help-49

[calm quest]

yayay!!

[burnt idol]

its more to present the work for the person grading it.

[tribal temple]

Of course, I don't think it's too difficult to see that 15 is greater than 1

But that's pretty much it

[calm quest]

lmao

thank you so much for your patience in helping me understand!!

i really appreciate it

[tribal temple]

It's always a pleasure to see and help you and hopefully that helped

[calm quest]

thank you it definitely did!!

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[twilit field]

I would like some help proving the divisibility rule for 11

[twilit field]

A number is divisible by 11 iff the difference of sum of the even digits and odd digits is divisible by 11

Any hints?

We first consider $11 \mid a_1a_2\dots a_n$\ implies 11\mid (11-1)^{n-1}a_1+(11-1)^{n-2}a_2+ \dots + a_n$

[
11 \mid a_1a_2 \dots a_n \implies 11 \mid (11-1)^{n-1}a_1 + (11-1)^{n-2}a_2 + \dots + a_n
]

[grand pond]

A dense set(Ping when reply)

[twilit field]

ooh

[steep falcon]

You can write the number as a_n*10^n and so on

[twilit field]

yeh, that's what I've done

[viral dagger]

if you expand doesent that imply $11\mid a_1(-1)^{n-1}+a_2(-1)^{n-2}+\dots+a_{n-1}(-1)^1+a_n$

[grand pond]

Skissue ping4response

[steep falcon]

whats 10 mod 11?

[twilit field]

right

[twilit field]

1

[steep falcon]

-1

not 1

right?

[twilit field]

10

sorry

,w 10 mod 11

[grand pond]

Results provided by WolframAlpha

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Requested by physicsrocks

[steep falcon]

-1 = 10 mod 11

[twilit field]

ah yes

this is what I was trying to arrive at

thanks

[steep falcon]

then you get a_n*(-1)^n, if n is even then its just summing up a_n if n is odd you subtract it

so you have addition and subtraction alternating

[twilit field]

yup got it

thanks

now the opposite direction

if $11\mid a1(-1)^{n-1}+a_2(-1)^{n-2}+\dots+a{n-1}(-1)^1+a_n \implies 11\mid a_1a_2\dots a_n$

[grand pond]

A dense set(Ping when reply)

[twilit field]

well, this sin't too hard

We add and subtract stuff

that's it

I take that back

what do I do

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

[south seal]

Do you need to prove!

[twilit field]

this

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[dry ore]

Ik how this works here but I don't understand why doesn't the part labelled Under attention works ? Why can't we js cancel them out

[surreal moon]

It's a nuance of derivatives. dy, dx, and dt are not variables that you can just cancel out like regular algebraic variables

$\frac{\dd y}{\dd x}$ is all one variable.

[grand pond]

SWR

[surreal moon]

The chain rule does not work because you are cancelling out dt. We just chose a notation that makes it look that way (for better or worse)

And it completely breaks down when you consider the second derivative, or when you consider partial derivatives

So the point the textbook is trying to make is that the chain rule looks like algebraic cancellation, and it can be handy to think of it as such, but that is not what is actually happening

[dry ore]

Hmm could you further explain this part if possible? What do you mean by completely breaks down

Hope u cld explain in terms of 2nd derivative I have learnt that not partial derivatives yet tho

[surreal moon]

If you think that chain rule is just an algebraic simplification, then you may be inclinded to believe that $\frac{\dd^2 y}{\dd x^2}=\frac{\dd^2 y}{\dd t^2}\frac{\dd^2 t}{\dd x^2}$. But that is simply not true

[grand pond]

SWR

[dry ore]

Js a random qn ig d^2y/dx^2 is d/dx(dy/dx) aren't we technically js expanding them like fractions?

[dry ore]

Oh

[surreal moon]

wdym? Could you explain more, or work out what you mean?

[dry ore]

Like this

Like technically aren't we js expanding it

[tawdry laurel]

Here we cant treat them algebraically

[tawdry laurel]

By your logic denominator of LHS is d^2 x^2

[surreal moon]

This is valid, yes, and you can get a sort of chain rule going here, but it's going to like quite different from the first-order derivtive's chain rule

[dry ore]

Ohhh

Ohh I understand now then thanks a lot!

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[vestal moth]

(I think this one is easy but can't figure it out, I am a beginner when it comes to vectors)

[serene cypress]

did u draw that diagram?

[vestal moth]

yes

this is what I tried:
M mid-point of CD
⇒MC⃗=1/2⋅DC⃗
=1/2⋅(DA⃗+AC⃗

MA⃗=MC⃗-AC⃗
=1/2⋅DC⃗-AC⃗
=1/2⋅(DA⃗+AC⃗)-AC⃗
=1/2⋅DA⃗+1/2⋅AC⃗-AC⃗
=1/2⋅DA⃗-1/2⋅AC⃗
=-k/2⋅DB⃗-1/2⋅AC⃗
=-k/2⋅DB⃗+k/2⋅CB⃗
=-k/2(DB⃗-CB⃗)
=-k/2DC⃗
=-kMC⃗

Now if we can show that MC⃗=-k⃗MB , we're done

[midnight plank]

@vestal moth Has your question been resolved?

[spring wave]

Try say A=0 B=1, and calculate the position of all other points, also in your diagram, you have drawn as if k is negative.

If k was positive, d would be between a and b

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[drifting root]

how do I prove by induction?

