help-17

[tough nymph]

should be 1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1

[full ferry]

why sum?

since they're independant events shouldnt you multiply them

[tough nymph]

did you learn indicator variables?

[full ferry]

nope

[tough nymph]

hmm ok

on the first roll

what is the expected value to get a face

u havent seen before

[full ferry]

1/1

.

[tough nymph]

<@&268886789983436800>

after u rolled the first dice

what is the expected value

to roll a new face

[full ferry]

5/6...?

[tough nymph]

no

ok lets suppose you rolled a 1

now what is the distribution

for the number of times you need to roll a face that is not 1

[full ferry]

there's 5 sides left among the 6 sides

sorry never heard these terms in english before

[tough nymph]

which word

[full ferry]

"distribution"

[tough nymph]

ok never mind it then

so 5/6

is the probability of rolling a new face right

[full ferry]

yeah

[tough nymph]

do you know how to find the expected value

using this?

[full ferry]

i do not😭

[tough nymph]

ok

use definition

E(X) = \sum_{i=1}^\inf P(X=i)*i

first few times are like

[full ferry]

ohhh

[tough nymph]

what is P(X=1)

?

[full ferry]

1

[tough nymph]

no

if it took me 1 roll to get a new face

what is the probability of that

[full ferry]

er

i dunno

[tough nymph]

5/6

do you get why

[full ferry]

wouldnt that be for x = 2?

or no

[tough nymph]

ok i shouldve been clearer

let the count start from 1 again

let X = the number of rolls to get the second new face

[full ferry]

oh

[tough nymph]

let Y = the number of rolls to get the first new face

so to get 2 new faces it is E(X+Y)

as you said E(Y) = 1

we now want to find E(X)

[full ferry]

E(X) = 5/6?

[tough nymph]

no

P(X=1) = 5/6

right?

[full ferry]

yes

[tough nymph]

ok whats

P(X=2)

[full ferry]

1 + 5/6?

or 4/6

[tough nymph]

neither

what is the situation

in order for X=2?

[full ferry]

2 rolls?

[tough nymph]

ye its 2 rolls

but why would you stop at 2 rolls

it means that you rolled a new face on your 2nd roll right

but you rolled the old face on your first roll

[full ferry]

yeah

[tough nymph]

so what is the probability

of that sequence

[full ferry]

could you tell me 😭

i have no idea

[tough nymph]

hmm let's use an example so its easier to understand

so suppose i roll a {1}

now I want to find E(X)

so if X=1 means, i roll {2,3,4,5,6} on first roll

X=2 means i roll {1} then {2,3,4,5,6} on second roll

P(X=2) = P(first roll is 1) * P(second roll is not 1)

= 1/6 * 5/6

make sense?

[full ferry]

ohh

then what's P(3)?

[tough nymph]

can you do P(X=3)?

[full ferry]

o

1/6 * 5/6 * 4/6..?

[tough nymph]

no

P(X=3) = P(first roll is 1) * P(second roll is 1) * P(third roll is not 1)

which is?

[full ferry]

1/61/6 5/6?

[tough nymph]

yep

so whats P(X=4)

[full ferry]

5/(6^4)

?

[tough nymph]

yep

so E(X) = 1/6 * 1 + 5/6 * 1/6 *2 + 5/6 *1/6^2 *3 + ...

do you know to find this sum?

[full ferry]

no

[tough nymph]

uh

it's this: https://www.cs.cornell.edu/courses/cs280/2008sp/280wk11_x4.pdf

[full ferry]

well i know how to find normal sums

[tough nymph]

its an inf sum

[full ferry]

oh yeah i know

[tough nymph]

E(X) = 6/5

ill leave you to derive the closed form solution of the inf sum

now say you want 3 new faces

[full ferry]

E(x) is sum of P(x)'s?

[tough nymph]

its the expected value

E(x) = 1 * P(X=1) + 2 * P(X=2) + ...

[full ferry]

ah ok

[tough nymph]

now we move on to 3 new faces

ultimately u want to get to 6 new faces

but its a slow build up and you see a pattern

3 faces is E(W+X+Y)

where W is 3rd new face

you found E(X) + E(Y)

so you need E(W)

say you rolled {1,2} already

the "new" faces would be {3,4,5,6}

and "old" faces would be {1,2}

so probability of rolling new face is 4/6

[full ferry]

yep

[tough nymph]

and you do the same thing as with E(X)

but this time

its 4/6 instead of 5/6

after doing some math

[full ferry]

6/4?

[tough nymph]

you get E(W) = 6/4

yea

so continuing this pattern

[full ferry]

ohh

[tough nymph]

E(6 new faces ) = 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1

[full ferry]

im probably cooked for this competition lmao

[tough nymph]

lol

[full ferry]

anyways thanks

[tough nymph]

np

[full ferry]

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[full ferry]

wait

.reopen

[vocal sleet]

✅

[full ferry]

once you get E(6 new faces), wouldnt that be a fraction?

[tough nymph]

ye

there is no problem with that

[full ferry]

the question asked how many throws it takes

so round up?

[tough nymph]

i dont see why you would round

it says on average

[full ferry]

well you could only throw something an integer amount of times

[tough nymph]

ok true

then ye round up

its ambiguous

i dont usually round with expected vlaue

[full ferry]

ok thanks :3

[full ferry]

oh right

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[snow umbra]

I'm unsure if this is correct because of the square root. Please help me understand. Operation of Functions.

[vale frigate]

u are correct

[snow umbra]

okay, Thank you!

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[fickle frost]

Can someone help me with explaining why its 270?

[vocal sleet]

@fickle frost Has your question been resolved?

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[bleak prawn]

The 'inner product' of 2 vectors means the dot product, right?

[bleak prawn]

Or is it something else?

This is the solution to a question I just wanna make sure

[magic wasp]

In Euclidean vector spaces, the inner product is the dot product, yes

[bleak prawn]

Awesome!

[bleak prawn]

And theyre saying because the dot product is = to 0 that that proves theyre perpendicular, riight?

