precalculus

[deft crow]

ohhh I see now lol

thank you

[viscid thistle]

@civic plaza Being taught things before knowing what they mean is bad, that kills learning.
It's pedagogically harmful

It's teaching students to rely on formulae instead of reasoning and intuition

[civic plaza]

You're suggesting teaching the cross product by formula though

Why not just teach them how to compute it and in later class when they learn it correctly they'll see it not for the first time

[proud raven]

ill look up determinants

i think i can memorize how to do it at least

well sort of

does that mean vectors of different sizes dont have cross products

like if you have 2 4d vectors

[civic plaza]

Bring taught before knowing in some instances is gonna be necessary because like I said, I was taught in mechanics because you use the cross product in that class fairly frequently

[proud raven]

or you just do something to split them into smaller vectors

[willow bear]

cross product is specific to R^3 yes

[proud raven]

🤔

[willow bear]

spaces of other dimensions just don't have it

[proud raven]

my first homework set is doing cross product of 2 dimensional vectors

maybe its just intuitive

the first one is like cross product ihat jhat

oh but 2d vector cross product exists

just not in 2 dimensions

[echo plaza]

I mean if you "try" it in 2d you'd just get a scalar

[proud raven]

how does that work

with only one column

[echo plaza]

not very well

just from generalising this

[proud raven]

oh its literally just sin 90

[echo plaza]

and I don't know of any use for the 2 dimensional one

[proud raven]

i guess itd always just be |u| |v| sin

itd always be k something

huh

[echo plaza]

you can't get a vector because you're stuck in R² and the vector would need to be perpendicular to both of them

[proud raven]

so theres no actual answer to this question but just an intuitive one

i can live with that 😄

[viscid thistle]

Knowing how to compute something that you don't understand is useless information that will confuse

@civic plaza

If you're mentioning the word "determinant" to students who have taken linear algebra in relation to the cross product, then they'll know how to use relevant theorems to determinants to help them solve problems
If you're working with students who haven't, then the word "determinant" is absolutely meaningless to them, just as meaningless as a simpler formula would be

[civic plaza]

Meaningless is an exaggeration.
Plenty of people do fine learning to compute the determinant in other classes before being formally introduced in a linear algebra class.

I used the cross product in three classes before taking linear algebra and I turned out okay.

I don't quite understand what the argument is really

[bright oak]

Actually, I strongly disagree with the idea that you can compute something without understanding it, and that is a bad thing... not in all cases

In fact, it can be a great learning tool, to use some magic formula to get the right answer, and then figure out why the magic formula works. Has happened to me plenty of times, I learned a lot after I got the answer, and then backtracked, figured out what's going on. Far more useful than just having an empty paper.

[proud raven]

20 minutes into another lecture i realized the point of the exercise was to show that a cross b doesnt equal b cross a

this is gonna sound misguided

is the cross product just something that happens to give useful results or is it a meaningful/intuitive operation on its own

[timber plinth]

it's useful and is meaningful

[autumn sonnet]

Is the correct way to solve log_(1/2)(log_(2)((1+2x)/(1+x))) > 0 to make both sides the exponent of 1/2? That way I got a solution but I'm not sure if it's correct

[lyric girder]

Pardon me if this doesn't belong here, but does anyone have any recommendations besides Khan Academy for an adult trying to re-learn Algebra and then learn pre-calc + Calc 1? I think I'm in the minority here, but I can't stand Khan Academy's UX and content organization.

If anyone has any book or other resource information I'll gratefully check it out. I'm considering signing up for the ThinkWell course at some point.

[limber bone]

dms

[proud raven]

@lyric girder you're not alone, i hate it too

if the question asks first whats the cross product

which i found, and then asks to confirm they are orthogonal

is there something special about the resultant vector that lets just the dot product work

or is that just the fact that the dot product can confirm if any two vectors or orthogonal

actually scratch all that i just realized how stupid that question is 😄

@lyric girder I told someone else that theres a book in #books-old that is listed as suggested for someone teaching themselves calculus

Calculus for the Practical Man (James Edgar Thompson)
Requires basic algebra, trigonometry
“A great book to learn calculus on your own.”

[torn swift]

multivar channel would work better

[turbid bear]

I have 3 questions regarding polynomials.

What is the actual point or goal of polynomials?
This question was sparked by my more detail-specific questions:

1. Allow division by a variable/indeterminate
2. Allow negative exponents/degrees of variables/indeterminates

These 2 rules seem arbitrary, but I suppose everything seems arbitrary if you don't understand the point of what you're doing.

[willow bear]

if you allowed negative exponents or division by the indeterminate you'd lose the guarantee that the polynomial can be evaluated at any point

and well... polynomials are basically abstractions of expressions that you can form with addition, subtraction and multiplication only

[turbid bear]

That definitely makes more sense now thank you

[viscid thistle]

hi I'm looking for precalculus textbook reccommendations.. one with less applications and more deep intuituve understanding and rigor

[limber onyx]

coolmath.com has a lot of precalc

https://coolmath.com

http://www.coolmath.com/precalculus-review-calculus-intro

[hearty ether]

lmao ive known about cool math games for like 10 years and never once considered there was an actual maths component on the site

[limber onyx]

Yep

[viscid thistle]

That's an interesting site gotta admit. It can be supplmentary with a good textbook besides it

And I actually never thought it did too @hearty ether

[hearty ether]

lmao

[limber onyx]

Probably you want a real answer

Why do you guys know coolmath games though

I know it’s popular for its games

[hearty ether]

idk

[limber onyx]

But I didn’t know it’s that popular across the nation

[hearty ether]

its the site we were allowed to play when we were done with our work

thats all

[viscid thistle]

I've seen my friends play on there once in a while when they're bored

[limber onyx]

The website also has real math games as well

Although the owner Coolmath Karen considers all games good for critical thinking iirc

I think I read literally everything on the website, including its financial advice

No joke

No joke that’s where half of the self taught in my username comes from

[viscid thistle]

coolmath was a big resource

did you learn mostly from it?