[drifting root]

For 5

Og

Nvm

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[river marten]

Looking to solve these but idk what to do with the 2

[plain osprey]

the horizental asymptote will use that 2

well tbf the 2 is a part of the function, u would have to use it

[midnight plank]

@river marten Has your question been resolved?

[river marten]

Do I

Multiply it to the numerator?

[plain osprey]

u can

but that will not help at all

ur given the expanded form of the function and the factored form of the function so that u can determine the asymptotes

and the x intercept(s)

[river marten]

Sooo how do I solve for the Horizontal asymptote then?

[plain osprey]

• If the degree of the numerator is higher than the denominator, then there is no asymptote
• If the degree of the numerator is equal to the degree of the denominator, then the asymptote is the ratio of the coefficients of the highest degrees
• If the degree of the numerator is less than the degree of the denominator, then the asymptote is y = 0

[river marten]

~>

I know that but

[plain osprey]

then apply it

[river marten]

I have 2 x's up top

Two below

Do I do...

[plain osprey]

u see that the numerator and denominator have the same degree right?

they both have x² with the highest exponent

[plain osprey]

u mean cancellation?

[river marten]

Is it gonna be like that?

[plain osprey]

not quite

u were supposed to look at the expanded form

[river marten]

Uhuh

And the top is same size sooo

[plain osprey]

no

look at the exponents of x

[river marten]

Sorry typo

It was typo ;~;

[plain osprey]

oh

[river marten]

Alright so so

Now wut?

[plain osprey]

they both have same degree 2

so divide the coefficients of x²

thats ur horizental asymptote

[river marten]

So

3/1?

[plain osprey]

dont forget that 2

[river marten]

So 6/1?

[plain osprey]

which is 6

yes

[river marten]

Oh

Oki thank you

[plain osprey]

if ur experienced i guess u can evaluate limit of that function

if u want to, ofc

[river marten]

@.@

I am definitely not experienced...

[plain osprey]

ok then

lmao

[river marten]

I actually know this now

[river marten]

Thank you a bunch ~〰︎~

Non-math question

Is season 3 out yet?

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[carmine hawk]

im not sure aboutb

b

i think a is unconnected

b idk i feel like its unconnected as well because theres no way to acces d

id assume c is strongly connectyed

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[runic hamlet]

x <= t <= x+1 and -1 <= sint <= 1

[subtle torrent]

Tysm

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[tidal turret]

let $\mathbb{S} = \langle (1,1,1,0),(1,0,0,-1) \rangle$ and $\mathbb{W} = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 - 2x_2 + x_3 + x_4 = 0}$ , $\mathbb{H} = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 - x_2 + x_4 = 3x_1 + 2x_2 - x_3 = 0 }$ Find if possible a subspace $\mathbb{T}$ that satisfies $\$ $dim(\mathbb{T}) = 2$ , $\mathbb{S} + \mathbb{T} = \mathbb{W}$ and $\mathbb{S} \cap \mathbb{T} \subseteq \mathbb{H}$

[grand pond]

Renato Chavez

[tidal turret]

<@&286206848099549185>

<@&286206848099549185>

<@&1280681159495385169>

,, S + T = { \mathbf{x} \in \mathbb{R}^4 \mid \mathbf{x} = \mathbf{s} + \mathbf{t}, , \mathbf{s} \in S, , \mathbf{t} \in T }.

[grand pond]

Renato Chavez

[tidal turret]

,, S \cup T = { \mathbf{x} \in \mathbb{R}^4 \mid \mathbf{x} \in S \text{ or } \mathbf{x} \in T }.

[grand pond]

Renato Chavez

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

<@&286206848099549185> <@&1280681159495385169>

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@tidal turret Has your question been resolved?

[tidal turret]

<@&286206848099549185> <@&1280681159495385169>

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@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[gilded spear]

hm

[gilded spear]

Are you familiar with parameterization?

@tidal turret

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[tidal turret]

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[flint anchor]

how can i solve this with limits?

[velvet oar]

This is just a function, what are you supposed to do with it? Can you post the full context with all instructions?

[flint anchor]

Differentiate the function.

thats what im asked

but using limits

[polar socket]

oh shart it’s transparent

https://mlqpfigbl8na.i.optimole.com/cb:SOI-.ac3f/w:606/h:361/q:mauto/f:best/https://apcalcprep.com/wp-content/uploads/2021/11/Limit-Definition-of-the-Derivative.png

using this right

[flint anchor]

i just need to know how do i determine the limit

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[last slate]

I have 4 different balls and 3 different boxes.Each box can hold any number of balls.
Find the total number of possibilities.

[last slate]

How to find it ?

[viral dagger]

well the firsr ball can go into 3 bins, the second can go into 3, and so on

[last slate]

oh wait

yeah

huh

3^4

ty
i confused this with smth else

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[knotty forge]

Stars and Bars

[last slate]

stars and bars?

[knotty forge]

.reopen

[midnight plank]

✅

[knotty forge]

you said any number of balls, does that mean that some boxes can be empty?

[last slate]

yeah

[viral dagger]

its distinct balls

[knotty forge]

then you want theorem two here

https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

$\binom{n+k-1}{k-1}$

[viral dagger]

"how many ways there are to put n indistinguishable balls into k distinfuishable bins."

https://brilliant.org/wiki/distinct-objects-into-distinct-bins/

[last slate]

stars are also different

[grand pond]

Melvin Eugene Punymier

[knotty forge]

(n indeistinguishible objects and k bins)

[last slate]

n is also distinguishable

[knotty forge]

It's called "Stars and Bars" because you can picture it like lining up some stars and partitioning them with bars

[knotty forge]

are they? The wiki should be updated if that is so

they are not distinguishible in theorem 1

[last slate]

no

like

my question

was for

[last slate]

different balls and different boxes

[knotty forge]

oh, hm...

