help-49

[steel crest]

sure

[cedar pawn]

Okay so we'll not get too abstract for now

Let's just approach the original question

[steel crest]

in your assignment though you needed to find (2A)^{-1} thats why I suggested it

[cedar pawn]

We want to find (2A)^-1

What equation does (2A)^-1 satisfy?

[last slate]

2A * (2A)^-1 = 1

[cedar pawn]

These are matrices now, so we get I, not 1

Do you know what I looks like, for a 2x2 matrix?

[last slate]

yes

[last slate]

yes

[cedar pawn]

What does it look like?

[last slate]

ill draw it

[cedar pawn]

Okay looks good, although we don't need to write out the whole matrix just yet

So we have this atm:

(2A) (2A)^-1 = I

[last slate]

yes

[cedar pawn]

Scalar multiplication of matrices is straightforward

It's just elementwise

[last slate]

yes

[cedar pawn]

We can take the 2 out of the 2A

And move it over to the other side

[last slate]

divide both sides by 2?

[cedar pawn]

Yes

(2A) (2A)^-1 = I
2 A (2A)^-1 = I
A (2A)^-1 = 1/2 I

Oki doki

Do you know about pre-multiplying and post-multiplying?

[last slate]

no

i assume not

[cedar pawn]

Do you know that matrix multiplication is not commutative?

So AB =/= BA

[last slate]

yes

i do

[cedar pawn]

Okay

[last slate]

is that always the case btw

[cedar pawn]

It's possible for AB = BA but most of the time it doesn't

[last slate]

oh ok

[cedar pawn]

It's not guaranteed to be the case like with numbers

For numbers, ab = ba

But for matrices, it's not true

[last slate]

yes

[cedar pawn]

This means that when we multiply both sides of an equation by a matrix, we have to be careful about whether we're multiplying at the start (pre-multiplying) or the end (post-multiplying)

If M = ABC:

Pre-multiply by D -> DM = DABC
Post-multiply by D -> MD = ABCD

[last slate]

okay

[cedar pawn]

A (2A)^-1 = 1/2 I

So we have this

How do we get rid of A on the left side?

[last slate]

divide?

[cedar pawn]

Divide by what?

[last slate]

A

[cedar pawn]

How do you divide by a matrix?

[last slate]

idek

you dont

[cedar pawn]

Yup

So what else can we do?

[last slate]

they gave me a formula

[steel crest]

think about the scalars

you can always multiply both sides of a matrix equation by a scalar

[cedar pawn]

You don't need a formula

What we've discussed so far is enough

[cedar pawn]

We're here at the moment:
A (2A)^-1 = 1/2 I

[last slate]

we want (2A)^-1

[cedar pawn]

Which means we need to get rid of A, right?

[last slate]

right

[cedar pawn]

Then we have (2A)^-1 = ...

How do we get rid of A?

[last slate]

i cant think of anything other than division 💀

[cedar pawn]

What does I (2A)^-1 = ?

[last slate]

idk

[cedar pawn]

What does multiplying by I do?

[steel crest]

what if the RHS was just I instead of (1/2)I?

[last slate]

give you the same matrix

oh

[cedar pawn]

^^^

[last slate]

so (2A)^-1

[cedar pawn]

Yep

So if we had I instead of A, we'd be done

[steel crest]

oh nm you did that step already

[cedar pawn]

How do we turn A into I?

[last slate]

multiply it by its inverse

[cedar pawn]

Yes!

[last slate]

yesir

[cedar pawn]

But we need to be careful which side we multiply it on

[last slate]

how do i know

[cedar pawn]

You can think of multiplying by the inverse as being the matrix equivalent of division, if it helps

A (2A)^-1 = 1/2 I

So we have this

If we post-multiply by A^-1, we get:
A (2A)^-1 A^-1 = 1/2 I A^-1

Because the A and A^-1 are split around the (2A)^-1, we can't combine them

Because matrix multiplication is not commutative

So we need to pre-multiply by A

[last slate]

yes

[cedar pawn]

Basically, if you want a matrix to go away, you need to put its inverse next to it

[last slate]

yes

does it matter next to it on the right or leftr

[cedar pawn]

So we need to pre-multiply by A^-1

No

Right and left are both fine

Because A A^-1 = A^-1 A = I

That's not a given since matrix multiplication is not commutative

But it can be proved and it is true

[last slate]

yes

[cedar pawn]

So A A^-1 and A^-1 A are both I

[last slate]

true

[cedar pawn]

And that means you can put it on either side

[steel crest]

I usually see it as the definition of inverse

for groups, matrices, etc, that it has to commute

[cedar pawn]

Potentially, but if we've stated that matrix multiplication doesn't commute in general, then we should prove it works in this case. If you want to define it that way, you have to prove it exists. Otherwise, you have to prove that it's commutative.

[cedar pawn]

So if we pre-multiply by A^-1, we get:
A^-1 A (2A)^-1 = A^-1 1/2 I

Right?

[last slate]

right

[cedar pawn]

Can you simplify that?

You can keep going

Yes!

[last slate]

oooooooooooh

[cedar pawn]

Although I would write the 1/2 before the A^-1

[last slate]

can i do this method whenever i wanna find any inverse

[cedar pawn]

Well

The reason this is useful is because in Part A, you found A^-1 already

If you didn't know A^-1 already, this wouldn't be as useful

But if you know what A^-1 is, and they ask you to find a different inverse that is related to A^-1, you can do this

I suppose you can also use this method to find any inverse you like, but that's simply where the formula they've given you comes from

[steel crest]

i dont jhavbe to prove it exists if it's defined as existing iff it's a right and left inverse

[last slate]

how about if i wanna find any inverse ?

