help-49

[wind oxide]

okay they did similar (-pressure) shenanigans in the next question

like

are we assuming the d_1 liquid puts pressure on the air or smth?

and its in the reverse direction.?

[modern sapphire]

the trapped pressure would also vary by height

[wind oxide]

but we dont have enough info to calculate that

[modern sapphire]

its same as any other liquid, the stuff above exerts pressure below

[wind oxide]

oh so we assume its sufficient enough?

[modern sapphire]

not really, you set up equilibrium equations for it, compare it with the pressure at bottom

and then you can calculate it properly

[wind oxide]

err sry i couldnt get this

like where do i make my eqm equations?

[modern sapphire]

you know how to get this right?

[wind oxide]

yes

[wind oxide]

thats why i asked where

and not how

[modern sapphire]

to the points inside the tube that traps the air

[wind oxide]

but we dont have the airs density

or the dimensions of the closed end of the tube

@modern sapphire ..?

[modern sapphire]

yea sorry, I was thinking there. I think in this case you cant really compute the pressure, except by assuming that there is a linear profile of the pressure distribution.

[wind oxide]

okay yes thts wt i was thinking too

we have to assume its consistent

[modern sapphire]

but I can again think of a contradication here too. The pressure near the cap is 0 for the right side

and yet, for the left side the p_atm component exists

[wind oxide]

well but they arent the same fluid

they are separated by 2 fluids

wait wtf m i saying

ignore pls.

my brains having a meltdown rn

okay so that means fluid problems are pretty much broken

[soft stone]

nah i get what u mean, but basically yeah we assume it’s already reached equilibrium that’s why the pressure difference makes sense in the first place

if it wasn’t the fluids would just move till they do so the equation only applies after that happens

[wind oxide]

ok

i amma move on

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[hollow bridge]

Hi Guys

[hollow bridge]

Can anyone prove 3 , 4 and 5 formulas

By diagram also

[keen wigeon]

by diagram is extremely complex

just use the formulas given

i assume formulas like sin(a+b)=sina.cosb+sinb.cosa is given

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@hollow bridge Has your question been resolved?

[hollow bridge]

Thank you

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[slow thorn]

find $\left[\sum_{k = 1}^{100} \frac{1}{\sqrt{k}}\right]$, where $[x]$ represents the greatest integer less than or equal to x

[lyric charm]

don't put plain text into dollars

and [] means what? floor?

[slow thorn]

yeahh

[soft stone]

you can use an integral to estimate it

[slow thorn]

yeah but im not sure how you would sneak the n properly into the summation though

like converting it in terms of 1/n and k/n

[soft stone]

yeah, you can rescale it into a riemann-sum form

[neat notch]

i think u have to break it

[slow thorn]

yeah but how

[dapper scarab]

something like this

[soft stone]

use 1/k = (1/n) * 1/(k/n)

Then Σ(k=1 to n) 1/√k = √n * (1/n) Σ(k=1 to n) 1/√(k/n)

[slow thorn]

oh shii nvmm

im dmub

1 min

[neat notch]

but wont the limits be 0?

both upper and lower limit

[soft stone]

not exactly 😭 since k starts from 1, the lower limit becomes 1/n, not 0

[slow thorn]

hmm not really helpful though

the limit sum is 20

but the answer is 18

[soft stone]

if you took 0, the integral of x⁻¹/² would blow up (it’s infinite there), so we start at 1/n instead to match the first term of the sum

[slow thorn]

even if i believe a little number is shredded off im not sure how to prove more than 1 is shaved off from 100 to infinity

[soft stone]

hmm

[dapper scarab]

@slow thorn consider this

[slow thorn]

thats... what anflo said

[slow thorn]

and this is the issue

[dapper scarab]

its not

if you calculate its <19

also you can get lower bound of sum the same way

[slow thorn]

ok hmm

um

[dapper scarab]

i think you integrated from 0 to 100

which results 20

[slow thorn]

oh yeaaa

what about the lower bound

[dapper scarab]

consider 1/sqrt(x+1)

[soft stone]

yeah, integrating from 0 is the mistake (∫₀¹ x⁻¹/² dx blows up)... since k starts at 1 the correct integral bounds are 1≤k≤101

[slow thorn]

aaarrggg

yeah im getting > 18... and < 19

thank yo

[dapper scarab]

nice

[slow thorn]

.close

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[slow thorn]

oh and

.reopen

[midnight plank]

✅ Original question: #help-49 message

[neat notch]

i tried anflo's method and its giving exactly 18

s

[slow thorn]

find $\left[\sum_{n = 1}^{\infty} \frac{1}{(n+1)\sqrt{n}}\right]$, where $[x]$ represents the greatest integer less than or equal to x

[slow thorn]

hum

[toxic prism]

ay that looks nice

[neat notch]

the limits will be 100/n and 1/n

[dapper scarab]

its from n to inf right

[slow thorn]

yeah so 18 + 1

[dapper scarab]

i think you can use same method as before

only problem is integrating ig

[soft stone]

can we use f(x) = 1 / ((x + 1)√x)

[dapper scarab]

yeah i just found it could be integrated

some substitutions

[soft stone]

yeah exactly, if you let x = t² then dx = 2t dt and it becomes 2 ∫ dt / (t² + 1) which gives pi/2

that’s the cleanest way to show the sum is between 1 and 2

[midnight plank]

@slow thorn Has your question been resolved?

[slow thorn]

wai how did u sub happen

[midnight plank]

@slow thorn Has your question been resolved?

[midnight plank]

@slow thorn Has your question been resolved?