[last slate]

wikipedia says "such as how many ways there are to put n indistinguishable balls into k distinguishable bins."

n indistinguishable

[knotty forge]

oh

would you just need to multiply by n! or something?

[last slate]

each ball has a chance to go to 3 boxes

we have 4 balls

so 3x3x3x3

[knotty forge]

ah

[last slate]

3^4?

[knotty forge]

I think you are right

[last slate]

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[last slate]

thank you

[knotty forge]

yeah I misread

[ebon mica]

wait i have a question

cant a box hold no balls too

[last slate]

it is possible

[viral dagger]

thats still accounted for

[last slate]

and it is considered in 3^4

[ebon mica]

how

[last slate]

.reopen

[midnight plank]

✅

[ebon mica]

doesnt 3^4 account for each ball going in either box 1 box 2 or box 3

[last slate]

a box having no balls is possible

[ebon mica]

yeah

[last slate]

all the boxing containing 0 balls is not possible

because we have to place the balls somewhere

[ebon mica]

yeah

[last slate]

well?

[knotty forge]

we do?

[last slate]

yeah

if we dont there would be 4 chances for each ball

[ebon mica]

i see

[last slate]

making it 4^4

[ebon mica]

i assumed that the ball didnt have to be placed somewhere

gotcha

[last slate]

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[half vigil]

<@&286206848099549185>

@grand pond

@normal raptor

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[stiff forum]

The red parts should cancel out in theory, but what happens to that absolute value?

(pls ping me when answering @stiff forum )

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[median solar]

How’d they go from the first step to the second?

[midnight plank]

@stiff forum Has your question been resolved?

[stiff forum]

I already claimed this channel, u gotta ask in one of the unclaimed ones

[stiff forum]

<@&286206848099549185>

[median solar]

I got help ty

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[humble wraith]

i dont understand how they got the first equation from substituing w for a +jb

[hard shard]

we know w is a complex number which can be written in the form a+jb

[humble wraith]

yea

[hard shard]

that way we can split the purely imaginary parts from the purely real parts

[humble wraith]

whered the j go though

[hard shard]

|c+jd| = sqrt(c^2+d^2)

magnitude of a complex number is a nonnegative real number

j is the imaginary unit

you people need to stop deleting your messages

you learned something

cool

be glad about it

[upper crag]

deleting messages is a cowardly act

[hard shard]

i wouldnt go that far

[humble wraith]

why not c^2 + d^2 ? since j^2 = -1

[gray mist]

I would write it as 8j instead of j8 so I was confused. Thanks anyways!

[hard shard]

if you draw the complex plane, the magnitude of a complex number is the distance from that complex number to the origin

you can use the distance formula

[humble wraith]

this one ?

[hard shard]

yup

[humble wraith]

then why is this a-12

why not a-4

[hard shard]

you need to split the real and imaginary parts

they kinda skip a lot of steps, so im gonna break it down

we start with |x/y| = |x|/|y| which we do the first one

then we multiply by the denominators to clear them

[humble wraith]

oh wait

[hard shard]

this gives you 3|w-12|=5|w-8j|

[humble wraith]

i forgot to multiply 12 by 3

[hard shard]

why?

[humble wraith]

in the first step where i multiply by denominators

i didnt

careles

[hard shard]

any other steps youre stuck on?

[humble wraith]

how did they derive eqn 2

?

[hard shard]

its a similar idea but with the second given equation

|w-4|/|w-8| = 1

[humble wraith]

ohhh alright

i got it

[humble wraith]

thanks man 👍🏻

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[twilit field]

Prove

[twilit field]

I was thinking of proving ab|lcm* gcd

and vice versa

Let the prime factors of a,b be 1,a_1,a_2,\dots, a and 1,b_1,b_2,\dots, b

let the largest common factor be some a_i and b_j

so gcd(a,b) = a_i

or wait

let gcd(a,b)=1

then lcm(a,b)=ab and we're done

let b=ka

then gcd(a,b)= a, and lcm(a,b)=b

so gcd*lcm=ab

let the prime factors of a = 1,a_1,a_2,\dots, a_n
\
Let the prime factors of b be b_1,b_2,\dots, b_m
\
we then have gcd(a,b) = a_1a_2 \dots a_k where thesre are the common prime factors

I'm lost

<@&286206848099549185>

Alternatively, I thought I could prove $lcm(a,b) \cdot gcd(ab) \mid (a,b)$ and vice versa

[grand pond]

A dense set(Ping when reply)

[twilit field]

.close

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[viral dagger]

for real number x and natural number n, prove that

[viral dagger]

topic is induction

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[whole yarrow]

yo is there a math bot here for homeworl?

[remote jasper]

i dont know

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[inland patio]

Let ${E_r}_{r>0}$ be a family of Borel subsets of $\mathbb R^n$. They are said to "shrink nicely" to $x\in\mathbb R^n$ if \

1. $E_r\subset B(r,x)$ for each $r$,\
2. there is a constant $\alpha>0$, independent of $r$, such that $m(E_r)>\alpha m(B(r,x))$.\

Let $U$ be a Borel subset of $B(1,0)$ such that $m(U)>0$. Then $E_r={x+ry:y\in U}$ shrinks nicely to $x$. How? My attempt: 1) is straightforward to verify, but I struggle with 2). $E_r$ is the translation by $x$ of the set $rU$, so $$m(E_r)=m(rU)=r^nm(U),$$but I don't know how to continue here.

