help-49

[sharp cairn]

but here it says n = 1 its correct

and im confused

do you guys understand whats my issue?

[undone ermine]

what have you calculated?

[midnight plank]

@sharp cairn Has your question been resolved?

[sharp cairn]

i mean i understand

but at the same time

they are not continous]

so it should have no extreme local points right?

[midnight plank]

@sharp cairn Has your question been resolved?

[midnight plank]

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[twilit field]

Prove that if $y,x \in \R, y-x>1$ There's an integer b/w $x$ and $y$

[grand pond]

wai

[twilit field]

I have no idea what to do

I can't use completness as I'm using this to prove completness

[snow ledge]

what is the longest possible interval that doesn’t contain an integer?

[twilit field]

1

[snow ledge]

Yeah

[twilit field]

I know that, but does that really work as a proof in analysis

[snow ledge]

Well what is the length of the interval (x,y)?

[twilit field]

|y-x|; |y-x|>1

oh

got it

thanks

[snow ledge]

You’re welcome

[twilit field]

wait, no

that doesn't prove there's an integer, does it

[snow ledge]

If the interval (a,b) doesn’t contain an integer, the maximum possible length of that interval is 1 (this seems obvious but try to prove this maybe)
Length of (x,y) = y-x > 1

I mean i kinda just reworded the question because the first claim still needs to be proven

[twilit field]

What I was trying was setting up an inequality that would force this

I've already proven that every field element is b/w two integers

[snow ledge]

Use the floor of y maybe?

Prove that floor of y is between x and y (it isn’t between them when y is an integer but in that case you can do the floor of y - 1)

[twilit field]

In that case won't I have to prove the floor funtion is well defined

[snow ledge]

I mean it is well defined

What in its definition seems uncertain

[ornate flame]

hello

[twilit field]

hi

[ornate flame]

grt and you

[twilit field]

Nothing, but I'm doing real analysis so I'll have to prove everything practically

[ornate flame]

yeah thas the spirit

[midnight plank]

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[chilly cobalt]

Translation : Let the 2022nd and 2023rd decimal of the sum […] be a and b respectively. Find the value of 10a + b (no calculator for this test btw)

[chilly cobalt]

i think i found the intended sol? (if im not wrong that is)

and we can just find the n = 2022, 2023 terms since the others dont contribute

is this right?

[noble vapor]

you might want to translate the question

[chilly cobalt]

i.. did...

does nobody read what i type

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

<@&286206848099549185>

[midnight plank]

@chilly cobalt Has your question been resolved?

[desert siren]

<@&268886789983436800>

[swift blade]

<@&268886789983436800>

[chilly cobalt]

anyways this is my qn ^

[formal blade]

I'll delete some of the irrelevant convo

[bright holly]

I tried a similar approach but the alternating signs are problematic

For example in case you wanna find the 4th term in an arbitrary series similar to above with alternating signs your term might be something like 0.0248-(16/(100000)+...) then it seems that your term will be 8 but any term after it may change the value and due to carry overs in subtraction you dk where to terminate the series

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[tired maple]

my teacher said i cant use this discord server anymore cuz then it proves that i need help of others to succeed in math and im not doing it myself so i wont be as active :( but i still got a question which i answered but need someone to check with.
its 6x-x^2 converted to m-(x-n)^2

[tired maple]

i got the answer

-(x+3)^2-9

so would the answer be -9-(x+3)^2??

[lyric charm]

your teacher forbade you from using this discord specifically bc it proves you need help from others and that's somehow a bad thing???

[lyric charm]

anyway double-check your arithmetic and/or working bc this is incorrect

[tired maple]

$6x-x^2\-x^2+6x\-[x^2+6x+(3)^2-(3)^2]\-[(x+3)^2-3^2\-(x+3)^2+9$

[grand pond]

Tan

[tired maple]

yes

[lyric charm]

oh but this is correct

[lyric charm]

your teacher is, to put it in technical terms, an idiot.

[tired maple]

because it implies that i cannot do problems by myself and have to rely upon others to give me the answers and when i gotta do the exams i will miserably fail

because i never did anything by myself

[lyric charm]

we don't give you answers

[tired maple]

and asked other people to do it for me

i knpw

[lyric charm]

relying upon HELP is not a bad thing at all by itself

[tired maple]

i tried telling him that

[lyric charm]

your teacher is, to put it in technical terms, an idiot.

[tired maple]

well to be fair

you would consider this basic right

[lyric charm]

anyway how is he going to enforce it

[lyric charm]

"basic" is extremely subjective

[tired maple]

well he just told my mom to look over me

and ensure im doing stuff by myself

[tired maple]

my teacher very boldly said to my mother "the stuff we are doing right now is very basic so if he is struggling at this, then he'll definitely have to struggle more in the future, so i suggest you put him into normal math instead of accelerated.

🤷

[lyric charm]

[runic hamlet]

I never imagined I would hear a teacher say that getting help is bad. what exactly do they think they are doing as their job

[tired maple]

i think he just thinks that u guys give me the answers

[lament knoll]

fr

[lyric charm]

to teachers like that guy, it must be systemic child abuse

[tired maple]

and that is obv bad

[lament knoll]

tell him to join the server lmao

[lyric charm]

well ok like he can go pound sand tbh

[runic hamlet]

well you did the good step of ignoring him anyway

[lyric charm]

anyway back to the problem at hand: this working is correct and gives the answer 9 - (x+3)^2 and i hereby VERIFY that the answer YOU obtained is correct.

[tired maple]

:D

anyways if i dont pass the next text tmrw im getting kicked out of accel

and i cant just pass

i have to ace it

[tawdry laurel]

o shit good luck

[tired maple]

ty

.close

[midnight plank]

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[tawdry laurel]

This is pure stupidity

sigh

gl

[midnight plank]

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[bright holly]

@chilly cobalt

I wrote the term as something/(1-x) which can be expanded into something*(1+x+x^2+...) and then I created factors pairs and computed 2^mew2(n) at those n values

I'm not sure if I'm correct but this seems correct tho it is a very laborious method so in case you find a shorter method or some slick trick do lmk

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[fossil ember]

So we have 3+d points that make up planar real number straight line polygons, lines, and points. We have eyes center point and a view plane center with actual XYZ being XYZ. Projection occurs by moving from polygon point to view plane on a line toward eyes point. What would be most efficient way to get any corresponding 2d view plane point and its corresponding z on each shape? Some math: p1+a(p2-p1)+b(p3-p1) defines planes and l1(1-c)+l2•c defines lines where abc are each single numbers. Linear Algebra projection works but is slow. X E.

