help-49

[visual tiger]

the derivative would be det(Row1',Row2,Row3) + ...

[novel flax]

yeye gotchu lemme try and let u know

ye got it

thx

ans is 8, (10 -2)

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[crude galleon]

O

[brazen palm]

Do you have a math question to ask?

[sudden yacht]

Then close the channel and don't spam

<@&268886789983436800>

[olive matrix]

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[dusty aspen]

This is my problem

[dusty aspen]

But I only want to know the second question

what are the nul-točke

solution is:

But why we only take the numerator's nul-točke?

And not the denominators?

because denominator's nul-točke is 1, -1

nul-točka: points where the parabole cuts the X axis

[visual tiger]

first of all

the denominator's null points

are the points where f is not defined

since

if denominator is zero

then f(x) would be .../0

[dusty aspen]

Ohhhh

[visual tiger]

division by 0 not allowed, doesn't exist

[dusty aspen]

Yes, I see now

[visual tiger]

a/b = 0 is the same as

a = 0 * b

which is a = 0

so

the denominator's null points are points outside the domain

the numerator's null points are the null points of the function

[dusty aspen]

👍

thanks again @visual tiger

.solved

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[proud mist]

Snitch.

[proud mist]

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[blissful totem]

💀

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[chrome ibex]

is this correct ?

[chrome ibex]

the text its not very important

this is a lotka volterra model

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[dusty aspen]

Hello, is this correct:
Injection: when no y element of Y doesn't have more than 1 x element X joined?

[dusty aspen]

Also, is this correct:
Surjection: when every y element of Y has at least 1 x element of X joined?

[visual elk]

this one is correct

[dusty aspen]

Injection: when no y element of Y has more than 1 x element X joined?

when none y has more than 1 x

correct?

[visual elk]

oh sorry

it's correct i was trippin

[dusty aspen]

okay 🙂

thank you! @visual elk

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[tepid breach]

Markov Chains question, I'm not entirely sure how to start

Maybe law of total probability at the start to introduce invariant dist

<@&286206848099549185>

[midnight plank]

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[tepid breach]

nvm I cooked

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[ocean osprey]

hihi

[ocean osprey]

problem:
y = -x^2 + 5x + 6

and we're also given this formula:

-b +- sqrt(b^2 - 4 * c)
/
2 * a

and you gotta like seperate them

it's a perabula thing

now, the answers page says that the correct answers are: (6,0),(-1,0)

but I can't get it.

<@&286206848099549185>

[fallow scarab]

!15m

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[grizzled merlin]

are you sure that formula is write

[ocean osprey]

yep

[grizzled merlin]

um

oh i supppose

it is in this case

do you know what a, b and c stand for?

[ocean osprey]

i THINK I do:
a = -1
b = 5
c = 6

right?

[grizzled merlin]

a is just 1

because the coefficient(number infront) of x^2 is 1

but obviously we dont write the 1

[ocean osprey]

but isn't it -1?

[grizzled merlin]

what makes you think that

[ocean osprey]

I wrote it wrong

mb

[grizzled merlin]

ohh ya its -1 then

[ocean osprey]

it's this:
y = -x^2 + 5x + 6

[grizzled merlin]

okay your formula is wrong then

you missed out the a in the discriminant (square root)

[ocean osprey]

whatttt??/

nnonoonn

x isn't equal to the formula

[grizzled merlin]

oh you need to factor it into two terms?

like (x-a)(x-b)

[ocean osprey]

it's +- because you divide it into two parts:
one for an exercise where you do subtraction
and one for an exercise where you do addition

and then you get two x's

[grizzled merlin]

do you understand why you need to do the plus and minus?

[ocean osprey]

yes

cuz you gotta do both

[grizzled merlin]

well why

[ocean osprey]

so you'll get the two points where the perabula touches the y axis

[grizzled merlin]

you mean x axis?

[ocean osprey]

yeah I mean x mb

[grizzled merlin]

okay well

the formula above is a general solution to equations in the form Ax^2 + Bx + C = 0

so in this case

-x^2 + 5x + 6 = 0 (because y=-x^2 + 5x + 6, and at the x axis the y value is 0, so you can replace the y with 0)

[ocean osprey]

yeah that's what I did

when I solved the square root of the formula, I got 1

[grizzled merlin]

aight hang on

i think you should understand why we take the plus and minus of the square root

[ocean osprey]

what?

I honestly have no idea why we do that

[grizzled merlin]

consider the equation x^2 = 1

[ocean osprey]

first time working with perabulas

[grizzled merlin]

solve that for x

[ocean osprey]

x = 1

you transfer the square root

and then it's x = sqrt(x)

which is 1

[grizzled merlin]

ok. what's (-1)^2

[ocean osprey]

1

[grizzled merlin]

hence, -1 is also a solution to x^2 = 1

[ocean osprey]

okay

so what now?

[grizzled merlin]

so wheneveryou need to take the square root to solve for a variable, remember you must include the +- to account for the negative root

i just felt that was quite an important piece of intuition for this topic

[ocean osprey]

okay

so what now?

[grizzled merlin]

alright so back to the quadratic

-x^2 + 5x + 6 = 0

we have it in the form ax^2 + bx + c = 0, so we can apply the quadratic formula to find our values for x that satisfy this equation

as we said,
a=-1
b=5
c=6

[ocean osprey]

okay,
So it's:

-5 +- sqrt(5^2 -4 * 6)
/
2*(-1)

right?

