help-0

[echo socket]

Where sec is certainly positive/negative

[covert agate]

oh

wait how did he multiply by sec/|sec| outside?

sec/|sec| is still a function

not a constant

<@&286206848099549185>

[lone heart]

@covert agate Has your question been resolved?

[little drum]

What's the issue Chromium?

[covert agate]

integrating |sec|

[covert agate]

?

[little drum]

hmm

[little drum]

Indeed, $\frac{\sec x}{|\sec x|}$ is still a function, that outputs the sign of sec x in the interval for which the integral is done

[ocean seal]

Ansh

[little drum]

the thing is, for whichever interval sec x is Riemann integrable, it's sign function remains the same throughout the interval

so the area under the curve is simply the area under the curve of sec x, with a sign function of sec x attached so that the interval decides on the sign of the area under curve

[covert agate]

i don’f get it

[little drum]

"the area is just the area under the curve of "sec x" with the sign of "sec x" in the interval

[covert agate]

wait so

if F is the antiderivative of f

[little drum]

we're not talking about shit in general (@_@;)
rn, we're talking about |sec x|

If you're soooo hell bent on generalizing though, at most you'd want to talk about some function that has an absolute value function involved

[covert agate]

yea

how do we find int |f|

in terms of F

or is there no general antiderivative expression?

[little drum]

We don't lol.. that's the point.. there's no general antiderivative expression as such
the most realistic approach is to break the |f| into intervals and add up to the respective areas to conclude

,w plot |x| + |x-1| + |x-2|

[ocean seal]

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[little drum]

consider this for example

[velvet pelican]

hmm that actually looks integrable

[covert agate]

it’s always bigger than 0 lol

[little drum]

🤦‍♂️

,w plot |x^3|

[ocean seal]

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Requested by Ansh#3502

[little drum]

theree

[velvet pelican]

iirc $\int |x| dx = x|x|/2 +c$

[ocean seal]

Frosty

[little drum]

hmm yep

[covert agate]

how..?

[little drum]

same thing, x > 0 => x²/2 ; x<0 => -x²/2

You really don't want to think about generalizing it tbh, or it'd be a mess if you start coming across integrals that'd be far easily dealt with by just being flexible with them, than being crude and using the generalized approach

[covert agate]

huh

[velvet pelican]

yea, im not sure whether you can generalise to |f|

[covert agate]

so how do we continue about sec

[little drum]

Start with integrating: $\int |x| \dd{x}$

[ocean seal]

Ansh

[little drum]

Note: $|x| = x\cdot sgn(x)$

[ocean seal]

Ansh

[little drum]

sign function - signum function, whichever you call it

[covert agate]

signum lol

sign ~ sine

not very good in saying out loud

[little drum]

Yeah $|x| = x \cdot \sin x$

Imagine

[ocean seal]

Ansh

[covert agate]

😋

[covert agate]

anyway, what then?

[little drum]

If you got how it works

$\int |\sec x| = \int \sec x \cdot sgn(\sec x)$

[ocean seal]

Ansh

[covert agate]

yea..?

[little drum]

$=sgn(\sec x) \int \sec x \dd{x}$

[ocean seal]

Ansh

[little drum]

that's it :|

[covert agate]

why can we just take it out???

[little drum]

:/

Did you even try the $\int |x| \dd{x}$?

[ocean seal]

Ansh

[covert agate]

yea

it requires ‘integration by parts’

not sure what that is

[little drum]

No it doesn't :/

[covert agate]

why can we take out the signum thing again?

[little drum]

Signum thing, within the interval of "x" is just the sign of the area

which is +1 or -1, i.e., a constant

[pulsar aspen]

Wait, the proof might be less obvious than you think. Imagine a continous but nowhere differentiable function.

[little drum]

:o

[pulsar aspen]

Splitting the case where f(x) = +1, f(x)=-1, or f(x)=0 might not work because it may create a nonmeasurable set.

[little drum]

why would differentiability affect the integral

[covert agate]

can this be rigorously proven

[pulsar aspen]

I'm just being careful. I might still be wrong, though. Just look what if f is a weierstrass function.

[covert agate]

int |f| = signum(f) int f?

[little drum]

the proof lies within the separation of intervals

[covert agate]

wait so this is true????

holy shit it is

differentiate the rhs

you get |f| right?

[little drum]

Well I prefer to think as integrals as the area under the curve rather than the "opposite of derivative" so, yeah that holds true for me

[covert agate]

wait

is signum differentiable?

[little drum]

:p

,w plot sgn(x)

[ocean seal]

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[little drum]

what do you think 👀

[pulsar aspen]

No, I mean, how does separation of intervals work on a function like Weierstrass function which is fractal

[covert agate]

isn’t it always 0..?

[little drum]

yha

[little drum]

hmmm.. abs(....) function basically flips the -ve graph w.r.t. the x axis :o
so whatever the interval is, say any arbitrary interval, when W was negative, is now positive, and the integral in the interval is still sgn(W) int W, in the interval

[covert agate]

so we can’t differentiate sgn(x) • f?

[little drum]

$\lim_{h\to 0} \frac{sgn(x+h)f(x+h) - sgn(x)f(x)}{h}$

[ocean seal]

Ansh

[little drum]

*y = |x| is not differentiable at x = 0

[pulsar aspen]

The problem is, the "interval" may not actually be interval. Is it always the case that {x | f(x) >0} can be decomposed into open intervals and countably many singletons if f is a nowhere differentiable function?