[steel crest]

more than one approach, but that's how it was taught to me and thats how I taught it to LNRD earlier

[last slate]

do i start with the fact that A * A^-1 = I?

[cedar pawn]

The formula they've given you is the result of that process

[steel crest]

for matrices 3x3 or larger there are algorithms, there aren't any formulations so nice and tidy as the smaller cacses

[last slate]

but isnt that only for 2x2

[cedar pawn]

Yes

You could attempt to try to solve that equation for larger matrices

But it would be excruciating

You'd have a system of 9 equations to solve for a 3x3

But for matrices larger than 2x2, there is an algorithm to follow

That evaluates the inverse

[steel crest]

there is a special case that is very easy and quick for large matrices, a very specific kind

diagonal matrices

[cedar pawn]

Finding matrix inverses is somewhat painful, so being able to do some matrix algebra is helpful if you have to do more than one of them

[steel crest]

that's when all the non-diagonal elements are 0

it's invertible if and only if none of the diagonal elements are 0, and the inverse is just you take the inverse of each number on the diagonal

[cedar pawn]

Going back to here:

It makes sense that (2A)^-1 = 1/2 A^-1

Because we can pull out scalar multiples from matrices

So if we have:
(2A) (2A)^-1 = I
Then we get:
2 A (1/2) A^-1 = I

And the 2 and 1/2 cancel out

So we should consider that to be the expected result

Also, matrices are generally used to represent geometric transformations

And multiplying by 2 is equivalent to an enlargement by a factor of 2

The matrix inverse is the opposite transformation

So that should involve an enlargement by a factor of 1/2

To shrink it back down again

[last slate]

oh ok

[cedar pawn]

Does that all make sense now?

[last slate]

i guess yeah

[cedar pawn]

Oki doki

[last slate]

i mean i understand how to find something in that particular position

but not generally

for any matrix

3x3 for example

[cedar pawn]

If you haven't covered that, don't worry about it

It's a complicated procedure

Hmm

Okay well you have to follow a procedure to do it

That's still 2x2

You could attempt to do that

[last slate]

can i do it to 3x3

[cedar pawn]

Yes, but it would be difficult

But you could if you wanted to

You'd end up with a system of 9 simultaneous equations

That would work but there is a faster method to do it

[last slate]

what is it?

[last slate]

damnn

[cedar pawn]

I can send you a link to another textbook that covers some matrix stuff if you like

If you don't like your current one

It isn't

But that is one method you can use

It's this

[last slate]

and this works for any 3x3?

[cedar pawn]

Yes

[last slate]

can we do this one

i made it up

[cedar pawn]

That's not 3x3

[last slate]

oh

yeah lolll

[cedar pawn]

Well they want you to use Gauss-Jordan Elimination for that one

But if you want to do it this way just to try, we can

[last slate]

does that method work for all 3x3?

[cedar pawn]

Yes

[last slate]

thank God

okay ill do it that way then since we have a test tmrw on it

[cedar pawn]

There are 3 methods for finding the inverse of a matrix, and all 3 work for all 3x3 matrices

[last slate]

ohh

[cedar pawn]

• Method of cofactors
• Gauss-Jordan Elimination
• Cayley-Hamilton Theorem

[cedar pawn]

This is the method of cofactors

[last slate]

life would be 10x easier

[cedar pawn]

The Cayley-Hamilton theorem one is a little bit more advanced so might not be useful to you at this point

The other two are standard methods for your level

If you have a test on Gauss-Jordan Elimination, it might be a good idea to do that one

[last slate]

how am i gonna just keep doing this

bro i cant keep doing that

[cedar pawn]

You're nearly done

[last slate]

bro idk

im giving up

it looks like theres no way im gonna finish that ever

[cedar pawn]

You're 2 steps away

[small jasper]

(don't let the numbers sucking scare you off)

[last slate]

bro i feel like numbers are just gonna keep coming

[cedar pawn]

You can use the middle row to finish the last row, and then the last row to finish the top row and you're done

[last slate]

youre right 😂

wait btw

are there 3x3 matrixes that arent inversable

invertable *

[cedar pawn]

Yes

[last slate]

how do you know

if they arent

[cedar pawn]

Any matrix with a determinant of 0 is not invertible

That's one advantage of the method of cofactors

Since Step 1 is to find the determinant, if it's 0 you can just stop there

[sick fossil]

the zero matrix

less trivially, any matrix which projects space onto a plane, intuitively because information is lost

[cedar pawn]

You've made a mistake in Line 3 x.x

-5 (-2) - 4 = 10 - 4 = 6

[last slate]

oh God

nooo

dman it

[last slate]

in general though do you always just use intuition to eliminate the numbers?

like is there not a pattern

[midnight plank]

@last slate Has your question been resolved?

[last slate]

how does that work

[spiral rock]

The system can be written as
Ax = b,
If you know the inverse then after mulitplying on the left by A^-1 you get
x = A^-1 * b

[last slate]

whats Ax = b

[spiral rock]

A is the matrix describing the linear system of equations

x is the vector (x1,x2,x3) i.e. your variables

b is vector of the constants in your equations
In this case b=(1, -2, 4)

[last slate]

vectors? all i know is a vector has a direction and magnitude, idk how this is a vector

and also, can all linear systems be rewritten like that? and was that the only way to solve for part b? is to rewrite it in that way specifically??

[spiral rock]

Yes!

