help-13

[fallow drift]

so like

is this right?

to use comparison theorem:

1. look at the leading degrees of xs in numerator and denominator
2. break the integral up so that you eventually get 1-inf for the bounds and a rational (consisting of leading degree xs of numerator and denominators)
3. use comparison theorem

?

how

[night wedge]

Well you can bound it by f(x)=1

[fallow drift]

hold on

[night wedge]

Integral of 1 over (0,1) is 1

[fallow drift]

if i integrate 1/x^2

and the bouns are 0-1

[night wedge]

It will not be convergent

[fallow drift]

yea.....

[night wedge]

It’s not like the other part of the graph which is unfortunately

[fallow drift]

s like

how would i do it instead

.close

[cedar kiln]

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[cursive shore]

Did I do the correct thing here?

i was a bit confused on how i would find the second unit vector

[hot cipher]

Have you learnt what the cross product means in 3D Space? (hope so)

[last apex]

The second vector wouldn't be a unit vector

[cursive shore]

yes i think so

[last apex]

Luckily there is a different constant multiple you can use

[cursive shore]

why not?

[last apex]

Mutiplying a vector by 2 multiplies its magnitude by 2

[cursive shore]

so youre saying it would just be the same vector with a larger magnitude, rather than a 2nd new vector?

[hot cipher]

Well, the vector P, result of the operation VxU, is perpendicular to both V and U
But Vector P isnt a unique vector, in fact, there are infinite Vectors that are perpendicular to both V and U

[last apex]

No, you had the correct idea, just used a constant that makes the vector no longer be a unit vector

[hot cipher]

Mostly because normality / perpendicularity is defined solely by direction and not magnitude.

[cursive shore]

ahhh i seee. so how do i find the second vector?

[cursive shore]

so how would i go about fixing it?

[last apex]

Use a different constant multiple.

[hot cipher]

We will go with the math later, but imagine that the purple vector is the unitary perpendicular you first found
And the orange line is the direction of all vectors perpendicular to U and V (3d space btw)

Do you see where the other unitary will land?

[cursive shore]

would it land going downwards?

[hot cipher]

Yes, and specifically, if Purple has magnitude = 1, what would be the constant you should multiply with to get the other vector?

[cursive shore]

-1

[hot cipher]

Perfect, the two unitary perpendicular vectors of any UxV have both magnitude 1, and are opposite to each other

If you find any vector result from UxV, then you can construct them both

Thing is, UxV will usually lead to a vector with magnitude =/= 1

[cursive shore]

ahhh i seee🤔

[cursive shore]

that makes a lot of sense

[hot cipher]

remember that a normalized vector and a normal vector are different concepts
ill take it you know how to normalize a vector and make it unitary

[cursive shore]

umm well maybe we use different terms at my school, but what do you mean by normalize?

[hot cipher]

example:
i give you the vector (1,5,2), whats the unitary vector with the same direction?

[cursive shore]

2(1,5,2)

[hot cipher]

do you know how to calculate the magnitude of a vector?

[cursive shore]

yes

[hot cipher]

What would be the magnitude of the vector i just gave you?

[cursive shore]

sqrt(1^2 + 5^2 + 2^2)

[hot cipher]

so, sqrt(30), right?

[cursive shore]

yes correct

[hot cipher]

Given that the magnitude >1, you have to find a number to divide the vector such that the magnitude becomes 1.

[cursive shore]

and that would be 1/sqrt(30)

[hot cipher]

Now, again, for (1,5,2), whats the unitary vector with the same direction?
-> absolutely awful numbers btw

[cursive shore]

it would be v/1sqrt(30)

[hot cipher]

yeah, you could go ahead and check on your own that the result in magnitude when you distribute the 1/sqrt30 will lead to the magnitude = 1

From this, you already got all the steps.

1. Do UxV = P
2. Normalize P = N
3. Use "N" and "-N"

Try and work the problem again to get the two vectors

[cursive shore]

ahhh okay thanks. Ill try that right now and show you what i get

[cursive shore]

okay i think i got it

[hot cipher]

You missed calculating the magnitude
-> and forgot a sign in -N

[cursive shore]

ohh i see the sign that i missed. But for the magnitude, do you mean i rounded it off incorrectly?

[hot cipher]

$\sqrt{(-1)^2+(-1)^2+5^2} = \sqrt{27} = 3\sqrt{3}$

[wraith dagger]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[hot cipher]

I dont really know where you got the 5.2 from

[cursive shore]

ohh i see what you mean

my calculator automatically turned it into a decimal number rather than 3sqrt3

but 3sqrt3 is more accurate thanks

[hot cipher]

Oohh, its the decimal aproximation of sqrt27, mb

[cursive shore]

no problem. Thank you so muuch for all the help! I have a much better understanding now!!😁😁😁

[hot cipher]

!done

[cedar kiln]

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

[hot cipher]

btw

[cursive shore]

oh sorry

.close

.close

[sacred iron]

Bot is broken right now. This channel is open.

[cedar kiln]

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[mighty shuttle]

so this is for pair of arbitrary sets?

Suppose There existed a uniform distribution on $\Omega$. Then $ \bigcup_{i=1}^{\infty}P({i}=1$. So $\lim_{n \to \infty} \frac{1}{n}=p$. So$p=0$, but $0≠1$.

[wraith dagger]

wai

[mighty shuttle]

I think the wording is horrible

[last apex]

Who said P({i}) is 1/n?