[midnight plank]

@fossil ember Has your question been resolved?

[fossil ember]

In case it helps, any 3 not collinear points are a plane no matter dimension. X E.

Barycentric does this sort of thing right? Could I use a bounding thing?

Any 4th point that is not in that plane could define a space. X E.

[fossil ember]

Could I possibly define a plane plane intersect and have same p1?

$p1+a(p2-p1)+b(p3-p1)=p1+c(p4-p1)+d(p5-p1)->a(p2-p1)+b(p3-p1)=c(p4-p1)+d(p5-p1)$, hmm, no. X E.

[grand pond]

maybeJosiah

[fossil ember]

Hmm, define p1 as 0, $a(p2)+b(p3)=c(p4)+d(p5)$. X E.

[grand pond]

maybeJosiah

[fossil ember]

Hmm, if we define l1 as 0, line defined as l2(c). X E.

So (p1-l1)+a(p2-p1)+b(p3-p1)=(l2-l1)(c) and a and b parts cancel. X E.

Still original. X E.

Let's see if I can derive a projection equation with t's. X E.

So we have t1-t3. For each point eyes point is l1, may as well set that as 0. So p1(t1)... . We want a point such that p4 on view plane is p4 on polygon plane. X E.

(p1(t1)+p2(t2))/2 does not equal (p1+p2)/2 but scaling does work though. X E.

In order to re-project, simply multiply by t1 and t2. t on line works. X E.

But then (p1(t1)+p2(t2))/3 would not work. X E.

We want p1(t1)(1-x)+p2(t2)(x)=p1(1-y)+p2(y). X E.

p3(y)=p1(t1)(1-x)+p2(t2)(x). X E.

y=(p1(t1)(1-x)+p2(t2)(x))/p3. X E.

That does not make much but p1-y(p2-p1) also is useful. X E.

p1(t1)(1-x)+p2(t2)(x)=p1(1-(p1(t1)(1-x)+p2(t2)(x))/p3)+p2((p1(t1)(1-x)+p2(t2)(x))/p3). X E.

p1(t1)(1-x)+p2(t2)(x)=p1+(p2-p1)(y), finding ratio of x and y. X E.

[desert siren]

X E.

[fossil ember]

So p1(t1)+(p2(t2)-p1(t1))(x)=p1+(p2-p1)(y)->p1(t1)-p1=(p2-p1)(y)-(p2(t2)-p1(t1))(x). X E.

(p1(t1)-p1)/(p2-p1)-y=-(p2(t2)-p1(t1))(x)/(p2-p1)->((p1(t1)-p1)/(p2-p1)/y-1)(p2-p1)/-(p2(t2)-p1(t1))=x/y. X E.

Well going nowhere, any ideas of how to find this stuff?

See pinned message for what I was originally trying to find. X E.

p1(t1)-p1=(p2-p1)(y)-(p2(t2)-p1(t1))(x), finding a ratio of x to y to satisfy this. If you know x or y it simplifies. X E.

So say x is 0.5, (p1(t)-p1+(p2(t2)-p1(t1))(0.5))/(p1-p2)=y. X E.

Problem partly solved. X E.

Lines solved. X E.

I can now get from on line more dimensional coordinates to 2d coordinates and back again. X E.

Should this process work for say like planar straight line real number polygon, just get from 2 edges corresponding then get from those edges corresponding stuff?

[fossil ember]

Actually those equations are mostly wrong. X E.

p1(t1)(1-x)+p2(t2)(x)=(p1+(p2-p1)(y))(t3), finding ratio of x and y. X E.

p1(t1)(1-x)+p2(t2)(x)=p3(t3), finding ratio of x and y. X E.

Dumb me. X E.

I guess I should vacate and figure myself. X E.

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[open swift]

Hey this is not a very serious concern, this is mainly recreational maths and wanted to know for what values the equation a^b+c^d‎ = e^bln(a)+ln(e^(dln(c)-bln(a)+1) holds (and for which it doesn’t), if you want the full derivation of this i wrote it down here (although I don’t know if it’s even correct)

[fallow scarab]

,rotate

[open swift]

my bad

lol

[fallow scarab]

you want solutions for the 4 variables (a,b,c,d) ?

[open swift]

mainly to verify if it’s an identity

like if it holds

and if so what values make this fail

i should’ve been clearer on the notation

if you see like a^blna+ whatever

that means whole thing is exponentiated

lmk

i shall go eat in 10 seconds

if anybody manages to figure out whether it’s an identity and for what a,b,c,d values this holds please let me know

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@open swift Has your question been resolved?

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[mint ravine]

hi

[mint ravine]

I have tried so many attempts on this problem but can't seem to solve it

First, I defined each pair of sibling into X, Y, Z

then i divide them up into x1 x2, y1 y2, and z1 z2 for each individual

and then did two possible cases

first case: first row has all three different sibling pairs (so x, y, z in any order)

second case: first row has two of the same siblings (e.g. X1, Y2, X2)

first case has 216 possibilities (6 * 4 * 3 * 3 * 2 *1)/2

second case has 48 possibilities (6 * 4 * 2)

then i added them up 216+48=264 cause they're mutually exclusive

please help!

[desert siren]

This is 6choose 3 * 3! * 3! correct?

Hm it isnt

[mint ravine]

no i don't think so

[desert siren]

Oh yeah this is not it anyways

[mint ravine]

i feel like im missing a case but i dont know what it is

[desert siren]

There isn't any other case

I would do this by complement btw, but let's do it with this method sure

[mint ravine]

yeah i was thinking about doing complementary counting but i suck at it

[desert siren]

Can you explain your thought process for these numbers btw

[mint ravine]

though if it's easier that way, i would prefer complement too

[mint ravine]

6 possibilities for first seat, 4 possibilities for next seat, 3 for the last seat on the first row

[desert siren]

In the first case your cases should be 36, idk how you are getting 216

[mint ravine]

how you get 36

[desert siren]

One of the rows can only have x1 y1 z1

While the other can only have x2 y2 z2

so 3! * 3!

Also, it doesn't seem like the rows are ordered btw

[mint ravine]

ok

[desert siren]

This also seems wrong.

