help-49

[fervent ember]

is 4 < 3?

you try to solve (x+1)/(x-2) < 3 and you get 2.5 < x as solution:

how does this match if you assume x = 3 and you get 4?

and explain how you get this:

@last slate ???

[last slate]

I splitted the inequality

[last slate]

no

[fervent ember]

so you should realize that your solution 2.5 < x cant be correct.

[fervent ember]

explain how you get this two lines out of the starting fraction.

[last slate]

I dont know why Im getting that for my answer

[last slate]

I multiplied x -2

[fervent ember]

what have you learned about multiplying an inequality?

[last slate]

if you multiply by negative you have to flip it

[fervent ember]

so, is x-2 positive or negative?

[last slate]

can it be both?

like two cases?

[fervent ember]

you have to look at two cases for each part of the inequalities.

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[wind oxide]

Two cars A and B are moving on parallel roads in the same direction. Car A moves with constant velocity 30 m/s and Car B moves with constant acceleration 2.4 m/s^2. At t=0, car A is 120m ahead of Car B and Car B is moving with velocity 12 m/s. Find the distance travelled by car B until it overtakes car A

[wind oxide]

two things that i have crossed out

i amma describe them digitally

[modern sapphire]

your amma so small she can be described digitally? /jk

[wind oxide]

1. Relative
   by ideas of relative motion, we can consider Car A having $a=-2.4$ and $v=30-12=18$ and find the time $t$ it takes to cover the $120$ metre gap

[grand pond]

rak³en

[wind oxide]

Then by $s = ut + \frac{1}{2}at^2$, we have $120 = 18t - 1.2t^2$

[grand pond]

rak³en

[wind oxide]

but the equation does not have real roots

So i thought maybe i blundered i went absolute distance approach

$x_A(t) = 30t$, $x_B(t) = 12t + 1.2t^2$, $x_B(t) + 120 = x_A(t)$

[grand pond]

rak³en

[wind oxide]

Simplifying this gives me the same exact equation

with no real roots

[slow thorn]

displacement is -120 if you say car b is fixed

[modern sapphire]

well, in frame of car b, the car a is 120 meters ahead

so s = -120m

[wind oxide]

OH

I FORGOT DISPLACEMENT IS RELATIVE TOO

THANK YOU

🫡

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[chilly cobalt]

uh.. how do i solve this?

[chilly cobalt]

heres what ive done

what do i do next? i lowkey forgor

[late violet]

Why not 2x - 2x = 0?

[chilly cobalt]

huh?

its dx and dy

[chilly cobalt]

it should be dy = vdx + xdv btw

[polar star]

combine denominators with v

cancel terms and then separate variables

[chilly cobalt]

is this correct?

[violet rivet]

shouldn't the first line be 2xy+x² there in the denominator?

[chilly cobalt]

wait

noooo

you're right😭

thanks, ill go do this now

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[silent torrent]

can someone help me ques 10

[silent torrent]

i think we just need to calculate the volume between green and red circal

i did a exercise in the past with the same 2 circals like that and i used polar form

i need your help

i just let f(red) = f( green) ( bc both of them =1) and i found x=y is there anything more

[wraith bay]

Do you have any idea how to find the volume between them?

[silent torrent]

use the polar form

i think so

so we hhave 1/r^3 dr dtheta

theta from 0 to pi/4 r from 0 to 2 costheta

is that right

[midnight plank]

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[midnight plank]

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[lavish venture]

identify the corresponding power series

[atomic brook]

i tried using sinx power series, but idk what to do now

i tried differentiabting it but i got (2n+1) instead of n

[nova yoke]

maybe try something like $\sum(2n+1) = 2\sum n + \sum 1$

[grand pond]

Bungo

[atomic brook]

will try thanks

thank you that worked.

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[sour sierra]

Uh I am stuck at drawing the golden theorem lmao

so ig I come to the math server for some umm help 🤓

[fallow scarab]

wut

[grim vector]

You want to make a spiral?

[sour sierra]

yeah

Those are all squares

[grim vector]

Get a compas

[sour sierra]

But the point that intersect the two line is uh

[grim vector]

Maybe verify the mesure of theses little squares

[sour sierra]

I just round off the values lmao

Let me just redo it with more precious measurements

[grim vector]

Then its normal to not have the same thing cuz rounding over rounding makes it changing too much sometimes

[sour sierra]

Ohh silly me

Let me try again

[sour sierra]

Thanks you

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[grim vector]

Gut

[sour sierra]

OPPs

Yes

.close

Sorry mod

I broke the channel

🙏

[grim vector]

Its already close

It just take a little time to reset

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[quasi thicket]

Let A, B, C, and D be four distinct points in the plane.
Which of the following statements, expressed using oriented angles, are always true?
Tick each correct option.

-If lines AB and CD are distinct and parallel, then
the oriented angle ABC is equal to the oriented angle DCB.

-If B lies on the segment [AC], then
the oriented angle DBA plus the oriented angle DBC equals 180°.

-If the oriented angle ABC plus the oriented angle BCD equals 0°, then
lines AB and CD are parallel.

-If the oriented angle ABC plus the oriented angle BCD equals 180°, then
lines AB and CD are parallel.

[midnight plank]

Please don't occupy multiple help channels.

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[sour sierra]

I think I need some help

[sour sierra]

.rotate

[pearl hull]

,rotate

[sour sierra]

I am not sure if this will work

I wonder if there is easier way

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[dreamy wolf]

1 is neither prime nor composite

[dreamy wolf]

right?

