help-23

[kind galleon]

But wait

If to ignore this domain from the question

And follow this it will get reflected over x axis still

[earnest nacelle]

y-axis*

[kind galleon]

Because I can put x = 1

And it will stay at x as my input, but the function will evaluate for -x

[earnest nacelle]

you can put x = 1, and the function will be defined at the input (-1)

[kind galleon]

Thus reflecting over x

[earnest nacelle]

but notice that for every positive x you put, the input will become the negative x

[kind galleon]

Yes

[earnest nacelle]

and vice versa

[kind galleon]

Yes

[earnest nacelle]

that's exactly what reflecting over the y-axis does

[kind galleon]

I don't understand

[earnest nacelle]

positive x-coordinates get sent to negative x-coordinates and vice versa

[kind galleon]

But why?

Graph is created based on what we put in x

[earnest nacelle]

I knew you would ask this, hence why I said this

[kind galleon]

Not what is in the parentheses

Graph is created in respect to original x value

[earnest nacelle]

oh goodness.

[kind galleon]

?

[earnest nacelle]

I mean the original graph is based on f(x), yeah

[kind galleon]

Im talking about the transformed one

[earnest nacelle]

but suppose you have some function f such that at x = 1, f(x) = 2
and at x = -1, f(x) = 4

now consider f(-x)

for x = 1, what is f(-x)?

[kind galleon]

-4

[earnest nacelle]

sure?

let's check

[kind galleon]

No

[earnest nacelle]

x = 1 makes the input to f(-x) (-1), so
f(-1) = 4

not sure where -4 came from

[kind galleon]

Im lost

[kind galleon]

Here 4

So it didn't reflect over y axis

x is 1 it stays in the second quadrant

[earnest nacelle]

but x = 1 produces an output normally reserved for x = -1

and vice versa

and that is the key here

if this doesn't help then I apologize and I'll probably step back for now

[kind galleon]

You know function transformations is probably the second hardest topic I've ever had in my life in Math

First one was complex numbers

[earnest nacelle]

I have heard of your struggles, hence why I said what I said at the start

[kind galleon]

What you said on the start

I don't remember

[earnest nacelle]

here.

[kind galleon]

I am the hardest student to teach to in my whole class

Nothing comes easy for me

And I am struggling

Wait

Is there a difference between x and f(this x)?

Because currently I have a perspective and x is different from the x inside f(x)

[earnest nacelle]

I'm going to try one last time

[kind galleon]

Why is this shit so confusing

[earnest nacelle]

for f(ax + b), think of x as some raw material, ax + b being a machinery that prepares this raw material for processing, and f being the actual processor that outputs stuff

[kind galleon]

Ok

[earnest nacelle]

so if x is modified in any way, then f takes the modified input

[kind galleon]

Yes

[earnest nacelle]

eg: f(2x) doubles x before using it as an input

[kind galleon]

but x is separate from 2x

[earnest nacelle]

f(x + 5) adds 5 to x before using it as an input

etc.

[kind galleon]

I mean the input as x

and the operation for x before input

[earnest nacelle]

the original input is always x. the modified input is whatever is in the parentheses as an argument to f.

[kind galleon]

And graphs are made in reference to the input as x

I mean the x cordinate

[earnest nacelle]

yes. but for f(x + 5) for instance, x = 2 gets you f(7)

[kind galleon]

Yes

So what is missing in my knowledge

I don't understand why any operation on the input is so confusing

[earnest nacelle]

I don't know. I'm not your teacher

[kind galleon]

My teachers doesn't teach shit anyways

That's why I have my book and this community

And internet

[earnest nacelle]

I'm sorry to hear that, but if I'm not going to be of use any more in this channel I'll take my leave first. I will say though, based on what I heard about you, you might find it beneficial to get a personal tutor

[kind galleon]

So the problem is that I was paying turbo big money for tutoring

Because I am on IB course, and the tutors cost at least 5x what the normal cost

And I realized that with good source and material I am able to do what we do on the lesson in 15 minutes by myself

[earnest nacelle]

then I invite you to ponder on possible reasons your lessons are taking longer than 15 minutes.

[kind galleon]

No I mean that if on the lesson I was able to sovle 3 questions, I can solve by myself 9 in the same time

[earnest nacelle]

but either way, I don't think a help channel is the place to discuss personal problems, so I will disengage for now.

good luck!

[kind galleon]

Yeah no personal problems, thank you

[austere goblet]

!done?

[safe radish]

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

[kind galleon]

No

Of course not

[austere goblet]

oh ok

[kind galleon]

Maybe you can help

Or already too tired

With me

I am the server destroyer, the help channel absorber

[austere goblet]

I don't know what else I can say that Towa hasn't already

[kind galleon]

WHy Is this literally so hard to understand for me

I don't get it

[earnest nacelle]

maybe go play with some graphs on Desmos/Geogebra or something

[kind galleon]

That's what I am doing currently

[earnest nacelle]

instead of having text explanation after text explanation chucked in your face

[kind galleon]

Doesn't really work

Wait

Why does it work like I described

y=f(-x)

[earnest nacelle]

then I'm not sure how else I can help you, sorry.

[kind galleon]

It literally works the way I described

Ok it doesn't nevermind

[shell torrent]

What does?

[kind galleon]

Ok I’m gonna close the ticket for now since I’m gonna be away from the book

[kind galleon]

Thanks for answering but I’m closing

Gonna come back today or tomorrow

.close

[safe radish]

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[safe radish]

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[radiant canopy]

guys when do u take +C when u integrate and when u dont

[quasi bison]

indefinite integrals require +C. definite integrals don't

[mighty mango]

quick answer:
+C when you get antiderivatives (indefinite integration)
and not when you find area under the graph (definite integration)

if you wanna know why, we could start with the indefinite integration case.