Euclidean vector spaces
No idea what this is btw 😝

[magic wasp]

Orthogonal, yes

[bleak prawn]

Orthogonal
Idk that either LMAO

[magic wasp]

Most likely all the vector spaces you've seen are Euclidean

[bleak prawn]

Ah oke

Thank you!

❤️

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[magic wasp]

Orthogonality is a generalization of perpendicularity; the former term is typically used with vectors in the context of linear algebra whereas the latter is used for lines in the context of geometry

(but for example two functions can be orthogonal, and it would be odd to call them perpendicular)

[bleak prawn]

Okkkkk makes sense

Thank you haha you explained it well

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[terse fiber]

How can I find out per graph if x depends on y, or if y depends on x or both?

[terse fiber]

<@&286206848099549185>

[flat osprey]

Do you mean y being a function of x?

or something similar

[terse fiber]

yes exactly

my bad for wording it poorly

[flat osprey]

So if y is a function of x, then there exists only one y for every x. vice versa for x being a function of y. Another way of saying is is that x maps to only one y

[full ferry]

maybe which graphs increase relative to x, or relative to y?

[flat osprey]

In the first image, y is a function of x as for each x, there's one y, same with graph 2

[muted valley]

thats why I was so confused- I mean what do u mean, I was having self doubt 😤

[full ferry]

changes*

[flat osprey]

Notice how in the 1st graph, x is also a function of y

[terse fiber]

I looked in the answers and saw that graph A is that x is a function of y and y is a function of x

[flat osprey]

But in the 2nd graph, there exists more than one x for some values of y (e.g. y=2) so x isn't a function of y

[terse fiber]

forgive me lmao, the question is in my own language so I kinda stupidly translated it literally word for word

[flat osprey]

You can apply these logics to graph c

[full ferry]

you'll see that if you draw a line perpendicular to the y axis, it only intercepts with one point per line

[terse fiber]

im not sure if its a great example but for graph A. if an airplane travels 2m per 2 seconds, that means y is a function of x right

[full ferry]

(except for graph C which y is a function of x)

[terse fiber]

but the height wouldnt be a function of the time

in C x is a function of y

I got that from my book fyi

[muted valley]

guys he means he needs a relation to express the graphs, I dont think just explaining him will work - he needs the proper answer

[flat osprey]

How come?

Wait which direction is the plane travelling

[terse fiber]

So to sum up, the question of the image I sent is

"Find out by looking at the graphs if x is a function of y, if y is a function of x, or both"

[flat osprey]

It's impossible without knowing the exact values of the extrema , the only double thing that I'd say js a)

I also feel like that wasn't the purpose of the exercises

[muted valley]

ik

[flat osprey]

is*

[terse fiber]

so for graph A. how could i see, by only looking at the graph if x is a function of y and vice versa

from what I see is that every 2 steps in x, y goes up

[muted valley]

wait I will try to solve it after I am free

[flat osprey]

This is also called the vertical line test

[flat osprey]

Refer to the video below

To check if x is a function of y, use the horizontal line

[terse fiber]

I'll check it out, I tried finding a video from Khan Academy, but didnt seem like he had on on the issues I have

[flat osprey]

I see

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[broken carbon]

How does Wolfram alpha even get this? Brute force?

[drifting jackal]

Tbh I prefer this site https://www.integral-calculator.com/

[vast shale]

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

this is one way

[broken carbon]

@drifting jackal oh wow, got it thanks

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[tulip temple]

I have a doubt when it comes to the solution of this question

[tulip temple]

the line (p+2q)x + (p-3q)y = p-q for different values of p and q passes through the fixed point?

this is the equation

when we simplify it:

here we get two equations

both of which should be equal two zero for the equation to hold true for all p and q

intrestingly enough

solving these two lines

gives us the point

why is it so?

[vast shale]

as you have mentioned, for the second equation to hold for all p, q, we need x+y-1=0=2x-3y+1. And by solving this system, you have found the point(s), at which it is true.

[tulip temple]

dang yeah

i have a new doubt

solve please:D

if i take too many channels they ban

points on the line x+y=4 that lie at a unit distance from the line 4x+3y-10 = 0

both the equations

ah nvm actually

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[lone quest]

This is from Hubbard's multivariable calculus book. He gave those definitions of a graph and a smooth manifold. in the third picture, where he talks about the xy=0, are the two axis combined not a manifold because we cannot represent them around 0 by a single graph that is a single function representing that graph around 0?

that is there is no neighborhood of 0 in which there exists a graph of that shape around 0?

My problem is around 0 I can have two functions one x = 0 other y = 0 and that would be okay, right? they make up the cross shape around 0, but if the definitions specify that there has to be only one function representing the shape in the neighborhood of some point then I can see why it is not a manifold

[white palm]

pretty sure it's one function expressing x in term of y or one function expressing y in term of x around 0.

one (exclusive or) the other not 2 functions to represent the whole loci

[lone quest]

Ok, I guess that makes sense because if it's flexible then we can represent any shape and that restrictions of the definition wouldn't have any meaning

[white palm]

yeah hubbard has a nice graphic illustrating this for a curve right after

(i'm reading this too at the moment)

[lone quest]

Oh then I must go right on hahaha been spending too much time looking at these definitions!

[lone quest]

Nice let's stay in contact if you want to! We may be of help to each other

Will send you a friend req. Thanks a lot for helping, do you have any thing to say before I close it?

[white palm]

nah i think you good to go

[lone quest]

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[dense sentinel]

Hello, I have almost graduated with a bachelors in applied math. I want to go back and build up my fundamentals again. I would like to start with algebra 1 and work my way up to calc3. Does anybody have any textbook or workbook suggestions? Thank you!

[vast shale]

Jean Dieudonné - Treatise on Analysis

[fallow remnant]

For Algebra 1 to pre-calculus, I suggest Stitz-Zeager, it's very long but it goes to very rigorous lengths and solidifies your fundamentals https://www.stitz-zeager.com/

And there's also Serge Lang's Basic Mathematics but people have mixed feelings about his books

[dense sentinel]

That looks great! Thank you 😊

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[quiet gazelle]

I got to (25/2)^-1/2

[quiet gazelle]

But I don't know how to proceed

[tranquil trellis]

you can just reduce that

[quiet gazelle]

How

I forgot how

[tranquil trellis]

$a^{-1/2} = \frac{1}{\sqrt{a}}$

[twin meteor]

Ari

[quiet gazelle]

Can you explain that

1 ÷ 5/√2?