You could just solve it normally

[spiral rock]

In linear algebra vectors have a generlzition which using vector spaces.
For this context you can think of them as R^n, i.e. n-tuples of real numbers

So here the vectors are 3-tuples. And you can extract direction and magnitude from them given you have a correct notion of them

[last slate]

literally "Ax =b"?

for all linear systems of equations

[midnight plank]

@last slate Has your question been resolved?

[bold peak]

Yes

[midnight plank]

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[midnight plank]

@white moss Has your question been resolved?

[midnight plank]

@white moss Has your question been resolved?

[midnight plank]

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[uneven relic]

Need help with exercice 2

[uneven relic]

Can translate if needed

[cedar mason]

what is this

dont get this either

ok ig its 'the diagonals are the same length'

[cedar mason]

i dont get this

[uneven relic]

Show the equivalence of the following statements

[cedar mason]

thats not very helpful

[uneven relic]

Yea

[cedar mason]

what does that even mean

[frozen talon]

that any parallelogram satisfying any one of (i), (ii), (iii) also satisfies all three

[cedar mason]

ah

ok

!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

[uneven relic]

1 but i missed the class where this was explained and its due in 1 hour soo

[frozen talon]

lol

[cedar mason]

well is it a test

[frozen talon]

usually in "the following are equivalent" proofs, you prove them in a little circle

(i) => (ii) => (iii) => (i)

[uneven relic]

No homework

But it counts

[cedar mason]

no cheating /j

[frozen talon]

so you only need to write three proofs, theoretically

[uneven relic]

Ig

[cedar mason]

start with i \implies ii

[uneven relic]

Do i jus say that since i says that its a rectangle ii is true because all angles of a rectangle and 90degrees

[frozen talon]

you can phrase it a little nicer, but yeah (i) => (ii) sounds pretty straightforward

[uneven relic]

Okok

[frozen talon]

... although how (i) saying "P is a rectangle" is equivalent to "one of the angles is right" is strange

[uneven relic]

Yea idk

[frozen talon]

it feels more like that blurb should be attributed to (ii)

or

idk

like the blurb of (i) should be the whole statement of (ii) and vice versa, that sounds slightly less trivial and more acceptable

[cedar mason]

it isnt

[frozen talon]

but whatever

[cedar mason]

a rectangle is literally just a ||gm with a right angle

[frozen talon]

sure but that at least requires like two whole lines of algebra to demonstrate

[cedar mason]

i think its a basic fact

[frozen talon]

it's basic but not immediate

[uneven relic]

You cant just write one line

[cedar mason]

in our geo courses we defined a ||gm first and defined a rectangle with that

[uneven relic]

You cant just sau because u know it

[frozen talon]

well provided there's a way to define rectangles independently of parallelograms, I'd say it's not immediate

[uneven relic]

Can i say that since we know atleast 1 angle is a right angle the rest are also right because the sum is 360

[frozen talon]

no, you also need to use the fact that P=ABCD is a parallelogram

[uneven relic]

🫩

[frozen talon]

you can make a quadrilateral with a single right angle that is not a rectangle

but if you ask that quadrilateral be a parallelogram, then it is also a rectangle

[uneven relic]

Huh

[frozen talon]

what's your definition of a parallelogram?

[uneven relic]

No but what did u mean by if u ask that """

[frozen talon]

I mean, if you subject a quadrilateral with a 90 degree angle to the additional constraint that the quadrilateral is also a parallelogram, then you can deduce that it is a rectangle

"ask that X has P(X)" means "if we (also) assume that X has property P"

sorry for the confusion

[uneven relic]

Im so lost rn

[frozen talon]

this is not sufficient, because this does not use the fact that your given object is a parallelogram

it is not sufficient because there is a quadrilateral with a 90 degree angle that is not a parallelogram nor a rectangle

[uneven relic]

Ahh

[frozen talon]

so somewhere in your proof, you need to include a fact about parallelograms

[uneven relic]

So ehat would i say for i

[frozen talon]

that's up to you

I can't tell you what to write, it's not my homework, and I'd be stealing the experience from you

my french is also terrible, I'm thankful the sentence structure is similar enough

[uneven relic]

Okok

[midnight plank]

@uneven relic Has your question been resolved?

[midnight plank]

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[midnight plank]

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[frozen talon]

. @cedar pawn in case you don't know, preface a message with a . to avoid provoking the bot

[cedar pawn]

Okay. The issue here was that some previous messages hadn't loaded yet

So I didn't realise there was a huge gap

But thanks

[midnight plank]

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[turbid karma]

hello can anyone help me find the period lol? i got 3 but i don’t think its right

[bold peak]

What is the period of a function

[burnt wraith]

Functions on LHS and RHS are different

which one are you asking for?

[turbid karma]

the right

[burnt wraith]

find period

[turbid karma]

2pi/3

[burnt wraith]

yes, how did you get that?

[turbid karma]

from looking at the graph

but when i do the period formula i get 3

[burnt wraith]

Can you tell us what you did?

About the formula approach

[turbid karma]

2pi/(2pi/3)

that’s what the examples say to do

[bold peak]

What is this period formula you speak of

Could you show an example perhaps

[turbid karma]

yes

[burnt wraith]

Have you gottten the funciton of this graph?

[turbid karma]

i have the amplitude and the midline

what i’m struggling with is the B value

[bold peak]

If you don't have the equation then you can't really use the formula

[burnt wraith]

B value can be acquired by knowing the period

[bold peak]

So answer my initial question

What exactly is the period

[burnt wraith]

but what you're literally looking for is the period

[bold peak]

What does it show

[turbid karma]

yes so the period is 2pi/3

but what i’m getting for the B value is 3

[burnt wraith]

Where did u get that?