[radiant canopy]

yes pls
and is definite migration only for area under graph and y

[quasi bison]

definite integration can be used for a lot of things

[mighty mango]

antiderivatives are used when you have a differentiated version of the function and you wanna find how the original one looked like

to show why we need +C

[radiant canopy]

oh

[mighty mango]

how about you differentiate
2x+2
2x+3
2x+4

what would you get?

for each, please.

[radiant canopy]

ok

1 sec

[radiant canopy]

oh differentiate mb

2

2

2

[mighty mango]

great

so now we have the differentiated version of every one of the functions

care to integrate it? (antiderivative)

integrate
2
2
2

and maybe you notice something interesting

[radiant canopy]

integration of 2 is 2x
but
we don't know the other part it could be 1,2,3,4, right ?

so we use C

[mighty mango]

exactly!!!!

[radiant canopy]

so there is no way to find that C

[mighty mango]

err you can, depends on the information given and the context of the question

but yeah, you cant if its just straight up "antiderivative this"

antiderivatives propose a family of possible original functions and the only difference between them is the C.

[radiant canopy]

oh

[radiant canopy]

and u only area under graph

[radiant canopy]

y*

[mighty mango]

as ann stated, definite integrals are used for many many many things

or wiat

i might be misunderstnading your statement

[mighty mango]

wdym y only area under graph?

[mighty mango]

our most famous use is the area under the graph shtick!

[radiant canopy]

oh

k

Y C doesn't matter in area under graph

[mighty mango]

@quasi bison can you take over? emergency struck

sorry 👋

[radiant canopy]

bye

[quasi bison]

i mean you can find volumes using definite integration

and also if you integrate speed of a particle you can find the arc length of its path

(while integrating velocity gives you displacement)

[radiant canopy]

Y C doesn't matter in the area under cirve

[quasi bison]

you know how $\int_a^b f(x) \dd{x} = F(b) - F(a)$ right

[flat frigate]

Ann

[radiant canopy]

yes

oh

SO C gets cancelled

got it

ig

[delicate shore]

the indefinite integral taught in highschool is an abomination

[radiant canopy]

my school has not taught integration (I'm not there yet)

[quasi bison]

yes, the C's cancel out

[radiant canopy]

k ty

.close

[safe radish]

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[safe radish]

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[fiery olive]

I need help understanding this function when it x goes to infinity. I understand arcsin goes to 0 which gives 0 but I don’t understand why arctan gives pi/2 when x goes towards infinity

[rigid inlet]

Perhaps you should recall how arctan is defined

[lone void]

this is graph of arctan

[desert pasture]

what is arctan(tan(x))

[worn pollen]

Tan inverse

[desert pasture]

no..

[worn pollen]

Oh

It depends

[rigid inlet]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[honest perch]

on x

[rigid inlet]

This isn't an open question, please go somewhere else if you aren't the one asking for help

[worn pollen]

I thought that dude was gone

[honest perch]

seems like it

[fiery olive]

.close

[safe radish]

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[safe radish]

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[stoic saddle]

<@&268886789983436800>

[safe radish]

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[tribal sparrow]

What is the downside to polynomial of best fit? I had an idea for a model I wanted to train with gradient descent but I realized I could just replace it with an n-degree polynomial bc of its power series expansion.

[marsh walrus]

its possible that your polynomial of best fit doesnt really represent the data

Finding the best fit for some given polynomial is easy but it can be harder to justify the degree of polynomial you choose

[tribal sparrow]

right i guess im asking how to deal with overfitting

[tribal sparrow]

Is there literature on selecting the right degree?

[marsh walrus]

whats the damn test called

[tribal sparrow]

I'm training this on asset price movements

[marsh walrus]

Some discussion here

https://stats.stackexchange.com/questions/159864/goodness-of-fit-how-to-evaluate-if-polynomial-of-order-n1-gives-statistically

Although we have some stats people here I think if you had more involved questions the stats server in #old-network is probably a better place to ask

[safe radish]

@tribal sparrow Has your question been resolved?

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[kind galleon]

For a transformation of function y=f(x), why does y=f(qx) for q < 0 reflect it over y axis?

[queen ingot]

Negative values of x become positive, and vice-versa

[kind galleon]

Yes but shouldn't that reflect it over x axis?

Since function will be evaluated let's say for example x = 1 then f(-(1))

[median vigil]

remember that the y-axis is the line x = 0

[kind galleon]

Yes

[queen ingot]

X is mapped to whatever -x is mapped to

[kind galleon]

Isn't x, and f(this x) different things?

[median vigil]

so going from x = 1 to x = -1 and so on should reflect across x = 0

[kind galleon]

Yes but the graphs are made in respect to x

Not to the operations done that will actually feed into the input

[median vigil]

if you start out with the point (x, f(x)) on the graph that gets sent to the point (-x, f(-x)) in the transformation

[kind galleon]

How is it?

shouldn't it be (x, f(-x))?

[pine mountain]

I'm assuming by graphs you mean transformations; since all transformations are done on x before the function f is applied to find the respective y, and no additional transformations are done afterward, nothing changes about the y-value for a certain x

[pine mountain]

From this reply

[median vigil]

if all you knew about the graph of y = f(x) is that it passes through the point (2,3), then the only thing you know about the graph of y = f(-x) is that it passes through (-2, 3)

[kind galleon]

Ok, for what value will function g be evaluted if g = f(-x) and let's say x=5?

[kind galleon]

Yes then I need to input -2 to get the same y output as before transformation

[median vigil]

yes. so if inputting the opposite x value gives the same y value as originally, that should look like a horizontal mirroring

[kind galleon]

Ok so I feel stupid right now?

