help-49

[upbeat yew]

how does it affect my first and second derivative?

[keen sinew]

what ?

[upbeat yew]

first derivative is for slopes by x

second derivative is for convex by x

[keen sinew]

oh yes

so you know that the slope from f(x) and f(y) is $\frac {f(x) - f(y)} {x-y}$

[grand pond]

robins

[keen sinew]

the derivative at point x is just the limit as $y$ tends to $x$

[grand pond]

robins

[upbeat yew]

yes that's the result of the slope

[keen sinew]

so that is the relation between limits and derivatives

\begin{document}

Let $a \in \mathbb{R}$, $f , g$ continuous on $\left[0 , a\right]$ and differentiable on $\left(0 , a\right]$ such that $f\left( 0\right) = g\left( 0\right) = 0$, with $g\left( x\right)$ and $g\left( x\right) \neq 0 , \forall x \in \left(0 , a\right)$.

Let $x \in \left(0 , a\right]$ and $h\left( t\right) = f\left( x\right) g\left( t\right) - f\left( t\right) g\left( x\right)$

First, lets compute $h'\left( t\right)$ :

[
h'\left( t\right) = f\left( x\right) g'\left( t\right) - f'\left( t\right) g\left( x\right)
]

Lets assume that : [
\lim_{x\rightarrow 0}\frac{(f'\left( x\right))}{(g'\left( x\right))} = l
]

According to Rolles theorem :

Since $h\left( x\right) = 0$ and $h\left( 0\right) = 0$, there exists $c_x \in \left(0 , x\right)$ s.t. $h'\left( c_x\right) = 0$

So $h'\left( c_x\right) = f\left( x\right) g'\left( c_x\right) - f'\left( c_x\right) g\left( x\right) = 0 \Longleftrightarrow f\left( x\right) g'\left( c_x\right) = f'\left( c_x\right)g\left( x\right)$.

Since $g$ doesnt cancel on $\left(0 , a\right]$, we can divide by $g\left( x\right)$. Now we have that :

[
\frac{f\left( x\right)}{g\left( x\right)} g'\left( c_x\right) = f'\left( c_x\right)
]

Similarely, lets divide by $g'\left( c_x\right)$ :

[
\frac{f\left( x\right)}{g\left( x\right)} = \frac{(f'\left( c_x\right))}{(g'\left( c_x\right))}
]

as $x$ tends to $0$, we have that $c_x$ also tends to $0$, so: [
\lim_{x\rightarrow 0} \frac{f\left( x\right)}{g\left( x\right)} = \lim_{c_x\rightarrow 0} \frac{(f'\left( c_x\right))}{(g'\left( c_x\right))} = \lim_{x\rightarrow 0} \frac{(f'\left( x\right))}{(g'\left( x\right))}
]
\end{document}

uhhh my translator made some mistakes.. wait a minute

[grand pond]

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

[midnight plank]

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[upbeat yew]

thank you very much!

[keen sinew]

no problem

have a good day

[upbeat yew]

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[keen sinew]

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[woeful turret]

i dont know how to start

[civic lynx]

• try solving for lambda first. how much of the given info do you need to do that?
• think about the properties of the adjoint (the adjoint is used in what formula?)

[woeful turret]

ye |A|=-1

i got lambda = 3

forgot to tell sorry

and could u say adjoint instead of adjugate would make it easier for me

[civic lynx]

sure :D

[woeful turret]

hmm the formula

A inverse = adj(A)/|A|?

[civic lynx]

now mess around with this formula a bit, see if you can find something that is contained in adj(A adj(A^2))

-# thanks mods i love you mods

[woeful turret]

uhh we need A^2 inside adjoint

so ig A-->A^2

[civic lynx]

that would help, yes

[woeful turret]

and then multiply by A

so we have A adj(A^2)/|A^2| = A(A^2)^-1

[civic lynx]

you are nearly there, but remember we are multiplying matrices

you have to multiply on the left for both sides, or on the right for both sides

[woeful turret]

oh right sorry

now?

[civic lynx]

now, what else can you simplify?

[woeful turret]

we can take |A^2| to other side

and take adj again?

so we get adj(|A|^2 A (A^2)^-1)

[civic lynx]

hmm, i don't see how taking adj would help in simplifying

we would hopefully like to get a simpler expression for B, so we can calculate it easily

[woeful turret]

then how to simplify 😔

[civic lynx]

like you wrote, |A^2| = |A|^2, and you know what |A| is, so we can ignore that

we are left with A, which we cannot do much about, and (A^2)^(-1). any properties spring to mind for this one?

[woeful turret]

(A^-1)^2?

[civic lynx]

yes :)

to convince yourself this is true, the inverse of A is adj(A)/|A|, and we can "pull powers" out of both of these things

mainly because adj(AB)=adj(A) adj(B), and |AB|= |A| |B|

so where can we go from there?

[woeful turret]

A A^-1 A^-1 = A^-1

so we have just B^-1 = adj(A^-1)

so B=adj(A)

now i have to calculate adj(A)?

[civic lynx]

yes, so you have two options

1. use the formula and hope to god you remember it on the exam
2. use the fact that A^(-1) = adj(A)/|A|, solve for adj(A) here. then use whatever technique to find A^(-1)

but i mean if you remember the formula easily then you can use that, no problem

[woeful turret]

which formula

[woeful turret]

wait why i have to calculate A^-1

[civic lynx]

$A_{ij} = (-1)^{j+i}M_{ji}$ where $M_{ji}$ is the $(j,i)$-th minor of $A$

[grand pond]

ηασιβ ♥

[civic lynx]

but if you haven't seen this then there's no point using it

[idle lance]

Wtf am I looking at

Holy

My brain is frying rn

[woeful turret]

ye i have..ive just never used it in a question

[civic lynx]

it is related to adj(A) via the formula above, so we can use it to solve for A^(-1)

[idle lance]

Can u explain what the fuck u just out

Put

[woeful turret]

bruh but why do we need A^-1?

[idle lance]

My friend

Stop ignoring me

[civic lynx]

it seemed easier to me, and i remember how to do it more often than i remember how to do adj(A)
but again, you can use the formula and get the same answer

[woeful turret]

bro open ur own help channel

[idle lance]

Brother I don’t even understand ur question

[woeful turret]

then y r u even here

[idle lance]

Idk I just look

[civic lynx]

this is math... in the math server
this is a help channel which is claimed by yoda, but feel free to discuss whatever in #discussion
if you are interested in what's going on, this is linear algebra, which you will likely come across in late secondary/early university

[woeful turret]

oh ok fine then

ty

[woeful turret]

im not in university :(

[civic lynx]

i have a very bad grasp on when people learn stuff, if im honest >.>

[woeful turret]

nah nwnw..ty again

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[frank sonnet]

Could anyone point me to a good source of learning vectors from the ground up, video or book. Im not really great at math but i enjoy and need math. I really want to learn vectors especially since its really usefull and i dont want just to google for solutions or ask here i want to understand what im doing.
Not sure if thats a valid reason for help thread 😄

[upper scroll]

try 3 blue 1 brown videos

they are great videos

[frank sonnet]

will i be able to learn from the growned up? does it give any definitions of things

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[safe onyx]

A = -8 j

I have positive, but the answer key says negative

Why did I get positive if it's supposed to be negative

[lyric charm]

you subtracted the wrong way around

also your handwriting is uh. 💀

[lyric charm]

you might wanna stop making your zeroes look like sixes btw

[safe onyx]

ooh the 0 came first

thank you

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[gloomy agate]

Help me with understanding why you cant find the surface area of a sphere by integrating the circunferece of the disks

[gleaming latch]

what did you integrate

[gloomy agate]

i mean, like, why can't you find the surface area of a sphere as the inetgral of the circunference of the disks

[gleaming latch]

you can

[gloomy agate]

u know, u devide the sphere in a bunch of disk. Why cant u find the surfacea rea if u integrate the circunference of each?