0 is even but 0 is neither positive nor negative

[inner gazelle]

yep

[dreamy wolf]

can you help me understand these

helpers

like we need to remember it or

[inner gazelle]

what dont you understand?

can you be more clear?

[dreamy wolf]

no no

like see

when we say a is divisible by b

when a\b = x where x is an integer right

?

[inner gazelle]

yes, that is true

[dreamy wolf]

ok then

[inner gazelle]

if you have 6/2 that gives you 3

then 6 is divisible by 2

[dreamy wolf]

0 is divisible by 2 means 0\2=0 wihich is an integere

acc to defination of even number i.e all integers which are divisible by 2 are even

this stisfies it

0 is divisible by 2 means 0\2=0 wihich is an integere
acc to defination of even number i.e all integers which are divisible by 2 are even
this stisfies it

[inner gazelle]

all numbers ending in 0 2 4 6 and 8 are even, yes

4 is even, 7838 is even, 678903 is not, 0 is too

[dreamy wolf]

but 0 does not count in odd numbers because odd numbers are those integers which are not divisible by 2 although 0 is divisible by 3 and all other numbers we dont count it here is bcuz it is also divisible by 2

[inner gazelle]

if you divide 0/2 you can make 0 pairs, without no element left

[dreamy wolf]

?

[dreamy wolf]

0 is too?

what too

[inner gazelle]

0 is even, 0 is even too

look

[dreamy wolf]

this understanding correct?

my teacher told this'

[inner gazelle]

yes, its right

[dreamy wolf]

ohh ok

btw

[inner gazelle]

if you divide 6 by 2 how many pairs can you make?

[dreamy wolf]

3 pairs

[inner gazelle]

is there any element left?

[dreamy wolf]

btw what you mean by pair here

[dreamy wolf]

reminder is 0

[inner gazelle]

a couple, a pair, two things together

what about 5, how many pairs can you make with 5 objects?

[dreamy wolf]

i know what is pair but 6\2=3(quotient) and 0 is reminder right why consider number "3" as 3 pairs

[inner gazelle]

its just an intuitive way of thinking

[dreamy wolf]

20?

[dreamy wolf]

can u help me with it

[inner gazelle]

you can make 20 pairs with 5 objects? but you only have 5

[dreamy wolf]

i though it like this may be wrong sorry :1,2,3,4,5 = (1,2),(1,3)(1,4)(1,5)(2,1)(2,3)(2,4)(2,5)

like that

[inner gazelle]

you're asking about an elementary concept, calm down ok, dont add more work

[dreamy wolf]

ohh

umm

sorry

yeah 2 pairs i didnt think that way

[inner gazelle]

there is an element left - 5 is not an even number

when you make pairs (divide by two) with even numbers you get no elements left (no reminder)

if you have 0 objects and try making pairs, will any element last?

@dreamy wolf

[midnight plank]

@dreamy wolf Has your question been resolved?

[dreamy wolf]

@inner gazelle

sorry i had powerc ut

[inner gazelle]

can you answer that question then?

[dreamy wolf]

wait a min

[dreamy wolf]

if we dont have any objects how can we make any pairs

we cant make nay

any*

[inner gazelle]

will any elements last tho?

[dreamy wolf]

nope

[inner gazelle]

so its even

[dreamy wolf]

nothing will last since we cant even make pairds of nothing

ohh nice

thnx makx cat

[inner gazelle]

ure welcome

remember to close the channel

[dreamy wolf]

hey i will close this

[dreamy wolf]

but i have dmed you

.close

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[regal barn]

2^m-2^n=56

[regal barn]

how can I solve it?

[empty cove]

are m and n natural numbers?

[regal barn]

yeah

[empty cove]

i suggest trying some small values of n

that will give you some intuition to what is going on

[regal barn]

2^6=64

so i will take 2^6-2^3

[empty cove]

yeah that's a solution

but im assuming you have to prove this is the only solution?

[regal barn]

ohh

there can be many?

how can I prove?

[empty cove]

this was a question

are you supposed to prove this?

[regal barn]

no

[empty cove]

in what context did this problem show up?

[regal barn]

only solving

[empty cove]

well then you're done

[regal barn]

i see

tq very much

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[lone hound]

.close

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[silent torrent]

can some one tell me what ques 3 is asking for , and guide me ques 5

[silent torrent]

for ques 3 i dont understand the question

ques 5 i divide into 2 cases

case 1 first time probal black 4/10 the lost 1 ball then second time Probal brown 6/9

case 2 first time probal brown 6/10 the lost 1 ball then second time Probal brown 8/12

so i add 4/10 * 6/9 + 6/10* 8/12 so i get the answer right

[sly burrow]

3rd question is asking that if there are 15 bulbs to be installed, and the ordering doesn't matter, the only thing that matters is number of yellow bulbs, red bulbs and blue bulbs. How many possible designs are there?
One of the possible design is: 5 yellow bulbs, 5 red bulbs and 5 blue bulbs.

[silent torrent]

can u check my work of ques 5

[silent torrent]

with ur idea ive solved ques 3 the answer is 14c2

[sly burrow]

yes, I checked it, it's correct.

[silent torrent]

lovely thank you

[sly burrow]

not exactly, here you're assuming that each color should appear at least one but one of the possible designs is: 10 yellow bulbs, 0 red bulbs and 5 blue bulbs.

[silent torrent]

so the answer is 14c2 +14c1 +3

14c2 3 colours 14c1 2 colours 3 1 coulor

[sly burrow]

still not correct 😅 , 14c1 should be multplied by 3c2.