[grizzled merlin]

you missed something in the square root

you just did 4c instead of 4ac

[ocean osprey]

what's ac

[grizzled merlin]

its a multiplied by c

so instead of 46, you would have 4(-1)*6

ok

the 4 is italisizing

[ocean osprey]

why 4(-1)*6 ?

where did that -1 come from?

[grizzled merlin]

its a?

[grizzled merlin]

read this formula

[ocean osprey]

but.. why is a there?

are you sure its supposed to be there?

[grizzled merlin]

100%

[ocean osprey]

it wasn't there in class, it only showed up as the 2*a

[grizzled merlin]

perhaps you were only dealing with quadratics where a=1

but tbh i have no idea why it wouldnt be there

[ocean osprey]

oh it is thereee

okay I see it

the square root says:
sqrt(b^2 -4 * a * c)

is this right?

[grizzled merlin]

yes

[ocean osprey]

ok lemme try to solve it now

i got 1

still

[grizzled merlin]

1 for what?

[ocean osprey]

i did:
sqrt(5^2 -4 * 1 * 4)

[ocean osprey]

for the square root

[grizzled merlin]

ok

double check all of your values

a=-1
b=5
c=6

[ocean osprey]

now I got 7 ???

oh it's cuz I used the (-1) instead of just 1

okay, is 7 right?

[grizzled merlin]

okay

so

7 is right

but

im assuming you mistyped and input 6?

[grizzled merlin]

as in: this has the wrong value for c

[ocean osprey]

ok

now I did this:
sqrt((5^2) -4*(-1)*6)

and I got 7

[grizzled merlin]

lmao i think your brain isnt working

5^2?

you wrote 5^4

[ocean osprey]

typo

I meant 5^2

[grizzled merlin]

so you have computed the square root but not the whole formula

remember

so rn you're at -b +-(7) / 2a

[ocean osprey]

I got
(6,0)
(-1,0)

[grizzled merlin]

that is correct

[ocean osprey]

alright thank you

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[half kiln]

The answer is y = a(5x+2), with a < 0
I'm trying to understand this but I simply can't

[surreal moon]

What is an affine function?

[half kiln]

f(x) = ax+b

[surreal moon]

Ah just a linear function

[half kiln]

Yes

[surreal moon]

Consider the point slope equation for a line. Do you know it?

[half kiln]

The "a"?

a = Δy/Δx

and
a = tg(θ)

[grim vector]

||(No difference in english? Cuz affine is ax + b and linear is ax to me)||

[half kiln]

Yesss

At least in my language it's like that

[grim vector]

Also its in fr and we have wierd enough stuff like that

[half kiln]

Portuguese as well...

I'll close this chat and try to search for my teacher to get a reply... because WHAT A QUESTION

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[sudden yacht]

It could simply be y = -(x+2/5)

[heady plume]

(x + 2/5)y < 0

Is the written condition basically, or you could manipulate it to be

y/(5x + 2) < 0

So just put it some constant C, with C < 0 and you're good to ho

Not ho, *go

[midnight plank]

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[mint ravine]

How do I make most of the terms here cancel out?

[mint ravine]

I think we could use the fact 1/(n*n+1)=1/n-1/(n+1) here

[surreal moon]

at first glance, it looks like you would want to consider telescoping series (but I could be wrong)

[surreal moon]

that looks useful

[lavish venture]

$\frac{1}{(2n+1)(2n+3)}$

[novel flax]

it cant hold even stuff

[lavish venture]

oh wait i see nevermind

this is only odd my fault

[novel flax]

2n+1 and 2n+3

[grand pond]

knief

[novel flax]

n from 0 to 99

[mint ravine]

oh ok

[lavish venture]

@mint ravine do you know partial fraction decomposition

[mint ravine]

no

[novel flax]

its coming

[mint ravine]

Am I supposed to?

[lavish venture]

$\frac{1}{(2n+1)(2n+3)} = \frac{A}{2n+1} + \frac{B}{2n+3}$

[lavish venture]

yes

[grand pond]

knief

[novel flax]

what i did was multiply and divide by 2

[lavish venture]

solve for A and B

[novel flax]

its simple so we can directly get it

(2n+3)-(2n+1)

---

2(2n+1)(2n+3)

[mint ravine]

sorry im having tech issues right now

let me restart my computer

[mint ravine]

A=(1-B(2n+1))/(2n+3)

[lavish venture]

nono

what you should do is multiply both sides by the denominator on the left

[mint ravine]

yeah that's what was i did

[lavish venture]

then set n = -1/2 to solve for A

and set n = -3/2 to solve for B

$1= A(2n+3) + B(2n+1)$

[grand pond]

knief

[mint ravine]

yeah i got that expression after multiplying both sides by the denominator on the left

[lavish venture]

did you do this

[mint ravine]

no

let me do that rq

[lavish venture]

do that

because it will make the other term be zero

so you can solve for A and B

[mint ravine]

oh cool

so a=1/2

[lavish venture]

yea

and B = ?