[mighty shuttle]

IT IS

[last apex]

Its right just you need to write it bettet

[mighty shuttle]

it's a uniform distribution

[mighty shuttle]

That's a relief lol, I'm HORRIBLE at probability

[last apex]

Your argument would be that there is no value you can assign to P({i})

[mighty shuttle]

how come

[last apex]

If you try to assign 0, then the sum wont be 1

If you try to assign a positive value, the sum will diverge

[mighty shuttle]

surely I can just use this axiom

as the sets {0},{1}... partion the space, the left is 1

[last apex]

Yes! This is what you contradict

[mighty shuttle]

the right would be $1= lim_{n \to \infty}np$

[wraith dagger]

wai

[mighty shuttle]

This would require p=1/n as n-> infty

[last apex]

I would phrase it as if P({i})=x>0 then 1 is a countably infinite sum of x

Which is clearly false

[mighty shuttle]

fair enough

tq

Another question

I can assume the transitivty of $\supseteq$ right

[wraith dagger]

wai

[last apex]

This is a property

No need to assume it

[mighty shuttle]

I think I'll do this later today

,ti

[wraith dagger]

The current time for math_rocks is 01:05 AM (IST) on Sun, 05/10/2025.

[mighty shuttle]

I'll sleep

tq

(a) is just induction though

Eh, I'll finish (a)

okay

done

thanks!

.close

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[cedar kiln]

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[robust pulsar]

Among all triangles inscribed in a circle with center O and radius R , which triangle has d(O,AB)^2 + d(O,BC)^2 + d(O,CA)^2 with the smallest value (use vector to solve)

[cedar kiln]

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@robust pulsar Has your question been resolved?

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[wind forge]

how to use l hopital rule in this

[wind forge]

??

[rapid flax]

oops

[past wave]

First check whether you can use lh or not

[wind forge]

oh

im getting

0/0

i thought we can use lh everytime we get 0/0

[past wave]

Yes correct

So differentiate the numerator

[wind forge]

thats the problem

i forgot power rule

can u pls simplify

[past wave]

$x^n -> nx^{n-1}$

[wraith dagger]

Duck Man

[wind forge]

can u pls use numbers

it would be much easier

i often get confuse when variables are used

[past wave]

Just think of it this way

The power comes down

And the new power is one less

[zealous plinth]

x²->2x, x³->3x², x⁴->4x³, etc

[wind forge]

dont we have to multply x^n ?

[wind forge]

i got more confused after seeing this

[zealous plinth]

What do you mean by this? We have something of the form f(x)/g(x) -> 0/0 as x-> 2, lhopital tells us this equals lim x->2 of f'(x)/g'(x). f(x)=3x²-x-10, g(x)=x²-4 here, so compute the derivatives here separately

Remember derivatives are linear, so $\frac{d}{dx}(3x^2) = 3\cdot \frac{d}{dx}(x^2)$ too

[wraith dagger]

Crackhex

[cedar kiln]

@wind forge Has your question been resolved?

[cedar kiln]

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[dire radish]

Can someone help me problem 22

[dire radish]

im unsure where to begin

to find the bounds for this problem

can sommeone guide me pls

[frank ice]

Yooo

Sooo

@dire radish identify the two circles

[dire radish]

x'2~y

bro

wait

i need to switch

elybaord

ok

x^2+y^2 = 1 minus the otehr one

on the inside

[frank ice]

((x-1)^2 + y^2 = 1), centered at (1,0) with radius 1

[wraith dagger]

YZYBlueBoy22

[dire radish]

yes

ok

and other one is centered at origin

rad 1

[past wolf]

i need help finding the x and y

[frank ice]

(x^2 + y^2 = 1), centered at the origin with radius 1.

[wraith dagger]

YZYBlueBoy22

[dire radish]

what the

how are you doing that without

the notation

[frank ice]

It's easy

Do you understand?

[dire radish]

ye

so

would my bounds be

0<r<1 and 0<theta<2pi

[frank ice]

Do you know how to subtract the second from the first to eliminate y/2?

[dire radish]

uhh

yea maybe i think

idk

💀

[frank ice]

[
x^2 + y^2 - [(x-1)^2 + y^2] = 1 - 1
]

[wraith dagger]

YZYBlueBoy22

[dire radish]

ohhh ok

[frank ice]

[
x^2 - (x - 1)^2 = 0
]

[wraith dagger]

YZYBlueBoy22

[dire radish]

[
x^2 - (x - 1)^2 = 0
]

bro

[frank ice]

You need to expand (x-1)/2

((x - 1)^2)

[wraith dagger]

YZYBlueBoy22

[dire radish]

2x+1=0

[frank ice]

[
x^2 - (x^2 - 2x + 1) = 0
]

[wraith dagger]

YZYBlueBoy22

[frank ice]

Yessir

[dire radish]

wjay would i do next?

im unsure

[frank ice]

[
2x - 1 = 0 \implies x = \frac{1}{2}
]

[wraith dagger]

YZYBlueBoy22

[frank ice]

You have to do this

[dire radish]

ok

[frank ice]

Now substitute (x = \frac{1}{2}) into Circle 2 to find (y)

[wraith dagger]

YZYBlueBoy22

[frank ice]

[
x^2 - x^2 + 2x - 1 = 0
]

[wraith dagger]

YZYBlueBoy22

[dire radish]

into x^2+y^2=1?

[frank ice]

No.

[
\left(\frac{1}{2}\right)^2 + y^2 = 1 \implies \frac{1}{4} + y^2 = 1 \implies y^2 = \frac{3}{4} \implies y = \pm \frac{\sqrt{3}}{2}
]

[wraith dagger]

YZYBlueBoy22

[frank ice]

This

Now, you have to convert intersection points to polar coordinates.

Bro

[dire radish]

yeah?

[frank ice]

Now, you have to convert intersection points to polar coordinates.

[dire radish]

yea dw

[frank ice]

[
x = r \cos \theta, \quad y = r \sin \theta
]

[wraith dagger]

YZYBlueBoy22

[frank ice]

Alright bro

[dire radish]

r = 1/costheta and r = (sqrt(3) /2) over sintheta?

