

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[sick garnet]

hello everyone, i sruggle getting good math grades can anyone gime me advice

[sick garnet]

struggle*

sorry for bad grammar

[visual elk]

That's an open ended question

What grade are you in

[sick garnet]

8

[visual elk]

Most of the time when i see someone who is bad at math, it is because they have poor foundations

[sick garnet]

i see

anyone advice for preparing on tests such as maths

any

[visual elk]

What are the things on your syllabus tho

[sick garnet]

vic curriculum Kambrya college

aus

state of victoria

[visual elk]

Can you share the syllabus

[sick garnet]

ok

https://f10.vcaa.vic.edu.au/learning-areas/mathematics/introduction

bye but pls share advice

[visual elk]

I see

I suggest that you should first master the properties of fractions. How they work and how they can be manipulated. Once you are comfortable enough with working with fractions, things are going to get a bit easier. Then you want to master the manipulation of algebraic expressions, factorization etc. These are the very basics

Most of the time people move onto doing higher topics without having a clear understanding of these.

Once you master fractions and algebraic manipulations, move on to exponents

Learn the rules of exponents in detail, and apply them to simplify large expressions

A big part of being good at math is being able to simplify expressions

Then you want to move onto solving equations

Solving equations the single most important thing you need in your level

Learn how to solve different kind of equations from the very start in detail

Like linear equations, quadratic ones, (cubic ones if you really wanna), systems of equations, exponential equations etc

When you master solving equations, you can combine that with your previous knowledge to solve all kinds of word problems

You just gotta practise and learn how to turn word problems into equations, simplify, and solve them

[midnight plank]

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[gaunt jetty]

Test

[gaunt jetty]

Double pin wtf

.close

[midnight plank]

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[bold peak]

Prepare for trouble and make it double

[midnight plank]

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[tall crag]

can u translate

oh u want the absolute maximum and min?

[tall crag]

ah

so

what is the problem?

did u take derivative?

@tidal turret

[tidal turret]

well i was trying to use lagrange multipliers method

[tall crag]

why

[tall crag]

just compute the partial derivatives and set them to zero

mh

if u like to use lagrange multipliers u can too

[midnight plank]

@tidal turret Has your question been resolved?

[tall crag]

u still have problem?

set up lagrange multipliers on circular boundary constraints

g_1(x,y) = (x-5)^2 + y^2 - 25 = 0

g_2(x,y) = (x-2)^2 + y^2 - 16 = 0

first for outer circle second for inner circle

then use lm

$\nabla f(x,y) = \lambda \nabla g(x,y)$

[grand pond]

Viͥђaͣnͫ

[tall crag]

then solve for (x,y) once for inner once for outer circle

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[twilit field]

wanna make sure I'm not doing anything wrong here

[twilit field]

Z=(Y^3+1)(X^2-1)

so E[Z]= E[Y^3+1]E[X^2-1]=E[Y^3]E[X^2]

[frail carbon]

this is not how expectations work, for the first equality you need independence (not the case), and the second isn't true ever since that's not how constants are handled

[twilit field]

so here I just brute force it?

[frail carbon]

I don't think there's a nice way around it

[twilit field]

got it

thanks

.close

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[paper prism]

.

[west ravine]

..

[icy island]

there r many indians here

[midnight plank]

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[prime star]

if $X_1, ..., X_n$ are iid with common mean $\mu$ and variance $\sigma^2$, show that $\frac{1}{n} \sum(X_j - \bar{X_n})^2 \to \sigma^2$ almost surely

[grand pond]

Red

[prime star]

i tried a whole bunch of things that did not get me closer to an answer

i have the strong law of large numbers on hand:

[potent knoll]

What are you confused about this, great bard?

[prime star]

right now i'm trying to show that a limit converges

but i dont think i'm supposed to do that

[prime star]

SLLN: $\frac{1}{n} \sum X_n \to \mu = E(\bar{X_n})$ almost surely

[grand pond]

Red

[coarse rivet]

need help

[prime star]

there are other help channels open, pls go there

[potent knoll]

Are you asking how to start?

[prime star]

yes

[potent knoll]

Let me see if I can assist the great bard

[potent knoll]

Well for starters you can decompose the sample variance

Express it into terms of the true mean and the sample mean

[subtle zinc]

maybe expand the square

[prime star]

where do you get the true mean part of it?

i have this:

[subtle zinc]

true mean = mu

[prime star]

$\frac{1}{n} ( \sum X_j^2 - 2 \sum X_j \bar{X_n} + \sum \bar{X_n}^2)$

[grand pond]

Red

[prime star]

then ??
$E(\bar{X_n}^2) - 2 E(\bar{X_n}^2) + E(\bar{X_n}^2) = 0$

[grand pond]

Red

[potent knoll]

Give me a moment

What is with this signal 😭

[prime star]

rip your wifi i guess :')

[potent knoll]

It looks good so far

[potent knoll]

Hold on, is this supposed to be the next part after the one you sent?

