help-10

[solemn burrow]

I don't understand the language written in it😭

[velvet radish]

You can still see the problems

[split sluice]

Question 2 part a is wrong 3x^2-12=0 the solution is plus or minus 2

[velvet radish]

It’s 2

[split sluice]

Plus or minus 2

Not just 2

-2 as well

Everything else is correct but that one question

3(2)^2+12=0
And
3(-2)^2+12=0

So 2 and -2 are solutions not just 2

[velvet radish]

Mhm

I corrected it though

Later

Though

I knew about this

[split sluice]

Btw u exposed ur name 🙏

[obtuse pebble]

@velvet radish Has your question been resolved?

[inner sierra]

🙏

[solemn burrow]

Brother do you need any more help?

[velvet radish]

Maybe

[obtuse pebble]

@velvet radish Has your question been resolved?

[iron edge]

wym maybe?

Do you have anymore question?

[velvet radish]

🤫

67

[mortal blade]

<@&268886789983436800>

[polar fossil]

if you have no further questions, i'm going to lock this channel

.close

[obtuse pebble]

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[eager sky]

I’ve just gotten this question wrong, does anyone know why?

[eager sky]

Oh wait hold on

Yeah looks a lot more pleasing when I send it like that

[stone prism]

right side is (-2)^2, not -2^2

[eager sky]

Oh thanks

But how does that change how I continue to do the equation?

[stone prism]

-4 and 4 is big difference

[eager sky]

Oh wait is it because when its in brackets it’s still a negative?

[eager sky]

Ok thanks

I removed the mistake but my answers still wrong

[mortal blade]

Cuz you made the same mistake, just eventually

[eager sky]

Oh what

Okay I see it

[stone prism]

what is (-2)*(-2)

[mortal blade]

Just start by writing (-2)² as its value

Don't kick it down the line

[eager sky]

4

I think

[mortal blade]

Good, now start with that

[stone prism]

yeah that one

[eager sky]

Alright

Nice I got the original answer

Thanks guys

.close

[obtuse pebble]

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[gusty garden]

What's 9 + 10

[atomic marsh]

Please do not troll in help channels

[gusty garden]

.close

[obtuse pebble]

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[neat sparrow]

So, suppose that I do a proof by cases, I write:
\

$\mathcal{A}1$: $P(s^c \mid f) \ne 0. \text{ [assumption for proof by cases]}$ \

However, now I want to name something that I derive using $\mathcal{A}1$. I can't call it a lemma, I guess, right? Because it's not something that I derived; I only derived it conditionally on $\mathcal{A}1$. So I guess I could write: \

L5$^{\mathcal{A}1}$: $EV(f \cap s) \ge EV(\neg s \cap f). \text{ [from L4, and $\mathcal{A}1$]}$ \

But idk that might not be very aesthetically appealing. Is there a standard word like 'lemma' that one can use here?

[warm shale]

! See my about me

[ember frost]

corollary?

[jaunty ocean]

I don’t think that works here. A corollary is result that follows immediately (or almost immediately) from a theorem or lemma that has already been established unconditionally.

[neat sparrow]

doesn't seem like it indeed https://math.stackexchange.com/questions/463362/whats-the-difference-between-theorem-lemma-and-corollary

[kind hawk]

whats wrong with calling it a lemma

every lemma has a list of conditions

[neat sparrow]

@kind hawk

[neat sparrow]

conditional lemma maybe?

[kind hawk]

are you working in some proof system where you have to be overly precise with everything?

[neat sparrow]

yes

[kind hawk]

otherwise there is nothing wrong with doing stuff like
case 1: assume m is even. we do some things. we arrive at a lemma
lemma 1: if m is even, then bla holds
then we do some other things
case 2: assume m is odd. we continue with other stuff
lemma 2: if m is odd, then ...

and you should be able to formally translate this into whatever specific restricted proof system you have

[obtuse pebble]

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[neat sparrow]

Suppose I know that \

1. $EV(f \cap s) = 0$ \ \
2. $P(s) = 0 \to EV(f \cap s) = undefined$ \

then I can derive $P(s) \ne 0$, right?

[warm shale]

! See my about me

[jaunty ocean]

Yes

The contrapositive of 2 reads if EV(f\cap s) is defined, then P(s) != 0.

[obtuse pebble]

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[cyan helm]

have i made any mistakes with this, thanks

[obtuse pebble]

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[muted sleet]

part b

[muted sleet]

i got the y(80-2y)

and then i got 40=y

but it doesnt add up cuz then 80 + x = 80

not sure where i went wrong

because im pretty sure everything i did was right

[blissful flame]

You could try a drawing, labelling the sides and see if you can spot your own mistake

[muted sleet]

yea i did

the shorter sides were y

and i did the one long side as x

try it

if y(80-2y)= area, then wouldnt y = 40?

[blissful flame]

why so

[formal island]

Do you understand where the 80-2y comes from?

And why we multiply by it to get the area.

[muted sleet]

because u just do the bracket = 0

[muted sleet]

i thought it came from the 2y+x=80

[blissful flame]

why would the Area be =0?

[muted sleet]

which was the total length

[muted sleet]

but i thought u always make the bracket = 0

to get a solution

[formal island]

Like, yes, it is, but do you know what it represents for the problem?

[blissful flame]

if you have 0 on the other side yes

but on the other side you have A, not 0

here its not really about math and equations that much, you need to make a clear drawing of the situation and try to understand where things come from

[muted sleet]

the total length for the fence

[muted sleet]

oh

then

well its not the perimeter

[formal island]

Lets try to break down the problem

If you need help figuring some part, diagram it, it should solve any doubt

What is a valid equation for the perimeter of fence?

[muted sleet]

2y+x=80

[formal island]

What do y and x represent there?

[muted sleet]

y represents the length

x is the width

[formal island]

We are looking for the Area of the garden, so, based on x and y, (whatever they are), whats the equation for it?