[frank ice]

[
x = r \cos \theta = 1 \implies r = \frac{1}{\cos \theta}
]

[wraith dagger]

YZYBlueBoy22

[frank ice]

Yo bro, I'm going to sleep

[dire radish]

okay

goodnight

thx

[frank ice]

Do you understand or?

[dire radish]

i got to here

[frank ice]

Appreciate it bro

[dire radish]

but imnot sure what to do with it

[frank ice]

Figure it out, tomorrow you are going to show your work to me.

We will check it together.

[dire radish]

uhh tomorrow i cant tomorrow

its ok but thank you

go to sleep

its ok

thx

..close

.close

[cedar kiln]

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[frank ice]

Wait

[dire radish]

?

go sleep bro its ok

i go tit

got

it

[frank ice]

We can finish that

I'm probably going to sleep bro

[dire radish]

ok

np

gn

[frank ice]

My apologies

[dire radish]

dont apologize

nw

thx

[cedar kiln]

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[frank ice]

I'm sure you will figure it by yourself ❤️‍🩹

[frank ice]

.close

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[hearty summit]

heyyy

[hearty summit]

idk how to start with this

[zealous plinth]

This looks like you want to use induction here

[hearty summit]

so maybe

for our base

set k = 0

n =0 sorry

[heavy lichen]

Oke so i learned wat Fibonacci was. Do you still remember that F(n)= F(n-1)+F(n-2)?

[zealous plinth]

Yea yea, thats the easiest base case to check

[heavy lichen]

Lemme open the ()s quickly

[hearty summit]

so ummm f1

= f0^2 + f1^2

1 = 1

[heavy lichen]

F2n + F2n + 2Fn + 1

[dull oxide]

you'll want to use strong induction, to be precise

[heavy lichen]

Im sorry if you didn't understand what i wrote

[zealous plinth]

Right, now assume it works for n (and all n up to it), and try to derive that it works for n+1

[hearty summit]

alr alr

give me a biot to try and figure it out 😭

[cedar kiln]

@hearty summit Has your question been resolved?

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[grand harness]

Can someone help me with this question on static equilibrium

,rotate

[grand harness]

Are the forces correct?

or am i missing anything?

[clear ember]

yeah

[grand harness]

alr. ill try doing more then ill send another pic

Do i start by cakculating torque? @clear ember

torque ccw = torque cw

[clear ember]

yes

well

,, \sum M_A = 0

[wraith dagger]

mmmm7

[clear ember]

where A is the point you take your moment about

[grand harness]

the pivot point?

the pivot point is where Ff is right?

force of friction

[clear ember]

sure

[grand harness]

alr

,rotate

[grand harness]

i feel like im doing this wrong

i need an angle

and i think the distances im using might be wrong as well

im so confused

@clear ember

[clear ember]

😭

where did you get the numbers from

like F_g * 4???

[clear ember]

correct

all of them should be in terms of theta

idk where you got 4 and 5 and all of that

when the ladder itself is only 3 meters

[grand harness]

it says the mass is 1m from the top. so thats 4m form the pivot. i think thats prob wroing tho

[clear ember]

but the ladder

is only 3 meters

[grand harness]

ohhhhh shoot. i misread it. i thiugh the ladder was 5m long. cause the questions i was doing befre this it was 5m lol. mb

can you do the equation part for me. the counter clockwise torque = clockwise torque part. cause thats what im stuck on

[clear ember]

yeah but still

what would pink be?

should be basic trig

[grand harness]

oh am i supposed to use the horizontal distance?

[clear ember]

yes

the "distance" has to be

perpendicular to the force

since the force is directed downwards

the distance is the pink

[grand harness]

i cant get the distance for pink

i need another angle or distance

it would be cos θ = x/2

but i dont have theta

[grand harness]

ohh i remember reading that. that makes more sense now

[clear ember]

well

u need to write it int erms of theta yeah

so for the block it's just the force the block exerts * 2 cos(theta)

[grand harness]

WAIT I THINK I GOT IT

is it Fgsinθ(2m)?

cause Fgsinθ would be the perpenducar force

[clear ember]

😭 no

[clear ember]

purple is the line of action of the force

so the pink is the "distance" we care about

which as you mentioned is 2cos(theta)

F_g (2 cos(theta)) is then what we should have

where F_g is the force exerted by the block's weighted

[grand harness]

wouldnt Fgsinθ times the hypotnuse distance be the same as this?

[clear ember]

Fg sin(theta) is purple though

we want pink

purple is the line of action of the force

the distance "vector"

must be perpendicular to it

pink is perpendicular to purple

[grand harness]

wait look at the triangle i drew on top of the box. is that correct? cause i remember my teacher doing something like that

i think theres two ways to do it. i remember my teacher saying that. ill just do your way then

[clear ember]

if u want to

move the triangle there

then u have to do some geometry

😭 to figure the angle out there

[grand harness]

the angle is the same as the bottom one isnt it?

[grand harness]

its fine. ill just use this

[clear ember]

this one is theta

also i don' tsee

[grand harness]

ill try doing the rest as well

[clear ember]

this one is 90 - theta

[grand harness]

ohh ok

[clear ember]

so not really no

but yeah

find line of action of force

[grand harness]

,rotate

[grand harness]

am i cooking now?

@clear ember

[clear ember]

no

😭

well the left side is fine

the right side no

[grand harness]

😭 . im cooked

[clear ember]

draw the line of action of F_w

u want the vertical distance now

since F_w's line of action is a horizontal line

[grand harness]

whats the line of action?

[clear ember]

take the force "vector"

and just draw a line along it

that's the line of action

of the force

line of action

[grand harness]

wait so if the force is in the x direwction i use vertical distance. and y direction i use horizontal distance?

[clear ember]

so the distance is a line from the pivot

perpendicular to the line of action

[grand harness]

oh ok

[clear ember]

so you want to find green here

[grand harness]

ohhh that makes sense

[clear ember]

this is also why F_f doesn't generate any moment about the pivot

[grand harness]

so 3sinθ?