[turbid karma]

and when i graph it on desmos im getting a different graph than the question

[turbid karma]

2pi/(2pi/3)

[burnt wraith]

The function can be shifted, the one in your question has been either left shifted or right shifted, depending the the magnitude

[turbid karma]

okay thanks

.close

[midnight plank]

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[pearl hull]

.reopen

[midnight plank]

✅

[pearl hull]

@turbid karma wait, lemme show you the function

[bold peak]

Huh

Lmao

I'm picturing you frantically trying to switch accounts to reopen the channel before the bot banishes it to hidden

[burnt wraith]

LOL

[burnt wraith]

@turbid karma If you're curious, this is what the function supposed to be after you shift it

[burnt wraith]

bro is going to modmail me 😭

[pearl hull]

.solved

[midnight plank]

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[burnt wraith]

Have a good one

[midnight plank]

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[lapis falcon]

Heya, I am trying to figure out an equation and solve for X. The information I have is that the total value is 100,000. But to get 100,000 I need to buy X amount of units. Each unit costs 1,750. But every time I buy a unit, the price of a unit increases by one. Like 1 unit is 1750, but two units is 1,751 and so on. Everytime I buy a unit the total value gained is 13.

[bold peak]

Well what have you tried so far

[keen widget]

i think we can use AP in this

summation of AP

[bold peak]

That is what we have to use yes

But I'd like to first know what OP has tried already

[keen widget]

OP?

[bold peak]

The person asking the question

[keen widget]

oh

ok

[lapis falcon]

I thought it might’ve been exponential

But I don’t know how to change e like value if it.

Because it won’t be a sharp slope for mx+b so there’s gotta be an exponent but I’m stuck in the exponent that increases every value if x

[bold peak]

Could you show us the original question

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[keen widget]

no i dont think it would be an exponential fn

u should use AP

[lapis falcon]

It’s not an original question, just a real world problem I’ve come up with

[bold peak]

Well it isn't very clear here, cuz you want to generate 100,000 value. And every purchase gives you 13 value

So that gives you x directly

And then if you want to calculate the total price, that's just the sum of an AP

[lapis falcon]

But every unit that gives me 13 value, I have to give 1,750 and every unit I buy of that 1,750 unit it increases by one. So like 1,750 X 1=13, then 1,751 x 2 =26 and so on until I get to 100,000

[midnight plank]

@lapis falcon Has your question been resolved?

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[twilit field]

$\alpha( \alpha(x_1)) = \alpha (\alpha(x_2)) \implies x_1=x_2$. It is this one-one.

[grand pond]

wai

[spiral rock]

its just that alpha is its own inverse

[brisk iris]

Can’t you say alpha is a bijection because there exists alpha^-1 which is alpha ?

[twilit field]

hmm, does that work?

[frozen talon]

as long as invertible functions are also injective and surjective, yeah

[twilit field]

One more question

Can I have a hint

Like I suppose I have to break it down

$\alpha^{5} = \alpha^{-1}$;
$\beta^4 = \beta^{-1}$

I suppose I can use this

[grand pond]

wai

[spiral rock]

You can use that sign is a homomorphism

[twilit field]

haven;t done that

$\alpha^{-1}\beta^{-1} \alpha^{-1} \beta^{-2} \alpha^{-1}$

[grand pond]

wai

[spiral rock]

The inverse of a k-cycle is a k-cycle

So it has the same parity

[twilit field]

yea

but I don't know if \alpha and \beta are disjoint

[spiral rock]

it doesn't matter!

[twilit field]

why not

[spiral rock]

just write each of them as a product of cycles

And the parity of the amount of cycles is the overall parity

[spiral rock]

We are not talking about order

[twilit field]

right $\alpha$ can be written as the product of 5 transpositions

[grand pond]

wai

[twilit field]

and \beta 4

[spiral rock]

so alpha is odd, beta is even

[twilit field]

yes

so it's even

[spiral rock]

Overall its odd

[twilit field]

how

(\alpha ^{-1} \beta^{-1}). is of even order

ooh

I get it now

[spiral rock]

Thats why I said to use that sign is a homomorphism.

It makes this very easy

[twilit field]

We haven't done homomorphims in class, so I can't use them😭

[spiral rock]

But you know the concept right?

[twilit field]

More or less

it's very similar to an isomorphism

[spiral rock]

homomorphism is an isomoprhism but not bijective

we usually start with introducing homomorphisms and then isomorphisms

[twilit field]

I did isomorphims in LA

that's how I know thjem

[spiral rock]

in LA?

[twilit field]

LIn ALg

[spiral rock]

yeah I know what you mean

but isomorphism are boring in linear algebra

since any two vector spaces of the same dimension are isomorphic

in groups they are a lot more interesting

wanna hear something cool?

[twilit field]

sure

[spiral rock]

you know An?

[twilit field]

The alternating group, yea

[spiral rock]

its a normal subgroup of Sn since its the kernel of the sign homomorphism. Now, for n>=5 its the only non-trivial proper normal subgroup of Sn

Its pretty annoying to prove this

do you have any more questions?

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

Not really

Thanks

[midnight plank]

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[slate ferry]

Its asking for the current (I) across the "45" resistance

[slate ferry]

what I tried was the following;

I_45 : 45 = I_t : R_t

current across 45 / 45 (the resistance) = total current / total resistance

[slate ferry]

reducing whatever the fuck this is, R_t = 45

I_45 : 45 = I_t : 45

I across 45 = Total I

total I = 180/45

I_45=4

nope. its wrong

what went wrong?

