help-13

[dire geode]

List 15.3.1 here
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/15%3A_Group_Theory_and_Applications/15.03%3A_Permutation_Groups

take the permutation group of {1, 2, 3} (which has cardinality 3! = 6) and calculate products of elements until you find one that doesn't commute.

[cedar kiln]

@tired ravine Has your question been resolved?

[tired ravine]

I don't have. I know how to do what's on the list 15.3.1. But I don't know how to use it to prove those statements.

[foggy merlin]

you just need to show that S_A is a group right ?

[dire geode]

@tired ravine did you do this yet?

[cedar kiln]

@tired ravine Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Again ?

[violet flume]

[crimson sedge]

Wasn’t this the same question u posted earlier

[bold hinge]

Ok this is simple! Use synthetic division

[crimson sedge]

Or long division

[bold hinge]

That works too

But I find it takes a lot longer

[leaden plover]

yes

[crimson sedge]

Oof

[bold hinge]

You see the only thing that really matters about polynomials is their coefficients

[crimson sedge]

TRU

[bold hinge]

I dont personally remember how to do synthetic division

If you actually asked me to do this problem rn I would multiply it out

[hollow osprey]

thats what im sying

this is literally the exact same question he asked before

[bold hinge]

Oh really? Was it not answered

[jaunty mural]

hi

@leaden plover do u not know how to do whatever division

[leaden plover]

synthetic division? no

i skipped one year of math

[jaunty mural]

ok let me demonstrate... again

this time no mistakes

$$f(x) = 2x^3 - 24x^2 + 44x - 40$$

[wraith dagger]

Shuri2060

[jaunty mural]

Now, I know x-10 is a factor

ok?

?

[leaden plover]

ok

[jaunty mural]

$$2x^3 - 24x^2 + 44x - 40 = (x - 10) (...$$

So I know it can be written like this

to start with

and I need to complete that right bracket

So in order to get 2x^3

What do I need for the 1st term?

[wraith dagger]

Shuri2060

[jaunty mural]

any clue?

[leaden plover]

nope

[jaunty mural]

a cubic

will be a linear times a quadratic

In order to get an x^3 term

it has to be something times the x

$$2x^3 - 24x^2 + 44x - 40 = (x - 10) (2x^2+...$$

[wraith dagger]

Shuri2060

[jaunty mural]

It's gotta be this

if that makes sense?

[leaden plover]

you know what, i appreciate your help, but i think i just need to rewind and figure out what i need to do first.

.close

[jaunty mural]

ok...

[cedar kiln]

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[bold hinge]

Lol

[cedar kiln]

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[crimson sedge]

How would I go about solving this? I have a formula that I think might work for this problem, however I am unsure of how I will find the derivative of V to help solve the question (if that’s needed).

[crimson sedge]

It is in the photo, I can type it out if you can't see it

[untold owl]

find r as a function of h using pythagoras theorem and differentiate wrt time
I think that should work

since you have dh/dt and h given

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

@untold owlWhat would r as a function of h using pythagoras theorem look like in this case? Because I know I can substitute r and h for a and b, but what about c? is that just a constant or??

[untold owl]

take the triangle above the water surface
it has sides (10-h), r and 10

[crimson sedge]

Why is (10-h) the side length?

Wouldn't it be h/2

[untold owl]

the line segment from the centre to the end of the sphere is the radius

which is 10

and we remove h from it since we need the segment till the surface

[crimson sedge]

ah I see

okay I will try and get back to you in a moment

@untold owl thank you!

[cedar kiln]

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[ripe valve]

Hi, I keep on getting 4.8 and not 10. Can I pls know where I went wrong?

[tropic oxide]

how are you getting a = 4.8? @ripe valve

[ripe valve]

333+1/3

divide 1/3

cube root the answer

oh shoot i see where i went wrong

thank you thank you

.close

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[jaunty mural]

Challenge:

================

Find all real solutions to $\cos x\tan x - \cos x = 0$.

$(\forall x\in\bR)$
$$\cos x\tan x - \cos x = 0$$
$\implies$
$$\cos x(\tan x - 1) = 0$$
$\implies$
$$\cos x = 0\lor\tan x = 1$$
$\implies$
$$x\in\left{\frac{\pi}{4},\frac{\pi}{2}\right}+n\pi, n\in\bZ$$

================

Find what is wrong with this proof.

[wraith dagger]

Shuri2060

[glad kestrel]

i know nothing of proofs but tan(x) is undefined at pi/2 lol

[tropic oxide]

[glad kestrel]

i also can't tell what you're changing every time you edit it

it looks the same

[jaunty mural]

rip, another time

.close

[cedar kiln]

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[south wren]

This is a linear algebra problem. I need to find for S and T and then make the Matrix Consistent.

I have found that the determinant of the Matrix is 0. So I know that the System has either an infinite number of solutions or no solutions.

However, I've played around with a few methods but I am not confident in what I have done.

[south wren]

Attempt 1

Attempt 2

Attempt 3

It was at that point I thought that something wasn't right and I threw the matrix into Excel to find the determinant and it came back 0

So now I don't really know where to go from here to find S, T or make the matrix consistent

So any and all guidance and assistance would be appreciated.

[fallen heath]

#❓how-to-get-help :o This channel is occupied..

[versed kayak]

@opaque ridge #❓how-to-get-help

[fallen heath]

Perform two individual operations $R_2$ <- $R_2 + R_1$ and factor out 2, and $R_2$ <- $R_2 - R_1$ :o

[wraith dagger]

Ansh

[fallen heath]

You'll directly get the values of s and t

[south wren]

😟

I have never seen the upside down !- operator before.

I'll look that up, read it, try your method - do you mind giving me a couple of minutes in case I have more questions, @fallen heath ?

[fallen heath]

Uhh it was "<-" the latex made it that way

[south wren]

Ahhh rock on.

