precalculus

[sacred eagle]

ok

what about for this 1

https://gyazo.com/7d7528edb7b363ac4c3b61aea177728c

i just gotta make sure all the negatives are right ll

[clever inlet]

Yeah

[sacred eagle]

that 1 is negative ?

[clever inlet]

Yep

[sacred eagle]

lol somehow im guessing the signs and its working

[hexed ermine]

No need to guess

Just learn it

Do you work better with radians or degrees?

[sacred eagle]

both

ooh

[clever inlet]

It ultimates comes down to X and Y coordinate of unit circle

[sacred eagle]

not sure

im wokring with both doe

[hexed ermine]

Well if I have a negative as a theta value, I would add to make it positive

since it's periodic it works like that

so -2pi/3 + 6pi/3 = 4pi/3

And I know that 4pi/3 is in quad iii, where x and y are negative

[sacred eagle]

how do u do this then

https://gyazo.com/3b4728587e5fbd8c47b8a9e45112f616

just 3/5 ?

[hexed ermine]

$$\frac {\sqrt{22}}{5}^2 + \cos^2 = 1$$

wtf

i'm having a hard time with this lol

[sacred eagle]

yeah idk

idek what its asking like

[hexed ermine]

anyways it simplifies to 22/25+cos^2=1

cos^2=3/25

So cosine is sqrt(3)/5

[sacred eagle]

oh what

i thought it was just 3/5

[hexed ermine]

No

[clever inlet]

Use the identity

[hexed ermine]

You have to take the square root

[sacred eagle]

ooh

i kinda dont ge it

[dense zealot]

?

Ur just plugging stuff in

U know cost = sqrt(22)/5

So u can make a right triangle since in the unit circle cos = x

And we want sin

We know the hypotenuse is 1

So 22/25 + (sint)^2 = 1^2

(sint)^2 = 3/25

sint = sqrt(3)/5

[sacred eagle]

https://gyazo.com/6ebbc7473a7492765ca13667d3c16274

hypotnuse is 9

[dense zealot]

?

Wdym

Hypotenuse is 1 in the unit circle

[sacred eagle]

ooh

how would u that that 1

8/9 + cos^2 =1 ?

[dense zealot]

64/81

sint = 8/9

So (sint)^2 = 64/81

[sacred eagle]

what is that

i dont get that 1

[hexed ermine]

$$\cos^2+\sin^2 = 1 \newline \Rightarrow \cos^2+(\frac{8}{9})^{2}=1 \newline \Rightarrow \cos^2+\frac{64}{81}=1$$

[dense zealot]

sqrt(17)/9

[hexed ermine]

cos^2=17/81

cos=sqrt(17)/9

So sqrt(17)/9

[thick raptor]

@hexed ermine wat?

cos^2 <= 1

[hexed ermine]

?

[thick raptor]

It should be cos^2 = 17/81

[hexed ermine]

Ues

Which simplifies to sqrt(17)/9

take the square root

[calm thicket]

I'm not sure how to represent the inverse functions geometrically so my usual tactic of a mental triangle isn't working.

[hexed ermine]

they are cofunctions

[calm thicket]

Oh my god

I'm stupid

Thanks

[hexed ermine]

LOL

np

[calm thicket]

Still tho is there a way to visualize inverses geometrically?

[hexed ermine]

No I don't believe

[calm thicket]

oh yikes

[hexed ermine]

oh fun I remember this

[calm thicket]

Yea

[hexed ermine]

I'm assuming you are given, unit circle right?

[calm thicket]

Correct

The first two I can do with that

and the third

and the last

But 4 and 5 are unique, which makes me theorize that there's a simpler way to do it

[hexed ermine]

WIthout calculator?

[calm thicket]

Yea

Wait, arctan(4/5) means we're asking for the angle whose tangent is 4/5...

So we can draw it as a triangle

with o=4 and a=5

[hexed ermine]

yes

[calm thicket]

Finding hypotenuse with pythag we get sin(that)=4/(sqrt(blabla))

Fourth one, we can do the same with adjacent and hypotenuse to get csc as h/o

[hexed ermine]

Nah, you have to use calculator

There is no way they are expecting you to find decimal precise, angle values

[calm thicket]

You sure?

[hexed ermine]

Well, do you know of any way to find the angle?

[calm thicket]

arcfunctions can be remodelled to decimal-able areas

We aren't concerned with finding the arctangent, we want the sine of the angle whose tangent is 4/5, right?