[mint ravine]

oh yeah i know

[desert siren]

Cool, work on it and come back if not getting

[mint ravine]

oka

okay

[mint ravine]

.close. X E.

[desert siren]

[fossil ember]

What was I called here for?

[mint ravine]

X E.

[fossil ember]

Okay, read bio, no links. X E.

[mint ravine]

please stop speaking in shakesphere

[fossil ember]

Bye now. X E.

[mint ravine]

man i have no idea

my brain is fried

[desert siren]

It's easy, for the second case you just choose one pair of twins, as $\binom{3}{1}$, then choose a person to sit with them, as $\binom{4}{1}$, and then arrange the other three as 3!

[grand pond]

Executor (ask on server b4 DM)

[desert siren]

Multiply all those for your second case

Lmk if you don't understand

[mint ravine]

im taking a break

i'll do this later

.close

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[short hill]

it's not a p series. ._.

[short hill]

https://tenor.com/view/cat-scream-gif-9259936078829692831

ping if respond

[midnight plank]

@short hill Has your question been resolved?

[modest star]

lol it is a "p-series" in the way that $\sum_{n=1}^{\infty} \f{1}{n^p} = \sum_{n=1}^{\infty} \f{1}{n^{-9}}$

[modest star]

but yeah the question is very weirdly worded

maybe they meant is it a convergent p series

[short hill]

so its -9 >_<

[modest star]

yea ig so

[short hill]

oke i will try dat

it worked >_< thx

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[subtle peak]

How can I show part a?

[subtle peak]

Can I let the constant c be equal to 0 and therefor v(t)=x(t)?

Also I don't know if I'm getting rid of a nontrivial solution by dividing both sides by u(t-c)

[steel crest]

well u(t-c) is zero half of the time

however

you can just split into cases: when u(t-c)=0, and when u(t-c)=1

and no you can't simply let c=0 because then the theorem isnt claiming anything

defeats the point of the theorem

[subtle peak]

Right

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@subtle peak Has your question been resolved?

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@subtle peak Has your question been resolved?

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[molten bay]

How can I start it?

[viral dagger]

the easiest way is to find an n that works and plug it into the 2n+5 mod 14

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

It looks tough

[viral dagger]

https://brilliant.org/wiki/chinese-remainder-theorem/

try to view the "system of congruences" parts

[misty cairn]

you can remove the constants

[molten bay]

Yes

6n-8=4mod13

6n+1=0mod 13

@misty cairn

[lyric charm]

$6n \equiv 12 \pmod{13}$ gives $n \equiv 2 \pmod{13}$ actually

[grand pond]

Ann

[molten bay]

Ohh right

n=13k+2

5n+7=2 mod 12

n=-1 means

n=11

and 3n-4=1 mod 11

3n=5mod11

@lyric charm

[lyric charm]

3n=5 mod 11.

[molten bay]

Opps yeah

So what should I do next?

@lyric charm

Sry for pinging

[lyric charm]

so we've reduced our modular congruences to:

• n ≡ 2 (mod 13)
• n ≡ -1 (mod 12)
• 3n ≡ 5 (mod 11)

for the last of these, multiply both sides by 4 so LHS becomes 12n and reduces to 1n

[molten bay]

12n=20mod11

[lyric charm]

yes now reduce.

and afterwards the system is in the right form to apply CRT.

[molten bay]

n=9 mod 11

n=11 mod 12

n=2 mod 13

11K+9=12L+11

11K=12L+2

L=9 satisfied

So i got n=9 mod 132

9L+132=13K+2
9L+130=13K

Thank you

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[frosty sable]

I need help with factoring

[frosty sable]

These questions are like impossible to me right now and I have to know this for my exam

I’m an idiot

[rare maple]

For part a which is that one term common to all

[frosty sable]

(a-1)^3

[rare maple]

Yes now how will the expression simplify

[frosty sable]

(a-1)^3x25x^2-5x-2

[rare maple]

Missing parenthesis

[frosty sable]

(a-1)^3 (25x^2-5x-2) ?

[rare maple]

Yes

[frosty sable]

Is that the final answer

[rare maple]

The quadratic can be factored too

[frosty sable]

I forget how to do that

[rare maple]

If we have a quadratic ax²+bx+c then we need to find 2 numbers p and q such that their sum is equal to b and product equal to ac

So in this question what will be those two numbers

[frosty sable]

I figured it out using the reverse box method
(a-1)^3(5x-2)(5x+1)
Is that correct?

Can you help me with e?

[rare maple]

Since you know how to factor quadratics

Is there a way to turn part e into one

@frosty sable

[frosty sable]

Idk

<@&286206848099549185>

[rare maple]

Can we write x² as k

If we do that it simplifies to 8k²+19k-27

[rare maple]

Which is solvable

[frosty sable]

Ok

I can’t find 2 numbers that multiply to -27 and add to 19 though

[rare maple]

We need them to multiply to -27*8

Also for these long factorizations it is better to learn the quadratic formula

That will save you a ton of time

[frosty sable]

We were never taught that

[frosty sable]

When do they usually teach that to you? I’m in grade 10

[rare maple]

i was taught this in 10th

[rare maple]

so i recommend learning it

This question becomes way easier with it

[midnight plank]

@frosty sable Has your question been resolved?

[frosty sable]

I’m just trying to factor it though, not solve for x

[rare maple]

If a and b are solutions of x then (x-a)(x-b) are its factors

What I am trying to say is that

You can use this in your rough work for now to find the numbers

Then show those numbers through factorization

[frosty sable]

Idk, I must be stupid or something, it’s getting late

.close

[midnight plank]

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[midnight plank]

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[quartz flame]

Hey all, I'm not quite sure how to intepret the solution to the last part of this question

[quartz flame]

Like in this part, why isnt it

$\int_{0}^{21.44}{e(t)-u(t)dt} + 5$

[grand pond]

giordan

[midnight plank]

@quartz flame Has your question been resolved?