[gentle thicket]

yes
and $\frac{1}{\frac{1}{a}}=a$

[twin meteor]

77²

[gentle thicket]

'kay?

[quiet gazelle]

That's actually the answer or I made something wrong?

[quiet gazelle]

Aight

[gentle thicket]

actually idk
you just said it that you calculated it so we assumed that it was correct

[quiet gazelle]

Well, I think that's it

Gonna try it again

Tyy

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[noble hornet]

I think that I haven't understood the question well, can anyone tell me what it asks? It says that there is always a set C where we can choose any 2 circles in it that intersect each other or not, but that is always true, isn't it? Any two circles in the plane intersects or don't intersect.

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@noble hornet Has your question been resolved?

[noble hornet]

<@&286206848099549185>

[red bear]

i think it is saying that there is a set of 27 circles with one of two properties, either every pair of circles is intersecting, or every pair of circles is non-intersecting

this seems like a pigeon-hole principle problem to me

[noble hornet]

Oh thank you!!

[noble hornet]

Yeah it seems like one of those

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[wary citrus]

Hello, this is my first post (I will try my best to adhere to the format!)

I have a question about square wheels. There are a bunch of resources for square wheels that roll 'smoothly' on bumpy surfaces. It seems like all of them use catenary curves for this effect. One side effect of using these curves is that if the square rotates linearly, the surface moves non-linearly, and vice versa. I was hoping to derive a formula for a curve/2D surface where the square rotates linearly and the surface/curve moves linearly.

I have been able to create a parametric function that I think satisfies this, but have been unable to convert it back into a simple cartesian function (this is important for a programming project haha). I will include the parametric function as an image here and link to the Desmos page I used to work it out, if it would be helpful to anyone.

Also, if the thing I am trying to do has been done better by someone hundreds of years ago (as is usually the case haha), please let me know!

TLDR: I need help converting the below parametric function into a cartesian one.

Desmos page: https://www.desmos.com/calculator/0mnbes4sno

Cheers!

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@wary citrus Has your question been resolved?

[wary citrus]

<@&286206848099549185>

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@wary citrus Has your question been resolved?

[vocal sleet]

@wary citrus Has your question been resolved?

[wary citrus]

There has to be a way to check if anyone else has replied to the question and auto not-resolve it with the bot, haha.

[sweet flower]

I think it has to do with timing out old questions. Because some are just too hard / off topic / not fun to solve and get timed out eventually

[wary citrus]

yeah that might be the case with this one...

ill give it till tomorrow then ill close it if no one seems to take an interest in it

[woeful igloo]

The best way to do this is find t = f(x) from the parametrization and substitute it into the second one y = g(t) as y = g(f(x))

And given the nature of the x = h(t), we can be sure a proper inverse function f does exist

[wary citrus]

yeah Ive been unable to figure that out unfortunately! I was hoping someone here might have some tips

[viral plover]

when you say rotates linearly do you mean with constant angular momentum

[wary citrus]

the second paramatarization simplifies a bit to -sin(t)+sin(2t)-cos(t)+1 if that is anything

[viral plover]

theres a great video series about all these types of curves, and if i recall correctly the only such solution where things move 'nicely' like you described is a circle

[wary citrus]

yes that is probably more precise

[wary citrus]

thats what I thought too, but I think my function does it? Unless I am sorely mistaken

if you look in here: https://www.desmos.com/calculator/gq3icuyafh the purple line seems to move linearly left/right and the green square seems to rotate with constant angular momentum

[viral plover]

https://www.youtube.com/channel/UCu7Zwf4X_OQ-TEnou0zdyRA
this is the channel btw

ill have quick scan see if i can find what im talking about/what i think im talking abt haha

[wary citrus]

oh hey, thats the channel that I used to get the catenary version of the function working :D

[viral plover]

i may well be misremembering

[wary citrus]

its a great resource

[viral plover]

i think whats happening here is that you dont see the slow down because it happens when the surface jumps back

again may be wrong

[wary citrus]

sorry for the awful organizing, but in SIMPLER, the rotation of the green square is set by r_ad and the translation of the parametric function is just an offset, so there shouldnt be a slow down. It should be linear

[viral plover]

is it possible that the simpler version you have is not correct?

i.e. that there is some sliding happening (rather than rolling)

[wary citrus]

https://www.desmos.com/calculator/szxlqmii2j

I tried to get rid of everything that wasnt needed

you can see here that only r_ad and o_ffset1/2 are changing

and they should be linear?

[viral plover]

im not saying thats wrong, my potential qualm is that it only LOOKS correct but it not actually true to what you are trying to model

[wary citrus]

yeah for sure. though I dont know how

Im definitely not smarter than mathematicians, and I do remember the video saying something about circles being the only shapes that can rotate/move like what I think my functions are doing...

so 🤷

[viral plover]

ah ok i found what i was after !

[wary citrus]

!

[viral plover]

https://youtu.be/ZWWgGk9JU0E
in this video at ~9m in theres the equations relating angular momentum and radius of the wheels

though i dont know how perfectly that translates onto the road-wheel situation

it is true that if both wheel have constant angular momentum then they must be circles though

[wary citrus]

oh you watched the next video in the series. I only peeked at the first one. Ill give it a look through

[wary citrus]

hmm. I wonder what the heck is happening in my functions to make me think that it has those properties then.

[viral plover]

well i dont think this translates over perfectly

[wary citrus]

maybe its microscopically intersecting somewhere

[viral plover]

yeah its possible

[wary citrus]

ill post this here in case anyone else wants to take a stab at it: https://www.desmos.com/calculator/bcmsnofw8m I would love to be able to describe C without relying on a parametric function

Q (black line) is the tangent point between C (purple line) and the square

is that what you mean?

[viral plover]

i realised i could just consult the equations for the first video haha

if we have both constant velocity and constant angular momentum then from this we must have that r is constant

and that the wheel is a circle

[wary citrus]

I mean that sounds right but Im staring at what seems to be a counter example rn

[viral plover]

seems to be being the key part of that phrase

near misses are rife in maths!

