precalculus

[patent beacon]

Bring the exponents down

It is now a linear equation, solve accordingly

[earnest nymph]

Or just 9/16 = (3/4)^2

[thick star]

i got it solved

thank you for your help though!

[heady jewel]

this is surely precalculus

[languid crane]

arithmetic = precalc

[prisma marten]

Calc is Arithmetic

[shrewd urchin]

[rigid beacon]

Calc is just doing infinite tiny arithmetic

[shrewd urchin]

Calc is arthinetic
If this statmemt is true. Then I was ready to study physics when I was in 2nd grade.

[prisma marten]

The hardest part of Calc is the arithmetic

[rigid beacon]

You aren't wrong

[steady jasper]

all math is fancy addition

edit: *most

[shrewd urchin]

[minor dagger]

Did I hear a vihart qoute? 😄

[steady jasper]

https://www.youtube.com/watch?v=N-7tcTIrers&index=8&list=UUOGeU-1Fig3rrDjhm9Zs_wg

[minor dagger]

Haha yes!

[sterile portal]

solve x^3-x<0

which is x(x+1)(x-1)<0 but idk where to go from there

[vernal rapids]

Um

Well, being under 0 means being negative

So one or all three are negative

if all three are negative, x is less then -1
If only one is negative, because x-1 is the smallest, it will be negative
Meaning x is under 1

x<-1
x<1

The union of these is x<-1

[sterile portal]

ok I understand that

these are the answers im given which is confusing me I gues

[vernal rapids]

Hmmm

draw a numberline for all of them underneath each other

Color it green where the individual terms are positive, and red when they are negative

and look for what x values are there 1 or 3 red lines

[sterile portal]

is this not what it's suppose to be?

[vernal rapids]

x<-1 AND 0<x<1

Meaning the last option

[sterile portal]

hmm I don't understand 0<x<1

[vernal rapids]

The product is negative if all three are negative, or only 1 is

in 0<x<1, only x-1 is negative

And so the product will be negative aswell

[sterile portal]

hmm ok

[sterile portal]

OHHH ok i understand now

thank you so much!

[sterile portal]

would this be the same as 2x?

[viscid thistle]

2^x

$2^x$

[obsidian monolith]

JustCurious:

[sterile portal]

ohh so the x stays in the exponent?

while e^ln(2)=2?

[viscid thistle]

yes

$x * ln2 = ln 2^x$

[obsidian monolith]

JustCurious:

[vernal rapids]

Please use \ln

And use parentheses

And \cdot

$x\cdot\ln2=\ln(2^x)$

[obsidian monolith]

DarK:

[shrewd urchin]

=tex

[frozen needle]

[shrewd urchin]

[rigid beacon]

[shrewd urchin]

[obsidian monolith]

Paul Bäcker:

[hoary yoke]

@proper nexus wat about it

[proper nexus]

Just testing how Tex works

[hoary yoke]

#bots

[proper nexus]

Ok, sorry about that

[timid barn]

uhhh

how do you find the center of this ellipse?

33x^2+13y^2-396x+26\sqrt{7}y+850=0

[pale kettle]

Ellipses don't have a single center

[timid barn]

i can choose between 4 answers

but i can only choose 1 of the 4

im doing an independent review test

https://www.varsitytutors.com/precalculus_diagnostic_1-problem-68314 @pale kettle

[pale kettle]

Ah okay I know what they mean

Well what have you tried

[timid barn]

i just finished calculus and i feel as if i forgot everything from precal :/

but i got a few of the structures of parabolas and circles alright

im studying precal cause im taking a test for college in 2 weeks

and it diagnoses my math skills

and ive been in summer mode for quite awhile 😦

[minor dagger]

@pale kettle they do have single center 😮

[next willow]

Wait did summer already begin?

[pale kettle]

@minor dagger yep mb was thinking about foci

[minor dagger]

Oh ok

[pale kettle]

the center of an ellipse is not really something that's talked about a ton

[minor dagger]

Actually it is important if you want to find the equation of the ellipse

@timid barn do completing the square for x and y

[timid barn]

ok ill try that

im tempted to use wolfram

[minor dagger]

No

[timid barn]

ok fine

[minor dagger]

Do you want me to remind you how to complete the square?

[timid barn]

i wish there was a role where i can freely do VC chats with any willing tutors here

and do screen share

yes @minor dagger

[minor dagger]

Isn’t that allowed here? But my pronunciation is exceptionally obnoxious

[timid barn]

ehh im used to it

i just finished high school

senior year was worst

my teacher whole taught cal had to learn cal on the way to teching it

[minor dagger]

Ok. So by completing the square, we want to write ax^2 + bx + c into (x + constant) ^2

[timid barn]

and her method of teaching is limited

[next willow]

Does wolfram complete squares? Wow

[timid barn]

yes

i remember that, i used it to factor

find factors

[next willow]

,w complete the square x^2-x+y^2+17y

[timid barn]

i mean roots

[obsidian monolith]

Display results online and refine query

Requested by Mat#9616

[next willow]

I am amazed

[timid barn]

but yea im still stuck

[minor dagger]

First, divide by a

Add and subtract b/(2a))^2

[next willow]

It also shows step by step solutions

[minor dagger]

Then factor it

[timid barn]

divide it all by 33?