Got it! ty for possibly looking and attempting to help 🙂

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[proud abyss]

can someone please tell me what is wrong with this? I am trying to multiply 11 and 13 using add-and-shift algorithm, but I am not getting the correct result

[lyric charm]

do you have a precise and exact writeup of the add-and-shift algo

idk why my question was so pleonastic but whatever

[proud abyss]

it is not given in the lecture material, but let me see if I can find it online

not sure if it counts, but it is the best thing i could find

basically, the idea is that we check q0, and if:

1. it is 1:
   we add M to AC and then shift C AC Q to right (by 1 bit)
2. it is 0:
   we simply shift C AC Q to right (by 1 bit)

[proud abyss]

also, it is Q here and not A

ah

figured it out

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[atomic notch]

hm?

[midnight plank]

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[sturdy junco]

misclicked sorry

[atomic notch]

sorry im new

[sturdy junco]

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[green dune]

Need help with 11

[green dune]

i tried doing the f(2+h) - f(2) / (2+h) - 2 thing for (a), i think im misleading myself tho

i know i need to find a triangle

i think

representing the difference between two points

2 and 2+h

[rare maple]

Have you learnt other ways of finding the derivative

Apart from this

[green dune]

i know the derivative rule, but it was from outside sources and not in class

a^3 = 3a^2

[rare maple]

Yes this will be useful here

I am assuming that we can use this here

[green dune]

the slope in other words

is kinda like the secant?

right

just needs to intersect the graph at two points

[rare maple]

It is the tangent

Slope that we measure

[rare maple]

Slope about a point of the graph is the slope of the tangent about that point

[green dune]

so then how do i find the slope

[rare maple]

To find the slope we first need the derivative

But we will also require the chain rule for this

[green dune]

whats that

[rare maple]

It is basically a way of finding derivatives of composite functions

Let y= f(g(x)) now
y' = f'(g(x)) * g'(x)

[green dune]

ok

i dont understand why exactly we're doing that tho

[rare maple]

Are you aware of the notation d/dx

[green dune]

a little

[rare maple]

Nevermind that isn't really a good method

[rare maple]

For now we just assume this is true

It might not be the best explanation so i would highly recommend looking it up online

[green dune]

aah alright

[rare maple]

After you have done the chain rule and power rule the question will be quite easy

[midnight plank]

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[sage zodiac]

I just need a key

[modern sapphire]

63+27 = 17+73 = 90

[lyric charm]

i have a feeling i have seen this exact question before

[gleaming hearth]

Use sin(90-Q)=cosQ
cos(90-Q)=sinQ
Sin²Q+cos²Q=1

[sage zodiac]

Yeah!! That's why I said I lost the key

[lyric charm]

you should recall the identities that jay just conveniently posted for you.

[sage zodiac]

That doesn't work @lyric charm had told me

[sage zodiac]

Yeah! I know the identities

[lyric charm]

ok so apply them

[sage zodiac]

Let Me do it

[lyric charm]

i also think that you may have misremembered me, or misunderstood me, or read me only partially, or all 3.

[modern sapphire]

wdym doesnt work? I am not saying numerator and denominator are 90. I am saying the angles are complementary, and hinting towards what jayy explicitly stated

[sage zodiac]

Ohh

[sage zodiac]

I tried but the irregular number of angles bothering me

[modern sapphire]

irregularity of the numbers is irrelevant here

[lyric charm]

ok lets start with just one application

and hope that you dont get an allergic reaction to it

focus on \textbf{only} the term $\sin^2(27\dg)$.

[grand pond]

Ann

[lyric charm]

i claim you can apply one of the co-function identities to it.

A) sin(x) = cos(90-x)
B) cos(x) = sin(90-x)

work out which one, then apply it and say what you get. ONLY this term, i repeat ONLY THIS TERM and don't go back to the full expression yet. only this term.

instructions clear? @sage zodiac

[sage zodiac]

Umm!! I'll try

[lyric charm]

this should not take you a lot of time.

[gleaming hearth]

The answer is 1 found it yet?

[sage zodiac]

I already know the answer

[gleaming hearth]

Can you do it now?

[sage zodiac]

Yeah!

But can I solve this by just using sin²A + Cos²A=1

And Sec²A-tan²A=1
Cosec²A-Cot²A=1

@gleaming hearth using the above identities?

[gleaming hearth]

Yesh

[sage zodiac]

How?

[gleaming hearth]

@sage zodiac use sin(90-x)=cosx
And cos(90-x)=sinx
First

[sage zodiac]

I get it

But I didn't want to use sin(90-x)=cosx identity becuz this topic is cutted out @gleaming hearth

That's why I was stuck

[gleaming hearth]

Not allowed to use sin(90-x)=cosx?

[sage zodiac]

Lol I don't know but is there a way to solve this without those identity?

[gleaming hearth]

I don't think so

[sage zodiac]

Hm me too

How many identities are there?

[lyric charm]

wait what

cut out?

as in your teacher specifically said it's haram?

or just never taught and somehow absent from syllabus?

[sage zodiac]

[sage zodiac]

That topic is cutted out

[sage zodiac]

@lyric charm tell me

[gleaming hearth]

This is all I have

π=180⁰ btw if you wonder

[sage zodiac]

cos(x) = sin(90-x) @gleaming hearth these type of

[sage zodiac]

Thanks for your hardwork I will save it

[gleaming hearth]

I sent it

[lyric charm]

trig identities are an open class

so there's no way to give you an exact count

especially because there exist plenty of identities that are learned in some regions/curricula and not others

[sage zodiac]

Oh ohk

[lyric charm]

i dont think anybody gives a shit about the identity for cos(5x) even though it definitely is out there

[sage zodiac]

Hm

Thanks guys

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[rare maple]

Consider a positive integer $a > 1$. If $a$ is not a perfect square then at the next move we add $3$ to it and if it is a perfect square we take the square root of it. Define the trajectory of a number $a$ as the set obtained by performing this operation on $a$. For example the cardinality of $3$ is ${3, 6, 9}$.
Find all $n$ such that the cardinality of $n$ is finite.