[silent torrent]

why =((

[sly burrow]

but there's an easier and general way, imagine if there had been 5 colors then you had to consider 5 cases.

[sly burrow]

because you also have to choose which two of these colors are appearing

[silent torrent]

ahh i see

cảm ơn

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[warm island]

I took the dervative and got that

Qx = b

i then said for the FONC to be satisfied there can be no directions which changes this.

So i then went Q(x + alpha d) = b implies Qx + alphaQd = b only when alphaQd is zero. There cant be any directions that violate the FONC.

Is this a vald approach?

[lyric charm]

what does FONC stand for?

[warm island]

first order nessecary condition

[lyric charm]

i am not sure i can fully grasp the logical structure of your argument here cause it is a little bit obscured

but presumably you would need to prove FONC=>min and also min=>FONC with clear separation between the two subproofs

[warm island]

okay thank you.

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[final shard]

prove that <HKE = <MKE

[unique juniper]

are there any relationships or something? like written down?

[final shard]

thanks for decided to help me with this messy diagram , but i just did that some mins ago so , anyways thanks

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[unique juniper]

happy mathing then

[final shard]

if you like proving geometry like this just inbox me and ill give you tons of them :)

[unique juniper]

i'm already exhausted with my national olympiads no more 😭

i'm already ignoring a lot of problems because i have stuff to do, don't wanna add more problems on my list

[final shard]

is that diagram i sent familiar to your olympiad?

[unique juniper]

maybe... i've forgotten a lot 😔

[final shard]

wow thats just the daily homework we have to do everyday

[unique juniper]

good luck

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[silent torrent]

can some one help me problem a

[silent torrent]

the problem is find CDF

my work, x+x^2/2 if x<0 , x-x^2 if x<1, 1 if x>1

i was thinking if x <1 it should be x+x^2/2 +x-x^2/2=2x because comunitive function so i think i will + 2 of them

[midnight plank]

@silent torrent Has your question been resolved?

[silent torrent]

help me

[humble wolf]

i suggest that you state the formula of CDF first

[midnight plank]

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[sand flume]

He says that 1+a)^2 > a. This implies that a+1 > a. But isn't it the other way around. a+1> a which is obvious so squaring it will also me it is greater than a. And for showing that the set is not empty how can he just say that the element exists. Doesn't he need to proof it somehow

[midnight plank]

@sand flume Has your question been resolved?

[midnight plank]

@sand flume Has your question been resolved?

[tribal temple]

It isn't the other way around, squaring a + 1 > a will give you (a + 1)^2 > a^2, but you don't know that a^2 > a (and if a is between 0 and 1, it isn't)
They also are not saying that "This implies that a+1 > a", they're saying that 1 + a is an upper bound for S, the set of all positive numbers x such that x^2 < a, if x is in S, then you have x < 1 + a because x^2 < a < (1 + a)^2

They also didn't just say an element exists for showing S is non empty, they gave you a particular element, that being a/(1 + a)

[sand flume]

Could you into more depth about this: x^2 < a<(1+a)^2

[sand flume]

But how did they produce it?

If that makes sense

[tribal temple]

The definition of the set S is all the [nonnegative] elements x such that x^2 < a, and you have a < (1 + a)^2 (you can e.g. expand out (1 + a)^2 = 1 + 2a + a^2, which as a is nonnegative, must be strictly greater than a), so you can chain those inequalities together for a given x in the set to give you x^2 < a < (1 + a)^2, to condense that down, x^2 < (1 + a)^2

[tribal temple]

For the thought process that got them to think about it, I wouldn't know what to suggest, but as for how they proved it was even a member, notice here: a/(1 + a) is nonnegative (a fraction of a nonnegative and strictly positive number), and (a/(1 + a))^2 <= a follows from rearranging a^2 <= a(1 + a)^2 = a^3 + 2a^2 + a (notice that a^3 + a is nonnegative also)

[sand flume]

One second let me read through this

[sand flume]

But how do we know that (a+1)^2> a

[tribal temple]

I said inside the message: "you can e.g. expand out..."

[sand flume]

Ok and then in his writing he says that since it is bigger than a. a+1 is an upper bound. Doesn't this word say that the first implies the second

[tribal temple]

It does, the implication is the fact that x^2 < a < (1 + a)^2

[sand flume]

Yes but what about the a+1

[tribal temple]

a + 1 is an upper bound of S because all elements of S are less than a + 1

[tribal temple]

That is what this shows

[sand flume]

So I understand why (a+1)^2> a and a+1>a is obvious since it is 1 greater than a. but the connection between the two I don't see. They are just two distinct upper bounds.

[sand flume]

So here your saying we start off from a^2<= a(1+a)^2 and rearranging this we get (a/(1+a))^2<a

[tribal temple]

a + 1 > a is irrelevant and isn't used anywhere

[tribal temple]

You could, sure

[tribal temple]

The main aim is to get some upper bound of a, which a < (a + 1)^2 allows you to deduce a + 1 is an upper bound for you

Then showing that the set is non empty allows you to consider the least upper bound of S

[sand flume]

Wait how can a+1 > a be irrelevant if that means a+1 is an upper blund

[runic hamlet]

a < a+1 < (a+1)^2. importantly a+1 is at least 1 so the last inequality holds. at least thats how I would argue

[tribal temple]

And, except perhaps in using it to prove said upper bound exists, you don't care that a + 1 is bigger than a

[tribal temple]

As per here, you don't need to use a < a + 1 to show the inequality

[runic hamlet]

essentially they are being smart and hiding a case distinction that way. the problem is that x<x^2 is not true if x < 1. so if a<1 then you could for example say that 1 is an upper bound for S. and if a >= 1 then you could say that a itself is an upper bound. it doesnt matter how your upper bound looks like, only that you have one. by doing it this way they combined both cases

and to show that S is nonempty, for a < 1 it holds that a is in S and for a >= 1 it holds that 1 is in S. so again thats some element in S. again it doesnt matter what element you have, only that S is nonempty

[midnight plank]

@sand flume Has your question been resolved?