[minor dagger]

Yes

[timid barn]

so it should look like x^2+(13/33)y^2-(396/33)x+((26\sqrt{7})/33)y+(850/33)=0

[minor dagger]

so it should look like $x^2+(13/33)y^2-(396/33)x+((26\sqrt{7})/33)y+850/33=0$

[obsidian monolith]

Compiler:

[minor dagger]

If you want the bot to use LaTeX, put dollar signs 😄

[timid barn]

oh.

i never used the math bots before

[minor dagger]

Yes you divided by 33

[timid barn]

and what do u mean by :
Add and subtract b/(2a))^2

i see what u mean if it was a 3 term equation

[minor dagger]

B= -396

A = 33

Now add and subtract b/2a raised to the 2nd power

[timid barn]

working on it lol

[minor dagger]

You can do that because you won’t be changing anything in the eq

Since adding a constant then subtracting it changes nothing

[timid barn]

except it's appreance as an equation

[minor dagger]

Even though, you can add 1 then subtract 1 whenever you feel like it

X = 5 is the same as X +1 - 1 = 5

[timid barn]

i got 36

[minor dagger]

Right, add and subtract

[timid barn]

ok...

[minor dagger]

Can you recognize x^2 - 12x + 36

Isn’t that the same as x^2 - 2(6)(x) + (6)^2

[timid barn]

yes

[minor dagger]

Does this mean anything to you?

[timid barn]

ur pulling a factor out of each individual one

[minor dagger]

And it is equal to?

[timid barn]

well yes they are the same equation

[minor dagger]

Uh

Actually

[timid barn]

i dont feel "hands-on" enough lol

[minor dagger]

x^2 - 2(6)x +36 = (x - 6)^2

[timid barn]

yes

because

oh i comepletely forgot about all of that

but what i normally do mentally

is that it adds up to b and it multiplies to c

[minor dagger]

Because in general (a + b)^2 = a^2 + 2ab + b^2

Now complete the square for y too by yourself and tell me what you get

[timid barn]

wait a minute, was i supposed to up the x and y terms on the opposite sides of the equal sign from each other....

[minor dagger]

No there is no need to

Just do exactly what you did but this tome for y

[timid barn]

@minor dagger ok i solved it already XD

thanks for the help

[minor dagger]

You are welcome 😄

Did you answer (6, -sqrt(7))?

[spark cliff]

yeah i have zero idea to do this :(

[torn swift]

it asks to find how long it takes for the cork to oscillate one cycle

[spark cliff]

yea o for it

[wanton cipher]

when in doubt, plot it out

[spark cliff]

i got it*

[wanton cipher]

https://www.wolframalpha.com/input/?i=0.2cos(20pi*t)%2B8+graph+frequency

[spark cliff]

.1 lol yeet

[lusty agate]

hello

[turbid bear]

What do you call a "function" in math with more than one input and more than one output? For example, if you have a pair as input and a pair as output, then are both pairs just viewed as one number? And if you have 2 inputs and 1 output... Is that just viewed as a formula and not a function?

[echo plaza]

the ordered pair is considered one input

[frozen needle]

if you input a pair, that still 1 input
same for the output

[rigid beacon]

Yea an ordered pair can be an input and an output

And then for the 2nd part where you have 2 inputs leading to 1 put out you could also have that and that's a function still I think

[minor dagger]

Yes, it is, you can have a function with an ordered pair as an input.

For the output part, yes you can, but how exactly could that be useful?

It would be like having two functions

Nothing useful

[pale kettle]

Yeah

It's still useful

[minor dagger]

Maybe, but I haven't stumbled on any uses, also can't think of any.

[pale kettle]

Functions from R to R^2 are essentially what you're describing

And they're extremely useful

For example, R could be like the time

and then R^2 is the path something takes as time goes on

[minor dagger]

You are right, didn't think about that 🤔

[pale kettle]

Tons of things like this come up in calc

[minor dagger]

Nice, I am thankful I rarely have to do any calculus

[royal gull]

S

[ashen ginkgo]

As a lesson learnt from programming, regarding a function f:AxB -> X as instead a function g: A -> X^B, by g(a)(b) = f(a, b). Obvious but useful

[grizzled orchid]

please use latex

please

[ashen ginkgo]

No

sry I don't remember how any more

[grizzled orchid]

google

[ashen ginkgo]

I don't care enough

[grizzled orchid]

it makes it much easier to read what u do

[royal gull]

umm, what whodatguy wrote is pretty easily understood in this case so not sure why are you saying this sloth

[grizzled orchid]

not when the font is tiny and ur glasses are outdated

[echo plaza]

yes but I really really want an excuse to use this command

$\fun{f}{A\cross B}{X}{(a,b)}{f(a,b)}\\fun{g}{A}{X^B}{a}{\br{\fun{h}{B}{X}{b}{f(a,b)}}}$

[obsidian monolith]

CaptainLightning:

[ashen ginkgo]

That's dope

[echo plaza]

tuong made it

it's too fucking good

[frozen needle]

Hi

[next willow]

Wow
Is it possible to align "f:" with domain and codomain line?

[frozen needle]

sure, you can put the f: in the same array line as domain -> codomain

and have an empty column or something at the second line

[next willow]

Awesome, thanks

[frozen needle]

🍻

[severe verge]

🍛

[spark cliff]

lol is the unit circle just straight memorisation or is there some way to make it easier

the units themselves, pi/6, pi/3 etc

[pale kettle]

It's just all the multiples of pi/6 and pi/4

[spark cliff]

what

how is 7pi/4 a multiple

that is one of them right

in the fourth quadrant

[pale kettle]

that's just 7 * (pi/4)

[spark cliff]

but like how am i supposed to know where tf they go on the circle

[pale kettle]

by thinking

you know where 0 is

you know where pi is

you know where pi/2 is

you know where 3pi/2 is

just compare whatever number you have to those numbers to figure out what quadrant you're in

[spark cliff]

lol i guess so 🤷🏻‍♀️

[lethal skiff]

is it still a vertical asymptote if the limit as that point is aproached is infinity from both sides.

@ me please im off to bed

[serene heath]

ye

@lethal skiff

[lethal skiff]

im still here thankfully

ty

wasnt sure if they where simply points you couldnt get a limit of

[spark cliff]

hello please help

[next willow]

This is a curious exercise

The answer is supposedly determined by just the first half of the first question

Would you be able to solve that half?