[little drum]

no (@_@;)

the point remains the same still though, although you can't really generalize.. flexibly checking on the intervals for your function should give you a thorough understanding of why your sgn(f(x)) is free to come out of the integral

Also @covert agate I found a decent site for you that I myself refer to when stuck with some questions
If you'd like to check https://tutorial.math.lamar.edu/classes/calci/ProofIntProp.aspx#Extras_IntPf_FToCII

[covert agate]

yea

thats what i use often

so

we assume sec θ > 0, then we’re done with the signum issue?

(will also need a caae where sec θ < 0, but that’s just sec θ > 0 negated)

[little drum]

mhmm

[covert agate]

so 2 answers lol

depending in θ

(depending on x*)

[little drum]

and sgn(sec theta) take cares to adjust that dependency

[covert agate]

original integral 1/√(x² + 1)

[little drum]

and for all practical purposes, you can assume that no-one's smart enough to ask you to integrate |sec x| in interval: say [pi/4, 3pi/4] (@_@;)

because there's a vertical asymptote at pi/2 right there

[covert agate]

wdym

[little drum]

the integral will be done in one of these intervals no?

the sgn(sec theta) derives the sign of the area accordingly and you're done

btw, ngl, the x = tan(u) substitution is hella dodgy too lol

[covert agate]

so is the sec u part incorrect

and should be |sec u|

[little drum]

nope it's correct

[covert agate]

!?

[little drum]

I'm only solely concerned about the x = tan(u) part

has something to do with how inverse trig works but nvm

[pale kestrel]

https://i.imgur.com/Lczqarh.png

Do you understand this?

if you do, then there is no modulus sign to worry about any more.

[little drum]

the sgn(secx) coming out was the issue

[covert agate]

(and i still don’t get it)

[pale kestrel]

For a general f, check if this is true, then.

[covert agate]

(ffs)

[pale kestrel]

by drawing a graph

[covert agate]

int |f| = sgn(f) F?

but sgn is not differentiable lol

[pale kestrel]

??????????

Convince yourself by staring at the area

[covert agate]

aren’t we working with antiderivatives

[pale kestrel]

wtf

[covert agate]

instead of area

[pale kestrel]

urghhhhhhhhhhhhh

[little drum]

hmmmmmmm

[pale kestrel]

FIRST convince yourself with an area argument

if that integral had definite limits

[little drum]

I read absolute value function has no antiderivative

[pale kestrel]

$$\int^b_a |f(x)| \dd x = \frac{f}{|f|}\int^b_af(x) \dd x$$

[hasty elk]

This is so much more entertaining as a lurker

[little drum]

Camilleone saves the day

[ocean seal]

Shuri2060

[little drum]

help

[covert agate]

👏

[pale kestrel]

This is like

You can see this is true right?

[covert agate]

uhh not really

[pale kestrel]

draw an example while I write some latex 💤

[covert agate]

yea i tried lol

[pale kestrel]

just draw a general f

that goes positive negative

then draw what the modulus version looks like

Then notice how the area is flipped by the f/|f|

[covert agate]

wait what even is modulus lol

[pale kestrel]

????????//

| . |

^absolute value signs

$$\frac{f}{|f|}(x) = \begin{cases}1&f(x)>0\\-1&f(x)<0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\int^b_a |f(x)| \dd x = \int^b_a\frac{|f(x)|}{f(x)}f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

@covert agate do u follow?

[covert agate]

yea

[pale kestrel]

But you wouldn't follow if I just took that fraction outside?

[covert agate]

i wouldnt

[pale kestrel]

Let
$$S_{>0}=\bigcup_i\left{(a_i, b_i):a_i<x<b_i, f(x)>0\right}$$
$$S_{<0}=\bigcup_i\left{(a_i, b_i):a_i<x<b_i, f(x)<0\right}$$

hmmmm

no.

[ocean seal]

Shuri2060

[echo socket]

Wouldn't it be easier to write {x : f(x) > 0} tho?

[pale kestrel]

I want to break the integral into bits

depending on the sign of f

So I need to involve the limits right?

ok no no no

I should just use the indicator function

$$\bd{1}_A(x) = \begin{cases}1&x\in A\0&x\not\in A\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\bd{1}_{\bR^+}(x) = \begin{cases}1&x\in \bR^+\0&x\not\in \bR^+\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\bd{1}_{\bR^+}(x) = \begin{cases}1&x>0\0&x\leq0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\bd{1}_{\bR^-}(x) = \begin{cases}1&x<0\0&x\geq0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\int^b_a |f(x)| \dd x = \int^b_a\frac{|f(x)|}{f(x)}\bd{1}{\bR^+}(x)f(x) \dd x + \int^b_a\frac{|f(x)|}{f(x)}\bd{1}{\bR^-}(x)f(x) \dd x$$

@covert agate ok ok this is the next step

[pale kestrel]

This is the indicating function

And I'm using it to indicate if f(x) is positive or not

[covert agate]

not good with union notation

[pale kestrel]

Chuck that

Just the indicating function is needed.

I'm setting A to be the positive reals and the negative reals in the 2 cases

$$\int^b_a |f(x)| \dd x = \int^b_a\frac{|f(x)|}{f(x)}f(x) \dd x$$

$$= \int^b_a\frac{|f(x)|}{f(x)}\bd{1}{\bR^+}(f(x))f(x) \dd x + \int^b_a\frac{|f(x)|}{f(x)}\bd{1}{\bR^-}(f(x))f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

I then split up the right integral into positive and negative area, essentially.