[gloomy agate]

according to gpt u can not thought 😭

and according to some other forums

[sharp coral]

when you want to integrate the size of the boundary of a region (which is arc length surrounding a 2D region and surface area surrounding a 3D region), it's important that the approximations you use for that boundary are tangent to the true boundary. Otherwise there is not necessarily any relationship between the approximations and the true value. This restriction doesn't apply to the interior (area of a 2D region, volume of a 3D region)

[fallow scarab]

!nogpt

[midnight plank]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sharp coral]

example of what goes wrong when your approximations are not tangent:

[gloomy agate]

loll i am aware, but i also found forums that stated the same

[fallow scarab]

show it

[gloomy agate]

https://math.stackexchange.com/questions/2205612/surface-area-of-a-sphere-with-integration-of-disks

i lowkey just don't undertsand what the responeses are talking about lol

[fallow scarab]

explanation is right there

[gloomy agate]

yeah yeah ig, but i dnt undertsand why the aproximation would be wrong though

[gloomy agate]

mmmmm but as it becomes smaller and smaller, wouldnt it aproximate lol?

to dx

[fallow scarab]

"it becomes smaller and smaller" is too handwavy

you need to be more precise in calculus

[gloomy agate]

calculus is hard lol

but yeah, its true, im going to have to change some stuff 😭

thanksss

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[sharp coral]

it's not sufficient for the approximating shapes to be getting smaller, you need the error to shrink as you do so.

[gloomy agate]

huh, thats a nice way to put it lol

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[lavish venture]

i don't see a question here

did you have a follow up

[silent dock]

blisapeared

[supple sigil]

beared

[near geyser]

okay so i want to understand this proof

<@&286206848099549185>

three lines in mod

and last two lines in absolute

i don't get it

[molten mauve]

a series corresponds to a sequence because you can take u_1, u_1 + u_2, u_1 + u_2 + u_3 etc.

so the idea is to apply cauchy's principle for sequences to this sequence

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[nova pike]

Can someone explain in detail how the first answer in the following mathoverflow link was derived: https://mathoverflow.net/questions/69700/are-there-any-solutions-to-frac3n-2n2k-3n-n?rq=1? The question is basically finding pairs (m,n) such that (3^n-2^n)/(2^m-3^n) is a natural number.

[midnight plank]

@nova pike Has your question been resolved?

[silent dock]

write d := 2^k - 3^n, then d | 3^n - 2^n = (3^n - 2^k) + (2^k - 2^n) = -d + 2^n(2^{k-n} - 1).. so d | 2^n(2^{k-n} - 1). we know that d is odd (so gcd(d,2^n) = 1), so d | 2^{k-n} - 1. A useful consequence is the size bound d <= 2^{k-n}-1 < 2^{k-n}..

Because N in mathbb N, we have
d <= 3^n - 2^n < 3^n. So 2^k - 3^n < 3^n ==> 2^k < 2 • 3^n and so 2^k < 3^{n+1}. That gives k < [(n+1)log 3]/(log 2).

Therefore, d < 2^{k-n} and k - n < [(n+1)log 3]/(log 2) - n.. so

|2^k - 3^n| = d < 2^{k-n} < 2^{[(n+1)log 3]/(log 2) - n}

That's it for the upper bound.. Now for the lower bound from
linear forms in logarithms..

Write the difference in a way that exposes a linear form in logs:

2^k - 3^n = 3^n(exp(klog 2 - nlog 3)-1).

Let

Λ := klog 2 - nlog 3.

If 2^k ≠ 3^n, then Λ ≠ 0. Baker's theory (more precisely, explicit lower bounds for linear forms in two logarithms) gives an explicit inequality of the shape

|Lambda| >= exp(-Clog B) with B ≈ max{k,n}

for some effective constant C.. this implies that 2^k/3^n cannot be too close to 1, hence |2^k-3^n| cannot be too small relative to min{2^k,3^n}.. so Bennet summarises the end product as

|2^k - 3^n| >= min{2^k, 3^n}^{0.9}

except for finitely many explicitly known exceptional (k,n) pair..

(background on what lower bounds for linear forms in logarithms are and why they exist is standard Baker theory: you can see this for more info https://pub.math.leidenuniv.nl/~evertsejh/dio19-5.pdf)

combining these bounds kills all large solutions..

damn

missing a ` somewhere i guess

[supple sigil]

just tex it

smh my head

[silent dock]

i'm on mobile

[supple sigil]

you don't even need to change much with FREEMATH

[silent dock]

still a pain to do on phone

[midnight plank]

@nova pike Has your question been resolved?

[nova pike]

I understand the first half but don’t really understand the second half (finding the lower bound by using Baker’s theory)…

[nova pike]

Thank you so much for the explanation though! I hope to learn more about Baker theory and what lower bounds for linear forms in logarithms are in the future

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[rain wasp]

wtf is this kind of rounding

[rain wasp]

how can we distinguish between 122 and 1224?

this is a stem-and-leaf plot

[quick creek]

i think they are not distinguishable

[rain wasp]

if so then this is kinda a bad way to plot statistics

[supple sigil]

122 would be 1 and 2

[quick creek]

oops yeah

[lyric charm]

remember that each one of these comes with a key

[rain wasp]

can one leaf in a stem have multiple digits?

[midnight plank]

@rain wasp Has your question been resolved?

[fathom onyx]

Not really, because the whole point is that each digit-leaf is... a leaf

And thus a single data point

[rain wasp]

mmm i see

[fathom onyx]

That being said

Not that it's common to do so in data handling - and in fairness it's a really niche case to be using these in the first place, but still - you can on occasion use more than one digit per leaf, as long as each individual leaf is markedly clear

For a real-world example, here's a timetable of train departures in Japan:

[rain wasp]

i see

what benefits does this plotting method provide?

[fathom onyx]

Functionally, you still should provide a key in any case, and in a sense this is one

[fathom onyx]

It's a very crude and fast way of showing the distribution of data

So for instance, with the above plot, I can tell that the trains are most frequent at around 17:00-19:00

[rain wasp]

alright thanks

so it's a quick way to intuitively the distribution of the dataset

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[rose eagle]

I'm trying to prove that n^7-n is divisible by 7 if n is an integer

Currently i have factorized to n(n-1)(n+1)(n^2-n+1)(n^2+n+1)

I don't really know what to do from here like what I know is that i probably have to show one of these factors can be divisible by 7 therefore the whole expression can, just dont know where to start

[small jasper]

[https://en.wikipedia.org/wiki/Fermat's_little_theorem]

[rose eagle]

im not supposed to use that just like i want to know why

like i was told to use casework for this and dont know where to start from what

[lyric charm]

are you allowed to use modular arithmetic

[rose eagle]

idk i was just told i can put it into like different cases based on a remainder or something and i just dont understand that

[lyric charm]

you can trudge through working things out when n=7k, 7k+1, 7k+2, ..., 7k+6 if you want to / are forced to go through that route

rather unpleasant though

should i understand that modular arithmetic is also off the table?