[hexed ermine]

I've always useda calculator when calculating arctan

[calm thicket]

So we can almost cheat

By drawing a triangle with opposite side of 4 and adjascent of 5

hypotenuse is easy

we know soh

sine=o/h

So we can skip the angle

[hexed ermine]

Oh yeah you are right

I'm not a pro at this, i'm in fact doing the same stuff as you lol

[calm thicket]

Similar tactic is usable for the 5th

and the rest we can just brute force anyways :P

And neato

this is just the review :p

Worst case we just end up using a taylor series xD

[hexed ermine]

Have no idea what that is

heard of it, haven't taken a class that involved it

[calm thicket]

ah

basically used to express functions as polynomials

[hexed ermine]

Oh I see

[calm thicket]

=tex \sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}...

[calm thicket]

Cosine is similar but with 2k instead of 2k+1

They're not hard to memorize and if you ever find yourself needing a decimal trig function they're your best bet to my knowledge :P

[hexed ermine]

You had to know that for your trig class?

[calm thicket]

Nah

I know it from calculus

[hexed ermine]

Oh you've taken calculus?

[calm thicket]

Nay

Will take it next year

3b1b taught 'em to me tho

[hexed ermine]

Oh, are you in 11th grade?

[calm thicket]

Aye

Same for you?

[hexed ermine]

Oh ok same

Yes

[dense zealot]

I'm in 1st grade!

[calm thicket]

I am the first grade

I'm so first that the school wants me to do a presentation for them

[calm thicket]

53 & 54 pls, not sure how to approach it, tried factoring, reducing to base terms and converting to respective tangents

[thick raptor]

hi √i

[calm thicket]

It says to factor but doesn't seem to work on those

Heyo

[thick raptor]

The solution is to rewrite csc = 1/sin = 2i/(e^ix - e^-ix)

[calm thicket]

[thick raptor]

anyways that looks kinda messy

[calm thicket]

I could try 1/sin^3(x)...

[thick raptor]

prob nah

oh wait we're stupid

=tex u^3-u^2-u+1=u^2(u-1)-(u-1)=(u^2-1)(u-1)

[thick raptor]

I'm not good at factoring by grouping

[calm thicket]

oo

[thick raptor]

I prefer depressing the cubic and applying Cardano's method

[calm thicket]

I could totally mess with my teacher by doing u-substitution

[thick raptor]

Well like do what I did

[calm thicket]

Yea

[thick raptor]

but replace u with csc

[calm thicket]

nah, gonna pull the "am smert" card

"Let u = csc(x)"

[thick raptor]

👀

[calm thicket]

👀

wish me luck

[thick raptor]

k

wishes √i all the bad luck

[calm thicket]

rip luck

[dense zealot]

Lol

[calm thicket]

66 and 69, it's the "not in fractional form" part that's throwing me off.

inb4 $$\frac{5}{\tan(x)+\sec(x)}\to 5\div(\tan(x)+\sec(x))$$

[calm thicket]

technically not a fraction

[grizzled hull]

Oh, those are cute.

They're hiding Pythagorean identities.

Lemme grab LaTeX.

Something like

=tex \begin{array}{rcl}
\frac{5}{\tan x + \sec x}&=&\frac{5}{\tan x + \sec x}\frac{\tan x - \sec x}{\tan x - \sec x}\
&=&\frac{5\left (\tan x - \sec x \right )}{\left (\tan x + \sec x\right )\left (\tan x - \sec x \right )}\
&=&\frac{5\left (\tan x - \sec x \right )}{\tan^2 x - \sec^2 x}\
&=&\frac{5\left (\tan x - \sec x \right )}{-1}\
&=&-5\left (\tan x - \sec x \right )
\end{array}

[grizzled hull]

For 66

@calm thicket

[calm thicket]

o

Neato

[thick raptor]

inb4 69 jokes

[calm thicket]

inb4 g_69

[thick raptor]

inb4 TREE(69)

[calm thicket]

Is that bigger than g_69?

[thick raptor]

Bigger is an understatement

[calm thicket]

..