[fallen heath]

Also... don't bother :o
Just look at your equations, no need of matrices
Add Eqn. 1 and 2 for s, and subtract 1 from 2 for t

[south wren]

That looks about right. Thanks for that! Couldn't see the forest through the trees.

[fallen heath]

Hey no >_< s should be 1 as x_1 + x_4 is s

t is right though

[south wren]

Aawwww man I suck. Thanks for the pick up!

I might go and have a cold shower or something. I'm 8 hours into trying to decipher my lecturers presentation. Thanks @fallen heath - You're a good egg!

!close

[cedar kiln]

@south wren Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Hello

[crimson sedge]

How to solve 7th question?

<@&286206848099549185>

.close

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[crimson sedge]

I believe (a) is true but (b) is false. But can we prove it mathematically?

[south tundra]

Show that the limit of (f(x + h) - f(x))/h doesn't exist at x = 0

So basically show that limit of (f(h) - f(0))/h doesn't exist as h approaches 0

[crimson sedge]

What about 2nd part? For f is integrable?

[south tundra]

Oh wait lol I read the question wrong

[crimson sedge]

Yea, limit exists.

[south tundra]

Hm, not really sure how to do that

[bright surge]

it's 0?

[south tundra]

nvm

[wraith crypt]

it's definitely differentiable

[bright surge]

it is, indeed lazy solution: where else would the f' in b come from

also, maybe look up the (first, whatever) fundamental theorem of calculus

[wraith crypt]

LOL

[elfin hemlock]

One would hope int f'(x) dx = f(x) wouldn't they?

that would be a nice property

[crimson sedge]

Oh yeah.

Thank you!

.close

[cedar kiln]

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[cedar kiln]

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[gentle lintel]

let’s say f(x) has a slant asymptote mx + b at +inf

[gentle lintel]

why is it that lim_(x → ∞) (f(x)/x) = m?

[gentle lintel]

this means lim_(x → ∞) (f(x) - mx - b) = 0 right

we can’t divide everything by x, so whats next

[fallen heath]

$\lim \limits_{x \to \infty} (f(x) - (mx+b) ) = 0$ is what actually is meant by slant asymptote at infty

[gentle lintel]

yea

[wraith dagger]

Ansh

[fallen heath]

so yeah, divide the L.H.S. by x :o

[gentle lintel]

??

[fallen heath]

what's wrong with it?

[gentle lintel]

how do we divide that into the limit

[fallen heath]

x is not infty... it's a number approaching infty

[gentle lintel]

i know

we get limit/x = 0

how do we bring x into the limit?

or we divide by lim_(x → ∞) x instead

[fallen heath]

hmm... that wouldn't work

[gentle lintel]

why

rhs = 0

[fallen heath]

Yes

okay so if you divide by it instead...

you get

$\frac{\lim \limits_{x \to \infty} (f(x) - (mx+b) )}{\lim \limits_{x \to \infty} x} = 0$

[wraith dagger]

Ansh

[gentle lintel]

yea

we use limit properties to rid b and bring m to rhs

i guess that’s how that works?

idk, unsure

[fallen heath]

the only doubt is if this is valid or not

oh wait.. right 😂 it's valid but rewriting it as:

[gentle lintel]

0/(big) = 0

right..?

[fallen heath]

$\lim \limits_{x \to \infty} \frac{(f(x) - (mx+b) )}{x} = 0$ is invalid

[wraith dagger]

Ansh

[fallen heath]

because... the quotient law won't allow it

[gentle lintel]

wait i dont think dividing by this is well defined

[fallen heath]

you need to have a finite limit of the denominator in order to be able to write it

[gentle lintel]

we can split and combine limits iff both limits exist

lim_(x → ∞) x apparently isn’t

[fallen heath]

Yes

[gentle lintel]

so how do we do this

[fallen heath]

lemme repost the question:

If slant asymptote means: $\lim \limits_{x \to \infty} (f(x) - (mx+b) ) = 0$, then how do we get to $\lim \limits_{x \to \infty} \frac{f(x)}{x} = m$

[wraith dagger]

Ansh

[gentle lintel]

yes

rigorously

(hopefully without epsilon delta)

a

[fallen heath]

Oh right :o

@gentle lintel

maybe like how $\lim \limits_{x \to 0} \sin x - x = 0$ can also be given as $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$

[wraith dagger]

Ansh

[fallen heath]

but :c sin x - ax would hab the same limit T_T

hmm!

[elfin hemlock]

difference 0 implies quotient 1

not the other way

[fallen heath]

then we're allowed to do it right?

[elfin hemlock]

Do what?

[fallen heath]

Writing $\lim \limits_{x \to \infty} \frac{(f(x) - (mx+b) )}{x} = 0$

[wraith dagger]

Ansh

[gentle lintel]

?

[fallen heath]

given, $\lim \limits_{x \to \infty} (f(x) - (mx+b) ) = 0$

[wraith dagger]

Ansh

[fallen heath]

wait lmfao 😂 why the heck not

[gentle lintel]

huh

[elfin hemlock]

I mean 0/inf is always 0

[fallen heath]

Yeah

[gentle lintel]

.

[fallen heath]

smh I must've banged my head

It does

[gentle lintel]

??

[fallen heath]

$\lim \limits_{x \to \infty} \frac{1}{x} = 0$

[wraith dagger]

Ansh

[fallen heath]

:o

[gentle lintel]

are we figuring out workarounds lol

[fallen heath]

No!

I mean

given, $\lim \limits_{x \to \infty} (f(x) - (mx+b) ) = 0$

[wraith dagger]

Ansh

[gentle lintel]

yea?

[fallen heath]

you can write $\lim \limits_{x \to \infty} \frac{(f(x) - (mx+b) )}{x} = 0$

[wraith dagger]

Ansh

[gentle lintel]

??????

what the fuck?

are there any in-between steps

[elfin hemlock]

.