[muted sleet]

oh

xy

[formal island]

Yes

[muted sleet]

oh damn

i see it now

[formal island]

Can we rewrite that equation into what they gave us?

[muted sleet]

make x the subject and then substitute

[formal island]

Thats where (80-2y) comes from

[formal island]

So, again, what does that part represent?

[muted sleet]

the 80-2y?

[formal island]

Yep, recall how you got to that using A = xy

[muted sleet]

the 80-2y represents the perimeter of the fence

[formal island]

Nope,

thats x

[muted sleet]

💀

[formal island]

since you found that x = 80 - 2y

What was X?

[muted sleet]

ohhhhhh

x was the width

[formal island]

yeah

[muted sleet]

80-2y represents the width of the rectangle

[formal island]

What we did is rewrite the equation of the area only based on the length of the fence

Since we know it adds up to 80

[muted sleet]

i see

[formal island]

Now, you got yourself a function of y

How do you find the maxima of a function?

[muted sleet]

dont think so

function like dy/dx?

[formal island]

Okay this might be a little uncomfortable, but when doing derivatives, variables are just random names

So we can do d/dy

[muted sleet]

wheres the x

[formal island]

Thats why i said its uncomfortable, lemme draw something up rq

gimme a sec

[muted sleet]

ok

[formal island]

I hope you agree with this part

this one is the obvious one

btw this is just a random example > hasnt have anything to do with the problem

[muted sleet]

how did d/dx become 3

only thing i didnt understand

[formal island]

I just took the derivative

[muted sleet]

ohh

[formal island]

the derivative of x^3 is 3x^2

[muted sleet]

yep

and that gets rid of the d/dx

but where did the y go now

[formal island]

im working towards that

X is a variable, right?

[muted sleet]

oh alright

[formal island]

just a name for some random thing which changes

[muted sleet]

yes because it can change

[formal island]

The name of the variable doesnt matter

You will later learn down the line that you can take derivatives over any variable, even if they are mixed, but thats beyond the point

For our function f(y) = y(80-2y)

We can take a derivative using y

if it makes it easier to do for you, just replace y with "x" for a moment, (with the consideration of not confusing yourself with the x in our original problem)

[muted sleet]

would it be d/dx? or d/dy

[formal island]

d/dy

if you took d/dx, then youd have no x in there

So it would become 0

[muted sleet]

so u will have y terms left

ok so d/dy y(80-2y)

u expand te brackets

the

[formal island]

yep, distribute

[muted sleet]

d/dy 80y-2y^2

you get d/dy = 80-4y

right?

did i do something wrong 💀

couldnt u find maximum point by doing turning point

<@&286206848099549185>

ok nvm

.close

[obtuse pebble]

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[rich star]

I'm trying to do this problem, but I'm struggling with how to do it without seeing the graph. Khan Academy's video on it had a graph, but this only lists the two points. Here's what I managed to get done.

Can I have some help with the rest?

[obtuse pebble]

@rich star Has your question been resolved?

[rich star]

Is the answer too straightforward?

I'll close this.

.close

[obtuse pebble]

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[lofty fable]

still unsure how to do this question

[lofty fable]

tbh i dont even know where to start

id assume maybe seperate ys and xs but idk

[native inlet]

this differential equation is not separable

but it is a linear first-order and those are always solvable via an integrating factor

[lofty fable]

and how exactly do they work?

[native inlet]

so if we have some differential equation (\frac{dy}{dx}+p(x)y=g(x)), right? Then we want to multiply by some function (typically called (\mu)(x)) to get (\mu(x)\frac{dy}{dx}+\mu(x)p(x)y) to be equal to(\mu(x)\frac{dy}{dx}+\mu'(x)y)

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[native inlet]

because the latter equation is exactly the product rule, so it simplifies to (\frac{d}{dx}[\mu(x)y])

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[native inlet]

the integrating factor is given by (\mu(x)=e^{\int p(x)dx})

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[lofty fable]

so do some of these factors just come from nowhere to make the question possible?

[native inlet]

it comes from the idea here

if we have (\mu(x)\frac{dy}{dx}+\mu'(x)y) to be exactly equal to (\mu(x)\frac{dy}{dx}+\mu(x)p(x)y) this happens iff (\mu'(x)=\mu(x)p(x))

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[native inlet]

the solution to that last differential equation with (\mu(x)) is exactly (\mu(x)=e^{\int p(x)dx})

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[native inlet]

so multiply that by everything in your equation and everything usually works out pretty nicely

[lofty fable]

so from the firs line i multiply everything by μ(x)?

[native inlet]

and in your equation what exactly is mu(x)?

[lofty fable]

arbitrary function i guess

[native inlet]

your equation is (\frac{dy}{dx}+5x^4y=e^{-x^5}\cos(2x))

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[native inlet]

and thus (p(x)=5x^4), so can you figure out what (\mu(x)) is?

[warm shale]

ΠαϳαμαΜαμαΛλαμα

[lofty fable]

so whats the line after this?

[lost zenith]

Use this

p(x) is always the x term that is multiplied by y

[lofty fable]

So is this right?

.close

[obtuse pebble]

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[distant ravine]

Hey, one of my courses suddenly brought back laplace transforms after I haven't seen it in a long time, and I'm having a hard time with it. I'm currently trying to figure out how they did this part.

[distant ravine]

How did they get from the y'''+6y''+11y'+6=1 to the equation below it?

I thought it might be the LT that's highlighted in red on the right

And also, what do they mean by "zero initial conditions"?