[midnight plank]

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[midnight plank]

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[tiny shale]

how do i get better at math? i do so many question yet feel like im not making much progress..

[tiny shale]

a level maths

[crimson swift]

?

[polar star]

these are not really good questions

u would need to provide far more detail

[fallow scarab]

it's also not a math question

#discussion or #serious-discussion or #math-discussion is better

[tiny shale]

.close

[midnight plank]

Channel closed

Closed by @tiny shale

Use .reopen if this was a mistake.

[crimson swift]

look like someone typed you

[tiny shale]

mb boss

[crimson swift]

your profile nothing
when you message it looks like someone typed your name and message

[crimson swift]

do more be patient try understanding questions than thinking about how many questions I did and how much I did not

[midnight plank]

Available help channel!

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[last slate]

Can someone help me

[last slate]

This is what i did so far

<@&286206848099549185>

[fallow scarab]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[last slate]

Oops sorry

[slow thorn]

pretty sure there has been some error in finding f(x)

you should get a ratio of two linear polynomials

[last slate]

Oh

What ratio?

Bit confused

Can u recognise where the error is?

[slow thorn]

it's not in the work you are showing

before that

in calculating f(x)

you could try something like this

or show what you tried to reach f(x)

[last slate]

Ok ill send

[last slate]

ok now simplify 2arctan using the identiy and convert it into a normal expression right?

[slow thorn]

well yes

but you still need a round of triangle drawing

[last slate]

This is What i did.

I

Sorry Its a bit messy

[slow thorn]

hmm

i cant see the last step

[slow thorn]

no like

all the steps you showed are correct for what you got

so it seems like the conversion of cos(arctan(..)) was where something happened

but idk what you did there

[last slate]

Ok gimme a min

Lemme re work it

I forgot to take the root of the hypotenuse

Is this correct nowM

[slow thorn]

cosine is not what you got

thats the sine of it

[last slate]

Ohh

OHHH

Ohh so its optuon B right

[last slate]

.

In this

[slow thorn]

yes

[last slate]

Oki thank u very much

Have a nice day

.close

[midnight plank]

Channel closed

Closed by @sick kelp

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[twilit sentinel]

An algorithm is a way to go from input set to output set right?

[twilit sentinel]

This needs the output set be well defined?

In that case how do we talk of approximation algorithms?

Because they simply pose a bound on how much worse it can be from the optimal

[fallow scarab]

can you show the context

[twilit sentinel]

For example for vertex cover problem approximation algorithm we have a 2x bound, it can end up with a cover upto 2x larger than the smallest, but it's simply an "upto" bound, not exactly 2x, so it seems to have a ill defined output set?

[smoky ridge]

an algorithm is just a list of steps, the input set doesn't need to be the same size as the output set

[surreal moon]

Outputs (aka ranges, images, or codomains), are usually defined to be a subset of some established set. And one of the axioms of set theory is that all subsets of a known set are well-defined.

[surreal moon]

The codomain would just be the set of all possible outputs. We don't really care about the size, so long as we know this set is well-defined (which it is)

[smoky ridge]

so for f(x)=0 you get dim(x)=∞ dim(f(x))=0

[twilit sentinel]

I'm still a bit confused

I'll try to state my confusion clearly wait

So in the ordinary vertex cover problem our domain is basically the set of all undirected graphs. And our co domain is the set of natural numbers.
And because we only want the smallest vertex cover, each graph will be mapped to a particular number for sure.

But in an approximation algorithm, that assignment of a number to each graph can be random based on what algorithm we use, it's just that each graph will be assigned a number in the range of [v, 2v] where v is the size of the minimum vertex cover

[surreal moon]

There's still no harm here. It's kinda like saying "some arbitrary number". We don't need th exact deets. We just work with the info we have

[twilit sentinel]

So somehow our algorithm is determining our function and not the other way round

[surreal moon]

I mean, an algorithm really is just a function.

[twilit sentinel]

An algorithm is the steps to compute a function, usually we have a well defined function, and then we can have various algorithms for going from an input to it's output

[twilit sentinel]

Umm

Not saying there's any harm but it seems to me that

In case of approximation algorithms we have the function undefined

When usually any algorithm we work with has a well defined function

And then we try to construct a algorithm which will achieve that

Here it seems we are defining a range of functions, and creating an algorithm that will end up being any one of those functions, that range satisfying the property of some bound over optimality

[surreal moon]

okay, let's scale the problem down. What do you mean by "approximation algorithm"? It doesn't return the most optimum result, just a best try?

[twilit sentinel]

Yea but it has a bound on how bad it's results can be

So for example if we want the smallest path from point A to point B it will at worst give a path that has length 2 times the actual smallest path

[surreal moon]

Kinda, but not really.