[grand pond]

psie

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[wanton sleet]

We consider the linear system of equations (1st image) We consider the linear system of equations Determine the coefficient matrix and the augmented matrix of this system of equations, check whether it has solutions and determine them if necessary.

My numbers look werid. I would be thankful, if you could look over it.

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[left meadow]

I am confused about eigenvectors. Lets say we have the matrix A, which has an eigenvalue -1 (with multiplicity 2). Now, when calculating eigenvectors, both $x_1$ and $x_2$ are free, so we get eigenvectors (1, 0) and (0, 1). This means that for $\lambda$ = -1, A(1, 0) = -1(1, 0), and A(0, 1) = -1(0, 1), (which is true). However, consider any random vector $\in R^2$, for example (8, 5). A(8, 5) = -1(8, 5). So shouldn't (8, 5) be an eigenvector? I understand that because both $x_1$ and $x_2$ are free, for every vector (a, b), A(a, b) = -1(a, b), so then why isn't every vector considered an eigenvector, since it satisfies Ax = $\lambda$x for $\lambda$ = -1?

$$A = \begin{bmatrix} -1 & 0\0 & -1 \end{bmatrix}$$

[grand pond]

Ezlanding

[exotic pelican]

they are eigenvectors, too. Maybe you might want to look up eigenspaces

to each eigenvalue, a whole subspace of eigenvectors corresponds to it

it's just that (1, 0) and (0, 1) form a basis for this eigenspace

(excluding (0, 0) which I think is usually not considered to be an eigenvector, but still contained in every eigenspace)

[humble torrent]

Yes or else any lambda would be an eigenvalue

[midnight plank]

@left meadow Has your question been resolved?

[left meadow]

So span{ (1, 0), (0, 1) } forms all the vectors that Ax = -1x, and eigenvectors are the just basis for these vectors

[exotic pelican]

eigenvectors are all these vectors except 0, not necessarily just the basis vectors

they are $\mathbb{R}^2 \setminus {0}$ while the eigenspace is $\mathbb{R}^2$

[grand pond]

rbit

[left meadow]

So you're saying the eigenvectors are the vectors such that span${v_1, v_2, ..., v_p}$ are the vectors such that satisfy Ax = $\lambda$x for the specific $\lambda$?

[exotic pelican]

you can ignore any spans, the eigenvectors are just all the vectors ${x \in \mathbb{R}^2, x \neq (0, 0): Ax = \lambda x }$

[grand pond]

rbit

Ezlanding

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[exotic pelican]

it's really just that (1, 0) and (0, 1) are "nice" eigenvectors, because they form a basis for this whole space (if we include (0, 0) again)

[left meadow]

So then why isn't every vector the eigenvector (such as (8, 5)) of $\begin{bmatrix} -1 & 0\ 0 & -1 \end{bmatrix}$ rather just (1,0) and (0,1)?

[grand pond]

Ezlanding

[exotic pelican]

yes every vector excluding (0, 0) is indeed an eigenvector

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[twilit field]

Show $e \mid a \land e\mid b \implies e\mid gcd(a,b)$

[grand pond]

A dense set(Ping when reply)

[strong lava]

"gcd"
greatest common divisor
for what order, one might ask, standard order or divisibility order?
Doesn't matter, how wonderful

[twilit field]

so $a=ek_1; b=ek_2 ]\implies ab=e^2k_1k_2\implies gcd(a,b) \cdot lcm(a,b) = e^2k_1k_2$

[grand pond]

A dense set(Ping when reply)

[dreamy lichen]

Was it an exercise in your book

or did you make that question yourself

[twilit field]

a question set

[strong lava]

I'd refer to the definition and see if it doesn't happen to be immediate

[dreamy lichen]

gcd(a,b) * lcm(a,b) = ab might be an overkill tbh

[strong lava]

yeah in no world did you prove this without proving that first

[twilit field]

That was the previous question

[dreamy lichen]

oh

well, then its fine

[dreamy lichen]

but this is still not it

[twilit field]

wdym

[dreamy lichen]

how does it show that e divides gcd(a, b)?

[strong lava]

$gcd(a, b) = \inf{}_| {a, b}$
If you ever encounter this, it might come up one day
That's what it's related to

[grand pond]

Bezier

[twilit field]

It doesn't

not directly

Well, this tells us $e$ is a common factor of $a$ and $b$

[grand pond]

A dense set(Ping when reply)

[twilit field]

so lcm{a,b}/e is an integer

so we have gcd(a,b)k_3=ek_1k_2

and we're done

[dreamy lichen]

does this suffice?

[twilit field]

oh no

[strong lava]

how unnecessarily complicated of a proof for that fact

[strong lava]

you think you are

[dreamy lichen]

You wont need any fancy theorems for this one

[strong lava]

but you still haven't understood what can and cannot pretend to qualify as a proof
go do some lean

[dreamy lichen]

Just few defns:

a is a divisior of b iff b = ka for some k
gcd(a, b) is the greatest common divisor of a, b

[dreamy lichen]

Lmao

You don't need to go to the axioms every time, but you should be confident that you would be able to trace your proof down to basic theorems and axioms. And if you're not, don't skip the steps

[twilit field]

I';m sorry, but when I see ENT, I just short circuit

[graceful drum]