[echo plaza]

why is that multiple choice

[odd dagger]

lmao

[spark cliff]

no tbh

@next willow i have no idea how to do this

and i have to submit in 10 minutes yolo i’ll just take the P

L

[odd dagger]

take the P

im pr sure its C

[next willow]

What's the initial horizontal velocity?

[odd dagger]

@spark cliff

do components bro

[spark cliff]

bro idk how to do this

i’m just guessing on like all of these

[torn swift]

not the best approach, I'm sure you realize that

[odd dagger]

well look at the y part of the parametric equations

its not gonna be A or D cuz they start with something that isnt 5

i mean they dont have a constant of 5 added for the initial height

[spark cliff]

i just don’t have a lot of time to do this

[prisma marten]

start earlier hopefully nexttime

[odd dagger]

im so glad i dont get questions like that

my course is nice and pure🥺

[spark cliff]

this is literally alg1 but i want to make sure i didn’t just mess up alg1

[prisma marten]

looks okay

[torn swift]

you can always check

will x=4 and y=6 give 4 in the first equation and 8 in the second equation?

if yes, then you are correct

[spark cliff]

it said use matrices to solve it but i literally didnt is that bad

[torn swift]

do you know how to set up the matrix equation for it?

[odd dagger]

lmao

[torn swift]

then I guess they might make you use Cramer's Rule?

I hope not though

[odd dagger]

@echo plaza 😂😂😂😂

[torn swift]

most likely they showed you how to solve it using the inverse matrix

[odd dagger]

using matrices to solve 2 unknown simultaneous equations

[prisma marten]

you could just RREF the matrix

no need for inverse

[odd dagger]

rref is just stylish elimination

use inverse

[torn swift]

yes, but in high school they don't show you that for some reason

at least in most HS algebra courses they don't

[odd dagger]

yeah

[torn swift]

they usually show off A^{-1} and Cramer's Rule

tempted to pin that "rref is just stylish elimination"

makes me laugh so much

[prisma marten]

I forget how to use inverse and cramers tho

[torn swift]

inverse of 2 x 2 is easy

[odd dagger]

Ax = B
x = A^-1B

[prisma marten]

half of my first year ChE courses were just do RREF

[torn swift]

which is why they show it to highschool kids

[prisma marten]

i understand what inverse means

but how to find A^-1

[torn swift]

Cramer's Rule is interesting but not worth the time investment

[odd dagger]

ohh

[torn swift]

I'll write it down for a 2 x 2 then

[prisma marten]

I have to review my lin alg notes so much in August

[odd dagger]

sign matrix n 1/D

u transpose and have other stuff too idr

[prisma marten]

like the inverse is just what you need to make the Identity

[torn swift]

$\begin{bmatrix} a&b\c&d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d&-b\-c&a \end{bmatrix}$

[obsidian monolith]

⚡Amphy⚡:

[prisma marten]

I remember that, but im so bad with the computational stuff

ad-bc is just the det(A)?

[torn swift]

yes

[prisma marten]

oh what is that matrix on the right

[torn swift]

the whole right side is the inverse matrix

for 2 x 2 only

[prisma marten]

I mean the [d -b; - c a]

[odd dagger]

@prisma marten here

[torn swift]

oh, that is nice formula for 3 x 3

except det(P) is a pain

[prisma marten]

why did a and d switch spots in the 2x2 example

[odd dagger]

matrix of minors

read the 3x3 method

the 2x2 is that but u can just remember it easily

[echo plaza]

adjunct I've heard it called

[torn swift]

you can use Cramer's Rule to get the inverse of the symbolic 2 x 2 actually

[odd dagger]

for 2x2 the determinants are of 1x1 matrices lol

[prisma marten]

oh the adjucate matrix

I remember this

I skipped this on the final exam and just wrote what I would do instead of doing the computations

[odd dagger]

minors are also called cofactors if u know thst

[prisma marten]

oh yea

okay so lost in translation

[odd dagger]

lmao

[prisma marten]

wow, I didnt even realize there was a whole different language out here for math

[odd dagger]

A^-1 = 1/|A| adj(A)
adj(A) = C^T

[next willow]

Bob Harris?

[odd dagger]

C is matrix of cofactors/minors

[spark cliff]

okay so in a cardioid and limacon graph i understand that when a>b it’s a dimpled limacon, and when b>a it’s a limacon with an inner loop

but what happens when a=b

a and b are supposed to be absolute values but i’m too lazy to type it out

[spark cliff]

never mind it’s just a cardioid lol

[viscid thistle]

A=LU is superior

do y’all end up learning linear algebruh in hs in America?

[torn swift]

not typically I think

to be more complete, it would be PA=LU

[viscid thistle]

permutations are irrelevant

only reduce good matrices m

[torn swift]

lol

the world isn't perfect though

[viscid thistle]

[torn swift]

we hope for Λ but often we can only get J

[viscid thistle]

lol

[wraith idol]

PALU

My favorite

[vernal rapids]

Le french interval 🥖

[obsidian monolith]

Whoever:

[wraith idol]

Bruh why am I so stupid

[heady jewel]

😂

[shrewd urchin]

Everyone is stupid

[hexed ermine]

😔

[wraith idol]

But I'm not everyone, I'm whoever

And it's different

[shrewd urchin]

Whoever $\in$ Everyone

[obsidian monolith]

Krishna:

[shrewd urchin]

[zenith sable]

lets learn something today

[shrewd urchin]

i is a devil's number

[zenith sable]

why is it a devil's number?