An approximation algorithm is still a computation, and computations are always well-defined

[twilit sentinel]

Yea that's what was causing the confusion

[surreal moon]

even if the approximation uses randomization, that just means that some seed value is a part of your well-define function

The random function, is well-defined because it has a seed input, and the same seed will always give the same output.

approximation algorithms will always represent a well-defined function

[twilit sentinel]

Let's look at sorting algorithms

It's domain is all finite lists of numbers and it's range is all ordered lists

And we have various algorithms that compute the function

Where input is mapped to it's own sorted self

The function is well defined

Then we create algorithms for ir

It

But in the case of approximation algorithms, the function is not well defined

How can you define the function

[surreal moon]

how not?

[twilit sentinel]

How will you define the function

[surreal moon]

However the algorithm is defined

[twilit sentinel]

Exactly

The algorithm is well defined

But not the function it computes

[surreal moon]

if the algorithm is well-defined, then the underlying function must be well-defined

[twilit sentinel]

That's what it is like for most algorithms but it seems not true for approximation algorithms

[surreal moon]

The definition may be a lot more complex, but it is still well-defimed

[twilit sentinel]

Hmm

True actually

Yea you are right

[surreal moon]

Do you have a specific algorithm in mind? Your vertex cover approximation algorithm, which algorithm are you thinking of exactly?

[twilit sentinel]

Yea

The vertex cover algorithm

Nah you are right

We can define the function using the algorithm itself

And since the algo is well defined

So is the function

It's just that the simple nice description of the function we had for the optimal case

[surreal moon]

exactly

exactly right on all points

[twilit sentinel]

Becomes a notoriously complicated descriptions

Yea

You're right

Thx a lot

[surreal moon]

np

[twilit sentinel]

.close

[midnight plank]

Channel closed

Closed by @twilit sentinel

Use .reopen if this was a mistake.

[surreal moon]

happy to help

[midnight plank]

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>.
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• Be polite and have a nice day!

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[odd lagoon]

is that true?

[fathom onyx]

yes

[odd lagoon]

okay

[fathom onyx]

The bit in the middle is a weird way to present this

[odd lagoon]

is this the linear transformation?

[fathom onyx]

Typically I've just seen this written as (first bit) = (last bit)

[odd lagoon]

alright

[fathom onyx]

For want of a better word, yes

If your mapping is, say, T

And that matrix is, say, M

[odd lagoon]

im not sure what mapping is, is mapping just, i take an input, and output it somewhere?

[fathom onyx]

Then T(v) = Mv is a way to define the linear transformation

[odd lagoon]

and for linear transformations

a geometric way to interpret that would be

i like shift the coordinate system?

and since our vectors

[fathom onyx]

Sorta, yeah

[odd lagoon]

or here just a vector

i guess

or maybe they are both vectors, i dont know sorry

[fathom onyx]

A linear transformation is a function where linear operations on vectors are preserved

[odd lagoon]

im having a really hard time understanding how it shifts at the moment is what i wanted to get at

like

[fathom onyx]

So

If v and w are vectors

and a is a scalar

Then T is a linear transformation precisely when T(v+w) = T(v) + T(w) and whn T(av) = aT(v)

[odd lagoon]

im sorry i cannot follow that

[fathom onyx]

Conveniently that can be condensed into

[odd lagoon]

im gonna try to reread it

[fathom onyx]

T is just a function

If it satisfies T(v+w) = T(v) + T(w) and T(av) = aT(v), we call this T a linear transformation

Such a function on vectors can be made by premultiplying a vector with a matrix

Hence their association with matrices

[odd lagoon]

im sorry im not trying to be difficult but i know i usually end up being difficult, i really appreciate you trying to help

im trying to find a way to put into words how i want help here if it can at all be provided

[fathom onyx]

This actually matches the matrix multiplication you gave

[odd lagoon]

im trying to understand this geometrically because i seriously don't seem to be able to understand math in any other way

[fathom onyx]

The matrix tells you where the unit vectors (1,0) and (0,1) go

[odd lagoon]

right i recently learned of them, they are usually called i hat and j hat?

[fathom onyx]

3Blue1Brown has a good series on this

[odd lagoon]

yes i actually

watched that

[fathom onyx]

They can be, yes

[odd lagoon]

im trying to think of this the way he described but right now

im having trouble seeign what is being scaled

so

wait let me try to phrase my confusion

accurately

i read this as "im linearly transforming a 2 row 1 column vector by some funny collection of numbers (i dont understand them yet in any other way)"

then i know of the definition

of multiplying a 2x2 matrix by 2x1 matrix

i dont actually understand why i get that, but i do

[odd lagoon]

now i try to understand why i get this

so im basically scaling my x

the horisontal composant

of this vector

with, another vector?

im scaling with another vector?

[fathom onyx]

Let's start with the indentity matrix

(1 0)
(0 1)

[odd lagoon]

right

and i just want to be clear i dont really understand square matrices

i understand somehow that can transform something

or

sorry

i did learn earlier

identity matrix

doesnt really transform

it keeps the thing the same?

[fathom onyx]

By our matrix multiplication deifnition (the first thing you've posted), if we multiply this by a vector (x,y)(col), it tells us that we add x lots of (1,0)(col) with y lots of (0,1)(col)

[odd lagoon]

im effectively multiplying by 1 when im linearly transforming something with the identity matrix

?

[fathom onyx]

"(col)" here just means "written as a column vector"

yeah

[odd lagoon]

ok sry let me read this

[odd lagoon]

this

[fathom onyx]

The Identity matrix is the "multiplying by 1" of matrix multiplication

[odd lagoon]

oh so when you type (x,y)(col), you're writing

x
y

?