[wary citrus]

yeah for sure. I dont have the understanding to prove it, but this whole tangent is also kinda besides the point for my question. even if it is off by -10^100 somewhere, I would still love to describe it in terms of x and y only to shove it into my programming project

[viral plover]

oh right yeah

its a catenary so its just gonna be y = cosh(x)

(scaled, translated and restricted as you please)

[wary citrus]

not the video formulas, the ones included in my desmos links/images at the beginning of the question

[wary citrus]

these guys

[viral plover]

oh i see

uh if i had to guess id say theres not gonna be a nice form in terms of just x and y

but why do you need it in such for a form a programming project?

surely the parametrised form is fine

[wary citrus]

I want to shove it in a shader for drawing

There are probably better ways to do it but I like the idea of having most of the art in a project I'm working on be infinitely scale-able

Ive got this little belt looking thing that I like the look of. The squares not rotating uniformly was bothering me though https://www.desmos.com/calculator/jjwdspqwvn

[viral plover]

Ok but why not just use cosh, it likely won't be perceivably wrong

(just forcing everything to move nicely like you want)

Then you get best of both worlds

[wary citrus]

I understand that I could do that, but I would like to do this other thing

with cosh, you can see that the squares dont rotate at the same rate all the time. Its distracting imo

[viral plover]

just make the squares rotate at a constant speed is what I'm saying ?

Or do you mean that when you do it's clear that it clips through

[wary citrus]

then the non uniformity would be moved to the belt

so with cosh, either the squares slow down/ speed up slightly or the belt does. This other formula doesnt have that problem, but I only have it in parametric form

[viral plover]

Right but just force the belt and the squares to move constantly like you did in desmos

And then just use the cosh formula for simplicity

Surely it doesn't matter that it's not perfectly accurate mathematically so long as it looks fine

[wary citrus]

I know I can fudge it. Im on a math discord to see if I dont have to. Its just a hobby project, so I can afford to be particular

[viral plover]

Well I bid you good luck

[wary citrus]

ha, thanks

[vocal sleet]

@wary citrus Has your question been resolved?

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@wary citrus Has your question been resolved?

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@wary citrus Has your question been resolved?

[wary citrus]

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[wary citrus]

.reopen

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✅

[wary citrus]

https://math.stackexchange.com/questions/4954974/converting-the-parametric-equations-x-y-cos2t-sint-cost-2t-sin

I was helped by a math wizard, in case anyone was curious :)

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[tough quarry]

can someone check my work

[tough quarry]

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[hard monolith]

Let $C$ a circle $C_1$ and $C_2$ two circles internally tangents to $C$ at $P$ and $Q$ respectively. $C_1$ and $C_2$ also are externally tangents to each other in $T$. Show that the perpendicular to $(PT)$ passing by $P$ cross $(QT)$ on $C$

[twin meteor]

UzuNGD

[hard monolith]

I tried to show that APQK where K is the 2nd intersection of PT and A the intersection of the perpendicular to (PT) and (QT) with C is cyclic quadrilateral but it doesn't seem to work
Also i noticed that A,K and O where O is the center of C are colinear. If i can prove it it would mean that C is the circumcircle of APK which would be enough to show that A is on C

Also I think inversion could be of some help but I don't know this thing

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[quasi kayak]

Am I supposed to be comparing the limit to 1 and show that it is bigger than 1?

[near tartan]

No

[woeful igloo]

you need to find the exact value surely

[near tartan]

In such limits you just need to divide the numerator and denominator by 3^n+1

[quasi kayak]

Well if I show that the quotient is bigger than 1 doesn't it become infinite?

[quasi kayak]

That is what I have meant my bad

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[empty laurel]

teacher gave us an assignment to answer f(x) + h(x), but she haven't taught us how to add the fraction one and the not fraction one (idk how to call them sorry 😭) can someone guide me how to solve??

[hard atlas]

how would you add $3 + \frac47$

[twin meteor]

Denascite

[empty laurel]

uhh thats 25/7 right??

where did the 4 come from tho?

[hard atlas]

yes. and how did you do that

random numbers

[empty laurel]

ah

wait so like

hang on

x+3 * 2-x

?

then i add it with x+7?

[hard atlas]

please dont forget to use brackets

but yes

$x+3 + \frac{x+7}{2-x} = \frac{(x+3)(2-x)}{2-x} + \frac{x+7}{2-x}$

[twin meteor]

Denascite

[empty laurel]

ohhh i kinda get it now

thank you!!

[hard atlas]

yw

[empty laurel]

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[hard atlas]

very well done

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[livid shuttle]

i got two simulatenous equations, but i got stuck idk how i can get each variable's values?

2 = a+b+c
1=2a+b

[outer warren]

how did you get those equatiuons

[livid shuttle]

i subbed x=2 into the derivative f'(x) = 2ax+b which got 1=2a+b

[outer warren]

don't see how you're arriving at
1 = 2a+b
from that

can you show full work

[velvet zephyr]

2ax + b = -9
8a + b = -9

is my relation correct?

rauqnwv?

[outer warren]

where's

at x=1, dy/dx = 1
coming from

are you doing a different question than the one you posted?

[livid shuttle]

wait no its

i posted the question before it

[outer warren]

ok, so you want when
x = 0, dy/dx = 1
you had that written at the top, but then changed it later

and do you know the slope at stationary points?

[livid shuttle]

derivative =0 at stationary points right

[outer warren]

yes

[livid shuttle]

so does that make b = 1?