[quartz flame]

help

[novel lion]

sec

[quartz flame]

what does that area in betwen the graph actualy mean tho

[novel lion]

amount of energy stored in battery

e(t) is being produced
u(t) is being used

so stored = u(t) - e(t) or e(t) - u(t) depending on part of the graph

[quartz flame]

@fiery night

[fiery night]

okay right

[fiery night]

does it matter what fucntion is above when finding the stored batter in this case

what is the difference when the usage is above than the production of energy compared to when the production is above

[novel lion]

yes

[novel lion]

when usage is above production that means there's net usage (battery drain)

and battery charge when usage is below production

[novel lion]

u can still write the second part as u(t) - e(t)

but

you will need to add a - sign

because u(t) - e(t) there would reverse charging and draining signs

so instead of making life harder with sign changes, just follow one convention (whichever one you decide, can be opposite signs or not)

[quartz flame]

so for 0<=t<=6.54

you'd find area in between graph

which is change in energy stored in battery

which upon interpretation is negative

so u just do negative

and then you add it to the secondone?

ok im wrong

[fiery night]

wait would the battery stored not be integral of e(t)-u(t) from 0 to 21 + 5

???

[quartz flame]

because in the first part how is the amount stored the amount used - produced

[fiery night]

produced - used is how much is stored

yeah

[novel lion]

that works too

i was just answering what they asked

[quartz flame]

but if i put this and this into the calc

its two different evaluations no?

like answers

[novel lion]

no

[fiery night]

astar, for the asnwer in red, is it written correctly, because when simplified it minuses the first integral

[novel lion]

yes

[fiery night]

right?

that gives a different value then

[novel lion]

huh

[quartz flame]

am i wrong

[desert siren]

Retiring sounds nice...

[grand pond]

giordan

[novel lion]

,w integrate 2cos((pi/12)(t-6)) + 2 -(3.5sin((pi/12)(t-6)) + 3.5) from t=0 to 6.54

[quartz flame]

yk what im saying

[grand pond]

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[novel lion]

,w integrate 3.5sin((pi/12)(t-6)) + 3.5 - (2cos((pi/12)(t-6)) + 2) from t=6.54 to 21.44

[grand pond]

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Requested by astar777

[novel lion]

,w integrate 3.5sin((pi/12)(t-6)) + 3.5 - (2cos((pi/12)(t-6)) + 2) from t=0 to 21.44

[grand pond]

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[novel lion]

,w 50.9532 - 12.1415

[grand pond]

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[quartz flame]

what is going on

[novel lion]

it is equal

[fiery night]

so my method works

goated

okay

[novel lion]

anyways

just read the question

[quartz flame]

thank you astar you're goated

[novel lion]

its asking the maximum amount of energy stored

this isnt the way

[fiery night]

then how

[quartz flame]

what are we calculating then 💔

[fiery night]

we found it occurs at t = 21

[novel lion]

thats final energy, no?

[fiery night]

no max

[quartz flame]

we found that the maximum amount of energy is stored when t=21.44

so what integral would we do to find the energy stored in the battery when t=21.44

[novel lion]

,w integrate 3.5sin((pi/12)(t-6)) + 3.5 - (2cos((pi/12)(t-6)) + 2) from t=6.54 to 12

[grand pond]

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[novel lion]

nvm yeah

im dumb

its correct

[fiery night]

yeah

so what is the final apporach

[novel lion]

both works

there's 1 mistake here

[quartz flame]

GHAZI BRO

IT SHOULD BE MINUE

RIGHT

[quartz flame]

^^^

like this

[fiery night]

yeah thats what i said above chat

it makes the same

as

[quartz flame]

YOUR THING IS WRONG GHAZI BRO

[novel lion]

this should be minus

[fiery night]

integal of e - u from 0 to 21.44

[quartz flame]

THATS WHAT I SAID

ghazi

pls

lock

in

[novel lion]

if you are doing e(t) - u(t) in first part of graph too then you dont need the -

[fiery night]

yeah

so integal of e-u from 0 to 21 + 5 woks

right

yeah

it does

[novel lion]

yes

[fiery night]

thanks astar

[novel lion]

both approach work

[quartz flame]

so the final answer is:

$\int_{0}^{21.44}{e(t)-u(t)dt} + 5$

[grand pond]

giordan

[fiery night]

[quartz flame]

[novel lion]

uhh

[quartz flame]

thank you mr astar

[novel lion]

refrain from using that gif

[fiery night]

yeah thanks a lot man

[quartz flame]

.close

[midnight plank]

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[fiery night]

okya good night then

[midnight plank]

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[mystic oar]

Hello, I'm currently studying the "surrounding of a point" (I'm not sure if its the correct English translation). I'm having trouble with this excercise because, from my understanding of the definition of "surrounding ", these sets can't be a "surrounding". Thanks in advance.

[lyric charm]

i think the word you're looking for is neighborhood

what is your native language?

[mystic oar]

Italian

[lyric charm]

and what is this I(2,1) notation

[crystal plover]

intorno è neighbourhood

[crystal plover]

I think it stands for the interval centred at 2 of radius 1, @inland dawn can you confirm?

[mystic oar]

It's this

[lyric charm]

ok but aren't we working in R^1

[mystic oar]

Yes but this is the general definition

[crystal plover]

frontiera poi è “boundary”

[crystal plover]

Having boundary points doesn't necessarily prevent a set from being a neighbourhood of some of its points, but sometimes this depends on the given definition

what definition of neighbourhood were you given?

[mystic oar]

The one in the second picture

[crystal plover]

Can you show the full picture?

[crystal plover]

isn't there anywhere else where it talks about when a set is a neighbourhood of a point?

Maybe when explaining what an interior point is

[mystic oar]

There are some comments but no alternative definition. But I understand the theory, I just have a hard time using it in this exercise

[crystal plover]

okay

So what's your solution for (1), what are the interior, exterior and boundary points?
(or what did you get stuck with?)

[mystic oar]

From what I understand, a point x is an internal one of a set E if the neighbourhood is all inside E. But I don't know what r is, for example for the number 2 I can't determine what is the neighbourhood with center in 2 because I don't have r so I can't tell if the neighbourhood of 2 is a subset of E

[crystal plover]

From what I understand, a point x is an internal one of a set E if the there exists a neighbourhood all inside E.

let's stick to nr 1 first

what do you think the interior points are?

[mystic oar]

I meant 2 as in a number that is inside the first set

So it could be an internal point

[crystal plover]

there's no “the neighbourhood”, you can take a lot of neighbourhoods of the same point, visually they are intervals centred at them
I(2,1) is a neighbourhood, but so is I(2,0.5), I(2,0.25), etc.

[mystic oar]

Ah so I chose the ones that fit?