[rose eagle]

why would n equal those like 7k, 7k+1, etc tho

oh wait are those like if you divide n by something that results in remainders

i havent learned modular arithmetic yet

[lyric charm]

those are the cases for the remainder of n when dividing by 7

[visual tiger]

Euclidian division says you must have n = 7k+r

[lyric charm]

a.k.a. modular arithmetic but shittier and uglier and 600 times more difficult

[lyric charm]

blehhh

[visual tiger]

Where r is the remainder, between 0 and 6

[rose eagle]

so like if i say n=7k+1 then n-1 is divisible by 7

and just have to rewrite the rest of the cases up to 7k+6

[rose eagle]

i dont even know what euclidian division is

😭

[west iron]

You could do this by induction too

It wouldnt be pretty, but it would be elementary

[rose eagle]

how would i prove it by induction

[west iron]

subtract the n+1 case from the n case, show it's divisible by 7

[rose eagle]

wouldnt that be annoying cuz i would have to expand something to the power of 7

[west iron]

Yes

Binomial theorem would be your friend there

[rose eagle]

man i got so much to learn this is my first week here 😭 i dont know anything

all these theorems and stuff

[gaunt jetty]

Do casework and consider each remainder from 0 to 6

[west iron]

its either that or learn some modular arithmetic

[gaunt jetty]

Most ooga way to do this

[rose eagle]

what does that mean 😭

[gaunt jetty]

Any integer divided by 7 will leave a remainder from 0 to 6

[rose eagle]

yea i was planning on doing that lol

[noble forge]

You can prove it by induction

[gaunt jetty]

You need to show that for every possible remainder, at least one of your factors becomes a multiple of 7

[noble forge]

base case n = 0

[rose eagle]

i was just gonna say if n=7k+1 then n-1=7k divisible by 7 and work the rest of the cases

up to 7k+6

[west iron]

there you go

there's the remainder 1 case

[gaunt jetty]

Ok well 0, 1 and 6 are all trivial

The rest are the ones you need to do some work with

[rose eagle]

ok ty ill work on it

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[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}

\begin{Definition}[Union of sets]
The \emph{union} of sets $A$ and $B$ is the set $A \cup B = { x : x \in A \text{or} x \in B}$.
\end{Definition}

\newcommand{\powerset}[1]{ \mathcal{P} (#1) }
\begin{Definition}[Power set]
The \emph{power set} of a set $A$ is $\powerset{A} = { X : X \subseteq A}$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
It's the case that $\powerset{A} \cup \powerset{B} \subseteq \powerset{A \cup B}$.
\end{Theorem}

\begin{proof}
Let $X$ be a set, $x \in X$ an element of $X$, and $X \in \powerset{A} \cup \powerset{B}$.
By definition of union, $$X \in \powerset{A} \text{ or } X \in \powerset{B}.$$
Then, by definition of the power set, $$X \subseteq A \text{ or } X \subseteq B.$$
So, by definition of subset, $$x \in A \text{ or } x \in B.$$
This means that, by definition of union, $$x \in A \cup B.$$
Then, by definition of subset, $$X \subseteq A \cup B.$$
Hence, by definition of the powerset, we have that $$X \in \powerset{A \cup B}.$$
Since $X \in \powerset{A} \cup \powerset{B}$ implies that $X \in \powerset{A \cup B}$, it follows that $$\powerset{A} \cup \powerset{B} \subseteq \powerset{A \cup B}.$$
\end{proof}

[grand pond]

Mor Bras

[fresh sparrow]

\newcommand{\powerset}[1]{ \mathcal{P} (#1) }

Let $X$ be a set, $x \in X$ an element of $X$, and $X \in \powerset{A} \cup \powerset{B}$.\\

erm what?

[grand pond]

artemetra

[fresh sparrow]

this is a bit backwards

maybe better to write\
Let $A, B$ be sets, and let $X \in \mathcal{P}(A)\cup\mathcal{P}(B)$, and let $x \in X$ be an element of $X$

[grand pond]

artemetra

[fresh sparrow]

here it would be great to clarify that "since x was chosen arbitrarily, this holds for any x. Thus, by definition of subset, ..."

[fresh sparrow]

other than that it's all good well done

[midnight plank]

@fickle sierra Has your question been resolved?

[midnight plank]

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[coral belfry]

,tex
Hi there. Is this proof of equal cardinality between Z×N and N×N correct? I have problems proving surjectivity sometimes and i wanna make sure

\vspace{1cm}
$\exists f: \mathbb{Z} \to \mathbb{N} $ , bijective. Then:

\vspace{1cm}
$\exists g: \mathbb{Z} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N} $ thats also bijective

\vspace{0.5cm}
Proof:

\vspace{0.4cm}
$g(x,y) = \left( f(x),y\right)
\
g(x_1,y_1) = g(x_2,y_2) \implies (f(x_1),y_1) = (f(x_2),y_2) \implies
\
(x_1 = x_2) \land (y_1 = y_2) \implies $ g is injective

\vspace{0.2cm}
Let $(f(a),b) \in\mathbb{N} \times\mathbb{N} $. We wanna show that: $$ \exists (a_0 ,b_0)\in\mathbb{Z}\times\mathbb{N} \ | (g(a_0,b_0) = (f(a),b)) $$

\vspace{0.2cm}
Indeed, for $a_0 = a\in\mathbb{Z} , \ b_0 = b\in\mathbb{N} : \ g(a,b) = (f(a),b) \square $
MEOW

[grand pond]

fijokazż

[lyric charm]

not enough words

g(x,y) = (f(x), y) -- you meant this to be the definition of g, yes or no?

[coral belfry]

yes

sorry im lazy on latex

[lilac hawk]

Apart from that it is correct though

[lyric charm]

ish?

[coral belfry]

where did i go wrong?

[lyric charm]

"let (f(a), b) in N × N" is sus

you are trying to invoke surjectivity without saying you invoke surjectivity

[lilac hawk]

Yeah the surjection part could be better formulated

[coral belfry]

i mean if i just write "proving surjectivity" on top would it be fine?

[lilac hawk]

That’s not the pb

[coral belfry]

we know f(a) and b are naturals. they are two random elements of the codomain of g

[lyric charm]

mrrgh

[lilac hawk]

Yes, but if you want that to prove that g is surjective, you need to have every natural to be in the form f(a)

[lyric charm]

it is bad to introduce an arbitrary element of (in this case) N as f(a) directly

[lilac hawk]

Which is true here but should be reminded

[coral belfry]

we know that f is bijective so we do have that

[coral belfry]

is bijectivity of f not enough then?

[lyric charm]

better is to say: let (a,b) in N × N. we wish to show there exists (a0, b0) in Z × N such that g(a0, b0) = (a,b).
since f is surjective there exists a0 in Z such that f(a0) = a...

[lilac hawk]

Say (a,b) in n, then assert that a is f(a_0) for some a_0

[lyric charm]

im taking issue with how you phrased this more so than with your logic

[coral belfry]

i see, yeah that makes more sense

okeoke ill have that in mind

thank you both

.solved

[midnight plank]

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[coral belfry]

,tex
Hi there. This proof of convergence for $a_n $ makes sense right? We just solve for n and choose any natural greater than that for our "base"

\vspace{0.2cm}
$a_n = \frac{n}{2n-1} $. proof that $a_n \to \frac{1}{2} $:

\vspace{0.3cm}
Let $\epsilon >0 $ fixed. We want $ \lvert a_n -\frac{1}{2} \rvert <\epsilon \implies \lvert \frac{2n-2n+1}{2(2n-1)} \rvert <\epsilon \implies $

\vspace{0.2cm}
$ \lvert \frac{1}{2(2n-1)} \rvert < \epsilon \implies 2n-1>\frac{1}{2\epsilon} \implies n> \frac{\frac{1}{2\epsilon} +1}{2} $

\vspace{0.2cm}
So, if we choose $n_0 = \ceil{\frac{\frac{1}{2\epsilon} +1}{2}} $ : $$ \forall n\geq n_0 : \lvert a_n -0.5 \rvert <\epsilon $$

[grand pond]

fijokazż

[runic hamlet]

havent checked whether the steps are correct but the general structure is correct

[coral belfry]

okeoke im pretty sure the rest is good

thanks

.solved

[midnight plank]

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[fickle sierra]

.reopen

Hello, I fixed the proof with the recommendations above.