[mint ravine]

B=1/-2

[lavish venture]

$\frac{1}{2(2n+1)} - \frac{1}{2(2n+3)}$

[grand pond]

knief

[lavish venture]

or

$\frac{1}{2}\left(\frac{1}{2n+1} - \frac{1}{2n+3}\right)$

[grand pond]

knief

[lavish venture]

so

$\sum_{n=0}^{99} \frac{1}{2}\left(\frac{1}{2n+1} - \frac{1}{2n+3}\right)$

[grand pond]

knief

[lavish venture]

so you can write out a few terms to see what’s going to cancel

and pull out the 1/2 tbh

[mint ravine]

kk

thank you goku

[dapper yoke]

Sorry this is much better

[mint ravine]

thanks naruto

[dapper yoke]

so we have a1 - an

n=100

[mint ravine]

yes

im just plugging in random values for n to see which one works

[dapper yoke]

so it's 1/3 - 1/201

[dapper yoke]

If you look at the formula the top of the summation is n-1

so in this case n-1=99

n=100

[dapper yoke]

That multiplied by 1/2

[mint ravine]

oh yeah

that saved me a lot of time

thank you!

[dapper yoke]

Oh and we have an issue

The summation on the formula starts at 1

But in the problem it starts at 0

Guess what to do now?

[mint ravine]

subtract by 1?

idk

[dapper yoke]

You have to add the first term to the formula

so replace by 0 and add it to what we have

so 1 - 1/3, all multiplied by 1/2

[mint ravine]

so 1/3

[dapper yoke]

Yes

That added to 1/6 - 1/402

And that's it

[mint ravine]

oh wow

thank you for helping

[mint ravine]

so if i did my math correctly

the answer should be 601/903

[dapper yoke]

Let me check

[mint ravine]

and you can't simplify it anymore since 601 is a prime

[dapper yoke]

This was equivalent to 0.49751243781

[dapper yoke]

This is 0.66555924695

[mint ravine]

you have to add 1/3 to it

[mint ravine]

nevermind

i see my mistake

thanks for helping everyone

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[grizzled merlin]

am i actually tripping or aren't there infinity many solutions to this

[heady plume]

!status

[midnight plank]

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5. I have a question about someone else's work/solution.
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[grizzled merlin]

shit i didnt realise that was from long ago

[heady plume]

Lmao

[grizzled merlin]

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[safe star]

hello ! I have a problem with partial derivative in a 3 D space !

[safe star]

i must find the partial derivative in

that's the proposition i have

but 0.7 is not the correct answer 😅

so my idea of how to fint the partial derivative might be wrong

i also have this graph in y=1

thanks y'all in advance !

[hearty rune]

this is the one to look at, the slope at x=1

[safe star]

so i suppose that's it's negative

but why looking at the graph with z ?

OH OKAY

THX

i just understood

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[leaden seal]

this is what i did

but i think its wrong

[heady plume]

Should be 5ms

[leaden seal]

working out?

[heady plume]

No m done. Just wondering the context of "first time waveform reaches 220V"

[leaden seal]

im not sure

i think it just means time

[heady plume]

Well anyways, the first time waveform reaches 220V is in T/4 time and T = 1/f

So 1/4f time or 0.005s = 5ms

@leaden seal

[leaden seal]

oh

so this is wrong?

[heady plume]

Naah that's just rigorous

[leaden seal]

okay

[heady plume]

π/2 is reached at T/4 time anyways. And if frequency is already given, it's better for mcqs to not overcomplicate

Plug T = 1/f, compute, onto the next ✓

[leaden seal]

right

[heady plume]

Btw I'm curious why your Vpeak = √2 • 220V

[leaden seal]

its a formula

[heady plume]

But Vpeak = Vrms √2.. it isn't mentioned in q that Vrms is given

Anyways, hope you got this

[leaden seal]

thanks

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[shut goblet]

hii, I still don´t understand the difference between asymmetry and antisymmetry. The following set {a,b,c,d} and relation R = {(a, a),(a, c),(b, b),(b, d),(c, c),(d, d)} were given. Why is the relation R not asymmetric but antisymmetric :((

[lunar ocean]

well, what's the definition of asymmetric, and what's the definition of antisymmetric?

[shut goblet]

the definition of asymmetry is: if x R y => y is not in relation with x

and antisymmetry means x R y and y R x => x = y

well it is antisymmetric because it isn´t symmetric

[lunar ocean]

...what?

[shut goblet]

but I don´t know why it is asymmetric

[lunar ocean]

ok wait what's the definition of symmetric

[shut goblet]

okk,soo it means that if x is in relation with y, then y needs to be in relation with x too for example

[lunar ocean]

well that's not the negation of antisymmetric

[finite karma]

that is symetric yes

[lunar ocean]

the relation R = {(a,b),(b,a),(c,d)} isn't symmetric, because cRd is true but dRc is false, and also isn't antisymmetric, because aRb and bRa but a isn't equal to b

[finite karma]

but to prove antisymetric the most used method is if you have xRy suppose yRx and try to prove that in this case that always x will be equal to y

[lunar ocean]

most used method
...well yes because that's the definition

[shut goblet]

okkk, does that mean that if we have something like

[lunar ocean]

and no that's not the definition of asymmetric

it's the definition of symmetric

[finite karma]

? xD

bro sry to ssay this but u gotta take back that claim if u not sure try to not share missinformation please

[shut goblet]

R = {(a,a),(b,b),(c,c)} this is symmetric, reflexiv, transitive and antisymmetric, right?