[fathom onyx]

yeah

[odd lagoon]

but with the

brackets/parenthesis

[fathom onyx]

it's just difficult to write that in a textbox lol

[odd lagoon]

yeah

i have

noticed and it has frustrated me

i would use latex but my silly laptop

the like slash

doesnt work

[fathom onyx]

oof

[odd lagoon]

and i cant be asked to copy and paste

that in every time

anyway

[shadow schooner]

\

[odd lagoon]

yes i cannot type that

[fathom onyx]

"it's too dangerous to go alone - take this with you" ^

[odd lagoon]

is that some kinda dark souls reference? xd

[fathom onyx]

loool

It's Legend of Zelda

It's from one of the first games

[odd lagoon]

ah alr sry havent played, havent played dark souls neither

[fathom onyx]

It's basically a staple line

[odd lagoon]

i dont understand what you mean by lots

[fathom onyx]

we compute x times the first column (which gives us one vector)

And y times the second column (giving us a second vector)

And then add the two together

[odd lagoon]

the add part i got from the definition, and ok i agree that's what we're doing by the second way you explained now

im with you so far

[fathom onyx]

For the identity matrix, this just results in having the same vector we started with

If instead we looked at
(2 0)
(0 1)
as our matrix

[odd lagoon]

yes

or wait

wait

[fathom onyx]

Then by the same token, our result would be x(2,0)(col) + y(0,1)(col)

[odd lagoon]

like x(1, 0)(col)

that's the same as x right?

[fathom onyx]

yeah. well, (x,0)(col)

[odd lagoon]

yeah ok

so we have

x + y

on this step

[fathom onyx]

We'd have (x,0)(col) + (0,y)(col)

[odd lagoon]

or is there a difference

between x and (x,0)(col)

[fathom onyx]

x and y are just ordinary numbers

we're trying to add vectors

[odd lagoon]

but isn't that vector like just smack down on the number line

[fathom onyx]

Basically coordinates except we're allowed to add them

[odd lagoon]

like a one dimensional line

(x,0)(col)

[fathom onyx]

Yh

It's just x along the "x-axis"

[odd lagoon]

so how is it different from just x, for the identity matrix i mean

[fathom onyx]

Because x is just a number

[odd lagoon]

and a number doesnt have a direction

[fathom onyx]

It's like how there's a difference between the number 5 and the vector (5,0)(col)

[odd lagoon]

im still confused but i cant sense this is not something useful to be confused about

okay so

for the identity matrix

we have got this

(x,0)(col) + (0,y)(col)

and it is different from x + y

i dont totally understand why

but i can accept it

[fathom onyx]

5 + 4 = 9

But (5, 0) (col) + (0, 4) (col) = (5,4) (col)

[odd lagoon]

okay actually

that cleared up my confusion

lol

[fathom onyx]

Ideally you should be familiar with vectors

lol

[odd lagoon]

it is the distance from origo

to a point

and i see now

the distance isnt the same as the number

[fathom onyx]

It's...

[odd lagoon]

oh no

wait no it isnt

like 9 isnt the distance

[fathom onyx]

yeeee

[odd lagoon]

between origo and the point (5, 4)(col)

[fathom onyx]

To compare (5,4)(col) with (x,y)(col), you've got x = 5 and y = 4

Don't wanna rush you @shadow schooner but this seems like a massive message lol

[shadow schooner]

Where I come from linear algebra is learned in 2 stage, first the often unrigorous version of the course playing with vectors in R^2 (plane) and R^3 (space) and an abstract second course defining everything abstractly. imo 3blue1brown is between the 2 and helps a lot to build intuition, but is not a substitute to these two course.

there you go

[fathom onyx]

oh it was a massive message

[shadow schooner]

I just want to give some context

[odd lagoon]

i have been watching 3Blue1Brown's first 2 or 3 videos on this

[fathom onyx]

question is tho is this Linear Algebra in the strict sense or is this HS maths

[odd lagoon]

and I noticed I had to go back to try to understand

what is happening

[fathom onyx]

oh wait it says you're an undergrad

[odd lagoon]

yes I am somehow undergrad

[fathom onyx]

I can go full lin alg on this shet

[odd lagoon]

bro bro dont

[shadow schooner]

lesgo

[odd lagoon]

talk to me like im 5 years old

[fathom onyx]

where's that other matrix I wrote...

[odd lagoon]

and i have not finished HS

[fathom onyx]

Ah here it is

[odd lagoon]

like i can find the determinant of a 4x4 square matrix but i dont understand this yet

someone taught me a method

it was easy to memorise

but i want to stop memorising

[fathom onyx]

Let's see what happens if we multiply this matrix to (x,y)(col)

[odd lagoon]

and understand

where this shti comes from

i am so tired

of just memorising

nonsense

i want to understand

you know the jungle book?

when that monkey sings about

wanting to be a man

i want to be able to understand this stuff

right now im just memorsing and it isnt fun

i have failed to understand the notation

now im tryign to understand geometrically

and i feel like i have a better idea of what is going on since i started doing that

[fathom onyx]

Can you see how if we multiply that matrix to that vector, how we get to (2x, y)(col)?

[odd lagoon]

i dont see it but with the definition i can get there

[fathom onyx]

By "see" I do mean by using the definition dw

[odd lagoon]

ok my goal is to see it see it

[fathom onyx]

I'm not expecting you to immediately see this lol

[odd lagoon]

and np i dont expect to either i just need help getting to the point where i start seeing it

like

this could take weeks

[shadow schooner]

yeah we have to agree on definitions then interpreting this geometrically is bonus.