\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}

\begin{Definition}[Union of sets]
The \emph{union} of sets $A$ and $B$ is the set $A \cup B = { x : x \in A \text{ or } x \in B}$.
\end{Definition}

\newcommand{\powerset}[1]{ \mathcal{P} (#1) }
\begin{Definition}[Power set]
The \emph{power set} of a set $A$ is $\powerset{A} = { X : X \subseteq A}$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
It's the case that $\powerset{A} \cup \powerset{B} \subseteq \powerset{A \cup B}$.
\end{Theorem}

\begin{proof}
Let $A$ and $B$ be sets, $X \in \powerset{A} \cup \powerset{B}$, and $x \in X$ an element of $X$.
By definition of union, $$X \in \powerset{A} \text{ or } X \in \powerset{B}.$$
Then, by definition of the power set, $$X \subseteq A \text{ or } X \subseteq B.$$
Since $x$ is choosen arbitrarily, this holds for any $x$.
Thus, by definition of subset, $$x \in A \text{ or } x \in B.$$
This means that, by definition of union, $$x \in A \cup B.$$
Then, by definition of subset, $$X \subseteq A \cup B.$$
Hence, by definition of the powerset, we have that $$X \in \powerset{A \cup B}.$$
Since $X \in \powerset{A} \cup \powerset{B}$ implies that $X \in \powerset{A \cup B}$, it follows that $$\powerset{A} \cup \powerset{B} \subseteq \powerset{A \cup B}.$$
\end{proof}

[grand pond]

Mor Bras

[civic lynx]

looks fine to me ^-^ for clarity, i might rewrite this sentence:

Then, by definition of subset
would become
Since this is true for all x in X, by definition of subset

but the proof makes sense to me without this

[cerulean oyster]

mb

stupid moment

[civic lynx]

divergence theorem you can do no wrong in my eyes

[cerulean oyster]

Tbh i was thinking about a sort of transitive idea for the same proof

Since The set A is always included in P(A) too.

[civic lynx]

ooh i do like how clean that is! but you may need to prove that A is the largest element in P(A) and contains all elements of P(A), which may break down in the infinite case (?)

would be very curious to see that tho

[fickle sierra]

Thanks for your responses!

.close

[midnight plank]

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[cerulean oyster]

Without dealing with the obvious complications of infinite sets, the general idea.
P(A U B) contains A U B and all its subsets
A U B contains A and B as "subsequences"
If you individually find P(A) and P(B), this now becomes the set of all sub-subsequences, which obviously has to be contained by A and B

Following back this logic backwards implies that P(A) and P(B) as set of sub-subsequences have to be also contained the original set of A U B, and also in P(A U B)

It might be flawed tho

nothing formal anyways

[fickle sierra]

P(A U B) contains A U B as an element or contains the elements of the union?

[cerulean oyster]

A U B as an element.

the power set contains all the subsets, so its a set of sets.

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[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}

\newcommand{\powerset}[1]{ \mathcal{P} (#1) }
\begin{Definition}[Power set]
The \emph{power set} of a set $A$ is $\powerset{A} = { X : X \subseteq A}$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $A$ and $B$ are sets. If $A \subseteq B$, then $\powerset{A} \subseteq \powerset{B}$.
\end{Theorem}

\begin{proof}
Let $A$ and $B$ be sets, $A \subseteq B$, and $X \in \powerset{A}$.
Then, by definition of the power set, $$X \subseteq A.$$
Since $A \subset B$, this means that $$X \subseteq B.$$
Thus, by definition of the power set, $$X \in \powerset{B}.$$
Since $X$ is chosen arbitrarily, this holds for any $X \in \powerset{A}$.
Since $X \in \powerset{A}$ implies that $X \in \powerset{B}$,
it follows, by the definition of subset, that
$$\powerset{A} \subseteq \powerset{B}.$$
\end{proof}

[spiral rock]

you only showed that one element of P(A) is in P(B)

you need to do show every element of P(A) is in P(B)

@fickle sierra

[radiant roost]

also, you probably shouldn't assume that A contains an element

[grand pond]

Mor Bras

[spiral rock]

this is better, but still not enough, as not every element of P(B), P(A) is a singleton

I would suggest that you let X in P(A) and show that X is in P(B)

[flint hare]

Just let X be an arbitrary subset. It's not useful to define X={x}, you're only accounting for singleton sets like that

[midnight plank]

@fickle sierra Has your question been resolved?

[fickle sierra]

I'll repost the proof

\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}

\newcommand{\powerset}[1]{ \mathcal{P} (#1) }
\begin{Definition}[Power set]
The \emph{power set} of a set $A$ is $\powerset{A} = { X : X \subseteq A}$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $A$ and $B$ are sets. If $A \subseteq B$, then $\powerset{A} \subseteq \powerset{B}$.
\end{Theorem}

\begin{proof}
Let $A$ and $B$ be sets, $A \subseteq B$, and $X \in \powerset{A}$.
Then, by definition of the power set, $$X \subseteq A.$$
Since $A \subseteq B$, this means that $$X \subseteq B.$$
Thus, by definition of the power set, $$X \in \powerset{B}.$$
Since $X$ is chosen arbitrarily, this holds for any $X \in \powerset{A}$.
Since $X \in \powerset{A}$ implies that $X \in \powerset{B}$,
it follows, by the definition of subset, that
$$\powerset{A} \subseteq \powerset{B}.$$
\end{proof}

[grand pond]

Mor Bras

[radiant roost]

good 👍

[fickle sierra]

<@&286206848099549185> Hello, could someone else check if this proof looks good?

[fickle sierra]

[midnight plank]

@fickle sierra Has your question been resolved?

[paper prism]

Its good yes

[fickle sierra]

Thanks for your responses!

.close

[midnight plank]

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[small warren]

Hiiii I'd love to get help on this if possible! 🩷

[small warren]

This is a challenging problem in similarity found in Khan Academy, from the picture I can see that we essentially have 4 different triangles and that I have to find out the length of segment CF.

[modern sapphire]

can you see something that relates CF the unknown with AB and DE, the knowns?

<@&268886789983436800>

[small warren]

I can prove similarity between triangles CBF and DBF by angle angle correspondence.

[modern sapphire]

this is good

what does that tell you?

[small warren]

Also, now that I look at it, I can do the same for the other 2 triangles

[modern sapphire]

mhm

[small warren]

That both big triangles are similar to their smaller counterparts.

[modern sapphire]

yes, thats true

[small warren]

But not necessarily all 4 are similar to each other

I can't prove that yet i think

[modern sapphire]

but you want something concrete that connects the results of the two similarities

[small warren]

Do you mean as in a two column proof or something?

[modern sapphire]

can you spot the common elements of the two triangles?

[modern sapphire]

no not really

[small warren]

Right angle and a shared angle at the end.

[modern sapphire]

there is one more thing

[small warren]

Yes?

[modern sapphire]

the side BE

[small warren]

Ah!!!!!!!