Im refering to this from my notebook on stretched and relfections page

if q will be q < 0 then f(-x)

Let's just stay with the reflection stuff

So if I put into this function x = 5, the input will be -5

[median vigil]

f(-x) is just the special case of q = -1

[kind galleon]

Yes

[kind galleon]

And the point will be drawn (5, f(-5))

[median vigil]

when you think about transformations, it's good to think about only one input-output pair at a time. so if you only know that the pair (5, f(5)) is on the graph y = f(x), then you don't know what the point (5,f(-5)) looks like on the graph y = f(-x), because you don't know f(-5), but you do know that (-5, f(5)) is and that it's also on y = f(-x)

[kind galleon]

But im assuming that you are explicitly saying that x is different from f(this x)?

And also the point (5, f(-5)) is like a reference to confirm on what the function will be evalueted for

Im just trying to work with it but I think I may end up confusing myself even worse

[median vigil]

like both (5, f(-5)) and (-5, f(5)) are on the graph y = f(-x), but only one of them can be found only from knowing the original input-output pair (5, f(5))

[kind galleon]

Yes

We don't know what will be (5, f(-5))

[median vigil]

so it would be more useful to think of (5,f(5)) as being "sent" to the pair you can find from that info, which is (-5,f(5))

[kind galleon]

But Im saying that this will be the result for y of this opeartion

[kind galleon]

But you know I still don't understand this reflection

Everything about the x variables seems super complicated

[median vigil]

when you think of function transformations, you should think of them as acting on pairs of input and output (x,f(x)) rather than just the output

[kind galleon]

Explain further

[median vigil]

like in this case we are seeing that the transformation y = f(x) to y = f(-x) is better understood as acting on the input x rather than the output f(x)

[kind galleon]

Yes but output is also very important here

Because it will be the final verdict of y cordinate

So for y = f(x), y = f(-x) will only reflect over the y axis if x < 0?

Because only then x will be plotted on the negative side of axis

[median vigil]

no, it reflects both

[kind galleon]

What

[median vigil]

the point (-5, f(-5)) from y = f(x) is mapped to (5, f(-5)) on y = f(-x)

[kind galleon]

Yes

[median vigil]

because -(-5)) = 5

[kind galleon]

Wait what

[kind galleon]

What it will be for the point (5, f(5)) for the reflection/

What would I do to reflect this over y axis?

[median vigil]

the point (5,f(5)) is on the graph of the original function y = f(x) but it is not on the transformed graph y = f(-x)

[kind galleon]

What would happen if I would feed x=5 into y = f(-x)

Would this reflect?

[median vigil]

y = f(-x) is just a function. the transformation (reflection in this case) is the process which takes pairs of input and output from y = f(x) and turns them into pairs of input and output from y = f(-x)

[kind galleon]

I don't understand

[median vigil]

like y = f(-x) has absolutely nothing to do with reflection on its own, it's only reflected in comparison to y = f(x)

[kind galleon]

Ok, let's say the function is linear, I feed 1 into x for f(x)

The point will be (1,1)

What wil happen for the reflection function?

[median vigil]

if you only know that the point (1,1) is on the original function, then the only information you know about the reflected function is that (-1,1) appears on it. so we say that the point (1,1) is sent to the point (-1,1) by the reflection

[kind galleon]

isn't x the same for both functions?

[median vigil]

think of the reflection as being a function which takes in pairs of numbers (input-output values of the old function) and returns pairs of numbers (input-output values of the new function)

[kind galleon]

I put x the same for both functions

But they return me different y

[median vigil]

but if you only know that it was reflected horizontally, it would be more useful to think of it as putting in different x to those functions to get the same y

[kind galleon]

OK I don't understand

I don't know what is happening

[median vigil]

if you only think about putting the same x in, you can't say anything about the relationship between the two functions

[kind galleon]

Why?

[median vigil]

like say you only know that (1,2) is on the original function graph. so you know f(1) = 2 and nothing else about the function. it's true that (1, f(-1)) is on the graph of the new function, but that tells you nothing because you don't know what f(-1) is, or even if it exists. the only thing you know for certain is that the point (-1,2) is on the graph of the new function y =f(-x)

[kind galleon]

So they match output?

I need to use different x to get the same output

As before transformation?

[median vigil]

yes

[kind galleon]

THat doesn't make any sense

What if I want to put anything that I want to x, for fun

And I know the range for this function

Ok you nkow what

Im gonna close this ticket

It's too late for this shit

.close

[safe radish]

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[tranquil geyser]

\textbf{Axioms - Separation Scheme:} \$\forall$ L-formulas, $\varphi(u, z_1,z_2,...,z_n)$ have an axiom $\forall z_1,\forall z_2,, ..., \forall z_n \forall x \exists y \forall w ( w \in y\iff (w \in x$ and $\varphi(w, z_1, ..., z_n)))$.

This is the exact definition i wrote down, I dont fully understand what its trying to say

[flat frigate]

toast

[tranquil geyser]

especially like what is u?

for all variables z1 to zn, and for all sets x, there exists a set y such that all elements of y are in x and satisfy these L formulas?

[safe radish]

@tranquil geyser Has your question been resolved?

[tranquil geyser]

@meager igloo

specficially what is $\varphi(u, z_1,z_2,...,z_n)$

[flat frigate]

toast

[tranquil geyser]

or how do i better interpret this? is $\varphi$ a function?

[flat frigate]

toast

[tranquil geyser]

oh its another L-formula

and ig (u, z1,, ...., zn) are free variables

[safe radish]

@tranquil geyser Has your question been resolved?

[wild cape]

@tranquil geyser separation says that we can take the elements of a set that satisfy a formula, and they will form another set

if you recall setbuilder notation from earlier math content, separation is what allows us to do that

[safe radish]

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[wild cape]

.reopen

[safe radish]

✅ Original question: #help-23 message

[wild cape]

by setbuilder notation, i mean from a set $A$ and statement $\phi$ concerning the elements of $A$, we form the set $\cbr{x\in A:\phi(x)\text{ is true}}$

[flat frigate]

ロケット・ジャンプ

[safe radish]

@tranquil geyser Has your question been resolved?

[safe radish]

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[wild cape]

.reopen

[safe radish]

✅ Original question: #help-23 message

[safe radish]

@tranquil geyser Has your question been resolved?