.. all this while i was thinking e divides a to the power e

and i was confused why you all said its so easy

[strong lava]

high schooler encounters logic notation in the wild, freaks out

[graceful drum]

according to definition of gcd of a and b, if d|a and d|b, then d| the gcd

try to use this

[dreamy lichen]

Hmm

that doesnt sound like the most common defn of gcd

[strong lava]

that's probably not his definition

[graceful drum]

oh thats the definition in my book

[twilit field]

$e=k* gcd(a,b)+ r \implies (e-r)/gcd(a,b) =k. 0\leq r<gcd(a,b). so r=0

[strong lava]

yeah
the standard definition afaik is
"Z is a principal ring so aZ + bZ = dZ for some d > 0. d is unique and is called gcd(a, b)"
Barely into the second month of undergrad

I miss this
Simpler times

[dreamy lichen]

The simplest defn is gcd(a, b) is the greatest common divisor of a, b imo ¯_(ツ)_/¯

[strong lava]

show it exists

[dreamy lichen]

one common divisor is 1

it's a bounded set of integers

[twilit field]

WOP

[dreamy lichen]

has a maximum

[strong lava]

ok fair

[dreamy lichen]

careful, WOP guarantees existence of minimum

not maximum

you actually need it to be bounded from above

[strong lava]

also WOP sounds like axiom of choice to me

[dreamy lichen]

which WOP doesnt require

[dreamy lichen]

That's weird

[twilit field]

Does this work?

[strong lava]

well ordering principle is kinda like the AOC thing

[dreamy lichen]

are you confusing it with WOT?

[strong lava]

what's WOP then

[twilit field]

Well ordering princple

[strong lava]

"N is well ordered" isn't a principle, it's a theorem
We don't do physics here, there are no principles beyond the axioms

[dreamy lichen]

hmm

k is gonna be 0

r is gonna be 0

is e 0?

[twilit field]

why does k have to be zero

[dreamy lichen]

WOP says that every non-empty set of naturals has a minimum

WOT says that every set can be well-ordered

[strong lava]

why on earth do you call that a principle

[dreamy lichen]

this is where you use choice

[dreamy lichen]

Why not

[twilit field]

oops

I see

[strong lava]

that's just disrespectful to the meaning of the word

,w principle

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by mateo713

[twilit field]

$gcd(a,b)=ek_1+r$

[grand pond]

A dense set(Ping when reply)

[strong lava]

none of those are even close to "theorem"

[twilit field]

I work using this

[dreamy lichen]

WOP is very commonly used for reasoning

you can prove stuff with it

[strong lava]

doesn't change the issue with its name

it's not an axiom

so why call it principle

[dreamy lichen]

principle isnt the same as axiom

[strong lava]

that puts it on the same level as principles in physics

which are kinda like axioms

that's why I never see the term principle in math

[dreamy lichen]

physics can go fuck itself tbh

[strong lava]

yes

[dreamy lichen]

okay lemme check now

[twilit field]

so $gcd(a,b)-r)/e=k$

[grand pond]

A dense set(Ping when reply)

[twilit field]

I now want to show r=0

if r were not zero, this would be negative

[dreamy lichen]

fractions rarely lead to anything

[dreamy lichen]

really? Why?

what if gcd(a,b) was 2 and r was 1

[twilit field]

e\geq gcd(a,b)

[dreamy lichen]

e is a common divisor of a, b that is greater than the greatest common divisor of a,b?

[twilit field]

oops

less

my bad

hmm

am I on the right track

[dreamy lichen]

not sure

my approach would be different

but i cant say with certainty that yours doesnt work

[twilit field]

I'm just not sure how to show r=0

[graceful drum]

how is gcd(a,b) defined in your book

[twilit field]

d \mid a \land d\mid b

[dreamy lichen]

greatest common divisor of a,b ig

[graceful drum]

bruh

[twilit field]

a d if e \md a and e \mid b then e\mdi d

wait

shit

oops

waitr

[dreamy lichen]

Wait is this your defn?

[graceful drum]

it becomes so simple then

[twilit field]

that was the den for integers

for naturals we used a different defn

d \mid a and d\mid b

[graceful drum]

i mean itll be the same

[twilit field]

if e | a and e| b then e \leq d

[dreamy lichen]

Btw there is an elementary theorem that uses gcd and that would be extremely helpful

[dreamy lichen]

its named after this guy

[graceful drum]

overkill

[twilit field]

guass

guas something

[graceful drum]

not gauss

hes bezout

[twilit field]

ah yes

[graceful drum]

so did u find the def

[twilit field]

I just told it

d | a and d|b and if e|a and e|b then e \leq d

[graceful drum]

ok, now do you realise that the conditions given in the question statement match the property of gcd given by the definition

[dreamy lichen]

I think this isnt the defn that was intended by the problem set

[twilit field]

yes

[dreamy lichen]

because then your task is literally the defn of gcd

[graceful drum]

prolly just a warmup

[twilit field]

I mean I can use gcd(a,b) * lcm(a,b)=ab

can I not

[dreamy lichen]

1 is an integer. Prove that 1 is an integer

that's literally what the problem would be asking

[graceful drum]

i mean we can see its important

cus op struggled

the authors of the book are beyond us

[twilit field]

No, I'm just bad at number theroy

[dreamy lichen]

using bezout is easier, bezout is probably more elementary and I still dont see how you would use that to prove it

[graceful drum]

ur just a beginner dont be hard on urself

[dreamy lichen]

Compare yourself with yourself few months ago

[twilit field]

I'm prepraring for my end sems lol

fair

[twilit field]

Prime factorisation?