The big triangles share that side, and also an angle

except Side-angle isn't enough to tell if they're similar

[knotty coral]

@small warren You first prove triangle ABE is similar to triangle CEF
Then prove triangle BDE similar triangle BCF

You will get two simultaneous equations of CF, solve it.

[modern sapphire]

the two triangles share the same side, and their smaller similar triangles take up the two parts (which add up to BE)

[modern sapphire]

!nosols pls

[knotty coral]

I was just giving a hint...

[small warren]

Oh

[modern sapphire]

its hardly called a hint when you outline the entire solution and detail the steps

[small warren]

True

[small warren]

Both smaller triangles share that side.

[modern sapphire]

they share the parts of the side, and the important thing is that these parts add up back to BE, which is the more helpful part

[small warren]

I meant CF is the opposite to both smaller right triangles.

[small warren]

Yep, segments BF+FE=BE

[modern sapphire]

ah yes, ofc

[small warren]

and BE is the opposite to both big right angled triangles

[modern sapphire]

opposite?

[small warren]

Yeah

Is this good terminology?

[modern sapphire]

eww, thats not good, its opposite to the angle

[modern sapphire]

nope

[small warren]

How so? As far as I'm aware this is widely used.

[modern sapphire]

its opposite to the angle A, we dont care about any other angle in this problem other than the right angle

[small warren]

ok then!

[small warren]

we can't prove similarity without using other angles though, can we?

[weak parrot]

why we beating around the bushes bro, the key is to translate the ratios of AB CF DE all onto BE

[modern sapphire]

not even close, its only used when you are talking specifically about some angle. Opposite side means its actually opposite to something. the word loses its context when you dont have anything like an angle in consideration

[modern sapphire]

we already did establish similarity tho...

[small warren]

Interesting, thank you.

[small warren]

DE/CF=AB/CF, is this ok? I haven't established similarity between all 4 triangles, however segment CF is related to both big and small triangles.

[knotty coral]

No

[modern sapphire]

nope

[small warren]

how come?

[weak parrot]

nope

[knotty coral]

That would mean DE = AB

[modern sapphire]

I haven't established similarity between all 4 triangles
yea, coz you cant, since its simply not true

[small warren]

Can you show me how it would mean that, please? I'm curious

[small warren]

Why?

[modern sapphire]

thats why you have to take help of the common side

[knotty coral]

DE/CF=AB/CF
Multiply CF on both sides, you will get DE=AB.

[small warren]

Was that not what I was trying here?

[modern sapphire]

no not really

[small warren]

Ohhh, and that's a false statement, 9 doesn't equal 12

[modern sapphire]

maybe you were trying, but there was no actual evidence to support that equal sign there

[small warren]

@modern sapphire do you know why?

[modern sapphire]

wdym? there is nothing there that supports any claim of similarity

[small warren]

Also, reposting image to make it easier to see the original question:

[small warren]

I see, ty.

[weak parrot]

now we onto burden of proof hhhhhhhhh

[modern sapphire]

you need to see if the angles are equal, or that all the sides have the same ratio, and we dont have ny info about that

[weak parrot]

actually if all four are similar you would arrive again at 9=12

[small warren]

If you don't want to help then I'm just going to ignore you buddy.

[modern sapphire]

only way the triangles can be similar is if BE were to be sqrt(108) long, but even then the similarity is flipped

[knotty coral]

No you are just diving in too much...

You first prove triangle ABE is similar to triangle CEF
Then prove triangle BDE similar triangle BCF

You will get two simultaneous equations of CF, solve it.

[weak parrot]

feel free. i find how the way this relates the mathematical proof very fun

[small warren]

DE/CF=EB/FB, AB/CF=BE/FE

[small warren]

Does this help me?

[quick creek]

if your still confused let BF:FE to a:1-a

[modern sapphire]

it does

[small warren]

Do you mean FE instead of FD?

[knotty coral]

Try to divide one equation with the other one maybe...you will get something if your proportionality statement is correct.

[quick creek]

sry wrong alphabet

[small warren]

NP i was a bit confused, i thought maybe I could construct another triangle and see how it related to the others or something

[quick creek]

use the 9 and 12 to get a
then youll be easy to get CF

[small warren]

BF = BE-FE

[small warren]

How do you do that?

[small warren]

What does : mean here?

[quick creek]

the ratio between the length

[small warren]

How does this look like with :? @quick creek

[modern sapphire]

yea, use this

[quick creek]

DE:BE=CF:BF
AB:BE=CF:FE

[small warren]

I can definitely see how this becomes easy if we know what BE is and how to relate the proportions of the given numbers to these sides to figure out the proportion or K scaling factor from segment DE to segment CF, but I'm not sure how to go about that.

[cerulean oyster]

Might I add?

[modern sapphire]

well, quite simply, BF/BE + FE/BE = 1

[small warren]

I don't see how to do that, I haven't used : before

[cerulean oyster]

9 fe/be = 12 bf/be
9 fe = 12 bf
3/4 = bf/fe

[small warren]

Yea sure

[cerulean oyster]

So theres a constant between the two lengths

And as so, the line CF lies always at the same proportion between the two

[small warren]

That makes sense.

[small warren]

That makes sense! Tysm

[cerulean oyster]

And wont bore you, but the if the base of the triangle doubles, then its slope halves.

Which implies that the AC isnt affected by BE

So AC is just a constant based on the two side lengths

[small warren]

unfortunately i don't understand, i don't get why segment AC is important

[cerulean oyster]

Oh, mb

Not ac

[small warren]

Ah

[cerulean oyster]

Cf

Lmao

[small warren]

Ahhhh

i see

you're good XD

I don't know how to move forward

[cerulean oyster]

This is the geometry counterpart to a main theorem for vector algebra applied to statics & stability, lmao.

[small warren]

Cool!!!

[cerulean oyster]

You know that BE doesnt matter

Which implies that you can choose your own BE

[small warren]

Well the legs of a right angled triangle can give the missing hypotenuse using pythagorean theorem

[cerulean oyster]

And you have a known ratio for CF placement

[small warren]

Interesting, that never occurred to me.

[small warren]

Yeah

[cerulean oyster]

You could have also applied calculus principles

[small warren]

Maybe so!

[small warren]

If BE=1, then BF/BE=BF, and FE/BE=FE?

[cerulean oyster]

More importantly, you get to keep the property of bf/fe = 3/4

But yes.

[quick creek]

thats what i was saying

[small warren]

Ooohhhhhhh

Thank you, but I haven't learned that property yet, if it's a property or method or something

so i was pretty confused

[quick creek]

the sum of the ratio(equals BE) is 1

[small warren]

Sorry don't you mean FE/BF=3/4?

[cerulean oyster]

bf is shorter

[small warren]

Yes but FE is similar to BE in ABE, and that triangle- wait if ABE isn't similar to DEB (the two big triangles aren't similar to each other), then how can we relate the proportions of 9 to 12?

[cerulean oyster]

Lemme draw a sec

[small warren]

Okay

[cerulean oyster]

This is not the algebraic proof yet, but just to show the fact that the height of CF is a constant

[small warren]

i see

[cerulean oyster]

Theres this fact about triangles
If you divide the base in half, the height halves

Supposing same slope

[small warren]

There's a dilation here after changing the lengths of the hypotenuses? I thought that it was mirrored

[cerulean oyster]

Which is basically the idea that if you go down the line, its the same as a linear function

[cerulean oyster]

Not even dilation, i constructed two different cases just sharing the sides 12 and 9

But yeah. geometrically it behaves like a dilation

[small warren]

Ty

[small warren]

Yeah because opposite over adjacent = tan which is the longest side in a right triangle which is diagonal to both legs right?