Find all the complex number z if $z^4=\frac{1-i}{\sqrt{2}\left[\cos(\frac{5\pi}{12})+i\sin(\frac{5\pi}{12})\right]}$

[spring thunder]

Find all the complex number z if $z^4=\frac{1-i}{\sqrt{2}}\left[\cos(\frac{5\pi}{12})+i\sin(\frac{5\pi}{12})\right]$

[zenith sable]

that cos and isin part is supposed to be on the denominator

[obsidian monolith]

emeric75:

[spring thunder]

ah

[zenith sable]

i dont get it

why is it failing

[spring thunder]

well you totally forgot to close the denominator

[zenith sable]

oh i see

[obsidian monolith]

Manovic:

[zenith sable]

okay yep this is the problem

$|1-i|=\sqrt{1^2+(-1)^2} = \sqrt{2}$ and $\arctan(1-i)=\arctan(\frac{-1}{1}) = -\frac{\pi}{4}$

[obsidian monolith]

Manovic:

[zenith sable]

$1-i = \sqrt{2}\left[\cos(\frac{-\pi}{4}) + i\sin(\frac{-\pi}{4})\right]$

[obsidian monolith]

Manovic:

[zenith sable]

$z^4=\frac{\sqrt{2}\left[\cos(\frac{-\pi}{4}) + i\sin(\frac{-\pi}{4})\right]}{\sqrt{2}\left[\cos(\frac{5\pi}{12})+i\sin(\frac{5\pi}{12})\right]}$

[obsidian monolith]

Manovic:

[zenith sable]

ok so now

[shrewd urchin]

$x^2=1$

[obsidian monolith]

Krishna:

[shrewd urchin]

Why I don't have white background

[frozen needle]

,tex --colour white

[zenith sable]

nobody gave help yet?

[frozen needle]

what's giving you trouble?

[zenith sable]

look up there's the question and what i have done so far

[frozen needle]

if you rewrite the thing into 1 complex exponential it'll look easier probably

[zenith sable]

$e^{\sqrt{2}i(\frac{-\pi}{4})}$

[obsidian monolith]

Manovic:

[zenith sable]

is this even correct

for the numerator

[frozen needle]

doesn't look like it is

[zenith sable]

$\sqrt{2}e^{i\frac{-\pi}{4}}$

[obsidian monolith]

Manovic:

[frozen needle]

now that's better

[zenith sable]

how would i write both the numerator and denominator in one exponential

$\sqrt{2}e^{i(\frac{5\pi}{12})}$

[obsidian monolith]

Manovic:

[zenith sable]

$\frac{\sqrt{2}e^{i\frac{-\pi}{4}}}{\sqrt{2}e^{i(\frac{5\pi}{12})}}$

[obsidian monolith]

Manovic:

[frozen needle]

now fuse them into 1 exponential

[zenith sable]

$e^{i(\frac{-\pi}{4}-\frac{5\pi}{12})}$

[frozen needle]

yeah

erm wtf is that - -

[obsidian monolith]

Manovic:

[frozen needle]

phew

[zenith sable]

double subtraction lol

[frozen needle]

so now you should be able to deal with it

[zenith sable]

not confident enough tho

ive never done a problem like this before i think

[frozen needle]

there's a formula for the nth roots of complex numbers

[zenith sable]

does it have a name

[frozen needle]

that should be somewhere in your notes

[zenith sable]

I got no notes

im literally just trying these problems out

havent taken a proper class to this

and never done a problem like this

but i guess this is how i learn, by doing them!

$z^n=r^n(\cos(n\theta)+i\sin(n\theta))$

[echo plaza]

\cos \sin

[obsidian monolith]

Manovic:

[zenith sable]

i found this formula

how do i continue lads

[frozen needle]

learn about nth roots first and then come back to the exercise

[zenith sable]

wait a second guys

$z^4=\frac{\sqrt{2}\left[\cos(\frac{-\pi}{4}) + i\sin(\frac{-\pi}{4})\right]}{\sqrt{2}\left[\cos(\frac{5\pi}{12})+i\sin(\frac{5\pi}{12})\right]}$

[obsidian monolith]

Manovic:

[zenith sable]

why did we even put this into exponential form?

but

this can easily be turned into:

$z^4=\cos(\frac{-\pi}{4}-\frac{5\pi}{12}) + i\sin(\frac{-\pi}{4}-\frac{5\pi}{12})$

[obsidian monolith]

Manovic:

JY1853:

[zenith sable]

$z^4=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})$

[obsidian monolith]

Manovic:

[zenith sable]

i know theyre the same but

isnt this easier to solve?

[frozen needle]

how so?

[zenith sable]

$z=\sqrt[4]{\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})}$

[obsidian monolith]

Manovic:

[zenith sable]

yeah z

[frozen needle]

ouch

[echo plaza]

[frozen needle]

don't do it like that

there are 4 fouth roots

when you write that, it looks like you assume uniqueness

[zenith sable]

$z = \cos(\frac{\frac{-2\pi}{3}+2k\pi}{4})+i\sin(\frac{\frac{-2\pi}{3}+2k\pi){4}$

[obsidian monolith]

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

[zenith sable]

ughh

well frick

[frozen needle]

$a^4=b$ does not imply $a=\sqrt[4]{b}$

[obsidian monolith]

Tuong:

[frozen needle]

it's not true in C, it's not true either in R

[obsidian monolith]

JY1853:

[zenith sable]

$z_k=\cos(\frac{\frac{-2\pi}{3}+2k\pi}{4})+i\sin(\frac{\frac{-2\pi}{3}+2k\pi}{4})$

[obsidian monolith]

Manovic:

[zenith sable]

for k = 0,1,2,3.

[frozen needle]

now that's better

[zenith sable]

$z_0 = \cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})$

[obsidian monolith]

Manovic:

[zenith sable]

$z_1 = \cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})$

[obsidian monolith]

Manovic:

[zenith sable]

$z_2 = \cos(\frac{5\pi}{6})+i\sin(\frac{5\pi}{6})$

[obsidian monolith]

Manovic:

[zenith sable]

$z_3 = \cos(\frac{4\pi}{3})+i\sin(\frac{4\pi}{3})$

[obsidian monolith]

Manovic:

[echo plaza]

not worth bothering

[rancid summit]

how would i be able to visualize the graph of x=(1/16)y^2+(1/4)y-(11/4)

[shrewd urchin]

Draw a parabola

Do you know how to graph it ?

