help-49

[obsidian cliff]

im honestly stuck i have not a single clue how to approach this problem (pings are welcome if im being unresponsive)

[obsidian cliff]

iim supposed to use intermediate value theorem somehow but idk how

[midnight plank]

@obsidian cliff Has your question been resolved?

[surreal moon]

First, i would consider f(1/2)

[obsidian cliff]

uhh

ok

im going to cry and think about it for a bit ig?

im not sure where im gonna go with that but lets see

[obsidian cliff]

also i did not mean to say cry that was a typo i meant just go

sorry if there were any alarms raised

[surreal moon]

There's two cases.\

1. $f(0)=f(1/2)$\
2. $f(0)\ne f(1/2)$

[grand pond]

SWR

[surreal moon]

Case 1 is easy. Case 2 is where you want ivp

[obsidian cliff]

ok

hmm

hmm

im just trying to articulate my words for ab it

ok now the question is

is there such a choice for L where i can get y = x + 1/2

(days like this i wonder if im good enough for a maths degree lmao)

uhh

maybe some kind of iterated process

[surreal moon]

That could work, but something simpler:\
let $g :[0, 1/2] \mapsto g(x)=f(x)-f(x+1/2)$

[grand pond]

SWR

[surreal moon]

What can you say about $g$?

[grand pond]

SWR

[obsidian cliff]

ok

well

im just gonna list off some things first

dk if they're relevant yet

g(0) = - g(1/2)

what are its bounds

[surreal moon]

Wdym?

[obsidian cliff]

ig thats impossible to determine since f doesnt have a defined range

[surreal moon]

Not relevant

[obsidian cliff]

yea ok

[obsidian cliff]

if i want to get to the end i need to show that

theres a point where

[surreal moon]

This is big though. How can you use this?

[obsidian cliff]

g(x) = 0

[surreal moon]

Exactly

How can you do that?

[obsidian cliff]

im going to think abt this

[surreal moon]

[obsidian cliff]

so

oh wait

so

g(0) and g(1/2)

have opposite sign

so

there exists $x_0$ with $x_0 \in (0,\frac 12)$ with $g(x_0) = 0 $

[grand pond]

lifelong dumbass

[obsidian cliff]

hence you've found it

ok how on earth does anyone come up with that

[surreal moon]

Why is this true?

[surreal moon]

Practice. Experience.

[obsidian cliff]

you can apply intermediate value theorem on g(0) and g(1/2) again

pick one of them to be >0, say g(0)

[surreal moon]

Why can we apply it?

[obsidian cliff]

because its a continuous function defined on a closed interval?

[surreal moon]

Good

Finally, why is g continuous?

[obsidian cliff]

because f(x) and f(x+1/2) are continuous

well

[surreal moon]

Good enough

[obsidian cliff]

yea it is for x in [0,1/2]

ok

uh

[surreal moon]

Everything seems to be in order now

[obsidian cliff]

sorry this just feels like i never wouldve gotten it

if it werent for the hint on g(x)

[surreal moon]

I had a problem like this back when i did topology. It's not something you can easily think of on your own. But once it's in your head, you just hope you remember how to use it when you need to

[obsidian cliff]

ok

(dreading uni in like a month)

hopefully it turns ok

im gonna try part b and c

with the following then

[surreal moon]

Good luck to ya

[midnight plank]

@obsidian cliff Has your question been resolved?

[obsidian cliff]

sorry im just having a bit of trouble

formulating the same argument for b

[obsidian cliff]

yea i got to $\frac{n-1}{n}$ distance apart but then having trouble like

[grand pond]

lifelong dumbass

[obsidian cliff]

covering every point

so to speak

im trying to consider functions $g_k(x) = f(x+\frac{k-1}{n}) - f(x + \frac{k}{n})$

[grand pond]

lifelong dumbass

[surreal moon]

This one is tricky. You can use induction to get you there. But the full path isn't easy to see

As an example, consider n=3. We know there is $x_2\in[0,1/2]$ st $f(x_2)=f(x_2+1/2)$

[grand pond]

SWR

[obsidian cliff]

mmm

ok

right

i agree with this

[surreal moon]

Now let $g(x)=f(x+x_2) -f(x+x_2+1/3)$

[grand pond]

SWR

[surreal moon]

The details here are where things get tricky. Anyway, good luck

[obsidian cliff]

ok

hmm

[obsidian cliff]

im gonna close this for now and save the msg thread for another time my brain is already mentally so tired

im sorry

.close

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[pseudo owl]

sinx+sin2x+sin3x+sin4x+sin5x=5
If x lies in the interval [0,2pi], how many values of x exist?

[pseudo owl]

Through bounding arguments, its clear that all of these terms equal 1

Does a value of x exist which satisfies this equation?

We want all these terms to be simultaneously 1

Feels as if there should be no solutions

[frozen talon]

you're basically there already

[pseudo owl]

I am unable to prove if none solution exists.

[frozen talon]

each sin(whatever) must equal 1, so in particular sin(x) = 1

there's only one x in [0, 2pi] that makes this work

[pseudo owl]

And it doesn't work for others..

Done lol

so 0 solutions

[frozen talon]

but you'll find that e.g. sin(2x) is not 1 for this fixed x

[pseudo owl]

I get it.

Thx.

[frozen talon]

ye

[pseudo owl]

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[ebon stone]

hello, im currently studying sequences for cal 3 and I have a question for the boundedness. is there no specific formula for it? like do i just have to think of 2 numbers that can bound a sequence?

[odd pivot]

Basically yes, you have to "find it yourself"

[ebon stone]

ah so there is no way to show it mathematically?

[odd pivot]

Ofc it depends on the sequence, do you have a particular example you would have help with ?

[odd pivot]

Oh yes there is

[lyric charm]

two different things at play here

[odd pivot]

As in, you have a definition of what is a bounded sequence by in general

[ebon stone]

uhhh sure, heres one of the problems in my set

[lyric charm]

1. why does your bound work?
2. how did you find the bound?

only (1) requires mathematical proof

[odd pivot]

But when it comes to finding the bound, it's case by case basically

[lyric charm]

(2) does not have a fixed formula or algo

[ebon stone]

if im asked to find its monotonicity, im probably good, but i still cant grasp boundedness properly

[ebon stone]

so i can just give a direct proof for it for any of these two?

[odd pivot]

Okay, so you want to show this sequence is bounded right ?
One thing you can do is to see if it converges to a limit, in which case it will be bounded

[subtle zinc]

is it increasing or decreasing or neither?

[odd pivot]

Are you required to find a specific bound ? Or just show it's bounded ?

[ebon stone]

alr lemme solve quickly gimme a sec

[subtle zinc]

oh nvm

i can't read

[sick fossil]

can you send more of the original question so we know what it’s asking?

[ebon stone]

in the set, just check for the bounded monotone theorem, but im preparing if the exams will have something like "check for bounded above / bounded below"

[odd pivot]

Okay, I see

[ebon stone]

if its just the bounded theorem i can just find the limit for the boundedness then just check for monotone then barabimbaraboom

but yk profs usually likes to throw curveballs

[odd pivot]

Okay so if you want to use bounded monotone thm it's good to practice finding bounds out of the blue indeed

One method that works for non-negative (≥0) sequences is to show they are decreasing

Because if it's decreasing and non-negative, it's bounded between 0 and the first term (cause the sequence never gets bigger since it's decreasing)

[ebon stone]

ah yeah makes sense

what about for alternating?

is it the same thing?

[odd pivot]

For alternating sequence you can look at the absolute value

[ebon stone]

then i can look from there?

[odd pivot]

Yep

The definition of a bound is a number which is always bigger than the absolute value of your sequence

A positive real number $M>0$ is a bound of $(u_n)_{n \geq 0}$ whenever for all $ n\geq 0$, $|u_n| \leq M$

[grand pond]

Twenty

[odd pivot]

So for alternating sequences, you can basically "forget" about the alternating part and just look at the absolute value

[ebon stone]

ah alright i actually didnt knew about that

thanks

uhhh can u also do a quick help for physics? im fine with it i just wanna clarify something

[odd pivot]

Sure

[ebon stone]

ah nvm my brain is too fried rn to explain the entire problem and process lmao

thanks agian

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[opal hollow]

Hi! I was trying to draft a lesson plan on multiplying fractions. I was told the answer is incorrect and that I messed up at alike bases. Can anyone explain this to me?

[runic hamlet]

having different denominators is completely fine for multiplying fractions

$\frac a b\cdot \frac c d = \frac{ac}{bd}$

[grand pond]

Denascite

[runic hamlet]

you are thinking of addition if I had to guess?

[opal hollow]

Yes but, why is it bad if i give it the same deno?

[runic hamlet]

$\frac a b + \frac c b = \frac{a+c}{b}$ where this time you need the same denominator

[grand pond]

Denascite

[runic hamlet]

you have to multiply top and bottom

not just top

so artificially increasing the denominators just makes every number bigger

for no benefit

[opal hollow]

I thought i did

[runic hamlet]

and later on you will cancel those numbers out again anyway

you didnt

[opal hollow]

Ohhh

The 10*10

I se

[runic hamlet]

writing $\frac25\cdot\frac32=\frac{4}{10}\cdot\frac{15}{10}=\frac{60}{100}=\frac35$ would be fine

[grand pond]

Denascite

[runic hamlet]

but really thats just way more complicated than $\frac25\cdot\frac32=\frac{6}{10}=\frac35$

[grand pond]

Denascite

[opal hollow]

Well, i wanted to teach alike bases so that they get in the habit of doing it when needed

But this helps

Thank you!

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[runic hamlet]

you should never do it for multiplication

it will just lead to errors

[modest oriole]

$\frac{2}{5}\cdot\frac{3}{2}=\frac{\cancel{2}}{5}\cdot\frac{3}{\cancel{2}}=\frac{3}{5}$

[grand pond]

Allen

[modest oriole]

this feels better

[runic hamlet]

yes but thats something you learn a bit later

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[twilit field]

I'm trying to compute this

[twilit field]

I was thinking It would have order (4), so it's the identity?

so (12345)

[sick fossil]

yes

[sick fossil]

3 is a fixed point of this

[twilit field]

Hmm?

[spiral rock]

The identity is not (1 2 3 4 5)

Its just the identity

[twilit field]

oops

right

so just (1)

[spiral rock]

I would notate it as e

But its just my preference

[twilit field]

(14356)=(14)(43)(35)(56)

Would that be right

and (d) would be

(17)(72)(25)(54)(14)(42)(23)(15)(54)(46)(63)(32)

[sick fossil]

Or you can multiply them first

[twilit field]

Or that

but yeah

I have to find this set

${(1324),(1342),(134),(132),(13)}$

[grand pond]

wai

[spiral rock]

(1 3) (2 4) is missing

[twilit field]

Maybe I'm mising something, but what does this mean

[zealous schooner]

Do you know what it means for a permutation to be odd or even?

[twilit field]

The parity of the number of elements in a transposition

[runic hamlet]

no

[twilit field]

Oh right

It can be expressed as the product of an odd/even number of transpositins

[runic hamlet]

ok

use that

[twilit field]

hmm

My concern is if $\alpha$ and $\beta$ are both even , we have a problem

[grand pond]

wai

[twilit field]

now this is true

ooh

Transpositions are commutatove

[runic hamlet]

no

[runic hamlet]

why

[twilit field]

Well, $\alpha^{-1}$ and $\beta^{-1}$ will be odd too

[grand pond]

wai

[twilit field]

I meant if both are odd

my bad

[runic hamlet]

and?

[twilit field]

Then Their product would be of odd order, would it not

[runic hamlet]

no

what happens to the number of transpositions when you multiply two permutations

[zealous schooner]

odd + odd = even

[twilit field]

We just add them

[runic hamlet]

yes

[twilit field]

and the sum of 4 odd numbers is even

Right I was taking their product for some reason

🤦

[zealous schooner]

Have you proved that the inverse of a permutation maintains the same parity?

[runic hamlet]

if not, look closely at this pic

[twilit field]

yes

Got it

Thanks!

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[median apex]

who is doing alevels...

[median apex]

😭

[vivid yoke]

Just send the question

who can help and want to help will help

[burnt wraith]

just send

[median apex]

;-;

ok

i ll do that later

cuz i need to find it ;-;

[burnt wraith]

It's alright

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@median apex Has your question been resolved?