[gentle lintel]

main concern is lhs

[gentle lintel]

this

[gentle lintel]

and this

[elfin hemlock]

0/inf is 0

agree?

that is all you need to use

you can't just divide by lim of a variable no

but no need to even do that

[gentle lintel]

yea

so how come spontaneously /x?

[fallen heath]

no.. screw that.. Just consider a separate function $g(x) = \frac{(f(x) - (mx+b) )}{x}$

[wraith dagger]

Ansh

[elfin hemlock]

do you agree with this?

[fallen heath]

and evaluate it's limit at infty :o

[gentle lintel]

yea

[elfin hemlock]

all you to agree with

[gentle lintel]

its what i thought

[elfin hemlock]

then clearly it is equal to 0

[gentle lintel]

sure

then?

[elfin hemlock]

Idk what the plan is with that but just saying that is valid

[gentle lintel]

lol

[gentle lintel]

please continue

[fallen heath]

chromium is confused about why is it valid

[elfin hemlock]

Is how I would do it

[gentle lintel]

no

[fallen heath]

no?

[gentle lintel]

i understand

im just confused about well defined ness

and weird bs

[fallen heath]

you do?

[gentle lintel]

not this

[fallen heath]

hmm

[gentle lintel]

just this

[elfin hemlock]

limit of num is 0

limit of denom is inf

[gentle lintel]

i wanna see your g(x) approach

[elfin hemlock]

0/inf=?

[fallen heath]

So... what's the lim g(x) as x tends to infty?

[gentle lintel]

depends on f

[fallen heath]

Is it an indeterminate form?

dude

[elfin hemlock]

no it isn't

0/inf=?

[fallen heath]

f is same from before

$\lim \limits_{x \to \infty} (f(x) - (mx+b) ) = 0$

[wraith dagger]

Ansh

[gentle lintel]

forget that

i suck

[elfin hemlock]

0/inf is what?

[gentle lintel]

alright so.

[elfin hemlock]

so clearly the limit is 0 like I have said 10 times now

[gentle lintel]

we have this

[fallen heath]

Yes we have

[fallen heath]

this

and I'm asking

[gentle lintel]

0..?

i’m not sure

at least i cant mathematically prove it

[fallen heath]

why are you not sure

[gentle lintel]

im thinking splitting it into

0 numerator

inf denominator

[fallen heath]

oh

[gentle lintel]

but that doesnt work

[fallen heath]

Why not?

0/inf

is ?

[gentle lintel]

???

lim (f/g) = (lim f)/(lim g) iff both limits are well defined

lim at inf of (x) is apparently not

[fallen heath]

Nooo ._. we're just trying to check if the limit even needs to be solved... It doesn't even have an indeterminate form

0 / inf is just 0...

[gentle lintel]

i know

i know

0/(BIG SHIT)

= 0

[gentle lintel]

but i cant prove it right here

[fallen heath]

so as x tends to infty... g(x) turns into 0/(BIG SHIT)

[elfin hemlock]

wdym you can't prove it?

you just said you agreed 0/inf=0

can you decide once and for all, do you agree with that YES or NO?

[gentle lintel]

how do you show that it’s 0 with limit properties or something

[elfin hemlock]

if YES then clearly that limit is 0

if NO then clearly say so

[gentle lintel]

i dont see it

[elfin hemlock]

denom approaches inf?

[gentle lintel]

yes, but not clear to me

[elfin hemlock]

num approaches 0?

[gentle lintel]

we can’t split it into num and denom with their respective limits

[elfin hemlock]

not clear that $\lim_{x \to \infty} x=\infty$?

[wraith dagger]

ScapeProf

[gentle lintel]

i dont think that’s well defined

∞

[elfin hemlock]

Yes it is

[gentle lintel]

so we can’t have lim (0/x) = 0/(lim x) = 0

[gentle lintel]

huh

[elfin hemlock]

what?

[gentle lintel]

we won’t consider it an existent limit then

a limit exists if it approaches some real L

[elfin hemlock]

you do normally

[gentle lintel]

∞ isn’t a real number

[elfin hemlock]

you might use a weird book that doesn't

[gentle lintel]

can you show the steps that make lim (g(x)) = 0?

[elfin hemlock]

[\lim_{x\rightarrow\infty}=\frac{f\left(x\right)}{h\left(x\right)}]
Knowing that $\lim_{x\rightarrow\infty} f(x)=0$ and $\lim_{x\rightarrow\infty} h(x)=\infty$ then the limit is on the form "$\frac{0}{\infty}$". This is NOT AN INDETERMINATE FORM
For example consider rewriting it as
[\lim_{x\to\infty} f(x)\frac1{h(x)}]
where $\lim_{x\to\infty} f(x) = 0$ and $\lim_{x\to\infty} \frac1{h(x)}=0$

[gentle lintel]

what is that first line???

[elfin hemlock]

for two arbitary functions where the condition on the next line is met

[wraith dagger]

ScapeProf

[gentle lintel]

????

[elfin hemlock]

.

Like I said

[fallen heath]

$\lim \limits_{x \to \infty} g(x) = \lim \limits_{x \to \infty} \frac{(f(x) - (mx+b) )}{x} = \lim \limits_{x \to \infty} ( f(x) - (mx+b) ) \cdot \lim \limits_{x \to \infty} \frac{1}{x} = 0 \cdot 0 = 0$

[wraith dagger]

Ansh

[elfin hemlock]

seems like you really need to recheck your limit algebras (and what a limit even is)

[gentle lintel]

oh

[gentle lintel]

i constantly am

[fallen heath]

Do you get that?

[gentle lintel]

yes

[elfin hemlock]

so do you agree with this now?

like I have said 15 times?