[ocean seal]

Shuri2060

[covert agate]

i don’t get this

[pale kestrel]

$$\bd{1}_{\bR^+}(f(x)) = \begin{cases}1&f(x)>0\0&f(x)\leq0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

^Do you get this?

Let's start with the definition of indicating function

Does that make sense

[covert agate]

yea

[pale kestrel]

Continuing on from this

$$\bd{1}_{\bR^+}(f(x))f(x) = \begin{cases}f(x)&f(x)>0\0&f(x)\leq0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

I'm multiplying by f(x) on both sides.

Does this make sense?

[covert agate]

yea

[pale kestrel]

$$\bd{1}_{\bR^-}(f(x))f(x) = \begin{cases}f(x)&f(x)<0\0&f(x)\geq0\end{cases}$$

[ocean seal]

Shuri2060

[pale kestrel]

And the same for the other case

Next I get to here

By 'splitting' my integral

The left term represents all the 'positive' area (above x-axis)

The right term represents all the 'negative' area (below x-axis)

The left term is the blue area

The right term is the green area

$$\int^b_a f(x)\dd x= \int^b_a\bd{1}{\bR^+}(f(x))f(x) \dd x + \int^b_a\bd{1}{\bR^-}(f(x))f(x) \dd x$$

Sorry to make things clearer

JUST consider this for now.

[ocean seal]

Shuri2060

[pale kestrel]

This is my claim for any general f

[covert agate]

yea

[pale kestrel]

So back to this

ahhh

1 sec

no ok i should get rid of this

Stick with this

$$\int^b_a f(x)\dd x= \int^b_a\bd{1}{\bR^+}(f(x))f(x) \dd x + \int^b_a\bd{1}{\bR^-}(f(x))f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\int^b_a |f(x)|\dd x= \int^b_a\bd{1}{\bR^+}(f(x))f(x) \dd x - \int^b_a\bd{1}{\bR^-}(f(x))f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

Next claim

Modulus sign on function means I add the positive area and subtract the negative area

^I can't take the fraction outside for a definite integral

The outside has no x's (my mistake)

[pale kestrel]

So instead, I will work with this

Ok now I'm suddenly unconvinced by this

🤔 🤔 🤔 🤔 🤔

@hasty elk @little drum is this actually correct?

[covert agate]

(wait i can just tag you guys???)

[hasty elk]

LOL

[little drum]

kek

[hasty elk]

Did you confuse yourself shuri

[pale kestrel]

yes.

im thinking antiderivatives and it makes sense (i think)

but when i go back to areas, i ?????

[hasty elk]

It makes sense

[little drum]

$\int^{\frac{3\pi}{8}}_{\frac{\pi}{4}} |\sec x|$... yeah makes sense
When you go to anti derivative, the antiderivative is piecewise defined

[hasty elk]

The idea is that ./|.| acts as a reflection of the negative part

[ocean seal]

Ansh

[pale kestrel]

ohhhhh ok.

no no, yes yes, godamnit.

[little drum]

[little drum]

what's the integral |sec x| from pi/4 to 3pi/4? would you just plug in the value in the anti-derivative and get F(3pi/4)-F(pi/4)?

[pale kestrel]

Ok, no, this is pretty much as far as we get for the definite integral

[pale kestrel]

Are you crossing poles? If yes, you can't

[hasty elk]

When you get down to it, it's abusing the hell to goddamn mars and back that the negative of the integral is the integral of the negative

[little drum]

ofc right, you're crossing the asymptotes

forget that, the integral itself is divergent when the upper limit is pi/2 :/ but that doesn't stop you from writing the integral in a finite interval as an antiderivative right?

[pale kestrel]

You can write down the antiderivative

Which is indeed an antiderivative

For the domain of the original function

However, if your definite integral spans across poles

It means you chop it up

[little drum]

hmm?

[pale kestrel]

(and it will be improper pieces)

[little drum]

yeah

[pale kestrel]

$$\int_{-1}^1 x^{-1}\dd x = \lim_{a\to0}\int_{-1}^a x^{-1}\dd x + \lim_{a\to0}\int_{a}^1 x^{-1}\dd x $$

[ocean seal]

Shuri2060

[pale kestrel]

Is how we interpret this, I believe. (if it doesn't exist, the integral doesn't)

[hasty elk]

Provided it exists

[pale kestrel]

So I am going to have to go back to unions

$$\int^b_a |f(x)|\dd x= \int^b_a\bd{1}{\bR^+}(f(x))f(x) \dd x - \int^b_a\bd{1}{\bR^-}(f(x))f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

$$= \sum\int^{b_i}{a_i}\bd{1}{\bR^+}(f(x))f(x) \dd x - \sum\int^{b_j}{a_j}\bd{1}{\bR^-}(f(x))f(x) \dd x$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\left(\sum F(b_i) - F(a_i)\right) - \left(\sum F(b_j) - F(a_j)\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

kinda the best I can do atm

For definite

[pale kestrel]

Split the positive parts into bits.