[outer warren]

yes

[livid shuttle]

so what would the 2 simulatenous equations be?

is it what i wrote as 2 = a+b+c and 1=-a+c ?

but with b=1

[outer warren]

where's

1=-a+c ?
coming from

[livid shuttle]

i got it from the second sentence 'when x=0, the slope of the curve is 45' so i thought that 45* is gradient of 1

so 1 = derivative where x=0

wait did i substitute it wrong

[outer warren]

you had

f'(x) = 2ax+b
applying when x=0, derivative is 1
gives you b=1

[livid shuttle]

wait no the 1=-a+c is when i minused the two equations i had cuz i didnt know what else to try

[outer warren]

knowing that (1,2) is on the curve gives

2 = a + b + c

you didn't manipulate the equation properly after subbing in b=1

and you've yet to apply

derivative =0 at stationary points

[livid shuttle]

is a = -1/2, b=1, c=3/2

[outer warren]

show your work

[outer warren]

looks ok

[livid shuttle]

why do i need to sub x=0 into the derivative when it the derivative is equal to 1

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[empty laurel]

hello i got confused again and got 3 diff answers each time i try to figure it out.

[empty laurel]

if i try to just cancel it out then the answer would be the question itself, so what i did is i multiplied first the x+3 (2-x)

then i added it with the x+7 so the answer would be x^2+10 / 2-x

but then i remember that theres a rule in standard form that says it should be ax^2 + bx + c

so i tried again and this is what i got

idk which one is right but i think it is the second one

im not sure

sorry if its too confusing to understand, im not very good at explaining in math terms

[hard atlas]

idk how you got x^2+10

[hard atlas]

this looks good

[empty laurel]

did i do it properly? or did i miss any step/got something wrong??

[hard atlas]

everything very nice

[empty laurel]

but i thought in standard form it has to be in a specific order (ax^2+bx+c)?? why is it ax^2+c

or is it because its a fraction?

[hard atlas]

well, b=0

[empty laurel]

ahh so it just becomes invisible if its 0?

[hard atlas]

yes

[empty laurel]

ok i see now

[hard atlas]

you could write it down as 0x

but that would just lead to unnecessary stuff

and could lead to errors later

[empty laurel]

oh okayy

thank you again for your help!!

[hard atlas]

yw

[empty laurel]

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[feral venture]

Is $[(x, y) \in f \wedge (x, z) \in f \Rightarrow y = z] \text{ equivalent to } [\forall x$ $\exists \text{ at most one } y$ $(x, y) \in f]$

[twin meteor]

Forsaken

[feral venture]

Or it has to be exactly one y? Because if there is no y such that (x, y) in R then R is the empty set so it is impossible for there to be no y, so it is automatically exactly one y?

Or it makes sense for there to be no y, it doesn't matter therefore that R is empty

[feral venture]

I think that is impossible, therefore it is automatically exactly one y

<@&286206848099549185>

[white palm]

is f a set theoretic function

or is it the generic cartesian product of 2 set

[feral venture]

How do you define a set theoretic function

[feral venture]

?

[white palm]

this

[feral venture]

I define it as a partial function for which the functional relation holds such that dom(f) = X (dom(f) is not a subset of X but is equal to it)

I don't know if these two are the same

[white palm]

i sure don’t know what a partial function is

[feral venture]

A partial function is a triplet (X, G, Y) where X is the domain, G is its graph and Y is the codomain where G is a functional relation

I hope this isn't convoluted

[white palm]

it is lol

[feral venture]

Is this from a book?

[white palm]

well yes

[feral venture]

Which book?

That is an aside by the way

[white palm]

Hubbard vector calculus it’s not a big boi set theory book it’s just the naive and usual definition i go to

[feral venture]

Okay

I think these definitions are equal

There is no difference

The one being a subset of the cartesian product and the other being the set theoretic function

I am just confused with some foundational aspects of relations and functions, I have them mixed up in my head

[white palm]

well the construction and definition in your book are clearly more formal, but i’m afraid i can’t help you much before i read part of your book lol

[feral venture]

It is okay, I have to go though

[white palm]

yeah np

[feral venture]

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[compact geode]

can someone explain how get we get to the second step

[compact geode]

like 2x/x^2+9

where did the 2 come from

and the 5/18

[vast shale]

5/9 = (5/18) * 2

[feral venture]

They multiplied and divided by 2

The denominator (the 1/2) went out of the integral so it became 18 instead of 9 and the numerator ( the 2) stayed in the integral making it 2x instead of x

[compact geode]

im so slow and confused

where did the 1/2 get taken out from w

[feral venture]

What w

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[worn temple]

can i post my maths thing

[worn temple]

here

can someone explain this // ANSWER : 5 mins

its really basic in a linear relationship when the staff doubles dont u like get half the waiting time?

logically

[fading island]

t = (s^2) ^-1 => t = 1/(s^2)

t = 1/((2s)^2) = 1/4s^2

time decreases by 4

for the doubling of staff

so if 4 staff takes 20 min

8 staff takes 5

[fading island]

no cuz theres a square

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[worn temple]

oh ok

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[fierce flame]

can somebody please solve this step by step

[fierce flame]

<@&286206848099549185>

[full forge]

I think a good place to start is determining how many square meters of surface area we're dealing with.

[bronze trellis]

Take base cubes top surface area to represent the sculpture all top surface areas as the subsequent placed top cubes exposes equal area as they block. So, top area is 4 by 4 by 0.5m

[fierce flame]

Step 1: Analyze the Sculpture Structure
The sculpture is made by gluing cubic boxes, and the image shows a pyramid-like structure. Each cube has a side length of 0.5 meters (half a meter).

Step 2: Calculate the Surface Area of the Sculpture
We need to calculate the total surface area that will be painted. The sculpture consists of layers of cubes:
• Bottom layer: 4×4 cubes = 16 cubes
• Second layer: 3×3 cubes = 9 cubes
• Third layer: 2×2 cubes = 4 cube
• Top layer: 1×1 cube = 1 cube

Step 3: Calculate the Surface Area of Each Cube
Each cube has 6 faces, and the area of each face is:
Area of one face=0.5×0.5=0.25 square meters Surface area of one cube=6×0.25=1.5 square meters

Then the faces that are to be painted according to me are 5 from top layer, 8 from 2nd, 12 from 3rd and 16 from bottom

[drifting mural]

Did you use chatgpt now? xD

[fierce flame]

yes but its not telling the correct solution

[outer warren]

figures

don't use got for math

just delete that

[full forge]

I agree 5 need to be painted from the top, I get 11 surfaces for the 2nd layer and 17 for the 3rd. The bottom layer you can avoid entirely, I think.