[limber onyx]

Yeah

btw, this is coolmathgames description from the owner: This is our brain-training site, for everyone, where logic & thinking meet fun & games. These games have no violence, no empty action, just a lot of challenges that will make you forget you're getting a mental workout!

You may not be doing math

But the games are often puzzles over action

They really are “math” games

In a loose sense

[viscid thistle]

These really excercise your brain for more mathematical thinking. I'm thinking of using a lot more now from that description

[limber onyx]

Maybe I should write a joke paper on the mathematics of puzzle flash games

I over analyze the mechanics, etc

But yeah, they have straight prealgebra too, although less known

I suspect the website is so popular because its unblocked on several school firewalls

And therefore people think the math is a facade

But it actually isn’t

[viscid thistle]

I would gladly enjoy that paper. Yeah schools usually allow puzzle games to make it more a fun way to learn for younger students. I have the same opinion on that as well.. math has a more beautiful aspect if you look deeper into it. The way they teach math now in schools isn't enjoyable which makes students forget the true beauty of math

[limber onyx]

Unfortunately puzzle skills don’t really translate over to traditional math

So it is a bit silly to call them math games

At least usually

You won’t learn factorization from 2048

[viscid thistle]

I'll call them puzzle skills for now

[kind pier]

I have no idea where to begin for these two questions

[timber plinth]

You want it to be that H(x) = f(g(x))

So, your answer was wrong because you did f(x) = sqrt(9), which is the constant function

instead, you might what something that isn't constant @kind pier

[kind pier]

root of x is not a constant

[timber plinth]

but sqrt(9) is

and f(g(x)) is constant if either f(x) or g(x) is

which i leave to you to figure out why

[kind pier]

so algebraically I want to functions to multiply together to get H(x)?

[timber plinth]

no

f(x) is a function, so you want f(g(x)) = H(x)

it's not multiplication, it's composition

$(f \circ g)(x) = f(g(x))$

[kind pier]

9+x?

[obsidian monolith]

Darkrifts:

[kind pier]

under the square root

[timber plinth]

Perhaps emil, try it

preferably on paper first

or any other such means

[kind pier]

its not actual homework

its infinite attempts practice

that didnt work though. I was thinking the parent function was square root of 9 + x and the other function was just the root of x

oh wait it did work nvm

[timber plinth]

nice

[kind pier]

oh ya the other on under it was |6-x| , x^5

thanks

[timber plinth]

Also emil, f and g aren't unique

there's more than 1 way to get it (even if you throw out the identity)

[kind pier]

I was thinking like graph transformations from what you said

the parent is always x by itself with the constants

and then whatever is on the x directly is the other function

[timber plinth]

I'm not sure that's the best interpretation, but i'm no teacher

You could also directly replace the occurrences of x with the function g(x) too

[kind pier]

this is work from like something we skipped and if you wanna learn it on your own you can hahaa so I got no notes

[timber plinth]

nice

So, that's "composition of functions"

It's rather nice, especially the decomposition, but they're obviously not exactly uniquely able to be decomposed, so that's a thing

it's basically take f after you take g

[kind pier]

so would |6-x^3| , x^2 have been right also?

[viscid thistle]

doesn't it have to be A

[timber plinth]

@kind pier no that would be x^6 in the middle

[viscid thistle]

becuase the other's change the conic, since they only have A or C so it would be a parabola

[viscid thistle]

Can I get help

If anyone is on

[torn swift]

Pablo from the Backyardagains

[royal aspen]

just post the question

[wary plover]

this problem is simple but I cant for the life of me remember quite what the idea is

is it E?

[short sorrel]

helps to draw a diagram first

this is how i'd interpret the question

the ramp being 210 ft long probably means the length you walk on

is 210 feet

so now, recall SOH CAH TOA

here, the angle with the ground is in the bottom-left

so the two sides we know are

the opposite

and the hypotenuse

follow so far?

as a result, we use the trig ratio with O and H

that happens to be SOH, for sine

so we have:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

[wary plover]

that's what I figured

[short sorrel]

er

[obsidian monolith]

Namington:

[short sorrel]

typo sorry

[wary plover]

im fine at trig idk why I couldnt remember how to do that

[short sorrel]

anyway, now we need to solve for theta

currently

theta has this nasty sine acting on it

to get rid of sine

we take the inverse sine

$\sin (\sin^{-1} (x)) = x$, after all

[obsidian monolith]

Namington:

[short sorrel]

so it "cancels" so to speak

(This is technically domain-dependent but dont worry about it)

but of course, what we do to one side

we have to do to the ohter

so take the inverse sine of both sides

and we get

$\sin \theta = \frac{31}{210} \
\implies \sin^{-1} (\sin\theta) = \sin^{-1} \left(\frac{31}{210}\right) \
\implies \theta = \sin^{-1} \left(\frac{31}{210}\right)$

[obsidian monolith]

Namington:

[wary plover]

cool, thanks

[kind pier]

if they are asking me to just interpret functions and sketch graphs without plugging in random numbers

is it easier to just see what a polynomial looks like when its (x+2)^2 - 3

or should I foil it out to see whats happening easier?

I also don't understand the impact of what the 4x does do the graph..

[short sorrel]

Vertex form is generally the easiest quadratic to graph

For a quadratic in the form

a(b(x-c))^2 + d

The vertex is at (c,d)

And the rest can be graphed according to a horizontal stretch (1/b) and vertical stretch (a)

Expanding a polynomial into ux^2 + vx + w tends to make it trickier to graph, as it's hard to get an intuition for what the first couple terms mean

Factored form: a(x-p)(x-q) also tends to be fairly easy to graph [at least for polynomials with real roots], as the roots are at (p,0) and (q,0), and the vertex is just in the middle of those roots and stretched by the a value

In your case, (x+2)^2 - 3 indicates that the polynomial's vertex is at

(-2, -3)

Note that the sign of the term inside the bracket is flipped, but the one outside is preserved

And, because the a and b values are just 1, the graph is shaped like your standard y=x^2

Just moved over to the new vertex (-2, -3)

So this one is quite easy to graph in vertex form, but potentially much more difficult as an expanded polynomial

@kind pier

[kind pier]

https://www.desmos.com/calculator/nnwceezs0v

is that still x^2? it looks like it is, but down 3 and shifted to the left 3

maybe it's wider though

[short sorrel]

Its still x^2, just translated.