[cerulean oyster]

If i go down BF / BE of the way (as if we had a certain % of the way down)

Then the height is equal to CF

which gives us the following equation

[small warren]

Are you saying BF = CF?

i'm not sure i follow

[cerulean oyster]

12 * (BF/BE) = CF

[small warren]

Oooo, how do I find this out?

[quick creek]

similar triangles

[cerulean oyster]

Random ass triangle i made

[quick creek]

lol

[cerulean oyster]

the red marks are at 1/2 and 1/3 of the base

[small warren]

idk what to do to get 12 * (BF/BE) = CF though

[cerulean oyster]

the height at 1/2 is 6/2 = 3
the height at 1/3 is 6/3 = 2

doesnt matter how long the base is

[small warren]

This is given? in this image?

[cerulean oyster]

yep.

[small warren]

Makes sense

[small warren]

Yea!

[cerulean oyster]

. this equation applies the same concept

instead of 1/2 or 1/3 of the way

we have bf/be

[small warren]

We have 12 over CF?

[cerulean oyster]

not yet

[small warren]

12/CF:9/CF?

[cerulean oyster]

CF is our tool to have both be equal

[small warren]

Okay

[cerulean oyster]

and we also have 9 * (FE/BE) = CF

Notice, both equations are equal to CF

[small warren]

Yep

[small warren]

So we can substitute then

[cerulean oyster]

so we have 9 (FE/BE) = 12 (BF/BE)

We are dividing in both sides by BE, and this is what tells us that BE doesnt affect the ratio

[small warren]

9/12=(BF/BE)/(FE/BE)

[quick creek]

that gives the ratio at the bottom

[small warren]

3/4=(BF/BE)/(FE/BE)

[quick creek]

BE cancels out

[small warren]

Interesting

3/4=BF/FE

[quick creek]

YES

[cerulean oyster]

A good choice for BE is 7

[small warren]

BF = 3 and FE =4?

[cerulean oyster]

because then BF is 3 and FE is 4.

[small warren]

That's what I was thinking

I wasn't thinking of like this strategy but I thought what if BE is 7

[cerulean oyster]

from there, just apply the same concept, choose any two sides, and move its corresponding ratio.

[quick creek]

now get CF

[small warren]

[quick creek]

lol

[small warren]

If BE=7, then BD = sqrt193?

[cerulean oyster]

A and D are just fluff here.

[quick creek]

no the ratio

[small warren]

Hmmmm

[cerulean oyster]

knowing the ratios, just apply this same idea

[quick creek]

BE=7 is just for easy ratio

[small warren]

CF=9 because 9=12*3/4?

[quick creek]

not that ratio

CF=BF/BE

[small warren]

Hmmm

[cerulean oyster]

You can do it one of two ways

[small warren]

Yea

[cerulean oyster]

9 * (4/7) or 12 * (3/7)

Notice, these are the same number

[small warren]

CF is 3/4ths the length of BE?

[cerulean oyster]

nope

[quick creek]

look

[cerulean oyster]

3/4 is the ratio of the short and long lengths

[small warren]

ohhhh

CF = 3/7!

[small warren]

i see

[quick creek]

uhhh
12×3=36

[knotty coral]

CF = 36/7?

[small warren]

ahhhh

Thank you all so much, I don't feel like I could do this on another problem or on this same problem on my own, I'm very stupid

[knotty coral]

You got all the process correct?

[small warren]

Yea!

Ty for the help!! @knotty coral @quick creek @cerulean oyster and others

.close

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[pearl wind]

Im blanking on how to do these can somebody check it

[fervent minnow]

the method and diagrams are correct, but you forgot the negative on the second fraction

[midnight plank]

@pearl wind Has your question been resolved?

[pearl wind]

Kk ty

Could somebody check and answer the questions I wrote down on these

[midnight plank]

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[pearl wind]

Ok my other channel got closed because I lost track of time. So my question is how to draw the triangle because it ends to be against the x axis for the one in the bottom left corner. They said drawing it like this means the same thing but I am confused. Wouldent that now but more than 300 degrees. And if now the angle does not include the measurements inside the turned wouldent it make every other one of my answers wrong

[pearl wind]

So I mainly don’t understand how to know if the angle is inside or outside the tingle and why it changes

[jolly pike]

the triangle foot is always on x axis

[pearl wind]

Yea I’m trying to fix it but I don’t know how to fix it without changing the angle

Because other people said they would still be the same and right but I don’t understand how it doesn’t change the angle

[jolly pike]

just draw a straight line from the point of circle and the xaxis

300=-60

[pearl wind]

Like I don’t understand which one of these is correct and how do I know if to include the angle with the triangle or not

[jolly pike]

the first one

[pearl wind]

so I’m just confused because I know all my other answers are right

Or are they not anymore

Because I’ve looked at my teachers answer keys not for this one specially but she includes the triangle angle

[jolly pike]

see the answer for c

-30=330

the same thing happens here

[pearl wind]

That one you are going clockwise instead of anti clockwise tho

But how can 300 degrees be the same thing as 340 or -30

[jolly pike]

not 340

[pearl wind]

Isn’t that where it ends up instead

[jolly pike]

no

300=-60

[pearl wind]

Because you are going 3 full quarters so 90•3 =270 then you are going two pi/6 aka two 30 degrees so 330 not 340

[jolly pike]

yes

[pearl wind]

So how would I get the answer for 300 degrees

I’m sorry I’m just very confused

[jolly pike]

see i tell u to rotate 270 in anticlockwise

where will u be if u start at origin

[pearl wind]

You would be at (0,-1)

I wouldent know how to write that as a triangle

[jolly pike]

yes but it is same as rotating 90 in clockwise right

[pearl wind]

Well that makes sense because I know how I would draw that one

[jolly pike]

so anti=+ clockwise=-

[pearl wind]

Yes

[jolly pike]

270=-90

[pearl wind]

But this has no -

Now I am confused I don’t know how to draw 270 either

[jolly pike]

• mean in clockwise -

[pearl wind]

Wouldn’t it be anti clockwise

[jolly pike]

it is just a line no triangle will come

[pearl wind]

But in our tests we have to draw it with a triangle

[jolly pike]

no u always rotate in anti so +

[pearl wind]

Yes

Isn’t that what I said?

[jolly pike]

u just think what happens if 300 was inside the triangle

[pearl wind]

I just don’t know how I am supposed to know on a test when I’m allowed to go inside the tangle or not. Because it feels like I am just guessing

and I don’t understand how both can be true at once and this not make all my other right answers wrong

To me now the triangle I would write for this not including this is now judt 360 but that does not make sense because that’s a full rotation aka zero

And I don’t understand how they can be the same answer if one is at pi/11 and the other is at pi/10 or 9

Wait

Could I draw it like this

Nvm

That doesn’t work

Like ona test I’m sure I’ll be able to figure it out but it won’t be because I think it’s true or I understand and it

[midnight plank]

@pearl wind Has your question been resolved?

[molten mauve]

what do you mean by this?

it doesn't matter how you draw the angle. if the point on the unit circle is below the x-axis, then the height of the right triangle is negative. if it's to the left of the y-axis, then the foot of the right triangle is negative

you can subtract or add 2π to the angle and all the trigonometric functions (sine, csc, cot, etc.) won't notice a difference

to check that the ray you drew is correct, you start in the positive x- direction and rotate ccw according to what angle amount the question asked for. cw if it's negative.