[pale kestrel]

Then you'll find the definite integral corresponds

[little drum]

Ahhhhhhh I seee how you were tryna put it >_<

honestly though, being flexible between switching to antiderivative and area arguments is for the best at hs/early uni level

[pale kestrel]

Sure. Well it's not bad being sus of stuff like this

although in calculus... well hmm

[hasty elk]

In calculus things tend to be sufficiently nice

[little drum]

no no.. I'm not trying to dissuade them here.. Just,

You'll get to know your stuff better as you go one step at a time

[pale kestrel]

It's a lot easier to look back and question later

[hasty elk]

You really only run into issues once you hit the more mindscrewy parts of real analysis, and at that point you have the tools to understand in far greater detail

[pale kestrel]

Like I'm doing rn lol, still not 100% convinced algebraically

[covert agate]

i’ve been watching and i understand 0%

👍🏻

[little drum]

The only use for the "antiderivative argument" which I see, at this level, is if you're proficient enough to assume a F and differentiate it, make some alterations and conclude your integral that way and correspondingly saves a hell lot of time than say, u-substituting/using IBPs etc. on several occasions

[pale kestrel]

ok ok 1 sec

I'm confused how the heck that antiderivative corresponds to the definite integral

Surely it doesnt and you have to chop that integral and then use that antiderivative

[pale kestrel]

I'm like hoping I'm wrong and just thinking about something wrong but

[hasty elk]

We're back to the same argument that the area is really specific to a...well, a specific antiderivative

[pale kestrel]

Am I wrong lol. If yes, just gonna figure it out haha

[hasty elk]

For modulus functions antiderivatives tend to require piecewise definitions

And that makes sense because there's no actual way to integrate |x| in one shot using only elementary functions without resorting to piecewise chopping

[pale kestrel]

Ok I am right

You have to take the definite integral piecewise

The area clearly isn't negative.

@covert agate sorry, does that make more sense?

[covert agate]

so that’s for the definite integral?

piecewise shit

[pale kestrel]

Back to the original problem

Your 'anti-derivative' IS gonna end up being piece wise

[covert agate]

uhh this is the og integral? (shouldn’t we end up with |sec u| instead of sec u)

[pale kestrel]

And not continuous in most cases

[hasty elk]

Er

Shuri

What

[pale kestrel]

what?

?

[hasty elk]

Your antiderivative is not gonna be continuous

Are you sure

[pale kestrel]

In this case it wasn't.

[hasty elk]

Desmos is actually screwing this up a bit

[pale kestrel]

huh?

You can create a continuous antiderivative

[pale kestrel]

But sure as hell not like this

[little drum]

:o

[hasty elk]

Yeah

[pale kestrel]

You want to shift the purple thingies up and down

to connect

So that antiderivative from the calculator is only valid over joined up bits

[hasty elk]

But saying antiderivatives are not continuous kind of breaks things very badly because antiderivates need to be differentiable

[pale kestrel]

So when you calculate the definite integral it needs to be broken up

[pale kestrel]

In that case the calculator approached wrongly surely

What it spat out wasn't an antiderivative

It was only one over piecewise domains

[hasty elk]

Calculators are dumb that way

:P

[pale kestrel]

If I took the purple graph mod 2

then it would be ok

[hasty elk]

Yes

[pale kestrel]

and also not ok

🤦

[hasty elk]

LOL

[pale kestrel]

This reminds me of winding numbers if we take mod 2

[hasty elk]

This reminds me of how integral of 1/x is not ln |x| + C

In my year 1 everyone went wtf at it

[covert agate]

?????????

wtf

[hasty elk]

^ there we go

[pale kestrel]

elaborate 😢

it is over the original domain

(but the C is not constant everywhere)

[hasty elk]

Precisely

[pale kestrel]

That's the special property of the constant of integration 😉

[hasty elk]

C can differ at less than 0 and more than 0, so the true integral is written piecewise

Very mindscrew

[pale kestrel]

OMG i did it

fuuuuuuuu

1 sec

Now we have a true anti-derivative 😏

[covert agate]

bruh

[pale kestrel]

@covert agate On a more serious note, I can't imagine anyone being asked to find the antiderivative of |f(x)| properly

Like where the hell did that question come from smh

[covert agate]

original problem

[pale kestrel]

oh so that was the problem.

well, big hmm.

[little drum]

=_= don't involve me smh

Your problem was: antiderivative = integral

maybe, I'm not even sure lol... that statement was only in concern of the substitution

[hasty elk]

I don't even know what the original problem was HAHA

[pale kestrel]

me neither

i thought it was meant to be integrate |sec x|

[tawny orchid]

I think?

I don't know

[hasty elk]

I just came here after finishing writing a four page latex on parameterizing a general circle in 3D

[hasty elk]

Saw entertainment

[covert agate]

3rd line should’ve been |sec x| no?

[tawny orchid]

Where's the 3rd line?

Blue ones?

[pale kestrel]

bottom.

[little drum]

the original problem in context 🤦‍♂️

[pale kestrel]

I'd honestly hate doing this kind of problem without limits

(i mean definite integral)

[hasty elk]

For me, it's tuesday

I deal with shady integrals all the time these days

[pale kestrel]

Wellll I don't find being able to do hard integrals by hand thaaattttttt important 😂 😂 😂
Realistically, most integrals are impossible to solve and we have computers 😏

[hasty elk]

Gotta keep on the ball if i don't want to do shady things

[hasty elk]

Two words: Pure. Math.

[covert agate]

wait what do you do in phd math lol

[pale kestrel]

cami is phd?

[covert agate]

yes

[pale kestrel]

😮

Must be working on riemann

[covert agate]

shuri are you bsc

[pale kestrel]

bsc vs msc

whichever is 4 yrs

think its msc

[hasty elk]

I'm working on functional analysis, harmonic analysis, and statistics all rolled in one

[pale kestrel]

and riemann

[hasty elk]

I don't even know how i got to this point

[pale kestrel]

👀

don't think ill get into phd

[covert agate]

how cursed is phd statistics like

[hasty elk]

It's all integrals

[pale kestrel]

that sounds nicer than normal stats

[covert agate]

multidimensional ones?