[carmine spear]

y''(0)=y'(0)=y(0)=0

and you're right they are using laplace transform here

yk the L{y'} = sL{y} - y(0) thing?

js spam that till everything is $\sum s^nY$ form

[warm shale]

donkey

[carmine spear]

and you'll get it

[distant ravine]

Think I got it

What about this one? I sat on this for a really long time and couldn't figure it out

The final part is the one where I get stuck

Where they go from the L^-1 to equation 3-37

I tried solving it but didn't know how to proceed from this part (blue question mark on the right)

$L^(-1)[(5/2)*(1/(5*s+4))]$

[warm shale]

Somalian Space Pirate 🇮🇱

[distant ravine]

I tried lol

Oh wait I see it now

[obtuse pebble]

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[spare fossil]

i have a question if i have two vectors and made the cross product and did a plane with the two vectors how can i calculate the length and the direction of any vector's shadow on that plane

[spare fossil]

using cross and dot product

[forest sinew]

https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/00e9c8f0eafedeab21a3d079a17ed3d8_MIT18_06SCF11_Ses2.2sum.pdf

maybe this is helpful?

At least it sounds like youre describing projection onto a subspace

[spare fossil]

yeah its helpful thank its in 2d but still the idea thanks

[worn yoke]

if you subtract off the projection of a vector onto the plane's normal vector, then you'd get the projection of the vector onto the plane

[obtuse pebble]

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[karmic hedge]

This is a step in a tableau for semantic logic, I dont understand it

Why are we creating an object Q from the universal quantifier

I thought only the existential quantifier could create objects

Oh wait its not creating an object (I think)

Cause theres no +

So idk whats happenning here in this step

nvm i got it

.close

[obtuse pebble]

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[finite terrace]

this is 9th grade and idk how to do this

we have to do half with calc and half without

can sm1 give me a exmaple of how to do c and d?

pls any1 chatgpt down bruh ;-;

[royal basin]

noai

!noai

[obtuse pebble]

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

[royal basin]

anyway c and d are kinda really different functions

c is linear, d is a reciprocal function

do you know how to graph linear functions in general y/n

[finite terrace]

uhm no

[royal basin]

mmm review that then

[final plaza]

is this a good time to mention desmos and wolframalpha

[royal basin]

cause everything else is gonna be out of reach

also yes desmos

desmos very good for anything graphing related

[finite terrace]

ik but i cant use it on test :sob

[royal basin]

tho it does that for you

so it's a learning tool not a cheating tool

[final plaza]

yeah play around with them, especially with desmos

it will help you get a feel for how graphing functions works

[finite terrace]

ok

[obtuse pebble]

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[upbeat sigil]

Help with Karnaugh Diagram
How do I combine this?

[obtuse pebble]

Please don't occupy multiple help channels.

[upbeat sigil]

is this pairing correct?

[serene elk]

We pair things in groups that are sized powers of 2

[upbeat sigil]

Is this correct

[serene elk]

So your group of 6 is weird

[upbeat sigil]

oh

[upbeat sigil]

this is referring to karnaugh graphs? or something more broad

[serene elk]

They general rule is make the largest power of 2 groups that you can

[serene elk]

Karnaugh

[upbeat sigil]

mhh how would you do that? I have the 4x1

should I split the other box into 2x2 and 2x1

im always thrown a bit off when im counting a cell double or thrice

[serene elk]

Looks good

[upbeat sigil]

wait Im unsure now, did you mean to say my original solution with the 3x2 is wrong? or just unusual? @serene elk

[serene elk]

Here you want 3 groups of 4

You need to work with powers of 2

[obtuse pebble]

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[upbeat sigil]

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[upbeat sigil]

@serene elk can I combine the 3 terms now or am I meant to just link them via "or"

[royal basin]

blue is $\overline{a} \land \overline{c}$ not $\overline{a \land c}$

[warm shale]

Ann

[upbeat sigil]

green too right?

[royal basin]

yeah the same goes for green

your red is actually correct

[upbeat sigil]

ok so im left with this

can I further reduce Y?

or is that how it should be

or rather *am I meant to reduce Y?

I dont see a common variable so Im unsure

[royal basin]

it's in disjunctive normal form already

so i think you're fine

[obtuse pebble]

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[rare sand]

🫩

[scenic zenith]

What have you tried

@rare sand

[rare sand]

i forgot how to do it

i know its like

the limit definition of a derivative or something

the

idk

idk

oh

[trim portal]

indeed

derivative of what function at what point?

[rare sand]

any point

root3

i think i figured it out

[trim portal]

and the function is arctan, yep

[rare sand]

do i js find the derivattive of arctan

[trim portal]

yes

[rare sand]

1/ (1 + u^2)

riht

[trim portal]

use parens

but yes

maybe it would be more appropriate to use x here, but thats just a detail

1 / (1 + x^2)

now you can just plug in root3 and done

[rare sand]

ohh ok

i had another question

i dont know how to approach related rates at all

[scenic zenith]

How do you find rate of change of a quantity ?

[rare sand]

the derivative

idk

[scenic zenith]

Correct

So it's asking us to find the rate at which fuel changes

[rare sand]

Oh

[trim portal]

fuel consumption to be exact, and make sure to differentiate it wrt the right quantity

[rare sand]

do i use quotient rule

when i like find the derivative of the exponent

would s become ds/dt

[trim portal]

you gotta use chain rule here

you need to calculate derivative of fuel capacity with respect to time

so dF / dt

by chain rule, that equals (dF / ds) * (ds / dt)

and now you can calculate dF / ds and ds / dt

from the equations given

[rare sand]

idk

is this wrong

[trim portal]

what is that supposed to be?

[rare sand]

dF/dt

[trim portal]

looks right then

[rare sand]

Omg fr

[trim portal]

though I dont understand how you got e.g. 20/400

like when you differentiate s/20, you should simply get (ds/dt)/20

but whatever you did, it apparently worked

[rare sand]

the formula our teacher gave us is

((bottom x derivative of the top) - (top x derivative of the bottom)) / bottom ^2

alr so for s i plug in 50 and ds/dt is 20

then thats my answer?

[trim portal]

Oh you used quotient rule

[trim portal]

yes

[rare sand]

was i not usppoed to

Supposed

[trim portal]

btw when it's s/20, it's the same thing as (1/20)s

so the derivative will be just (1/20)ds/dt

[rare sand]

oh

💀

[trim portal]

its linear

quotient rule is kind of an overkill, but it works

[rare sand]

😭i forgot

okay

okokok

thank you

.close

[obtuse pebble]

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[slate sandal]

How to solve 2x+3=11

[slate sandal]

I forgot how to do these

[timber mesa]

Subtract 3 from both sides then divide by 2

[slate sandal]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

okay thanks

[slate sandal]

Though what about when it's 5x-1=9?