[clear ember]

yes

F_w (3 sin(theta))

[grand harness]

alr

lemme try solving it now

gimme a min

to find Fw, would i do sum of all forces equals 0?

[clear ember]

yes

it's alway sstandard

to do all 3

sum of force along x = 0

sum of forces along y = 0

and sum of moments about a point = 0

[grand harness]

alr

[clear ember]

anyway i have to go

but i think it's really

all just math from here on out

u can probably do it

[grand harness]

yah i think i got it. gimme a min and ill send u a new pic

@clear ember what do i now?

,rotate

[grand harness]

im stuck at the bottom

wait if i make the sinθ into cosθ by doing 90-θ

so 618.3(3cosθ-90)

would that work?

wait but then i couldnt really factor the cosθ still

[clear ember]

,, \cos( \theta) k = t \sin(\theta) \implies \tan(\theta) = \frac kt

[wraith dagger]

mmmm7

[clear ember]

for some real numbers k and t

provided theta is not 90 degrees

but that's obviously not the case here so we're good

[grand harness]

so what do i do?

at the bottom

[clear ember]

?

😭

[clear ember]

^

this

move the numbers to one side

get tan(theta) on another side

by dividing by cos(theta)

so you'll have tan(theta) = something

for which u can solve for theta

[grand harness]

oh ok

,rotate

[grand harness]

like this? @clear ember

[clear ember]

yes

[grand harness]

ohh ok tysm for the help

[clear ember]

well

it's tan(theta) = number

not tan^(-1) (theta) = number

also i would recommend

writing out the g's

and then computing at the very end

that way you don't propogate uncertainty

throughout your work

[grand harness]

oh ok

[grand harness]

but then i get a super small number like 0.0169. To get the angle wouldnt i have to do inverse tan?

[clear ember]

i mean this is wrong

,, \tan(\theta) = t \implies \theta = \tan ^{-1} (t)

[wraith dagger]

mmmm7

[grand harness]

oh wait the theta is the fraction

[clear ember]

in your case at least cuz we only care about one solution

[grand harness]

ok i get it now

thx

[clear ember]

yeah it should be tan(theta) = number

not tan^(-1) (theta) = number

[grand harness]

alr i fixed it. tysm for the help

[clear ember]

[grand harness]

cya

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[pseudo merlin]

why do they dot product it

[pseudo merlin]

nvm..

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[fathom coyote]

Hello everyone, I am Turkish and I am in my last year of middle school, that is, in the 8th grade. Can someone write me a dm and explain the EBOK topic?

[crimson delta]

well not sure what EBOK is cause google is unhelpful but this isnt how the server works

ask questions on here

[cedar kiln]

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[fathom coyote]

no 🙁

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[livid tundra]

How can I approach the following problem :
If L2 is regular, L1 U L2 is regular, and L1 n L2 is regular, then L1 is regular.

[wise quail]

What is L_x?

[chrome elk]

Languages

[chrome elk]

You have NFAs to generate those 3, could you modify them somehow to generate L_1?

[cedar kiln]

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[buoyant jungle]

v is the "mid point of the segment" yeah?

[split ice]

which segment exactly?

[digital cliff]

i dont think theres anything to indicate its the midpoint of anything at a glance

[carmine bronze]

If it were, there would probably be tick marks to indicate equivalent segments.

For sr, t is a midpoint because it has tick marks.

[buoyant jungle]

the question is asking me which one of these points if any is a mid point of a segment

[buoyant jungle]

huh

[digital cliff]

youve been spoiled then

the tick marks indicate equal lengths

[carmine bronze]

The tick marks on the line segments for st and tr.

[buoyant jungle]

i see

okok i got it ty

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[edgy sun]

how can you do this without saying x≠0

[edgy sun]

this is a bad example but any time you use u sub, why wouldn’t you have to do that since you are dividing by x

[trail sparrow]

you already kind of assumed x≠0 when you subbed in dx

sub x²dx directly perhaps

[edgy sun]

wdym

[trail sparrow]

you had du=3x²dx and you then used dx=du/3x²

since you divided by x there anyway, you already assumed x≠0

cancelling after isnt the issue

[edgy sun]

right that’s what i’m asking about

how is that valid even if you don’t know that x≠0

[trail sparrow]

how about you club the x²dx

and write it as du/3

avoiding the division

[edgy sun]

right but then you have to plug it back in

[trail sparrow]

what do youmean

[edgy sun]

idek what i’m asking

like when you take that step of dividing du by 3x^2, would you not have to say that x≠0? Otherwise you would be dividing both sides by zero

[trail sparrow]

yes but you dont have to divide

you're not using dx=du/3x²

you're using x²dx=du/3

x² stays with dx

[edgy sun]

OOOOHHHH

I c

doesn’t this logic fail when nothing cancels out though?

[trail sparrow]

it holds here

but you're right about it might failing

[edgy sun]

wdym

[trail sparrow]

theres no error here if you let the x² stay with dx

im not sure about any other cases

cant really comment without seeing it first

[edgy sun]

,,\int(cos(\frac{x^2}{2})dx

err

[trail sparrow]

that doesn't look elementary bro 💔

[edgy sun]

wdym not elementary

maybe that’s better?