[delicate sequoia]

hey what it would be the arithmetic sequence of 1,0,-1,0

[torn swift]

correct

you plugged in g(x) where ever you saw an x in f(x)

now just have to expand and simplify

[wary plover]

I keep forgetting how to do this kind of problem

is it something to do with logs?

[native sequoia]

you can write $3^{-7x+5}$ as $3^5\cdot(3^x)^{-7}$

[obsidian monolith]

EpicGuy4227:

[warm prawn]

Answer 'e', maybe?

[wary plover]

how does that help me

[torn swift]

you can substitute in what 3^x is

[wary plover]

oh yeah

I can do it then

[warm prawn]

I wrote X as log3(2) 🤷

[torn swift]

that's one way, but it was easier to manipulate the expression

[modest cobalt]

I just don't get it, the software wasn't very helpful

[viscid thistle]

I’m taking precal H next year, any suggestions on what to review/learn?

[timber plinth]

precal

[viscid thistle]

I heard the first topic is limits and continuity but that’s it

[glacial island]

quadratics, forumlas, polynomials

[timber plinth]

so uh, Precal H?

honors?

[viscid thistle]

@timber plinth i think it’s the same curriculum as calculus a

Yes

[timber plinth]

Uh, in that case

learn what a limit means, and a bit of the stuff about them

Make sure you know the stuff from prerequisite courses ofc

[viscid thistle]

Will I be doing integrals this year?

[timber plinth]

idk

[viscid thistle]

Doesn’t derivative come first

[timber plinth]

typically, since it's a more "fundamental" concept maybe?

[viscid thistle]

Oh

[timber plinth]

idk, it's certainly simpler for more people to get at least

Definitely typically taught first

[viscid thistle]

1 more thing, will I need to know trig stuff(unit circle, identities, etc)

[timber plinth]

You know, idk about the fundamental part really, but it's more simply interpreted with prior knowledge

and yes ofc

[viscid thistle]

Do you expand on the stuff from precal

[timber plinth]

well yeah

that's the point of precal

[viscid thistle]

True

[timber plinth]

But uh

$\lim_{x\to x_0} f(x)$ means stuff as you approach x_0 (or go to infinity)

[obsidian monolith]

Darkrifts:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

I watched a few videos on it

Seems fairly straight forward

[timber plinth]

and the derivative has like 5 different ways to notate and idk what your teacher will use

[viscid thistle]

Isn’t it to find the slope at a specific point

Or something like that

[timber plinth]

More or less yeah

That's one way to do it by constructing tangent lines

[proud raven]

find a set of parametric equations through the point parallel to the given vector

point is 0,0,0 and vector is 3,1,5

how do you begin with something like this

isnt that vector already going through that point

i mean isnt a vector already going through every point

is the question asking to turn the vector into parameters and then find an equivalent set of parameters?

<@&286206848099549185>

[hard hornet]

I would probably build the equation for the line first, but there are multiple approaches

[proud raven]

is the line the direction vector

[hard hornet]

no a line with function r(t)

Like a line line

$r(t) = r_0 + t<x,y,z>$ where $r_0=<x_0,y_0,z_0>$

Lemme dix

Fix

[obsidian monolith]

MemesPlease:

[proud raven]

so this is the same as asking to find a line thats parallel to a line

then you just express it as a vector

[hard hornet]

Line parallel to a given vector

Parallel to your directional derivative

Once you built the line

Everything comes together

Your x components belong with x, y components belong with y, z components belong to z

[proud raven]

so the question is asking you to find the t part of your equation

with r0 being the 3,1,5

[hard hornet]

Yes

[proud raven]

😮

[hard hornet]

You can find the equation

With the equation of a line

[proud raven]

so

we arent given a line here though

[hard hornet]

We can build one

Makes it easier for me imo

Build the line

[proud raven]

the line is just any line parallel to the vector?

[hard hornet]

Not any line, its a line parallel to the given vector, AND given an initial point

That is, <0,0,0>

[proud raven]

are you able to walk me through the problem i think i might be retarded

or ill try to find more youtube vids

[hard hornet]

Ok lets go through some basics

Given a directional vector and an initial point, how do you build a line with respect to t, time

Lets say

I wanted to build the equation of a line parallel to some vector, and passes thru a particular point

[proud raven]

just place the vector there

[hard hornet]

How though,

Think about it

[proud raven]

put the head there

or tail

[hard hornet]

I want a line that is parallel to <5,8,1> for example, and passes through the point <2,1,2>

[proud raven]

just put the beginning of the vector on that point

[hard hornet]

give me the equation

[proud raven]

oh

youd just add

uhh

<5,9,3

[hard hornet]

An equation that expresses the line

@proud raven

[proud raven]

sorry im trying to figure it out

[hard hornet]

Give it ri me in the form i posted above

Ah ok take ur time

$r(t) = r_0 + t<x,y,z>$ where $r_0=<x_0,y_0,z_0>$

[obsidian monolith]

MemesPlease:

[proud raven]

im confused because you only have one point

[hard hornet]

Yup

Thats perfectly fine

[proud raven]

do you have to go through all that angle nonsense then

[hard hornet]

For line building

You only need a point and a direction

[proud raven]

wait you only have one vector

[hard hornet]

Or two points

No need angle nonsense

[proud raven]

am i using this

[hard hornet]

Noooo dont use any formula

Think about it

In order to build a line

You need some sort of initial point

And how it moves parallel to the direction derivative

[proud raven]

do i have to take a derivative?

[hard hornet]

How do you build a line in 2 dimensions

Given an initial point and a slope

[proud raven]

use an equation

(y-y1)=m(x-x1)

[hard hornet]

Initial point

[proud raven]

x1,y1

[hard hornet]

Yup

That is right

Or

y=mx+b right?