[crystal plover]

if you draw the the set (1,3], you can highlight the point x=2 and draw a little interval centred at 2 inside the whole set (1,3]

[crystal plover]

yes

[crystal plover]

since you can do this, we say that the point x=2 is an interior point for (1,3]

[mystic oar]

But like, in theory could I not choose an infinite amount of interior points?

[crystal plover]

there are infinitely many interior points, but I am not sure that's what you meant

[mystic oar]

How do I give an exact answer?

[crystal plover]

you can describe them using interval notation

can you spot which points are internal and which external?

[mystic oar]

I mean that between 1 and 3 in R^1 there are infinite numbers so potentially infinite internal points

[crystal plover]

we are looking of what's inside A = (1,3] to start
A has infinitely many points, so it makes sense to possibly have infinitely many interior points and infinitely many exterior points

[mystic oar]

Would something like this make sense to define the internal points?

[crystal plover]

not clear on what's going on there

how about we approach this problem visually, then we describe the points using some notations

can you draw the interval (1,3] in ℝ?

[mystic oar]

Like this?

[crystal plover]

if you pick a number like x=2, you can draw a little interval containing 2 living inside A = (1,3]

do you agree?

Do you think you can do the same thing for all points in (1,3], or does it fail for some?

[mystic oar]

The way you phrased the question I feel like some should fail but I don't see it

[crystal plover]

what would you say about x=3

[mystic oar]

Ok I see it

[crystal plover]

so which points are internal?

[mystic oar]

(1,3)?

[crystal plover]

yes

do you think any points outside A can be internal?

[mystic oar]

No

[crystal plover]

good

so interior points are done

do you remember what an exterior point is?

[mystic oar]

Its neighbourhood is outside A

[crystal plover]

a point is not a neighbourhood

[mystic oar]

Ik?

[crystal plover]

I asked you what an exterior point is and you answered “it's a neighbourhood [...]”

[mystic oar]

Its not it's

[crystal plover]

still, there is no “the neighbourhood”

a point has infinitely many neighbourhoods

so “its” doesn't make sense, which one?

because I can assure you, if you pick any point not in A, I can find a big enough neighbouhood containing all of A

[mystic oar]

There is at least one neighbourhood that has no elements of A?

[crystal plover]

this is better

now look back at the drawing

pick a point not in A, for example x = 0

can you find a small enough neighbourhood of x=0 having no elements in common with A?

[mystic oar]

[-1,1]

[crystal plover]

neighbourhoods are open intervals in your convention

and [-1,1] isn't one

I was asking more for a visual check, not necessarily with specific values

[mystic oar]

(-1,1)

[crystal plover]

yes, but we can also take is smaller, say (-0.1, 0.1)

but yes, so 0 is an exterior point

do you think all points not in A are external?

[mystic oar]

No

[crystal plover]

elaborate

[mystic oar]

I'm getting confused

[crystal plover]

you said not all points not in A are external

which are they and why is that?

[mystic oar]

so the boundary points are 3 and 1 because they have necessarily elements both in A and outside. All the other points outside of A are external

?

[crystal plover]

yes

except add the word neighbourhood somewhere

with the correct quantifier

[mystic oar]

3 and 1 are boundary points because they have at least 1 neighbourhood with elements of A and of outside A. (3,+inf)U(-inf,1) external because they have at least 1 neighbourhood that has only elements outside of A. ?

[crystal plover]

at least is not enough, every point has at least 1 neighbourhood with points inside and outside A

[mystic oar]

The smallest neighbourhood?

[crystal plover]

what's “the smallest neighbourhood”?

think about it

can you find any neighbourhood of 3 that has only points inside A or only outside A?

[mystic oar]

No

[crystal plover]

so what distinguishes boundary points is that every neighbourhood has that property

[mystic oar]

Ok

[crystal plover]

is it clear now?

[mystic oar]

So once I determine the boundary points using that property I distinguish internal and external points by whether or not they are inside A?

[crystal plover]

well yeah, points are either boundary, internal or external
boundary points can be inside or outside the set, internal only inside and external only outside

[mystic oar]

Ok I understand now. Thank you very much : )

[crystal plover]

prego

ora fai il 2

[mystic oar]

Boundary: 1,2
Internal: (1,2)
External: (-inf,1)U(2,+inf)
?

This channel is still in use...

[crystal plover]

yes

[mystic oar]

Thank you : )

.close

[midnight plank]

Channel closed

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Use .reopen if this was a mistake.

[serene pivot]

Can I ask my question now?

[crystal plover]

this channel is not available yet, it takes some time

you can use an available channel

[serene pivot]

oh sorry im new to this server

[crystal plover]

you can see the available channels in the list

and ask your question there

[mystic oar]

Read #❓how-to-get-help

[midnight plank]

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[mint ravine]

Can someone please help; I don't know how to get started

[mint ravine]

What I tried so far: So I know there are 16 choose 8 ways to place the dots, but that's overcounting

[lyric charm]

i think there's a kinda clever way to encode an arrangement of counters into a sequence of numbers

and then count the sequences of numbers instead after figuring out what constraints they must satisfy

[mint ravine]

can you give an example?

[lyric charm]

well i was about to explain the encoding that i came up with

if we imagine placing the first counter at some position $(x_1, y_1)$ for example (where $x_1$ and $y_1$ are both between 1 and 4) then we know that there has to be another counter at some position of the form $(-, y_1)$ (because column $y_1$ has to have exactly 2 counters)

[grand pond]

Ann

[lyric charm]

so we could call the x coordinate of that counter x_2

and then in its row there has to be another counter -- (x_2, y_2)

and so on

and so we get a sequence of eight coordinates $[x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4]$

[grand pond]

Ann

[lyric charm]

which represents placing counters at $(x_i,y_i)$ for $1 \leq i \leq 4$ but also $(x_2, y_1)$, $(x_3, y_2)$, $(x_4, y_3)$ and $(x_1, y_4)$

[grand pond]

Ann

[lyric charm]

... wait.

no hold on i dont think that works exactly.

we might "loop back" before having "covered" all the counters this way.

my bad

[mint ravine]

oh ok

[lyric charm]

hm

not ready to say how to account for all possible permutations

[mint ravine]

I think you could try finding the possibilities for every row and column, and multiply

[dreamy lichen]

why not just make sure you dont loop back

by ensuring that x1, x2, x3, x4 are different

and similarly y1, y2, y3, y4

so you'd get 2 interconected permuations of 1-4

i think

[lyric charm]

ok but hold on

xx..
xx..
..xx
..xx

this arrangement just ends up unaccounted for doesn't it

[dreamy lichen]

oh you're right, it looked so nice and promising

oh we could prolly just consider all the points

(x1, y1)
(x2, y2)
...
(x8, y8)

[desert siren]

This seems like something it'd be easier to do by hand

[dreamy lichen]

now there must be exactly 2 of 1,2,3,4 in x's and in y's

[mint ravine]

isn't that overcounting?