[drifting mural]

it isnt good for maths in most cases more for like programming. Tho this task is fairely easy even if you would just count them by hand

[bronze trellis]

Then, find the sides surface areas. Possible by counting, remember some are on the other side of the sculpture. I got 40. Multiply by 0.5. Add answers then divide by 4. You should get E

[fierce flame]

yeahh, its too inacurate

[outer warren]

consider a top down view
and the side view

[full forge]

I counted by imagining the boxes moving on top of eachother to align perfectly

The area shouldn't change

So instead of a pyramid it's more like a corner staircase from minecraft

[drifting mural]

This.

[full forge]

Yes this. You don't need to imagine it as a staircase if you do this.

[drifting mural]

@fierce flame Come on lets try together. Lets start with the top how many surfaces can you see from the side and from the top?

[fierce flame]

I think 5 surfaces from top layer then on 2nd layer there are 8 complete faces plus the exposed faces 4 - 1 (the top cubes area) so 11. then third layer is 12 + 5.

[outer warren]

not what we're hinting at

[full forge]

damn that's what I got too lol

[fierce flame]

i think bottom can be skipped as its written in question

still the options can incorrect, its coming in decimal

[drifting mural]

i think with the base they mean the area touched by the floor?

[outer warren]

imagine you're in a plane from above or a bird

what would you see?

[fierce flame]

so then bottom layer can be added as 16+ 7

[full forge]

Oh do they want the bottom layer painted as well? But like the top part of the bottom layer?

[drifting mural]

You didnt get the second layer right first of all

[drifting mural]

Try and imagine this

If you get the idea he is presenting you will even see the formula for all bases

But you got the side ones right its 4, 8, 12, 16 cause its 4*n, n=1,2,3,4

tricky part is the top

[outer warren]

(if you add/subtract properly you'll get the same result, it's just very tedious)

[full forge]

The sides he suggested are right I think

[drifting mural]

Im not a native speaker but i think they want all the surfaces "you can see"

[full forge]

Oh wow

I totally misunderstood the question then, i thought they wanted everything except for the part touching the floor

But either way I believe I had the wrong amount of surface area

idk if this helps or not:

[fierce flame]

shouldn't it be 1, 3, 5, 7

[full forge]

That sounds right from the top

[drifting mural]

jup

[full forge]

Using their method you don't have to count

Which is nifty

[fierce flame]

then the total surfaces become 56x0.25

[drifting mural]

the formula is n^2-k where n is amount which you count the surfaces and k the amount of surfaces on top of those

but this isnt finished yet ; )

[full forge]

If you do end up counting this made it easier for me at least

[drifting mural]

jup this

[full forge]

If you end up with a decimal it might not be the end of the world because of how the question is formulated

[fierce flame]

also then 14 becomes the total surface area divided by 4 = 3.5 paint cans.

[drifting mural]

so how many cans would she need?

[fierce flame]

which is not in the options but options can be incorrect i think

[fierce flame]

3.5

[drifting mural]

yeah but imagine you cant buy half a can from the market

so you buy a full can instead for the half one

that means?

[full forge]

If you can answer this you're home

3 cans wouldn't be enough

[fierce flame]

4 cans

[drifting mural]

bingo

[fierce flame]

got itttt!!!!!

[full forge]

Sorry for not helping much

[drifting mural]

*if i understood the assignment

[fierce flame]

no you did help

[drifting mural]

dont worry i think you were

[full forge]

I misunderstood the question and even then I still made mistakes by the way I interpreted it

Cool problem though

Is there a way to avoid counting the side surfaces?

[drifting mural]

I mean i could be the one who got it wrong so

[drifting mural]

here is the formula for the sides

[full forge]

So you still need to count the sides

[drifting mural]

uhm only how many boxes are on one side since its n^2

but its a pyramid

[full forge]

Sorry can you remind me what n was again

[drifting mural]

its this (for pyramids)

[drifting mural]

n is the amount of surfaces on one side

[full forge]

yeah so you'd have to count them? Or do you solve for n somehow?

I might have missed the solving for n part

[drifting mural]

I mean depends you have to know how many floors there are

you know there are 4 floors for this pyramid

[full forge]

ah but here we know the amount of floors

okay yeah that's what I missed

then I am with you

[drifting mural]

hence why the sum goes from 1 to 4

[full forge]

ah ok

[drifting mural]

this sum

oh wait no my bad

its not k^2 its 4*k

this is the sum

see i make mistakes too

[full forge]

caught in 4k

[drifting mural]

this is the sum for how many blocks there are in total

[full forge]

literally

[drifting mural]

this is for the second part (the top view)

[drifting mural]

dont judge me 😭

[full forge]

I just imagined the top view as a square

[drifting mural]

yh there you can do n^2-k

[full forge]

for the sides I had to count them, for a very tall pyramid that is probably not ideal

[drifting mural]

no you dont have to count

[full forge]

no no I get it but

I meant I didnt know that

so I counted them

[drifting mural]

ah

now i get it lol

[full forge]

but just multiplied the number of visible surfaces from the top

i suppose

i guess i did count there

too

[drifting mural]

yes also @fierce flame you got it right?

[full forge]

but not all of them just the sides

[drifting mural]

dont wanna block this channel lol

if yes close it : )

[fierce flame]

yes yess can I ask you guys another question?

[drifting mural]

https://tenor.com/view/pon-gif-15097379091171620462

maybe

[full forge]

[fierce flame]

11. The circles of the figure should be numbered from 0 to 10, each with a different number. The five sums of the three numbers written on each diameter must be odd numbers. If one of these sums is the smallest possible, what will be the largest possible value of one of the remaining sums?