[timber mesa]

Add 1 to both sides and then divide by 5

[royal basin]

ok this is crossing into nosols territory

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[slate sandal]

Oh

My bad

[royal basin]

https://www.youtube.com/watch?v=Z-ZkmpQBIFo

[royal basin]

not at you, at @timber mesa

[obtuse pebble]

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[blazing tangle]

Need to use cylindrical coordinates to find the volume

[blazing tangle]

I can visualize the solid

But its hard to understand the cylindrical part

normally integral is set up as dzdrd(theta)

and usually I write it in rectangular form then convert to cylindrical

this will be the x-y plane

will this give the limits for x and y?

is this the correct setup in rectangular form?

Not sure why desmos sees the triple integral as a parabola, but thats less important

wow there a lot occupied

hmmmmm

The jacobian!

wish it was this easy on ti84

anyone know much about cylindrical coordinates

[hardy widget]

this is fine

[obtuse pebble]

@blazing tangle Has your question been resolved?

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[night plume]

would someone look at this pls thanks

[jolly raptor]

There's one mistake in the adjacency matrix. You can note that adjacency matrixes are always symmetric across the diagonal

which should make the mistake easier to spot

[obtuse pebble]

@night plume Has your question been resolved?

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[wise echo]

can someone please help me with my calc hw

[wise echo]

I know that the integral of 2x + sinx is x^2 -cosx

and then I tried plugging in -2

and it didn't work

[wary badger]

where is that coming from?

did you find f'(2)

i don't know why you're integrating h

[wise echo]

I thought I had to use g(2)

because of the integral?

[wary badger]

f is the integral of g

why are we integrating h

[wise echo]

oh

because h is the reverse derivative of 2?

[wary badger]

what?

stop looking for reasons to integrate it

find f'(2)

[wise echo]

ok

[wary badger]

then set h = f'(2)

[wise echo]

ok

[wary badger]

this is a calculator question yea

so you can just graph it

and solve for the intersection

are you using a ti-84

[wise echo]

ok so if f = 3

f(2) = 3 *

then i just graph that and h(x)?

[wary badger]

where did you get 3?

f'(2)

not f(2)

[wise echo]

oh ok

so it's -2

[wary badger]

yes

[wise echo]

wait but then when I graph it im not getting one of the answer choices

I might be cooked

[wary badger]

no i definitely get something

that is one of the answer choices

let me see

what you did

[wise echo]

did you put in the x^2 - cosx or 2x+ sinx

[wary badger]

why would you put x^2 - cosx?

it says h(x) = f'(2) = -2

so 2x + sinx = -2

2x + sinx + 2 = 0

[obtuse pebble]

@wise echo Has your question been resolved?

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[reef spoke]

i am able to understand the situation but not able to put it in mathematical terms

do i use triangles by any chance?

to trace the path of the shadow

.close

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[proven glade]

use the equation of motion..... @reef spoke

[obtuse pebble]

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[reef spoke]

what

[lone echo]

Does this help

the idea is computing the position of the shadow as a function of t.
The speed is the derivative of the position wrt time

[obtuse pebble]

@reef spoke Has your question been resolved?

[reef spoke]

i understood this

but how do we get the speed of the shadow as a function of t?

im only able to get values of it when t is given

do i just take a random instant t

and get its

[lone echo]

well, first off, you get the position of the ball as a function of t

you should be able to do that with the equations of motion, for the ball itself ignoring the light and the shadow

[reef spoke]

its just 50-16t^2

[lone echo]

okay

now, from that, you construct the triangles

you will have a small right triangle, with base 30 ft, with height 16t^2 ft, and hypothenuse function of t

[reef spoke]

oh ye

[lone echo]

you will have a similar triangle to that one, with height 50 ft, base of the position of the shadow as a function of t (in my diagram, starting from the left)

[reef spoke]

so we just ignore the value of t and get the shadows speed in t itself?

then put t value later

[lone echo]

no, this is for the POSITION, not the speed

[reef spoke]

yea i understood

[lone echo]

now, having the position, you can differentiate wrt t to obtain the speed

[reef spoke]

ohhhh i see

dang

i understood

tysm

[lone echo]

np

[reef spoke]

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[scenic zenith]

\textbf{Check convergence for } $\displaystyle \frac{1}{x^{n} + x^{-n}}$.

[warm shale]

FishTank

[scenic zenith]

This q was asked in my test today, when it ended i asked if my answer was correct, he just said "no"

Can anyone see my work and check what have i done wrong?

[lone echo]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[scenic zenith]

I don't have the exact question with me

It was this expression and we had to check if this series is converging or not

[obtuse pebble]

@scenic zenith Has your question been resolved?

[obtuse pebble]

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[jagged fulcrum]

why don't the photons/ELECTRO MAGNETIC WAVES travelling through light years miss out our electrons? or get accepted by some one else in between?

and I've heard electrons are so small, then why don't these waves sort of travel past us

[hardy widget]

discord.gg/physics

[formal island]

#serious-discussion, close this

[jagged fulcrum]

kk

.close

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[steel night]

Im confused with why there is both bio and chem

[steel night]

for the venn diagram

the example says study either biology OR chemistry

not bothh

[worn yoke]

in math, "or" means "one, the other, or both"

[steel night]

why both?

[tardy epoch]

https://en.wikipedia.org/wiki/Logical_disjunction

[steel night]

ah makes sense

thanks

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[void zephyr]

find the equation of the elipse with the foci (4.0) (-4,0) guys plz help

[obtuse pebble]

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

[void zephyr]

7

[obtuse pebble]

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

[void zephyr]

I can't start i think it is incomplete given

[hardy widget]

yeah you're right about that - the question doesn't allow for an incomplete answer

btw, in the future, please specify stuff like this when you first ask the question

gives more context and prevents a game of 20 questions

[void zephyr]

Sorry 😔

So it can't be solved right

[hardy widget]

not uniquely

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

@void zephyr Has your question been resolved?