[lunar ocean]

yes, assuming it's on the set {a,b,c}

[shut goblet]

yeah okkk

thanks a lotttt

[lunar ocean]

you're the one who's spreading misinformation actually

[shut goblet]

why what was wrong?

[finite karma]

can you read my emssage again and tell me what is wrong "undergrad?"

[lunar ocean]

the message you were replying to here is the definition of symmetric

you said it's the definition of asymmetric

or, well, "assymetric", which would be a pretty strange way to misspell "symmetric"

[finite karma]

anyways i'll break it down to you easily . to prove symetric you need to have is when xRy you need to prove that yRx aswell . for antisymetric you need to prove that for xRy then y sin't in Relation wtih x ( as i said the best method for this is to suppose xRy and yRx and prove that x=y always in this case ) ; for transative u need to prove that when you ahve xRy and yRz prove that xRz

[lunar ocean]

for antisymetric you need to prove that for xRy then y sin't in Relation wtih x ( as i said the best method for this is to suppose xRy and yRx and prove that x=y always in this case )
false

that's the definition of asymmetric

it is true that all asymmetric relations are antisymmetric, but not all antisymmetric relations are asymmetric

[shut goblet]

xRy and yRx => x = y is antisymmetric

[finite karma]

yes

[finite karma]

try to focus in this

[shut goblet]

why are all asymmetric relations antisymmetric?

[lunar ocean]

well suppose R is asymmetric, and consider two elements x,y such that xRy and yRx

[finite karma]

assymetric means if xRy then Y can never bever in relation iwth x

[lunar ocean]

this situation is a contradiction because R is asymmetric, therefore by the principle of explosion x = y

which proves that it's antisymmetric, because we showed that if xRy and yRx then x = y

[shut goblet]

waitt

if it is asymmetric, that means that xRy but yRx is not in the relation,right?

[lunar ocean]

exactly

[shut goblet]

and it is antisymmetric because xRy and y is not in the relation => x is not equal to y ???

[lunar ocean]

R is antisymmetric if whenever xRy and yRx, [something else]
but because R is asymmetric, "xRy and yRx" just never happens, so it's true vacuously that R is antisymmetric

[shut goblet]

ohh,just because it never happens?

[lunar ocean]

no, it's antisymmetric because if xRy and yRx, then x = y

but yeah basically it's just, "xRy and yRx" is impossible, so it's saying "every time [this thing that never happens] happens, x = y"

[shut goblet]

and that looks like (a,a),(b,b) for example?

[lunar ocean]

or a different way to look at it: there can't be a counterexample, which would be two x,y such that xRy and yRx, and yet x != y, because the "xRy and yRx" part is already impossible

[lunar ocean]

well no, the relation {(a,a),(b,b)} is antisymmetric but not asymmetric

because if you take x = a and y = a, then xRy and also yRx

because aRa and aRa are both true

[finite karma]

i'll give you an exemple take the relation < . if xRy then x < y and y can never be < x

[shut goblet]

wait, i need to reread your explanation

[finite karma]

so taht means if x R y then y can never be R x

[shut goblet]

okk

right

[finite karma]

@shut goblet i believe this exemple should be very easy

[shut goblet]

ohhh,rightttt now I got it I guess

[finite karma]

np :)))

[lunar ocean]

and to be clear, while "symmetric" does sort of feel like the "opposite" of "antisymmetric"/"asymmetric", they're not actually negations of each other

the relation {} is both symmetric and asymmetric

again for vacuous reasons, "symmetric" and "asymmetric" both start with "if xRy, [...]" and that never happens with the empty relation

[shut goblet]

THANKS A LOT, I really thought that it was just the opposite of symmetry

[finite karma]

well depends on the prespective

[lunar ocean]

there are also a lot of relations that are, intuitively, "sometimes symmetric and sometimes asymmetric", and end up not being symmetric or asymmetric, which both assert that it always happens a particular way

[shut goblet]

ohh, do you know how I can practice this topic?🥹

[lunar ocean]

{(a,b),(b,a),(c,d)} isn't symmetric, because the (c,d) part makes it not symmetric, and isn't antisymmetric, because the (a,b) (b,a) part makes it not antisymmetric

[shut goblet]

right, because after definition a should be equal to b but it can´t be

right?

[lunar ocean]

yep

[shut goblet]

can I ask you another question?

it is about transitivity

[lunar ocean]

and if you had {(a,a),(c,d)}, then that is antisymmetric, but isn't symmetric (the (c,d) still breaks it) and isn't asymmetric (because of the (a,a))

[shut goblet]

omg, thanks

there was the following relation given:

R = {(a,a),(a,c),(a,d),(b,b),(c,a),(d,a)}

Why is R transitive?

I thought that it is not possible to be transitive bc {(c,c),(c,d),(d,c),(d,d)} are missing

[lunar ocean]

...oh, huh, good point, i missed that lol

yeah you're right, this isn't transitive

cRa and aRc should imply cRc, if R was transitive

[shut goblet]

yeahh

my teacher wrote in the solutions that it is transitive :c

[lunar ocean]

yeah i think your teacher just didn't notice that

which tbf i didn't either until you pointed it out

[shut goblet]

does my solution makes sense?