[safe radish]

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[pastel pebble]

Hi

[pastel pebble]

How to I differentiate

Ln(2-sqrtx)^2

I got to 2ln(2-sqrtx)

What is a for a/a +b

[molten acorn]

i think you forgot to apply the chain rule

[molten acorn]

wdym

[pastel pebble]

O

Wait how do i apply chain rule to ln

[molten acorn]

do you know what the derivative of ln is

[pastel pebble]

1/x

Or

A/ax+b

[molten acorn]

ive genuinely never seen this form before

[pastel pebble]

Its what my textbook says

[molten acorn]

but for this ques youll have to apply chain rule twice btw

lemme clarify what i meant:
let x^2 be f(x)
ln(x) = g(x)
2-sqrtx = h(x)

the function you want to diff is f(g(h(x)))

[molten acorn]

just use this form tho, no need to use A/ax +b

[pastel pebble]

Wheres x^2 come from

[molten acorn]

this is the original func right

it has a ^2

[pastel pebble]

O

Can i use inside differentiation

[molten acorn]

wdym

[pastel pebble]

Like

F ‘ x/ fx

Also how do i differentiate (2-sqrtx)^2

Do i expand

Or

[prime shard]

that or chain rule

[safe radish]

@pastel pebble Has your question been resolved?

[safe radish]

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[surreal kayak]

x sqt - y sqt = 30

[earnest nacelle]

uh... picture please?

and what are you asked to do

[surreal kayak]

s

[earnest nacelle]

otherwise I assume it's $\sqrt{x} - \sqrt{y} = 30$ with no instructions

oh it's a square

[surreal kayak]

y

[plucky elk]

Factor your expression on the left

[earnest nacelle]

have you heard of the difference of two squares?

if you have, you can do what riemann asked to

(also I'll step down and let riemann take over if needed)

[surreal kayak]

ok

[mossy lotus]

merely factorizing 30 would be sufficient, especially for this problem (not a good advice in general for such problems however)

[earnest nacelle]

that would work too, yeah. perhaps a better method than what I have in mind

[mossy lotus]

yea, only works if you know some basic nt

[quasi bison]

you should use the symbol ^ for exponents, and write this as x^2 - y^2 = 30

otherwise nobody will understand you

[mossy lotus]

which is sadly not a school syllabus subject for most part

[earnest nacelle]

I was going to have OP list factors

then compare the factors of the LHS to the factors of 30

[surreal kayak]

ok

[safe radish]

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[autumn sundial]

Hi! Is anyone here familiar with desmos? Im a highschool student and i genuinely need help with my "desmos art project" they said we can make anything and i'm making a guitar, but im geuinely struggling to figure out how to make the body of the guitar and google and youtube arent helping much

[fleet willow]

like how

ig what do you want it to look like?

a good place to start is polynomial curves i think

[earnest nacelle]

perhaps start by sharing the image of the guitar you are using for reference and what you have tried so far

[hard crest]

i think elliptic curves could have a useful shape for a guitar body

[fleet willow]

yeah

[autumn sundial]

Heres the thing this is kind of a research project we havent really worked with graphing curved lines so im tryna research that rn

[fleet willow]

or polynomial curves in general 😛

[autumn sundial]

This is kinda what i figured out so far but im not happy with it

[hard crest]

try throwing random stuff into desmos with sliders and see where that gets you

[autumn sundial]

Thats one way to go ig

[fleet willow]

also

[hard crest]

sliders are great for adjusting shape

[fleet willow]

xx = x^2

[autumn sundial]

Ik

[fleet willow]

oh

why didnt you type x^2

[mortal sandal]

As in ellipses or as in y²=x³+ax+b lol

[hard crest]

the second one

[autumn sundial]

It wouldnt let me😭

[hard crest]

but also sure conic sections, try those

[fleet willow]

caret ^?

[autumn sundial]

???

[fleet willow]

use a caret ^

shift + 6

[autumn sundial]

Im on phone💀

[mortal sandal]

Bezier curves are cool too

[quasi bison]

desmos' phone keyboard has a dedicated "x^2" key does it not

[fleet willow]

here's an example elliptic curve: https://www.desmos.com/calculator/sltf4mulpe

they look very guitar-like 🙂

[autumn sundial]

Oo nice

[fleet willow]

you would have to modify the top tho

[mortal sandal]

https://www.desmos.com/calculator/bgvoyeztw6 here's a project with bezier curves if you want to see how they look

[fleet willow]

since it diverges out

[autumn sundial]

Currently lookong for it

[autumn sundial]

Thats so cooll

[mortal sandal]

The "code" isn't exactly a shining example of how yours should work probably lol

[autumn sundial]

Nvm i found i

[mortal sandal]

It's a bit scuffed in order to be able to make 100 of them

[autumn sundial]

Yeah

[fleet willow]

oh yeah i also have a project on bezier curves

https://www.desmos.com/geometry/d3d7e3b2b8

i did it because my calculus textbook told me to

[autumn sundial]

I need to do some huge research💀

[fleet willow]

ok so

[mortal sandal]

Ooh nice

[fleet willow]

do you know the binomial theorem

[autumn sundial]

No....

[fleet willow]

oh

do you know pascal's triangle

[autumn sundial]

Where i am in grade 10 they teach like- domain and range and linear graphs mostly

[fleet willow]

some of the motivation for bezier comes from the binomial theorem i think

[mortal sandal]

This project would be good for @autumn sundial to play around with just to see the kind of shapes you can get out of bezier curves

[mortal sandal]

You can probably get away with just not expanding the binomials

[fleet willow]

i had a project where you could do up to 10 vertices but it's lost to time 😔

[mortal sandal]

or hmm

maybe they actually naturally arise

and it's factoring that gets you the binomials

[fleet willow]

lol idk

i didnt really study beziers that much 😛

[mortal sandal]

Anyways if I have 2 points a,b

The natural way to have a path between them is to start at a and move towards b at a constant speed

[autumn sundial]

Yea

[mortal sandal]

So if t=0 I should be at a, and t=1 I should be at b

I can parametrize my line as (1-t)a + tb

[autumn sundial]

I think i get it?