[odd lagoon]

longer maybe

[fathom onyx]

They do have a definition

It's more we''re trying to make sense of it, iirc

[odd lagoon]

i had the determinant

explained to me geometrically

and that was nice

i still remember most of it

i can still see

most of it

[fathom onyx]

Let's try then the following

Let's take a generic 2x2 matrix (such as the one in the definition)

[odd lagoon]

ok

[fathom onyx]

What do I get if I muliply this to the vector (1,0)(col)?

[odd lagoon]

i get (a, c)(col)

[fathom onyx]

yeee

[odd lagoon]

and that's it i think

[fathom onyx]

Notice how that's also just the first column of the matrix?

[odd lagoon]

yes i do notice that

and i see we multiplied the other column away by 0

[fathom onyx]

yeee

So that means our î vector got moved to that vector

[odd lagoon]

i have to admit i didnt understand the basis vectors very well

[fathom onyx]

(because î by definition is (1,0)(col))

ah

[odd lagoon]

in uni i was introduced to them as the |1|

[fathom onyx]

Well there's one

[odd lagoon]

but in 3Blue

they were just 1

[fathom onyx]

Ah

[odd lagoon]

at least during the first 3 videos

[fathom onyx]

A key property of a basis vector is that it has length 1

I think that's what you're on about

[odd lagoon]

maybe, like

i get that

they dont change the vector (i think) other than direction

but in 3Blues video

i wasnt able to understand

what they were used for

they seemed important

[fathom onyx]

Ah

[odd lagoon]

he literally called them

"two very special vectors"

not just special

very special

[fathom onyx]

They technically are

I'll do the 1-D case cos it's simpler (even if a little dull)

the number 1 is our basis "vector" here

[odd lagoon]

i like 1-D

[fathom onyx]

Because every other number we can make just by it being some multiple of 1

hehehehe

e.g. 35258 is 35258 times this basis vector

[odd lagoon]

but it is so counterintuitive for me to understand

because multiplying by 1

changes nothing

[fathom onyx]

yeee

[odd lagoon]

i haven't gotten any new information

i don't understand what that means

[fathom onyx]

2-D is where it gets a little more fun

Suppose instead I want to describe the vector (14,5)

[odd lagoon]

and i agree with this

[fathom onyx]

(gonna skip using (col) all the time just for this)

[odd lagoon]

ok i will assume col

[shadow schooner]

the idea is to represent any element in R^2 as as sum of 2 vector stretched by adequate scaling factors

[fathom onyx]

I could probably do something similar - but I can't use "1" on its own because there's no vector called 1

It'd help if I had a vector of size 1 though

Well, let's see - ik (1,0) has size 1

Could I multiply this by anything to get to (14,5)...?

[odd lagoon]

no

[fathom onyx]

Right

Because I'm only going right

[odd lagoon]

yes

[fathom onyx]

There's no up-ness here

[odd lagoon]

i agree

[fathom onyx]

So while (1,0) is perhaps a useful basis vector (cos I could get the 14 bit there) I can't only use this vector

[odd lagoon]

yes

that makes sense

[fathom onyx]

Well by the same logic, (0,1) is also a vector of size 1

[odd lagoon]

we need to add a dimension

[fathom onyx]

yeeeeeee

Now - I have two vectors to play with

How many multiples of (1,0), and of (0,1), will I need for (14,5)?

[odd lagoon]

14 of i-hat and 5 of j-hat

[fathom onyx]

yep

Now it turns out that if I want to get to any vector (x,y) I just need x lots of i and y lots of j

[odd lagoon]

okay that's nice

[shadow schooner]

btw formally the dimension is defined as the number of vector you need to decompose any vector in your space R^2, R^3, R^n in a unique way.

[odd lagoon]

that's what you meant by lots

[odd lagoon]

im sorry but that is really hard for me to understand

[fathom onyx]

So these two vectors - (1,0) = i-hat and (0,1) = j-hat - form a "basis" to get to any vector

[odd lagoon]

im 5 years old

[odd lagoon]

OKAY

that's nice

now that feels

nice because now they have a reason

to be called basis vectors

[fathom onyx]

That's what we mean by basis vectors

Similarly, we say that vectors form a basis in a space if, by the method we've gone through, we can use these vectors to make any vector in that space

i.e. add a bunch of them together

[odd lagoon]

so like

when i have a 4 by 4 matrix

with a variable in there

[shadow schooner]

they allow you to write any vector in R^2 as a a * i-hat + b * j-hat in a unique way

[odd lagoon]

and i want its determinant

if the determinant isnt 0

i can make any vector

and that's why we have a bunch of solutions there

[fathom onyx]

So, a 4x4 matrix is helpful with regards to a 4-D vector

dw about actually thinking in 4 spatial dimensions

[odd lagoon]

i actually

already worried

about that

and now i can

write two planes

[fathom onyx]

Just think of it as "there's just 4 coordinates"

[odd lagoon]

and draw lines between the two planes

i find roots

[fathom onyx]

So (1,2,4,6) is a 4D vector

[odd lagoon]

in the domain

ok yes

[fathom onyx]

In the 2x2 case - we've basically got down to "the first column tells us where i-hat goes and the second column tells us where j-hat goes"

This extends upwards by dimension

[odd lagoon]

wait

let me think about that

because i didnt make that connection yet

im just writing down my thought process

that we have two basis vectors that

if scaled

can give us any vector

we would like

tha tis

2d

for i-hat and j-hat only

i know there is k-hat but

whatever

[fathom onyx]

that'll be for 3D etc

yh

[odd lagoon]

yeah ok im still

right so

1, 0
0, 1

the identity matrix

or 2x2 case

for i-hat and j-hat?

is that the same thing?