[dreamy lichen]

I wouldnt go into that

you would need to define gcd using prime factorization

etc

[twilit field]

I see

okay

so what do I do then

[dreamy lichen]

bezout

[twilit field]

the $\alpha a+ \beta b = gcd(a,b)$ one?

[grand pond]

A dense set(Ping when reply)

[dreamy lichen]

yeah

[dreamy lichen]

so you would need to prove that e divides that

[twilit field]

hmm

pretty easy

e|a

and e|b

so e|\alpha a ; e| \beta b

[dreamy lichen]

Yeah

bezout is a pretty elementary theorem so I think that you can use it here

and bezout is only proved with stuff like induction, euclidean algorithm etc which are like the most elementary it can get

[twilit field]

fiar

thanks

I didn't use it because this version of bezeouts was on the next page

but it will fly in the exam

Thanks!

good night!

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[wanton sleet]

The task is:

Determine the coefficient matrix and the augmented matrix of this system of equations, check whether it has solutions and determine them if necessary.

I would be glad if you could help me, because I am not sure about my result. The numbers look ‘funny’.

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[nova wharf]

Hi there I was tring to do this eccercise . THought of adding the director vector u = 3 4 2 and trying to find another called v = [ x , y , z ) such that u . v = 0 and then i got stucked

[midnight plank]

@nova wharf Has your question been resolved?

[nova wharf]

<@&286206848099549185>

<@&286206848099549185>

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@nova wharf Has your question been resolved?

[nova wharf]

@gray widget No c como seguir

[nova wharf]

<@&286206848099549185>

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[stiff terrace]

is the answer A?

[surreal moon]

!show

[midnight plank]

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

[stiff terrace]

i just want to verify if my work is correct

this is what i have

[midnight plank]

@stiff terrace Has your question been resolved?

[surreal moon]

@stiff terrace I'm still reading it fyi

[stiff terrace]

okk thank u

[surreal moon]

Okay your $\lambda_1=1$ and $\vec{\alpha}$ look good so far

[grand pond]

SWR

[surreal moon]

And your $\lambda_2$ and $\vec{v}_2$ and $\vec{v}_3$ look good.

[grand pond]

SWR

[surreal moon]

So yeah, I would say A is correct

[stiff terrace]

i have a question

does the order i put v1 v2 v3 matter?

[surreal moon]

(I do not know if the other options would all be wrong though. I doubt any of the other options would be correct. And even if they were, the problems says to find a matrix, not all matrices, so choosing A should be sufficient)

[surreal moon]

No. The order will simply determine the order of eigenvalues along the diagonal matrix, but it'll be diagonalized nonetheless

[stiff terrace]

ok 😭

ill try A

omg

it was right phew

thank uu

can u help me with this problem too? im not sure how i should approach it

[surreal moon]

will take a look in a few min

@stiff terrace $P_3$ is the space of all quadratic polynomials?

[grand pond]

SWR

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[jolly roost]

can someone explain how in the cofunction identity and angle can be negative?

[jolly roost]

Like this one

how does it make sense for the top angle of the triangle to negative

[sharp coral]

that would indicate that we are dealing with unit circle trigonometry rather than right triangle trigonometry

[jolly roost]

i see

.close

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[iron flame]

I might be stupid, but why is

[iron flame]

not equal to sec(x^4)?

Mind you, both sec(x^4) and 4x^3 * sec(x^4) (the right answer) seem very reasonable, it's just that I cannot pin down why the first one is wrong.

[lavish venture]

FTC sir

how do you think they got 4x^3

[iron flame]

I don't know, it just seems reasonable, is all.

[lavish venture]

what’s the derivative of x^4

[iron flame]

4x^3, granted.

[lavish venture]

yep

[iron flame]

But FTC states that g'(x) = f(x).

[lavish venture]

i’ll write it for you

[iron flame]

f(x) is not 4x^3 * sec(x^4).

[lavish venture]

$\frac{d}{dx} \int_{h(x)}^{g(x)} f(x) dx = f(g(x)) \cdot g’(x) - f(h(x)) \cdot h’(x)$

perhaps to see why we need the chain rule here it would be best to consider an example

other than the one given of course

how about

$\frac{d}{dx} \int_{0}^{x^2} \cos(x) dx$

let’s save the derivative until after

and first evaluate the integral

lemme change the lower bound actually

[iron flame]

sin(x^2), no?

[grand pond]

knief

[iron flame]

Wait, was it supposed to be pi or 0?

[lavish venture]

yes but about about the lower bound

[lavish venture]

i changed it to zero

[iron flame]

Ah, gotcha.

[lavish venture]

anti derivative of cos is sin

so now what

[iron flame]

You would do F(b) - F(a), hence sin(x^2) - 0, no?

[lavish venture]

yes exactly

now

let’s take the derivative of it

what’s the derivative of sin(x^2)

[iron flame]

Which gives us sin(x^2), which gives us cos(x^2) * 2x, granted.

[lavish venture]

not quite

not * x^2

yep

do you see now

[grand pond]

knief

[lavish venture]

do you see where the chain rule comes in

[iron flame]

I mean, I can see why it is needed, but I am struggling to see how it being needed doesn't contradict the FTC rule of g'(x) = f(x).

[lavish venture]

that’s only when the upper limit of integration is simply x

FTC is more commonly written as

$\frac{d}{dx} \int_a^x f(t) dt = f(x)$

[grand pond]

knief

[neat coral]

is that

is that

[lavish venture]

this is because we’re differentiating with respect to x

[neat coral]

the man

the myth

[lavish venture]

yep

[neat coral]

the legend

the one who helped me friend?