[mortal sandal]

to multiply a point by a real number, just multiply each component, to add points, just add components

[fleet willow]

ohh i see the motivation behind bezier curves now

[mortal sandal]

It's shorter notation so I don't have to write out the individual coordinates when I'd just be doing the same thing to both of them

[fleet willow]

so you know how drey talked about linear interp

you extend it quadratically

i'll explain in detail

[autumn sundial]

[fleet willow]

so you have 3 points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$

[flat frigate]

nadat12

[fleet willow]

you want to take these points

so that there's a curve that gets from $(x_1,y_1)$ to $(x_3,y_3)$

[flat frigate]

nadat12

[fleet willow]

so obviously, if our polynomial is B, then B(0)=(x1,y1) and B(1) = (x3, y3)

the fun part is this:

so from point 1 to point 2

lineraly interpolate

$(tx_1+(1-t)x_2,ty_1,(1-t)y_2)$

and from point 2 to point 3

linearly interpolate

$(tx_2+(1-t)x_3,ty_2,(1-t)y_3)$

[flat frigate]

nadat12

[fleet willow]

so we have those two points

[flat frigate]

nadat12

[fleet willow]

NOW

linearly interp from these two points

that's where it gets fun!

$(t(tx_1+(1-t)x_2)+(1-t)(tx_2+(1-t)x_3),t(ty_1+(1-t)y_2)+(1-t)(ty_2+(1-t)y_3))$

[flat frigate]

nadat12

[fleet willow]

this formula is pretty complicated

so we can simplify

[mortal sandal]

Here's a picture btw

[fleet willow]

that's what i described looks like

[autumn sundial]

Ohhh

[fleet willow]

for cubic beziers

it's the same

linearly interp once

twice

[autumn sundial]

This calrified it sm thank you😭

[fleet willow]

and thrice

in general

[fleet willow]

sorry for the very vague explanation

it's hard to explain

[autumn sundial]

No no that was actually a rlly good explanation

[mortal sandal]

Yeah that was good

[autumn sundial]

Thank you both

[mortal sandal]

No problem

[fleet willow]

the general bezier form follows $B(t)=\sum_{k=0}^n\binom nkt^{n-k}(1-t)^kP_k$ if you're wondering

[flat frigate]

nadat12

[fleet willow]

where Pk are the points

[mortal sandal]

4 points is usually used because it's the minimum number needed to get interesting behavior

All the 3 point examples kind of just look like bendy quadratics

[fleet willow]

the weird parenthesis thing is called the "binomial coefficient"

[mortal sandal]

And if you have too many points it becomes pretty hard to understand the behavior

[autumn sundial]

Ooo

[fleet willow]

the bezier curve animations as linear interps over linear interps is satisfying

[mortal sandal]

Long curves are usually made out of sections of multiple 4-point bezier curves

[mortal sandal]

Yeah

[fleet willow]

check out this

[mortal sandal]

Wow

[autumn sundial]

Woahhh

[fleet willow]

linear interps over linear interps

[mortal sandal]

Reminds me of the fourier drawing animations almost lol

[fleet willow]

anyways wish you luck on your project! hope this enlightened you!

[mortal sandal]

There's some clever way to chain together bezier curves so that the first derivatives always match

but idk it

[fleet willow]

yeah, i think that's what my calculus book was studying 🤔

[mortal sandal]

id have to do it by hand

[autumn sundial]

Yes thank youuuu i think i can get somewhere w this info

Thank you too dreyyy

[mortal sandal]

No problem!

[mortal sandal]

Out of curiosity what math are you doing right now?

[autumn sundial]

Did they leave

💀

[mortal sandal]

Idk ig lol

[autumn sundial]

Ig not

[fleet willow]

im still here just looking at the wikipedia page for beziers

[mortal sandal]

Ig not lol

[fleet willow]

uh im doing abstract algebra

[mortal sandal]

Ooh nice ok

[fleet willow]

i took some multivar calc but it was confusing 😵‍💫

[mortal sandal]

That one's fun

[fleet willow]

as in read it from a textbook

[mortal sandal]

Multivar calc has a lot of unmotivated stuff that gets motivated later

[mortal sandal]

As in like, in later courses lol

kinda like teaching high schoolers determinants and matrix multiplication and stuff

Or even how those are computed in a college intro linalg course

When I first took it I forgot stokes/greens/etc the second the final was over lol

[fleet willow]

lol me too

no one remembers all those nasty confusing theorems

[mortal sandal]

That's why differential geometry is so OP

Turns out they're all the same theorem

[fleet willow]

stokes?

[mortal sandal]

Generalized Stokes yeah

[fleet willow]

nice, good thing i got a textbook a few weeks ago

[mortal sandal]

though I still don't actually understand how you would prove generalized Stokes

Ironically it's the definition of boundary which is the hardest part to understand

something about cochains 💀

[fleet willow]

jordan curve theorem but generalized?

oh no

[mortal sandal]

Probably not theres some weird algebraic notion of the "boundary" of a...

[fleet willow]

,ti

[flat frigate]

The current time for nadat12_50450 is 01:17 AM (PST) on Sun, 15/02/2026.

[fleet willow]

yeah i gotta sleep 🙁

night

[mortal sandal]

I don't even remember what the object is called

Rip

Good night

[safe radish]

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[sick wyvern]

is this supposed to be solvable? chatgpt made it and cant answer it. i got t = 17.6381s
and 784 m

[quasi bison]

!noai first and foremost

[safe radish]

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

[quasi bison]

even for generated problems

[marsh gorge]

ill check if there is a solution

[quasi bison]

that said, i would make velocity-time graphs here

[sick wyvern]

i did not graph anything i just used the 4 or 3 kinematics equations

im in ee 1st year 2nd sem btw

[marsh gorge]

wait

[quasi bison]

ok, my suggestoin stands though

hey hang on

once A and B both go into constant speed, isn't A running at 18 m/s while B is running at only 8 m/s?