[dreamy lichen]

why?

oh

permutations

(1, y1)
(1, y2)
(2, y3)
(2, y4)
...
(4, y7)
(4, y8)

how about this?

[mint ravine]

Still overcounting, I think

[dreamy lichen]

how

[mint ravine]

y1-y8 can be any number from 1-8, right?

[dreamy lichen]

nope, it certainly must be between 1 and 4

and not any number

you have to place constraints on that

in such a way that you're guaranteed to have 2 counters in every row

try figuring out what the constraint should be

[mint ravine]

if y1-y3 are all 3

then they're in the same column

[dreamy lichen]

*in the same row

[dreamy lichen]

but yeah, they cant be all 3

so what are the constraints?

[mint ravine]

Exactly 2 counters in each row/column so max is two counters for each y-coor from 1-4

and we already defined the x-coor so we don't need to worry about that

[dreamy lichen]

Exactly

so we are looking for number of permutations of the string
11223344 basically

[mint ravine]

8!/(8-2)!

or just 8!/6!

which is 56

[dreamy lichen]

why?

[mint ravine]

oh wait permutations mb

8!/(2!*4)

I think that's right

cause there are 8! ways to place the numbers

but then you have to account for the overcounting cause each digit repeats twice

and there are 4 cases of that (1, 2, 3, 4 all repeat twice)

so 8!/(2!^4)?

[visual panther]

2!^4 not 2!*4

[mint ravine]

but then you have to divide by 4 cause the square could be rotated 4 different ways (90°, 180°, 270°, 360°)

oh wait no

they'll be different cases for each

[desert siren]

I don't think rotation's a thing here.

[mint ravine]

but this case could be rotated all 4 times and still look the same

[desert siren]

Why did you suddenly start thinking about rotation? The question doesn't specify that the square can be rotated.

[mint ravine]

ok then nvm

[mint ravine]

so the answer is 8!/(2!^4) or 2520?

but i feel like that's still overcounting but idk

[visual panther]

Are these "counters" distinguishable or do they all look the same?

[mint ravine]

they're indinguisable

so we're def overcounting

[visual panther]

By a lot!

[desert siren]

I still think doing it by hand is the best option.

[mint ravine]

maybe but i don't wanna restart all over

[visual panther]

Starting with rowa 1 and 2, I see three cases:

First: Counters only in two colums: 6 possibilities

Second: Counters in 3 rows: 4 * 3 * 2 possibilities (4 for the double and 3, 2 for the singles)

[mint ravine]

How? I thought each column has exactly 2 counters

For case 1

[visual panther]

Name colums a,b,c,d. 1st row a,b; 2nd row: a,b

This leaves us with no choices in the 3rd and 4th row, so a total of 6 possibilities.

Second case: We have 2 counters in one column (4 poss.) and 2 coulumns with one counter (3*2 poss.) -> 24 arrangements in rows 1 and 2.

Now in row 3, we have to pick the last coulmn and one of the columns with one counter -> 2 poss.; 4th row, no choice. Total: 48

[mint ravine]

what doess poss stand for

possibilities?

[visual panther]

possibilities, or should I write arrangements

[mint ravine]

oh ok

[mint ravine]

but you can have 2 counters in one column in 4 choose 2 or 6 ways though

[visual panther]

If it is only one column, then it is 4 chose 1 = 4

[mint ravine]

but there are 4 boxes in a column and you are choosing 2 of them to be a counter?

sorry if im sounding demanding by any chance

[visual panther]

Pick your column with 2 counters in row 1 and 2: 4. Then pick an extra column in row 1: 3, then pick another column in row 2: 2 -> 4 * 3 * 2

[mint ravine]

ok

[visual panther]

In this case, because we have to put 2 counters in the remaining column, we have to pich the remaining coulmn in row 3 and one of the coulmns with one counter. So only 2 choices.

[mint ravine]

wait i think you're getting column and row mixed up

[visual panther]

NO, when I pick 2 in a row, I am talking about their distribution in the columns.

[mint ravine]

oh ok

[visual panther]

Let me draw a picture. One sec.

[mint ravine]

these questions are so hard bro

[desert siren]

They're meant to be right?

What website you been doing them from

[mint ravine]

aops

or art of problem solving

[desert siren]

Yeah so its meant to be hard lol

Preparing for an olympiad?

[mint ravine]

amc 8 and amc 10

maybe olym afterwards

[visual panther]

Like this.

[mint ravine]

Oh I understand

So now we add up those cases

so 6+48+36=90

[visual panther]

Sure!

It would be nice, if they were distinguishable. 90 * 8! = 10!

[mint ravine]

yeah that would be a better answer

thanks for helping nevertheless

[midnight plank]

@mint ravine Has your question been resolved?

[midnight plank]

@mint ravine Has your question been resolved?

[mint ravine]

oops forgot to close it

.close

[midnight plank]

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[hybrid crow]

We have to prove both implications

Let V1 and V2 be subspaces

where V1 is contained in V2

then the union of V1 and V2 = V2, and V2 is a subspace

Now, assuming that the union of V1 and V2 is a subspace

can someone help me show that this would imply that one of the subspaces contain the other?

I understand intuitively why this is true, but can't exactly prove it.

y = x and y = -x can both be seen as subspaces of R^2

yet the union of them isn't closed under addition and is not a subspace

and one of them doesn't contain the other, so this example is consistent with the proof

I just added it in case it would be helpful.

[midnight plank]

@hybrid crow Has your question been resolved?