[lavish venture]

indeed

[neat coral]

HOOZAH

wait

go to sleep

[lavish venture]

lol

[neat coral]

you dont have exams tmrw

[lavish venture]

nah

i have to hand in homework for one class tomorrow

then i’m off for the rest of the week

[neat coral]

then do that now

[lavish venture]

and finals thereafter

[lavish venture]

it’s in person

he’s the one professor who wants paper homework

[neat coral]

teleport

[neat coral]

lmao

[iron flame]

By all means, if you have something you need to do, knief, please don't let me keep you from that.

[neat coral]

yo wait

knief

[lavish venture]

nah i don’t

[neat coral]

what are you majoring in?

[lavish venture]

pure/applied math double major

[neat coral]

sick

cya

gn

[lavish venture]

does this make sense to you

the upper limit is no longer x, it’s x^2 so the derivative with respect to x is no longer 1

[iron flame]

Do you mind if I write how I'm sort of picturing it in my head in iMath and then upload the picture in just a bit?

[lavish venture]

sure

@iron flame

maybe i can mention something else which may give you more of an intuition for this

perhaps it’s because we’re writing x which is very common to use

but for instance we could just as easily write u or x^2

like

$\frac{d}{dx^2} \int_0^{x^2} f(t) dt = f(x^2)$

[grand pond]

knief

[lavish venture]

if we were differentiating with respect to x^2

then yes it would be true

but we’re differentiating with respect to x

not x^2

so chain rule applies

[iron flame]

I think I somewhat get it.

the derivative of f(t) from 0 to x is F(x) - F(0). If we take the derivative of that, it is quite clearly going to be d/dx F(x) - d/dx F(0). Assuming F(0) (and therefore it's derivative) is zero, for conveniences sake, the derivative of F(x) will be d/dx F times d/dx x - and should x not simply be x, the chain rule will apply.
Regardless, however, we can still say that F'(x) = f(x), because f(x) does not refer exclusively to f(X^1).

Something along those lines, I think.

[lavish venture]

F(0) need not be zero

F(0) will always be a constant

hence its derivative will be zero

[iron flame]

Hence why I said for conveniences sake, since I did not want to think about it.

[lavish venture]

but the rest is good

[iron flame]

Alright, thanks!

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[twilit field]

Here, I wrote a=nk+b

[twilit field]

feel that's overkill

so like I got gcd(nk+b,b) ; gcd(b,n)

easy enough to show they're equal

[misty gorge]

using the definition of equality mod n isn’t overkill

[twilit field]

Thanks

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[low tiger]

Help with question 8 please

[rose trout]

What do you think?

[rose trout]

It is indeed not surjective, since 6^x and 6^|x| are both positive for any x

Now is it injective?

[low tiger]

Idk...

I think it is one-one

[rose trout]

Let's say you're given some value y, like 12.
Can you solve the equation $6^x + 6^{|x|} = 12$?

[grand pond]

Azyrashacorki

[rose trout]

Try and split into cases

[low tiger]

??

I don't understand

Oh

Okk

[rose trout]

If you solve that equation I just gave and end up finding multiple solutions, then the function can't be one-one

[low tiger]

So not one one

Thx!

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[vocal swan]

https://youtu.be/Xzv7citDgsE?si=asXg3c1U3obWbxKA

[dawn dagger]

real

[feral sedge]

complex

[junior flower]

what the actually fuck

[brazen moth]

Cool

[serene cypress]

ok.

[brittle shoal]

laylaaa :3

[junior flower]

hi swerriee ily

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[brittle shoal]

ily moree 💞

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[last slate]

guys

[last slate]

help me again

a cicular racetrack has a radius of 100. if a car drives halfway around the circle what is the distance traveled?

is it 50 or 200

or wait no

100

[brazen moth]

100 pi

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[last slate]

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[graceful ferry]

\lim_{n \to \infty} \frac{1 + \sqrt[n]{n} + \sqrt[n]{n} + \dots + \sqrt[n]{n}}{n}\

[graceful ferry]

😭

[last slate]

$\lim_{n\to\infty}\frac{1+\sqrt[n]{n}+\sqrt[n]{n} +\dots+\sqrt[n]{n}}{n}$

[grand pond]

roach

[graceful ferry]

Thanks

[last slate]

is that it though

[lethal path]

I don't think this is what you mean though

[graceful ferry]

Yes

[last slate]

what

[graceful ferry]

I mean this is something I wrote

Wait

To solve this (the upper one)

Honestly i have no idea if this is right

I'm tryig to use squeeze theorem

,rotate

[last slate]

replace the 1 with a \sqrt[n]{n} too maybe

[graceful ferry]

?

[last slate]

wait no idk what i wrote there

[last slate]

oh yeah for the LHS just replace every root with 1 since theyre greater and you get n/n = 1

[graceful ferry]

Oh okay

Is the RHS okay ?

[last slate]

yeah you got sqrt[n]{n} its limit is 1

[graceful ferry]

And by squeeze theorem the middle converges to 1 too right ?

[last slate]

yep

[graceful ferry]

$\lim_{n \to \infty} \left( \left( 1 + \frac{1}{1} \right)^{1} + \left( 1 + \frac{1}{2} \right)^{2} + \left( 1 + \frac{1}{3} \right)^{3} + \dots + \left( 1 + \frac{1}{n} \right)^{n} \right)$

[grand pond]

prograce

[graceful ferry]

This cinverges to 2 ?

Or e ?