[lunar ocean]

i think i would have given more detail to make it more directly related to the definition of transitivity, instead of just listing the pairs that were missing and leaving it to the reader to work out what exactly is wrong with them

[lunar ocean]

like i did here

but you definitely had the right basic idea, because those are the pairs that should be there but aren't

[shut goblet]

like cRa and aRc => cRc ∉ R?

for example

or like it should imply cRc, if R was transitive

[lunar ocean]

...i mean that's kind of a pile of symbols that doesn't exactly make grammatical sense, specifically "cRc ∉ R" is a bit odd, but yeah basically

"cRa and aRc are both true, so if R was transitive, cRc would also be true, but it's not, therefore R is not transitive"

[shut goblet]

yeah, you are right it is a bit of odd I should write it like you did it

makes things for everyone clear

i am really,really thankful for your patience and explanations

thanks a lotttt!!

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[grim monolith]

how did this get simplified

[grim vector]

$\sqrt[n]{x} = x^{\frac{1}{n}$

[grand pond]

Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[grim vector]

And

[grim vector]

So

[grim monolith]

i se

tahnks

[midnight plank]

@grim monolith Has your question been resolved?

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[slim portal]

Can anyone proof to me why does odd root can have a negative number inside of it?

[dapper yoke]

what number multiplied by itself 3 times is equal to -8?

[slim portal]

-2?

[dapper yoke]

Yes, now can you find a number that multiplied by itself is -4?

[sage helm]

I mean, if you think about it, (-1)^3 =-1 so (-1)^(1/3) would be -1

same thing with other odd roots

[dapper yoke]

This happens because negative times negative times negative is still negative, while there is no sign that multiplied by itself is negative

On real numbers

This is the signs rule

[slim portal]

I think there are no number to it

[dapper yoke]

So for square roots you have only two options: negative x negative or positive x positive

Both of them will give you positive

But what happens with cubic roots? You have negative x negative x negative or positive x positive x positive

[blissful totem]

are u asking for intuition or proof

[dapper yoke]

In the first case, from the picture above, negative x negative is positive, and that times negative will give you negative

In the second case, positive x positive is positive, and that times positive will give you positive

So for the cubic root you may find the root for both negative and positive numbers

But for square roots you may only find the value for poditive numbers

[slim portal]

OOOOO

I seeee

Basically the square of any number is positive so the value inside of the square root must always be positive, while on the other hand, odd root can have negative number like (-2)³ is -8 or 2³ is 8. therefore it is acceptable to have negative number inside of the square root

Is this correct?

[blissful totem]

yeah

[slim portal]

Thanks anyway, now I'm clarified

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[last slate]

I'm unable to understand the 7th question, that when we take coefficient of highest power as unity, theres only a single answer?

[dapper yoke]

Fundamental Theorem of Algebra

[winged wharf]

juaincha do you know wtf "highest power of the unknown be taken equal to unity" means

[dapper yoke]

Yes

[last slate]

How does it apply here?

[dapper yoke]

Maybe it helps to reduce the amount of solutions

[last slate]

But we need complex coefficients for that?

[dapper yoke]

If the highest power of the unknown is 4, then the amount of solutions are 4. Now if it is not 4, the solutions decrease

[dapper yoke]

I don't think

[last slate]

It says coefficient of the highest power, so if it was 4x⁴ now it's x⁴

[dapper yoke]

Ohh okay

So no Theorem of algebra

I'm going to sleep

Goodbye

[last slate]

um

[nimble iron]

It's bc you can just simplify things

[winged wharf]

is it asking why is there only one polynomial that satisfies ex 1 - 4

the grammatical errors are throwing me off in the question sorry

[nimble iron]

She said it's bc why do we use (x-r1)(x-r2)=0 instead of 3(x-r1)(x-r2)=0 let's say

But it's bc they're equivalent

[last slate]

Ohh

[nimble iron]

Moreover, they ask for "a" equation

[last slate]

okay I think got it

[winged wharf]

its the same thing as writing 100 x 0 or 1214124 x 0

they both become 0 anyways

doesn't matter what number I put before it

[last slate]

I understand, thank you!

[nimble iron]

Np

[last slate]

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[mint ravine]

Does anyone know of any physical calculators that are affordable and can be used for the SAT?

[mint ravine]

.close

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[mint ravine]

What is Simon's Favorite Factoring Trick and how does it correlate with the Diophantine Equation?

[mint ravine]

Can someone please explain it to me

<@&286206848099549185>

[fallow scarab]

Show where you're reading about Simon's favorite factoring trick

[midnight plank]

@mint ravine Has your question been resolved?

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[fallow scarab]

!15m

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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.

[fallow scarab]

!15m

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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.

[fallow scarab]

Why urgent

What did you attempt for the last 3 days

Jeez I'm reading

[zealous schooner]

You don't have to be unpleasant, people here are just volunteers

And you're not supposed to get help on these problems

[fallow scarab]

So this was a lie

[midnight plank]

@gentle horizon Has your question been resolved?

[misty gorge]

.close

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[south sparrow]

is it sufficient here to say that when T = 0 (the zero map), p(T) = 0 so the set becomes just {0} which is not L(V)

[south sparrow]

im not sure if it means these are not equal for ANY T or there is some T where they arent equal

im leaning towards the former

[rough birch]

what does $\mathcal{L}(V)$ mean again?