[polar star]

we should try contradiction

ill give u the hypothesis:

suppose that V1 and V2 are s.t neither subspaces contain eachother, thus we have

x in V1 \ V2 and y in V2 \ V1

now use closure in addition to establish a contradiction

consider what happens if x+y in V1 and if x+y in V2

[midnight plank]

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[hybrid crow]

thanks, i already did it in the other channel

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[sudden bronze]

yo'

[sleek cloud]

No

[sudden bronze]

yes

how do i do directed numbers

the one with negatives and positives

[midnight plank]

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[hybrid crow]

Let the union of V1, V2, and V3 be V4 (a subspace)
then assume that there isn't a subspace that contains the other two, such that there exists a in V1 only, b in V2 only, and c in V3 only
then (a+b), (b+c), (a+c), and (a+b+c) must all be elements of V4 by closure under addition
We would then be forced to say (a+b) is in V3, (b+c) is in V1, and (a+c) is in V2 to avoid obvious contradictions ((a+b) in V1 would imply b in V1, which is one of these obvious contradictions.
and then I have no idea what to do from here...nor do I know if this is even the right approach

Could also include that (a+b+c) is an element of V1, V2, and V3 (idk if this helps).

[wheat nymph]

u r assuming that none of the subspaces contain another subspace
but theres also the case where one contains another

[pine wave]

You have to use the fact that F has at least 3 elements

If your proof isnt doing that, it cannot work

[hybrid crow]

sounds about right given what he wrote under it

although i'm still not sure where to go with that 😭

[wheat nymph]

why the thumbs down

[hybrid crow]

i don't think that case changes anything

my hypothesis still works for a proof by contradiction

[wheat nymph]

u could have $V_1 \subseteq V_2$ but $V_3 \subsetneq V_2$

[grand pond]

Bob the Builder

[wheat nymph]

then u cant assume a exists

either way the 2nd case isnt too difficult

[hybrid crow]

huh

y can't i assume a exists

its just in v1 but not in v2 or v3

[wheat nymph]

but V1 is in V2, so a is in V2

[hybrid crow]

V1 doesn't contain V2 and V3. V2 doesn't contain V1 and V3. V3 doesn't contain V2 and V1

so no, V1 is not in V2?

maybe my wording was a bit inaccurate

thats what I meant tho

[wheat nymph]

well it seems to me like u r proving the converse

[hybrid crow]

if this statement is wrong then there exists a subspace that does contain the other two

which is what i'm trying to prove

via contradiction

[pine wave]

Suppose not, WLOG start with
V1 has a but not b
V2 has b but not a
neither have a+b so V3 must have that
neither has a+2b so ...[assumption used: 2 is not 0]

Maybe generalize this?

[wheat nymph]

yes i know
im saying that u r trying to show that the complement of that statement is false, and the complement is that either

1. none of the subspaces contain another subspace
2. the case im describing

[hybrid crow]

so a + 2b is in V3

so b is in V3

[wheat nymph]

but yes this works

[hybrid crow]

I'm trying to show that if the union of V1, V2, and V3 is a subspace, then this would imply that one of the subspaces contains the other two

[pine wave]

Needs a bit of cleaning up i guess

[wheat nymph]

...

[hybrid crow]

I'm using contradiction Bob, so i'm saying to first assume that V1 does not contain V2 union V3, etc.

[wheat nymph]

anyways u can use element's method

[hybrid crow]

and showing that this will lead to contradiction

[hybrid crow]

uh

i don't really get it

b is in V3 i guess?

[pine wave]

We start with an easily analysed condition: V1 and V2 contain stuff the other doesnt contain.

[hybrid crow]

a is in V1 but not in V2, b is in V2 but not V1

[pine wave]

If there is no such pair of spaces then you can easily check cases to find that theres a space containing the other two

[hybrid crow]

$a \in V_1 \cup V_2$ such that $a \notin V_3$

[grand pond]

Évariste

[hybrid crow]

i guess i'll think on it tonight

atleast i sorted out my mistake

with a set containing the other two

or not

u can close channel

[midnight plank]

@hybrid crow Has your question been resolved?

[midnight plank]

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[meager ore]

Let $G$ be a group of order $p^n$ (where $p$ is a prime). Prove that $G$ has a normal subgroup of order $p^{n-1}$.

[grand pond]

Halex

[meager ore]

Anybody got a clue

[pine wave]

Well known that these groups are nilpotent, but that might be too high-powered

Another way might be through Sylow

[meager ore]

Yes, this exercise is proposed in the Sylow's Theorems Section of my classnotes. Could you elaborate?

[pine wave]

Does you sylow section mention anything about p-subgroups?

[meager ore]

Yes

[pine wave]

like how different sizes of p subgroups relate under conjugation

[meager ore]

2nd Sylow Theorem?

Any two Sylow ( p )-subgroups of ( G ) are conjugate.

[grand pond]

Halex

[pine wave]

Not that useful because theres only one sylow p subgroup

[meager ore]

Hm

If ( G ) has a unique Sylow ( p )-subgroup ( P ), then ( P ) is normal in ( G ), i.e., ( P \trianglelefteq G ).

[grand pond]

Halex

[pine wave]

Another way im thinking of... is to induct by order, then for the general case notice that you have nontrivial center, so you want to split it like this and combine back somehow.

[pine wave]

Here in this problem P=G

To show you have nontrivial center, consider class equation.

[meager ore]

What does nontrivial center mean?

Like I know that |Z(G)| = p^k for some k>=1

[pine wave]

Yeah k>=1

That exactly

[meager ore]

And also |Z(G)| >= p

[pine wave]

nontrivial, as in not the trivial group

[meager ore]

if we go by induction on n, how would we proceed?

[pine wave]

Then I think you can write $G=C\times G'$ then if $G'$ trivial this should be easy, else by induction choose a normal subgroup $H'$ of $G'$, and you should be able to finish from there.

[grand pond]

Element118

[pine wave]

Hmm... weird something off

I suspect G=C\times G'

[meager ore]

why can we write G = C x G'?

[pine wave]

Defo can write G/Z(G) and consider a normal subgroup of that

[pine wave]

I think we cant

[meager ore]

we can choose $H \subseteq Z(G)$ (as $Z(G)$ is nontrivial)

[grand pond]

Halex

[meager ore]

This means $H$ is normal in $G$

[grand pond]

Halex

[meager ore]

We can apply inductive hypothesis here right?

Can we take |H| = p?

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

@meager ore Has your question been resolved?

[feral sedge]

since Z(G) is a p-group, by cauchy's theorem it has an element of order p

and you are able to continue from there

[meager ore]

By induction: we will prove that for every group $K$ of order $p^k$ with $k < n$, there exists a normal subgroup of order $p^{k-1}$.