B is literally too slow to catch up @sick wyvern

don't trust gpt to generate problems

[sick wyvern]

ahhhh

wait you're right

[marsh gorge]

its not possible for b to ctahc up

i checkedd every interval

a achives a speed more than twice of b

and has coverd more dist

[sick wyvern]

okay thanks, but i need to ask a question

in any case there's a catch up problem, and let's say b starts 5 s later, should i write it as t - 5 or t + 5?

[marsh gorge]

if u want to try these types of questions i know some great books

[sick wyvern]

i would love that

[marsh gorge]

in these types of problems when someone starts late

calculate the distance the other person has covered'first

in the time it takes for the other perso to start

then when they both are in the same timeline

use kinematics from there

so like lets say a started 5 seconds early

distance travvled by a in those 5 seconds would be 37.5 meters

add this to the equation and then start them at the same time

[sick wyvern]

and if a has like, accelerated for a time 3 s, what should i do? should i add 5?

[sick wyvern]

or like constant velocity before stopping

for a given time

[marsh gorge]

so if has accelerated for 3s at 2ms^-2 for example in 5 seconds they would have travelled 25 meters

speed would be 10ms^-1

from there on uu can write x(a)=25+10t

and if be starts with a velocity of like 4

write b as x(b)=4t

[sick wyvern]

what did you use as t for this one

vi = 0

[marsh gorge]

yep

[marsh gorge]

5 second head start for a

[sick wyvern]

so what if like a accelerates for a given time say 3 s, then stops but still moves at a constant velocity, and b starts 5 s later

calculate the distance travelled by a in 5 seconds??

[marsh gorge]

yep

[sick wyvern]

oh okay

what about b then

[marsh gorge]

then check whats up with a after 5 seconds

so like after 5 seconds treat the distance travelled by a as a funtion of time t and add the distance travelled before as a constant

and for b

take distance as funtion of the same time t

[sick wyvern]

and then equate them?

to solve for t

[marsh gorge]

yep tp find t

if u get no solution then its obvious that they wont cross

[sick wyvern]

alright thanks dude

[marsh gorge]

try that book

its great

look at this question

from the kinematics section

[sick wyvern]

i should expect problems like this to appear tomorrow

[sick wyvern]

what the

[marsh gorge]

Idk

[sick wyvern]

ahh okay

i understand

[marsh gorge]

It’s not difficult

U can check the solution

[sick wyvern]

you can use displacement using average velocity equation?

[marsh gorge]

For his question?

U will have to

Split it up

Into different parts

Like when b leaves the cycle

A picks it up

And stuff

[sick wyvern]

im going to try and solve it

[marsh gorge]

Yea check the solution in the book later

If u want to do better try the book

[safe radish]

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[desert pasture]

I would like to prove this

Here's what I've done so far

I would like to add, here $f$ is continuous

[flat frigate]

wai

[desert pasture]

here's the OG problem

[desert pasture]

Let the total bounded variation on $[a,c] \subseteq [a,b]$ be $M$. Let $(a_n)$ be a strictly increasing sequence converging to $c$ such that $ \lim_{n \to \infty} \sum_{i=1}^{n-1} \abs{ f(a_{i+1}) - f(a_i)}≠M$. There thus exists a strictly increasing seqeunce $(b_n)$ converging to $c$ such that $ \lim_{n \to \infty} \sum_{i=1}^{n-1} \abs{ f(a_{i+1} - f(a_i)}<\lim_{n \to \infty} \sum_{i=0}^{n-1} \abs{ f(b_{i+1})-f(b_i)}≤M$.

I'm trying to get a contradiction here, not sure how to

[flat frigate]

wai

[safe radish]

@desert pasture Has your question been resolved?

[desert pasture]

.close

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[runic ruin]

Can someone please explain why I’m getting F wrong

[runic ruin]

The pink version is what I’m doing and the black version is the model answer

Think they did a different approach but shouldn’t we get the same answer? Thanks

[safe radish]

@runic ruin Has your question been resolved?

[glass carbon]

the problem is that you treat the force R as a component, and this is the result, if you want the point P to be in equilibrium, the force R would have to be applied in the opposite direction

So, the equation should be F - 30*sin(60 deg)= Rcos(60 deg)

[runic ruin]

So for every resultant force I change the sign or something because the arrow goes the opposite way?

I mean in these kinds of problems

[glass carbon]

you have two options: the force R is the result of adding the vectors of the other two forces

so you can add these two forces (taking into account their signs) and equate them to the force R (on the other side of the equation)

or you can imagine that you apply force R in the opposite direction (i.e. you balance the forces acting on point P) and then the system is in equilibrium and the sum of forces on each axis is zero

[runic ruin]

Oh I see

Thank you!

.close

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[grim plover]

shortest approach to compute $\sum_{n=1}^{\infty} \frac{n^2+6n+10}{(2n+1)!}$

[flat frigate]

rak³en

[grim plover]

the given solution is to let m = 2n+1 and then compute the sums m^2/m!, m/m! and 1/m!

over m = 3,5,7,..

which is quite long and tedious

[quasi bison]

i mean you're gonna need to reduce this to sums of reciprocal odd factorials somehow right

let me cook

[grim plover]

well thats what i am asking is there a shorter way to reach that?

maybe some weird level 1000 algebra foresight split

[low canyon]

That's what I would do too

(I mean, the solution)

[grim plover]

(duh who else could you possibly be referring to)

[low canyon]

?

[low canyon]

To your very last message?

[quasi bison]

this is the beginning of what i would do

[grim plover]

hmm put n = 2,3,4 to find A,B,C?