[hardy widget]

what else do you need help with?

reposted: #1440891248956407939 message

.close

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[proper beacon]

If graph A is m-colorable and graph B is n-colorable (A and B have the same vertex set V). How many colors do I need to color their union C?

[proper beacon]

This isn't a test question nor homework btw, just my curiosity

My current guess is that it's mn

But I can't prove that

[trim portal]

if the 2 graphs are exactly the same, then you'd need only m colors

so it cant be as simple as that

[proper beacon]

mn is at least the upper bound though, right?

[trim portal]

a lower bound is certainly max(m, n), and this is strict

[trim portal]

yeah, that sounds right

[proper beacon]

Can this be generalized for more graphs?

Upper bound is always just the product, right?

And lower bound is just the max of numbers of colors needed for the subgraphs

[trim portal]

yeah, sure

[proper beacon]

How do I prove the upper bound is tight?

The way i think of this is that the maximum is ofc reached when the union graph is complete

[trim portal]

you'd have to construct an nm-colorable graph which can be divided into 2 graphs, one of which is n-colorable and the other one is m-colorable

[trim portal]

This is a pretty interesting idea actually

if we take a complete graph with n vertices, it's gonna be n-colorable

and we can make m copies

nvm, my idea wont work

actually it will

[trim portal]

now from each node from each copy, make an edge towards every other node from different copies

[proper beacon]

My brain accidentally played 5D chess when trying to visualize that

[trim portal]

the union of
the graph of m copies of complete graph with n vertices
and
all edges from each node from each copy towards every other node from different copy

is gonna be the complete graph of nm vertices

[trim portal]

ill draw and take a pic

[proper beacon]

I'm following well, yeah

[trim portal]

this will be the first graph. 12 vertices, 4-colorable

the second one would be the complement of this (so that their union would be complete)

the complement will be 3 colorable. You can simply color each of the 3 "components" with different colors

so this construction should prove the tightness

[proper beacon]

Very nice example

How do we prove the upper bound in the first place?

[trim portal]

if u take the set of colors, you can do cartesian product ig

[solar meadow]

Whoo! Good thing I'm colorblind

[trim portal]

then if the vertex was assigned color i in n-coloring and color j in m-coloring, its gonna get colored with (i, j)

try proving that this works

[proper beacon]

Let i and j correspondingly be proper colorings for A and B. If u~v is an edge in C, then it must be an edge in either A or B (or both). If it's an edge in A then i_u≠i_v and if it's an edge in B then j_u≠j≠v so (i_u,j_u)≠(i_v,j_v)∀u~v. Then if we assign a color k for each element of {(i_v,j_v)|v∈V}, we'd get a mn-coloring (because there are m values of i and n values of j)

@trim portal is this correct?

Sorry for the ping

[trim portal]

Yep

[proper beacon]

Thank you so much

[trim portal]

np

[proper beacon]

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[austere whale]

hello there

[versed stratus]

is this a test?

[austere whale]

no.. its a module exercise

its weekly assignments in a nutshell

[versed stratus]

got it

[austere whale]

so basically

sin (2theta) = 2sin theta cos theta

what should i divide by 2...

[solar meadow]

$\sin (2\theta) = 2\sin \theta \cos \theta$

[austere whale]

the whole sin or the angle only?

[trim portal]

\sin and \cos

[warm shale]

Erebus

[trim portal]

what are u trying to get to?

[austere whale]

the sine value

[trim portal]

you want sin(theta)cos(theta) on the RHS?

[austere whale]

nope, basically were simplifying it

[trim portal]

yeah, i mean what are u dividing by 2, why are u dividing it?

[austere whale]

so for example, if this was 2sin(theta/3)cos(theta/3), the simplification is sin(2theta/3)

[trim portal]

yep, good

so
2sin(theta/3)cos(theta)/3) = sin(2theta/3)

divide both sides by 2

and u get what you need

[austere whale]

both sides here, 2sin(theta)cos(theta)/ 2 = sin(theta)cos(theta)

[trim portal]

and u are dividing the whole side, meaning the whole sin

not just the angle

[austere whale]

why are we not dividing the angle only?

[trim portal]

it's an equality, we have to do the same thing to both side

so if we wanna divide one side by 2 (the 2sin(theta/3)cos(theta/3)), we gotta divide the other side as well

the other side is sin(2theta/3)

and when we divide that by 2, we get sin(2theta/3) / 2, not sin(theta/3)

$2\sin\left(\frac{\theta}{3}\right)\cos\left(\frac{\theta}{3}\right)=\sin\left(\frac{2\theta}{3}\right)$

[warm shale]

MathIsAlwaysRight

MathIsAlwaysRight

[trim portal]

we must do the same thing to both sides

[austere whale]

i see i see

I asked the question regarding why not just dividing the angle by 2 because I assumed that it would arrive at the same answer

[trim portal]

$\sin\left(\frac{\theta}{3}\right)\cos\left(\frac{\theta}{3}\right)=\sin\left(\frac{\theta}{3}\right)$

[warm shale]

MathIsAlwaysRight

[trim portal]

you would arrive at obvious non-sense

[faint talon]

I don't understand what you are searching for here

You know sin(2x) =2sin(x)cos(x)

So sin(x)cos(x) =½sin(2x)

[austere whale]

i was confused

i thought sin(2x/2) = 1/2 (sin(2x)) lmaoo

[faint talon]

Et c'est ok

[austere whale]

thanks guys !!