[fathom onyx]

same to what?

[odd lagoon]

is the identity matrix

telling me where

i-hat and j-hat goes

[shadow schooner]

yep

[fathom onyx]

yeah the first column says where i hat goes

the second, where j hat goes

[shadow schooner]

they stay where they are

[fathom onyx]

This matrix specifically doesn't really move those vectors, so we call this the identity matrix

[odd lagoon]

okay

now we have those

and

we can scale them with a number a to get a point a for the respective variable that the column takes care of

a, b

like

i suck at explaining

in math

[fathom onyx]

So if I have the matrix
(a,b)
(c, d)
the first column tells me my i-hat will move to (a,c)(col)

[odd lagoon]

like the first column was i-hat, conventionally horisontal

so in my mind, that takes care of x

[fathom onyx]

And my j-hat, from the second column, becomes (b,d)(col)

[odd lagoon]

the x axis

because that's conventionally horisontal

the 2nd column the y axis

[odd lagoon]

i will read from here

[fathom onyx]

This is a shitty chat-up line, btw, do not use

[shadow schooner]

maybe a more concrete example what do you think this matrix / linear map one does geometrically

[odd lagoon]

idk what chat-up means

so i wont be able to use it

[odd lagoon]

so

my first impression

[fathom onyx]

If it helps, where're you from if I can ask?

[odd lagoon]

was that this is the inverted identity matrix

but that is probably not true

and the only reason i thought that was for meta reasons

[odd lagoon]

sweden

im from sweden

but ok

so i-hat and j-hat

you know

like you can kind of think about them, i guess

[shadow schooner]

from the previous discussion the colum of the matrix are where (1,0) and (0,1) are sent by the transformation

[odd lagoon]

a square matrix

that just says

my vector remains the same

so

now to the scary part for me

if i have

[fathom onyx]

well according to wiktionary the closest meaning to "chat-up line" = "pick-up line" is "raggningsreplik" (not that that's remotely the point 🤣)

[odd lagoon]

(2, 0)
0, 1

dont tell me

so

now i have doubled

i-hat

?

[fathom onyx]

yee

[odd lagoon]

now tell me is that right

ok

then

wtf

happens here?

(1, 1)
1, 1

[fathom onyx]

well j doesn't change

It goes to (0,1)

[odd lagoon]

i could see that too so dont worry

yeah i didnt do anything

to j-hat

[odd lagoon]

when i look at this

now i can see

the basis vector

[fathom onyx]

So that matrix essentially stretches the coordinate grid by 2 along the x-axis

[odd lagoon]

in the horistona ldirection

doubling

[fathom onyx]

yeeee

[fathom onyx]

Try this matrix now

(pinging is fun /s)

[odd lagoon]

i will try that matrix

since yo uasked me to

oof

mm

i dont like that

i dont really understand what is going on

[fathom onyx]

Draw the vectors out

[odd lagoon]

i cant see it

this is why i made the thread so

[odd lagoon]

i can draw this for you

100%

[fathom onyx]

Only draw what i becomes and what j becomes

[odd lagoon]

i is 0 though

or wiat

are

making i vertical

[fathom onyx]

So step 1 - what does i become

[odd lagoon]

?

[fathom onyx]

yeee it's vertical

[odd lagoon]

bro

[fathom onyx]

pointing up

[odd lagoon]

ok

let me draw it

[fathom onyx]

What does j become?

[fathom onyx]

yeee

er

[odd lagoon]

😦

[fathom onyx]

your label for j is on the wrong side

[odd lagoon]

no

[fathom onyx]

It's pointing left now, so it should be on the left

[odd lagoon]

that's positive j

[fathom onyx]

Ah

[odd lagoon]

so it is going negative j

[fathom onyx]

then your i and j need to be swapped

if those are axes

[odd lagoon]

no

[fathom onyx]

yeah

[odd lagoon]

realyl?

😦

[fathom onyx]

i = (1,0)(col) - i.e. 1 along the x-axis

[shadow schooner]

with the formula compute where (2,2) gets sent could be nice maybe

[odd lagoon]

i was vertical

?

[fathom onyx]

i is vertical after applying the matrix

because it then becomes (0,1)(col)

[odd lagoon]

oh no

[fathom onyx]

If you swap the labels for i and j here then you're fine

[odd lagoon]

you wanted me to

multiply two square matrixes

[odd lagoon]

multiply the identity matrix

with that one

[shadow schooner]

it’s ine way to do that computation

[odd lagoon]

this one would be most left

[fathom onyx]

(if it helps - it's the same as multiplying by 1)

(i.e. it doesn't change the matrix at all)

[odd lagoon]

yes but

i dont understand

2x2 matrices

so

i dont understand this

i dont understand what the hell

this means

the identity mtrix

matrix*

is nice

[shadow schooner]

but we want you to see geometrically how a matrix can be viewed as a function that takes in vectors and spits out vectors

[odd lagoon]

i understnd that now

i didnt before

you explained it well

and i could even see

(2, 0)
0, 1

that it doubles i-hat

and i kind of think of that as the gaps becoming twice as big

on a cartesian plane