[grand pond]

Triaengle

[rare maple]

I am unable to prove that n=1mod3 do not work

[midnight plank]

@rare maple Has your question been resolved?

[edgy pond]

n=1mod3 does work as then n^2=1mod3 too

so there exists a way to continue to add 3 into n so that n becomes n^2, and the cardinality becomes finite

[rare maple]

With the examples I have tried it doesn't seem possible

I have tried a lot of values and the pattern I am observing is that it of of the square roots is always =-1mod3

[edgy pond]

hmm

[rare maple]

What I was thinking of trying was that the assuming we encounter no number n = (3k+2)²

The function is decreasing

And it will at some point go below 16

From where I can prove that it doesn't work

[midnight plank]

@rare maple Has your question been resolved?

[edgy pond]

I think we can definitely prove that with any n=(3k+2)^2 in the set the cardinality would be infinite

as k^2=1 or 0 mod3 only

[rare maple]

Yes

[edgy pond]

We can use induction to prove that if the chain start with 3n+1 wil result in (3k+2)^2

then (3(n+1)+1) will result in a (3k+2)^2 to

[rare maple]

How do I do that

[midnight plank]

@rare maple Has your question been resolved?

[midnight plank]

@rare maple Has your question been resolved?

[rare maple]

<@&286206848099549185>

[rough fox]

every number 0 or 1 mod 3 fulfills finite cardinality

[rare maple]

Pretty sure not every 1 mod 3 does

[rough fox]

hmm

you're right let me think about it more

[rare maple]

Infact I am inclined to think that none of the 1mod3 work

[rough fox]

you are right

it's just all the numbers 0 mod 3 then

[rare maple]

I am unable to prove the 1mod3 part

[rough fox]

"observation"

well

let's start with the fact that 4 and 16 are infinite cardinality

Since the only 1 mod 3 square number with finite cardinality from 1 to 16 is 1

That means every 1 mod 3 number up till 16^2, excluding 1, is infinite as well

And since the only finite cardinality square number from 1 to 16^2 is 1

that means every 1 mod 3 number up till 16^4 is also infinite

that's the idea at least

can write it formally but too lazy

Do you follow?

[rare maple]

how do we conclude that

[rough fox]

Let's say you have any number 16 < n=3k+1 < 16^2

The procedure is to add 3 to it until it is a square number

but this square number is less than or equal to 16^2 (It is guaranteed to hit 16^2 or square number below)

[rare maple]

now the number enters the interval

1,16

[rough fox]

yes

and that is infinite

so the range is extended to 16^2

and then can induct

[rare maple]

thanks for the help ]

[rough fox]

[rare maple]

have a nice day

.solved

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[lyric charm]

<@&268886789983436800> scam

[jaunty ivy]

yay casinos

[midnight plank]

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[tidal ice]

Are these right? Specifically the restrictions.

[tidal ice]

For the first graph I have my domain from (-inf,inf) and range (-inf, 7) and for the second graph (-inf,6] and range (-inf, 9)

[ashen jasper]

why does the domain of the second function stop at 6?

[tidal ice]

The closed circle at 6,9

It doesn’t continue after the fact no?

If there was an arrow it would be infinite

But it doesn’t have that

[ashen jasper]

if that's the case, your piecewise function should probably state the upper bound of the last definition.

[tidal ice]

Sorry can you dumb that down for me

I kind of notice the last restriction is x greater than/equal to 3

So the domain would be negative inf to inf right?

[ashen jasper]

if you intend for the last definition to be applicable only up to x = 9, then the upper bound of the last definition should indicate that (right here).

[tidal ice]

The stuff on the sharpie is what the teacher wrote so now I know it won’t be applicable to just x = 9 but beyond that

So it would be (-inf, inf) for the domain right?

[ashen jasper]

that depends. are you defining the function based on the graph, or are you drawing the graph based on the definition?

which one is given first - the graph, or the definition?

[tidal ice]

The graph

[ashen jasper]

then that means the graph takes priority, and you should respect the domain set by the graph.

[tidal ice]

So would the domain be from negative infinity to infinity

[ashen jasper]

I don't see anything on the graph to the right of x = 9.

[tidal ice]

so negative infinity to 9?

[ashen jasper]

yes, looks like it.

[tidal ice]

Thank you

Is the range good

I used a (-inf,9)

[ashen jasper]

that's not a square bracket for the upper bound, right?

[tidal ice]

Correct. @ashen jasper was

[ashen jasper]

in that case, I don't see anything wrong with the range.

[tidal ice]

Thank you.

Have a good day.

[ashen jasper]

hang on.

[tidal ice]

..?

[ashen jasper]

your first question's domain looks to be incorrect.

[tidal ice]

Talk to me

What’s wrong?

[ashen jasper]

look at the definitions of f.

[tidal ice]

Ah I see my mistake

[ashen jasper]

pay attention to the first one, and especially notice that it has a lower bound.

[tidal ice]

Domain: [-8, inf)

[ashen jasper]

yeah. range looks good.

[tidal ice]

Thank you boss take care

[ashen jasper]

likewise.