[rancid summit]

from looking at desmos yeah but

my teacher wants me to explain in words my thought process

to graph it

hmm so its a parabola thats perpendicular to the y axis

[shrewd urchin]

Graph of y=f(x) is set of all values of x for which y is defined .

So I would say

Pick some real values for x

Then find corresponding value of y

Then you can graph it

I mean you can't just think of a parabola without drawing it .

I am sure The first person who drew parabola did the same way

[rancid summit]

so pick some points and show how i got them, then explain its a parabola

[shrewd urchin]

Yes

Are you introduced to parabola before @rancid summit

[rancid summit]

yeah its like x^2 or y^2 if its a func in terms of x

[shrewd urchin]

Hmm

[rancid summit]

y i mean

im guessing since my function is in terms of y, ill say that thats why its rotated

[shrewd urchin]

Do you know the maximum and minimum point on a parabola

[rancid summit]

yeah

[shrewd urchin]

Yea y=f(x) because x is a free or independent variable

If x^2 is the quadratic term

Ok

I am going right now I have to sleep . I woke up early this morning.

[rancid summit]

thats a good step actually, i forgot about that being the easiest to base the graph off of

[shrewd urchin]

Yes

Find the min value of quadratic

[rancid summit]

thank you, have a good rest

then explain it keeps going on, to infinity

[shrewd urchin]

Assuming you dealing with a>0

Hmm

Because

X is a free variable it can take on any value

[rancid summit]

but its range is still limited and cannot go below its minimum value

[shrewd urchin]

Yew

Yes

I told x can take on any value

Not "y"

[rancid summit]

yeah domain is infinite

[shrewd urchin]

Hmm

[rancid summit]

[-infin, +infin]

[shrewd urchin]

Ree

(.....)

Not [...]

[rancid summit]

ohhh yeah

forgot about that if its infinity

[shrewd urchin]

Infinity is a hard number to define

Lol

[rancid summit]

im guessing thats why infinity/infinity is a indeterminate form

[trail badge]

(Domain isn't "infinite", but yes R)

Domain: R

😛

[rancid summit]

thanks for the help, i think i can describe it now

[trail badge]

U're welcome

[shrewd urchin]

I mentioned above

X can't take on any real

[trail badge]

I didn't say that X can take any real name, I said Domain is R and not Infinite, I just made a language correction

[rocky bison]

The Real numbers are an uncountably infinite set though @trail badge?

So whilst you can't have a value of infinity (In a way this is almost nonsensical anyway) you can having values that diverge TO infinity

[shrewd urchin]

Ya

[trail badge]

Ohh

I thought that they are talking about domain and not "lim" (or some like that)

[rocky bison]

Don't think of infinity as a value

Because that's what's most likely throwing you off

The real domain extends to infinity

But infinity isn't an element of the set

We also call it uncountable because between 1 and 2 there's infinitely many numbers

So we can't count through it, unlike the natural numbers for example

That also extends to infinity but it's countable

[trail badge]

yeah, I know

[shrewd urchin]

Infinity is just a result if the fact that there is no such thing a biggest number or a smallest number.

[serene heath]

wrong

5 is the biggest number

[shrewd urchin]

You are arrested for doing illegal math .

[echo plaza]

saying the naturals are countable isn't saying much

yes there is an injective function from the naturals to the naturals

[trail badge]

Nope

The biggest number is ||coming soon...||

[frozen needle]

5 = +∞

that's what my physics teacher told

[heady jewel]

well technically true as infinity just means a very big number

it can be 500 or 5 million

[viscid thistle]

lol

[royal gull]

[vernal anchor]

5 is the biggest number i ever saw

[royal gull]

https://www.youtube.com/watch?v=7Xon9kOEhl4

[shrewd urchin]

Panda cop is finding the people who say 5 is the biggest number .

[trail badge]

The biggest number is 5!

Who understood my joke?

[shrewd urchin]

I will sue you 5mil$

[trail badge]

5! = 5x4x3x2x1 = 120

[shrewd urchin]

Bad joke

[trail badge]

[shrewd urchin]

under arrest and sued 5! Million dollars

And yes 5! = 5x4x3x2x1 = 120

[thick star]

https://media.discordapp.net/attachments/197558652029239297/603068063478317076/unknown.png

does this become 7x-3 </= e^3
or 7x-3>/= e^3?

[native sequoia]

since $e^y$ is increasing, $a\le b\implies e^a\le e^b$

[obsidian monolith]

EpicGuy4227:

[thick star]

i'm confused

[native sequoia]

let a=3 and b=ln(7x-3)

[thick star]

e^3 >/= e^7x-3

?

i dont get it

[native sequoia]

$
3\le \ln(7x-3)\implies e^3\le e^{\ln(7x-3)}=7x-3$

[obsidian monolith]

EpicGuy4227:

[thick star]

ok so
7x-3>/= e^3?

[native sequoia]

ye

[thick star]

how about log_3(2x) </= 4
would it be 3^4 </= 2x

?

[native sequoia]

ye

[thick star]

oh i sorta meant the other way

but if it is

i dont get the connection

~-~

[native sequoia]

oh ye it should be the other way

[thick star]

ok

[willow bear]

@thick star just write <=

</= looks dumb

[viscid thistle]

Or just leq

[tropic crown]

does (lnx)^2 = lnx^2?

[torn swift]

(lnx)^2 =/= ln(x^2)

[tropic crown]

ty

[serene heath]

what if x=1

[frozen needle]

^ and that's an example of why one shouldn't confuse f and f(x)

[tidal sequoia]

I'm currently learning sum + difference, double-angle, and half angle identities

What's a good way to be good at them and remembering the identitities

[prisma marten]

Practice

And derive them yourselves

Double angle is just sum

Sin(x+x)

[tidal sequoia]

whats the point of a double angle though? wouldn't somebody just write down Sin(90) instead of Sin(45+45)

or insert

whatever

[royal gull]

sometimes its easier to compute, when the angle is a variable for example

[short sorrel]

Moreover

It's often useful in actual problems

If you know one angle is twice another (say, you have a large angle and you have the angle after bisection)

Then you can establish a relationship using a double angle identity

This becomes relevant for related rates problems in calculus in particular.

As well as for trigonometric proofs

[tidal sequoia]

So I'm doing a half-angle identity problem

https://gyazo.com/a9e4aa2353afbfe6a6678ba3c051924e

I got to a point where I'm at

(1-2cos(12x)+cos(12x)^2)/4

I'm not sure what to do to get it to this specific form

that is being asked

[tidal sequoia]

<@&286206848099549185>

[proud raven]

$\frac{1-2cos(12x)+cos^2(12x)}{4}$

[obsidian monolith]

jan Niku:

[proud raven]

like this?

[tidal sequoia]

yeah

[proud raven]

is that what you start with?

or did they give you something else

[tidal sequoia]

That's what I come up with from (sin(6x))^4

[proud raven]

oh

[tidal sequoia]

using half angle identities

but it's supposed to go into that one form

[proud raven]

sorry im trying to track it

[tidal sequoia]

I'm like 99.9% positive that the form I have it in is still true, I just dont know how I'm supposed to manipulate it into the screenshotted format

[proud raven]

oh wait

they just get rid of all exponents

so its flat powers

and only 2 different forms of x

[tidal sequoia]

I dont follow

:/

where does the 1/8cos(___x) come from

[proud raven]

im just looking like

i figure every time you apply that reduction formula you get another power of 2

so it probably means they reduced 3 times

which makes sense

or they like

power reduction, half angle, power reduction

each time adds another 1/2 factor

i think i get it

[tidal sequoia]

OH OK I think I get it. I can't reduce the whole thing so I have to break it apart, and then I can reduce cos^2(12x)/4 to 1/8cos(24x)

[proud raven]

youre almost there the problem is just fucky

yea

then the first box is a constant

then your 2cos(12x) cancels with the 4

you just need to reduce one more time

$\frac{1-2cos(12x)}{4}+$\frac{1+cos(24x)}{8}$

texit doesnt like it

lol

youll have a constant come out of each fraction when you have all the powers to 1

[tidal sequoia]

I'm sorry wdym by that

[proud raven]

do you have all of powers to 1

no cos^2 or anything

[tidal sequoia]

This is hw for an online class, and it comes with a video, but the video doesn't explain anything well

😦

https://gyazo.com/e09d2dad9f9b8ff1110b83912a6287e9

[proud raven]

okay

but you should have like

i think 1+cos(something)

in the numerator of each of those functions

right

[tidal sequoia]

yeah

[proud raven]

$\frac{1-2cos(12x)}{4}$

[obsidian monolith]

jan Niku:

[proud raven]

is my first one

so split it

$\frac{1}{4}-\frac{2cos(12x)}{4}$

[obsidian monolith]

jan Niku:

[proud raven]

you have to do that with the other fraction as well

does that make sense?

[tidal sequoia]

I did, that how I reduce the other one to 1/8

[proud raven]

but you still have numbers

1/4 and 1/8

i think thats what they want in the first box

theyre asking you to combine all of the parts that aren't a trig function into a single term

[tidal sequoia]

ok so the constant is just supposed to balance out the trig functions as if they didn't have the fractions

?

https://gyazo.com/4657c7a3722b8c3e37def7219c9f8685

[proud raven]

the numbers are part of the function

yea balance out i guess

you have to account for them

[tidal sequoia]

so cuz I'm a bit dumb right now, and bad at math. Where do the numbers come from that I need to start accounting for them

like they must "dissappear" or "get lost" I would assume

[proud raven]

i mean

theyre just ones usually

when youre dealing with these identities

you can derive them if you want

if you think that would help you understand

https://www.themathpage.com/atrig/double-proof.htm

they all come from sum and difference identities

and unit circle stuffs

like how you can change sin^2+cos^2 to 1

[tidal sequoia]

ok. I guess I'll have to just spend some time on the sum/difference identities, and the double and half angle identities

They are still relatively new, I'm sure if I keep doing stuff and reading and watching videos I'll get it

[proud raven]

if you can memorize the sum/difference, you can derive the rest in a pinch

oh yea if theyre new, theyll come to make sense for sure

[marble basalt]

hmm... 🤔

[proud raven]

its super confusing when you first learn it

i still have my notes from trig when i learned all this and the whole page is just big question marks over notes about the proof

[tidal sequoia]

any tips for breaking down and digesting proofs easier

[proud raven]

do em yourself 😄

[tidal sequoia]

I don't have much experience with any kind of proofs or derivations

lol

[proud raven]

idk if these are proofs im talking about

derivations yea

i mean do the cosine double angle

we can do like

you need your cosine sum identity

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

[obsidian monolith]

jan Niku:

[proud raven]

so you have cos(2a)

which is really the same as a+a

so you can just plug in a for a and a for b

$cos(a+a)=cos(a)cos(a)-sin(a)sin(a)$

[obsidian monolith]

jan Niku:

[proud raven]

and then you have products you can simplify down

$cos(a+a)=cos^2(a)-sin^2(a)$

[tidal sequoia]

ok

[obsidian monolith]

jan Niku:

[proud raven]

tada 😄

sorry if that was overboard lol

i like trig

[tidal sequoia]

no thanks I appreciate it very much

I need and like things broken down as much as they can be. I need to fully understand why something occurs or why a process is what it is. Like for example when my teacher in the past stated you can't divide by 0, and didn't explain it, it bothered me bc I needed to know why it's not allowed. (and eventually watched a cool ted-ed video)

so that helps

[proud raven]

oh!

you know before my trig exams i usually didnt even do problems

i used to just rederive everything

im the same way lol

if i get it i get it i dont need formulas or anything

i never did make sense of where the sum/difference formulas THEMSELVES come from but

thats a rabbit hole ill dive into some day

[pale bison]

you can probably google the proof of that

[tidal sequoia]

sum/difference are probably derived by the normal trig identities somehow

[proud raven]

if you get really comfortable with your trig stuff

you can figure out most of your trig stuff in calculus on the fly too

😄

in case you need motivation its gonna save you many headaches

[tidal sequoia]

good to know

[viscid thistle]

hey guys

i had a problem here

I really have no clue how to find the value of the second one

What does i equal exactly

[willow bear]

it doesn't equal anything

it's the index variable (or dummy variable or counter) of the summation

[viscid thistle]

Ah I see

But I still don't get it

Because if it were the same as n, the answer would be 0

The difference of the both of them would equal 0, 24-24, but when I did it, 0 was wrong

[short sorrel]

this is a notational thing

$\sum_{n = 1}^{4} 2n + 1 = \left(\sum_{n=1}^{4} 2n\right) + 1$

[willow bear]

do you know how summation notation works

[obsidian monolith]

Namington:

[viscid thistle]

I just learned about it slightly today

So not that much, no

[short sorrel]

it's like the difference between 4*2 + 1 and 4*(2+1)

[viscid thistle]

oh.....

[short sorrel]

in the right sum (the one that uses i as the index), the +1 is included in the sigma sum

in the left one, it's not

[viscid thistle]

So in my case, it would be 2(1) + 1... etc on the left and on the right it would be 2(1+1)... etc

Right?

[short sorrel]

...not quite

[viscid thistle]

Agh :/

[short sorrel]

in the left sum

$\sum_{n=1}^{4} 2n + 1$

[obsidian monolith]

Namington:

[short sorrel]

this means

$\left(\sum_{n=1}^{4} 2n\right) + 1$

[obsidian monolith]

Namington:

[short sorrel]

in other words, the term you're adding up over and over again is 2n

[short sorrel]

and then you add on +1 after everything else

[viscid thistle]

Then if that's the case, wouldn't it also be the same process for the one with the 2n+1

I'm sure it would be different, right

I mean, I know the value of the left one with the 2n+1 = 24, but according to that explanation, the (2i+1) would also equal 24

Which is wrong because their differences aren't 0 (I checked answer)

[short sorrel]

$\sum_{n=1}^{4} 2n + 1 = ((2 \cdot 1) + (2 \cdot 2) + (2 \cdot 3) + (2 \cdot 4)) + 1$

[obsidian monolith]

Namington:

[short sorrel]

the +1 isnt included in the term that's added up each time

because its not in the brackets

[viscid thistle]

OOOOHOHH

My goodness

I get it now

Hahahaha

Thank you for the help

Really appreciate it!

[short sorrel]

👍

[viscid thistle]

@short sorrel

What about this one

Any clue on that? :/

[short sorrel]

same deal

the term included in the sum is the n^2

the +3 is outside of the sum

[viscid thistle]

So would it be A?

Actually

No

Ugh

[short sorrel]

A seems right

[viscid thistle]

Oh, ok

Yup! Correct

Appreciate it

Again haha

[silk sequoia]

how do you do this one?

ping me

[serene heath]

simplify each fraction

[silk sequoia]

how do you do that

idk what to do with the factorials

[viscid thistle]

n! = 1 x 2 x 3 x ... x n

^ so just write out this expansion for each factorial

[short sorrel]

That's

[viscid thistle]

It should be clear where to go from here

[short sorrel]

Terribly inconvenient

Consider

n! = 1×2×3×...×(n-1)×n

Now, if you look at the part to the left of n

1×2×3×...×(n-1)

This is just (n-1)!

And that's multiplied by n

So n! = n(n-1)!

This identity will allow you to simplify and solve those fractions.

This should make intuitive sense, too

Consider 5! = 1×2×3×4×5

That's just 4! × 5

Since 4! = 1×2×3×4

So 5! = 4! × 5

And n! = (n-1)! × n

Use this to simplify the fractions, then add.

[silk sequoia]

oooooh okay that makes a lot more sense

so when you simplify you get 1/n + (n+1)!

right?

[serene heath]

why n+1!

[silk sequoia]

because on the right side, on the numerator (n+1)! = n!(n+1)

if im understanding this right

[short sorrel]

Right

So where's the factorial come from

[silk sequoia]

oh crap

there shouldnt be a factorial

oops

[short sorrel]

There we go

So you get 1/n + n + 1

Now, rewrite that in a form that fits one of the answers

[silk sequoia]

n^2+n+1/n

[serene heath]

[short sorrel]

Brackets around the numerator

But yes

,w (n^2 + n + 1)/n = (n-1)!/n! + (n+1)!/n!

[obsidian monolith]

Display results online and refine query

Requested by Namington#1983

[silk sequoia]

apologies, it slipped my mind

but i understand the problem now

ty!

[shrewd urchin]

Aaw

[viscid thistle]

Hey, I just had a question here

I'm not sure how to really solve this, wondering if anyone could help

<@&286206848099549185>

[pale kettle]

!15m

[brave topaz]

@viscid thistle , They are asking for you to find the sum of the first nine terms of what they gave you. The part highlighted is just saying that you sum the terms starting at k = 1, and ending at k = 9.

The sequence they gave you, 2(4)^k-1, is a geometric sequence. Your first term, when k = 1, is just going to be 2.