The first term is the average of the squares. The second and third terms involve the square of the average. I believe these are two different things

Instead of them all cancelling out to 0, should they not be converging to different limits?

When you subtract them, they shouldn't cancel to 0. Shouldn't they cancel to the variance formula?

[potent knoll]

So far so good, the latest one you sent seems off tho

[midnight plank]

@prime star Has your question been resolved?

[prime star]

oops just got back to this

[prime star]

yes they're definitely supposed to :')

[potent knoll]

I think that should help you correct your current step

[prime star]

so, where is mu supposed to appear?

[potent knoll]

Well, were following your solution so, we don't need it that (for now, I believe)

[potent knoll]

But anyways! Here's what I think is wrong with this step

Why did it do that 😭

Landscape, not portrait bruh

[fallen sparrow]

,rccw

[potent knoll]

Thank you 🥲

[prime star]

wait i think i got it

$\frac{1}{n}\sum X_j^2 - 2 \bar{X_n}^2 + \bar{X_n}^2 = \frac{1}{n}\sum X_j^2 - \bar{X_n}^2$\
$= E(X^2) - E(X)^2$\
$=Var(X) = \sigma^2$

[grand pond]

Red

[potent knoll]

That's right

[prime star]

probability...

[prime star]

yayy thank you so much

[potent knoll]

Np, it was a pleasure helping the great bard

Do you need any more help? If no more, you can .close

[prime star]

.close

[midnight plank]

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[hallow meteor]

After SVD we have X = U D V^T

[hallow meteor]

Why are the Principle components the eigenvectors of V and not the eigenvectors of U

Like why are eigenvectors of V the principle components in the first plac

Place

<@&286206848099549185>

[potent knoll]

It's because they are directly related to the covariance matrix of your data

[hallow meteor]

Oh right

And V is the eigenvectors of the covariance matrix

Right?

[potent knoll]

It's not U because it contains the eigenvectors XX^T that are related to the covariance across the sample/rows. not X^TX which is related to the covariance across the features/columns

[hallow meteor]

Ohh

[potent knoll]

Yes, that is right

[hallow meteor]

So what does U give us

[potent knoll]

Left singular vectors

[hallow meteor]

Yeah but

What's the interpretation

Cuz like V gives the directions ordered by variance along which features have max variance (right?)

Then what does U give

[potent knoll]

Yes, in a way

[potent knoll]

The coordinates

[hallow meteor]

The coordinates... Of?

[potent knoll]

Your original data points

Well after they have been projected to the new variance-maximizing axes defined by V

Think of it like this basically

V is the map directions
U is the new addresses (scaled by D)

[hallow meteor]

Wait lemn3

Lemme try make sense of this

[potent knoll]

It's okay, you can do this 😭

[hallow meteor]

So it's the data but written in terms of the new basis?

New basis being the pca axes?

[potent knoll]

Exactly

[hallow meteor]

Thanks

[potent knoll]

Np, anything else you need help on?

[hallow meteor]

I will be needing help, im Learning like svms, perceptrons, anns, SVD, pca, bagging amd boosting from scratch

👶

Can I @ u if I need help?

[potent knoll]

Go ahead! I might be a bit delayed tho but feel free to:) (also studying 😭)

[hallow meteor]

Thanks

.close

[midnight plank]

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[coral belfry]

so our prof asked us to evaluate cos(3θ) using De Moivre's rule, which im not sure how to implement here. I solved it like this, using two or three easy identities. How would we go about doing it the way my prof intended?

[coral belfry]

,rccw

[fallow scarab]

well first of all you're supposed to use demoivre

,tex .demoivre

[grand pond]

riemann

[coral belfry]

yes

[fallow scarab]

use n=3 and equate the real and imaginary parts of both sides

[coral belfry]

one side is raised to the n tho how do i find the real part?

oh wait i can just do the identity

[gaunt jetty]

u gotta do binomial expansion

[coral belfry]

hahah okay i thought there was a nicer way

thank you

.solved

[midnight plank]

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[midnight plank]

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[tired sonnet]

im having a hard time understanding this topic

[rose trout]

Any part in particular doesn't make sense to you?

[tired sonnet]

wait

finding the values of a and b

in both questions

[visual tiger]

Do you know what's the range of y = asin(bx)?

At the very least y = sin(x) to start

[tired sonnet]

do i know the range of that format?

[visual tiger]

You know what the range of a function means right?

[tired sonnet]

yes

[visual tiger]

So, do you know what the range of y = sin(x) is

[tired sonnet]

no

oh

wait

-1,1

?

maybe

[visual tiger]

[-1,1]

[tired sonnet]

yes

[visual tiger]

And so the amplitude is...?