[gentle lintel]

yes

this was all i needed lol

[fallen heath]

Lol

[gentle lintel]

now how do we obtain m

[fallen heath]

Now

$\lim \limits_{x \to \infty} \frac{(f(x) - (mx+b) )}{x} = 0$

[wraith dagger]

Ansh

[fallen heath]

$\implies \lim \limits_{x \to \infty} \frac{f(x)}{x} - m = 0$

[wraith dagger]

Ansh

[gentle lintel]

oh

[fallen heath]

now ask me why $\lim \limits_{x \to \infty} \frac{f(x)}{x}$ is valid?

[wraith dagger]

Ansh

[gentle lintel]

i get that

so in general

mx + b can be a parabola

or any weird function

idk

in terms of asymptotes

[fallen heath]

mx+b can never be parabola (@_@;)

[gentle lintel]

jdjdjsnddnono

no

fuck

[fallen heath]

what are you on...

[gentle lintel]

here, mx + b is an asymptote right

[fallen heath]

Yeah

[gentle lintel]

said asymptote can be anything

parabola or something else

and we apply similar techniques

[fallen heath]

No, like an asymptote can only be a line

[gentle lintel]

huh

oh

non-line asymptotes are too unimportant?

[fallen heath]

oh right... okay srry

Yes, so hmm..

[fallen heath]

Yes

But you gotta be reasonable with each step (@_@;)

[gentle lintel]

im about to start integrals and here i am failing simple limits lmfao

anyway, thanks for bearing me

.close

[cedar kiln]

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[fallen heath]

@ moderators

;-;

[cedar kiln]

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[crimson sedge]

Is there any formula for $\int \frac{dx}{\sqrt{x^2-a^2}}$?

[wraith dagger]

!!!

[crimson sedge]

Or do I have to solve these type of questions using substitution?

[south tundra]

I mean you could derive the formula yourself using the substitution

[crimson sedge]

Is this result true? $\int\frac{dx}{\sqrt{x^2-a^2}} = \ln(|x + \sqrt{x^2 - a^2}|) + c$

[wraith dagger]

!!!

[south tundra]

Yeah seems to be correct

[fallen heath]

Yes it is

:o

[crimson sedge]

Ok

.close

[cedar kiln]

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[cedar kiln]

@stable kelp Has your question been resolved?

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[steel ledge]

When changing a basis of matrix, when do we simply multiply the matrix by the change-of-basis matrix, and when do we use the formula ?

[steel ledge]

This formula:

[jaunty mural]

If I have transformation T

Multiplying by the transformation matrix will send vectors in basis A to basis A, let's say.

$$M_{AA}x_A=T(x)_A$$

[wraith dagger]

Shuri2060

[jaunty mural]

Not standard notation by me.

If you want to input a vector in basis A but output the vector in basis B

or the other way round

I think this is what you're referring to?

$$M_{BA}x_A=T(x)_B$$

[wraith dagger]

Shuri2060

[jaunty mural]

Lets say we use this notation to represent that

Then you just need to transform basis once

[steel ledge]

to give more context to my question, my current exercise is asking me to express T from the canonical basis of B (start) into the basis of B (arrival) ? and here I only used the transform basis matrix

but in a previous exercise, expressing T in regards to a basis, I had to use a formula

I just don't see the difference between those questions

[jaunty mural]

Can you show me the original french

[steel ledge]

yes okay

[jaunty mural]

but i doubt it helps me

[steel ledge]

haha

Given T : R^3 -> R^3 the linear application defined by (...)
Given B_can the canonical basis of R^3 and B the canonical basis of R^3 given by (...)
Give the matrix that represents T in regards to the bases: (...)
départ : start / arrivée : arrival

[jaunty mural]

Yes

[steel ledge]

What I did here: I expressed T in the canonical basis to get its associated matrix. Then I calculated the changing matrix from B (start) to the canonical basis. I originally wanted to use the formula, but we only multiply T and the changing basis matrix

[jaunty mural]

$$M_{BA}x_A=T(x)_B$$

[wraith dagger]

Shuri2060

[jaunty mural]

So this is what you need

the starting vector is written in basis A

and then it needs to end in basis B

yes?

[steel ledge]

yes

[jaunty mural]

So your original matrix

will transform x written in A to x written in A

Then you need to transform the final to x written in B

So you apply change of basis once

[steel ledge]

this makes sense to me, but then when do we need to use the formula?
In the previous exercise we did, they asked us to "express the linear application T in regards to the basis B". We also expressed T in the canonical basis there. Technically, doesn't that also mean T is in a basis (canonical) and needs to end in basis B ?

[jaunty mural]

Before, we had matrix for x_A to (Tx)_A

We need to change the matrix so it is

x_B to (Tx)_B

We need to move x_B to x_A first

Then use our matrix for x_A to (Tx)_A

Then we need to move the result from (Tx)_A to (Tx)_B

x_B -> x_A -> (Tx)_A -> (Tx)_B

This is our process

We do this because our matrix is written in basis A

[steel ledge]

Thank you! this makes a lot of sense. I just don't think I will be able to recognize this pattern myself. How do I know when I need to move x_B to x_A ?

[jaunty mural]

You need to think about what the starting basis and end basis

x_A -> (Tx)_A -> (Tx)_B

x_B -> x_A -> (Tx)_A

So these 2 cases are possible like in your question

You need to know which basis your matrix works in

[steel ledge]

okay maybe I'm wrong but I think this is also what you're saying... x_B -> x_A is when we express the vectors of the first basis depending on the vectors of the second matrix right?

[jaunty mural]

no

x_B is expressed in basis B

x_A is expressed in basis A

or did i misunderstand u

[steel ledge]

i mean the transformation from x_B to x_A !

[jaunty mural]

I guess yes actually but uhh

xB -> xA

You multiply by P_AB matrix

from your notes

so yes

You write the new vector as you said.

[steel ledge]

x_A -> (Tx)_A -> (Tx)_B
x_B -> x_A -> (Tx)_A
Then in the second one wouldn't we just have x_B -> x_A without needing (Tx)_A ?