[humble torrent]

Why do you believe it converges in the first place?

[graceful ferry]

The first value is 2
Second is 9/4+2
Third is 16/9+9/4+2
Just intuitevely I feel like it convereges to 2 by how it behaves

I'm not sure, Maybe it goes to infinity becayse it keeps getting bigger

[humble torrent]

The terms of this sum, that is (1 + 1/k)^1/k

What happens to these as k->infty?

[graceful ferry]

->e

[humble torrent]

Not quite, but that’s sort of irrelevant

What we do know is that it doesn’t go to 0

[graceful ferry]

Okay, yes

[humble torrent]

So you’re taking a sum of terms that don’t go to zero

That’s a problem

[graceful ferry]

Oh

[graceful ferry]

By the way this the lim (the second one) I'm just looking at the denominator for now but this might be a wrong thing to do

[humble torrent]

Oh you wrote it wrong

[humble torrent]

This is not the denominator

[graceful ferry]

Ohhh

[humble torrent]

Of the thing you’re talking about over here

[graceful ferry]

Yes my bad

[humble torrent]

In any case, yes in that case the terms go to e

But I would not recommend checking it like this

The denominator will indeed diverge

But that might not be a bad thing, since your original problem is the reciprocal of that

[graceful ferry]

Diverges meaning approaches infinity ?

Then my expression converges to 0 ?

[humble torrent]

Yes in this instance

[humble torrent]

Maybe, maybe not

You’ll have to analyze it more directly

[graceful ferry]

Okay let me see

Any hints on how to change the expression or play with it?

[humble torrent]

One possible route is to bound it below and above

[graceful ferry]

?🙂

But here the LHS converges to infinity and RHS converges tk 1/e

[humble torrent]

Is this really true?

[humble torrent]

Yes for the RHS that’s in the right direction

Wait

No, sadly you’ve manipulated the inequality inappropriately

The RHS must be smaller the way you’ve written it

So it’s a lower bound

Not a upper bound

[graceful ferry]

Okay

[humble torrent]

Now since you’ve got 1/e as a lower bound

Try and see if you can find 1/e as a upper bound

Or atleast as the limit of one

[graceful ferry]

Ok I will try

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[twin kernel]

For polar equations am I supposed to make a table or not

[twin kernel]

And if I used a table how much is the theta or rsupposed to go up by

[nova yoke]

can you show what you are trying to solve

[twin kernel]

Any polar equations

Like this

But some videos don’t use tables so I’m confused

[nova yoke]

well if you're able to graph it without making a table, that seems fine unless the problem explicitly says to make a table

[twin kernel]

If I’m using table how much is the theta or r suppose to be going up each time

[twin kernel]

Like this one is 0,pi/6,pi/3 etc

[nova yoke]

it's up to you, unless someone asks for a particular step size

whatever helps you to draw the graph properly

if there are trig functions involved and you have to compute by hand then there are only a few angles you can do easily anyway

pi/6, pi/4, pi/3, pi/2 etc

[twin kernel]

So am I suppose to graph the points like I would (x,y)

Like the first one is (2,0)?

[nova yoke]

yea except recognize that's in polar coordinates

meaning: radius = 2 and angle = 0

it's not the point (x=2,y=0)

(well in this case it is)

(but in general it won't be)

[twin kernel]

Wdym

Like the second one is (sqrt3,pi/6)?

[nova yoke]

well consider the last entry in the table: (r,theta) = (-2, pi)

that's not the point x=-2 and y=pi

you have to do the conversion (either explicitly or mentally) if you want to plot the right point

[nova yoke]

yea, if you wanted to write this as an (x,y) point, i.e. rectangular coordinates, do you know how to do it?

[twin kernel]

How r these theta with pi graphed on the x axis

[twin kernel]

Yeah but what if I just wanted to graph it like that

[nova yoke]

wdym

you mean without converting?

just start at the origin, and mentally draw a line in the direction of angle theta, of length r.. your desired point is at the other end of that line

if r is negative then you go the opposite direction from theta

[twin kernel]

U mean I just draw these points in random spots?

[nova yoke]

the origin is a specific spot, not random

it's the point where the x and y axes cross

[twin kernel]

Origin is (0,0)

In x,y

?

[nova yoke]

yes

and in polar it's r=0 with theta undefined

(or theta can be anything)

[twin kernel]

I don’t understand how u graph the pi/6 ,pi/3 and stuff tho

Or the sqrt3

[nova yoke]

well pi/6 is the angle

aka 30 degrees

measured counterclockwise from the positive x axis

and sqrt(3) is the length

to get the x y coordinates you would do:
x = r cos(theta), y = r sin(theta)

where cos(theta) = sqrt(3)/2 and sin(theta) = 1/2

[twin kernel]

So I always have to convert them to Cartesian?

[twin kernel]

I still don’t know how I would graph those points with the x y coordinates

[nova yoke]

the same way you graph any x,y points

[twin kernel]

So I always have to convert them into x,y first before plotting?

[nova yoke]

that's one way

the other way is to do what i said earlier

start at the origin, and draw an invisible line in the direction of angle theta, of length r

and when you reach the end, that's your point (r,theta)

[twin kernel]

I don’t understand what that means

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[barren onyx]

i thought answer was m+nC2 -mc2-nc2

[rough birch]

um

namely when we form a triangle we are choosing 3 points not 2 points

so replace your 2 with 3 and this fact is correct

you additionally will have a bit more complicated casework when you include A

[barren onyx]

oh ye

[barren onyx]

but the solution is different