[midnight plank]

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[azure bolt]

not sure how to do this

i constructed a diagram

so we have

oh and DC = x

forgot to label

oh well

,,\sin C = \frac hb \ \implies h = b \sin C

[grand pond]

calvin

[azure bolt]

then

[grand pond]

calvin

[azure bolt]

wait i feel like we needed h

maybe

,,(b\sin C)^2 + x^2 = b^2

[grand pond]

calvin

[azure bolt]

and [h^2 + x^2 + 2ax + a^2 = c^2]

[grand pond]

calvin

[azure bolt]

so then

[azure bolt]

here

,,x^2 = b^2 - (b\sin C)^2

[grand pond]

calvin

[vivid yoke]

the area of ABC is just h*a/2

[azure bolt]

yeah im not sure

[vivid yoke]

<@&268886789983436800>

[azure bolt]

<@&268886789983436800>

[surreal charm]

@modulators

[azure bolt]

🔥

[surreal charm]

lance react me

[azure bolt]

yeah but im not sure how to continue from here

[azure bolt]

what about it

Ohhhhh i get it

abc = ha/2

h = bsinC

abc = absinC /2

br

bRUhh

gg

im cook

thank you alexis

.clsoe

.close

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[azure bolt]

not sure how to do g, h and i

[hybrid widget]

what is $\sin(360^{\circ}-\theta)$?

*degree measure

[velvet compass]

^{\circ}

[azure bolt]

i got it thanks

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[lyric charm]

\sin

[hybrid widget]

kk

[grand pond]

Annie Maqionde

[clever sedge]

Cool idea for a bot

[midnight plank]

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[shut canyon]

What is the difference?

(Zc → Gc) ⊕ (Gc → Zc)
(Zc → Gc) ∨ (Gc → Zc)

Glossary:

• Gx: x is good
• Zx: Zeus says it's good
• c: cake

[shut canyon]

It seems subtle I don't yet understand it.

[hollow obsidian]

in general, A v B is also satisfied when both A and B are true, whereas its symmetric difference (or XOR) requires that exactly one of A or B is true

[olive matrix]

A v B means or, not and

[hollow obsidian]

yah i'm saying that A v B is also satisfied when both are true

[olive matrix]

ahh ok

[hollow obsidian]

the subtlety between XOR and OR is that OR is weaker in the sense that at least one proposition needs to hold, whereas XOR requires that exactly one needs to hold

[shut canyon]

and with these conditionals it matters?

[shut canyon]

Thank you for your help!

[hollow obsidian]

well, what you have is a set of propositions

[midnight plank]

@shut canyon Has your question been resolved?

[midnight plank]

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[leaden seal]

can someone explain using the log rules (which rule) has been used in the bit in red

[surreal charm]

log_c (a) - log_c (b) = log(a/b)
nlog(a) = log (a^n)

[leaden seal]

thanks

.close

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[river robin]

If my chain costs 200k and watch half a million and her smile 100k. What’s my networth?

[white cloak]

Wtf is this

[pearl hull]

👀

[slender walrus]

Don't use math help channels for stuff like that.
Take it to #chill

.close

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[river robin]

Wow… abuser

[woeful turret]

LMAO

the audacity

[river robin]

You should remain in Star Wars only.

[woeful turret]

damn

[pearl hull]

nah, you need an actual, legit math question to use a help channel

[river robin]

My question is good.

[lyric charm]

ayo maybe don't dilute the meaning of that word

[pearl hull]

not for help channel but pretty decent for #chill

[surreal charm]

you became green

yay

[woeful turret]

:)

[midnight plank]

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[coral belfry]

Hi there. Does anyone know a good place to learn abt sequences with outputs in Q, and how theyre connected to functions in the reals? I have some parts in my notes but i wanna see practice questions abt using them to determine continuity of a function and stuff

[fallow scarab]

https://mathcs.org/analysis/reals/numseq/sequence.html

this one is a little more modern looking https://intro-real-analysis.github.io/

[coral belfry]

okayokay they have what i need

thank you

.solved

[midnight plank]

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[urban sentinel]

I already tried like just solving f(x)=x^2+3^2

But I dont think that’s what I’m supposed to do

[ruby geyser]

the thing i can tell u is that the zero here is -3

[urban sentinel]

Yeah that’s the answer

Dunno how

[mystic condor]

let (x+3)^2=0

[urban sentinel]

Ohhh

Ohhhhh okay so that’s all it is

What about the equation of the line of symmetry

[mystic condor]

well think about what the graph looks like

[urban sentinel]

How am I supposed to know what the graph looks like

[mystic condor]

its a quadratic

if it was just x^2 what would it look like

[fallow scarab]

<@&268886789983436800> spazzing creep

[urban sentinel]

Wtf was that 😭

[mystic condor]

hes typing again

[ruby geyser]

i meant the constant just lifts the graph

[mystic condor]

we all know you didnt mean that

[ruby geyser]

it just slipped my mind

[tribal temple]

You are aware that we can see messages you've deleted, right?

[ruby geyser]

sorry

[tribal temple]

That was really strange, don't do that again

[ruby geyser]

i just dont know whats happening to me these days

its just that my pills are stuck in customs

[mystic condor]

@ruby geyser youre not helping go someplace else for this

[tribal temple]

If you want to talk about that, do so in #chill or something, though I would strongly suggest against trolling

[mystic condor]

anyway what do you have with the graph of x^2

[ruby geyser]

X=-3 is the line of symmetry equation sorry

[urban sentinel]

I know I just don’t know how

[ruby geyser]

sure man

[mystic condor]

the function is exactly the same as x^2 except x is shifted by the 3 right

like for each x value you take the function of whatever is 3 ahead of x

[neat rapids]

hello im new

im just confirming by under graduate math

[fallow scarab]

#discussion

[neat rapids]

what does that mean

oh sorry

[urban sentinel]

Oooh

Sorry I took a call mb

[ruby geyser]

i hope this helps

[urban sentinel]

Ohhh

I used the wrong technique

Anyway thank u @ruby geyser @mystic condor

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[limber sinew]

just tell me if it’s supplementary or congruent PLZ

nvm!

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[midnight plank]

Please don't occupy multiple help channels.

[fickle sierra]

<@&268886789983436800>

[midnight plank]

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[flat geyser]

can someone help me get an idea of what a Z-module homomorphism $\phi:\mathbb{Z}/n\mathbb{Z}\rightarrow\mathbb{Z}/m\mathbb{Z}$ would be like

[grand pond]

JamR_71111

[flat geyser]

To show that $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\cong\mathbb{Z}/(m,n)\mathbb{Z}$ would we need any more info/intuition about the things in that hom set other than exactly what the definition of an R-module homomorphism would tell us ?

[grand pond]

JamR_71111

[flat geyser]

rn, I can only see that it would be the case that $\phi(\overline{x}+\overline{y})=\phi(\overline{x+y})=\phi(\overline{x})+\phi(\overline{y})$

[grand pond]

JamR_71111

[flat geyser]

of course with the other multiplication part but aside from those two things I don't know what else (if anything) i need

[midnight plank]

@flat geyser Has your question been resolved?

[visual tiger]

That's one rule, you can also find the rule $\phi(k\overline x) = k\phi(\overline x)$ interesting

[grand pond]

Rafilouyear2026

[flat geyser]

ye thats what i mean by this. if you mean that there's a key to it in there, i don't see it very clearly yet

[visual tiger]

And so you don't have a lot of freedom for what a homomorphism can be

[visual tiger]

If you pick some strategic value of x for example

[flat geyser]

like an identity ?