[jaunty mural]

Im giving different examples

If they tell you the transformation matrix in the new basis. And want you to give answer in new basis

But you need to start in old basis

That is the 2nd one

[steel ledge]

okay just a sec i'm wrapping my head around it 😆

okay i have a question. Is the canonical basis considered a basis like B or C or is it "neutral" ?

[jaunty mural]

It is like any other basis

[steel ledge]

Right. So when we express T in canonical basis, aren't we moving T from a basis to another ?

to get the matrix

[jaunty mural]

what do u mean

[steel ledge]

if we are given a linear transformation of T, and we use the canonical basis vectors to get the matrix of T

[jaunty mural]

$$M_{EE}x_E=T(x)_E$$

[wraith dagger]

Shuri2060

[jaunty mural]

But this applies to any basis

To get the matrix of T in a basis A

you check where T sends the basis vectors

Yes, you can consider the transformation T to be 'moving basis' if T is bijective, I suppose.

But I'm not sure if this necessarily extends outside of R^n

T is a linear transformation
I would consider it different from changing basis? perhaps

Even if it is the same for vector spaces, it may not be the same for generalisations of vector spaces.

[steel ledge]

okay I see thank you! I think the thing I was confused by was what exactly do we do when we say x_B -> x_A and T(x)_B -> T(x)_A .

We have said x_B -> x_A is doing (x_B)*P_AB = x_A.
What is T(x)_B -> T(x)_A ?

sorry i will learn to use latex

[jaunty mural]

You left multiply by P_{AB} to go from basis B to basis A

T(x) is just a vector

like x

$$P_{AB}[x]_B = [x]B$$
$$P{AB}[T(x)]_B = [T(x)]_B$$

[wraith dagger]

Shuri2060

[jaunty mural]

This is formal notation I have seen from my own notes

It may differ from lecturer

[steel ledge]

okay ! so this is to change vector from basis B to basis A

right ?

[jaunty mural]

yes

you see the concept of vector x

exists outside of writing it down in coordinate form

When we write it down in coordinate form, we are using a basis to express it

$$x$$

this is a vector

[wraith dagger]

Shuri2060

[jaunty mural]

$$[x]_B$$

[wraith dagger]

Shuri2060

[jaunty mural]

This is vector x written in coordinates in basis B

[steel ledge]

ohh

[jaunty mural]

when i say x = (1 2 3)

this is technically improper

i should be saying [x]_B = (1 2 3)_B

but obviously there is no need to do this all the time if it is clear

In other subjects everything is euclidean, so why bother.

[steel ledge]

yeah!! that's all very clear to me . but then I'm wondering why is there a need for this formula at all if the changing basis matrix alone does the job of changing a vector's basis ?

[jaunty mural]

I am starting x in basis B

Then I change to basis C

[steel ledge]

yes

[jaunty mural]

Then I apply the matrix which is from basis C to C

[steel ledge]

why do we need the last one at all

[jaunty mural]

Then I change the result from C to B

Because [T]_B

represents a matrix

such that you input a vector in basis B

and output a vector in basis B

[steel ledge]

okay

if we were only to multiply T_C by the changing basis matrix from C to B, we would get T_C expressed in basis B but it wouldn't give us an output in B ?

[jaunty mural]

ehhh

right idea but i dont think u got your letters right

[steel ledge]

oh that's possible

wait i think i get it

[jaunty mural]

$$[T]{C}:=[T]{CC}$$

[wraith dagger]

Shuri2060

[jaunty mural]

Lets say we define this shorthand

[steel ledge]

Original:
T(bonjour) = T(au revoir)

Multiplying by the changing basis matrix
T(hello) = T(au revoir)

Using the formula
T(hello) = T(goodbye)

is that it ?

[jaunty mural]

this is good analogy but ehhh haha

[steel ledge]

haha

[jaunty mural]

yes

[steel ledge]

okay oh my god i get it thank you so much. So when they ask us to express T in another basis, they don't care about the output and that's the difference

[jaunty mural]

$$[T]{C}P{CB}:=[T]_{CB}$$

[jaunty mural]

ehhh no

[wraith dagger]

Shuri2060

[jaunty mural]

When they say to express T in basis A

they want $[T]_{AA}$

[wraith dagger]

Shuri2060

[jaunty mural]

or $$[T]_A$$

[wraith dagger]

Shuri2060

[jaunty mural]

same thing.

Your question is different

Your homework

it is telling you specifically

what inputs and outputs it expects

[steel ledge]

oh i understand this! i meant in general

[jaunty mural]

In general when you write T in basis A it means this

(Sorry, T needs a basis. The matrix does not. The matrix represents T in a basis.)

[steel ledge]

okay !

but i understand now, i'm pretty sure

so thank you so much for taking the time

[jaunty mural]

👌

[steel ledge]

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[crimson sedge]

is my book wrong here? in the third image i quote "Thus the cycles of theta are (1, 2), (3), (4, 5, 6)". i think the author confused cycles with equivalence classes. all the cycles of theta would infact be (1, 2), (2, 1), (3), (4, 5, 6), (5, 6, 4), (6, 4, 5) since while equivalence classes have to be disjoint from each other, to differentiate two cycles, the order simply has to be different (e.g (1, 2) ~= (2, 1))

[crimson sedge]

fyi orbits are equivalence classes

cycles however i dont think so

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[jaunty mural]

what are equivalence classes in this context

[jaunty mural]

k

[crimson sedge]

S subscript n is a permutation group

[jaunty mural]

i think its just lazy writing

[crimson sedge]

in other words the author is wrong?

lol

[jaunty mural]

{1, 2}, {3}, {4, 5, 6}

are the classes

typo or abuse of notation

[crimson sedge]

ik that but what would the cycles of theta be

[crimson sedge]

in this example

[jaunty mural]

what???