[nimble wyvern]

Beautiful handwriting damn

[grim plover]

the equations aren't very pretty though >_<

[grim plover]

okay ig

[low canyon]

What do you want? A two line solution? You won't get it!

[grim plover]

well no something that involves less calculation i suppose

guess i wont find one

ty

[low canyon]

I don't think you can find one due to that factorial, unless there exists pre-made formulas for very specific problems like this, which would be useless in the first place

[grim plover]

.close

[low canyon]

Due to that odd factorial*

[nimble wyvern]

Hm

Lemme see if I can make up something shorter

[grim plover]

the best you could do is learn formulae for odd/even factorial sums of the form m^k/m! for some integer k

right

[nimble wyvern]

Probably not but worth trying

[grim plover]

dm me if u find smth

.close

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[quasi bison]

you can do it in many ways

if you are gonna put values of n i recommend 0 and ±1/2

.reopen

[safe radish]

✅ Original question: #help-23 message

[quasi bison]

the key fact to use afterwards will be that the sum of all reciprocal odd factorials is sinh(1)

@grim plover can you share the given sol that you were looking at

[safe radish]

@grim plover Has your question been resolved?

[safe radish]

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[slender steeple]

how should i go about solving this?

[earnest nacelle]

I would probably expand the LHS, pull everything to the LHS, and then treat this as a quadratic in x

there's very likely a smarter way to do this but that's my first instinct

[slender steeple]

oh yeah ignore x=e in the top

thats from a previous problem

[earnest nacelle]

yea I wasn't considering that unless you say so

[winter whale]

i think quadratic is the smartest way to solve this

[slender steeple]

hm

is this right

[winter whale]

looks about right

[earnest nacelle]

hm how did you get this

quad formula?

[slender steeple]

quadratic formula for when b is even

[earnest nacelle]

oh you skipped the factoring step and the cancellation against the 2 in the denominator

[slender steeple]

yeah theres a shorter formula for when b is even

[earnest nacelle]

fair

[winter whale]

damn

[slender steeple]

i believe its uh

[earnest nacelle]

in that case the only thing I would suggest here is to write a brief explanation for why the negative solution is rejected

[slender steeple]

ill just write it down

[earnest nacelle]

cuz you dw to give examiners a chance to put a big fat "why?" on your paper

[winter whale]

tbf i thought you js factorised the 4 and cancelled the 2 out

[earnest nacelle]

same, I thought they did that too

[fair helm]

Are you sure its rejected? 🧐

[winter whale]

tbf that formula is probably just a simplification of the quadratic formula for a special case

[earnest nacelle]

OP seems to have rejected it

[earnest nacelle]

unless I'm a dumbass and there's another reason why the \pm became just a +

[winter whale]

tbf i would still stick to the original formula, cuz simiplification is easy

[earnest nacelle]

not beyond me

[slender steeple]

oh i rejected the minus sigh cause

[fair helm]

I never use the formula personally

[winter whale]

you self depricate a lot eh? chin up king

[slender steeple]

1 < √(e^3+1)

[earnest nacelle]

I speak only the truth

[fair helm]

So

[earnest nacelle]

if you have a justification best to write it on paper anyhow

[winter whale]

nah you're goated dw

[earnest nacelle]

but uh hm

[slender steeple]

honestly idk how to justify the removal of the minus sign

other than that

[fair helm]

Does it say there should only be one sololtioun?

[slender steeple]

it does say that problems can have more than 1 solution

i don't think this is one of them

[fair helm]

Why

[earnest nacelle]

eh you cannot reject this

[slender steeple]

cause the original problem was ln(x) + ln(x+2) = 3

[earnest nacelle]

so yes I'm a dumbass

[fair helm]

Oh yeah then you can reject it

[earnest nacelle]

oh wait so there's more context

[slender steeple]

yeah

[earnest nacelle]

argh the lack of original context here is killing all of us

[fair helm]

Say ln function is not defined for x<0 or something

[slender steeple]

aight

[slender steeple]

my bad

[junior smelt]

(I had a strong feeling that was the case )

[slender steeple]

i try to include the og problem

my sleepy self must've forgotten

[earnest nacelle]

I had thought so too but I do not like to assume

[slender steeple]

,ti

[flat frigate]

The current time for google7133 is 11:07 PM (+08) on Sun, 15/02/2026.

[slender steeple]

yeah its like 11pm

[earnest nacelle]

but I guess my brain didn't think that that x(x-2) inside the ln was sus

[earnest nacelle]

it's 11 here too

[slender steeple]

dang

[earnest nacelle]

anyway ok my bad for spewing a bunch of bs and confusing you probably

anything else though?

[slender steeple]

none as of rn

[earnest nacelle]

aye

all the best

[slender steeple]

ill probably leave the channel open unless the last question offers me no difficulties

[earnest nacelle]

sure thing

[slender steeple]

does this just have no solution?

oh wait nvm

i forgot exponents can be negative

actually

is there no solution?

[earnest nacelle]

let's see

there are two solutions I believe

[mighty mango]

I believe so too.

[slender steeple]

i think this is correct?

[mighty mango]

indeed.

[vale shard]

ooo good job :3

[slender steeple]

welp that's my last question for the night

[vale shard]

night ?

oh

[slender steeple]

thanks

.close

[safe radish]

Channel closed

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[earnest nacelle]

well done!

[vale shard]

w speed!

:D

[safe radish]

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[tranquil geyser]

Tysm this makes alot of sense

[tranquil geyser]

I’m still getting used to the notion of a formula ngl

is there like a good analogy for L, L-alphabet, L-terms, L-atomic formulas, L-formulas, L-sentences and L-srrucures

[safe radish]

@tranquil geyser Has your question been resolved?

[safe radish]

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[vague rain]

I am alr?