.close

[obtuse pebble]

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[torn sandal]

Exercise 2 limit of ln(1+x) when x approaches 0

[torn sandal]

Idk how to continue

Should i add the absolute value to the inequality

I have a feeling the answer has something to do with bringing a whole new value out of nowhere

[ruby orchid]

what if you just let x = 0

[torn sandal]

Why

[ruby orchid]

the limit defines what the function approaches as x approaches 0

[torn sandal]

Yes

[ruby orchid]

that includes what would happen if x was 0

[torn sandal]

If x was 0 then the function is also 0

What are you trying to say

Did you mean the other x

X_0 from the definition

Yeah idk what im doing

I tried to prove that e^-eps -1 is less than -e^eps +1

So that i can insert an absolute value inside

[obtuse pebble]

@torn sandal Has your question been resolved?

[hexed surge]

it's pretty good actually

you can't take the absolute value like that

[hexed surge]

don't overcomplicate it

hint: ||just take the max of both values||

[torn sandal]

we kinda can't since -e^eps + 1 is less than 0

shouldn't both values be positive

[hexed surge]

sure

how would you adapt?

[torn sandal]

flip the inequality on the lhs?

[hexed surge]

kinda

[torn sandal]

@hexed surge

[hexed surge]

sure about this?

one term in the max is negative

[torn sandal]

wait

yeah i accidentally wrote e^-eps instead of e^eps

is that it?

[hexed surge]

👍

[torn sandal]

alright

moving to the second limit now

I have a strong feeling that it's wrong

Ok it's wrong

Delta can't be negative

ok i thought it was +inf

I only had to change the characterisation

[torn sandal]

Just to top it all off with the 3rd limit

I think im done here

.close

[obtuse pebble]

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[rigid shadow]

any super duper math nerds able to help me with a custom formula i want to make with lots of dynamic variables and data points/sets where the formula can be used in increasing or decreasing order -- ideally this should be able to be made into a google sheets formula as well???? i have tried really hard on my own to make the sheets formula, have chatGPT make the results as well as the formula. I just cant find a good result for what i am trying to do i think.... details super complex including variables for things like this: noble position (top mid and low), noble rank names, percentages per positions assigned to ranks (55% to top, 30% to mid, 10% low, and 5% misc/other), a starting min and ending min for the initial data to create the formula with, set assignment ratios on the top end, subordinate counts for each noble rank to divide the total territory per rank into territory from noble rank into single individual territory at rank, and a few other parameters.
anyone able/ willing to help as AI is dropping the ball and i am failing in some of the logic
the going up initial formula works with 'galaxies' as the territory but inversing the formula should work in reverse dividing a single galaxy into fractions and even further down it should divide a planet into chunks using the same methodology
i am using a form of google sheet table for displaying data an image attached shows some of that to help clarify formatting of stuff: territory numbers for galaxies are messed up due to confusion with chatgpt trying to provide data, total members is based on the subordinates column downwards, territory per individual is another messed up data set currently (please let me know if other info needs clarifying)
examples of common errors or misunderstanding:
emperor = 160,000,000,000,000 galaxies
archon = emperor territory / 4 --> 4 archons
praetor = archon territory / 15 --> 15 praetor per archon OR emperor territory / 60 as there are 60 total praetors in the empire
then you have noble positions:

PART 2 BELOW

[rigid shadow]

for grand noble you have grand moff, moff, and governor representing the top middle and lower ranks of that position/tier than the following applies:
there are 20 grand moffs per praetor for 1.2K total in the empire, there are 3 moffs per each grand moff, and 5 governors per each moff
say that there is 800 billion galaxies distributed to the Grand noble tier than the following would apply == 55% of it goes to a the grand moffs, 30% to the moffs, 10% to the governors, and 5% to imperial knights serving those ranks or other misc things like merchants etc. --- what this also means is that the 55% of 800 billion is split between 1,200 grand moffs, the 30% is split between the 1,200 grand moffs * 3 subordinate moffs for 3,600 moffs total so 3,600 moffs receiving 30% of 800 billion, this continues down until the 100% is distributed
moving onto high nobles we have a repeat of situation: say there are 500 million galaxies to distribute to all high nobles, than the archduke gets 55% split between the total archduke count (i believe there are 108k of them), dukes receive 30% and there are 540k of them etc etc downwards repeating this cycle
at the bottom the imperial baronet should have a fixed number of 10,000 galaxies, imperial baronise oversee 100 galaxies under a baronet, and imperial knights oversee 10 galaxies. the total number of galaxies the emperor rules over can be adjusted to be higher to meet the requirements
this equation should also fit a fraction form-factor for the galactic section where each noble rules a portion of a galaxy fraction wise with the same %'s but with different subordinate counts as seen in the image
the equation should have a way to deal with there being less ranks in the planetary section where there is only a top and bottom rank, the percentage will be 55% top and 35% bottom with the remainder being the misc %

i know this is a crazy amount of stuff and ideally has some form of google sheets formula too but i do need the help. i more or less of the logic down but need help with the calculations and formula itself as i want the numbers for per individual at each rank and how much territory each rank has. also i would ideally like each rank to be smoothed out, so like what that means is say a rank comes out to 1,601,709 then the indicidual would actually rule 1.6 million and the 1,709 extra galaxies would be post calculation given to the misc category slightly altering the 5% but should only show this result in a sepereate column of the calculation results in a table.
please let me know how much if any of this is possible, i will accept help or answers for this

[obtuse pebble]

@rigid shadow Has your question been resolved?

[elfin cosmos]

this isnt a math question

[rigid shadow]

pure curiosity in what way?

[elfin cosmos]

you're asking how to format an excel spreadsheet with a bunch of random quantities. nobody here is going to be able to help you with that

what is this even for?

might as well just ask chat gpt

[rigid shadow]

for it what it is for i would say story building/novelization/personal as for the spreadhseet that is an added bonus really but the real thing i want is the formulation to solve what is essentially a LARGE word problem with data set values that i am struggling with

yes well AI services are struggling too

[elfin cosmos]

im sorry but this is too hyper specific

maybe try to clarify what you're asking

[obtuse pebble]

@rigid shadow Has your question been resolved?