[hasty elk]

^ that's me daily

[pale kestrel]

$\int\int\int\int\int\int\int\int\int\int f(A,B,C,D,E,F,G,H,Z)$

[ocean seal]

Shuri2060

[covert agate]

i can’t be bothered to count how many integrals there are

sweet jesus

[hasty elk]

One dimension, two dimensions, n dimensions, countably many dimensions...

All sorts of shady things appear

[pale kestrel]

Sick

countably many

[hasty elk]

Phd courses: here have n dimensional integrals for free

[covert agate]

in fucking statistics??

[hasty elk]

Phd research: here have countably many integrals for free

Yes, in fucking statistics

Tbh i like it better this wY

I actually understand things

Not like bullshit shady hypo testing

[sterile idol]

no thanks :)

[hasty elk]

Give me countably many integrals any day

[sterile idol]

do we even have a choice

[pale kestrel]

@covert agate Ok back to the original problem, when you do your substitution I think you're meant to separately consider the integral over different domains

If you do sub in sec

[covert agate]

wdym

[pale kestrel]

$\int\frac{1}{\sqrt{1+x^2}}\dd x$

It was this right

[ocean seal]

Shuri2060

[covert agate]

yea

[pale kestrel]

oh it was a tan sub

$$x = \tan u$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\dv{x}{u} = \sec^2 u$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\dd{x} = (\sec^2 u)\dd u$$

[ocean seal]

Shuri2060

[pale kestrel]

Ok, so whenever we substitute for indefinite integrals

We have to be able to undo the substitution

[covert agate]

yea

[pale kestrel]

to get the final answer in terms of x

Therefore, we only consider bijective versions of tan

[covert agate]

bijective??

[pale kestrel]

Yes, we only let u go from -pi/2 to pi/2

uhhh you know what bijective means or not?

[covert agate]

no

[pale kestrel]

D:

[covert agate]

not surjective

[hasty elk]

uh oh

[covert agate]

maybe injective

[pale kestrel]

Ok you really need to know

at this stage

[covert agate]

isn’t it analysis

[hasty elk]

NO

[pale kestrel]

Not really?

Its just like basic functions stuff

Elementary set theory???

Like something you should know anyways.

[hasty elk]

bijective functions is like everyone should know

[covert agate]

(can barely do sets, fuck)

[pale kestrel]

what do u mean barely do

if u can union intersection complement

you're good to go for now lol

[covert agate]

set theory is uhh

translated version of logical operators?

idk

[pale kestrel]

the heck?

[hasty elk]

wat.

[pale kestrel]

$A\cap B$

[ocean seal]

Shuri2060

[pale kestrel]

A intersect B

D:

[covert agate]

yea

[pale kestrel]

anywayyyyyyyysssss

back to functions

[covert agate]

cup as to or
cap as to and

[pale kestrel]

$${(x, y)}$$

[ocean seal]

Shuri2060

[pale kestrel]

Curious, have you seen functions denoted like this

[covert agate]

no

[pale kestrel]

as a set of ordered pairs (x, y)

[covert agate]

cartersian product shit??

✝️

[pale kestrel]

hmm

I think you should read up

but idk what to suggest

$f: A \to B$

[ocean seal]

Shuri2060

[pale kestrel]

This notation is familiar?

[covert agate]

yea

[pale kestrel]

f is a function that maps to domain A to codomain B

[covert agate]

codomain is superset of range

[pale kestrel]

yes.

[covert agate]

when codomain = range

f surjective?

[pale kestrel]

yes

[covert agate]

huh so

all supersets of range = range

then f surjective????

[pale kestrel]

$f(x) = f(y) \implies x=y$

[covert agate]

range = universal set

wtf

[ocean seal]

Shuri2060

[pale kestrel]

this is injective

[covert agate]

injectivity

[pale kestrel]

what do you mean surjective

superset

what

anyways, range = codomain

means surjective

[covert agate]

superset is like anti subset?

[pale kestrel]

look, i never use that word

[hasty elk]

i have a feeling i need to track down a good book

[pale kestrel]

almost

ever

$${(a, f(a)) : a\in A}$$

@covert agate so this is an alternate way to represent functions

[ocean seal]

Shuri2060

[covert agate]

huh

[pale kestrel]

This represents f

For example, if f(x) = x^2 for real x

(1, 1) is in the set
(2, 4) is in the set
(sqrt2, 2) is in the set
(-sqrt2, 2) is in the set

[hasty elk]

think of it in terms of a graph, the set of all coordinates of the function f is exactly that set

[covert agate]

oh yea coordinates

[pale kestrel]

Why I bring this up

[sterile idol]

Oh I know this y = f(a) so (x, y) and a is in some set?

[pale kestrel]

Well if a function has an inverse

[covert agate]

inverse iff injective

yea?

[pale kestrel]

no

[covert agate]

wait, how do we define codomain lol

[pale kestrel]

bijective = injective AND surjective

[pale kestrel]

yes.

its just a superset of the range

[sterile idol]

Some of this is recognizable from haskell

[pale kestrel]

lel

[hasty elk]

@covert agate https://www.math.tamu.edu/~florent/teaching/lecture_notes/220lecture_notes.pdf

[pale kestrel]

Inverse exists iff bijective

[hasty elk]

it's got nice colours

[pale kestrel]

If inverse exists, then the inverse function will be all those coordinates but swapped

And you kinda should be familiar with this stuff

before doing idk

calculus

When substituting in indefinite integrals, you need a bjective substitution

or you can't undo it (since the final answer needs to be in terms of the original variable)

[hasty elk]

before doing any kind of mathematics at the high school level even

[covert agate]

bruh

i dont get it

[pale kestrel]

which bit.