For $n=1$, $G$ has order $p$ and the trivial subgroup ${e}$ is normal of order $p^0=1$.

Assume that every group of order $p^k$ with $k<n$ has a normal subgroup of order $p^{k-1}$. Since $G$ is a $p$-group, $C(G)$ is non-trivial. Let $H$ be a subgroup of $C(G)$ of order $p$. Since $H\subseteq C(G)$, $H$ is normal in $G$. By Lagrange's Theorem, we can conclude that:
\begin{align*}
|G/H| = \frac{|G|}{|H|} = \frac{p^n}{p} = p^{n-1}.
\end{align*}
By the inductive hypothesis, there exists $\overline{K}\trianglelefteq G/H$ of order $p^{n-2}$. By the correspondence theorem, there exists $K\trianglelefteq G$ such that $H\subseteq K$ and $K/H = \overline{K}$. Then:
\begin{align*}
|K| = |K/H|\cdot|H| = p^{n-2}\cdot p = p^{n-1}.
\end{align*}
Therefore, $K$ is the normal subgroup of order $p^{n-1}$ we were looking for in $G$.

[grand pond]

Halex

[feral sedge]

looks roughly correct

[meager ore]

What does it need to be correct?

[feral sedge]

i say "roughly" not because it is wrong but because i just saw it had all the right ideas

[meager ore]

So, is this proof correct?

If there's something I am missing, please let me know

[feral sedge]

the proof is fine but i would delete that first line "By induction: ..."

or at least rewrite

since that looks like the induction hypothesis, not what you are trying to prove

[meager ore]

You are right

Just that part?

[feral sedge]

yeah

if this is part of a graded hw assignment you could specify why H exists (cauchy's theorem) but well the proof is correct regardless

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

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[midnight plank]

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[winter stone]

Hello, I'm having a lot of trouble achieving this simplification, I feel there is some idea I'm missing

[velvet compass]

Multiply by conjugate and apply j^2 = -1

Then do the common denominator and add

[austere drift]

Hello

[winter stone]

Ya I did that but I can't see how I arrive at the final formula

[velvet compass]

I got

[grand pond]

Modus

[velvet compass]

Then, after the addition

[winter stone]

This looks a lot simpler than what I've got

I got a bunch of things squared in both parts of the fraction

[velvet compass]

Well, I still haven't multiplied by conjugate

[winter stone]

Oh right, that way is probably easier

I multiplied by the conjugate right away

I'll try it like this

[velvet compass]

Yeah, then it works fine, I already see

Because in the denom we are going to get what they have

[winter stone]

Got it, thanks

.close

[midnight plank]

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[sudden holly]

Hello, I am struggling to solve this ³√(10-x) +³√(3-x)=1. I've tried to square both sides of the equation and got 6 as an answer, but it can't be right, can someone please explain what is the correct way to solve this.

[civic gazelle]

let cbrt(10-x) = a and cbrt(3-x) = b then solve the system of equations

[oak nymph]

Did you mean cube root or sqaure root? Because if its cube root squaring is not helpful

[sudden holly]

Oh sorry square

[civic gazelle]

should've questioned that as well

[sudden holly]

Thanks a lot

[civic gazelle]

change that to square root yeah

and make sure x<=3

[oak nymph]

Squaring is the correct method. If you got 6 as the ONLY answer after squaring, that means the original equation didnt have any solutions since we need x<=3

Its not necessary that every solution after squaring must be a solution before squaring

And it is true that there is no solution

[civic gazelle]

wouldn't it make a big mess tho

actually by observation, x<=3 so sqrt(10-x)>= sqrt(7) >1, so the equation has no solutions

[sudden holly]

I am confused.

If I graph it there will be an approximate solution

[lyric charm]

wait can you send a picture of the equation as written originally? im a bit confused about whether the roots are square or cube or what

[desert siren]

So its cube.

[civic gazelle]

oh cubed root

[civic gazelle]

^

[lyric charm]

yeah then @civic gazelle's suggestion is good

[civic gazelle]

how do you even type cube root on a mobile keyboard tho

[sudden holly]

Ok thanks

.close

[midnight plank]

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[lyric charm]

(stuff)^(1/3)

[midnight plank]

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[civic gazelle]

1^(1/3)

[civic gazelle]

.close

[midnight plank]

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[midnight plank]

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[narrow edge]

can anyone explain the second page?

bruh

why did everyone stop typing

😭

[sudden yacht]

Why do you ask again??

[normal prawn]

yea

[sudden yacht]

I mean, why did you close the other channel?

[oak nymph]

What are you having trouble understanding?

[lyric charm]

!noclopen

[midnight plank]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[narrow edge]

i asked 30 minutes ago so the other channel will probably be ignored

[sudden yacht]

You were being helped there, weren't you?

[sudden yacht]

You were being helped!

[desert siren]

You didn't you just posted the page and skedaddled

And then when you asked

K was helping

[narrow edge]

yes my laptops battery died

[narrow edge]

sorry he didnt message for a minute

i wanted immediate help

[desert siren]

You problem

[sudden yacht]

Having said that you won't!

This is not how to ask for help, you know?

[narrow edge]

sorry

:(

[oak nymph]

Can you explain what problem you are having in the second page?

[narrow edge]

"there are 5 ways we can choose 1 y and 4 x"

doesnt this mean that we are going calc a lot of combinations for x and y?

[oak nymph]

You can cleverly count it by choosing which bracket y is picked from

after picking one bracket for y, the rest must all be x, and since there are 5 brackets y can be picked from

There are 5 such terms

In general, we give a name for the number of ways to pick r things from n things, we call it $\binom{n}{r}$

[grand pond]

Herzog

[narrow edge]

i got the first para of the second page

what does it mean by (5 2) ways

[oak nymph]

(5 2) is the number of ways to pick 2 things from 5 things

[lethal path]

so it's $\frac{5!}{2! (5-2)!}$ ways

[grand pond]

south

[lethal path]

using the definition of nCr

[narrow edge]

yes but why do we want 2 ys

y's

[oak nymph]

We want to find the coefficient of the y^2x^3 term

The goal is to expand (x+y)^5

[narrow edge]

y^2x^3 what

[oak nymph]

y^2x^3 means y²x³

[narrow edge]

i read that as y to the power of 2x to the power of 3

hmm

i think im getting it now

ill try to derive that expansion myself

why cant there be an x^5 * y^5 term

the degree of the term has to add up to 5?