[grand pond]

Arnavutköy

[south sparrow]

its the set of operators on V

aka Hom(V)

[rough birch]

ok i will leave this for somebody else to help you with then

[sharp coral]

p(0) is not necessarily 0, though

[south sparrow]

oh wait ur right what am i talking abt

would it be p(T) = aI for some scalar a

I is identity

[zealous schooner]

But that should still qualify as a counterexample

[south sparrow]

yea it would

[south sparrow]

my main issue is just with this wording

[zealous schooner]

They mean it can't be equal for any T

So your argument doesn't prove it for all T

[south sparrow]

yea

i think i can figure it out for all T, i just was unsure abt the wording

ty!

,close

oop

.close

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[south sparrow]

ok i think i got it

.reopen

[midnight plank]

✅

[south sparrow]

the dimension of the set here is equal to dim V but dim L(V) = (dim V)^2. since (dim V)^2 < dim V (we are given dim V > 1, which guarantees this), we cannot have equality

i think..?

[hard umbra]

why is the dimension equal to dim V?

[south sparrow]

or i guess its less than or equal to dim V

wait

hold on my brain isnt working rn

[hard umbra]

this exercise appears before chapter 5B so presumably you should not know that fact yet

[south sparrow]

yeah i knew it because im just reviewing stuff but i was trying to think of a way to prove it without relying on stuff thats not in the book yet

[hard umbra]

why doesn't it work for dim V = 1

[south sparrow]

then dim L(V) = 1 so every nonzero T is a basis of L(V) and we can construct any other linear operator U as a multiple of T

[hard umbra]

there is a simpler explanation

[south sparrow]

o

without rlying on 5.B stuff?

[hard umbra]

yes

[south sparrow]

hmm

[hard umbra]

what is L(F)

how do i write down an element of L(F)

[south sparrow]

a linear map on F

so like z -> az for some a in F

?

[hard umbra]

yeah

so just a number

[south sparrow]

yeah

[hard umbra]

the number a

what is {p(T) : p in P(F)}

lets say T is just the number a

[south sparrow]

b0 * I + b1(az) + b2(az)^2 + ...

?

[hard umbra]

which also happens to be a number right

[south sparrow]

oops

yes

[hard umbra]

so L(F) is just F

and {p(T) : p in P(F)} is also just F

theyre both F

[south sparrow]

ok

makes sense

[hard umbra]

whats good about F

remember we have a bunch of polynomials in {p(T)}

[south sparrow]

uh could you be a bit more specific

dimF = 1?

[hard umbra]

well lets say ive got some polynomials p(x) and q(x)

[south sparrow]

im not sure what youre getting at

[south sparrow]

mhm

[hard umbra]

and i go like

p(x) q(x)

[south sparrow]

ok

[hard umbra]

theyre polynomials so theyre nice

you're used to doing arithmetic with them

[south sparrow]

mhm

[hard umbra]

now i take some S, T in L(V)

and i go like

S T

hmm

[south sparrow]

oh we have commutativity?

[hard umbra]

we do don't we

numbers commute

[south sparrow]

mhm

[hard umbra]

S and T don't really

what can you say about {p(T) : p in P(F)} for any choice of T

[south sparrow]

uh im not sure what's useful to say about it

[hard umbra]

hint: you've already said it

[south sparrow]

everything in it is just numbers

[hard umbra]

lets go bigger now

any V

[south sparrow]

its just elements of V ?

or isomorphic to elements of V

[hard umbra]

perhaps the example with V = F was confusing

lets take V to be any vector space now

not necessarily F

so T is a linear map V -> V

[south sparrow]

mhm

[hard umbra]

and p(T) is some expression like T^2 + 5T - 2I

[south sparrow]

yeah

[hard umbra]

what can you say about p(T) q(T)

[south sparrow]

oh

its some expression like that

with the maximal power doubled

[hard umbra]

something like that

but also

[hard umbra]

remember this

[south sparrow]

mhm

[hard umbra]

p(T) q(T) = q(T) p(T) still right?

[south sparrow]

yes

[hard umbra]

polynomials still commute

what about L(V)?

if you take some random elements there

[south sparrow]

they dont typically commute

[hard umbra]

they surely don't

so can {p(T) : p in P(F)} = L(V)?

[south sparrow]

no

[hard umbra]

do you see the argument now?

[south sparrow]

yes i do

[hard umbra]

and when V = F, as mentioned earlier, everything is just numbers

so you just have commutativity always

thats why it fails

[south sparrow]

and dimV > 1 because when dimV = 1 the operators do commute?

[hard umbra]

ye

[south sparrow]

mhm

i see

that makes sense

[hard umbra]

nice

[south sparrow]

ill try and use this to make a complete proof, tytyty <3

[hard umbra]

[south sparrow]

.close

[midnight plank]

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[hard umbra]

i suppose an unbased solution would be to look at the invariant subspaces

[south sparrow]

so my proof was to let U,W be arbitrary operators on V that do not commute (which is always possible with the guarantee that dim V > 1). then for sake of contradiction assume {p(T): p in P(F)} = L(V), which lets us pick polynomials p, q with p(T) = U and q(T) = W. then p(T)q(T) = q(T)p(T) but UW != WU, a contradiction

[hard umbra]

seems good to me

mfw UWU

[south sparrow]

UWU

crazey

[hard umbra]

alternative solution btw

[south sparrow]

this is probably the indended route lol

i like this proof though :>

[hard umbra]

the eigenvectors of T are eigenvectors of p(T)

so you just find S with different eigenvectors

hm no this might mess up when you get weirdness like p(T) = I

then everything is an eigenvector

commutativity is definitely more based

[midnight plank]

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[jovial crest]

i need to prove AY is the diameter of (ABC)

[jovial crest]

if we take quadrilateral BHCY its a paralellogram right

since the diagonals bisect eachother

i just wanted to know for any quadrilateral if the diagonals bisect eachother we can say its a parallelogram

[spring wave]

ok

first how did you prove that X lies on the circle?