[covert agate]

what surjective means

lol

[pale kestrel]

Readup on chapter 5

[covert agate]

?

[pale kestrel]

cami sent

[hasty elk]

https://www.math.tamu.edu/~florent/teaching/lecture_notes/220lecture_notes.pdf

go

page 46

[pale kestrel]

$$f:\bR\to\bR$$

[ocean seal]

Shuri2060

[pale kestrel]

We usually do stuff like this

even though f might not map to all of R

Just for convenience/generalisation

[sterile idol]

Wait what's the difference between codomain and range

[pale kestrel]

Codomain can be any superset of the range

The range is everything f can map to

$$\Im(f) := f(X) := {f(x) : x\in A}$$

[ocean seal]

Shuri2060

[pale kestrel]

Where A is the domain

Image of f is the same as range of f.

[sterile idol]

So super set of Z is R?

an example

[pale kestrel]

Not relevant to this, but yes?

$$\bR\supset \bZ$$

[sterile idol]

Yeah it wasn't supposed to be

[ocean seal]

Shuri2060

[sterile idol]

What's Im here

[pale kestrel]

Image of f

[sterile idol]

and :=

[pale kestrel]

= Range of f

:= defined to be

[sterile idol]

image?

[pale kestrel]

🤷‍♂️ look it up, but that's how it's defined

It's another way of saying range

[sterile idol]

Ah ok

[covert agate]

alright then?

[pale kestrel]

so we restrict the domain of tan

to

(-pi/2, pi/2)

that makes tan bijective and therefore invertible

We only consider the middle curve this way

Now we have a substitution function that is invertible

This is important for indefinite integral substitution. After substituting and integrating, you need to undo the substitution to have the final answer in terms of the correct variable

Definite integral substitution --- you don't necessarily need a bijective substitution.

[sterile idol]

because you're constricting to a range of values right?

in a definite integral

[pale kestrel]

You also convert the limits of the integral when you substitute for definite.

So there is no need to undo the sub

[sterile idol]

Oh ok

I have only reached till limits in school till now

so I don't know any of this

Good day

[lone heart]

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[oblique vigil]

I’m absolutely clueless as to how and why exactly this happened. Thank you to whoever helps me!!!!

I need it explained like I’m 5, because I’ve been literally trying to understand how and why for the past 2 days

[lone heart]

@oblique vigil Has your question been resolved?

[sage bronze]

(-log(3) -4log(8))x = -2log3
(-1)[log(3)+4log(8)]x = (-1)(2log3)
divide both sides by -1

[oblique vigil]

Do you know why we divide both sides by -1?

[placid zinc]

"Divide both sides by -1"
Is a way to change the sign of both sides

If -x = -3
Then x = 3

If (-log3 - 4log8)x = -2log3
Then (log3 + 4log8)x = 2log3

You can also think of this as "Multiply both sides by -1", same thing.

[oblique vigil]

Can we do that because the entire equation is negative on both sides?

[placid zinc]

You can just always do that

You can multiply both sides by 2 any time you want. You can multiply both sides by -1 any time you want

That is, you can flip the sign on both sides any time you want

[oblique vigil]

Ok, now this is making a lot of sense!!!!!!!!!!!

So no matter what the equation, this isn’t special to logs or anything

[placid zinc]

Ye the logs aren't doing anything to make this happen

[oblique vigil]

Thank you so incredibly much!!!!!

[placid zinc]

Np, feel free to ask if you have anything else.

[oblique vigil]

Thank you!!! :))))

.close

[lone heart]

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[warped spruce]

is it the answer the this question?

just wanna make sure

you need to simplify

?

[alpine sable]

Yes

[warped spruce]

are you sure bro?

[alpine sable]

Yeah

[warped spruce]

thank you my friend

[alpine sable]

Np

[warped spruce]

have a nice day!

.close

[lone heart]

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[alpine sable]

U too

[lone heart]

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[near nest]

hello

[glass lichen]

You got help with this already.

[near nest]

another problem I have

how did the power change from ^1/4 to ^1/2

[glass lichen]

Explained already.

Refer back to like 10 minutes ago when I explained it to you

[near nest]

okay

.close

[lone heart]

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[astral wasp]

Can someone tell me the steps to solve it?

[gray isle]

what are the options?

[gray isle]

start by explicitly marking the lengths on the sides of the triangle
as well as marking the location of the angle

then based on the relative positions of the sides you know to the angle you want to find
identify the appropriate trig function to use and set up your equation

[astral wasp]

I don't understand how to identify the appropriate trig function

[gray isle]

did you do that

start by explicitly marking the lengths on the sides of the triangle
as well as marking the location of the angle

[astral wasp]

I did the 1st part I don't understand the 2nd part

[gray isle]

wdym by first part

[astral wasp]

I marked the lengths on the sides of the triangle.

Do I put hypotenuse where I marked AB and 90 where C is?

[gray isle]

can you show me how you marked your diagram

Do I put hypotenuse where I marked AB and 90 where C is?
you can if you want

[gray isle]

can you identify whether BC (the side of length 13) is opposite or adjacent to angle A?