How does that work intuitively though,

Explain to me how that equation works

[proud raven]

i mean

i guess multiplying x by your slope just gives you y

that doesnt sound right

sorry i know what each individual part is

i dont really know what you want me to say 😅

[hard hornet]

In slope intercept, you have some constant value, this is where the line is "nailed at"

and then you have the slop

that dictates how slanted the line looks

Makes sense, you only need two components for a line

where it is "nailed at", and how slanted it looks

if you have these two components, you can model infinitely, and all possibilities of lines

makes sense right?

[proud raven]

yes

[hard hornet]

now lets up it one dimension, and introduce vectors!

This picture might help

Now, lets say we wanted to construct a line L, that goes through a particular point P, and is parallel to the vector v

in the same way we did it in two dimensions

in three dimensions

there is an initial point, and a "slope", actually not really called slope, but the directional derivative that dictates which direction the line goes

so again I ask you

what is the equation of the line, if it is parallel to <5,8,1> and passes through the point <2,1,2>

$r(t) = r_0 + t<x,y,z>$ where $r_0=<x_0,y_0,z_0>$

[obsidian monolith]

MemesPlease:

[proud raven]

<5,8,1> + t<2,1,2>?

[hard hornet]

you got the order a bit off

the parallel or directional derivative goes with t, because any t value would make it parallel to that vector

and your initial point is <2,1,2>

[proud raven]

what is r0 in the example you gave

in the graphic i mean

[hard hornet]

initial point

[proud raven]

so r0 is a vector which goes from the origin to the initial point

[hard hornet]

that is correct, but i wouldnt want you to think of it that way

[proud raven]

im trying to parse the image

[hard hornet]

but yes you are basically right

https://cdn.discordapp.com/attachments/363224154469826562/588231695283650581/unknown.png

[proud raven]

that was in the lecture i watched too and i was struggling with it

[hard hornet]

notice how

[proud raven]

what is v

[hard hornet]

we only get the line L, when we add the vectors r_0, and a, in which a is your directional derivative multiplied by t

v is your directional derivative

thats what your line is parallel to

[proud raven]

so our v is the <5,8,1>

[hard hornet]

I know this is very hard to picture

[proud raven]

so were trying to find r

[hard hornet]

but imagine vector r, in the picture, is like moving back and forth on the line

[proud raven]

so when we add them we get L

[hard hornet]

where the end is at the origin, and the tip is moving back and forth on the line L

we are trying to find the equation of a line in three dimensions, parameterized with t

Read what I wrote about r

r is not stationary

it moves back and forth depending on what t value you give it

thus it traces out a line

I don't know how to animate, so I can't really show you

[proud raven]

i mean in this example we are doing

which is just at a moment in time right

[hard hornet]

we essentially graphed an equation of the line

r(t) = <2,1,2> + t<5,7,1>

if t was 0

we get the vector <2,1,2>

if t was 1, we get <7, 8, 3>

if t was 2, we get <12, 15, 4>

etc etc

notice

that depending on your t value

we trace out infinitely many vectors

and those infinitely many vectors arrows, trace out a line

thus thats why the equation of the line in three dimensions

is an initial point + t multiplied by the vector it is parallel with

it takes time to process, don't worry if you don't get it at first

it took me a while too

[proud raven]

im upset we have homework on something we didnt even cover lol

im trying to find a graphing app to visualize this

[hard hornet]

o wait what did you guys cover so far

[proud raven]

its hard to say

i think he actually assumed we already knew this

but i cant be sure

[hard hornet]

what topic

just give me anything

[proud raven]

calc iii

[hard hornet]

"its hard to say" doesn't offer anything

alright

can you be a bit more specific

[proud raven]

weve been reviewing vectors for the past 2 weeks

and reviewing calc i stuff

but the homework has been mostly on parameter stuff like this and planes

[hard hornet]

oh

[proud raven]

today we reviewed the chain rule and the product review

product rule

so the answer is an equation in that form

two vectors added together and a scalar

[hard hornet]

could you post your example again

we are going to go through this together

[proud raven]

through point 0,0,0 parallel to 3,1,5

[hard hornet]

alright

so what is your equation

using what i showed you

[proud raven]

so that means our line is just our vector

or

with the scalar

[hard hornet]

your line is essentially infinitely many vectors, in which the line is traced out by connecting all these vectors

[proud raven]

but its just scalar*the vector

[hard hornet]

IM SO FUCKIN MAD I CANT SHOW YO

[proud raven]

t<3,1,5>

[hard hornet]

aaaaAHAH yup

[proud raven]

lol im mad i cant figure out how to graph any of this

[hard hornet]

yup yup

Alright

[proud raven]

id really like to play with this

[hard hornet]

knowing that r(t) = t<3,1,5>

[proud raven]

but its impossible in 3d

[hard hornet]

no initial point

that means

[proud raven]

desmos is great for 3d stuff

[hard hornet]

x corresponds to x, y corresponds to y, z corresponds to z

so

x = 3t
y = t
z = 5t

t<3,1,5>

3 = x, 1 = y, 5 = z

thats your parametric equation

for symmetric equation, just solve for t

[proud raven]

if its asking

symmetric equations of the line through the point parallel to the given vector

is that just the same thing

[hard hornet]

I would build my line first

figure out parametri

then do symmetric

alright

I gotta go soon, ping me back later and I can show you more examples

[proud raven]

okay

😄

[modest cobalt]

I don't really know how to do this at all, I don't know how to start or anything

[proud raven]

try pinging helpers

idk or id help you 😦

[modest cobalt]

<@&286206848099549185>

this one's been killing me for a while now

[fast oriole]

idk what text channel to put this. b ut its just a simple equation which im confusing myself on

z^1/3 OVER 2

im trying to flip this

ive got 1 OVER 2 x cuberoot(z)

Is this wrong?

[proud raven]

flip?