[jovial crest]

i took angles and proved ABXC is cyclic

[spring wave]

ok

tak the line segment XY

try showing it is parrallel to BC

[jovial crest]

ok

[jovial crest]

i wanted to know if this is sufficient to say HBYC is a paralellogram

[spring wave]

HYBC is a parralelogram and H is the orthocenter is sufficient

[jovial crest]

yeah yeah i just wanted to know if saying the diagonals bisect eachother is enough to prove HBYC is a parallelogram

[spring wave]

yes

that is enough

[jovial crest]

thanks!

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[gusty falcon]

hi, for a triangle with fixed perimeter, the area is maximized when it’s an equilateral triangle right?

[gusty falcon]

Also what about if two sides are fixed? then maximum area is when it’s a right triangle?

[rough birch]

yes

[gusty falcon]

okay sure, nice

[gusty falcon]

what about this?

[rough birch]

you mean when the perimeter is not fixed but we know 2 of the side lengths? yes then the right triangle where the two sides are the legs

[gusty falcon]

i guess yeah

[gusty falcon]

why?

also this person is wrong then right?

[rough birch]

well one way to think about it is the area of a triangle is $\frac{ab\sin(C)}{2}$

[grand pond]

Arnavutköy

[rough birch]

if two of the sides are fixed and the perimiter is not fixed then yes it is when its a right angled triangle

[gusty falcon]

hmm fair

[gusty falcon]

so the person that responded is wrong right?

[gusty falcon]

also no trig allowed 😭

[rough birch]

yeah

[gusty falcon]

any geometric proof?

[rough birch]

base times height over 2

[gusty falcon]

uhm like i guess for a fixed perimeter a square has the greatest area so we want it to be as close to a square?

[rough birch]

so take one of the edges

and make that the base

the height will always be less than or equal to the second side

the case of equality only happens when the second side IS the height

which implies that both sides form a right angle since they are the base and height

[rough birch]

for fixed perimeter, the maximum area of an n-gon is the regular n-gon

[gusty falcon]

why?

but okay i see the intuition nonetheless

if u have two sides and if u fix one as the base

the area is maximized when you make the height orthogonal to it

[gusty falcon]

given 2 sides why can’t u like try to make it regular?

why right triangle and not “as regular” as possible

[midnight plank]

@gusty falcon Has your question been resolved?

[midnight plank]

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[barren onyx]

Convert this to parabolic equation

[barren onyx]

i took 4 common

4(y^2-5y) = -(12x+67)

[twilit field]

||4(y^2-5y+6.25)-6.25=-12x-67||

[barren onyx]

6

25/4?

[zealous schooner]

should be -25 outside the brackets

[barren onyx]

why 25/4

[twilit field]

-25?

[twilit field]

6.25

(2.5)^2

[barren onyx]

ok in our syllabus they reduced perfect square for 10th lol

so i don't know what method you're using

[twilit field]

completing the square

[barren onyx]

ye but why (2.5)^2?

[twilit field]

https://youtu.be/prx_Bf2hakw?si=guhMxX92T90aUkGy see this to understand the method

[barren onyx]

thnx

[zealous schooner]

the bracket is being multiplied by 4

[barren onyx]

oh ok

so its (5/2)^2

[zealous schooner]

it might be easier to think about it as $4y^2-20y+25-25=-12x-67$ since $4y^2-20y+25=(2y)^2-2(2y)(5)+(5)^2=(2y-5)^2$

[grand pond]

kheerii

[barren onyx]

(y-5/2)^2 = (-12x-67+25)/4

[zealous schooner]

no

[barren onyx]

why

[zealous schooner]

+25, not +6.25

[barren onyx]

25/4 is same 6.25

oh i get

It

[zealous schooner]

but what you wrote has a term of 25/16, not 25/4

[barren onyx]

so taking 4 common u get

bruh what to do next

do i take 4 common

[zealous schooner]

you're done, you can simplify the fraction

[barren onyx]

Ok

taking 4 common u get 4(-3x-42/4)

but this is incorrect

it should form (y-5/2)^2= -3(x+(7/2))

?

@twilit field ?

[barren onyx]

its not giving the same answer dude

[twilit field]

sorry, was enrolling myself in a course

yeah, one minute

where are you stuck

[midnight plank]

@barren onyx Has your question been resolved?

[midnight plank]

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[pure wraith]

The extended Euclidean algorithm is an efficient way to find integers u and v such that au + bv = gcd(a, b). Using two prime numbers p = 26513 and q = 32321, find the integers u and v such that pu + qv = gcd(p, q).

[pure wraith]

I'm not quite sure how I'd do this, all I really know is that gcd(p, q) = 1

[brave merlin]

Good start. I would do pu + qv = 1

[buoyant yoke]

start with $1\cdot p + 0\cdot q = p$ and $0\cdot p + 1\cdot q = q$

[grand pond]

Bonk

[pure wraith]

What does "start" mean here exactly?