[scenic stirrup]

@rigid shadow that's WAY to complicated and I don't think nobody here is can make this to you directly. What you want is a database with Tiered Distribution and Dynamic Rounding. This would be very difficult to achieve with a single cell formula (lambda, etc.), so instead you should create a row-by-row 'Waterfall Model'.

Maybe I couldn't help you either, but you can probably create something yourself using these topics.

[rigid shadow]

okay thanks

.close

[obtuse pebble]

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[nova parcel]

do u think this is the most optimal way to pack my cocoa snails?

[lost acorn]

the most optimal way to pack circles on a plane is via this hexagonal lattice pattern

[lost acorn]

those aren't perfect circles but I'm sure this would be a good approximation for the best packing of your cocoa snails

[nova parcel]

thanks, this helps. i think i did this intuitively in the middle, but becuase of the limitations of my baking tray at the edge i had to opt for this different design

[obtuse pebble]

@nova parcel Has your question been resolved?

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[potent pendant]

Hi, I am faced with an interesting combinatorics problem: you are given the string “oof” and you can do the following operation 3 times. insert an alphabetical lowercase character anywhere in the string. Find the amount of distinct resulting string. Calculators allowed. How to solve?

[inner sierra]

assuming we are working with the english alphabet?

[potent pendant]

Yes

[inner sierra]

we can do the operations 3 times

so we need to consider each time, how many possible cases are there

so for the first character, how many characters can you choose

@potent pendant?

[potent pendant]

Yes I see

The problem is that you can overcount

[inner sierra]

nah, i don't think theres's overcount for this one

assuming X is your character, so you can only insert it at Xoof, oXof, ooXf, oofX

oh wait

X could be o tho

[potent pendant]

Ya

[inner sierra]

and X could be f

[potent pendant]

<@&286206848099549185>

[obtuse pebble]

@potent pendant Has your question been resolved?

[potent pendant]

<@&286206848099549185> pls help 🙏

[wary lintel]

okay, tried to solve it

Start with the string "oof" (length 3) and insert a lowercase letter anywhere 3 times.

[potent pendant]

Ok I’m following

[wary lintel]

After 3 insertions, the string length is 6.

Treat "oof" as a fixed subsequence

choose positions for the original 3 letters in the 6-length string (
((6/3) = 20) and fill the remaining 3 positions with inserted letters (26 choices each, 26^3 = 17576)

Multiply to get the total number of distinct strings: 20x17576 = 351520

[potent pendant]

How to remove duplicates?

Wait

That still includes duplicates though

[wary lintel]

Ah, you mean how to make sure we only count distinct strings, i.e., avoid duplicates caused by repeated letters in the original string like "oof"?

[potent pendant]

Sorry, I must have worded my question inaccurately

I am looking for unique strings

Fixed the question

[wary lintel]

ohh okayy

wait let me solve rq

[warm shale]

YOLO SO DO WHAT YOU LOVE

[wary lintel]

@potent pendant

okay this should be right

[potent pendant]

Hmm, lemme see if I understand

@wary lintel how did you account for oof when f is placed before the f in string and after?

[wary lintel]

The two "o"s in "oof" are identical, so swapping them doesn’t create a new string, but the f is unique, and its position relative to inserted letters always creates a distinct string. Inserted letters in different positions also produce distinct strings. Therefore, we only divide by 2! for the repeated "o"s, and the total number of unique resulting strings remains 175760

[potent pendant]

I believe that there is still over counting

For example when when you can do only one operation on oof:

In the case when you can only add f

there are only 3 ways. The fourth way is a duplicate

Also when you insert two numbers to an empty string, the answer is not 26^2. Resulting strings like “aa” and “bb” are repeated. What do you think @wary lintel

[wary lintel]

mhm

let me think

i think inserting letters in multiple ways can lead to the same final string, so counting insertion sequences overcounts

to get the true number of distinct results, you must deduplicate the final strings

so I was wrong

after doing that, the total number of unique final strings is 919496.

maybe I'm wrong again idk

[potent pendant]

May I know how you found 919496 please? Thank you so much for your time!

[wary lintel]

well I used python

if you want I can paste the code

[placid bone]

Becuase you can add o and f for the 3 random characters, you can't do $26^3$ blindly it will overcount.
I'd first calculate without o and f for the extra 3 and then add them in to the original known to get the answer.
Basically o2fX3 + o3fX2 + o4fX + o5f + o2f2X2 + o3f2X + o4f2 + o2f3X + o2f4 (I think, if I got it all right) which I am pretty sure you can write concisely as a summation and use formula to calculate.

[potent pendant]

Uhhh, surprisingly i am trying to use my knowledge I gain from this problem to solve a coding problem. The string can have a length of up to 1e6 so likely, the Python code will be too slow if not optimized.

[warm shale]

@placid bone

[wary lintel]

Wait you can also do it with the inclusion-exclusion formula

do you know it?

[placid bone]

I think I forgot o3f3 but it should give you the idea

[potent pendant]

Kind of

[potent pendant]

May I ask what 2 means in the string?

[placid bone]

Count of previous character

o2fX3 = oofXXX, i.e. the characters you're permuting

$X \in \Sigma_{Eng} / {o, f}$

[wary lintel]

Let &(S)& be the set of the $(\binom{6}{3}=20)$ index triples $(T=(p,q,r))$ with ($0\le p<q<r\le5)$. For each triple $(T)$ let $(A_T)$ be the set of length-6 strings with positions $(p,q,r)$ equal to o,o,f respectively. We want $(\left|\bigcup_{T\in S} A_T\right|)$.