[small jasper]

$$I=\int^{\infty}{1} \frac{1}{(x+1) \sqrt{x}} \dd{x}$$
Substitute $x=t^2$. Then, $\dd{x}=2t \dd{t}$.
$$I=\int^{\infty}{1} \frac{1}{(t^2+1) \cdot t} \cdot 2t \dd{t}=2 \int^{\infty}_{1} \frac{1}{t^2+1} \dd{t}$$

[grand pond]

Civil Service Pigeon

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[night gyro]

idk how to start

[near cliff]

well,
$$\vec{BA} = \vec{BO} + \vec{OA}$$

[night gyro]

shouldnt they give a diagram or i have to make my own?

[near cliff]

you dont need a diagram

you can do this purely with calculation

[night gyro]

is it better to draw?

[near cliff]

if you want to... it's faster to do with just calculation imo

[night gyro]

how do i even draw this

its a triangle but everything is given

all the vectors

[near cliff]

i dont think you should draw in this case

@night gyro do you know how to add vectors?

[night gyro]

i mean with the triangles one yh

but this 1 i hv never seen

can u walk me through?

[near cliff]

sure

wait a sec

[night gyro]

ok

[ashen kite]

i don't think BO and OB is the same vector

[night gyro]

idk i need help

[near cliff]

One can ``split'' up a vector $\overrightarrow{XY}$ like so:

$$\overrightarrow{XY} = \overrightarrow{\fbox{$X$}\color{red}{A}} + \overrightarrow{{\color{red}A}\fbox{$Y$}}$$

Notice that the start of the first vector is $X$ and the end of the second vector is $Y$, and that the letters in the middle are the same. These can be any two letters.

E.g.
\begin{align*}
\overrightarrow{XY} &= \overrightarrow{XA} + \overrightarrow{AY}\
&= \overrightarrow{XB} + \overrightarrow{BY}\
&= \overrightarrow{XC} + \overrightarrow{CY}\
& \vdots
\end{align*}

Just as long as the start and end correspond to the original vector, and the middle two letters are the same

[night gyro]

yes i see

[near cliff]

aight

so can you think of ways to express $\overrightarrow{BA}$ as the sum of two vectors?

[night gyro]

Ba can be anything no? like BA=BY+BA

[near cliff]

no the middle two letters must be the same, and the start of the first vector must be B, and the end of the second vector must be A

<@&268886789983436800>

[night gyro]

i did tht?

[near cliff]

you said BY + BA

which is

not following that rule at all

[night gyro]

middle 2 ltters r the same B

and and withs A

[near cliff]

the middle two letters in BY + BA are Y and B...

(or A and B, if you swap the ordering)

[night gyro]

so how should it be like?

[near cliff]

you can do BA = BO + OA

then that the two middle letters are O, the start is B and the end is A

[night gyro]

ohh

[night gyro]

so for this it should be BY+YA?

[near cliff]

yup

[night gyro]

okk

[near cliff]

anyways, another thing you need to know is that

$$-\overrightarrow{XY} = \overrightarrow{YX}$$

[near cliff]

so with that you should be able to solve the problem

[near cliff]

no you cant do that since the start of the first vector of O, when it should be B.

[night gyro]

i used this

[night gyro]

i see

[near cliff]

you can't switch the orderings of letters like that

$$\overrightarrow{OB} \neq \overrightarrow{BO}$$

[near cliff]

this is correct

[night gyro]

i used it i got the ans correct

okk

[near cliff]

If you want to switch the order, you'll have to take the negative of the vector:

$$-\overrightarrow{BO} = \overrightarrow{OB}$$

[night gyro]

yess got bothh

[night gyro]

yh

tyy

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[spiral rock]

fast mods

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[inland patio]

I'm studying the Poisson process. We have that $U_1,U_2,\ldots$ is a sequence of exponentially-distributed random variables with parameter $\lambda$. Then $T_n=U_1+\cdots+U_n$ and also, for every $t\geq0$, $N_t=\sum_{n=1}^\infty \mathbf{1}{{T_n\leq t}}$. We discard the null set here where $N_t$ is infinite. $(N_t){t\geq0}$ is then called the Poisson process.\

Moreover, it is a fact that if we fix $t>0$ and define for every $r\geq0$, $N_r^{(t)}=N_{t+r}-N_t$, the collection $(N_r^{(t)}){r\geq0}$ is another Poisson process and independent of $(N_r){0\leq r\leq t}$. The r.v. $N_t$ also has a Poisson distribution with parameter $\lambda t$.\

The claim is now that for every $j\in{1,\ldots,k}$, $N_{t_j}-N_{t_{j-1}}$ is independent of $(N_{t_1}, N_{t_2}-N_{t_1},\ldots,N_{t_{j-1}}-N_{t_{j-2}})$. I understand this. But why does this then imply the independence of $N_{t_1}, N_{t_2}-N_{t_1},\ldots,N_{t_{k}}-N_{t_{k-1}}$?

[grand pond]

psie

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[errant nebula]

I've got the polynomial but I'm not sure what to put down for the lagrange error formula. I have something written but am not sure if it's correct

[lavish venture]

3b?

[errant nebula]

uh

2 b

[lavish venture]

why are you doing f^4(2)

[errant nebula]

that.. isnt supposed to be there

I think i just wrote it wrong

but I figured out that f^2(z) = 6

since thats the highest bound

[lavish venture]

f^2?

[errant nebula]

fuck

f^4

😭

I'm just wondering what to fill in for x now

2.9 or 3?

[lavish venture]

does it matter?

[errant nebula]

I think it does right?

[lavish venture]

they are both equidistant from 3

[errant nebula]

if I fill in 3 it equals 0

[lavish venture]

oh

you said

3

not 3.1

[errant nebula]

but if I will in 2.9 then it is slightly differnt

[lavish venture]

im dyslexic

😭

[errant nebula]

😭

[lavish venture]

ya

2.9 then

[errant nebula]

but why not 3?

[lavish venture]

well i mean are we sure they even want that filled in?

[errant nebula]

idk bro

[lavish venture]

instead of leaving it as x?

usually we plug in a value for x when we want the error bound for a particular value

[errant nebula]

I thought I was supposed to fill in the x value that gave the highest possible value

[lavish venture]

not an interval

[errant nebula]

so like I filled in 4 for 1b

[lavish venture]

well i mean that would give error 0 because the taylor polynomial always matches the function at the center

[errant nebula]

the fifth derivative is 24/x^5

[errant nebula]

makes sense

but then what do I do?

[lavish venture]

yea then just choose the point furthest from the center i guess

if they want an upper bound on the error

[midnight plank]

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[errant nebula]

.close

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[tacit kelp]

Can someone help me solve the following cubic equation please

[tacit kelp]

No idea how to start

[crude idol]

U can test rational roots

[tacit kelp]

Is there anything besides trial and error

[hearty rune]

one of the usual suspects is a root

[tacit kelp]

Like a way that will give me an answer straight away

[hearty rune]

not really, the rational root theorem is a good start
Once you have once root, it'll become simpler to find the others

unless you want to use the ungodly formula of which we do not speak

[crude idol]

Which is...

[midnight plank]

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[floral ruin]

I tried lhopital in this one

differentiating wrt t

is that correct or we have to differentiate both x and t?

[lavish venture]

🤔

why would you differentiate wrt x

[floral ruin]

i was just confused

we just differentiate wrt t right?

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[little cipher]

does anybody know how to solve this calculus problem?

[lyric charm]

have you heard of the mean value theorem?

[little cipher]

yes

[lyric charm]

ok

can you apply it here in such a way that talks about f(6)-f(1)

[little cipher]

could it want me to do something with the formula f'(c) = (f(6)-f(1))/(6-1) ?

[little cipher]

honestly, idk

im just guessing

[lyric charm]

yes

so $\frac{f(6)-f(1)}{5} = f'(c)$

[grand pond]

Ann

[sharp mauve]

How to solve it

[lyric charm]

...you are not OP, are you?

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[hoary drift]

I simplified this to exp(integrate(ln(1 + x²))) but how do I do that integration

[hoary drift]

This

Nvm I got it

I had to use integration by parts

.close

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[edgy crater]

hi, can i check for part (i), f(x) = 1 is valid, but also non constant functions such as sin(2pix) + 2 would also work, since, sin(2pix) is periodic every 1? im i bit stuck on part (ii), im not sure how to find the integral of f(x) (if i am meant to), although i presume it has something to do with part (i) and the fact that F(0) = 0, ty

[runic hamlet]

you know that if you integrate f(x) from 0 to n, you get F(n)-F(0)

if you integrate f from 0 to 1 you get F(1)-F(0)

you are supposed to make a connection between F(n) and F(1)

[edgy crater]

thanks il keep going

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[near geyser]

option 4 is wrong to me

but why they corrected?

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[spiral rock]

Z12 x Z30 is abelian

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[coral belfry]

hello, so im tryna prove that (A\B)×C = (A×C) \ (B×C) and bc im lazy i used only iffs instead of proving both ways separately. is there a point where we should have an "=>" that i missed? if i were to guess, id say its where i found that
(x,y)\notin B×C but im not sure

[coral belfry]

now that i think abt it, it might be that they are related by subset not equality

but still idk how to prove either one

[wary epoch]

Unfortunately, ${(x, y) | x \not\in B \land y \in C}$ is not a valid set

[coral belfry]

nono, by using brackets i dont mean a set

its just a logical sentence

{x not in B }
{and y in C}

not as a set, but just as a sentence that comes from (x,y) not in B×C, or smth like that

[grand pond]

Xwtek

[lone herald]

thank you mr "bot"

bruh

[coral belfry]

anyways, i think i proved both ways so we dont have to analyze it further

but thanks

.solved

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[lone herald]

bro how? i was trying the same equality method and.. i think they r not equal in the sense...but by some relation to subset..(sorry for my bad eng)

[lone herald]

@coral belfry

[coral belfry]

i took the first thing i showed earlier, but replaced one iff with =>, and then i went backwards and it worked idk

[lone herald]

oh...so u did like this, p=>q & q=> soo p <=> q lol ... easy

[coral belfry]

yeah. but im not 100% sure p=>q is entirely correct

[lone herald]

i mean it should be ... cuz we did prove by one directonal method ...well ig equality doesnt work in general for this type of scenarios

solved?

[coral belfry]

yesyes

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[old sandal]

Anyone help me solve this question?

I assume the right hand side become v^3 + v but what does the left side do

[eager drift]

so like in this image that i shared he seperate E into 2 componentts E_x and E_y and then he just draw the axis on E_x one time and E_y the other time that means when we draw the axis on E_x the E_x and the sigma_x are just a vector and the sigma_y just becomes 0 and same for the other cases so it just becomes 2 vector cases instaed of a single tensor case right?

[old sandal]

im unsure what tensor is this is just an inital value problem

[floral apex]

what have you tried?

it looks separable

[old sandal]

it asked to use the substittution v=y/x

[floral apex]

sure, you could do that. do you understand how to use it?

also just since the bot isnt working can you ping if you reply

[finite jungle]

I understand the sandwich theorem and all that but why is it between |x| and -|x|

[vocal warren]

-1<=sin(1/x)<=1
if you were to multiply this inequality by x and x was negative you'd have to change the inequality signs, so to simplify things you just multiply the abs of x so you know it's positive

[mint dagger]

Am I doing it wrong?

A calc differentiation question

[compact pivot]

Need help?

[mint dagger]

Yes peis

Pris

[bleak parcel]

can anyone help me find videos on each of these questions? i'm not too sure what each question is called so for a lot of them, i'm not sure where to look. thank you for your time. i know how to do the fractions in the beginning

[compact pivot]

Wait

[mint dagger]

🥹

[compact pivot]

The function to be differentiated is:(f(x)=\ln (1-x^{2}+x^{3})\sqrt{1+x}) ? ? Is it

[grand pond]

Celestial

[mint dagger]

Yesyes

I think i did it correct but my friend’s answer differed from mine

[compact pivot]

there is an error in the final calculation when evaluating the function at (x=1).

[grand pond]

Celestial

[mint dagger]

Where exactly?

[compact pivot]

The problem asks to differentiate the function (f(x)=\ln (1-x^{2}+x^{3})\cdot \sqrt{1+x}) and then find the value of the derivative at (x=1)

[grand pond]

Celestial

[compact pivot]

?

[mint dagger]

Yesyes

For f^1(x) then f^1(1)

[compact pivot]

so we use the product rule for differentiation: ((uv)^{\prime }=u^{\prime }v+uv^{\prime }). Let (u=\ln (1-x^{2}+x^{3})).Let (v=\sqrt{1+x}).

[grand pond]

Celestial

[compact pivot]

❓❓⁉️⁉️

[mint dagger]

??? 😭😭

[compact pivot]

To find (u^{\prime }), we use the chain rule for the natural logarithm: (\frac{d}{dx}[\ln (g(x))]=\frac{g^{\prime }(x)}{g(x)}).Here, (g(x)=1-x^{2}+x^{3}), so its derivative is (g^{\prime }(x)=-2x+3x^{2}).Therefore, (u^{\prime }=\frac{3x^{2}-2x}{1-x^{2}+x^{3}}).

[grand pond]

Celestial

[compact pivot]

To find (v^{\prime }), we use the power rule and chain rule for the square root: (\frac{d}{dx}[\sqrt{g(x)}]=\frac{g^{\prime }(x)}{2\sqrt{g(x)}}).Here, (g(x)=1+x), so its derivative is (g^{\prime }(x)=1).Therefore, (v^{\prime }=\frac{1}{2\sqrt{1+x}}).

So you correctly found these right

[grand pond]

Celestial

[compact pivot]

Screw it

The correct value for (f^{\prime }(1)) is (\sqrt{2}).

[grand pond]

Celestial

[mint dagger]

Yesyes so my answer for f^1(x) is right

[compact pivot]

Why r u saying your wrong tho

[mint dagger]

But f^1(1) is wrong right?

[compact pivot]

I need to do my English homework rn

[mint dagger]

Because my final asnwer for that is 1

😭 okay..

[compact pivot]

Sry

[mint dagger]

Ty for the help

[compact pivot]

No

Prob

[mint dagger]

👋

[compact pivot]

Btw

Can u solve this

[mint dagger]

What even is that 😭😭😭

[compact pivot]

Uh

.

Btw gtg

Cya

[mint dagger]

Good luck

[compact pivot]

👋🏻👋🏻

[orchid fossil]

yea good luck

[compact pivot]

AHH HELP HELP HELP HELPNPLSMSNSNSNSNSNS

[echo nest]

Ignore the first definition and the range of u. I want to know if you guys agree there is an error on this article because if you count from k.u until (k+1).u - 1 you will get u different numbers, but the article talks like there are k different numbers. I think what the authors meant was to count to k.(u+1) - 1

What do you guys think?

[sacred folio]

.help

hmm

[twin kernel]

how u do this

[twin kernel]

for b and c

[midnight plank]

@twin kernel Has your question been resolved?

[midnight plank]

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[sly cape]

In linear inequalities, whenever multiplying both sides with -1, do I have to change the sign? Also why?

[spiral rock]

yo do change the sign

[sly cape]

Why’s that?

[spiral rock]

there are few ways to see that,
if x < y then you can add -y-x to both sides to get -y < -x after simplification

[shell wigeon]

,, a < b \implies a-a-b < b-a-b \implies -b < -a \implies -a > -b

[grand pond]

Nel

[sly cape]

Also I can either multiply by -1 or divide by -1

It works the same?

[shell wigeon]

Dividing by -1 is multiplying by 1/(-1) = -1, so...

[sly cape]

Also

The

X€R

Means

X equals to all real number?

[shell wigeon]

No, just that X is a real number

[sly cape]

€ is

Equals

[spiral rock]

$\in$ means "in"

[grand pond]

ExpertEsquieESQUIE

[spiral rock]

so $x \in \bR$ means $x$ is in the set of real numbers, so $x$ is a real number

[sly cape]

R is real numbers?

[grand pond]

ExpertEsquieESQUIE

[spiral rock]

yes

[sly cape]

Real number is all numbers

Also in square root

And power

And fraction

[spiral rock]

and a lot more

for example $\pi$

[grand pond]

ExpertEsquieESQUIE

[surreal moon]

What constitues real numbers, while well-defined, is quite hard to fully appreciate

[sly cape]

Also with this I understand

Suppose a < b

And a is 5 and b 10

And if we multiply by -1

Then -5 is bigger than -10

So a>b

[midnight plank]

@sly cape Has your question been resolved?

[midnight plank]

@sly cape Has your question been resolved?

[tall lintel]

-a > -b yes

Remember originally that a was 5

[midnight plank]

@sly cape Has your question been resolved?

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[rigid turret]

conceptually i understand the forces. each has gravity down and each has normal force perpendicular to ramp. A has force upward from the car which exerts that same force on B which exerts same force equal magnitude opposite in direction

but

how is block A at constant velocity

it has net force going up the ramp

[formal orchid]

The overall moving force is constant, right?

Isn't it more intitive that Block A should also have a constant velocity?

[rigid turret]

yes

but free body diagram shows that block A is not net 0 force

wait

nevermind

it might just not be to scale

[formal orchid]

Why does the net force need to be 0?

Won't the object be at rest then?

[rigid turret]

no

an object in motion stays in motion unless acted upon by another force

it just means constant velocity

[formal orchid]

Oh yeah, my bad

Can you show me the other options as well?

[rigid turret]

na thats the correct answer

was just looking for a breakdown

but im good now

.close

[midnight plank]

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[midnight plank]

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[rich yew]

this one too?

[midnight plank]

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[slate ferry]

<@&268886789983436800>

[midnight plank]

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[slate ferry]

seriously how do people even believe that scam

its so popular for some reason

[midnight plank]

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[last slate]

hi

[last slate]

idk how to interpret this

and i dont know why it isnt a function either

Also what is the output

[bold peak]

So relations aren't functions

Relations are, well, subsets of the Cartesian product

Let's start very very simple

Equality is a relation

[last slate]

even when they give me y = something x?

thats not a function either?

[bold peak]

Functions are also relations

A function is a relation where you have "unique output" in some sense

Like for one input, you only have one output

That's a function

But relations are more general

[rough plover]

you don't necessarily have to understand what a "relation" is to solve this problem

[bold peak]

One thing can be related to a lot more

[rough plover]

the definition of related is given in the problem

[last slate]

the whole section is called “properties of relations”

[bold peak]

You need to understand what transitivity is

Also it's better to actually understand what you're dealing with

Maths is about more than solving problems

[last slate]

so all functions are relations but not all relations are functions?

[bold peak]

Yes

You can think of a relation as a bunch of arrows

[last slate]

right

[bold peak]

A relation is a function if every element has only one outgoing arrow

Ig I should make it a bit more precise

A relation from A to B is a subset of A x B

[last slate]

true

[bold peak]

A relation is a function if, for every a ∈ A, ∃! b ∈ B st. (a,b) ∈ R

Note that two things can still map to the same thing on the other side, that's still allowed

[last slate]

Yes

[bold peak]

Now back to relations

[last slate]

But it wouldnt be invertable

unless its a bijection

[bold peak]

Relations have three properties that we usually care about

[bold peak]

Correct

[last slate]

yes?

[bold peak]

Actually four sorry

The first is reflexive

Actually wait before that

Right now we are looking at relations from A to A

This notion doesn't make sense otherwise

[rough plover]

that would be for equivalence relations, no?

[bold peak]

I never said the properties need to hold

These are the properties we care about. As in usually given a relation, you want to check if these properties hold

[last slate]

yea

[last slate]

so reflexive is first?

[bold peak]

Yes

So given a relation from A to A

We say it's reflexive if every element is related to itself

We don't care what else is related

Just that everything is related to itself

[last slate]

What if some where and others werent?

that would be neither right

[rough plover]

then the relation wouldn't be considered reflexive

[bold peak]

That's not reflexive no

[rough plover]

but you could say it's reflexive for a subset

[last slate]

And neither anti reflexive

[bold peak]

It needs to include every pair (a,a)

[bold peak]

Anti reflexive isn't a thing we care about usually

So much so that I've never found a mention for it

[last slate]

Book talks about it so

[bold peak]

Fair enough

[rough plover]

can you tell if your given relation is reflexive?

[last slate]

can i think of xRy as just x maps to y ?

[bold peak]

Is related to is better wording

Mapping usually implies functions

[last slate]

I cant even understand the inequality

[bold peak]

walltile let's slow down

Alr moving on

Symmetry is simple

If a is related to b, then b is related to a

If this is true for all a and b, your relation is symmetric

God my phone keeps autocorrecting relation to relationship

[last slate]

a and b being any random element

[bold peak]

Yes

[last slate]

or any ordered pair

or

[bold peak]

a and b are elements

[rough plover]

it's good to distinguish random and arbitrary. if you take probability or statistics in the future, random means something specific

[bold peak]

True

[last slate]

so arbitrary?

[bold peak]

Yes

[last slate]

and when we talk about these properties of relations, and we examine xRx or xRy or anythingRanything

These are always arbitrary values?

[bold peak]

I gotta log off cuz class, all yours walltile

[last slate]

from my set of the list of ordered pairs?

[rough plover]

yeah, if you want to say that a relation R is symmetric/reflexive, you need to show that all arbitrary pairs have the property

[last slate]

Its always gonna be a set of ordered pairs that i take elements from right

[rough plover]

yes, they would be ordered pairs

order matters bc that makes symmetry a notable property to check for

[last slate]

and same thing consequently applies for the transitive property right

[rough plover]

indeed

you can see all elements in the relation as ordered pairs

[last slate]

can you please explain this to me

[rough plover]

so, you say that x is "related" to y, if it satisfies some property

in this case, the property is y>=x+1

for example, we can say that 1 is related to 2, bc 2>=1+1

however, we don't say that 1 is related to 1 (itself), bc 1>=1+1 isn't true

[last slate]

when we say “related”

whether it be with this or a function

what does it really mean

[rough plover]

here, it is some arbitrary relation the problem defined

[last slate]

i think of it as an input from the domain set giving an output from the target set and that arrow is the relation

[rough plover]

you can see it as a relation with the specific inequality given

[rough plover]

yeah that works. and you would draw arrows for pairs that satisfy the inequality

[last slate]

how do i know when im dealing with a function vs a relation that isnt one

[rough plover]

in this case, it doesn't make a difference. but to answer your question: a function needs to pass the "vertical line test", i.e., every domain element only has one outward arrow

[midnight plank]

@last slate Has your question been resolved?

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[last slate]

thank you bro

[midnight plank]

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[last slate]

hi

[last slate]

another question

i need help understanding this

[lyric charm]

which part?

this is an outline of a bunch of basic vocabulary to do with directed graphs.

[lyric charm]

be more specific. which part(s) of this confuse you?

welp. op offline.

[last slate]

back

[last slate]

everything tbh

(V, E) = ( {verticies} , {edges} )?

[lyric charm]

sure.

[last slate]

yeah that’s confusing like how am I supposed to format this

[lyric charm]

have you seen a picture like this before? as in, have you ever seen any depiction of a graph in the graph-theory sense?

[last slate]

Seen something like this related to properties of relations but nthn else

[lyric charm]

ok like

"graph" in this sense should be understood to mean "a bunch of nodes connected by a bunch of edges"

this (V,E) stuff is a more formal way to say the same thing

but if you think that everybody writes down those formal notations every time they think of or present a graph you are quite wrong

[last slate]

nodes are just elements?

[lyric charm]

nodes can truly be whatever the fuck you want

[last slate]

what do we mean by nodes exactly

alright

[lyric charm]

they're depicted as points or little circles but like

they are just things

say, does your city have an underground rail transit or metro system

[last slate]

yes

it does

[lyric charm]

<@&268886789983436800> scamtroll

[lyric charm]

wonderful. the map of your metro is a good example of a graph that you may deal with on your daily commute

when using the metro you dont care that much about where exactly the stations are geographically but rather how they are connected to each other

[last slate]

true, but its a subset of which V x V??

[lyric charm]

you're thinking in the wrong way imo.

but if you insist, then V would be the set of all stations and E would be the set of all rail corridors between each pair of adjacent stations.

though a metro system is an undirected graph cause you can travel both ways on it.

[last slate]

I just don’t understand the V x V here

[lyric charm]

i really really caution against bogging yourself down with formalism and/or taking it too seriously

read and understand this:

an edge in a directed graph is specified by where it begins (a vertex) and where it ends (another vertex).

[last slate]

so a node is the vertex?

[lyric charm]

"node" and "vertex" are synonyms yeah

[last slate]

i see

whats the degree in degree out thing about

[lyric charm]

in-degree = "how many edges are going INTO this vertex?"

out-degree = "how many edges are going OUT OF this vertex?"

[last slate]

ah i see

btw

is there a particular formula to calculate a power set

[lyric charm]

completely unrelated question?

wdym by "calculate a power set"

like write a power set down in full?

[last slate]

Yea, actually just forget it for now i dont wanna create confusion

i have a related question tho

how do you define a walk in a directed graph

[lyric charm]

do you want me to somehow regurgitate the formalism, or do you want to know what a walk actually is?

[last slate]

no actually

the book already does a great job at making things complicated

[lyric charm]

"no"? so which one is it? neither of the 2 things i said?

[last slate]

no, the actually part

[lyric charm]

ok

dont answer either-or questions with yes nor with no in the future

anyway a walk is a route you can follow in the graph if you always walk along the edges in the direction of their arrows

i.e. it's a sequence of vertices you visit one directly after the other (and so any two adjacent vertices in that sequence must be actually joined by an edge in your graph for the walk to be valid)

[last slate]

So how would you describe an example walk here

[lyric charm]

here's one: 0 → 2 → 4 → 5

a couple more examples:
0 → 3 would be illegal in a walk because there's no edge going directly from 0 to 3
4 → 0 would be illegal because you're following an edge but you're going the wrong way

[last slate]

Can you also say 0 -> 2 -> 0 is illegal because you need to follow a tail to tip rout?

if im not wrong

[lyric charm]

2 -> 0 cannot be part of a walk in this graph, yes.

(because you're going through an arrow the wrong way)

[last slate]

i see

so you must follow the arrows but also the ones that connect

[lyric charm]

i mean. yeah. thats what a walk is

idk like, imagine you're moving through a part of your city where all the streets are one-way

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[rustic owl]

d/dx of x(-x^2+16)^1/2

[rustic owl]

i got -x(-x^2+16)^-1/2 + (-x^2+16)^1/2

i dont know how to simplify it more

[spiral rock]

Its -x^2(....

[chrome swift]

U know what chain rule is

[rustic owl]

why

[lyric charm]

first term seems wrong to me

you need chain and product here

[rustic owl]

i did that

let me do again

[chrome swift]

Naw

[rustic owl]

im posotive this is right

positive

[chrome swift]

Naw man

Tell me what ur doing in ur first step

Like how are you applying the chain rule and product rule

[rustic owl]

x* 1/2 (-x^2+16)^-1/2 * -2 + (-x^2+16)^1/2 * 1

[lyric charm]

derivative of -x^2+16 is not -2

[rustic owl]

ooo

-2x

oop

so -x^2(-x^2+16)^-1/2 + (-x^2+16)^1/2

[chrome swift]

I don’t see the issue with that first term

[rustic owl]

so how do i simplfy now

because i need to set this whole thing set to 0 and dne to find the critical numbers

[chrome swift]

Oh

I mean

Whats wrong with multiplying with the common denominator to preform subtraction

[rustic owl]

huh

i dont understand

explain further

i thought i had to factor

but didnt know how

also for the problem x^2/x+1 its decreasing from (-2,0) right

becuase -1 isnt a critical number because its not in the domain

[midnight plank]

@rustic owl Has your question been resolved?

[tall lintel]

Yep, excluding x = -1.

[midnight plank]

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[cloud adder]

For B do I need to find the position of x(2) and then the velocity x'(2) and compare my answers to see if it's moving away or towards the origin?

[tall lintel]

Explain :)

How would we determine whether its moving away or towards the origin using these two?

[cloud adder]

the signs

of the andwers

[tall lintel]

Wdym

[tall lintel]

Good, I agree!

In other words, the particle is currently left of the origin (x(2) < 0), but is moving to the right (x'(2) > 0)! Hence, it is going towards the origin.

Wair sorry, other way around

[cloud adder]

it's currently left of the origin

[tall lintel]

Yep.

[cloud adder]

But it's moving towards the right

Which is towards the origin

That makes sense

[tall lintel]

Indeed :)

[cloud adder]

.close

[midnight plank]

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[near geyser]

total elements are n^2

a11+a22+...ann

[west iron]

do you know what the dimension would be without the constraint?

[midnight plank]

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[near geyser]

yeah

[fervent slate]

Bruh are you liek you inyear 12 or uni or idk

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[wheat tiger]

guys i just had my gcse math exam this was one the questions can u check if it is right

[wheat tiger]

basically there a 15 counter - pink - 2 green -3 blue -10 in a box greg takes 3 counters calculate the probability that one pink still remains in the box

i wrote 34/35

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[coral belfry]

,tex
Hello, so we have that: \
T: the set of all circles in $\mathbb{R} ^2 $ with center at $ \left( 0,0\right) $ and \ g: $ T\to \mathbb{R} _{>0} $ the function of the value of the circumference of every such circle. \
\
To prove that g is surjective, can we just say that $\forall y\in\mathbb{R} _{>0} \ \exists r\in T $ s.t. $g(r) = y $ \
\
and that this is true because for any choice of circumference in the reals, there will exist a circle of radius r that has that circumference?

[grand pond]

fijokazż

[coral belfry]

(i skipped over the fact that g(r) = 2πr but idk if we need it. maybe if we used it, we wouldve had to prove that y/(2πr) is an element of T right? which id have no idea how to prove)

[oak kelp]

<@&268886789983436800>

[coral belfry]

lol

[blissful pier]

sure, just pick a circle with radius of y/2pi, then what is g(y/2pi)?

[coral belfry]

y

wait what we can just do that

(?)

[blissful pier]

sure, for an arbitrary y in R+ we can always produce a circle such that the circumference is y

is that not the definition of surjectivity?

[coral belfry]

lol trueee

thank you very much

.solved

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[proud jacinth]

how do i find the 5% and 95% interpercentile range of a data set

[midnight plank]

@proud jacinth Has your question been resolved?

[proud jacinth]

question 3

<@&286206848099549185>

can anybody help me please?

[glass canyon]

Did you try using a percentiles formula for grouped data?

[proud jacinth]

no i haven’t tried anything idk what to do

[glass canyon]

https://youtu.be/eceK3eFyS8Y?si=2yMYWnzz-GoTSsDg

Perhaps have a gander at this for a

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[ionic sapphire]

proof

i got the method while typing out my thought process(bruh moment)

.close

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[safe urchin]

I have a quaternion representing the orientation of one object relative to its parent, and the parent has an orientation relative to the origin. What operation do I use on the two quats to get the first object's orientation relative to the origin? Is it just addition, or something else?

[onyx tide]

Hi

[pearl idol]

it sounds like you're just trying to compose two rotations in which case the answer is just quaternion multiplication :)

[safe urchin]

awesome, thanks!!!

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[pearl idol]

you're welcome

[safe urchin]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[pearl idol]

be careful about the order of multiplication tho since it's not commutative

[safe urchin]

How should I do it then? Object * parent1 * parent1parent?

or in reverse?

[pearl idol]

so if q represents rotation from origin to parent and r represents rotation from parent to object, then rq would be rotation from origin to object bc you read the rotations from right to left

[safe urchin]

rq would be rotation from origin to object
Meaning the rotation you apply to the origin to get the object? Like saying "the angle you apply to a vector to get the new vector"?

Just want to make sure I'm understanding the terminology

I've never worked with quats before

[safe urchin]

If I have an arbitrary object's "orientation" stored on the object as a quaternion, would that be called the "rotation from the origin to the object"?

[pearl idol]

yeah exactly

[pearl idol]

That'd be my typical assumption yeah, make sure to double check the conventions in whatever code you're using

[safe urchin]

Thanks

[safe urchin]

And I'm guessing if I were to compose from an arbitrary start, I'd want it to be the identity quat? So it ends up as Identity * object * parent?

oh wait ok I get the right-to-left thing

thanks

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[vocal flame]

Can someone help me find where I went wrong on this problem

[vocal flame]

Like ig I could have said pi/6 instead of the long decimal for 1 but idk what I did wrong

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[trim sun]

Idk what equation to use to find y

Is it y-y1=m(x-x1)?

[shell wigeon]

Yes it is

[trim sun]

Am i plugging in the numbers correctly??

[shell wigeon]

,rcw

[shell wigeon]

Why 12?

[trim sun]

I plugged in 1 into 18x-30

[shell wigeon]

Uh

18*1 - 30 = 12?

[trim sun]

-12?

😭

[shell wigeon]

Yes

[trim sun]

Okay i got it thanks 😊

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[fossil knot]

Is tg tangent?

[quartz vale]

yeah was going to ask too

european thing

@shut canyon remember tangent is an odd function

[shut canyon]

tan

[quartz vale]

then you can use small angle approximation

then cancel x since x ≠ 0

you have your answer after lol

no need for l'hôpital

[shut canyon]

limₓ→0 (tan(-5x) ÷ 2x) = limₓ→0 (-tan(5x) ÷ 2x)

[quartz vale]

yes then treat tan(5x) ~ 5x

[shut canyon]

Thank you!

[quartz vale]

my pleasure!

[quartz vale]

but of course you can still fallback to this if you get stuck

[shut canyon]

tan(5x) ~ 5x as x→0, so limₓ→0 tan(-5x)/2x = -5/2

[quartz vale]

nice 🔥

[shut canyon]

In general when you see these limit exercises what do you look for?

[quartz vale]

ofc indeterminate forms like that

0/0 and ∞/∞ for example

also with trig functions (like i did), note even/odd properties

the algebraic tricks like factorization, rationalization (like multiplying by a conjugate), or rewriting expressions to cancel terms

for behavior at infinity, check dominant terms in polynomials or exponentials

[fallen sparrow]

Theres certain standard limits

[quartz vale]

these are good to know too

like sin(x)/x and tan(x)/x

and e 😮‍💨

[fallen sparrow]

Try to algebraically manipulate the given limit to find several of these standard limits

Then you can apply them

[shut canyon]

tysm🐈‍

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[near geyser]

n(A/B) means?

[small jasper]

number of elements in A but not in B

[near geyser]

So i will find what they in intersection

Then remove it

[small jasper]

yes it does involve subtracting the number of elements in $A \cap B$

[grand pond]

Civil Service Pigeon

[near geyser]

8

One more question

[small jasper]

good

[near geyser]

What does this + mean?

[small jasper]

,rotate

[near geyser]

What is direct sum?

[small jasper]

in $A$ or $B$, but not both

[grand pond]

Civil Service Pigeon

[small jasper]

aka the symmetric difference

[near geyser]

A+B- intersection?

Thanks

[small jasper]

this counts the number of elements in the union of A and B (aka in at least one of A and B)

that's not what the symmetric difference is

[near geyser]

I got 55

Thanks

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[near geyser]

.close

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[wraith notch]

So this comes from the openstax book about trig substitution, and I'm confused how it goes from step 2 to step 3. My sqrt algebra is a bit weak so I don't really know how the 9 got on the outside there

[wraith notch]

It's a bit strange how the 9 gets out unscathed.... and its not really distributed...

I'm not sure you can factor out of a square root, I don't think thats a thing..

so its $$ \int \sqrt{9 - 9 sin^{2}} \theta 3 cos \theta d \theta $$

breh

[grand pond]

Alphurion

[plucky furnace]

okay so you're actually "factoring out" a 3 here

[wraith notch]

Oh

[plucky furnace]

which multiplies with the 3 already outside

you can factor 9 out inside the square root

and since sqrt(ab) = sqrt(a) * sqrt(b), you end up with 3*sqrt(1-sin^2 theta)

[wraith notch]

and then that multiplies with the 3 from the 3 cos \theta?

[plucky furnace]

yup

[wraith notch]

Ohhhhh okay thank you

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[near geyser]

I didn't understand the last step

[lyric charm]

mmmm wonderful misuse of \otimes.

[near geyser]

\otimes?

[lyric charm]

when you say last step do you just mean the example that they give of one multiplication calculated within the group?

[lyric charm]

$\otimes$

[grand pond]

Ann

[lyric charm]

this symbol

[near geyser]

Yeah

(2,7)(3,11)

[lyric charm]

yeah

in a product of groups the multiplication is done coordinate-wise

[near geyser]

I have read direct sum

But this is new to me i meant multiplication

And what they did in next step?

[runic hamlet]

the point is that in each entry you are doing the operation of the original group

so in the first entry you do the operation from U(5) which is multiplication

and in the second entry you do the operation from U(12) which is also multiplication

[visual tiger]

yeah, so to compute (2,7) x (3,11) in U(5) x U(12)

you compute 2 x 3 in U(5)

and you compute 7 x 11 in U(12)

[midnight plank]

@near geyser Has your question been resolved?

[near geyser]

They have multiplied all the elements

But why they write again these two?

(2,7)(3,11)

[runic hamlet]

its an example

of how the group operation works

in that new group

the element (2,7) combined with the elemnet (3,11) gives the element (1,5)

[near geyser]

I didn't understand the next modular part

Could you simplify please?

2.3, 7.11

(6,77)?

Ohh wait

I got it

Mod 5 and mod 12

Why they multiplied only these two?

Will we multiply (1,5) with others?

[runic hamlet]

its an example

[near geyser]

So they didn't mention

[runic hamlet]

do you want them to write out the entire 16x16 multiplication table?

[near geyser]

We will multiply all Of them

They should mention but