[astral wasp]

Opposite

[gray isle]

and you've also already identified that AB (the side with length 16) as the hypotenuse

are you familiar with
soh cah toa?

[astral wasp]

No

[gray isle]

its a mnemonic to help remember what sin,cos,tan do in a right triangle

and it would be helpful to look that up

if you don't know it

[astral wasp]

Okay

Thanks

[lone heart]

@astral wasp Has your question been resolved?

[lone heart]

@astral wasp Has your question been resolved?

[lone heart]

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[strong fractal]

Does this look right?

[strong fractal]

anyone/

[echo socket]

You wrote that intersection of A and B is {1, 3, 9}?

[strong fractal]

I wrote A'∩ B is {1, 3, 9}

[echo socket]

What's A' ?

[strong fractal]

lol

it means compliment of A

(NOT A)

[echo socket]

Ah

Yeah it's {1, 3, 9}

I think iii is wrong

It should have elements that are in A or are not in B

So that'd be every element except 1 3 and 9

[strong fractal]

Yes well spotted thanks

[lone heart]

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[lone heart]

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[fresh relic]

Let f be a function, a in the domain of f .
State the definition of f continuous at a.

[fresh relic]

Can anyone explain to me what this question is telling me to do?

[wanton junco]

it just tells you to say what does it mean for function f to be continuous at a

[fresh relic]

It has no holes or jumps?

[echo socket]

Have you been introduced to definition of continuity at a point?

[wanton junco]

at point a

[fresh relic]

Not yet

[echo socket]

Well you need that

[wanton junco]

and you're being told to state the definition ?

a bit weird

[fresh relic]

It just says state the definition of f continuous at a

[pale kestrel]

the limit of f as x goes to a is equal to f(a)

[fresh relic]

thats the answer?

[pale kestrel]

Might as well tell you. Yes.

$$\lim_{x\to a}f(x) = f(a)$$

[ocean seal]

Shuri2060

[tacit arch]

Sometimes the questions are really that easy

[fresh relic]

wow

it is an easy question but I would have never known it lol

[tacit arch]

It's probably in your textbook

[wanton junco]

oddly enough on one of the exams this answer was not accepted because it was not rigorous enough

they expected an epsilon delta definition instead

[tacit arch]

https://c.tenor.com/Gg6DzBNQ2VEAAAAM/table-flip.gif

[pale kestrel]

?

[tight locust]

??

[pale kestrel]

Well whatever your exam wants.

[last tendon]

I don't really buy that, there is nothing to prove when asking for a definition

[pale kestrel]

But the left hand thing

is defined by epsilon delta

[wanton junco]

I corrected it

[pale kestrel]

I can write it out in full if I want

[tacit arch]

I assume if OP knew the epsilon delta definition, they wouldn't ask

Sorry for side railing. Feel free to .close @fresh relic

[fresh relic]

Oh no worries. Appreciate the help from everyone

.close

[lone heart]

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[inner sentinel]

how do i go go about doing this one

[pale kestrel]

'best' 🤔

I think what they want you to do is draw it

and decide which way of chopping it up is smarter

[inner sentinel]

i drew it but idk how to understand which way to chop it

[pale kestrel]

can u show diagram

[pale kestrel]

eh?

you should get used to just sketching on paper but ok

You should shade in which region is of interest

[inner sentinel]

the y-axis about the x-axis?

[pale kestrel]

?

no clue what you mean

shade in the bit which is relevant

for integration

[inner sentinel]

the question says rotated about the y-axis about the x-axis

[pale kestrel]

the part of the solid you're rotating

[pale kestrel]

wrong.

They give you 2 equations

what is the 3rd equation which forms the last side of the triangle

[inner sentinel]

y = x ?

[pale kestrel]

where is there any mention of that

[inner sentinel]

wdym 3rd equation

[pale kestrel]

If those brackets and highlighting make it easier to read

[inner sentinel]

ohhhh

i misread i think so is it saying like y = -x + 8 and x = -y + 8?

[pale kestrel]

huh?

y=3x

y=-x+8

y-axis

Are the 3 lines that bound your region

ie. triangle.

[inner sentinel]

the third equation is the y axis?

[pale kestrel]

The 3rd equation is x = 0 (the y axis)

So now shade in the correct region

[pale kestrel]

no.

[inner sentinel]

uhhhh

[pale kestrel]

yes.

Now this thing is going to be rotated about the x axis.

[inner sentinel]

ya

[pale kestrel]

Visualize this solid

then figure out which answer

[inner sentinel]

how do i visualize it

like i can only visualize it being rotated around the red line function

[pale kestrel]

??

no, rotate this thing around the x-axis

Use a 3d graphing program if necessary

/draw the thing on paper

and try

[inner sentinel]

is it shell with respect to x

[pale kestrel]

wait a sec

Draw what you mean

like a visualisation

cus I don't think so.

[lone heart]

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[lone heart]

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[alpine sable]

I asked this question last night, but apparently it was unsolvable. I'm going to add some more information now and ask if it's solvable.

[alpine sable]

So, I want to solve for function f.
I know that the curve has no inflection points in range [0,8], that it is concave-down and increasing, and that it passes through 2 specific points.
With this information, is it possible to solve for f?

[ebon condor]

No

I’m sure you can find a polynomial that solves that

But imagine you had something like 1/100000 x^6

You can just add a bunch of insignificant terms that change the curvature slightly

[alpine sable]

What if there were a stationary point in [1,8], and I knew where it was?