[atomic hinge]

!done?

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[tidal ice]

Yes sir.

.close

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[grim coyote]

Does anyone have a video or a text about 2d shapes that look like 3d when viewed from a specific point in a 3d system?

[midnight plank]

@grim coyote Has your question been resolved?

[grim coyote]

like this

[cedar mason]

uh

so

projections?

[midnight plank]

@grim coyote Has your question been resolved?

[grim coyote]

yeah i think so

But of a 2d figure that looks 3d, optical illusion

[median apex]

anyone here play kirka./

[atomic hinge]

!occupied

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@grim coyote Has your question been resolved?

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[hoary hamlet]

in the last line, y is substituted into y - x. since y is maximixed when x - q = 1, shouldn't y = p + pq be subbed in, rather than only pq being subbed in, as shown in the solution?

[small jasper]

You suggest y=p(q+1), but q+1 is already x

I’m not sure where you’re seeing pq being substituted in

[hoary hamlet]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

i didnt see that

i thought i saw pq/1 rather than px/1

thank you

😭

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[distant gulch]

what exactly is math?

Cause there are so many concepts, but what's their unifying definition?

[fallow scarab]

https://en.wikipedia.org/wiki/Mathematics

[distant gulch]

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[balmy quartz]

how do you do this fast without a calculator

should take 1.5 min

[mortal mirage]

well first, which two can you eliminate immediately based on tom's conditions?

[slender walrus]

know your multiples of 3

[balmy quartz]

first and last

but then u still have to multiply the price of each one by 3

[mortal mirage]

not just 3, i think

[balmy quartz]

and then find out if he has enough for the supplement

[mortal mirage]

its per person per day

so you multiply by...?

[balmy quartz]

what is

the cost per person?

[lost sphinx]

cost per person doesnt depend on days

[mortal mirage]

the cost for one person for one day is, say for the second one, 10 dollars

[lost sphinx]

i suppose you can eliminate B in the next 5 seconds when you see 1200x3 is already 3600, he cant pay for his supplements

yeah its not just 3 but its mainly 3

[mortal mirage]

so the cost for 1 person for 7 days is?

[lost sphinx]

its cost for tom, not cost per person

his family doesnt need to pay for this

[mortal mirage]

ah icic

[balmy quartz]

.close

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[hard shard]

left ghg^-1 be k in H

[twilit field]

oh

then $gh=kg$

[grand pond]

wai

[twilit field]

I think double inclusion might be easier here

Let $x \in gH$. Then $\exists h_1 \in H : x=gh$. As $ghg^{-1} \in H$, it follows $xg^{-1} = h'$ for some $h' \in H$. So $x=h'g$. So $gH \subseteq Hg$. Similarly, $Hg \subseteq gH$. Thus we have $gH=Hg$

[grand pond]

wai

[dreamy lichen]

looks pretty good, just watch your indices

you write h1 and then proceed to write plain h everywhere else

[twilit field]

Got it.

Thanks!

Let $x \in g( H \cap K)$. Then $\exists h \in H \cap K: x= gh$. $h \in H \cap K \implies h \in H \land h \in K$. So $gh \in (gH \land gK)$. So $g(H \cap K$. Thus $g(H \cap K) \subseteq gH \cap gK$.
\
Let $y \in gH \cap gK.$ Then $ \exists a \in H : y=ga$. And $y \in gK \implies \exists b \in H : y=gb$. Then $yg^{-1}=a; yg^{-1}=b \implies a=b$.
\
Then $(a=b) \in H \cap K$. So $(ga=gb=y) \in g(H\cap K)$.So $gH \cap gK \subseteq g(H \cap K)$.
\
Thus $gH \cap gK = g(H \cap K)$

[grand pond]

wai

[twilit field]

Is the formatting and eveyrthing fine?

[spiral rock]

can be improved

you say $gh \in (gH \land gK)$ which should be $gh \in gH \land gh\in gK$

[grand pond]

ExpertEsquieESQUIE

[spiral rock]

Also some missing parentheses

[atomic hinge]

This first one would be fine if it was $\cap$, since you can’t and sets

[grand pond]

BBMaths

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

.close

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[twilit field]

Thanks everyone!

[midnight plank]

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[fossil creek]

540

[burnt wraith]

bruh

[pearl hull]

.solved

[midnight plank]

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[sharp coral]

i don't see where it says you can assume dim W > dim V

[midnight plank]

@digital basalt Has your question been resolved?

[midnight plank]

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glad cypress]

yo can someone help w me with this question

[tacit jungle]

yah

[glad cypress]

can we talk in dms if thats fine w u

[tacit jungle]

no

[glad cypress]

alr

well this is quite unique

i been trying for a bit

[tacit jungle]

okay find the pattern

[subtle blaze]

What do the circles mean?

[glad cypress]

pretty sure its the number of times its used in the sequence

[glad cypress]

?

if it were that simple

[subtle blaze]

Where did this come from

[glad cypress]

i wouldve done it

[glad cypress]

uh

[pastel sentinel]

i think you start with 2 and then find what numbers 2 points to that adds up to the next number if that makes sense

[bold peak]

That cannot be the case cuz you'd run out wayyyy before 1000

[wooden summit]

Yo yall i need help with understanding fonction polynomes

[scenic wyvern]

!occupied, sorry

[bold peak]

2 points to 23 but the next term is 21

[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).

[pastel sentinel]

21 - 2 is 19
2 and 19 point to each other

[glad cypress]

might be the nodes then

[wooden summit]

what channel do i use for help

[scenic wyvern]

#help-46

[bold peak]

Any of the available ones

[wooden summit]

ok

[glad cypress]

this is for something

the answer to this question anyways

[bold peak]

Hmm could be smth like you move to the one with the most circles and add that

[glad cypress]

ye

[bold peak]

Yeah that tracks actually

[glad cypress]

u can do this for 11 sequences

but the 12th one

[bold peak]

Start with 2, move to the one with most circles that 2 points to and add that

[glad cypress]

im a bit stuck on it after that

[bold peak]

So we're looking for a loop

[pastel sentinel]

qait maybe

[glad cypress]

yea

[pastel sentinel]

the next one would be 29

which would have the most amount of circles, which is the same as 2, but u js were at 2 sooo

[subtle blaze]

I mean you can follow the nodes 2 19 29 23 but then there isn’t a 48 node

[bold peak]

Problem is, this doesn't work for 29 to 23

[glad cypress]

list all the nodes first

[hoary anchor]

Did u got this from Matrix67?

[pastel sentinel]

maybe all the numbers it gives you is one full loop?
so once u know the loop u can figure out the whatever number in the sequence

[glad cypress]

it loops after the 12th im pretty sure

[subtle blaze]

It’s also not as if we’re looking at the greatest sum for n step path

[glad cypress]

not from anyone here

and if i solve it i get smth

[bold peak]

Yup, definitely not

[subtle blaze]

Where did you get it from

[bold peak]

That breaks down almost immediately

[glad cypress]

someone

[bold peak]

What do you get lol

[subtle blaze]

That’s not helpful

[glad cypress]

its not someone from around here

[glad cypress]

hmm if we were to convert to usd id say about 200-300?

[subtle blaze]

That’s not the point

[glad cypress]

but thats if u convert

im not rlly looking to do that

[subtle blaze]

Like from what source

[glad cypress]

discord? ig?

if thats your question

[subtle blaze]

Oh

[glad cypress]

idk where he got the question

[bold peak]

Yeah no someone isn't just handing out money for a problem that's easy to solve

Or even is reasonable

[subtle blaze]

So what do the circles mean

[glad cypress]

i got very close

and im ass at math

[glad cypress]

i think they're the nodes

i said this

[subtle blaze]

It was a worthy try

[glad cypress]

not money but an item worth around that much

[glad cypress]

the reason it doesnt work is cause

u cant get the 12th term

ill do the addition

wait

[glad cypress]

Observe it well and list the nodes

wait ill send

19+29=48
48+23=71
71+48=119
119+10=129
129+44=173
173+17=190
190+5=195
195+57=252
252+73=325
325+17=342
342+2=344 12th term
344+19=363
363+29=392
392+23=415
415+48=463 16th term

@subtle blaze @bold peak

and what i been trying is to

multiply the 12th term with an amount of cycles (which i cannot figure out)

[glad cypress]

maybe check the terms i gave and compare

cause ik this part of my working is right

[glad cypress]

from what i know this is something called a prime node transversal or smt

hold on i mightve solved it

[pastel sentinel]

ok so if you added all of the differences of the first 12 terms together 1000 times, i think ud get 344000

possibly

mayhaps

[glad cypress]

i just got 31200 smth

[pastel sentinel]

😭😭

[glad cypress]

youre rushing it

i rushed it too and got 18000 smth

💀

[pastel sentinel]

well is that the answr?? its under 30k

[glad cypress]

no

that was wrong too

[pastel sentinel]

awe

[glad cypress]

im assuming its between 20k to 30k

lol

probably on the higher end

so 25k+

but below 30k

but estimating isnt gonna get me the answer is it 😭:

[pastel sentinel]

guess every number and eventually ull get it right /j

[glad cypress]

"is it 25001"

"what about 25002"

LMAOO

[pastel sentinel]

😭😭😭😭

[glad cypress]

I dont think its hard

because

u can do it if youve done algebra 2

💀

thats what he told me anyways

[pastel sentinel]

i didnt understand much in my alg 2 class 😞😞

[glad cypress]

relearn i need an answe

💀

[pastel sentinel]

💀💀

[glad cypress]

shit atp ill find some math professor on discord

i want my items 💀

[pastel sentinel]

lmaoo good luck

[glad cypress]

have u seen one full cycle

from the start node

the arrows direct u

[glad cypress]

then u get this

ive even derived a formula

[pastel sentinel]

yeah but like
the numbers u gave are diff from the nodes

[glad cypress]

which seems to fail

[pastel sentinel]

or wtvr

[glad cypress]

u gotta add it

[glad cypress]

ive been given hints

and im still failing

😭

i thought someone here could do this

cause

i think it takes 10 mins

[pastel sentinel]

bro needs to email a math professor

[glad cypress]

No i just need to find someone who can do prime node transversals

whoever made that name cooked

[pastel sentinel]

lmao

i would learn it cuz this problem is interesting but i have my own problems to figure out 💔💔💔

[glad cypress]

💀😭

find me youtube tututorla

tutorial

[pastel sentinel]

🫡u got this chat

[glad cypress]

what chat

they all took one look

and abandoned me

💀

[pastel sentinel]

yea cuz like
what... in the world is that

[glad cypress]

Guess what

its a prime node transversal

💀

whatever the fuck that even means

[pastel sentinel]

ywah idk bro

[glad cypress]

perhaps smth to do w prime numbers

i havent seen it integrated into any steps yet

[pastel sentinel]

probably

[glad cypress]

and idt any of these numbers are even primes

oh some

not all

[pastel sentinel]

huh

interesting

[glad cypress]

its fun but

i have a time limit

of 24 hrs