[earnest nacelle]

yes.

you can also check to be sure

according to you, Cheryl is 18 now.
5 years ago she's 13, and in eight years she would be 26, exactly double that of her age 5 years ago

so yeah

[vague rain]

Thx

[earnest nacelle]

anything else you wanna ask?

[vague rain]

No, is all

[earnest nacelle]

alright, you may close the channel then. all the best!

[pine mountain]

!done

[safe radish]

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[sleek trellis]

Hi

[sleek trellis]

I need some help with a proof

[nimble wyvern]

Hm

[brave wolf]

What exactly do you need help with

[sleek trellis]

A step by step analysis of the whole thing actually

To specify, the 4th last line

how do we arrive at 0 < x < pi/2

[brave wolf]

What's the theorem they are proving?

i suppose its part of the squeeze for lim sinx / x as x -> 0

[brave wolf]

okay, yeah, there is your answer

[sleek trellis]

through the sandwich theorem

[brave wolf]

0 < |x| < pi/2 is in the assumption of that theorem

[sleek trellis]

Okay okay oh sorry

Sorry for wasting your time

My bad

[brave wolf]

Yeah, the limit only cares about the values which are close to 0

[sleek trellis]

Common sense

[sleek trellis]

yeah

[brave wolf]

np, we sometimes miss the obvious

[sleek trellis]

Thanks

.close

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[fresh grove]

Source: Madhava Mathematics Competition

There is solution paper available. But I want to do this myself

[fresh grove]

By the way is it wasting your time?

I kind of thought it's fine

[fleet willow]

ok so i dont really know how to solve it

but maybe it relies on the fact that f attains all natural numbers

and g(a) and g(b) are distinct for all distinct a and b

[fresh grove]

Ahh

[fleet willow]

so for any n, let f(a) = n for some a

then n >= g(a)

right

[fresh grove]

yes

but there might be a b such that f(b) = n ?

[fleet willow]

yeah

ok

maybe assume there's two a and b such that f(a)=f(b)=n

[fresh grove]

Ok

[fleet willow]

and then n>=g(a) and n>=g(b)

but they can't be the same

[fresh grove]

yep

[atomic bison]

Since f is one to one, it means that f(n) = 1 for some n in N.

Does that give a bit of a hint?

[fresh grove]

how is f is oneone?

you mean g?

[fleet willow]

onto*

but yeah

ok i see

[atomic bison]

Does it mean surjective?

[fleet willow]

ok hint: ||use induction||

[fleet willow]

no, one to one = injective

but onto = surjective

[atomic bison]

One to one should be bijective

[fleet willow]

no injective is right

[atomic bison]

But still, since f is surjective, f(n) = 1 for some n in N

[fresh grove]

oneone onto is bijective?

[fleet willow]

one to one + onto = bijective, yes

[fleet willow]

using this, use ||induction|| to prove it for all n

[fresh grove]

i haven't looked at this. i hope that's fine

[fleet willow]

it's ok

but nico's hint is very powerful

ok

if f(n) = 1, what's g(n)

[fresh grove]

definitely 1

since it can;t be 0

[fleet willow]

what if f(k) = 2

what must g(k) be

[fresh grove]

wooow

2

wooow

this is soo good

[fleet willow]

so now

you can use ||induction||

[fresh grove]

yes yes

I loved this

[fleet willow]

this problem is very nice

[fresh grove]

I was going to sleep

I think now i'll get a good sleep

.close

[safe radish]

Channel closed

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[fresh grove]

thankyou @fleet willow @atomic bison

I really enjoyed the conversation

when you both helped me soo much

tysm

[safe radish]

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[sour bolt]

can someone help explain this to me

idk the way my teacher is explaining it is confusing me

[simple gazelle]

how would you go about multiplying them

[unkempt geyser]

First remember FOIL
First
Outside
Inside
Last

[sour bolt]

ye ik that

but i’m confused because im getting these crazy answers

[unkempt geyser]

You will come up with a large equation once you have finishing FOILing

[simple gazelle]

show what you tried

[unkempt geyser]

Then, find common terms, and simplify

[sour bolt]

would it be 0.5x^2-1.3x+0.67x?

[simple gazelle]

how did you arrive at this conclusion

[sour bolt]

foil

2x times 1/2x

1x^2

-1 times 1/2x

[unkempt geyser]

Easiest way to check is to pull up Desmos and see if you get the same curve

[sour bolt]

-0.5x

wait no

i get it now

hang in

on

is it x^2-11/6x+2/3?

[pine mountain]

use parentheses around (11/6) but yes

[sour bolt]

okay

ty

.close

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Channel closed

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[safe radish]

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[stoic raptor]

whats the point of higher derivatives, like i just learned it and its so easy i just take derivative of a derivative etc. but i want to know what applications it might be used in in the future so can someone give me some insight thanks

[brave wolf]

One cool application is e.g. taylor series

[stoic raptor]

oh iheard of that

its used in physics

[brave wolf]

You know how first derivative is used to find tangents?

[stoic raptor]

yeah

[brave wolf]

What if insteand of a tangent line, we tried to make a tangent parabola? Or a tangent 3rd degree polynom, or a tangent 4th degree polynom...

it would approximate the function better and better

[stoic raptor]

what the hail

bro jsut use tangent

its close enough

[brave wolf]

$x-\frac{x^{3}}{6}+\frac{x^{5}}{120}$

is it this close?

[flat frigate]

MathIsAlwaysRight

[stoic raptor]

oh wow

that is beautiful

[brave wolf]

and the cool thing is, that if you take an "infinite degree" polynomial, also known as an infinite series, it actually becomes exactly the function (often for nice enough functions, sometimes it doesnt)

[stoic raptor]

isnt that common sense though

[brave wolf]

Maybe, but it doesnt always work

For example do you think that the following is true:
If a function has all derivatives = 0 at x = 0 and f(0) = 0, then it's just f(x) = 0

[stoic raptor]

yeah

[brave wolf]

it's not true

[stoic raptor]

wait what