[timber plinth]

@fast oriole yes

[willow bear]

this person posted in 3 channels and i happened to respond to them in #prealg-and-algebra

[timber plinth]

acceptable

[fast oriole]

cant hit if you dont miss a few times

[timber plinth]

but uh the reciprocal of $\frac{z^{1/3}}{2}$ is just $\frac{2}{z^{1/3}}$

[obsidian monolith]

Darkrifts:

[timber plinth]

which is what I assume you mean by flip

[fast oriole]

so the initial question is

integral z^1/3 OVER 2

z^-1/3

so i flipped it to 1/2 * cuberoot(z)

1 OVER***

But i want to try pull a constant out of the integral

[timber plinth]

you can pull the 1/2 out yes

[fast oriole]

why is it a half

[timber plinth]

because a/2 = a*1/2

[fast oriole]

because 2 on the demoninator

so i dont just flip the 2

so its 2/1

[timber plinth]

i dont know

[fast oriole]

(flip upside down)?

[timber plinth]

i don't know what the problem even is

the necessity of the flip at all is lost on me

[fast oriole]

Im just trying to flip it so i can isolate the variable

[timber plinth]

you don't need to flip anything

[fast oriole]

I dont know how to use your fancy picture thing

[timber plinth]

just take a/b = a * 1/b

[fast oriole]

and if b is negative xponent, flip it to positive

[timber plinth]

yes

[fast oriole]

$\frac{z^{-1/3}}{2}$

[obsidian monolith]

FTJ:

[fast oriole]

that is the thing im triyng to integrate

[timber plinth]

yes yo can put that on the bottom

put the z on the bottom, take the 1/2 out

[fast oriole]

right

so it leaves me with

1/2 pulled out, and 1/cuberoot(z)

right?

[timber plinth]

yes

so integrate

[fast oriole]

ok awesome

ty for ur help

[proud raven]

,rotate

😦

,rotate

,rotate

[proud raven]

on number 25, is there a faster way to do it than turning it into vector point form

or would you just use a fast way to get it into that form

[willow bear]

is it really that time-consuming to write it in vector form

[proud raven]

like a minute or so for each line

no not really

i was just curious

[willow bear]

$L_1 : (x,y,z) = (6,-2,5) + t(-3,2,4)$

[obsidian monolith]

Ann:

[proud raven]

oh wait

yea its faster on these

[willow bear]

what did you think it was

[proud raven]

i thought there was additional steps

[willow bear]

what steps

[proud raven]

idk im still trying to sort out the steps for these problems

looking at i thought i had to do some weirdness combining lines

to figure out points

but the vector tells you if theyre parallel anyways

idk how to figure out if theyre identical and not just parallel but ill figure out tomorrow

thanks for answering my stupid question 😄

[gloomy shard]

hey, how come
4 log 4(x)
becomes 2 log 2(x)

[willow bear]

$ 4 \log_4(x) = 2 \log_2(x)$? is that what you meant?

[obsidian monolith]

Ann:

[gloomy shard]

yes

it doesn't make sense to me

it should increase to 8 i think

in my head that's what should happen if x is 4 lets say

[kind pier]

did they make it 2^2?

[willow bear]

$4^{4 \log_4(x)} = x^4 \ 4^{2 \log_2(x)} = (2^2)^{2 \log_2(x)} = 2^{4 \log_2(x)} = x^4$

[obsidian monolith]

Ann:

[gloomy shard]

math

this rule, right...

why does it contradict this

2^2 = 4
4 ^ 1 = 4

[viscid thistle]

what contradicts what

[gloomy shard]

so whenever i say, that 4 is in fact 2 ^ 2

nvm logarithms don't work like that

[kind pier]

I'm having trouble drawing the max and mins using the number line

Im pretty sure the first one is right, but then I don't know

[gloomy shard]

i just explained that to myself why @obsidian monolith is correct

[kind pier]

for the interval notation how do you know which parts get the bracket to be included?

[viscid thistle]

find the derivatve, find its roots, those are your extrema

the sign of the derivative determine increase/decrease

@kind pier

[kind pier]

the top got cut off

its the original function

its x^4 - 7x^3 - 13x^2

[viscid thistle]

i thought 4x^3-21x^2-26x is the function

[kind pier]

for the intervals I also made a mistake for decreasing its (-infin,-1.03) U ( 0,6.2)

but i dont know when to use brackets

who gets to claim the cv's?

[viscid thistle]

for the brackets you find the limits

its always open at -inf and + inf

[kind pier]

right I mean for the cv's though

who gest to claim them

[viscid thistle]

what is

a cv

[kind pier]

our max and min x values

[viscid thistle]

you mean f(x) when x is an extremum?

0,0 is a local maximum

the minima are horrendous looking

[kind pier]

so im suppose to use interval notation to write out the intervals of increasing and decreasing areas of the graph

do the increasing or decreasing intervals get to claim the cv's

so like (-infin,-1.03] or (-infin,-1.03)

[viscid thistle]

i think it doesnt make a difference whether you include the extremum in the interval

you can see it as decreasing on (-inf,-1.03) and reaching a minimum at -1.03 then starting to increase

you just need to make sure the union of all intervals is R since your function is continuous

[kind pier]

do I just multiply the y values by 3? for part b I mean

[vernal rapids]

$$(x^{2^{2^{2^2}}})' ?$$

[obsidian monolith]

DarK:

[pale kettle]

chain rule

[vernal rapids]

$$2(x^{2^{2^{2}}})\times 2(x^{2^2})\times 2x^2\times 2x$$ Is this true? lol

Lol. I messed up the formatting

[echo plaza]

wha

[obsidian monolith]

DarK:

[echo plaza]

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

[obsidian monolith]

CaptainLightning:

[pale kettle]

Yeah it's weird, because

[echo plaza]

power rule

[vernal rapids]

$2^4(x^{2^{2^{2}}})(x^{2^2})(x^2)(x)$

[obsidian monolith]

DarK:

[pale kettle]

$x^{2^{2}}$ is defined as $x^{(2^2)}$ rather than $(x^2)^2$

[obsidian monolith]

Zopherus:

[vernal rapids]

Oh.

Well Im too lazy to put 100 ()

[ancient stirrup]

Tetration?!