{1, 2}, {3}, {4, 5, 6} are the classes

so the cycles are those in some permutation

[crimson sedge]

.close

[cedar kiln]

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[cedar kiln]

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[uneven epoch]

!help

[uneven epoch]

If $R_1$ and $R_2$ are two irreps of $\mathrm{SU(N)}$ and $T^a_{R_j}$ the generators in a given rep, is the following true? How?

$$\mathrm{Tr} (T^a_{R_1 \oplus R_2}T^b_{R_1 \oplus R_2}) =\mathrm{Tr} (T^a_{R_1}T^b_{R_1} + T^a_{R_2}T^b_{R_2})$$

[wraith dagger]

Siupa

[dire geode]

maybe try #groups-rings-fields

[uneven epoch]

ok, thank you

.close

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[jaunty mural]

sketch it.

[smoky stump]

I am very confused on how to sketch it?

[jaunty mural]

check desmos

and see how functions

and their inverses

are related graphically

[smoky stump]

but what did the solution do?

[jaunty mural]

it explains?

[smoky stump]

I find it od how it takes it's lower limit -1

[jaunty mural]

it figures out donain and range

[smoky stump]

and then subs into the same equation and call it interesected

[jaunty mural]

they notice graphically -1 -1 is an intersection

idk

[smoky stump]

but then how would you draw that graph?

[jaunty mural]

first they work out inequality

the graph is given

[smoky stump]

okay and they align the inequality with the grraph gievn

[jaunty mural]

yh

[smoky stump]

okay thanks

.close

[cedar kiln]

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[glad kestrel]

learn on the interwebs

[jaunty mural]

god just stop.

This isn't google

[dense wing]

As I stated already.... discord isn't for lessons

Google, put some effort in if you want to learn tbh

[cedar kiln]

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[echo fjord]

Hi, I am modelling a gun for a game, when I started to model the bullet of the gun, I didn't found a lot of blueprint for the bullet.

The only one I found is the following. The problem is that some distances that are missing.

I am wondering is it possible to calculate a distance between 2 point with only the data I have ?
I am sorry for my lack of knowledge. Thanks you in advance < 3

[echo fjord]

In short: How to find the length of the Green ? Please help < 3

[jaunty mural]

look up what the sin, cos, tan

ratios are

[echo fjord]

What do you mean ?

[jaunty mural]

https://www.mathsisfun.com/sine-cosine-tangent.html

[echo fjord]

Thanks !

But is it possible to calculate ?

I mean, here is only 2 values

[jaunty mural]

Yes, otherwise I wouldn't have suggested.

[echo fjord]

Hooo, alright thanks ! I do not know anything abouts these, but I will try to solve it my myself then ty (way more easy than without knowing where to search) < 3 (Problem solved)

[cedar kiln]

@echo fjord Has your question been resolved?

[echo fjord]

I can't manage to find a logical result, I don't understand what I did wrong. Somehow I only find value close to 0.27 (which is not okay.

If you look the blueprint, the height of the side of the angle (0.27) is not even visible because how small it is.
The result is most than probably more than 10x 0.27) (all result I found are not what the result that should be). I am confused.

c is what I am looking for, I have only A and a. I even tried to put the value in a calculator but I guess I am doing something fundamentally wrong.

I hardly imagine the length been 0.97
What am I doing wrong....

Am I misreading units ? And the result is 9.7 ?

[obsidian coral]

I got 0.97

With that given info

[echo fjord]

I am so lost

[obsidian coral]

Using this, I also got 0.97

[echo fjord]

I am obviously doing something wrong

I don't know what

[obsidian coral]

What value are you getting?

[echo fjord]

I need to find the length of the green line

Basically I did 9.85 -9.58 (in red) to get 0.27 of difference

I have only 0.27 and 16.24°

as values

[obsidian coral]

Can you mark exactly where 16.24 is in that picture? In this picture

[echo fjord]

I'll zoom that part, 1 sec

[obsidian coral]

I mean in the part that has the rectangle around it

[echo fjord]

hum ?

basically the part i zoom is like this one (I suppose ? Maybe it's here I fkg up ?)

The black lines don't exist

it's to represent

[obsidian coral]

You see the rectangle stuff here? Where does 16.24 point to?

[echo fjord]

with zoom the angle since it's very small

[echo fjord]

it's the slope ?

I guess

Is it not ?

I supposed it is

[obsidian coral]

Because that gray line looks angled, so whatever triangle you tried to draw, in here, doesn't match with the rectangled stuff

[echo fjord]

I know how to read mechanical schematics but I learned it like 10 years ago, so....

[echo fjord]

So I am misreading the schematics ?

That would explain a lot of stuff

But that seems illogical too

tbh, omg my brain

[obsidian coral]

Give me a few minutes and I'll circle the areas that I'm referring to

[echo fjord]

Thanks you so much !!! < 3

I am very lost lol

I got it

I am actually reatd as fuck

it's 0.27 / 2

that's why

because it's both side

=.=

but still the result is wrong, wtf

[obsidian coral]

That line there looks sloped

[echo fjord]

haa

this line have no mean sorry

it was to represent that is was an angle

[obsidian coral]

So the black line on the left is 9.85mm and the black line on the right is 9.58mm?

[echo fjord]

yes

but it was 0.135 actually, since it was for both side

0.27 / 2 = 0.135

but still with this value it don't work

The drawing is an exaggerated view since the slope is so small, but now I am starting to doubt the angle

I think the schematics I have is a troll

It's not possible that the angle can ever be 16° with a height of 0.27

[obsidian coral]

It's hard to tell

Sorry

[echo fjord]

Don't worry, I am the one sorry

I think the schematics is wrong

16° angle is not possible with such light slope

math never lie

I just got trolled

[echo fjord]

Sorry, thanks you so much for the much tho !! < 3

[obsidian coral]

Well, I tried

[echo fjord]

16.24° angle, lol

Why I didn't saw that from the start that it was not logical

My issue is solved

[cedar kiln]

@echo fjord Has your question been resolved?

[cedar kiln]

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[ruby fulcrum]

x² + 8c + ▭
so here ik you gotta order it so its
x² + ▭ + 8c
in the question it says u gotta find the number inside the rectangle to make it a trinomial perfect square
but i dont know what to do because the 8 doesnt have an exact square root

[hollow osprey]

well we know that a square is in the form

(x+a)^2 = x^2 + 2xa + a^2

so if we let a^2 = 8c, then what would a be?

[ruby fulcrum]

so like the square doesnt exist its just where the number to make it a trinomial perfect square be

but as i said in the question i cant manage to find it because 8 doesnt have a perfect root

[hollow osprey]

8 doesn't need a perfect root?

you just need the x^2 + rectangle + 8c to be in the form of (x+a)^2

do we agree that:

$x^2 + 2\sqrt{2}x + 2$

is a perfect square?

[wraith dagger]

Yottachad

[ruby fulcrum]

yes it is a perfect square

[hollow osprey]

okay so then just apply the same pricinple as i said before

2 doesn't have a perfect square root but it still works right

[ruby fulcrum]

yea

thx

.close

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[steady wolf]

A teacher bought disposable masks to be distributed to students when they return to face-to-face activities. consider
there will be a total of N masks and K students. Also consider the following restrictions:

• Each student must receive at least A masks
• Each student must receive a maximum of B masks
• It is not necessary for all purchased masks to be distributed

Develop a mathematical expression that represents in how many ways such a distribution can be made, considering
as many restrictions as possible at the same time.

[steady wolf]

Im just straight up lost

<@&286206848099549185>

.close

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[white barn]

Can we ask about physics on here? Or is there a better place

[glossy halo]

Physics discord has question channels too

[white barn]

K

.close

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[wheat silo]

what do I do after this?

[jaunty mural]

expand abs u

[wheat silo]

ok am I on the right track? I simply can't see how I will get a vector

[jaunty mural]

well theres some things to notice

There are 2 possible directions for u

u can be scaled by any (positive?) amount.

not sure if youre allowed to scale negatively

depending on what angle between means

[wheat silo]

something like this

or the other way around

[jaunty mural]

Do you see the problem?

you have an infinite number of answers

Either set u_1 or u_2 to be 1

or continue what you're doing, but rearrange to find the ratio between the 2

[cedar kiln]

@wheat silo Has your question been resolved?

[wheat silo]

thanks

.close

[cedar kiln]

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[cedar kiln]

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[lethal vale]

where is 24, 28 coming from

err

idk why i said that lol

[ember geode]

doing this matrix multiplication

[lethal vale]

1 sec

lemme do it

[ember geode]

[lethal vale]

yep

got it

tyvm

really on point today for the assistance

good sir

.close

[cedar kiln]

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[kind stirrup]

Could someone explain to me number 3? I'm really stuck on average rate of change

[glossy halo]

What have you gotten so far?

[cedar kiln]

@kind stirrup Has your question been resolved?

[glossy halo]

So far so good!

[kind stirrup]

Is this right????

[glossy halo]

Looks good, not sure about the "3(h+6x)" since it would be "3h+6x" or "3(h+2x)"

[kind stirrup]

Oh yeah that part was just me writing i wrote the answer below it

Thank you!

[cedar kiln]

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[glossy zenith]

https://cdn.discordapp.com/attachments/919906407292227604/933608951512170506/unknown.png

[glossy zenith]

what i got so far

ik 729^5/6 is 243

but how do i solve g and y

[glossy halo]

you dont - you need r and s which are the exponents of g and y

[glossy zenith]

oh

oh

so i just do

243, g^10/3, y^5/2

after i simplify

and those are my answers

[glossy halo]

Looks like it

[glossy zenith]

would it be

y^4*2/7

[glossy halo]

should be

wait

[glossy zenith]

im thinking 4*2/7 because

the y in the numerator should cancel one y out

so instead of y * y * y* y * y

its just y * y * y * y

[glossy halo]

ye 8/7

[glossy zenith]

..

isnt ^0 always 1

[glossy halo]

Yes, but it asked what the exponents are

[glossy zenith]

oh

what happens to the exponents then

they become 0?

4*0=0

and that for the rest

[glossy halo]

Yeppers

Best of luck with the rest - I am going to sleep

[glossy zenith]

alr thanks for the help

[cedar kiln]

@glossy zenith Has your question been resolved?

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[gray vine]

how do i determine $58\sin \left(x\right)\cdot \cos \left(x\right)$ without calculator, if $2\sin \left(x\right)=5\cos \left(x\right)$

[wraith dagger]

Theophania

[gray vine]

we know that sin2x=2sinxcosx, but maybe that's not relevant here

[tropic oxide]

you're right, it is not relevant here.

you might want to find tan(x)

[gray vine]

$\frac{2\sin \left(x\right)}{5\cos \left(x\right)}=1$

[wraith dagger]

Theophania

[gray vine]

like this? i don't know how we can find it otherwise

[tropic oxide]

that is a bit indirect but what you've just written amounts to $\frac{2}{5} \tan(x) = 1$

[wraith dagger]

Ann

[mighty mortar]

... I feel like you're supposed to use the Pythagorean identity...

[gray vine]

$\frac{2}{5}\tan \left(x\right)=\sin ^2\left(x\right)+\cos ^2\left(x\right)$

[tropic oxide]

i mean, yes, but there's nothing wrong with doing a little preparatory work here

[mighty mortar]

Square the second equation and use sin^2 + cos^2=1

[wraith dagger]

Theophania

[mighty mortar]

I guess.

[gray vine]

like this tsumiki?

[tropic oxide]

this will do you no good.