[tired sonnet]

1

[visual tiger]

Yes

If instead of sin(x) we had sin(bx)

What changes?

[tired sonnet]

the period

[visual tiger]

Yes

[tired sonnet]

and the amplitude i think but idk why

[visual tiger]

The amplitude does not change

[tired sonnet]

oh

so just the period

[visual tiger]

sin(bx) changes the speed of the input inside the sin

So it fluctuates faster/slower

But the range, and so the amplitude, don't change

[tired sonnet]

ok

[visual tiger]

And so what's the period of sin(bx)?

[tired sonnet]

360

360degrees or 2pi

[visual tiger]

That's the period of sin(x)

[tired sonnet]

divided by b

?

[visual tiger]

Yes

[tired sonnet]

okay

[visual tiger]

So

Finally

Let's multiply by a in front

y = asin(bx)

What changes compared to sin(bx)?

[tired sonnet]

its the midline

[visual tiger]

The midline?

[tired sonnet]

y=a is the midline

[visual tiger]

Unfortunately no

The midline didn't change

If the range of sin is [-1,1], what's the range of sin multiplied by a

[tired sonnet]

-1a,1a

[visual tiger]

If a is positive (let's assume so for now), yes

So range is [-a,a]

The midpoint of that range is still 0

[tired sonnet]

okay

[visual tiger]

What about the amplitude?

[tired sonnet]

of what

[visual tiger]

Of y = a*sin(bx)

[tired sonnet]

is it not just a

[visual tiger]

It's a if a is positive

[tired sonnet]

ok

so for 4(i) the value of a is 3

[visual tiger]

If a is positive, a is 3

There's also a possibility that a is negative

In that case a is ...

[tired sonnet]

would it not ask for -a then

[visual tiger]

No?

It asks for all possible values of a and b

[tired sonnet]

oh

[visual tiger]

So a,b can be any real numbers that satisfy the conditions

[tired sonnet]

so a can be 3 and -3

what about b

fwait

360 divided by pi

or 2pi divided by pi

[visual tiger]

The period is 2pi/b (2pi/|b| if b is any real number)

So you need 2pi/|b| = pi

[tired sonnet]

b = 2?

[midnight plank]

@tired sonnet Has your question been resolved?

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[night gyro]

hello i have 2 questions

[night gyro]

how does that red bit not turn into 2 pi

also

is it right to do 4x^3/3pi

[tribal temple]

(assuming that you have some form of "ice cream cone", with half a sphere stuck to the end of a cylinder?)

[night gyro]

wait

[sudden yacht]

What you wrote in red is exactly what's written already

[night gyro]

ill show full pick

i jst had questions abt wht i posted

[night gyro]

they wrote 4/3 x^3 pi

[sudden yacht]

You too 😅

[night gyro]

ok so wht i wrote works

[sudden yacht]

Maybe I'm not understanding your doubt atp

[tribal temple]

(as long as you're clear enough by how you mean (so e.g. have ) then you're fine )

[night gyro]

?

[night gyro]

i wnted to know if the both i wrote and wht they wrote mean the same thing is ok

[night gyro]

also how does tht not turn into 2 pi

[tribal temple]

To clarify, do you mean in a similar sense as to how pi + pi would make 2pi?

[night gyro]

yh

[sudden yacht]

Well, it does

2π/3 + 4π/3 = 6π/3 which is indeed 2π

[tribal temple]

(I would write out what I mean in TeX but I'm on my phone so can't really )

[sudden yacht]

Where are you seeing it not happening?

[midnight plank]

@night gyro Has your question been resolved?

[night gyro]

hello sorry i was trying the question

i still dont know

yoooo

nvm i got it

[night gyro]

thank you

.close

[midnight plank]

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[night gyro]

hello why is this not correct

[gusty falcon]

10 cm greater than the area of the rectangle

if you added 10 to the area of the rectangle then that'd be equal to the area of the triangle

[night gyro]

hm

but it says 10 cm greater

u add 10 to the other side to balance it out?

[unborn stone]

If you're 10 cm taller than your brother, he's the one you're gonna give boots to get taller

not you

It's the same thing here

If the triangle area is greater than the area of the rectangle, you're gonna add 10cm² to the area of the rectangle to balance it out

You need to give this 10cm² to the smaller one to balance it out, not the bigger one

[night gyro]

ooooo

okk i see

[night gyro]

thank you

.close

[midnight plank]

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[midnight plank]

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[scarlet flower]

this is my teachers solution

when it says find all vectors why couldn't you just show the final answer as the vector product multiplied by some scalar k

[fallow scarab]

You do need the scalar k

You should tell your teacher

[scarlet flower]

lol i will thx

.close