👀

[vernal rapids]

$2^4(x^{2^{2^2}+2^2+2+1})$

[ancient stirrup]

I have been summoned

[obsidian monolith]

DarK:

[ancient stirrup]

I think you need to chen lu the chen lu for the tetration term

[vernal rapids]

$2^4(x^{2^{2^2}+2^2+2+1})=16x^{15}$

[obsidian monolith]

DarK:

[ancient stirrup]

Still nani tf but that's better

[vernal rapids]

2^4(x^{8+4+2+1})=2^4x^{15}=16x^{15}

$$2^4(x^{8+4+2+1})=2^4x^{15}=16x^{15}$$

[obsidian monolith]

DarK:

[ancient stirrup]

Them exponents

what was the original

[vernal rapids]

We could have just calculated the exponent and use $(x^n)'=nx^{n-1}$..

[obsidian monolith]

DarK:

[vernal rapids]

$$(x^{2^{2^{2^2}}})' $$ original

[obsidian monolith]

DarK:

[vernal rapids]

it is not tetration
I'm just lazy to write a lot of ()

[ancient stirrup]

Ohhh

I'm confused so is it

((((x^2)^2)^2)^2) '

or grouped the other way

[vernal rapids]

$((((x^2)^2)^2)^2) '=2^4(x^{2^{2^2}+2^2+2+1})=16x^{15}$

[obsidian monolith]

DarK:

[echo plaza]

(((x^2)^2)^2)^2= x^16

[ancient stirrup]

[vernal rapids]

It feels good when the correct solution comes out even when you try to oversimplify it

[short sorrel]

$\pi^{\pi^{\pi^{\pi}}} \in \bZ$?

[obsidian monolith]

Namington:

[viscid thistle]

No

[short sorrel]

Prove it

Then publish a paper

[viscid thistle]

No

[short sorrel]

Because it's an open problem

[viscid thistle]

Number theory is boring

[short sorrel]

(though it's an integer with probability 0)

(but it's technically possible!)

[viscid thistle]

You can use the term almost surely to sound cool

[vernal rapids]

I dont see why proving this would help society

[short sorrel]

Welcome to mathematics.

[timber plinth]

i mean, proving it wouldn't help society directly, but 1) it'd be nifty because math 2) it'd provide useful information which may incidentally help society outside of math on accident

and lets be real, if you're doing math you're probably going for 1) more than 2)

[modest cobalt]

I really need help with this one, I don't even know where to begin

[modest cobalt]

<@&286206848099549185>

Is there a better place for this question? No one's answering, this is my third try

[proud raven]

i can try to help i guess

but it might be a sec

[modest cobalt]

ah thank you

[proud raven]

is this statistics 😦

[modest cobalt]

yup lol

it's for summer school, I have to learn all of it in 3 days

[proud raven]

im just guessing, and i dont know

but im honestly just trying to find a formula that has all of the key words in your question and then solving the inequality

so i might steer you wrong lol

[modest cobalt]

ah it's alright, at this point anything will help

[proud raven]

if i can parse it

[modest cobalt]

maybe it'll give me an idea of what to do

[proud raven]

i gotta give up 😦

unless you have a formula you are using for this sort of thing

[modest cobalt]

nah the software wasn't very helpful

It's alright though

Thank you for trying anyway

[cursive estuary]

try #probability-statistics

I haven't done statistics for a while so I'm not entirely comfortable doing it without research and I don't have time atm

[midnight fulcrum]

Can anyone help on this? if i have 2 sets A and B and A*B = ∅

A and B must be both ∅?

[echo plaza]

A=∅ or B=∅

[stark inlet]

If I had y=2(5(x-2))+3 would my horizontal stretch factor be 1/5?

[steel granite]

would you say the same if you wrote it at y=10(x-2)+3

[ancient stirrup]

omg LMAO im sorry I laughed when i saw this

[torn swift]

Lol

Need ... though

[ancient stirrup]

beats my infinite integrals i did a few nights ago

for real tho you could eZ prove it

[short sorrel]

go.

and enjoy your tenure as a mathematics professor

[torn swift]

Zeta is a professor?

[short sorrel]

(probabilistic proofs dont count, its trivial to show it's false with probability 1)

[ancient stirrup]

Step 1:
REEEEEEEE

[short sorrel]

nah @torn swift its just that this is a fairly well-known open problem

[ancient stirrup]

Okay now that we have our axioms of choice, REEEEEEE, let's begin

[short sorrel]

i mean i say "open problem"

its not really

its obviously not an integer

but no one's proved nor calculated it

[ancient stirrup]

that's the meme

So pi^pi^pi^pi is about 2.598 quadrillion

[short sorrel]

?

[ancient stirrup]

obviously it's an integer /s

[short sorrel]

yer a bit off

pi^pi^pi is already 1.34 * 10^18

[ancient stirrup]

pi^(pi^pi)

[short sorrel]

and exponents grow, well, exponentially, so pi raised to that power would be absolutely massive

[ancient stirrup]

I'm talking about (pi^pi)^pi

[short sorrel]

no one writes $\pi^{\pi^\pi}$ meaning $\left(\pi^\pi\right)^\pi$

[obsidian monolith]

Namington:

[ancient stirrup]

the latter is too thicc to analyze

[short sorrel]

the convention is to treat exponents as though they have implied brackets

[ancient stirrup]

ye

[rocky bison]

Thats incorrect

That latex IMG atleast

Unless that was example of incorrectness

[short sorrel]

well ok

no one should write that

[ancient stirrup]

technically

fixes professor glasses

[rocky bison]

Those two things are NOT equivalent

[short sorrel]

yes

[ancient stirrup]

^

[short sorrel]

thats my point

no one writes the former

meaning the latter

[ancient stirrup]

exactly, so when you talk about pi^pi^pi^pi you have to specify

[short sorrel]

no

its unambiguous

[rocky bison]

It is tbh

Atleast when not in any mark down

[short sorrel]

or does $\int_{-\infty}^{\infty} e^{-x^2} \dd{x}$ wake you up at night?

[obsidian monolith]

Namington:

[ancient stirrup]

that infty tho

What's that, that doesn't look like anything I learned in precalc

[short sorrel]

you right, this is prealg 101

[ancient stirrup]

but yes that does keep me up at night

[short sorrel]

i mean, it doesnt even use algebra

so yeah its prealg

[ancient stirrup]

exponents trigger my OCD

they need to get in line

[short sorrel]

petition to rename real analysis "prealgebra"

[ancient stirrup]

LOL

"today we're going to talk about these integrals-- no algebra necessary!!!"