Each successive term is going to be 2 multiplied by a power of 4. 4 is your common ratio, or the amount that each term gets multiplied by to get to the next term

The formula for finding the sum of the first n terms of a geometric sequence is:

Where a1 is the first term, in this case 2, and r is the common ratio, in this case 4. N is the number of terms you want to sum up to, so in this case that's nine. Plugging in those values into the above formula, you'll find that one of your answer choices is the correct one.

Let me know if this helps 😃

[viscid thistle]

@brave topaz Ty so much! Got it right 😃

[brave topaz]

Awesome, nice job!

[viscid thistle]

😃

[tropic crown]

how can we prove this limit using the definition?

[next willow]

To start, do you have a candidate value for the limit in mind?

[tropic crown]

no i dont

kinda confused on that part

when i did practice questions, they usually gaev a limit

[next willow]

I'd suggest you to draw the function to get a possible idea on that

Also, do you know what is a continuous function?

[tropic crown]

Possibly 3?

[next willow]

Why?

[tropic crown]

The way I got this number is probably not the right way of thinking things

But they make the function = 0

[next willow]

Well they are asking you to evaluate

$\lim_{x->1} f(x)$

[obsidian monolith]

Mat:

[next willow]

Intuitively, if x approaches 1, what value does f(x) tend to assume?

Maybe look at your graph

[tropic crown]

If x approaches 1, it will = -2

[next willow]

Ah good

f(1)

[tropic crown]

Lemme give u a pic of what I've done so far

Not sure what to really make of it...

[next willow]

This is one definition of a continuous function, have you already seen something related?

[next willow]

x^2-3x is continuous

I think there are some problems, because after you figure out the value of the limit is -2, you should start with

$|x^2-3x-(-2)|<\epsilon$

[obsidian monolith]

Mat:

[tropic crown]

Okay, let me give that a try

sadly, i was not able to prove it

can u help identify what i did wrong?

@next willow

[next willow]

The procedure seems correct but at the beginning you took -2 out of the absolute value, when it was inside

It is not possible to do it

$|x^2-3x-(-2)|<\epsilon$\
$-2-\epsilon<x^2-3x<-2+\epsilon$ \

[obsidian monolith]

Mat:

[tropic crown]

okay, let me try -2

and see the result

but im curious as to why its - (-2)

[next willow]

You can also simplify the minuses and write |..+2|
I wrote it like that because
|f(x)-a| is the distance of f(x) from a

In our case a=-2

[tropic crown]

hmm, i still get the same result

not sure what im doing wrong

[next willow]

Let's see, I didn't read all before

Ok we can continue from

$\frac{1}{4}-\epsilon<(x-\frac{3}{2})^2<\frac{1}{4}+\epsilon$

[obsidian monolith]

Mat:

[next willow]

To solve this you can write down the system:

[tropic crown]

oh

i think i see where i went wrong

lemme check

[next willow]

$\begin{cases} \frac{1}{4}-\epsilon<(x-\frac{3}{2})^2 \ (x-\frac{3}{2})^2<\frac{1}{4}+\epsilon \end{cases} $

[obsidian monolith]

Mat:

[tropic crown]

i believe its suppose to be (&-1/2)^2

[next willow]

Mm, I think you should solve the system anyway

Because it is not correct to simply apply the sqrt to the left and right

Solve separately the inequalities above, then intersect solutions

Also, you have to claim that you consider
ε<1/4

[tropic crown]

how do we determine which piecewise equation to use?

[next willow]

You mean of the two intervals we get as a solution to the system?

[tropic crown]

yes

[next willow]

You choose the interval which contains 1

It is the one you need to prove the limit definition

As we need to find a neighbourhood of 1

[tropic crown]

thanks a lot mat, i really appreciate the help

[next willow]

np

[tropic crown]

im not even sure how to approach this question

im not familiar with E - N definition of a limit at infinity

im mostly familiar with epsilon - delta

[willow bear]

the underlying idea is mostly the same

here's the definition

$\lim_{x \to +\infty} f(x) = L \iff \forall \varepsilon > 0, \exists N : \forall x > N, |f(x) - L| < \varepsilon$

[obsidian monolith]

Ann:

[willow bear]

@tropic crown also maybe this should go in #calculus

[viscid thistle]

This is taught in precalculus

Limits

But it’s better to be in #calculus

[short sorrel]

i'd be very surprised to see a class called "precalculus" that teaches epsilon-delta limits

anything's possible, I suppose

but many calculus classes don't even touch on epsilon-delta

much less precalc

[shrewd urchin]

[viscid thistle]

They used epsilon delta?

Yeah nvm

I would be surprised too

I thought it as a regular limit question

my hs maff teach didn't know what epsilon delta was

and squez theorm

lol

she be like stop making up fancy things

[languid crane]

Lmfso

We don’t even touch the Mean Value Theorem on calc

[viscid thistle]

reeip

[languid crane]

Yep

[viscid thistle]

We’ve already established how your calculus sucks @languid crane

[languid crane]

Yes I know

My mental health is shit so pls just ignore

[serene heath]

[languid crane]

😭

[tropic crown]

u guys are actually right, im currenty in a calc coursse thats passed precalc

[prisma marten]

I feel most Calc courses are past pre Calc

[shrewd urchin]

🤔

[tropic crown]

i just thought it be appropriate to ask here because "the precise definition of a limit"

was in the earlier chapters of my text

[prisma marten]

Episilon Delta definition?

Personally haven’t used it other than Midterm and Final, but I’m also just in engineering so it might be relevant elsewhere

[tropic crown]

i just woke up, i still dont know exactly what

https://cdn.discordapp.com/attachments/363224154469826562/603845588634697734/183668144404037632.png

this is suppose to say when answering

https://cdn.discordapp.com/attachments/363224154469826562/603832414136958987/unknown.png

[prisma marten]

Yea that’s Epsilon Delta it seems