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[grave kernel]

True or false and why: If a and b are irrational,
then a^b is irrational.

here, i was considering a^b as rational taking it as p/q
and then i have no idea
took log and im stuck

[tough shale]

a^b may be rational may be irrational

[lethal path]

are you allowed to use the fact that a = sqrt(2)^sqrt(2) is irrational?

[grave kernel]

well

i dont know

nothing is stated

[lethal path]

it's quite nontrivial (it's from the Gelfond-Schneider theorem)

but it's true

[tough shale]

for example, if you take $\sqrt{2}^\sqrt{2}$, it is irrational
but if you take $(\sqrt{2}^{\sqrt{2}})^\sqrt{2}$, it is rational

[grave kernel]

hmmm

but like

sqrt(2)^sqrt(2)
isnt that what im already supposed to prove tho

[grand pond]

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

[lethal path]

no

[lethal path]

which class is this for?

[grave kernel]

this is an exercise on methods of proving
induction, contradiction and stuff

[lethal path]

oh, like an easier example is e^(ln 2) = 2

[lethal path]

yeah so for this class you just have to state the contradiction

[grave kernel]

right

[lethal path]

you don't need to prove that ln 2 is irrational or anything

yeah if it's about intro to proofwriting or similar

[grave kernel]

right right

that makes much more sense honestly

thank you so much

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[tidal turret]

hey guys, I need some help with combinatorics
The question is
Let f in {h : {1,2,3,4} -> {1,2, . . ., 50} | h is injective} be defined as f(x) = x for 1 <= x <= 4. Find how many functions g in {h : {1,2,3,4} -> {1,2, . . ., 50} | h is injective} satisfy that #(Im(f) \ Im(g)) = 0 or (Im(f) \ Im(g)) = 4

[tidal turret]

what I have done so far is

knowing that f is injective, then its image its of cardinality 4, #(Im(f)) = 4

same for g, we know #(Im(g)) = 4

so theres two possibilities right, that the set Im(f) and the set Im(g) are disjoint is one possibility

the other possibility is that they are not, in that case specifically Im(f) = Im(g)

[radiant roost]

yes

[tidal turret]

so, first of all, the number of injective functions f, such that the image has cardinality 4, in the interval 1 <= x <= 4 is something along the lines of 4!
because say for example
f(1) = 1
f(2) = 2
f(3) = 3
f(4) = 4

like order doesnt matter
f(2) = 2
f(3) = 3
f(4) = 4
f(1) = 1
is still describing the same function

[radiant roost]

yes

[tidal turret]

but, after this is when it starts getting messy

when we try to find the number of possible gs

[radiant roost]

you're picking 4 from 46

these are "permutations"

[tidal turret]

wait a second dude

I got lost how we figured the number of possible fs are 4!

order doesnt matter, so its a combination of picking what ?

[radiant roost]

it's not a combination

it's 4P4

[tidal turret]

the domain of f is {1,2,3,4} but his codomain is {1,2, . . ., 50}

[radiant roost]

we're interested in the ones whose image is {1,2,3,4}

[tidal turret]

how many of those there are?

[radiant roost]

4P4 = 4!

[tidal turret]

but can you show a little example of fs that work?

[radiant roost]

f(n)=n

[lunar ocean]

well for instance f(1) = 1, f(2) = 2, f(3) = 4, f(4) = 3

[radiant roost]

f(n) = (n%4)+1

[tidal turret]

the problem is dat

[tidal turret]

no

[lunar ocean]

?

[tidal turret]

that doesnt follow f(x) = x

[lunar ocean]

that's an injective function and its image is {1, 2, 3, 4}

[tidal turret]

right?

[radiant roost]

what do you mean "follow"?

[lunar ocean]

that is indeed not the function f(x) = x

[radiant roost]

oh, right

[tidal turret]

but we needed that f(x) = x

[radiant roost]

i don't think so

oh... because f was defined in the problem

well call it g then

[tidal turret]

well first I was trying to find the number of f

[lunar ocean]

the number of f that satisfy what, exactly?

[tidal turret]

then we can get to calculate the number of g

[radiant roost]

there is only one f

[tidal turret]

yes, I think so too

[tidal turret]

just that f in the set of {h : {1,2,3,4} -> {1,2, . . ., 50} | h is injective} and that f(x) = x

[tidal turret]

what about g?

[radiant roost]

g with image {1,2,3,4} right?

[tidal turret]

Im(g) has cardinality 4

we know know exactly his image

[radiant roost]

yes Im(g) has cardinality 4

but you had mentioned two cases

[tidal turret]

yes

either Im(f) and Im(g) are disjoint or Im(f) = Im(g)

[radiant roost]

yes

[tidal turret]

so either Im(g) = {1,2,3,4} or Im(g) does not intersect {1,2,3,4}

[radiant roost]

yes

[tidal turret]

this is where it gets tricky

[fathom onyx]

Not really?

[radiant roost]

note that g cannot satisfy both conditions

[fathom onyx]

(also, that's not what f is)

(f is just defined as f(1) = x1, f(2) = x2, f(3) = x3, f(4) = x4, where x1 to x4 are distinct whole numbers between 1 and 50 inclusive)

[radiant roost]

no, f(x) = x

[tidal turret]

ye

[fathom onyx]

oh shit yh it's defined there

Then this is even simpler

You're then to determine how many injective functions g there are from {1,2,3,4} to {5, 6, ..., 50}

This should be simple to calculate

[tidal turret]

how

[fathom onyx]

Okay, first, do you understand why, when im(g) = im(f), how there were 4P4 such functions g?

[tidal turret]

I divided into two possibilities, that Im(g) = {1,2,3,4} or that Im(g) subset {5,...,50}

[fathom onyx]

yh ik I have read the above

ANd you've already done the first case

But do you understand WHY the first case's result is 4P4 = 24 such functions g?

[radiant roost]

do you feel comfortable with the first case?

[tidal turret]

kinda, yeah, since we are not restricted by g(x) = x
we can do
g(1) = 1
g(2) = 2
g(3) = 4
g(4) = 3
like I dont understand it completely, but I sort of have some intuition

[radiant roost]

another example...
g(1)=3
g(2)=2
g(3)=4
g(4)=1

[tidal turret]

yeah, but how 4p4

[fathom onyx]

(Right, so you didn't understand it?)

[tidal turret]

yeah

[fathom onyx]

so why claim you had an intuition

We have to pick four numbers for each of g(1), g(2), g(3) and g(4)

Starting at g(1), we have four such choices (in the case that im(g) = im(f))

Once it is picked, then we've only got three choices for g(2)

Then once that is picked, we've got two choices for g(3)

And by this point, we've only got the one remaining choice for g(4)

@tidal turret Can you see how this gives us 24? (there's a multiplication to do)

[tidal turret]

Yeah, sort of gettt it, is kind abstract though

[fathom onyx]

It really isn't that abstract

It's akin to the question: "How many ways can you seat 4 people on a bench?"

In the first space (e.g. leftmost) on the bench you have 4 choices of person to seat

Then in the second, you have 3 choices left

Then in the third, 2 choices left

Then in the last seat, whoever's remaining

[tidal turret]

yeah

yeah, so it sort of means 24 seats, but then it gets messy as hell, when Im(g) subset {5,...,50}

46 P 4?

[radiant roost]

it's exactly 24

not "sort of"

yeah 46 P 4 is right

[fathom onyx]

idk why you keep saying "sort of means" here

[tidal turret]

I mean, 4 ways of sitting the first dude

but why we multiplicate them

why not just sum them?

[fathom onyx]

-# (the verb is "to multiply")

[tidal turret]

because they are all happenning simultaneously?

[radiant roost]

there are 4 ways of choosing f(1). for each of these, the number of ways N to pick the remaining three values is the same. so the total is 4N

[tidal turret]

there is an multiplication principle

[radiant roost]

in other words there are N ways with f(1)=1, N ways with f(1)=2, N ways with f(1)=3, and N ways with f(1)=4

so 4N, and not 4+N

[tidal turret]

we can make like a tree diagram out of this

coutning really is hard

[tidal turret]

4N = N + N + N + N

[radiant roost]

yes

[tidal turret]

then for f(2) we have 3N seats

because one is picked already

[radiant roost]

sure, maybe you should change the variable

[strange notch]

Can anyone help with 13 and 14

[fathom onyx]

!occupied

BOT DOWN WTF

[radiant roost]

try one of the available rooms, like #help-31

[fathom onyx]

"Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see https://discord.com/channels/268882317391429632/488120190538743810⁠ for instructions)."

-# thanks, me; have a cookie, me

[tidal turret]

how many ways of arranging ABCD there are?
4 ways of arranging A, 3 ways of arraining B, 2 ways of arranging C , 1 way of arranging D

sort of like that, all them happening simultaneously

[radiant roost]

it's the same problem

[tidal turret]

I mean, kinda get it, is tricky tho

[radiant roost]

yeah

[fathom onyx]

Genuinely slightly concerned what here you're finding "tricky"

How they're the same problems? or the computation itself?

[tidal turret]

counting is tricky

[tidal turret]

the counting itself

4 ways of arranging the first letter

3 ways of arranging the second letter

[fathom onyx]

https://brilliant.org/wiki/combinatorics/

Highly recommend you read through this then

[tidal turret]

this explains it

of the why its multiplication

[radiant roost]

that is so vague

[fathom onyx]

yeh that''s not more explicatory than what we've said

[tidal turret]

I mean, this multiplication principle

[fathom onyx]

That's like if we told you 8+3 = 11; and you said "I understand it now because the calculator said 8+3 = 11"

It's honestly not functionally different here

[radiant roost]

i think this is a little different from the permutations

[tidal turret]

its the events happening one after the other

[radiant roost]

it's like saying how many ways can you arrange ABCD on one bench and EFG on a second bench

it's 4! times 3!

[tidal turret]

yeah I sort of get it now

but as I said, the counting itself, not about the number of ways to seat x people, but like counting in general is tricky

[radiant roost]

yep

i've had the moment of like "wait why doesn't it work if i calculate it this way?"
and seen others ask the same thing

[tidal turret]

well multiplication principle and addition principle is useful for simple counting

but then, if Im(g) subseteq {5, ..., 50} theres like 4! x 46P4 ways of this shit happening

[radiant roost]

no, it's addition here

4! + 46P4

[tidal turret]

you mean 4! for Im(g) = {1,2,3,4} and 46P4 for Im(g) subseteq {5, ..., 50}?

[radiant roost]

yes

[tidal turret]

that sort of tracks, but as I said, counting is tricky dude

[radiant roost]

no argument here

[tidal turret]

,w 1 + 4! + 46 Pick 4

[grand pond]

Results provided by WolframAlpha

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

Requested by renatoxbz

[tidal turret]

wait a second

4! ways of picking f(x) = x

yes

but

[tidal turret]

no dude

its 4! + 4! x 46p4

[fathom onyx]

huh

[radiant roost]

no

[tidal turret]

because every f gives another

[fathom onyx]

No

You have a FIXED f

[radiant roost]

yeah there is only one f

[fathom onyx]

You said it yourself right here

[tidal turret]

the issue is that

[radiant roost]

is the problem mis-stated?

[tidal turret]

46 pick 4 is happening simultaneously with 4!

[fathom onyx]

Do you understand what "nPr" means?

[radiant roost]

g is either of the first type or the second type

so the total number of functions g is the sum

[tidal turret]

yes maybe

[tidal turret]

right

Im(g) = {1,2,3,4} or Im(g) subset {5,...,50}

[fathom onyx]

Right, so this rule applies

You have to add them

There's no further multiplication at this step

[tidal turret]

I might have messed up the translation

[fathom onyx]

!xy is what I would say, but the bot is down

[tidal turret]

haha

Ejercicio 1. Sea $\mathcal{F} = {h : {1,2,3,4} \to {1, 2, \text{ } \dots \text{ } , 50} \text{ } / \text{ } h \text{ es inyectiva }}$. $\$ Definimos en $\mathcal{F}$ la relacion $\mathcal{R}$ como $$f \mathcal{R} g \quad \text{si y solo si} \quad #(\operatorname{Im}(f) \setminus \operatorname{Im}(g)) = 0 \text{ o } 4.$$ a) Analizar si $\mathcal{R}$ es una relacion reflexiva, simetrica, antisimetrica y/o transitiva. $\$ b) Sea $f \in \mathcal{F}$ definida como $f(x) = x$ para $1 \leq x \leq 4$. Calcular cuantas funciones $g \in \mathcal{F}$ satisfacen $f \mathcal{R} g$

[grand pond]

Renato

[tidal turret]

this exercise was b) of this original problem

[fathom onyx]

The translation then is correct

[tidal turret]

I mean, yes and no

because I proved the relation is symmetric, reflexive, and that is not transitive and not antisymmetric

but assuming that R is symmetric and R is reflexive

[fathom onyx]

And that has to do with anything related to part b, why?

[tidal turret]

fRf is possible and fRg => gRf

[fathom onyx]

A if and only if B

So any time you see the statement A, you can replace it with B

So you replace "fRg" with the definition there, # (Im(f) \ Im(g)) = 0 or 4

I see no problem, translation-wise

[tidal turret]

yes but if fRg holds then gRf aswell

[fathom onyx]

In both senses of "translation" (mathematical and linguistic)

[fathom onyx]

why are you bothering yourself with gRf

[tidal turret]

so #( Im(g) \ Im(f)) = 0 or 4

[fathom onyx]

Again who cares

You're told to count the number of functions g that satisfy fRg

[tidal turret]

I think you are right

[fathom onyx]

i.e. the number of functions g that satisfy # (Im(f) \ Im(g)) = 0 or 4

That's literally what you translated it to

Why are you doubting this

[tidal turret]

this is still the same two possibilities that Im(f) = Im(g) or Im(g) subseteq {5, ..., 50}

[tidal turret]

because

this is what someone did

[radiant roost]

the answer doesn't seem right

[tidal turret]

it is not, but there is a note from the one who corrected it

[fathom onyx]

[the green writing is the marking; it has the correction]

[tidal turret]

you and she is saying that the events are disjoint, but like the corrector is saying that each permutation gives a different function

Is tricky

no idea why is that even the case

but I think our analysis is kinda correct, like the number of possible fs is 1, and theres two possibilities that Im(g) = {1,2,3,4} or that Im(g) subset {5,...,50}

[tidal turret]

.close

[tidal turret]

mother fucking bot

the bot has me trapped as a hostage

anyone seeing this?

[tidal turret]

.close

.close

.close

.solved

.close

.solved

.close

mother fucking bot dude

.close

[tidal turret]

our reasoning is correct dude

the answer is just 4! + Permutation(46,4)

that is equivalent to 4! + 4! * combinations(46,4)

[tidal turret]

the answer that the dude who took this exam gave is wrong because they did 4! + combinations(46,4)

but like, 4! * combinations(46,4) = permutations(46,4)

I was just overcomplicating the problem, and I didnt explained myself correctly but you guys were right from the beginning

and the answer from the guy who corrected is right aswell

both of us were right

because 4! * combinations(46,4) = permutations(46,4)

generally speaking: k! * combinations(n,k) = permutations(n,k)

this was the huge misunderstanding

in this case the one who took this exam didnt recognized that order matters

order matters because we can have f(1) = 5, f(2) = 6, f(3) = 7, f(4) = 8, and have f(1) = 8, f(2) = 7, f(3) = 6, f(4) = 5 and both be different functions

both are valid permutations

anyways I appreciate the help as always

this one was a hardone, because I havent reviewed the basics of combinatorics

but in fact it shouldnt have been too hard

is just that I am behind on the material

I appreciate it

[modest heart]

The bot is down

[tidal turret]

no shit

[modest heart]

[This channel is available to anyone needing help now]

[civic gorge]

Hi

i'm confused on why combinations with repetitions and stars and bars problems have different formulas

[pearl hull]

I’ll pin this question, consider it a temporary occupation.

[civic gorge]

tyty

Say I have 5 different types of candies and i want to buy 10 candies

the number of ways to do that is 14 C 4

isnt this the same as saying numbers of ways to choose 10 numbers from 0-4 when repetitions are allowed

i have realized

i might be stupid

k in stars and bars is the types of candies

k in combinations with replacement is 10

they are equivalent

god bless me

.closed

[pearl hull]

Are you done with the question?

[civic gorge]

yep

[pearl hull]

Alright

Available

[tidal turret]

3. Dada la funcion
   $$f(x,y) = \begin{cases} \frac{x^2\sin(y)}{\frac{1}{3} (x^2 + y^2)} &\text{ si } (x,y) \neq (0,0) \ 0 &\text{ si } (x,y) = (0,0) \end{cases}$$ (a) Hallar todos los $v \in \mathbb{R}^2$ de norma 1 tales que $\pdv{f}{v}$ $(0,0) = 3\$
   (b) Decidir si $f(x,y)$ es diferenciable en $(0,0)$.

[grand pond]

Renato

[tidal turret]

I need some help with b)

how do I check if this piecewise is differentiable at (0,0)?

[tidal turret]

sky you dont need to do this 😭 tho I appreciate you being a manual bot

[pearl hull]

LOL

[tidal turret]

,align &f(x,y) = \begin{cases} \frac{x^2\sin(y)}{\frac{1}{3} (x^2 + y^2)} &\text{ si } (x,y) \neq (0,0) \ 0 &\text{ si } (x,y) = (0,0) \end{cases} \ &\lim_{(x,y) \to (x_0, y_0)} \frac{f(x,y) - f(x_0, y_0) - \nabla f(x_0,y_0) \cdot (x - x_0, y - y_0)}{|(x-x_0, y - y_0)|} = 0 \ &\lim_{(x,y) \to (0, 0)} \frac{f(x,y) - f(0, 0) - \nabla f(0,0) \cdot (x - 0, y - 0)}{|(x-0, y - 0)|} = 0 \ &\lim_{(x,y) \to (0, 0)} \frac{f(x,y) - \nabla f(0,0) \cdot (x , y )}{|(x, y )|} = 0 \ &\nabla f(x_0, y_0) = \left(\pdv{f}{x} \ (x_0,y_0), \pdv{f}{v} \ (x_0,y_0)\right) \ &\nabla f(0, 0) = \left(\pdv{f}{x} \ (0,0), \pdv{f}{v} \ (0,0)\right) \ &\pdv{f}{x} \ (x_0,y_0) = \lim_{h \to 0} \frac{f(x_0 + h, y_0) - f(x_0, y_0)}{h} \ &\pdv{f}{x} \ (0,0) = \lim_{h \to 0} \frac{f(0 + h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{f(h, 0)}{h} = \lim_{h \to 0} \frac{\frac{h^2 \sin(0)}{\frac{1}{3}(h^2 + 0^2)}}{h} = \lim_{h \to 0} \frac{3h^2 \sin(0)}{h(h^2 + 0^2)} = 0 \ &\pdv{f}{y} \ (0,0) = \lim_{h \to 0} \frac{f(0, 0 + h) - f(0, 0)}{h} = \lim_{h \to 0} \frac{f(0, h)}{h} = \lim_{h \to 0} \frac{\frac{0^2 \sin(h)}{\frac{1}{3}(0^2 + h^2)}}{h} = \lim_{h \to 0} \frac{3 \cdot 0^2 \sin(h)}{h(0^2 + h^2)} = 0 \ &\nabla f(0, 0) = \left(\pdv{f}{x} \ (0,0), \pdv{f}{v} \ (0,0)\right) = (0,0) \ &\lim_{(x,y) \to (0, 0)} \frac{f(x,y)}{|(x, y )|} = \lim_{(x,y) \to (0, 0)} \frac{f(x,y)}{\sqrt{x^2 + y^2}} = \lim_{(x,y) \to (0, 0)} \frac{1}{\sqrt{x^2 + y^2}} \cdot \frac{x^2\sin(y)}{\frac{1}{3} (x^2 + y^2)} = \lim_{(x,y) \to (0, 0)} \frac{3x^2\sin(y)}{ \sqrt{x^2 + y^2} (x^2 + y^2)} \ &\lim_{r \to 0} \frac{3(r\cos(\theta))^2\sin(r\sin(\theta))}{ \sqrt{r^2} (r^2)} = \lim_{r \to 0} \frac{3r^2\cos^2(\theta)\sin(r\sin(\theta))}{ \sqrt{r^2} (r^2)} = \lim_{r \to 0} \frac{3\cos^2(\theta)\sin(r\sin(\theta))}{r} = \lim_{r \to 0} 3\cos^2(\theta)\sin(\theta) \cdot \frac{\sin(r \sin(\theta))}{r \sin(\theta)} \ &3\cos^2(\theta)\sin(\theta) \neq 0 , \forall \theta

[grand pond]

Renato

[steel crest]

!nosols

!noans

.nosols

[grim vector]

This one shouldve work

[steel crest]

bot is broken it seems

[tidal turret]

I need some help with this

4. Sea $\varphi(x,y) = \left(\frac{1}{\pi} \tan\left(\frac{\pi y}{x}\right), \sqrt{x^2 - 3\cos(\pi y)}, x + 4y\right)$ \\ (a) Calcular $D_{\varphi}(1,-1)$.\\ (b) Sea $f : \mbb{R}^3 \to \mbb{R}$ una funcion diferenciable tal que $f(0,2,-3) = 1$. \\ Sabiendo que: \begin{itemize} \item $\text{La derivada direccional de } f \text{ en la direccion } \\ v = \left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right) \text{ es } \frac{7}{\sqrt{2}}$ \item $\text{La derivada direccional de } f \text{ en la direccion } \\ w = \left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} , 0 \right) \text{ es } \frac{-13}{\sqrt{10}}$ \item $\pdv{(f \circ \alpha)}{t} \ \left(\frac{\pi}{4}\right) = -3 \text{ para } \alpha(t) = \left(\cos^2(2t), 2\sin(2t), -3\tan(t)\right)$ \end{itemize} Calcular la ecuacion del plano tangente de $f \circ \varphi (1, -1)$

[grand pond]

Renato

[steel crest]

what's sea

[tidal turret]

let

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

,w partial derivative with respect to x of (1/pi)*(tan((pi * y)/x))

[grand pond]

Results provided by WolframAlpha

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

Requested by renatoxbz

[gleaming jasper]

Hi iam a teacher in math

Who want a Learn a little

[midnight plank]

@tidal turret Has your question been resolved?

[bold peak]

!occupied please claim an available help channel

[midnight plank]

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

[midnight plank]

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[burnt wraith]

,rccw

[grand pond]

Couldn't find an attached image in the last 10 messages.

Sep 17, 2025, 1:39 PM

[burnt wraith]

,rccw

[rich yew]

much better

[urban needle]

wait

nth

answer is 15 solution

but i got only 14

.close

[midnight plank]

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[polar nexus]

Need help

[lyric charm]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[polar nexus]

I am stuck

[lyric charm]

stuck where?

[polar nexus]

Between

[lyric charm]

between what?

[polar nexus]

Like i am halfway

But I am getting it wrong

Repeatedly

[lyric charm]

ok, so show your work.

that way we can see where you are messing up.

[lyric charm]

,rcw

[lyric charm]

last term

$\frac{5}{2} \cdot 5 \neq \frac{10}{2}$

[grand pond]

Ann

[lyric charm]

and first term, the ^2 is nowhere to be seen.

the first term should be $2 \times \paren{\frac52}^2$ \ and not $2 \times \frac52$

[grand pond]

Ann

[lyric charm]

finally in the second term i am seeing an extra five pop out of nowhere

$\sqrt{3} \times \frac{5}{2} \neq 5 \sqrt{3} \times \frac{5}{2}$, surely.

[grand pond]

Ann

[lyric charm]

@polar nexus these are your mistakes.

are you fixing them?

[midnight plank]

@polar nexus Has your question been resolved?

[midnight plank]

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[last slate]

yo

[pearl hull]

yo yo yo

Any questions today?

[midnight plank]

@last slate Has your question been resolved?

[rich yew]

My guy do you have a question?

[cobalt rover]

Q 48

[rich yew]

THIS ONE AINT OPEN

it’s occupied

#help-3

[midnight plank]

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[last slate]

Hello. Could someone please assist me to solve part a) ?

[last slate]

This is what I did, but I am not sure if I answered the question

Hello. I need help please

Thank you. I have to run

.close

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[tidal turret]

how do I prove this limit is 0

like, I have tried multiple ways

nothing really is solid

[forest bolt]

what is sen

[full leaf]

sine (in some languages)

[tidal turret]

I mean I have some ideas

dude I think I got it

[twilit jetty]

That works, nice job

btw you also could use sin(w^3) <= 1 instead of sin(w^3) <= w

[tidal turret]

ye

[twilit jetty]

Oh wait nvm you can’t do that

[tidal turret]

nice observation

?

[twilit jetty]

you need w^3 so that it cancels with the w^2 on the bottom

[tidal turret]

shit is so nasty

[twilit jetty]

That’s why half of it didn’t matter

[tidal turret]

any tips for basic calculus?

[twilit jetty]

You’re doing good on basic calculus already

[tidal turret]

wdym

[tidal turret]

exam is in a week dude I'm cooked

[tidal turret]

care to elaborate?

[midnight plank]

@tidal turret Has your question been resolved?

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[open dew]

im stuck at the end ive looked it up and watched a video I just dont understand it

[slender walrus]

ur so close, this is stuff you've most definitely done before

how would you solve something like
5p = 7

[open dew]

divide by 5?

[slender walrus]

both sides, yes

[open dew]

yea

[slender walrus]

exactly the same idea/principle applies here

[open dew]

so id divide by the -2x+4?

[slender walrus]

something was multiplied to the p you wanted to isolate,
and that's what you divided by
same idea for the y

yes

[open dew]

would I distribute the negative to the 1?

[slender walrus]

wdym

[open dew]

like when I put the y under the left side theres a negative does it turn the whole 2x+1 into -2x-1?

[slender walrus]

can you write down the whole equation you're getting

[open dew]

yeah

[slender walrus]

before the attempt at further simplification

[slender walrus]

that'd be wrong

[open dew]

okay so theres no negative 1 its just a positive

[slender walrus]

would still be wrong

[open dew]

oh

[slender walrus]

note what you're actually supposed to divide by

so id divide by the -2x+4?

[open dew]

okay

[slender walrus]

do that first, because you're trying to do multiple things at once and ended up treating that as
-(2x+4)

[open dew]

yeah okay ill try that

this look right?

[slender walrus]

no

same issue

you're still treating it as -(2x+4)

[open dew]

keep the minus in the denominator?

[slender walrus]

yes

[open dew]

okay

[slender walrus]

the - in its current state applies only to the 2x

[open dew]

alrighttt

[slender walrus]

to be safe, chuck whatever you're dividing by in the denominator
worry about what to do after

[open dew]

okay what now then?

[slender walrus]

what do you have now?

[open dew]

yeah

[slender walrus]

that wasn't a yes/no question

[open dew]

I dont know would i have to subtract the numerator 1 over then divide the 2?

[slender walrus]

no

show me what you have after that division only

do nothing else

[slender walrus]

and ur done

[open dew]

this is it?

[slender walrus]

yes

[open dew]

ohhh okay thank you for your help

.close

[midnight plank]

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[slate ferry]

this guy

so if A (area) is so small (like in the wire) shouldnt R (resistance) be so large

[eternal pawn]

Because rho (P) is very small

[slate ferry]

oh...

hm

so rho p in every wire is so small that dividing it by a super small value like A doesnt make it large

ig that makes sense

lets say we increased L ----> 10L

A will decrease to 1/10 A to make Vol constant

R got 10 times larger

or wait no

10/(1/10)

it got a 100x larger

so how is smth that is 100x larger still.... so small? is resistance in real life examples just multiples of 0.0000000000000000001 or smth

[robust isle]

depends on the material, but yeah for metals it's pretty f-in small (table from wiki)

[slate ferry]

yeah i see that makes sense

[eternal pawn]

You don't have .1 mm wires do you

[slate ferry]

I... dont think i do 😭

[vivid crow]

wires like copper or aluminum have an extremely low resistivity

even though the wire is thin the p of copper is so incredibly small that it keeps the overall resistance (R) of the wire close to zero

[near cliff]

In general for experimental setups the resistance of the wires can pretty much be neglected

[slate ferry]

I seeee

[near cliff]

For larger wires like power lines though, resistance starts to become an issue

[slate ferry]

so i was just- confused

alright this is a lot clearer now

[slate ferry]

ig we didnt get there yet

ty everyone!

.close

[midnight plank]

Channel closed

Closed by @slate ferry

Use .reopen if this was a mistake.

[midnight plank]

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[last slate]

yoo

[last slate]

whats an identity matrix

I

denoted by capital i

[small jasper]

[https://en.m.wikipedia.org/wiki/Identity_matrix]

[last slate]

how do you turn a matrix to its identity matrix

[small jasper]

What do you mean “its identity matrix”? There’s only n by n one identity matrix for a given n.

[midnight plank]

@last slate Has your question been resolved?

[small jasper]

?

https://math.libretexts.org/Bookshelves/Linear_Algebra/Fundamentals_of_Matrix_Algebra_(Hartman)/01%3A_Systems_of_Linear_Equations/1.03%3A_Elementary_Row_Operations_and_Gaussian_Elimination

Are you trying to refer to this?

or ...

[last slate]

i think i just didnt word it properly

i need help with part b

i dont understand what 2 is here

[steel crest]

2 is 2

[main current]

2 is an integer. You can multiply an integer by a matrix, which gives you a new matrix

Take A. You just multiply all of the values in the matrix by 2, and you have 2A

[last slate]

i just multiply by the inverse?

or by A then inverse it?

[steel crest]

you didnt ask what it means to multiply by 1/(ad-bc)

why is that more obvious?

[last slate]

its the parenthesis

the parenthesis make it seem like you need to multiply 2 first

[steel crest]

sure

[last slate]

but the question says to use the answer in part a

so if i use the answer in part a then i just mulyiply the result by 2

[steel crest]

you can prove a theorem that

(cA)^-1 = c^-1 A^-1

want to prove that?

[brisk iris]

If m is a scalar and M a matrix.
(mM)^x = m^x M^x

[last slate]

what even is c^-1

[steel crest]

whats 2^-1 ?

[last slate]

i thought that theroem applies to two matrices

not a scalar

[steel crest]

2 is a matrix

[2]

1x1 matrix

[last slate]

oh

[steel crest]

if you wish to think of it that way

you dont need to think of it that way but it works

of course we have that

[2]^-1 = [2^-1]

[last slate]

so that would be [2] * x = I?

[steel crest]

not sure what youre asking

[last slate]

I being the identity matrix

wait nevermind

im confusing stuff

in general how do you find the inverse of any matrix?

other than that formula

[steel crest]

there are algorithms

it's not straightforward

[last slate]

my book is terrible at explaining

is there any place i can learn them

[steel crest]

plenty,but, what is your questiin here, youre a bit all over the place

[last slate]

i just dont think i understand the concept itself

[steel crest]

which concept

[last slate]

inverse of matrices

[steel crest]

let M be an nxn matrix

suppose there exists an nxn matrix N with the property that

MN=NM=I

then N = M^{-1}

that's the definition

same as the definition of a multiplicative inverse of a number x

suppose there exists a number y such that xy=yx=1

then y = x^{-1}

or "1/x"

[last slate]

why n x n and not m

it has to be a square matrix or else its non invertible?

[steel crest]

correct

otherwise MN != NM

[last slate]

interesting so you can only invert square matrices

[steel crest]

you can have right-inverses and left-inverses of non-square matrices but they're not really talked aboutr as much because theyre not so useful

or important

[last slate]

so if you have an nxm matrix you cant find an inverse for it correct

[steel crest]

if m!=n

[last slate]

right

[last slate]

and this applies for any nxn matrix right

any matrix thats able to be inversed

[steel crest]

not all square matrices have inverses

[last slate]

oh

[steel crest]

but it is the definition of inverse for all invertible square matrices

for example, look at the formula you have for a 2x2 inverse

if ad-bc = 0 then there is no inverse

division by 0

[last slate]

right

[cedar pawn]

Ehhhhh

That would violate dimensionality conditions

[warm rivet]

a 2x2 matrix represents a transformation of 2 dimensional space (by matrix vector multiplication). The inverse is just the transformation you need to "invert" the operation done by A

[steel crest]

nu nu

[warm rivet]

if you first apply A and then scale by 2, to invert you must apply A^{-1} and then scale by 1/2 to counteract the 2 (or you can use the formula)

[steel crest]

unless the matrix is also 1x1 🧑‍🔬

[cedar pawn]

But it's 2x2 in this case

[steel crest]

nu nu

you're correct

[cedar pawn]

I agree it would be nice if everything worked nicely if you did it like that

But sadly it doesn't

[steel crest]

so up to the person to prove that (cA)^{-1} = c^{-1}A^{-1}

[cedar pawn]

Do you want them to prove that?

[steel crest]

theyre not even here anymore i think

but i dont have a particular desire either way, theyre not MY student, but i think it would be a good exercise

[last slate]

bro whats the inverse of numbers

whats the inverse of 2

[upper cedar]

Depends on the operation you're inverting. Think about it: the inverse for addition and multiplication are going to be different.

[last slate]

my book fing sucks

the way it has things written just pisses me off

[cedar pawn]

The "inverse of a matrix" is more specifically the multiplicative inverse

[steel crest]

I wrote this above

[cedar pawn]

The inverse A^-1 of a matrix A (more specifically the multiplicative inverse of A) is the matrix such that A A^-1 = I, the identity matrix.

[upper cedar]

-# Oh sorry I should have inferred multiplicative inverse from the context.

[steel crest]

have you never seen the notation ebfore, like

[last slate]

what does multiplicative inverse mean

[steel crest]

$9^{-1}$

[grand pond]

gfauxpas

[steel crest]

?

[cedar pawn]

So imagine we have a number x

[last slate]

yeah but what is that

[cedar pawn]

We multiply it by 2

[last slate]

ok

[steel crest]

you didnt learn negative exponents?

[cedar pawn]

The multiplicative inverse is the number that we need to now multiply it by to get back to x

[last slate]

the book says inverses arent exponents

[cedar pawn]

Which is 1/2

[steel crest]

this is a number

[last slate]

forgot if its the book or what

[cedar pawn]

The multiplicative inverse of a number x is 1/x

[steel crest]

9 is a number

[steel crest]

here, did you miss this?

what I wrote in this part?

[last slate]

i just dont understand anything rn its that moment where the frustration is happening

[cedar pawn]

Okay let's just pause for a sec

Let's start with numbers

[last slate]

okay

[cedar pawn]

We have a number x

[last slate]

ok

[cedar pawn]

The multiplicative inverse of x is the number we need to multiply x by to get 1

So in other words, xy = 1

y is the multiplicative inverse of x

We can write y = x^-1

For numbers, the multiplicative inverse is just 1/x

So y = x^-1 = 1/x

And we have x (1/x) = 1

Which is correct

Happy with that?

[last slate]

1 is our goal for all inverses?

[cedar pawn]

1 is the multiplicative identity

So yes

An identity is something that leaves the original object unchanged

[last slate]

okay so an inverse of a number is a another number you multiply by the original number to obtain 1

okay

[cedar pawn]

Yes

[last slate]

got it

same thing in matrices?

[cedar pawn]

Well

It's slightly different with matrices

Because you can't do 1/matrix

But the principle is the same

So now we have a matrix M

We want M M^-1 = I

[last slate]

okay

[cedar pawn]

I is the identity matrix

Which is the matrix equivalent of 1

[last slate]

right

[cedar pawn]

For numbers, we have:
x(1) = x

For matrices, we have:
MI = M

[last slate]

ahhh

[cedar pawn]

So I leaves the original matrix unchanged

So it's like multiplying by 1

[last slate]

right

[cedar pawn]

Multiplication is the primary operation you do with matrices

So an "inverse matrix" is the multiplicative inverse

[steel crest]

in fact matrices were invented in order to do matrix multiplication

historically

[cedar pawn]

And the identity matrix is the multiplicative identity

The multiplicative bit is just dropped because that's always the case

[last slate]

right

like x = x*1

[cedar pawn]

Yep

[last slate]

M = M * I

[cedar pawn]

Multiplying by 1 leaves x unchanged

[steel crest]

best not to use a symbol for matrix multiplication

[cedar pawn]

Matrix multiplying by I leaves M unchanged

[last slate]

right

[cedar pawn]

So for numbers, we have x x^-1 = 1

For matrices, we have M M^-1 = I

Is that okay?

[steel crest]

Greenie I think hes just about ready

[last slate]

yes

[cedar pawn]

What is he ready for?

[steel crest]

to prove:
if c is a scalar
and M is an invertible matrix
then
$(cM)^{-1} = c^{-1} M^{-1}$

[grand pond]

gfauxpas

[steel crest]

non-zero scalar*

[last slate]

i dont understand that

[steel crest]

well let's break it down

[last slate]

can we continue from the og pase

pace*