[mighty mortar]

nOT THERE

[tropic oxide]

what i was going to suggest next if i wasn't interrupted

was to write $\sin(x) \cos(x)$ as $\frac{\sin(x)}{\cos(x)} \cdot \cos^2(x)$

[wraith dagger]

Ann

[gray vine]

$$58\sin \left(x\right)\cos \left(x\right)=58\frac{\sin \left(x\right)}{\cos \left(x\right)}\cdot \cos ^2\left(x\right)$$

[wraith dagger]

Theophania

[tropic oxide]

and $58 \frac{\sin(x)}{\cos(x)} \cdot \cos^2(x)$ is in turn equal to $58 \tan(x) \cdot \frac{1}{\tan^2(x) + 1}$, and the application of the pythagorean identity lies (somewhat indirectly) in transforming $\cos^2(x)$ into $\frac{1}{\tan^2(x) + 1}$

[wraith dagger]

Ann

[tropic oxide]

this is the identity im talking about: $\tan^2(x) + 1 = \frac{1}{\cos^2(x)}$

[wraith dagger]

Ann

[gray vine]

i'm not familiar with $\cos ^2\left(x\right)=\frac{1}{\tan ^2\left(x\right)+1}$ and $\tan ^2\left(x\right)+1=\frac{1}{\cos ^2\left(x\right)}$

[wraith dagger]

Theophania

[gray vine]

so i will try to see how we get them step by step

[wraith dagger]

Ann

[gray vine]

ah okay.

$$2\sin \left(x\right)=5\cos \left(x\right)$$
$$\frac{\sin \left(x\right)}{\cos \left(x\right)}=\tan \left(x\right)=\frac{5}{2}$$

[wraith dagger]

Theophania

[gray vine]

now that we know what tan(x) is, we should use it here, i assume?

[tropic oxide]

no reason to assume anything

as i explained earlier you have $58 \sin(x) \cos(x) = \frac{58 \tan(x)}{\tan^2(x) + 1}$; you may reread what i've written to explain to yourself why this is true if you feel it's necessary, but said explanation need not (and in fact probably should not) rely on the known value of $\tan(x)$.

so you can (and should) insert the known value of tan(x) directly into the expression whose value you seek

[wraith dagger]

Ann

[gray vine]

$$58\sin \left(x\right)\cos \left(x\right)=58\frac{\sin \left(x\right)}{\cos \left(x\right)}\cdot \cos ^2\left(x\right)=58\tan \left(x\right)\cos ^2\left(x\right)$$
$$\cos ^2\left(x\right)=\frac{1}{\tan ^2\left(x\right)+1}$$
$$58\sin \left(x\right)\cos \left(x\right)=58\frac{\sin \left(x\right)}{\cos \left(x\right)}\cdot \cos ^2\left(x\right)=58\tan \left(x\right)\cdot \frac{1}{\tan ^2\left(x\right)+1}$$

[wraith dagger]

Theophania

[gray vine]

$= 58\cdot \frac{5}{2}\cdot \frac{1}{\left(\frac{5}{2}\right)^2+1}$

[wraith dagger]

Theophania

[gray vine]

$=29\cdot 5\cdot \frac{1}{\frac{25}{4}+\frac{4}{4}}$

[wraith dagger]

Theophania

[gray vine]

(145/1)/(29/4) = (145*4)/(29)

= 580/29 = 20

,w 58sinxcosx, 2sinx=5cosx

[wraith dagger]

Results provided by WolframAlpha

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Requested by Theophania#1704

[gray vine]

thanks

.close

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[cedar kiln]

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[tender warren]

 Study the monotonicity of the sequence of partial sums for a series
with positive terms.

Isn't it always increasing because the terms are positive?
and so Sn+1 - Sn > 0

[crimson sedge]

Partial sums?

[tender warren]

yes

for SPT

[cedar kiln]

@tender warren Has your question been resolved?

[tender warren]

@crimson sedge

[crimson sedge]

Maybe they mean the term itself

Like 1,1/2,1/3,1/4...

I honestly don't know

[tender warren]

mhm

ok

thanks

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[crimson sedge]

how does this work?

u = e^t

ohh nvm

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[jaunty mural]

lol

[crimson sedge]

where did the u^2 on the denominator come from?

[jaunty mural]

dt

du

probably

or u mean that step??

its just algebraic manipulation

[crimson sedge]

[jaunty mural]

putting 2 things under one denom

[crimson sedge]

could u show me an example pls?

I don't understand

it should be

(u^2 - 1)/u

sorry not u^2

u

[jaunty mural]

honestly, if you rearrange the left side

its hard not to get the right side

they just distributed the square over the fraction

[cedar kiln]

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[crimson sedge]

What formula do I use to find part b?

[crimson sedge]

I already found answers for part a.

[obsidian coral]

Heron's formula would work

[obsidian coral]

Also $Area = \frac{ab}{2} \sin \theta$ is a valid equation too

[wraith dagger]

dldh06

[crimson sedge]

ok

.close

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[proven sail]

Find the boundaries of the rows

<@&286206848099549185>

might not be the best translation of the instructions

[crimson sedge]

I think rationalisation would work

Multiply both numerator and denominator with $\sqrt[4]{n^2+n+1} + \sqrt[4]{n^2-n+1}$

[wraith dagger]

!!!

[dawn jetty]

I think you need 2 additional terms

since its a 4th root not a 2nd

but yes a conjugaet should work, yielding an O(1.5) function in both the nominator and denominator

a^n-b^n=(a-b)(a^n-1+b^n-1+ba^n-2+ab^n-2 . . .)

wild guess I think the limit if 1/2

[proven sail]

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[short heath]

60$ 30$ 17$

[crimson sedge]

You can create linear equations out of the given information.

[short heath]

What do you mean?

Its just easy math

[crimson sedge]

For the first case, assume car as c and ball as b. You can make a linear equation as 2c + 2b = 60. Now you can make linear equations from other two cases and solve for the variables.

[crimson sedge]

It's fundamental method.

[short heath]

Car=17
Ball=13
Bow tie=4

Is it right?

[crimson sedge]

Perfect

You can verify it by just substituting back the variables in your equations.