[warm shale]

YOLO SO DO WHAT YOU LOVE
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

@placid bone

[wary lintel]

$[
\left|\bigcup_{T\in S} A_T\right|
= \sum_{\varnothing\neq J\subseteq S} (-1)^{|J|+1}; \big| \bigcap_{T\in J} A_T\big|.
]$

[warm shale]

YOLO SO DO WHAT YOU LOVE
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[potent pendant]

Hmm I think I am starting to understand. Does X mean any letter that is not o or f

[placid bone]

Yes

[potent pendant]

What if X was s placed between two other character? Right now I think you have only calculated when x is at the end

[placid bone]

Those are only the counts, I haven't calculated anything yet

Also just place the known characters first and fill the extras in the last and you've fixed the problem

I think the formula should look something like this

[scarlet laurel]

wth

are all these symbols

😭

[placid bone]

$\sum_{\substack{o + f \le 6 \\o \ge 2 \\f \ge 1}} 24^{6 - o - f} \cdot \frac{P(6, o + f)}{o! \cdot f!}$

[potent pendant]

May I know what the p function is? In the P function, o + f = 6, so can the stuff inside the p function be simplified to p(6,6)?

[placid bone]

P(n, r) is the number of ways to arrange r objects in n spaces

Also, I made a mistake

[warm shale]

@placid bone

[potent pendant]

Thank you so much! I think I understand the equation :). Just wondering, can this problem be calculated using principle of inclusion and exclusion? Mostly I am just worried about the number of terms in the tedious summation.

[wary lintel]

yes

[wary lintel]

.

[potent pendant]

Oh sorry about that! I thought that was a general formula mb. I am not good at set theory and am a bit confused with notation. May I know what the absolute sign and the brackets at the start and end of the equation is?

[obtuse pebble]

@potent pendant Has your question been resolved?

[potent pendant]

<@&286206848099549185> apparently this is the formula that I need to use to solve the original problem. However I do not know what any of the variables mean. Can anyone help? Thank you

[potent pendant]

<@&286206848099549185> pls help 🙏

[obtuse pebble]

@potent pendant Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[gleaming gazelle]

.

[lyric valley]

what r u having trouble with

[elfin cosmos]

is this latvian?

[gleaming gazelle]

yes

[gleaming gazelle]

How do you solve that

[elfin cosmos]

anyway multiply sqrt(10) out to 8 and 6*sqrt(10)

[gleaming gazelle]

what does that mean

how?

why?

[elfin cosmos]

do you know how to do 3 * (8 + 6)?

[gleaming gazelle]

I guess?

[elfin cosmos]

do you know what √ means?

[gleaming gazelle]

Yes

[elfin cosmos]

so you have √10, you multiply 8 and 6√10 by √10

8 x √10 + 6√10 x √10

[gleaming gazelle]

Why is it si random

Wasnt ( ) first??

math is confusing

[elfin cosmos]

if you have

3 * (8 + 6)
= 3*8 + 3*6

right?

[gleaming gazelle]

8+6=14
3x14=42

I dunno

[lyric valley]

instead of doing that

u can multiply 8x3 and 6x3

and sum those results

ull also get 42

[elfin cosmos]

but that works, so for the problem:

√10 * (8 + 6√10)
= √10 * 8 + √10 * 6√10

right?

you multiply to each term

[gleaming gazelle]

You removed the () but why?

[lyric valley]

because he multiplied both terms by square root of 10

thats how parenthesis work

[elfin cosmos]

yes, you multiply both terms (8 and 6√10)

it is the same

[gleaming gazelle]

huh

[elfin cosmos]

how do you add 8 + 6√10?

you cant

so you multiply each term

[gleaming gazelle]

Cant it just be 14 squareroot 10

[lyric valley]

no

if u sum 8 bananas with 6 apples do u have 14 bananas?

[gleaming gazelle]

no

[lyric valley]

same logic

[gleaming gazelle]

But its not 8a+4b its the same thing by your logic 2+2 isnt 4

[lyric valley]

yes it is

[elfin cosmos]

8 does not have √10, so you cannot combine

[lyric valley]

and 8 isnt a square root

its just a numeral

[elfin cosmos]

if it was 8√10 + 6√10 you would be correct

but it is just 8, not 8√10

[gleaming gazelle]

ohh

But still im confused by the question

I dont get it

[elfin cosmos]

so, because you cannot add, you have to multiply each term individually

[gleaming gazelle]

How do i solve that mid test?

[elfin cosmos]

like

a * (b + c)
a*b + a*c

[gleaming gazelle]

So squareroot 80+ 6 squareroot 10?

[elfin cosmos]

then you have

8 * √10 + 6√10 * √10

what is √10 * √10?

[gleaming gazelle]

Squareroot 100

[elfin cosmos]

which is?

[gleaming gazelle]

10

[elfin cosmos]

yes, so it is

8√10 + 6 * 10

= 8√10 + 60

that is final answer

[gleaming gazelle]

Im gonna kms

Wth

How do you instantly go from that to the answer

[elfin cosmos]

wdym?

you understand

a * (b + c)
= a*b + a*c

right?

[gleaming gazelle]

Alright?

[elfin cosmos]

so for the problem

√10 * (8 + 6√10)
= √10 * 8 + 6√10 * √10
= 8√10 + 6√100
= 8√10 + 6 * 10
= 8√10 + 60

8√10 is simplest form

you multiply, then simplify

[gleaming gazelle]

I kinda get it but i wouldnt be able to do that normally.

[elfin cosmos]

you have the two numbers in the parentheses

you multiply the number outside (√10) to both, and add

[gleaming gazelle]

So you remove () and add a x square root 10?

[lyric valley]

just do this

[elfin cosmos]

yes

[gleaming gazelle]

Is there a diff like question you could give me so i could see if i understand

[lyric valley]

√5(4 + √15)

[gleaming gazelle]

Why is square root 10 times 8 8 square root 10

[lyric valley]

do that

[elfin cosmos]

you understand why √10 * √10 = 10, right?

that is all you need to know

[gleaming gazelle]

yes

because its squareroot 100 which is 10

[fossil pendant]

-# roots are very ugly values, they dont simply simplify with anything! if you write 8 * sqrt(10) on the calculator, it wont give you a clean value so it just says 8sqrt(10).

[elfin cosmos]

yes, 8√10 is simplest form, since √10 is final

√100 is not final, since √100 = 10