What I'm confused about it how is the Euclidean algorithm relevant at all

[buoyant yoke]

this is the method i learned

you start off with those two equations

write them above each other

and then subtract them from one another until you get to 1

ill show the first step

one second

\begin{align*}1\cdot 26513 &+ 0\cdot 32321 &= 26513\0\cdot 26513 &+ 1\cdot 32321 &= 32321\(0-1)\cdot 26513 &+1\cdot 32321&=5808\-1\cdot 26513 &+1\cdot 32321&=5808\end{align*}

[grand pond]

Bonk

[buoyant yoke]

do you see what i did here?

[midnight plank]

@pure wraith Has your question been resolved?

[midnight plank]

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[mighty tree]

,rotate

[mighty tree]

Hi guys any one have any idea how to do 13

[buoyant yoke]

!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

[mighty tree]

I do not know how to start at all

I thought of integration

But my sch hasn't went through it yet only differention

My only thought process is coordinate q and r have same graident maybe we can link it to that

[buoyant yoke]

first try making an expression for the area of the rectangle

[mighty tree]

Hmm I'm not very sure how to do that

[buoyant yoke]

also, notice that it is symmetric around the x-axis

[royal sable]

yea that makes it much easier

[buoyant yoke]

let x be the width of the rectangle

[mighty tree]

hmm yes so points q and r have same y coodinate right

[buoyant yoke]

then what would be height of the curve at that point?

[royal sable]

no, x, but i think you get the idea

[mighty tree]

hmmm x/2?

not sure if thats a square if we js count the upper part

[mighty tree]

oops mb

[buoyant yoke]

where are you getting that from?

[mighty tree]

cutting the rectangle into 2 at the x axis

[buoyant yoke]

yes but we have the point x

on the x axis

what is then the value of the curve on the point x?

[mighty tree]

wait which point x?

if its my marking ignore it i kinda did it wrongly

[buoyant yoke]

you know functions right?

[mighty tree]

yea

[buoyant yoke]

f(x)=x^2+4 or smth

for example

[mighty tree]

hmm yea

[buoyant yoke]

so this is a parabola

and the y value depends on x

now we have the curve x=y^2-4

then what is the value of y?

[mighty tree]

oh 2?

[buoyant yoke]

what?

how is y a constant

x is a variable

y is also a value

and dependent on x

[mighty tree]

im not sre i understand what do u mean by value y the value of y that makes x=0 or smth?

[buoyant yoke]

we want to find the area of the rectangle right?

[mighty tree]

hmm yes

[buoyant yoke]

so we need the width and heigh

we set the width equal to x

then what is its height?

[mighty tree]

im still not sure

[buoyant yoke]

what we want is a function of the area of the rectangle in terms of x

[mighty tree]

am i supposed to express the height in terms of x?

[buoyant yoke]

yes

[mighty tree]

i cant find any relation between the width and height tho

[buoyant yoke]

how not?

[mighty tree]

or is PS twice the length of QP

[buoyant yoke]

the height is bounded by the curve

right?

[mighty tree]

hmm but it only touches a side of the rectangle tho

[buoyant yoke]

so how do we involve the curve here

[buoyant yoke]

yes?

[mighty tree]

ur letting QP/RS be x right so im supposed to express PS in terms of x

[buoyant yoke]

yes

and QR=PS

[mighty tree]

so are u trying to releate QR to the eqn of the curve?

[buoyant yoke]

yes

because thats how QR is defined

[barren onyx]

are u free now

[buoyant yoke]

!occupied

[midnight plank]

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

[barren onyx]

ye im gonna open a new channel

just asking if he's free or not

[mighty tree]

hmm is it sqrt(x+4)

[buoyant yoke]

indeed

so then what is the area of the rectangle?

[mighty tree]

x sqrt(x+4)

[mighty tree]

so ur saying as long as 2 points with the same x coordinate that touches this curve here it is always reprsented by sqrt(x+4)

[buoyant yoke]

yes

and then how do we maximise it?

[mighty tree]

diffrenatite the thing

er lemme try

[buoyant yoke]

wdym?

[mighty tree]

lets say p and s touches the curve too is the distance between them also sqrt(x+4)

[buoyant yoke]

yes

anything on the curve

[mighty tree]

ohh

[buoyant yoke]

for any pair (x,y) on the curve: x=y^2-4

[mighty tree]

i got d/dx=2x+4/2sqrt(x^2+4x)

[buoyant yoke]

or you can see it as: for any x and y that have the relationship that x=y^2-4, is on the curve

[mighty tree]

oh

[buoyant yoke]

formally: $C = {x,y\in\mathbb{R} :x=y^2-4}$

[grand pond]

Bonk

[buoyant yoke]

anyway

[mighty tree]

er yea idk what this is

i have not learn set union notation yet and stuff

[buoyant yoke]

,w d(x*sqrt(x+4))/dx

[grand pond]

Results provided by WolframAlpha

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

Requested by .bonk__

[mighty tree]

eh wat

oh wait

nvm careless mitake mb

er wat

i got

2x+9/2sqrt(x+4)

[buoyant yoke]

!show

[midnight plank]

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

[buoyant yoke]

you forgot an x on the 3rd line

[mighty tree]

huh where did the x come from

isiint d/dx x=1

or d/dx (x+4)=1

[mighty tree]

oh my handwrigint mb