[full forge]

Okay so by 5 sums I think they mean 5 of those circles because after that it is just the same sum again

since they are connected I mean

then one of the sums will be some combination of 1 2 3

[drifting mural]

I dont even get the problem

[full forge]

let me see if i can explain it with a picture

assuming i understood it at all

i hope that makes sense

after sum 1 they just repeat

so there will be 5 sums

1 3 2 might not necessarily be the right order but since the goal was to get the final sum as large as possible i thought it was appropriate to place the 3 there

2 might end up getting a larger sum though

i doubt 1 would get the largest

[drifting mural]

nah still dont get the problem

ima try translate it

[full forge]

are you with me when i say 1 + 3 + 2 is sum 1

so 6 is sum 1

in some order

our goal is that one of the other sums is as large as possible, i think

totally possible i misunderstood though

wait

[drifting mural]

i think i got the problem now

[full forge]

sorry 0-10

ignore that

[drifting mural]

its only the red line you connected

[full forge]

it should be 0 1 2

my bad

[drifting mural]

but all the variations of that red line

[full forge]

im sorry 😦

yeah the red line for another sum should be as large as it can be

oh

wait now i get more

[drifting mural]

@fierce flame you here?

[fierce flame]

yes but this question is too tricky for me

[full forge]

if the number in the middle was 1 instead you could get 1+10+8 but that is smaller i think

they wanted a sum to be as large as possibe

i might have misunderstood still

[drifting mural]

uhm 🤓 why did you solve it for her?

[full forge]

I DONT KNOW IF IT IS SOLVED

[drifting mural]

following the logic thats the answer

[full forge]

im sorry i got carried away

this is not how you help people with math

[drifting mural]

cause first you looking for the "smallest possible solution" which is 0+1+2 (in any order)

after that you look at the biggest numbers left which is 9+8 now which of those 3 numbers (1,2,3) makes 9+8+x= max?

obviously the 2 thats why you put it in the middle

[full forge]

im sorry i have failed you

can you forgive me

[drifting mural]

theres something cooler here tho

the difference between max and lowest number is always the same for any combination

[full forge]

that's neat

is there something that is still confusing you emm

[fierce flame]

lol but how did you arrive that this solution

[full forge]

yes good question

and honestly it was a little trial and error but also not

are you with me that the question asks us to consider the lowest sum possible

[drifting mural]

https://tenor.com/view/crying-omg-omg-omg-omg-crying-face-oh-my-god-wow-gif-8679087715539115273

[full forge]

and are you following what the sums are

[full forge]

yes this

tron put it in easier to understand terms and frankly more concise

basically by the way the question is formulated

you need to consider a sum of 0 + 1 + 2 in some order, the order is important because only one number can be in the middle of the circle

[fierce flame]

okay so you started with the smallest

[full forge]

yes because the question asked us to

kind of like this image

but i missed that it was 0-10

i thought it was 1-10

that's why i put 1 3 2

but after realizing my error i put 0 2 1

it could have been 1 2 0 as well, that doesn't matter

then the question wants us to get as big of a sum as possible

[fierce flame]

got ittt

[full forge]

so i put the 2 in the middle for that reason

really sorry for giving away the answer

i thought putting the 2 might get the largest number because 2>1>0

[fierce flame]

yes plus then you get all odd sums

[full forge]

oh yeah so about he odd sums, yeah i used some trial and error there

i cant explain it, but there is logic there too i think

i just tried to consider what the biggest odd number i could get

[drifting mural]

omg nah

[full forge]

where one number has to be 2

[drifting mural]

i forgor about those odds

[full forge]

wait did i do it wrong

[drifting mural]

yh just a small thing

[full forge]

i cant see it can yo upoint it out

[drifting mural]

The five sums of the three numbers written on each diameter must be odd numbers

[full forge]

i cant find the even sum

[drifting mural]

means very a+b+c=2x+1

[fierce flame]

all 5 are odd

[drifting mural]

nah you right

i just didnt check for odds

[full forge]

Tron is putting this into math terms though

I don't think in math

This is a problem for me

[drifting mural]

you already put the odds on side and the even on the other

[full forge]

Yeah but that was intuitive for me

[drifting mural]

did you do it by accident or you figured it out?

[full forge]

no I figured it out

[drifting mural]

smart

[full forge]

i like to think sometimes i can be smart

but i cannot be math smart

[drifting mural]

its a good start

anyways @fierce flame you got the idea?

[fierce flame]

yes i got it thank you so much guys

[full forge]

if you got all the help you needed just type .close

[drifting mural]

https://tenor.com/view/handshake-gif-20334449

[full forge]

and again sry for giving away the answer

[fierce flame]

alright thanks guys, it was fun and informative:)

[fierce flame]

no worries, you actually helped by giving away the solution first, i would not have been able to figure it out otherwise

.close

[vocal sleet]

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[vocal sleet]

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[obsidian basalt]

Qn.

[thin vale]

Do you know the theorem, dim(U+V) <= dim(U)+dim(V) ?

[obsidian basalt]

To start of, I did U + V, which results in (x, x, x, 0). Now, there were several approached that I could think of:

1. It should be R3 because we have only 3 entries (I think this is incorrect)

Yeah, I briefly read about that theorem.

[thin vale]

regardless of the amount of entries with an x, the vector is in R4

it has 4 entries

So if dim(U+V) <= dim(U) + dim(V)

What is dim(U) and dim(V) in this problem?

[obsidian basalt]

That's a column vector right, so obviously 1x4 (right?)

[thin vale]

Not the dimension of the vector, the dimension of the subspace

[obsidian basalt]

Do you mind if I also share the other approaches

[thin vale]

U and V are subspaces of R4

[thin vale]

yes I do actually

I'd like to just teach you how to do it

[obsidian basalt]

I assume they are not given to us

Am I supposed to assume it by default?

[thin vale]

Do you know how to find the dimension of a vector space?

[thin vale]

It is clear by how they are defined that they satisfy the properties of a vector space

but you could check them manually aswell

[obsidian basalt]

Are you asking about checking the dimensions or checking if they are a subspace? I understand they are subspace by verifying the scalary multi and vector addition properties.

[thin vale]

I was talking about checking if they were a subspace since I wasn't sure if you knew why that was true or not

but now that you know, do you know how to find the dimension of them?

[obsidian basalt]

Not really.

[thin vale]

Do you know how the dimension of a vector space is defined?

If not, I can tell you

[obsidian basalt]

Yes please!

We've just started with this course so there are a lot of gaps in my understanding

[thin vale]

the dimension of a vector space, is the number of linearly independent vectors needed to span the vector space

Does that sound familiar

[obsidian basalt]

Alright yes, I've definitely heard about it. Btw, I have a strong feeling that the professor wants us to make use of the concept of "Sum of subspaces" in this question. Does that linear independent vectors thing, relate to that?

[thin vale]

Yes

We're getting there.

you need to find the dim(U) and dim(V) to use that concept

[obsidian basalt]

Alright, awesome!

[thin vale]

Here's a hint for finding dim(U). Does the vector (1,1,0,0) span U?

[obsidian basalt]

Yeah, I think it does

Is the dim 2?

Maybe??

[thin vale]

Well, if (1,1,0,0) spans U

then you have 1 linearly independent vector

that spans the vector space

so what is its dimension, based on what I said above about the definition of dimension

[thin vale]

[obsidian basalt]

1

[thin vale]

Yes

Just because there are two entries filled up in (x,x,0,0) does not mean that it will have dimension 2

all of the vectors inside of U, are of the form (1,1,0,0) or (2,2,0,0) or (18,18,0,0)

[obsidian basalt]

Ah okay, they are the same variable (they have to be the same), so that's technically only 1 independent variable. Is my understanding close to being correct?

[thin vale]

You can think of them as all being scalar multiplies of the vector (1,1,0,0), forming a line in R4

[obsidian basalt]

Right!

[thin vale]

Yes

So similarly, can you now tell me dim(V)?

[obsidian basalt]

yeah that's 1 as well

[thin vale]

Okay so here is where the sum of subspaces concept comes in

If R4 = U+V

then what must the dimension of U+V be?

if it were true that they were equal

[obsidian basalt]

This was the point of confusion for me. The x (although named the same variable), belong to the different sets. So can they be called "x", yet have different values?
For ex: (1,1,2,0)

[thin vale]

Yes, (1,1,2,0) can be a vector in U+V

since (1,1,0,0) in U and (0,0,2,0) in V

so their sum is a valid vector in U+V

you could replace the definition of V to be

$$V=\left{ \begin{bmatrix} 0 \ 0 \ y \ 0 \end{bmatrix} \mid y\in \mathbb{R} \right}$$

[twin meteor]

Austin

[thin vale]

if this helps you not be confused, it means the same thing as how V was previously defined

I will be back in about 5 minutes, but to continue working on the problem,
Think about if R4=U+V, then what must be dim(U+V)?
Additionally, recall that dim(U+V)<=dim(U)+dim(V) and we know what dim(U) and dim(V) are.
This should be enough to solve it, but I can help if you're stuck in 5 minutes

[obsidian basalt]

I'll be back and respond

[vocal sleet]

@obsidian basalt Has your question been resolved?

[obsidian basalt]

@thin vale sorry, I had to go for lunch.

So yes, based on your extremely helpful comments above, the dimension should be 2. Is that correct?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obsidian basalt]

.reopen

[vocal sleet]

✅

[thin vale]

dim(U) + dim(V) = 2

But R4 had dimension 4

So..

[obsidian basalt]

U + V = R2?

[thin vale]

No

The problem asks you to determine if U+V = R4

[obsidian basalt]

Yes

[thin vale]

The vectors in U+V are still in R4, so this can’t be true.

[thin vale]

If U+V=R4
Then dim(U+V)=dim(R4)
Recall that dim(U+V) <= dim(U) + dim(V) = 2

Can something less than or equal to 2, be equal to 4?

[obsidian basalt]

Oh!

Nope

[thin vale]

So U+V ≠ R4

[obsidian basalt]

Ahhhh!

[thin vale]

Additionally U+V ≠ R2

U+V is a 2-dimensional sub space of R4

[obsidian basalt]

It makes a lot more sense now!

Alright. So the counter example can be taking U =(x, x, 0, 0) and V = (0, 0, y, 0)
And then I just pick up two random X and Y values, to indicate that it is a 2D subspace, not a 4D.

[thin vale]

No

There is no “counter-example” in this problem, I’m not sure why they phrased it asking for one. What we just did is a direct proof that U+V ≠ R4, which you could think of as a counter example of you’d like

I guess proving U+V ≠ R4 is a counter example to the statement U+V=R4 😂

[obsidian basalt]

Yeah, that sounds reasonable enough.

Okay, let me read more on this. I think I understand the approach now.

Thank you so much for your help!

[thin vale]

No problem

[obsidian basalt]

.closed

.close

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Use .reopen if this was a mistake.

[vocal sleet]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[stone birch]

This may be a very dumb question but my brain is a bit fried, does this work just cause if I'm pulling out an e^-t from the bottom term, because that bottom term is being squared it's actually e^-2t?

[civic drift]

Hmm

I don't think it's exactly how you imagine

$({e^{-t}(1+e^{2t})})^2$

[twin meteor]

ColdTe²

[stone birch]

okay is it like... ((e^-t)(1+e^2t))^2 = (e^-t)^2(1+e^2t)^2

[civic drift]

$e^{-2t}(1+e^{2t})^2$

[twin meteor]

ColdTe²

[civic drift]

Factored out e^{-t}

[compact zodiac]

yes

[civic drift]

Remember not to directly pull it out

Knowing how it was done is essential

Cause in a different case it might be incorrect

[stone birch]

alright, that makes sense, I just needed a bit of a refresher

thanks!

.close

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[vocal sleet]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[vast shale]

Yo

[vast shale]

How do I do this

[radiant bone]

Do you know the definition of a function?

Or when something isn’t a function

[vast shale]

Yeah

If you give me a min to search it up

[radiant bone]

lol ok

Something isn’t a function if it has more than one output per input

Basically, you can’t put it one number and get two out

So your current answer is incorrect because you have two outputs for one input (1->3&5)

[vast shale]

@vast shale

Ok

[vast shale]

yep I’m here

[vast shale]

Ok so the first one is (-7,6)

[radiant bone]

Yep

[vast shale]

That’s what I got

I guessed

Oh

[radiant bone]

Oh

[vast shale]

😭

Genuinely thought I was tweaking

Ok the next one is (6,1) right

Lol

No it’s not

[radiant bone]

Why would it be that

You can’t just guess cuz we’re here