[lone heart]

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[alpine sable]

I'm stuck trying to find A B C and D

not sure what I did wrong

[pale kestrel]

Cx + D

should be the numerator on top of the quadratic denominator

[alpine sable]

Because of x^2?

[pale kestrel]

the numerator should always be 1 degree less than the denominator.

[alpine sable]

oh so it's not the form (ax^2 + bx + c) but degree?

I thought we were only allowed to use Cx + D if the denominator's in ax^2 + bx + c form

[placid zinc]

It isn't only about degree, as something like x/(x + 1)² has a different form

But you do have the form ax² + bx + c there

[pale kestrel]

True true

squared (or more) factors have special case

[alpine sable]

that looks painful

guys I can't solve these equations

I think I'm stupid

sigh

@pale kestrel help pls

[pale kestrel]

have u considered cover up method?

or substituting like x = 1

x = -1

[alpine sable]

oh right, now that I have a new equation I should try that

so what I do is find the values that make each of these undefined and substitute them in right?

[pale kestrel]

uh.....

after you multiply the denominators out

[alpine sable]

yeah

so

ohh

alright

I managed to find A and B but not C and D

A = 1/4 and B = -1/4

oh 1 sec

Found C using x = 0

but if I sub plus or minus 1 for D I get either A or B?

so what should I do in this case? @pale kestrel

C = -1/2

[frigid geyser]

when is $x^2 +1 = 0$?

[ocean seal]

xdk1235

[alpine sable]

no it's (x^2-1)

[frigid geyser]

huh this has +1

[alpine sable]

something must've happened while I was rearranging

no nothing went wrong

[alpine sable]

am i supposed to go by the fractions or the resulting equation? I'm pretty sure it's the resulting euqation because that's what i see on my textbook

[frigid geyser]

x^2 + 1 is right

are you not using the cover up method?

[little drum]

$\frac{1}{4}(x+1)(x^2+1) - \frac{1}{4}(x-1)(x^2+1)+C(x^3-x)+D(x^2-1) = 1$

[alpine sable]

These two are the only methods I've been taught

[little drum]

You could totally compare the coefficients at this point after getting A and B-

[ocean seal]

Ansh

[alpine sable]

for D?

[little drum]

For both C and D

[alpine sable]

I already found A, B and C

[little drum]

oh

[alpine sable]

C = -1/2

but I ndo what to put in for D because it becomes a zero

oh my God

it could be 0

why did I think it couldn't be zero?

oh no wait

C's not zero

[little drum]

tbh, I don't see how you got C = -1/2

[alpine sable]

it says D is -1/2

not C

https://tenor.com/view/dh9511dh-head-banging-on-the-wall-gif-13355825

[little drum]

:p right? so you get C = 0, D = -1/2

[alpine sable]

ye

right thank you again so much

.close

[lone heart]

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[sullen trout]

Hey friends. Starting taking Calculus II this semester, and I haven't been in a calculus course in a few years. I get the priviledge of uncovering all of the gaps in my mathematical knowledge.
I'm stuck on finding the integral of a function. I feel like I'm overthinking the u-substitution.

[alpine sable]

have you tried u = x^2 - 1?

uhh 1 sec

[sullen trout]

I have u = (x^2) + 1.
du = 2x

And my trouble is substituting this back into the equation

[alpine sable]

i missed the integral sign between 1/2 and 1/sqrt(u)

[little drum]

Coffee's translation: Put $u= x^2 - 1$ and the expression converts to- \
$\int \frac{1}{2\sqrt{u}} \dd{u}$

[sullen trout]

What is being substituted in the place of x in the numerator?

[ocean seal]

Ansh

[sullen trout]

I also saw that calculation on an external website. What I'm tying to figure out is where the x in the numerator goes.

[sullen trout]

Sorry, I'm also working while doing homework.

I'm confused why dx was substituted into the equation when we're solving in terms of u.

why would it not be sqrt(u-1) / sqrt(u)

[alpine sable]

oh because we can't have u inside the integral when we have dx or vice versa

I learned that here

[sullen trout]

also good to know, thank you

i'm thinking more so of how dx = (1/2x) was substituted into the function when du is already present

it looks like dx (du)

It makes sense how the x's cancel out to leave 1/2

but why was thewait

WAIT

it clicked

holy heck

I see it now. The u-substitution was distributed to du/2x, which is du (1/2x).

because you're replacing dx in terms of du, the du is also accompanied by 1/2x

cool. Thanks for the help yall!

[alpine sable]

POG

@sullen trout could you type .close please?

@pale kestrelwanna see something I solved on my own?

Can I dm you? It's not a question

[pale kestrel]

go for it

[sullen trout]

.close

[lone heart]

Channel closed

Closed by @sullen trout

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Available help channel!

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lucid wadi]

Hi

[lucid wadi]

How is it 7-x

isnt it supposed to be x-7

[pale kestrel]

$$x = 7 - A$$

[ocean seal]

Shuri2060

[pale kestrel]

$$A = 7 - x$$

[ocean seal]

Shuri2060

[lucid wadi]

I still dont get it

[pale kestrel]

Add A to both sides.

Then. Take away x from both sides.

[lucid wadi]

wait

can't i just do x-7

then divide by negative 1

= -x+7

[pale kestrel]

whatever floats your boat.

/........

[lucid wadi]

oh i see

thank yoy

you*

have a good day

.close

[lone heart]

Channel closed

Closed by @lucid wadi

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[astral solstice]

I want to prove $2^n>n^3$ for n grater or equal than 10