[trail sparrow]

you're missing an x and a dx

maybe

[edgy sun]

just a dx

[wraith dagger]

∫ cj ☠

[trail sparrow]

theres no closed form for that i believe

[edgy sun]

idk i’m trying to think of a nice example where it doesn’t apply

,,\int (x^2+3)^8dx

[wraith dagger]

∫ cj ☠

[edgy sun]

i think that works

[split ice]

you could check the bounds and hope that $x \neq 0$ over the interval, as a start
since it does, however, how about this? \as $x \to 0,$ we can take the limit of the integrand to be $$\lim_{\substack{x \to 0 \ u \to -2}} \frac{\cancel{x^2}u^3}{3\cancel{x^2}}$$ which exists and is equal to $\frac{-8}{3}$. i.e. it doesn't matter because you're cancelling, as caspian put it

[wraith dagger]

ηασιβ ♥

[trail sparrow]

what substitution do u plan on using anyway

[split ice]

if there are no cancellations, however, you'll likely have an integrand like $\frac{1}{x^2}$, meaning it should be an improper integral near 0. i suppose in that case you are not evaluating this "limit" i proposed, you're just carrying it through to check for convergence

[wraith dagger]

ηασιβ ♥

[edgy sun]

is there not an issue in this equation?

the domain isn’t restricted here

[trail sparrow]

the division by zero is our algebraic manipulation

that we do symbolically

its not a part of the integral and doesn't affect the continuity

[edgy sun]

How can you count it as a valid method if carrying it out through algebraic manipulation is illegitimate

[trail sparrow]

you dont really even write du=3x²dx

its du/dx=3x²

but we treat it like fractions

we're just

mnemonically expressing

the new differential

does that make sense..

[edgy sun]

how would you get that in a form such that we can utilize it in the original equation?

[trail sparrow]

its not formal algebra

but the result holds

so we use it like that

its like a convenient abuse of notation

[split ice]

the same reason you can divide/multiply by dx, even though it's infinitesimally small
essentially, what we are doing is actually a simplification of some deeper construct that just so happens to work out like this, and we formalize it properly in real analysis/diff geo. i must confess i am not adept enough at real analysis to explain a proof to you

it's a little bit like how you're taught a seemingly magic method of multiplying numbers in primary school, then you can go back in high school and justify it

[edgy sun]

oh damn

[trail sparrow]

not adept enough in real analysis either

but yes it works

[edgy sun]

“trust me bro”

i lowkey wanna see a proof just to see if i could reason it out

[split ice]

we can also think of the u-sub in the multivariate sense, i.e. we are taking the xy-coordinates and stretching them into uy-coordinates via a linear transformation. there is some byproduct to "balance out" the stretching; in multivariate, it's the determinant of the jacobian matrix, and i think if you worked it out for the 1x1 case you would get the u-sub back

perhaps i will do that as an exercise later

[edgy sun]

do you happen to know anyone who could show me a proof?

[trail sparrow]

ther was this calculus manifolds something book

it has the rigor behind it

i think

not too sure

[split ice]

trying to find one bc im curious myself now - calculus of manifolds probably has one because its on the cusp of diffgeo, i think i saw a proof at one point that treated dy/dx as a quotient of differential forms

[edgy sun]

i need someone to be able to explain it to you then dumb it down for me lmfao

[barren iron]

altering a function at finitely many points doesnt change its integral over an interval because it is equivalent to the area. So even if you're using a substitution that isnt defined at a finite number of points in the interval it doesnt matter? i think this makes sense?

[edgy sun]

yah but if there is a fundamental issue with how you go about u substitution, how do you know it’s valid? (I.e getting du in terms of dx)

i’m not sure how exactly that applies to the area

[trail sparrow]

trust me bro ahh

[edgy sun]

this is why i hate how they teach math in school

it should follow logically instead of making assumptions to fill in later

[trail sparrow]

atleast its not like physics

[split ice]

i have definitely felt this sentiment, but if you keep going with it then you are going to start elementary-school math with "we are working in the semiring N, which satisfies these ring axioms but not these ones...."

[trail sparrow]

they be treating the dx with such disrespect

[edgy sun]

physics is even worse

[edgy sun]

Well i belive it to an extent. If the teacher is like “We are learning 1 2 3s,” and a student is like “What’s in between 2-3” i don’t think that knowledge should be withheld from the until later ykwim

[trail sparrow]

this man wants to teach real analysis to 3 year olds

[barren iron]

if the u substitution is invalid only at a finite number of points in the interval, you can just split the integral into parts at those points and add it, integral at a particular point is just 0, Like say the interval is [a,b] and ure using a substitution thats not defined at c,d in [a,b] (a<c<d<b) then just integrate over (a,c) , (c,d) , (d,b) where it is valid and add them

[trail sparrow]

"nooo don't treat du/dx as a fraction it only works 100% of the time" ahh

[edgy sun]

I c what you are saying, but this whole thing seems like a HUGE generalization. Plus it’s not necessarily saying the function isn’t defined at x, it’s saying that the way we manipulate the function makes the computations flawed, since randomly we find that x≠0, despite the fact that the function may be defined at zero

Like the thing with math is every step shown should logically follow from the one before it. If we divide by zero, how do we keep going without saying the method of solving it is illogical

[barren iron]

We never divide by zero; if a substitution would do so at isolated points, we split the interval there and apply the substitution only where valid. This is justified because modifying or excluding finitely many points doesn’t change the integral.

[trail sparrow]

that kind of sounds like cheating but i dunno how to put it

[split ice]

This isn't the best answer, but:
Consider $$\int 2x^3 ; \dd x = \frac 12 x^4 + C.$$ We're going to use the reverse chain rule, i.e. $$\int f\big(u(x)\big) \cdot u'(x) ; \dd x = F\big(u(x)\big),$$ where $F$ is the antiderivative. \
Split the integral into $\int 2x^3 ; \dd x = \int 2x(x^2) ; \dd x.$ In our case, we will let $f(\alpha) = \alpha$ be the identity, and $u(x) = x^2.$ Then $f\big(u(x)\big) = x^2$ and $u'(x) = 2x$, so $$\int 2x(x^2) ; \dd x = \int u'(x) \cdot f\big(u(x)\big) ; \dd x = f\big(u(x)\big) \cdot \underbrace{u'(x) ; \dd x}_{\dd u} = F(u(x)).$$ Now, $F(\alpha) = \frac 12 \alpha^2 + C,$ so $F(u(x)) = \frac 12 x^4 + C$ as desired. \
; \
This is the machinery behind the notational steps. As I said, we refer to $u'(x) ; \dd x$ as simply $\dd u,$ thus it's easy to see that $\dd u = \dv{u}{x} \dd x$ - again, a notational "representation of the truth." Then when we subbed $u$ into the integral, we needed to find a $u'(x) ; \dd x$. Instead of inspecting around for one, in practice we simply sub in an expression that works for almost all $x,$ i.e. $\dd x = \frac{\dd u}{2x}.$ This allows us to "pick out" the $u'(x) ; \dd x$ expression easier, which we simply replace with $\dd u,$ but \textbf{no actual division is happening.}