[short sorrel]

just do integrals

and never simplify

[ancient stirrup]

lmaoooooo

call it advanced geometry

[short sorrel]

because havent learned that yet

[ancient stirrup]

they legit won't know the difference

if you call it geometry

area of a shape = area of a shape

[gloomy shard]

how to find lg5
if lg2 = a

[willow bear]

what is lg?

decimal log?

consider that $\lg(2 \cdot 5) = \lg(10) = 1$

[obsidian monolith]

Ann:

[willow bear]

@gloomy shard

[gloomy shard]

lg is log base 10

[willow bear]

yes, ok, decimal log

that's what i thought

and what i wrote still stands

[gloomy shard]

alright, thanks, will try to figure this one out :D

how it even works

wait how do you just multiply them

it's still not the answer of lg5

expressed as a

[willow bear]

well yes duh i didn't just give you the answer ofc

[gloomy shard]

ah i see :P

right, common log functions with the same base, addition 🤦

thanks so much!

[full palm]

Can x^3 + x^2 + x be factored completely? I applied the quadratic formula and got 0 and (-1 +/- 3i)/2 as zeros. When I try to factor completely and then expand to check my work, I get a different polynomial. Does anyone know where I went wrong?

[wind igloo]

That function has only 1 real root.

Not sure what you mean by 'factor completely'

[full palm]

Factor into linear factors with complex numbers. Fundamental theory of algebra. @wind igloo

[wind igloo]

Should work, then.

Want to show your work?

[wind igloo]

Looks like you did something wrong when expanding to check your work.

Everything works out for me.

[full palm]

Doesn't quite work out

[wind igloo]

Hmmm.

How very odd.

Oh. Duh.

It's sqrt3 i

not 3 i

$$ \sqrt{-3} = \sqrt{3} i \neq 3i $$

[obsidian monolith]

samanthaCS:

[wind igloo]

@full palm

[full palm]

Substituting that gives me something even more hairy. Can you show me how you arrived at your solution? (btw, I'm practicing this cause I like it, not for a test).

[wind igloo]

Hm?

It's the quadratic formula.

b^2-4ac = 1^2 - 4(1)(1)

So the radical is -3

So the root is $$ \frac{-1 \pm \sqrt{-3}}{2} $$

[obsidian monolith]

samanthaCS:

[wind igloo]

And when you do the expansion, the numerator is $$ (-1 + \sqrt{3} i )(-1 - \sqrt{3} i ) = 1^2 - \sqrt{3} i + \sqrt{3} i - 3i^2 = 4 $$

[obsidian monolith]

samanthaCS:

[full palm]

That seems an odd way of doing it. In all other problems in the exercises, I just factor into conjugate pairs when there's a square in a factor. This still doesn't make sense since I cannot expand back into x^3 + x^2 + x. I'm not sure if something just isn't clicking or I misstated my problem.

[wind igloo]

How is it an odd way of doing it?

And how does it not expand back into the original polynomial?

I mean, yes, I'm being more explicit than necessary.

But this is exactly the method you used, isn't it?

[full palm]

I didn't learn to use the numerator like that.

The problem was to find all zeros and then factor completely.

[wind igloo]

That's just shorthand.

Sorry, I didn't want to type the whole term.

So I just did the 'tricky' bit.

Here, let me do it properly. One sec

$$ x^3 + x^2 + x = x(x^2 + x + 1) = 0 $$
$$x^2 + x + 1 = 0 \implies x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i*\sqrt{3}}{2} $$

[obsidian monolith]

samanthaCS:

[wind igloo]

$$ x (x-\frac{-1 + i\sqrt3}{2})(x-\frac{-1-i\sqrt3}{2}) = x(x^2 -x(\frac{-1 + i\sqrt3}{2} -x(\frac{-1-i\sqrt3}{2}) + (\frac{-1 + i\sqrt3}{2})(\frac{-1-i\sqrt3}{2}) $$

[obsidian monolith]

samanthaCS:

[wind igloo]

Hmph.

Is that truncated for you too?

[full palm]

yes

[wind igloo]

Bleh.

[full palm]

Goddamn bot

[wind igloo]

$$ =x(x^2 -x(\frac{-1 + i\sqrt3}{2} -x(\frac{-1-i\sqrt3}{2}) + (\frac{-1 + i\sqrt3}{2})(\frac{-1-i\sqrt3}{2}) $$

[obsidian monolith]

samanthaCS:

[wind igloo]

Better.

$$ =x(x^2 + \frac{x}{2} -\frac{xi\sqrt3}{2} + \frac{x}{2} + \frac{xi\sqrt3}{2} + \frac{(-1+i\sqrt3)(-1-i\sqrt3)}{4} $$

[obsidian monolith]

samanthaCS:

[wind igloo]

$$ = x(x^2 + x + \frac{(-1)^2 + i\sqrt3-i\sqrt3 - (\sqrt3)^2 i^2}{4})$$

[obsidian monolith]

samanthaCS:

[wind igloo]

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

[obsidian monolith]

samanthaCS:

[wind igloo]

Does that all make sense now?

[full palm]

Yes thank you

You really went all out for me, thank you

[wind igloo]

You're welcome

Eh. It's good practice.

I've gotten rusty with Latex.

[ruby otter]

samantha familiar name

[wind igloo]

I haven't been around in a while.

[ruby otter]

you ain't familiar the name is!!

[wind igloo]

Ok. It's a common name

Nope

[ruby otter]

that pic is 3.84mb it took me time to upload it !!

[wind igloo]

Shoulda just linked it

[ruby otter]

most people don't trust links they think that's sketchy

[wind igloo]

🤷

[ruby otter]

btw is that name common coz i never thought so

[wind igloo]

It's pretty common in the US

[ruby otter]

It's origins??

[wind igloo]

No real idea beyond what I can quickly google

[ruby otter]

Are you a native american??

[wind igloo]

I was born in the US, but I am ethnically Chinese.

[ruby otter]

Any connection with India??