Hmm

[visual tiger]

Ever heard the term "generator"?

[flat geyser]

yes

[visual tiger]

Ok

[flat geyser]

1 generates both z/nz and z/mz

[visual tiger]

Both those "1"s are different

But the 1 from Z/nZ generates it

And same thing for the other

[flat geyser]

ok

[visual tiger]

So

If you know phi(1)

[flat geyser]

Hmm so phi(1) has to generate z/mz?

[visual tiger]

No not necessarily

[flat geyser]

rip

[visual tiger]

If x = 1 generates Z/nZ

That means every element of Z/nZ can be written as kx

[flat geyser]

right

[visual tiger]

So...

[flat geyser]

O

[visual tiger]

Once we know phi(1)

We know phi(anyone)

[flat geyser]

Ok so the elements of the hom set depend entirely on what you send 1 to?

[visual tiger]

Well a homomorphism depends entirely on what it sends 1 to

Yeah

[flat geyser]

Nice okay

[visual tiger]

So

What are the elements you can send 1 to?

[flat geyser]

only things that preserve the multiplication rule $\phi(r\cdot\overline{1})=r\cdot\phi(\overline{1})$ ?

[grand pond]

JamR_71111

[visual tiger]

Yeah, it's a bit vague for now but sure

Maybe you can find useful values of r to plug in

[flat geyser]

n ?

[visual tiger]

Yeah let's try

[flat geyser]

ok let me see

$\phi(n\cdot\overline{1})=\phi(\overline{0})=n\cdot\phi(\overline{1})$

[grand pond]

JamR_71111

[visual tiger]

Phi(0) being?

[flat geyser]

0 in the codomain right

awesome

[visual tiger]

So what can you say about the order of phi(1)

[flat geyser]

a divisor of n maybe

[visual tiger]

Exactly

[flat geyser]

Nice

sorry im slow with these things, is this enough to make a map for the isomorphism

and im thinking the map is just sending the homomorphism with $\phi(\overline{1})=y$ to $y\in\mathbb{Z}/(m,n)\mathbb{Z}$

[grand pond]

JamR_71111

[visual tiger]

Not quite, but close

y doesn't live in Z/(m,n)Z, but in Z/mZ

[flat geyser]

right right

Maybe it sends that homomorphism to the remainder of y when divided by gcd(m, n)

[visual tiger]

Let's let $\overline k = \phi(1)$, we'll find a way to express the condition for k

Forget about the order thing for now

[flat geyser]

ok

[grand pond]

Rafilouyear2026

[visual tiger]

So $m\mid nk$

[grand pond]

Rafilouyear2026

[flat geyser]

I see that $nk\equiv0(modm)$

Yeah

[visual tiger]

Mod m but yes

[grand pond]

JamR_71111

[flat geyser]

ya

[visual tiger]

You can divide both sides by d = gcd(n,m)

$\frac md \mid \frac nd k$

[grand pond]

Rafilouyear2026

[visual tiger]

I'll call m' = m/d and n' = n/d

So $m'\mid n'k$

[grand pond]

Rafilouyear2026

[visual tiger]

And now, we can use [???]

[flat geyser]

No clue lol sorry

Wait

(m',n')=1 ?

[visual tiger]

Yes

[flat geyser]

Meaning $m'\vert k$

[grand pond]

JamR_71111

[visual tiger]

Exactly

[flat geyser]

So k is a multiple of m/(m,n)

[visual tiger]

How many k like that are there in {0,...,m-1}

[flat geyser]

(m,n) solutions

Hah awesome

So we send each map that sends 1 to some l-multiple of m/(m,n) to l in Z/(m,n)Z ?

[visual tiger]

Wait

[flat geyser]

uh oh

[visual tiger]

You're saying l = phi(1)?

[flat geyser]

No, l * (m/(m,n)) = phi(1)

And we send phi to l

[visual tiger]

That looks better

[flat geyser]

is that a valid map for it

[visual tiger]

I mean we just did half of the stuff, that if phi is a homomorphism, then phi(1) is a multiple of m/(m,n)

Now, you need backwards implication

If k is a multiple of m/(m,n)

[flat geyser]

But if we have an element of $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}),$ doesn't that element have to be a homomorphism

[grand pond]

JamR_71111

[visual tiger]

Is the app defined by phi(x) = kx a homomorphism

[flat geyser]

using the thing that sends each map to whatever multiple of m/d it sends 1 to turns out to suffice for an isomorphism between that hom set and Z/dZ

thank you so much for your help in basically all of the set-up, i REALLY need to get used to this kind of stuff

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[half glen]

Hi ik this isn't a math question but i'm struggling to understand what this question is asking for bc what does it mean by between terminals a and b

[velvet compass]

Here you have a parallel and series connection, do it step by step (simplifying the circuit each time) and you will get the result

[quick creek]

yeah

[half glen]

so is it the same as just finding the Rt total resistance?

[quick creek]

i think so

[velvet compass]

this is the resistance that the meter (or source) sees if I connect it exactly between these two points

In this case there is no a big difference between those two, but generally there is

e.g. two resistors connected in series, if you take a meter and measure at points A and B (at the terminals of the first resistor), the resistance will be different than the resistance of the entire circuit between points A and C (i.e. equivalent)

[half glen]

ohh

okay thank you it makes more sense to me now!

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[wanton spade]

no clue how to do part A

[wanton spade]

i know (b) follows from (a) by rank nullity

[runic hamlet]

you have to construct functions T such that ST=0

so in particular im(T) is in null(S)

and as usual for these problems: pick a good basis

[restive sleet]

I’m not sure either but maybe try thinking about what it means for A(T) = ST to be zero

\begin{itemize}
\item First, remind yourself what \textbf{null space} means here. in other words:
``For which linear maps $T$ does $ST = 0$?''
This question defines the null space of $\mathcal A$.

\item Next, translate $ST = 0$ into plain language:
this means \emph{every vector in $V$} gets sent by $T$ into something that $S$ kills.

\item Ask yourself: ``Where must the outputs of $T$ live for this to happen?''
(Hint: think about the null space of $S$, not of $\mathcal A$.)

\item Now switch perspective: instead of thinking about operators, think about
\textbf{choosing images of a basis} of $V$.
For each basis vector of $V$, where is $T$ allowed to send it?

\item Once you see that, count degrees of freedom:
\begin{itemize}
\item How many independent choices do you have for each basis vector?
\item How many basis vectors are there?
\end{itemize}

\item Finally, remember that linear maps are completely determined by what they do to a basis.
\end{itemize}

[grand pond]

Anon

[supple sigil]

don't need to overcomplicate it

null A is precisely those in T in Hom(V, V) where im T ⊂ null S

[restive sleet]

oh yeah that makes sense

[supple sigil]

ok I don't mean it that way your exposition is helpful

I was saying that in reference to some deleted messgage

[restive sleet]

oh okay

[midnight plank]

@wanton spade Has your question been resolved?

[wanton spade]

Tv_i for a basis vector v_i will be a combination of dim null S vectors, and V has dim V vectors

wait so that should give me dimV dim null S choices?

[supple sigil]

yes

[wanton spade]

hold on would it suffice to show null A = L(V, null S)?

[supple sigil]

yes that's just a basis-free way of thinking about the same thing

[wanton spade]

ohh

[supple sigil]

(the fact that dim Hom(V, U) = (dim V)(dim U) for finite-dimensional spaces is why they specified V is finite-dimensional)

[wanton spade]

i see

thanks guys

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[rain wasp]

so i want to construct a karnaugh map of this