[wraith dagger]

ηασιβ ♥

[trail sparrow]

yes theres no

division

the notations are used that way

[split ice]

I guess your main concern, however, was that this division doesn't work for 0, so why use it? difficult question to answer, but because it works for finitely many points, it is good enough of a "net" to catch this expression out, then what we actually end up with is $f\big(u(x)\big) \big(u'(x) ; \dd x\big)$, or simply $f(u(x)) ; \dd u$. If you actually wanted to worry about the validity, though, you would fall back to the intial "machinery"

[wraith dagger]

ηασιβ ♥

[split ice]

in hindsight, should've just taken a pic of my chalkboard 🙃

[edgy sun]

but if we apply that couldn’t we end up with a lot of wacky answers?

like if you are doing trig proofs, and you come across algebraic manipulation that is impossible, then it’s probably not correct.. If we use this logic and just say it’s “good enough” doesn’t all of that also fall apart?

[trail sparrow]

its not algebra, we aren't dividing anything, the basic principle falls back to the reverse chain rule as he said

we're using the notations to figure out the function involved in the anti derivative

[split ice]

the function we find is still satisfied along all of R, just not at these finite points. it can be formally shown that no function exists that is "wacky enough" to give you a wrong answer, given that constraint

shown how? take a limiting process over the rectangle $f(x_i) \Delta x_i$ underneath that point, or split the integral up

[wraith dagger]

ηασιβ ♥

[split ice]

and use the fact that a function that is discontinuous at countably many points is Riemann integrable

[edgy sun]

are we also not multiplying anything when we go from du/dx=2x to du=2xdx

[trail sparrow]

i think we somehow employ the chain rule, du/dx isnt a fraction but we're allowed to treat it that way through the chain rule

integral of f(y) * dy/dx wrt x can be shown equivalent to the integral of f(y) wrt y

but it looks as though we cancel the dx

[edgy sun]

maybe i should just trust bro 🥀

[trail sparrow]

it can be shown that integral[f(y)*dy/dx dx] is equal to integral[f(y) dy]

we never cancel the dx

[trail sparrow]

🥀

[split ice]

we are multiplying limits, i suppose:
if $\dv{u}{x} = 2x$ was our integrand, then $\dv{u}{x} \dd x = 2x \dd x$ is the integrand next to $\dd x.$
$\dv{u}{x} \dd x = u'(x) ; \dd x,$ which we note as $\dd u.$ Then $2x \dd x \coloneq \dd u$ is a sort of notational defintion

[wraith dagger]

ηασιβ ♥

[split ice]

so maybe not multiplying limits, actually - i think you could argue, with a bit more justification needed, that $$\dd x = \lim_{x \to x_1} (x_1 - x),$$ or simply $$\dd x = \lim_{\norm{\Delta} \to 0} \Delta x$$ where $\Delta$ is a discrete difference.

[wraith dagger]

ηασιβ ♥

[split ice]

hence why $\lim_{\norm{\Delta} \to 0} \sum_{\Delta_i} f(x_i) \Delta_i x = \int f(x) ; \dd x$

[wraith dagger]

ηασιβ ♥

[edgy sun]

so is there a way to do it without needing to use this notation?

[split ice]

well, "this notation" is part of the formal definition of the integral (the sum i just wrote), so probably not

unless you would like to jump straight to measure theory and define the Lebesgue integral

[flat mica]

you can use sup and inf instead, but you'd need to.do a little resear ch because its not usually taught in calc i. it isnt more difficult though

[edgy sun]

sup and inf?

[flat mica]

supremum and infimum, least upper bound and greater lower bound

[edgy sun]

Idk what’s happening

i think i’ll prob just trust

[abstract breach]

You should

[flat mica]

as i said you'd need to look up those concepts if you wanted to use them

good news is

mathematicians have ironed out all the handwaving and contradictions from calculus

it took a while, and wasnt done by the inventors of calculus, but we did it

also, to your original question

the value of the integrand at a single point never changes the value of an integral, this can be proved

so you can just ignore x=0

if you'd like

[edgy sun]

right but i was arguing that if there is a point in the computation where you have to do something sketchy, that the whole process would have to be reviewed as true. For example if I had a binomial function that couldn’t be factored with the quadratic formula, the quadratic formula would have to be re-evaluated as true

[flat mica]

every quadratic is factorable with the quadratic formula if you use complex numbers, they just don't like using them in calc i curricula

maybe i missed your point

look though, you had

x²dx=du/3

theres no division here

replace x²dx with du/3

[edgy sun]

I’m just showing that if a formula doesn’t work for one instance, it’s truth has to be re evaluated

[flat mica]

which formula are you suspicious of

[edgy sun]

are there not instances where this doesn’t work tho? Where x doesn’t cancel out in any way

[edgy sun]

None. Just generally

I’m pretty sure a fluid dynamics formula got proven wrong the other day

[flat mica]

sure, then integration by substitution is unhelpful. but it isnt false

[edgy sun]

mmm

ty

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[wary forge]

Do I need to prove that each term on the right is less than or equal to the term beneath it?

[dire geode]

yes

[wary forge]

I'm having difficulties with the second inequality

I can show that liminf(x_n-y_n) </= limsup(x_n)-liminf(y_n)

haven't been able to do it for the added terms though

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[naive hill]

Could someone help me with the table, i don't understand it

[naive hill]

inequalities

[dire geode]

which part exactly?

[naive hill]

like how the table is structured like that, why it's a - - on the left side and ++ on the right side, etc

[dire geode]

the table shows sign changes

when numbers x are less than 3, the sign of x-3 is negative

another way to say that is if x < 3 (numbers x are less than 3) , then x - 3 < 0 (sign of x-3 is negative)

try the rest of the row for (x-3) and the next row (x-4)

[cedar kiln]

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[naive hill]

ohh i see

[cedar kiln]

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[pseudo merlin]

How do i do c

[violet flume]

do you understand what its asking

[wraith dagger]

jan Niku

[cedar kiln]

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[mighty light]

clearer photo:

Hello I'm a little stuck on this question a) could I have some hints please?

just confirming if it's okay to do this and substitute then solve?

[calm sierra]

yes

[mighty light]

okay thank you

.close

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[faint dirge]

given this g(y) (sawtooth from 0 to 1 period 1, up for 0.9s and down for 0.1s), how would I solve
$\int\frac{1}{-y+4g(y)}dy$
?

[faint dirge]

a sawtooth like this (without the offset by y-0.2)

[wraith dagger]

Astral

[faint dirge]

oh wait I may be dumb

.close

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[gray island]

Is this reduced to the exact form of differential eqn or it can be reduced furthermore?

[past wave]

this is correct

[gray island]

.close

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[tight widget]

Can I get some hint please. My smooth brain can't figure out.

[crimson sedge]

it's b

[tight widget]

How lol?

[crimson sedge]

okay so

using power of circles, we get rp=rp'=half of pp'

let me write how

rp²=ra×rp
rp²=rp'²
rp=rp'

then ofc pp' is the sum of both the two

so since theyre equal and add up to it theyre half of pp'

and since most options have pp and rs

your best bet is to go with rs²-ab²

then use the formula

and chop up rs into it's constituents

and it'll become ra=bs

try it

[tight widget]

Ok

[tight widget]

I didn't saw this part. Thanks for it.

[crimson sedge]

np now continuing from there

now repeat with the opposite side

sq²

youll find ra=sa

[crimson sedge]

now since ra is bs, put the value of ra as bs

in the same chopped up equation

youll find 4(ra×rb)

now think of what ra×rb can be written as

at least guess😭

[cedar kiln]

@tight widget Has your question been resolved?

[tight widget]

My brain hurts lol. But this is the right path for solution so marking it solved.

[cedar kiln]

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[soft egret]

i might be wrong here

[soft egret]

but for b i strongly believe that u can solve it 100% graphically

because the car is at constant velocity for 0.505 sec

so you need at least basic rudimentary calculations to find out when it will stop

i.e 15-6(x-0.505) = 0

you need at least this basic calculation to actually find out the time for stop

how can u do this graphically

at best u can only get approximate ans

[cedar kiln]

@soft egret Has your question been resolved?

[cedar kiln]

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[raven kayak]

how do i sketch graph of y = f(x)

[neon void]

graph y = x for x >= 2 and graph y = 2 for x < 2
its just 2 different graphs stitched together

[raven kayak]

like this?

[neon void]

yes

[raven kayak]

oh

so what was the point of doing the f(-1) stuff

those dont correlate to the graph?

[neon void]

what f(-1) stuff

[raven kayak]

oops not -1

but i found f(1) f(2) f(3)

[neon void]

those are just meant to help you find points on the function i guess?

[raven kayak]

okayy

tyy

.close

[cedar kiln]

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[cedar kiln]

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[honest jasper]

how do you start this?

[quiet plover]

differentiate the equation given to you

you might get a Df / f'(x) term

[honest jasper]

okay ill start with that

thx

.close

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[limpid apex]

How

[limpid apex]

42×tan(38) isn't equal to 32.8

[past wave]

,w 42 * tan(38 deg)

[wraith dagger]

Sending query to Wolfram Alpha, please wait.

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported and should be fixed soon.
If the error persists, please contact our friendly support team at our support guild!

Exception

aiohttp.client_exceptions.ClientConnectorError: Cannot connect to host public5c.wolframalpha.com:443 ssl:default [Name or service not known]

[past wave]

How do you know this?

[limpid apex]

Calculator

[oblique flare]

calculator is in radians

[limpid apex]

Oh shi

Rookie mistake

[oblique flare]

either change to degrees or multiply the argument of tan by 180/pi

[crimson sedge]

He’s looking for a length no

[limpid apex]

.close

[cedar kiln]

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[honest jasper]

if i have to partially differentiate f(x-t) wrt x, how do you do it?

[past wave]

Just differentiate normally and put dt/dx = 0

[honest jasper]

sorry idk

i want to show lhs = rhs

[tropic oxide]

treat the t as a constant

[honest jasper]

but how do I show it without given a specific function

[tropic oxide]

well f and g are singlevar functions

how would you differentiate e.g. f(x-7)

[past wave]

Tell me what the derivative of f(x) is wrt x

[honest jasper]

f'(x)

[past wave]

Now answer ann

[honest jasper]

f'(x)

[past wave]

Nope

What is the derivative of (x-2)³

[honest jasper]

youre right therell be extra terms

idk sorry 😭

[jade blaze]

try by chain rule

[jade blaze]

take x - t = u and x + t = v

[honest jasper]

ok ill try this

[past wave]

That is not chain rule lol

[jade blaze]

no but we will use it after

like taking u and v

[past wave]

Kk

[honest jasper]

is it f'(x-t)(-1) for f(x-1) wrt t

[past wave]

No -1

Other than that it's correct

[honest jasper]

Even if it’s wrt to t?

[cedar kiln]

@honest jasper Has your question been resolved?