[wind igloo]

Nope

[ruby otter]

So that name must be common as you said

[wind igloo]

According to Wikipedia it was in the top 10 girls' names between 1988 and 2006

In the US

[ruby otter]

Maybe in India that samantha came from some other country

[wind igloo]

Maybe

[ruby otter]

Talking about India do you know my boi??

Ramanujan

[wind igloo]

I have heard of him.

[ruby otter]

-_- you have heard of him??

[wind igloo]

Well, I have heard of a famous Indian mathematician of that name.

[ruby otter]

https://en.wikipedia.org/wiki/Srinivasa_Ramanujan

almost everyone here know him if I'm correct

[echo plaza]

[ruby otter]

[viscid thistle]

lmao look at him

[ancient stirrup]

whO'S thAt?!111111

[full palm]

Ramanujan was incredible, he just didn't like proving his work.

Why does one use a natural logarithm to get the inverse of 2^2x?
It seems counterintuitive since my first instinct is to use log base 2.

[torn swift]

Log base 2 isn’t programmed into your calc typically, but log10 and ln are

ln occurs frequently so its values are known

[short sorrel]

yep

you can absolutely use log_2

but thats usually more awkward

ln is very well-behaved, and log_10 is commonly used, so those two are more common

(you'll see what i mean by "ln is very well-behaved" if you ever take a calc course)

they try and drill the habit into you of using ln

because it's generally easy to work with

[full palm]

I'm training to be a computer engineer so I will be taking Calc and Discrete Math.

[kind pier]

r(x) =
x3 − 1x2 − 2x
x − 2

how do i find the range of that ?

I think im suppose to build the inverse of it, but when I do I simplify it down to y(y+1)=x and then can't do anything else

[kind pier]

<@&286206848099549185>

[limber bone]

which 1

x^3-x^2_2x?

[kind pier]

its a fraction

[limber bone]

x^3-x^2-2x?

oh

x^3-x^2-2x/(x-2)

well

just graph it lol

not easy to find its inverse here

[willow bear]

$r(x) = \frac{x^3 - x^2 - 2x}{x - 2}$

[obsidian monolith]

Ann:

[willow bear]

is this the thing

[limber bone]

u can factor stuff

[kind pier]

ya, I don't understand where they are getting these fractions

[limber bone]

there is a comon factor ig

common*

[willow bear]

if not, do you have a picture of the thing

[kind pier]

they are simple like -1/4

[willow bear]

@kind pier do you have a picture of the problem

before we can start doing the thing or helping you with the thing we need to know what the thing is

[kind pier]

the second picture is the first part and the one with the red x is what I can't solve. I got the inverse down to y(y+1)=x with y cannot equal 4, but I don't think even that is right.

[limber bone]

graph it on desmos

or on ur calculator

and see max and min y values

[kind pier]

isn't there a way to get an exact value though? the other answers are exact values

[limber bone]

wdym exact

[kind pier]

if i graph it on my calc im going to get a decimal string right?

[kind pier]

ok well i figured it out

how they were getting those fractions

that is a stupid amount of work

[viscid thistle]

Guys

so the answer is C

but can't it also be A?

[vital siren]

@kind pier remainder and factor theorems?

[kind pier]

i was trying to find the range of a rational

and they want exact value

[vital siren]

aouf

[kind pier]

so I think i had to just use calculus

[vital siren]

what's the question

[kind pier]

I didn't know how to do that when I asked I just watched youtube, but the guy above me had a question

[vital siren]

oh

[formal wave]

if anyone free why does the sign flip when square rooting

oh nm

[void coral]

hey can i get some help

with this function

apparently its horizontal asymptote is y=0

but i dont understand how it crosses y=0

[pale kettle]

Review what an asymptote is

[sour quiver]

@void coral HA is the general behaviour of a graph

it can cross over

[willow bear]

@void coral do you mean you don't understand how y=0 manages to be an asymptote yet be crossed by the graph or how the crossing happens in the first place regardless of y=0 being an asymptote?

[shrewd urchin]

@formal wave ?

$20-4x^2<0$ $\implies$ $20<4x^2$ $implies$ $5<x^2$

[obsidian monolith]

Krishna:

[willow bear]

why do many dollar signs

$20 - 4x^2 < 0 \implies 20 < 4x^2 \implies 5 < x^2$

[obsidian monolith]

Ann:

[willow bear]

$ marks the beginning or end of math mode in latex

your message is like
"Start math mode. Type 20 - 4x^2 < 0. End math mode. Start math mode again. Type \implies. End math mode. Start math mode again..."

[shrewd urchin]

$20 - 4x^2 < 0 \implies 20 < 4x^2 \implies 5 < x^2 \implies x>\sqrt{5}$ or $x<-\sqrt{5}$

[obsidian monolith]

Krishna:

[shrewd urchin]

,w factor x^2-11/12x+1/6

[obsidian monolith]

Display results online and refine query

Requested by Krishna#4971

[kind pier]

unsure how to do that

[midnight summit]

if 1+i is a zero for Q then i-1 is also a zero for Q

you should know this

and also if (1+i) is a zero for Q then $ x-(1+i)$ is a factor for Q

@kind pier

[obsidian monolith]

mohad12211:

[vernal rapids]

Are we having complex numbers?

[midnight summit]

yeah?

[kind pier]

I mostly dont understand how to distribute imaginary numbers

when we have three terms inside

(x-(1+i))(x+4)^2 right?

i need the polynomial

[midnight summit]

well no

it's

$(x+4)(x-(1+i))(x-(1-i))$

[obsidian monolith]

mohad12211:

[midnight summit]

if 1+i is a zero then 1-i is also a zero

[kind pier]

i thought the point of making the numbers imaginary was to force a difference of squares

how come you have both imaginary terms -

[midnight summit]

i have 1+i

and 1-i

[kind pier]

but I was looking at that as lets say the entire b in a^2-b^2 = (a+b)(a-b)

[midnight summit]

yeah

[kind pier]

but we have double (a-b)'s

[midnight summit]

next step

$(x-4)(x-1-i)(x-1+i)$

[obsidian monolith]

mohad12211: