help-23

[opaque fern]

Hey, Im so confused by this but how do you partition this into Voronoi regions?

[brave wolf]

is it this thing?

cant u just draw lines inbetween every pair of dots which are "close"

and then draw the boundary from those lines

(drawn in black here)

[brave wolf]

(to be a bit more exact, draw bisectors between the pairs of points)

[opaque fern]

What i don't grt is thst it should be gor every near pair right

[brave wolf]

well, yeah, but i think it can be done intuitively

just draw them until they intersect with sth else

[opaque fern]

Wait what is the purpose of thr diagonal line

[brave wolf]

its the bisector between the diagonal pair

[opaque fern]

Oh yeah

[brave wolf]

i marked those points red here

[opaque fern]

Ok

So basically draw a bisector for every pair of adjacent points and call it a day, yes?

[brave wolf]

yeah, that's how I'd do it

including the diagonal adjacency

and then you have to pick which parts are actually the boundaries

[opaque fern]

Ok ty!!

.close

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[wispy berry]

where is your banner from

[wispy berry]

NOOOOOOOO

.close

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[naive dragon]

Lmao

[wispy berry]

wow

[quasi timber]

[opaque fern]

💀

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[latent mountain]

I tried solving this with phasors but Im getting Pie as the answer

[latent mountain]

but answer is 5pie/3

this is my working

[dark elm]

Phase 2 isn't 180+60

Since it's moving right

[latent mountain]

Yes but its measured from +ve x axis right?

[dark elm]

If it was +180 it would also be moving away from the centre

[latent mountain]

Im sorry i didn't understand

[dark elm]

So on your diagram the y axis is where the block is horizontally

And the x axis is the speed

[latent mountain]

no

its just the phasor diagram

[dark elm]

But it is doing that

Your phase 2 should be in bottom right quadrant

[latent mountain]

why?

[dark elm]

Okay so at 0 degrees the block is at the origin moving quickly to the right

[latent mountain]

yes

[dark elm]

So what would it be like at -60 degrees

-90 is having the block on the left side with 0 speed btw

[latent mountain]

at the midpoint between amplitude and origin moving towards left

[latent mountain]

wait how

[dark elm]

Wait I'm confused hold on

[latent mountain]

shouldnt it be at the origin moving towards left at -90

[dark elm]

Actually I got it wrong the whole time :P

It should be in the top left

Your graph's x axis is the position and the y is the speed

[latent mountain]

No man it does not represent speed and position

if u graph speed and position in shm youll get a kind of cosine wave

[dark elm]

Yeah, and that's what a circle is

(cos(t),sin(t))

[latent mountain]

nooo

v = -Awcos(wt + phi)

[dark elm]

That's with time on the x axis

I'm saying position is on the x axis

[latent mountain]

okayyy

wait then you get an ellipse

v = w*sqrt(A^2 - x^2)

reagraning u get an ellipse

[dark elm]

But you can rescale to get a circle with the phase as the angle along the circle

<@&268886789983436800>

(scam)

[latent mountain]

lmao

[dark elm]

So in your diagram 0 degrees is like (1,0) where it is all the way to the right and 0 speed to the right

[latent mountain]

the diagram which I drew is a represntation on shm in 3d

https://tenor.com/view/fasores-gif-21369082

[dark elm]

And 60 degrees is like $(0.5,\frac{\sqrt{3}}{2})$ where it's 0.5 to the right of the centre and with speed $\frac{\sqrt{3}}{2}$

[latent mountain]

yes correct

[flat frigate]

BBMaths

[latent mountain]

idk abt the speed tho

how did u calculate tht

[dark elm]

The speed is the y position (divided by A)

[latent mountain]

k

[dark elm]

So what does 120 degrees look like on your graph

I'm a bit worried because I don't even see how it's 5/3pi

I get 4/3pi

[latent mountain]

wait i got it

[dark elm]

Oh wait it's 1/6

nvm

yeah it is 5/3pi

Very tired today lots of mistakes lol

[latent mountain]

phase diagram looks like this

k thanks for your time

[dark elm]

Yeah that's where you want it

so the angle between the two is 60 degrees which is also 300 degrees=5pi/3 radians

[latent mountain]

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[distant chasm]

I have to prove this thingy. It says that it's enough to take n! out of the brackets on the left. I tried doing that and it was n!(1+[1/n] * n^2) which ended up being n!(1+n)
but that's not the same as (n+1)! I think? Did I make a mistake while taking it out of the brackets?

[finite yacht]

It is the same

[distant chasm]

oh

yeah my brain might be done for today, thanks though!

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[feral condor]

how can we solve 4z²-10iz-7-i=0 in C

[feral condor]

?

[split kayak]

quadratic formula

and possibly throw in eulers identity too

[feral condor]

how so

[feral condor]

whats that?

[burnt mantle]

use coefficients 4, -10i, and -7-i as your a,b,c

[feral condor]

ok ok oik

tnx

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[plush fossil]

Hello can someone help me out with this question, I don't understand it but our teacher said that we need to apply Pythagoras theorem somehow. Any help is appreciated.

[vague zinc]

Start out by drawing a diagram

[plush fossil]

like this ?

[vague zinc]

Yep, that's good

[vestal zephyr]

No we dont

[cerulean nimbus]

no we dont bud

sarj aleti do not lie

[vague zinc]

Are you all doing this problem together😭

[vestal zephyr]

Maybe

[vague zinc]

You guys know the british flag theorem?

[cerulean nimbus]

no

[vague zinc]

It's based on the pythagorean theorem

[cerulean nimbus]

oh wait thats helpful ty

[plush fossil]

thank you

[vestal zephyr]

Thank you so much for this assistance, saviour.

[vague zinc]

Thank the lord

[solar hazel]

[cerulean nimbus]

who is this

[vestal zephyr]

ありがとうございます

[vague zinc]

You probably shouldn't use something you haven't seen in class though

[solar hazel]

wtf is happening in this channel

[vestal zephyr]

Our teacher doesnt teach

Hes mute and didnt show up for a month

[vague zinc]

Three people trying to solve the same problem in class

[cerulean nimbus]

yes

[vague zinc]

I assume

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[raw sedge]

Hi, could anyone explain where the 2 comes from in the second expression? I think it has to do with the negative exponent but I just can't understand it somehow. Some other examples or maybe a break down would be great

[meager jacinth]

lemme try ts one sec

is it whole root or nah

whole power

or jst x

which 2

[raw sedge]

Oh, srry

it's the coefficent 2

[meager jacinth]

oh ye

[raw sedge]

Where it says 2x^1/2 in the denominatoir

[meager jacinth]

so its x^-1/2/2 right

to start

[floral yew]

power rule of differentiation

[meager jacinth]

x^-1=1/x

to the power of neg 1 flips it

btw why tf are you asking this while doing trigg?

[raw sedge]

im doing chain rule

differentiation of cos is -sin

[meager jacinth]

no i do not

I thought you said smth else

anyways

[raw sedge]

Im just not the best at Algebra

[meager jacinth]

Alr so lemme help

so the 2 was already there right

[tender horizon]

is this derivative of (1/2)x^(-1/2)?

[meager jacinth]

so its 1/2*1/x^1/2

[floral yew]

are you differentiating sqrt(x)?

[raw sedge]

i should have said that for context, mb

[meager jacinth]

basically

yall got any tips for the amc8

this is my last attempt

[floral yew]

then your answer is correct its power rule x^(-a) can be written as 1/x^a

[raw sedge]

Yea it's right, I just want to know how to simplify it

Because the answer choices don't include the unsimplified version

[floral yew]

you got 1/2sqrt(x) right.. what more simplification do you want?

[tender horizon]

just send the question so we know what you did wrong

maybe it's not a simplification problem

[floral yew]

$\frac{d}{dx}x^{n}\ =n\ \cdot x^{n-1}$

[flat frigate]

Xerxes

[floral yew]

in your case n = 1/2 now try it

[raw sedge]

$\frac{1}{2}x^{frac{-1}{2}$

[flat frigate]

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

[raw sedge]

[floral yew]

i get what youre trying to say

[raw sedge]

Ok, so I know how to do the power rule

That's fine

[floral yew]

then there's your 2 in the denominator which is 1/2 from the power rule

[raw sedge]

Ah

This might be a stupid question by why does it go from 1/2 to 2?

[floral yew]

what?

[raw sedge]

From the 1st to the 2nd expression in this photo

The coefficent of 1/2 goes to 2

[floral yew]

1/2 does not equal 2

$\frac{1}{2\sqrt{x}}\ =\ \frac{1}{2\ }\cdot\ \frac{1}{\sqrt{x}}$

[flat frigate]

Xerxes

[floral yew]

the 2 is in the denominator its still 1/2

[raw sedge]

Ok I understand it now

[raw sedge]

Yeah I was just being dumb

I forgot it's in a fraction

[raw sedge]

This is exactly what I asked for, basically a breakdown of what it is so I can understand where things are coming from. Srry if it wasn't clear. Ty though!

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[lone void]

can someone help w this

o wait

can we write [x]=x

no nvm

[winged flare]

squeeze\
$x[\frac{n}{x}] \in (x(\frac{n}{x}-1), x(\frac{n}{x}))$

[winged flare]

very sorry for the badtex

[onyx tiger]

sandwhich

[lone void]

ye i thought of squeeze but i got a bit confused..
x-1<=[x]<=x
does this hold if we put x-->n/x?

[onyx tiger]

how ever one eway is to split the gif as fractional part + integer

[winged flare]

anything as long as both ends approach a finite value

simplifying gives (n-x, n)

[lone void]

ya ok got it then thx

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[small epoch]

What are the positive integer pairs (m,n) such that (3^n-2^(n-1))/(2^m-3^n) is an integer? So far what I found is that there are some that work for n=1 and n=2 but cannot seem to find ones for higher n

[small epoch]

I also asked this question in #help-7｜zen1thxyz for anyone interested

[gray mulch]

m = n - 1 always works I think

[small epoch]

Other than that cause that’s trivial

[safe radish]

@small epoch Has your question been resolved?

[charred summit]

The only other possible cases are 2^m - 3^n = +- 1

[small epoch]

Can you prove how?

[charred summit]

Nevermind I found a counterexample

[small epoch]

What’s the counterexample?

[charred summit]

(4, 2)

[small epoch]

Oh, other than that

[charred summit]

This is with 1 =< n =< 500 and 1=< m =< 5000

And for n >= 3 the only solution is m = n-1

Obviously not a proof but strong evidence

[small epoch]

Thanks, that’s still very helpful

I have another fraction I want to check when it is an integer. It’s (3^n+2^(n-2))/(2^m-3^n)

[charred summit]

m max 50000 and n max 5000

[safe radish]

@small epoch Has your question been resolved?

[small epoch]

Another fraction too: (13 * 3^(n-2) - 2^(n+1))/(2^m-3^n)

[small epoch]

That’s what I got too by checking small cases

[safe radish]

@small epoch Has your question been resolved?

[small epoch]

I have to go now but thanks for your help

.close

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[vestal onyx]

can you give me a bidmas question

[vestal onyx]

i need help with bidmas

[plucky elk]

✨ google ✨

[vestal onyx]

ok

[quasi bison]

op left

.close

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[harsh crow]

hey, i need some help with the b question, saying:

show that |arctan(u)|<=|u|

that i could do it

but i dont get how to prove it is continuous in 0

which is the 2nd line of the question

f(0)=0 as said in the thing, and then yeah i dont think you need more data

the question in itself is "deduce that f is continuous from the right, as it's continuous in 0"

[plucky elk]

you just insert the inequality |arctan(u)| <= |u| into the integrand in the right side of f(x)

and use f(x) <= |f(x)|

[harsh crow]

like i do the integral of arctan(u)/u ??

[plucky elk]

yes that's the definition of f(x)

[safe radish]

@harsh crow Has your question been resolved?

[harsh crow]

but i dont get how to make it countinuous in 0

like how can i know it is continuous in 0

[fathom adder]

Tu as réussi pour la continuité à droite?

[plucky elk]

yes you asked that already

[plucky elk]

did you do this yet

[harsh crow]

nan je comprends pas comment on trouve comment c'est continu

[fathom adder]

Fais ce que dit riemann

[harsh crow]

yeah i tried but i dont understand what to do just after as i cant do limits inside an integral

[fathom adder]

You don't have to

Just bound it for now

[harsh crow]

im sorry im lost lmao

[fathom adder]

May we see what you have done

[harsh crow]

yeah wait

i gotta re-wright it as i crossed it

[plucky elk]

do this integral

[harsh crow]

but why would it mean that, if this goes to 0, the other integral that is f(x) will be continuous ??

[fathom adder]

Also 0 <= f(x) as x > 0

[fathom adder]

Continuity definition + gendarmes theorem

[harsh crow]

i think i dont get the definition of continuity then

but i just need to do the limit of the integral and it will be fine??

[fathom adder]

lim x -> 0+ f(x) = f(0) is the right side continuity in 0

[harsh crow]

i mean the integral gives x so

it goes to 0 if x goes to 0

[fathom adder]

Yes so ?

Thats what you want right ?

[fathom adder]

See

And the exercice gives you f(0) = 0

[harsh crow]

yeah okay

it's just that i was doubting about how it shows it's continuous

[fathom adder]

And you can do the same logic in x -> 0-

[harsh crow]

i get that yeah it goes to 0 but ig im just bad with continuity

[harsh crow]

no need it's an odd function

so it's logical

[harsh crow]

but okay thanks a lot

[fathom adder]

Oh yeah indeed

[harsh crow]

thanks lmao it's just that i was idk dumb or thinking too much

i know what's continuity but i wasnt understanding how i was showing the continuity of a function using another

[fathom adder]

Know that its just a question of limit here

[harsh crow]

yeah

anyways thanks

i can do the rest

but i may come back later for polynomials lmao

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[hearty spear]

I’ve recently got a book about electronics and I have few questions
Is d just delta
Is Q charge and what is q
And what is U (not union)

[light shoal]

can you show a screenshot?

[hearty spear]

K

[hearty spear]

I can send u link if u want

[light shoal]

nah just paste it here

[hearty spear]

Pasted everything

[median vigil]

they are using Q to represent the total charge which has flowed through a circuit, whereas q is representing the charge of a hypothetical individual particle

U is potential energy, so U is the electrical potential energy of that a point charge of charge q would have at a given point with voltage V

i'm not sure where d is showing up other than in the derivative

[hearty spear]

Derivative?

[median vigil]

have you taken calculus before?

[hearty spear]

Nope

[median vigil]

ok. have you encountered limits yet?

[hearty spear]

Limits of what

[median vigil]

of functions

[hearty spear]

I’m just a student at a school

[hearty spear]

No

I’m in hear 10

Year

[last heath]

I think d is derivative and Delta is change in a quantity

[hearty spear]

*9

[hearty spear]

Ye Ik delta

[last heath]

ok im a bit late but what are you asking rn?

[hearty spear]

He answered all my questions

[hearty spear]

Apart from small d

[median vigil]

ok so the idea of $\odv Qt$ is basically to look at $\adv Qt$ for various sizes of $\adif t$, and notice that as you make $\adif t$ very very close to $0$, a pattern emerges where $\adv Qt$ appears to approach a definite value, which we call the derivative $\odv Qt$

here is a table calculating it for t = 0. You can see that it is approaching 1 = 1 * cos(0)

[flat frigate]

cloud ☁

[median vigil]

this is something you would learn in a calculus class

[hearty spear]

.solved

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[true tree]

i need help with my delta math work

[true tree]

like i dont know how to say it exactly, and no its not a test/exam

[quiet plume]

[safe radish]

@true tree Has your question been resolved?

[safe radish]

@true tree Has your question been resolved?

[azure dew]

use ASA

look carefully at AC and BD

[spice canyon]

do you know isosceles property? Hint: ||(IT has something to do with the angles CAB and DBA being equal)||

[true tree]

never mind guys i got help by my best friends to help me

solve.

[spice canyon]

so it's dome?

do uh

.close

if it's solved

[true tree]

yes

[azure dew]

type .close to end

[spice canyon]

do .close

[true tree]

.close

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[cloud hound]

I didn't read it was a boolean matrix at first so I designed this for standard multiplication

Okay so my proposed algorithm for this is

Supplied with an algorithm multiply(A,B) in O(f(n)) for multiplying nxn matrices,

ALGORITHM power(A, k)
// input: nxn matrix A and power k to raise it to
// output: matrix A^k

B = I // output (identity matrix)
D = A // dummy
p = powerpart2(k)
for i <- 1 to p[0]:
  D <- multiply(D,D)
  if i in p: B <- multiply(B,D)
return B


ALGORITHM powerpart2(k)
// input: positive integer k
// output: array of indices of 2 that partition k, descending order

// determine the highest power
i <- 1
while 2^(i+1) < k: i <- i + 1
P <- [i]
s <- 2^i
// determine lower powers
while i > 0:
  while 2^i > (k - s): i <- i - 1
  P[] <- i
  s <- s + 2^i
  if k = s: break
return P

[cloud hound]

Here's my analysis which I would like to be evaluated:

First I analyse powerpart2 which is executed once
The basic operation is comparison, which occurs log2(k) times in the first while loop and log2(k)-1 times in the second, so we can write that powerpart2is in O(log(k))

Now the main algorithm, multiply is the basic operation, which is executed 2|p| times in the worst case (when all i in p) further, in the worst case, k = 2^m - 1, so |p| = log2(k+1)
Therefore power is in O(log(k)*f(n)) + O(log(k)) = O(log(k)*f(n))

[safe radish]

@cloud hound Has your question been resolved?

[safe radish]

@cloud hound Has your question been resolved?

[ripe wharf]

well there is a more convenient way of exponentiation
https://en.wikipedia.org/wiki/Exponentiation_by_squaring

nvm
I also wonder about "will boolean matrices have more efficient power algoritm"

[safe radish]

@cloud hound Has your question been resolved?

[safe radish]

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[oak cobalt]

I dont know if this is the right server for me to be asking this but...which sci-calc should i choose?
On one hand i could choose the
Caiso FX-570MS which from what ive seen IS allowed on board exams/licensure exams
OR
Casio FX-570ES plus
Which ISNT allowed on board exams and stuff but it looks more.... sophisticated? The orientation atleast.

Are they almost exactly the same?
I couldve sworn the screen is different on the two.

[oak cobalt]

Like look at that, the screen on the ES version looks just slightly bigger and the numbers and stuff looks cleaner. But somehow its not allowed on board/licensure exams.
Do you guys know any calcs (from casio) that has almost the same screen as the 570ES but is allowed on exams and stuff?

[peak estuary]

the only reason to have a calculator is for exams

why else would you get one

[oak cobalt]

Yeah but like i also want to have satisfactory experience while using a calculator.

[azure dew]

i like the first one

[oak cobalt]

The smaller screen?

[azure dew]

im using it for 5 years and got no problems

[oak cobalt]

Do the numbers not looks weird to you?

[azure dew]

the calculation shows □×10ⁿ thing when it goes big

but its convenient

[azure dew]

pretty clean

[oak cobalt]

Oh.

What does MS and ES even mean?

[azure dew]

the integrals just look not familiar compared to plus

[oak cobalt]

Also am i not seeing it but does the ms version not have a = at the top of CALC?

[azure dew]

idk that sry

[oak cobalt]

I can see the solve at the top of calc but not =

[azure dew]

no it has

the alpha mode has =

the picture just shows shift mode

the only thing is MS's integral not look familiar

if your doing a lot of it use plus

[oak cobalt]

Welp, got no choice i guess

Ill just go for the MS one. These exam calculators are very picky.

.close

[safe radish]

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[neat finch]

I need herp for Integration by Substitution plz

[onyx tiger]

take t = 3x+1

[neat finch]

i got this integration, i got the formule but can't understand how to decide which is v or u

[onyx tiger]

can you show the formula

[neat finch]

it is something like that : v'(u(t)) dt = [v(u(t))] ab

to start i heard i should find the v' and the u and then do this part [v(u(t))] ab

[onyx tiger]

🤔

[neat finch]

maybe i don't go to the right direction

[onyx tiger]

oh

you mean

[neat finch]

i don't know if it's the right one we just start this chapter

[onyx tiger]

it is like $\int v(u(x))\cdot u'(x)dx = \int v(t)dt$

where $t = u(x)$

[flat frigate]

scoob

scoob

[onyx tiger]

if you set t = 3x+1 here

what is dt/dx =?

[neat finch]

i don't know if i could answers that, it's what is on my chapter but if i got t=u, what should i do after ?

[onyx tiger]

method you are using is you derivative of something u(t) in the integral you have

your u(t) here is 3x+1

[neat finch]

does that mean v'=1/x?

so i got u=3x+1 u'=3 v= 1/x v'= -1/x^2

[onyx tiger]

mb

[onyx tiger]

you are right

v' = 1/x

[neat finch]

u=3x+1 u'=3 v=ln(x) v'=1/x

[onyx tiger]

indeed

[neat finch]

okey i start to get it but now i got that i should do this part ?

[onyx tiger]

find v(u(t))

then apply limits

[neat finch]

okey so i end up with that

[onyx tiger]

a is lower limit b is upper limit
you have to substitute v(u(upper limit)) - v(u(lower limit))

now substiute

v(u(b)) - v(u(a))

[neat finch]

ln(10)-ln(2) ?

[onyx tiger]

@clever gale what's wrong ?

[onyx tiger]

yes

[neat finch]

so i didn't need to do u' v' or find primitive ?

[onyx tiger]

wdym?

[neat finch]

that the final answerd for this and i don't need to do more

[onyx tiger]

you can do log(a) -log(b) = log(a/b)

[neat finch]

i mean whith ipp wee have to find primitive and i was thinking wee should do the same for the Integration by Substitution, i find 1.609 at theend

[onyx tiger]

yes apart from simplifying you are done

[onyx tiger]

i'm not sure what do you mean by primitive

[onyx tiger]

but yeah this is answer

[neat finch]

ho sorry maybe it's not the same terme, by primitive i mean for exemple a=x what can found a when i derivate it, like x^2/2 -> ' -> = x

[onyx tiger]

that is integrating f(x) = x

may i ask where are you from ?

[neat finch]

france

[onyx tiger]

ok

[neat finch]

welp thanks for help 😄

[onyx tiger]

no worries

[safe radish]

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[cloud hound]

Yeah I just didn't get to sit down and run the calcs to check

[cloud hound]

Anyway I'll repin the original question which was about the anlysis

[safe radish]

@cloud hound Has your question been resolved?

[cloud hound]

no goddamnit

[safe radish]

@cloud hound Has your question been resolved?

[open wedge]

uhhh

do you have a question?

[cloud hound]

Pinned

The question is mostly if the analysis is correct (although if the algorithm won't work that would be good to know first 😆)

[split kayak]

I think the analysis is correct, althought, specifically for boolean matrices, there is a neat optimization

Since the act of "dotting" rows and columns in boolean algebra is basically

Once you reach at least 1 succesful AND operator you can stop the entire operation

Which would reduce the avg time.

Althought there is surely a faster method

[safe radish]

@cloud hound Has your question been resolved?

[safe radish]

@cloud hound Has your question been resolved?

[gray mulch]

There are 2 problems with your algorithm

1. your while loop starts at i = 1 and skips 0 completely, which will be problematic for odd k

[cloud hound]

Ah good point

[gray mulch]

2. even if it did start at 0, you'd get a wrong result, because you square before checking if i in p

It would use a value that overshoots the index by 1

[cloud hound]

Yeah I'd slightly change it in that

Case

[gray mulch]

So for i = 0, D would be A² Instead of A
For i = 1, D = A⁴ instead of A² and so on

[cloud hound]

I'll make the changes (when I get back to computer) and get back to you 👍 thank you! (Ofc if you have anything else to mention you can continue, I'll see it in about two hours)

[gray mulch]

You gotta put it after the if statement (and outside of it of course)

Everything else including the analysis looks good

[cloud hound]

Okay here is the improved algorithm

ALGORITHM power(A, k)
// input: nxn matrix A and power k to raise it to
// output: matrix A^k

B = I // output (identity matrix)
D = A // dummy
p = powerpart2(k)
for i <- 0 to p[0]:
  if i in p: B <- multiply(B,D)
  D <- multiply(D,D)
return B

Since I (somehow) got the worst case analysis right I will also try best case analysis

Ah I see powerpart2 is inefficient (currently the inner while loop executes all the way before checking if k = s)

So it's best case is probably also in omega(log(k))

But we can fix that

ALGORITHM powerpart2(k)
// input: positive integer k
// output: array of indices of 2 that partition k, descending order

// determine the highest power
i <- 1
while 2^(i+1) < k: i <- i + 1
P <- [i]
s <- 2^i
// determine lower powers
while i > 0:
  if k = s: break
  while 2^i > (k - s): i <- i - 1
  P[] <- i
  s <- s + 2^i
return P

Just a lot of simple things with moving the checks in the loops up

Now
In the best case, k is a power of 2, in which case the comparison in the first while loop of powerpart2 executes log2(k) times and the second short circuits

Ah well it's still in $\Omega(\log(k))$

[flat frigate]

Coolempire93

[cloud hound]

So that means $\texttt{powerpart2}(k) \in \Theta(\log(k))$

[flat frigate]

Coolempire93

[split kayak]

Also, are boolean matrices diagonizable?

or some of them at least

[cloud hound]

Now to the other algorithm

[cloud hound]

I'm going to try to write an algorithm specifically for boolean matrices next

I was trying to avoid diagonalizability (in both cases really) since I'm not familiar with it

As before, powerpart2 is executed once

And because k is a power of 2, array p only contains one element
regardless the multiplication operation occurs between p[0]+1 and 2(p[0]+1) number of times, so we are still in log(k)

In conclusion

$\texttt{power(A,k)} \in \Theta(\log(k) \cdot f(n))$, where $A$ is an $n \times n$ matrix

[flat frigate]

Coolempire93

[cloud hound]

Now to see if there's anything special for the boolean case (but if anyone comes by feel free to check my analysis once again)

[split kayak]

On a completely unrelated note, i learn that boolean (logic) matrices are isomorphic to binary relations

And matrix product is isomorphic to composition

[cloud hound]

I mean boolean vectors are isomorphic to sets [in a powerset] so this makes sense

But yeah you can use boolean matrices to check properties of relations

Like the product to check transitivity

Reflexivity along the diagonal and symmetry by symmetry

or antisymmetry

[split kayak]

anyways

Tbh i havent came up with much past the obvious improvement of the "stop evaluating past a succesful AND"

Which as far as my understanding goes reduces avg time from n^3 to n^2.5

[cloud hound]

How did you obtain that analysis

[cloud hound]

Hmmm

In analyzing by hand I obtain an unexpected result

I think it must be wrong

I get $\mqty[A & B \ C & D]^2 = \mqty[A + BC & B(A + D) \ C(A + D) & BC + D]$ which looks good, but then I get $\mqty[A & B \ C & D]^4 = \mqty[A + BC & B \ C & BC + D]$

[flat frigate]

Coolempire93

[cloud hound]

Which would mean that you end up with the same A + BC , B, C, BC + D matrix when you raise the original to the 16th power, 64th, etc

[split kayak]

yeah i think something went wrong

[cloud hound]

Let's check the second term

[split kayak]

the ^2 version is correct if thats what you mean

[cloud hound]

\begin{align*}
A^3B + 2AB^2 + BD^3 + ABD^2 + A^2BD + 2B^2CD &= AB + AB + BD + ABD + ABD + BCD \
&= B(A + D + AD + CD) \
&= B(A + D)
\end{align*}

[flat frigate]

Coolempire93

[cloud hound]

that's weird how did I get just B

[split kayak]

Oh, boolean, mb

anyways

[cloud hound]

Wait B(A + D) was what was in the ^2 matrix

Surely it can't stabilize after just one composition

O.o

[split kayak]

personally i found some matrices that switch between 2 states

,tex
$\begin{bmatrix}
0&1&0\
1&0&0\
1&1&0
\end{bmatrix}$

[cloud hound]

Yeah but an aribitrary one?

[split kayak]

I just tested a few and the two did it

I wouldnt be surprised

Althought i ought to believe some approach a full 1 matrix and others a full 0

[cloud hound]

Yeah some do

I think partial orders approach all 1's

[cloud hound]

Oh this is also interesting the bottom element is just BC

Okay well I guess since any direct rules are difficult let's see what I can do for actually doing the calculation

[split kayak]

becomes

1. a+abc+bcd -> a + (a+d)bc -> a + bcd
2. ab+bc+bd+abd -> b(a+c+d+ad) -> b(a+c+d)
3. bc+ac+cd+acd -> c(b+a+d+ad) -> c(a+b+d)
4. abc+d+bcd -> d + (a+d)bc -> d + abc

[cloud hound]

Yeah that's what I got as well

Somehow just messed up the multiplication by abcd again

[split kayak]

1. a+abc+bcd+abcd -> a(bc+bcd)+bcd -> a+bcd

Lmao.

[cloud hound]

Is this one the 5th power?

[split kayak]

yep

Yeah they seem cyclic

[cloud hound]

Which doesn't make sense because if you have a total order it won't be AH

I see

it's because it's 2x2

$\mqty[1 & 1 \ 0 & 0] \odot \mqty [1 & 1 \ 0 & 0] = \mqty[1 & 0 \ 0 & 0]$
$\mqty[1 & 0 \ 0 & 0] \odot \mqty[1 & 1 \ 0 & 0] = \mqty[1 & 1 \ 0 & 0]$

[flat frigate]

Coolempire93

[cloud hound]

R = {(0,0), (0,1)}
Compose
R(R) = {(0,0)}
Compose
R(R(R)) = {(0,0), (0,1)}

But

R = {(0,0), (1,2), (2,3)}
R(R) = {(2,3)}
R(R(R)) = {}

The eigenvalue of $\mqty[1 & 1 \ 0 & 0]$ is $\mqty|1 XOR \lambda & 1 \ 0 & 0 XOR \lambda| = (1 XOR \lambda)(0 XOR \lambda) \vee 0 = 0 \vee 0 = 0$ supposing these act like standard eigenvalues

[flat frigate]

Coolempire93

[cloud hound]

ALGORITHM booleanmultiply(A, B)
// input: nxn matrices A and B
// output matrix C = AB

// row-by-column multiplication
// let the matrices be indexed A[row][column]
for row <- 1 to n:
  for col <- 1 to n:
    C[row][col] <- 0
    for k <- 1 to n:
      if AND(A[row][k], B[k][col]): 
        C[row][col] <- 1
        break
return C

I suppose the brute force method looks something like this

[cloud hound]

Taking the basic operation to be the AND operation,
In the worst case we never short circuit (all 0's matrix) so $\texttt{booleanmultiply}(A,B) \in O(n^3)$ and in the best case every value short circuits (all 1's matrix) so $\texttt{booleanmultiply}(A,B) \in \Omega(n^2)$

[flat frigate]

Coolempire93

[cloud hound]

Thus we have $\texttt{power}(A,k) \in \Omega(n^2\log(k))$ and $\texttt{power}(A,k) \in O(n^3\log(k))$

[flat frigate]

Coolempire93

[cloud hound]

Can anyone verify if this looks right?

[cloud hound]

In the meantime, I'll now attempt the optimization based on boolean matrices as relations and matrix multiplication as composition

Recalling that (AB)[i][j] = 1 when there exists 1 in A[i][k] and 1 in B[k][j], i.e., i ~ k ~ j (i relates to k under A, k relates to j under B)

So we can calculate the values in A^k directly from A simply by following along this path (and we will see what the efficiency is like)

Actually now that I begin to write this out I realize this quickly devolves into a graph search

The composition essentially creates a digraph where we start at the ith root and check each path of length k to see if it ends at j

A* won't work here because there's no heuristic for how 'close' or 'far' you are from j

I assume DFS would work better than BFS but I'm not sure

In terms of time both have O(V + E) complexity

In the worst case, we would have an all 1's matrix

So each i relates to every j, and so on

Gives us $n\binom{n}{1} = n^2$ edges and $k*n$ vertices so $\in O(n^2 + kn) = O(n^2)$

[flat frigate]

Coolempire93

[cloud hound]

Which is faster than the other algorithm in the best case

[cloud hound]

Used your idea to come up with an even better algorithm

Essentially looks like

[split kayak]

Is there any actual meaning for the det of a boolean matrix?

[cloud hound]

No clue

I guess for 2x2 it tells you if it's reflexive or if 1~2 and 2~1 (the opposite ig xd), and then for higher using the minors if it's all reflexive or 1~2 and 2~1 and 3~2/2~3 and 3~1

Ah

It tells you if it has a permutation embedded 😂

Useless info

At least for a relation

Knowing if it's 1~1 and 2~2 (det 1) vs 1~2 and 2~1 (also det 1) is important

And it could even be a combination

I guess in general it tells you if the digraph of the relation has all elements involved in (some form of) cycle

But it doesn't inform transitivity because you need all of the edges to form cycles

And det 0 doesn't mean it's antitransitive either

Not even that it's not transitive

Wait I'm thinking of symmetry

I'll have to think about transitivity in a secon

Let me finish the algorithm first 😂

ALGORITHM booleanpower(A, k)
// input: matrix A, power k to raise it to
// output: matrix A^k

// to determine the value of A^k[i][j] 
// we attempt to find i ~ l1 ~ l2 ...  ~ l(k-1) ~ j
// storing relationships in an array along the way for cache access in later steps (this represents each 'layer' of the graph)

G = // all 0's kxnxn matrix
B = // nxn output matrix

// DFS
for i <- 1 to n:                // row in A^k
  for j <- 1 to n:              // col in A^k
    M = // k-length array of 0's
    M[1] = i
    l = 2                       // relationship #
    while 0 < l <= k:
      for m <- M[l]+1 to n:     // search until find i ~ m
        if A[M[l]][m] == 1:
          G[l][M[l]][m] <- 1    // mark M[l] ~ m at lth step
          M[l+1] <- m
          break
      if M[l+1] == 0: l <- l-1  // M[l] relates to no elements
      else: l <- l + 1
      if l = k+1:
        if M[l] = j: B[i][j] = 1
        else: l <- l - 1        // go back, path not found

return B

Okay I just quickly finished the algorithm without using the graph idea (pretend those lines aren't there for now, since the graph is never read)

Essentially brute force DFS reading through the matrix A every time

Supposing as usual that the algorithm is correct (hopefully it's stll readable enough for you to debug), our analysis proceeds

Basic operation is checking if A[M[l]][m] == 1 (that is, if M[l] relates to m under A)

Using the all 0's matrix as a worst case, we can see that the inner inner for loop runs the max amount of times

Because no path will every be found

No

That's not necessarily worst, it'll short circuit after the first round

Not sure it's possible to engineer a matrix A such that the inner inner for loop runs the maximum possible amount of times every time (that is, it has to check every single path and the one to reach j is always last)

But supposing it is

The maximum number of times would yield $\frac{n(n-1)}{2}$ basic operations, the while loop would run through all $k-1$ values , and then we multiply by $n^2$ for the outer for loops, yielding $\texttt{booleanpower}(A,k) \in O(k\cdot n^4)$

[flat frigate]

Coolempire93

[cloud hound]

But again I doubt this is possible

It's funny because both the all 0's and all 1's matrices for this are great cases

For the all 0's, you get n^2*k operations, and for the all 1's, you get n^2*k operations

Damn that's still worse than the other one though

Have I determined these two correctly?

[safe radish]

@cloud hound Has your question been resolved?

[safe radish]

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[lone void]

i got the stars and bar method confused

[lone void]

for
x1+x2+x3+x4 = 5

is it 5+4-1 C 4-1 or 5+4-1 C 5-1

[naive dragon]

the first one

[lone void]

ok so in the formula n+r-1 C r-1

r is no. of objects?

[fiery stratus]

yes

[naive dragon]

yeah

you can just derive the formula

[lone void]

alr got it thanks a lot

[naive dragon]

I never memorize it tbh

[lone void]

NO THANK YOU GOOD SIR

[naive dragon]

lmao ok

[lone void]

try taking chemistry

.close

[safe radish]

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[lone void]

also thanks for the help

[safe radish]

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[covert gale]

Would it not be √0?

[covert gale]

√(ii) + (11) = √-1+1 = √0

[cloud hound]

There's a video on this

[covert gale]

Yes, that's where I got the problem, but I still don't understand how it is √2

[cloud hound]

Ah

Well that would be because we're putting it down into real space

"i" isn't the length of the side

|i| = 1 is the length of the side

Consider if you drew the triangle in the opposite direction

The sides would be -1 and i

But the length of -1 is |-1| = 1

And the length of i is |i| = 1

Now you can apply pythagorean theorem

[covert gale]

Ah, I understand now. Thank you

.close

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[prisma wren]

<@&268886789983436800>

[safe radish]

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[proven halo]

I need help with this question here;
Factor the given polynomial completely

• I'm stuck at trying to find the greatest common factor and cannot get to the next step.
• Second problem is factoring afterwards

[dry forum]

just use the quadratic formula

[proven halo]

I did but I got stuck at a certain point

Let me do it rq

[onyx tiger]

Don't you see a common factor

In coefficients

[proven halo]

ohh wait 3

yipee

[dry forum]

ncie

[proven halo]

i just need to apply it mainly

the way i wouldve done it is using the AC method but factoring gets more annoying

270 from multiplying and wouldve had to find the gcf from that (my main point where I'm stuck at)

[onyx tiger]

Where is 270

[proven halo]

x^2+51x+270

[onyx tiger]

Another question?

[proven halo]

i think im just mixing different methods together and getting confused

no

its 18x^2+51x+15

18*15

[onyx tiger]

What is AC method

[onyx tiger]

Factor 3 first

[proven halo]

ok

so i wouldnt multiply anything yet?

[onyx tiger]

What is the factored polynomial

[proven halo]

im at 3(6x^2+17x+5)

it woul be 6x^2+17x+5

we ignore the 3 for now

[onyx tiger]

Solve quadratic inside bracket

[proven halo]

so division again?

wait nvm

im stuck

ohh ok i see

its what i had done earlier

but NOW i do have to the AC method

ok ok so uh

i know to multiply a*c which is 30 and match b
which is 15+2 = 17

mb i put 7 instead of 2

[proven halo]

and 15*2=30

(2x+5)(3x+1)

got it

:D

wait

Shoot

i fogro the 3

YAYYY

.close

[safe radish]

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[shy skiff]

Hi I currently have a math homework on Forces and Friction, I wasn't in for the lesson as I had an exam and my maths teacher gave us this homework today to give in Tomorrow but I have no idea what I'm doing since I wasn't in the lesson. Could anybody help me? I'm okay with the force pushing up the hill but once the force is horizontal I have no idea.

[dry forum]

can you rotate it please

[shy skiff]

yes ofc one second

[cloud hound]

,rccw

[shy skiff]

oh thank you!

[dry forum]

soln or explaination

[shy skiff]

wdym?

[dry forum]

should i give you the solution or the explaination @shy skiff

[safe radish]

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

[shy skiff]

I need more of hybrid, I need to understand it or I won't be able to do it in exams

[dry forum]

ok so vc or chat ?

[shy skiff]

uh is chat okay?

[dry forum]

okay sure

[subtle tartan]

I bet you are familiar with coordinates and stuff, so here is a hint
(plus you should apply the sum of forces according to each axis)

don't forget the weight btw

[shy skiff]

yes right now I have up to that point, I'm now trying to work out the line perpendicular to the slope. I cant figure out if im supposed to add 10cos30 and 12sin30 or to subtract them

Im not that sure what im actually working out, I missed this entire chapter due to an exam overlapping

[dry forum]

,rccw

[subtle tartan]

well, it all depends of the orientation of your axis
if it's in the same direction then it's +, if it's the opposite it's -
and always friction is the opposite direction as movement so if it's a hill, and the object is falling, the friction will try to get it up and avoid its fall

[dry forum]

@shy skiff does this help?

[shy skiff]

so are the axis relative to the slope? if so how do i know if its in the same direction? the horizontal force is going diagonal relative to the slope's axis

[shy skiff]

umm, not really Its kind of like what my teacher does, i dont really understand the math

[subtle tartan]

you are free to choose whatever direction, after all since it's all relative, then if the arrow is directing up, then everything will change, each + become a - and vice-versa
but in the end you should find the same result,
just make sure the x axis is parallel to the slope so you can do the stuff with cos and sin and so weight doesn't cancel out so you can see its effect really on both axis
you are actually very close to solving it
if the arrow of the force is pointing in the same direction as the axis then it a + else it's a -

[shy skiff]

😭

I genuinely can't understand what im doing

i have my R

[subtle tartan]

dw you are doing great

[shy skiff]

but when it comes to the resolve going up the slope

i dont know what to do, I have an example using a smooth surface

but that was before the friction chapter

This horizontal force is throwing me completely off, I've never encountered it

I think my diagram might be completely wrong honestly

[subtle tartan]

just add a vector f_x and f_y to each equation
and since you will suppose it will stay in its position you will take each equation = 0
and you find f_x and f_y
it is in equilibrium after all

[shy skiff]

Uhmmm... Vector f_x? Idk if that's just another format of writing vectors or something...

I think I may have learning difficulties 😭 I have never been tested for it though because I got through GCSE with 7s and 8s so there was no need but at A level I'm just a D student, I can't do the exam questions because they aren't anything like the examples in lessons

[subtle tartan]

it's smth like this

well I might have smth wrong because I am not totally focusing on it

those are your forces

and you got your axis

do the projections according to x and y

and don't forget you can move your vectors however you like if it helps with project
a vector got an infinite number of representatives you you can move them all to the center, the intersection of the two axis

it might help a bit with that 12N force

and about angles, draw triangles it will help

see? you can calculate each angle

hopefully it helps

[safe radish]

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[pallid karma]

tutorial

[dry forum]

less go

[edgy breach]

look for similar angles

or supplementary

use them to relate the two angles together

[plucky elk]

!nosols

[safe radish]

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

[edgy breach]

its wrong anyway

[dry forum]

i was just confirming if i was right

[plucky elk]

#❓how-to-get-help

[pallid karma]

the answer is 129 but idk how to show my work

like where do I start

I’m allowed to like plug it into desmos and stuff

If that’s needed

[timid ridge]

basically you write words how you get 129

thats geometry

[pallid karma]

hey can I get some real help thanks

[timid ridge]

im being deadass? describe how you got 129

[edgy breach]

relate the angles together

[pallid karma]

The answer was obviously fucking given now the essential reason I made this is to figure out how to SHOW MY WORK

Like I said already

[cerulean berry]

example can be u can point out which angles are complementary

[pallid karma]

anyways can I get some real help from a actual kind individual.

[edgy breach]

you can label some of the angles

[cerulean berry]

and which are like alternate, exterier etc

[pallid karma]

I know the type of angles but what do I do with that

Yea

[edgy breach]

and determine if they are corresponding to each other or supplementary

[pallid karma]

Angle cbe and angle feh are corresponding

But I’m not sure what I do from that

[cerulean berry]

okay so u can say that since cbe = 14x +17 then abe must be 180-cbe

then mention the property used

[edgy breach]

<@&268886789983436800>

[cerulean berry]

and then do that for every step

just write that down till u reach the answer

[pallid karma]

Aren’t they congruent

[cerulean berry]

wdum congruent

[pallid karma]

They both would be equal to each other

[cerulean berry]

abe and cbe?

[pallid karma]

Cbe and feh

[cerulean berry]

yes and how did u figure that out?

[edgy breach]

✅

[pallid karma]

wait I’m smart bye thank you so much.

.close

[safe radish]

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[cerulean berry]

yay np

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}
\begin{Lemma}
Let $a$, $b$, and $c$ be integers, and let $p$ be a prime.
\begin{enumerate}
\item If $p \ \cancel\mid \ a$, then $\gcd(p,a) = 1$.
\item If $a \mid bc$ and $\gcd(a, b) = 1$, then $a \mid c$.
\item If $p \mid bc$, then either $p \mid b$ or $p \mid c$.
\end{enumerate}
\end{Lemma}
\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}
\begin{Definition}[Intersection of sets]
The \emph{intersection} of sets $A$ and $B$ is the set $A \cap B = { x : x \in A \text{ and } x \in B}$.
\end{Definition}
\begin{Theorem}
Assume $A = {n \in \mathbb{Z} : 2 \mid n}$, $B = {n \in \mathbb{Z} : 9 \mid n}$, $C = {n \in \mathbb{Z} : 6 \mid n}$.
Then, $A \cap B \subseteq C$.
\end{Theorem}

\begin{proof}
Let $x \in A \cap B$.
Then, by the definition of the intersection, $$x \in A \text{ and } x \in B.$$
This means that $$2 \mid x \text{ and } 9 \mid x.$$
By the definition of divisibility, we have that
$$x = 2k \text{ and } x = 9l$$ for some integers $k$ and $l$.
So $2k = 9l$. This means that $2 \mid 9l$.
Since 2 is prime, by lemma 2.c, either $2 \mid 9$ or $2 \mid l$.
Because $2 \ \cancel\mid \ 9$, we have that $2 \mid l$ is true.
Then, by definition of divisibility $l = 2m$ for some integer $m$.
So, $$x = 9l = 3(3)(2l) = 6(3l)$$.
Since $l$ is an integer, so is $3l$.
And, by definition of divisibility, $6 \mid x$.
Since $x$ was chosen arbitrarily, it holds for any $x$.
Thus, $x \in C$.
We shown that $x \in A \cap B$ implies $x \in C$, therefore, by the definition of subset,
$$A \cap B \subseteq C.$$
\end{proof}

[flat frigate]

Mor Bras

[safe radish]

@earnest mirage Has your question been resolved?

[earnest mirage]

<@&286206848099549185> 👋

[earnest mirage]

<@&286206848099549185>

[safe radish]

@earnest mirage Has your question been resolved?

[earnest mirage]

Hello

<@&286206848099549185>

[azure dew]

i think its good
im not a helper but

[earnest mirage]

Thanks for your response!

[earnest mirage]

<@&286206848099549185> Could someone else check the proof please?

[solar hazel]

yes it's fine

[fluid ember]

Looks good

[granite quartz]

can u help me

[prisma elk]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[earnest mirage]

Thank for your responses!

.close

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[uneven shale]

Can someone help with this circuit design i js have no idea where to start

[last heath]

did they give you a schematic or did you build this by yourself?

also this isn't really a math question I'd recommend asking elsewhere

[uneven shale]

that image is the only schematic/requirements the teacher gave us

[last heath]

where is the schematic diagram

[uneven shale]

yeah ive asked in a physics server but got no response 😭

[last heath]

that would be helpful

[last heath]

try https://discord.gg/engineeringstudents

[uneven shale]

we don't have a diagram on that

[uneven shale]

okay i'll try asking

[last heath]

also u have to be patient cuz I understand not many ppl responds right away

I asked a circuit question before and they haven't responds for 3 months lol

[last heath]

honestly it seems a bit complicated without knowing the schematic

[uneven shale]

yup 😭😭

[safe radish]

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[safe radish]

@uneven shale Has your question been resolved?

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[shut gust]

HELLO

[shut gust]

ANYONE ONLINE

[light shoal]

do you have a math question?

[solar hazel]

many are

[still charm]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[rigid blaze]

question?

where the question

[safe radish]

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[shut gust]

Anyone online

[shut gust]

Hey

Anyone Oneline

[cerulean berry]

yes?

[shut gust]

Good MOrining

[proud tree]

do u have a question

[still charm]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[shut gust]

I Need Help Solving Binary Subtraction

Like this question

00000011-01100100

Can Anyone guide me how to solve this 00000011-01100100

[quasi bison]

do you know how to do column subtraction in base ten?

[shut gust]

yeah

[quasi bison]

ok and in base two?

[shut gust]

the problem is that the zeor have no carry digit

yeah inew

[quasi bison]

well your binary numbers are meant to be 2's complement signed integers right

[shut gust]

the problem is that the zeor have no carry digit left an leftmost

yeah it is

[quasi bison]

just imagine there is a "1" to the left of the 0000 0011 (the top number)

[shut gust]

-128 64 32 16 8 4 2 1

ok

then

[quasi bison]

then subtract as normal but borrowing from this phantom 1

[shut gust]

Wait What How will it work will it not ruin the answer

[quasi bison]

that, or turn 0110 0100 into its negative & then add to 0000 0010

[quasi bison]

it won't.

[shut gust]

Ok

[quasi bison]

it'll give you the correct signed binary representation

[shut gust]

Thx

Your Thousand Thank you can you accept my friend request

[quasi bison]

you're welcome but no sorry

[shut gust]

ok no problem see yaa

i just asking for help using it but if you dont want its ok bye buddy

what is the denery value of 1111000 in 2s complement

.close

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[shut gust]

.close

.close

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[safe radish]

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[oak cobalt]

When you get the value of an intergral, do you get the radii of the sum of all of the dx right??

[oak cobalt]

Like you dont get the whole circle thing just the radii of it.

[burnt notch]

What do you mean with radii of the sum?

And what circles are you talking about??

[oak cobalt]

Like do you only get this

Or do you get this whole thing

[burnt notch]

With what integral?

The usual integral of a function from a to b?

[oak cobalt]

Yeah

I think

Wait

[burnt notch]

Then yes, you get this. The area between x=a, x=b, the function and the x-axis

[burnt notch]

????

[oak cobalt]

Oooh, for a second i thought you get like the whole thingy majiggy

[oak cobalt]

Thats what i used

[burnt notch]

Yeah but that one is a totally different integral

And still, why were you talking about circles?

[oak cobalt]

Maybe advanced studying isnt for me i think i mightve skipped a few too many

[burnt notch]

Yeah I have such a hunch too

[oak cobalt]

Hold on can i send links here

[burnt notch]

Cover the basics well, then you can go more advanced

[opal fractal]

shouldn't pi be multiplying the function squared for it to be an integral of volume of revolution?

[burnt notch]

Yea

[onyx tiger]

that would be rotating it by pi with respect to x axis and adding

[oak cobalt]

This is what i based it off of, thought i understood it and i tried to recreate the graph they showed as best as i can.

https://vt.tiktok.com/ZS5Tcy2wN/

[oak cobalt]

To be fair...i wasnt even sure if what theyre saying is correct or not.

[burnt notch]

Yeah ok I get now what you were talking about

I can't open the link cause I don't have tik tok unfortunately

But I believe they were saying something correct

[burnt notch]

Also notice this @oak cobalt

[oak cobalt]

So its just what i created on desmos that was wrong?

[burnt notch]

Without knowing what you wanted to do, it's hard to say

[opal fractal]

the definite integral you wrote on the first line will just give you a number, not plot the cone

if you wanted to plot the cone, you should use a 3d graphing calculator

[oak cobalt]

Do you know any?

[opal fractal]

desmos has one

[oak cobalt]

What i was trying to know was if THIS was getting the correct value of THAT function

[oak cobalt]

Man, maybe im not on that level yet...

[burnt notch]

Well...

The first one is a NUMBER

The second one is a FUNCTION

So they're totally different things

[oak cobalt]

Well then...what should i put to get the area under than function?

[burnt notch]

And that integral neither gives you the area enclosed nor the volume of the solid obtained by revolving f around the x-axis

[burnt notch]

Simply the integral from a to b of f(x)

[oak cobalt]

So like this?

[opal fractal]

yep

[burnt notch]

Exactly

[oak cobalt]

Is this still correct? Or do i change something on the first one?

[opal fractal]

the first line will give you the area between 0 and 5 for any function f(x) you write, so you don't need to change it