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[shut blaze]

<@&286206848099549185>

[shut blaze]

,rccw

[wraith dagger]

Couldn't find an attached image in the last 10 messages.

Oct 7, 2025, 5:42 PM

[shut blaze]

,rccw

[shut blaze]

Need help with 96

,,The arithmetic progression a(n) difference is 3, it's given that S13-S12= 44.

[wraith dagger]

supanova (Oh Ay)

[shut blaze]

Find the arithmetic progressions first number a1

[split ice]

please wait 15 minutes before pinging helpers in the future :)

[unique copper]

d=3
s13-s12=44
a=?

use formula for sum of n terms of AP, do you know what that is?

[split ice]

if i wrote $S_{13} - S_{12}$ as
\begin{align*}
&(a_1 + a_2 + \dots + a_{12} + a_{13}) - \
&(a_1 + a_2 + \dots + a_{12})
\end{align*}
then can you see what this evaluates to?

[wraith dagger]

ηασιβ ♥

[cedar kiln]

@shut blaze Has your question been resolved?

[shut blaze]

Wait

[shut blaze]

This is what I wrote

[shut blaze]

Also I'm confused wdym,,before" before what? Before another person's questions is answered or? (Gen question)

[unique copper]

no! a+(n-1)d is the formula for nth term of AP

S_n means the sum of all terms upto the nth term of AP and it has a different formula
for example S_3 = a1+a2+a3

[shut blaze]

Ohh

[unique copper]

He meant that after you post your question, you have to wait for 15 min and after that you can ping helpers

[shut blaze]

Ohhh

Okkie

Sorry didn't know

[unique copper]

Once you have posted your question in a channel, the channel is reserved for you. No one will post their questions in your channel so you don't have to worry about that.

[unique copper]

now, do you happen to know the actual formula or should i tell you?

[shut blaze]

Plz do bc I'm lost in these formulas even if there aren't alot of them

[unique copper]

Sum of first n terms of an AP,
Sₙ = (n/2) [2a+(n-1)d]

[shut blaze]

Okk

[unique copper]

now apply that there

[shut blaze]

How? I don't know a

[unique copper]

assume it variable 'a'

put it in equation and that's how you will get value of 'a'

just like you put 'x' in equation and find x's value

[shut blaze]

Oh so find it?

[unique copper]

yes that's what's asked

[unique copper]

here, what do you think is a1?

a=a1 they are different ways to write same thing

it's clear now?

[shut blaze]

No

[unique copper]

do it again with correct formula and see if you get first term a1 or a

[shut blaze]

Okay I'll send if I'm doing right

[unique copper]

what exactly you don't understand

[shut blaze]

What to do with the given formula

[unique copper]

you know d=3
and S13-S12 = 44
use the formula to find values of s13, s12 and substitute in this equation

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[shut blaze]

I tried am I going in the right direction or?

[unique copper]

no S_13 ≠ (a1+a13) × 13 / 2

[shut blaze]

.reopen

[cedar kiln]

✅ Original question: #help-13 message

[unique copper]

Sₙ = (n/2) [2×a1+(n-1)d]

S13 = (13/2) [2×a1+(13-1)3]

you get how S13 is found here?

is this formula even taught to you? If not, we can do the question a simpler way if you want

[shut blaze]

Yea wait

[shut blaze]

Yes I was just sick that day

Got the notes tho

[unique copper]

oh you wrote that actually mb

[shut blaze]

Oh, thought my internet was stuck or smth

But okay

So

[unique copper]

yes it's correct

now find S12

[shut blaze]

Okkie

[unique copper]

then put their values in the equation you know

[shut blaze]

Okkie

[unique copper]

ping me if you need me

[shut blaze]

I have answer can I send?

[unique copper]

yes

[unique copper]

correct

[unique copper]

now, before closing I want you to understand this aswell

There are 2 ways to do this: 1 st by formula which is direct and will always work, other is by observing and this will work in many cases but not always but when it does work it saves lot of your time

and 2nd method does work in this case, it was given that S13-S12=44

It's literal meaning is that sum of first 13 terms minus sum of first 12 terms is equal to 44

so that will translate to
(sum of first 13 terms) - (sum of first 12 terms) = 44
(a1+a2...a12+a13) - (a1+a2...a12) = 44

[shut blaze]

Yes I get that much

[unique copper]

do you see most of the terms cancel out like this?

[shut blaze]

Yes actually wait that's smart

[unique copper]

and you directly get a13=44 and with simpler formula you get a+12d = 44

so you save lot of stress with this

[shut blaze]

Is it okay that I took a ss?

[unique copper]

yes it's public channel no problem

[shut blaze]

Okie thx have a nice day :DD

[unique copper]

ok bye

[shut blaze]

.close

[cedar kiln]

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[upbeat lotus]

Hey just out of curiosity I made the graph of factorial , but I don't understand why it is defined for negative numbers, why it is 0 in the negative x axis , why it is undefined at negative integers , i don't understand.

[tropic oxide]

look up gamma function

this thing actually never equals 0, it just approaches 0 really really really closely the further left you go

[upbeat lotus]

I was actually solving some problems on the gamma function can you enlighten me like how should I think

Should I put n= some random integer in the gamma function

[tropic oxide]

god idk, Gamma(z+1) = z*Gamma(z)? that's how you extend it to the left.

[upbeat lotus]

Then see what happens

[tropic oxide]

gamma, not grammar.

gamma, like the greek letter.

[upbeat lotus]

Yes sorry auto correct I'm typing with my mobile

[upbeat lotus]

Yes so if I put some random negative rational which is not equal to an integer , then I should get something approaching to 0

That's for giving me a direction I'll try it

.close

[cedar kiln]

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[wild belfry]

In combinatorics how do you tell if order is important or repetition is allowed? Is there a rule or guidelines I can follow to figure that out about a statement?