What is the tree function

how does it work

[thick raptor]

In fact it actually makes me kinda sad that people don't understand the difference in magnitude between TREE(3) and Graham's number

[calm thicket]

I've never heard it ;0

[thick raptor]

@calm thicket requires graph theory. I only know how to recursively approximate it

[calm thicket]

o

I'll take the recursive then

[thick raptor]

Okay

So this is gonna be super loose and bad

Define the following:

a+0 = a
a*0 = 0
a^0 = 1

a*(b+c) = a*b+a*c
a^(b+c) = (a^b)*(a^c)

Operations not associative

In a+b, a*b, and a^b, I say b is to the right

We also use infinities

_>

[calm thicket]

uhuh

[thick raptor]

=tex H_0(n)=n\H_{a+1}(n)=H_a(n+1)\H_{(\infty\text{ is all the way on the right)}}(n)=H_{\text{(replace last }\infty\text{ with }n)}(n)

[thick raptor]

So some simple examples

=tex H_3(2)=H_{2+1}(2)=H_2(3)=\dots=5

[thick raptor]

=tex H_\infty(2)=H_2(2)=4

[thick raptor]

=tex H_{\infty+2}(2)=H_{\infty+1}(3)=H_\infty(4)=H_4(4)=8

[thick raptor]

i.e. pop off the last num until it ends with $$\infty,$$replacing the $$\infty$$with $$n$$

[thick raptor]

=tex g_n\approx H_{\infty^{\infty+1}}(n)=H_{\infty^\infty\cdot\infty}(n)=H_{\infty^\infty\cdot n}(n)

[calm thicket]

uhuh

[thick raptor]

It gets pretty big

TREE(n) is bigger than $$H_{\infty^{\infty^{\infty^{\dots}}}}(n)$$

[thick raptor]

With n powers of infinity

Btw

This H probably grows faster than anything you can think of

[true vigil]

is TREE(4) infinite

[thick raptor]

@true vigil not according to ZFC

[true vigil]

what do you mean

[thick raptor]

ZFC can prove TREE(n) is total

I.e. finite for all n

[true vigil]

nice

[thick raptor]

ZFC has a PTO much larger than TREE(n)'s corresponding ordinal in the H function

Meaning it can pull off enough induction over graphs to prove TREE(n) is always finite

[true vigil]

is it big

its ordinal

how do you construct it

[grizzled hull]

Amusing that this is a precal topic.

[thick raptor]

Well it requires some understanding of ordinal notations

Fairly decent sized notations actually

@true vigil biggest countable ordinal you know?

[true vigil]

uhhhh I dunno, is that important? I know much bigger ordinals in ZFC like the ordinal of R, of 2^R, of 2^2^R, ...

[thick raptor]

Oh wait, were you asking about ZFC's PTO?

[true vigil]

yeah

do you just need a big countable ordinal to prove TREE(n) is finite?

[thick raptor]

TREE(n)'s PTO much smaller =/

Not quite

you need to do induction up to some ordinal

Also ZFC's PTO > all computable ordinal notations

Except maybe Tarnovski's C notation, but that's pretty complicated

Oh yeah, back to the question...

@true vigil idk how to calculate PTOs and it's actually pretty hard

Also the proofs tend to be scary lookin'

You can get lower bounds by taking theories ZFC proves consistent

And finding their PTOs

[true vigil]

[viscid thistle]

Why was this in precalc tho

[thick raptor]

why not?

seems close enuf

[timid jacinth]

I have the standard form of the equation of a circle; (x-1)^2+(y-2)^2=9, I have to find the intercepts algebraically

I'm trying to find the y-int first

so I set x = 0 and so far I have this

(0-1)^2+(y-2)^2=9

1 + y^2-4y+4 = 9

y^2-4y+5=9

y^2-4y=4

y(y-4)=4

not sure where to go from here

[tired cedar]

why did u expand the y bracket tho

[timid jacinth]

figured it'd be easier

[tired cedar]

well

[timid jacinth]

should I not?

[tired cedar]

hm i kinda skipped out on precalc stuff

but u should have 2 y intercepts right

[timid jacinth]

yep

[tired cedar]

so its gonna be +- something + some random number

[timid jacinth]

Right now I just need to isolate the y

[tired cedar]

um i haven't done a lot of precalc, but try without expanding the y bracket

[timid jacinth]

oh nvm

you're right

I shouldn't have expanded

one sec

[tired cedar]

you already have the solutions to Y as of now but it's not right

yeah work with the bracket

[timid jacinth]

so

if I do it this way

I have

1 + (y-2)^2=9

(y-2)^2=8

y-2 = +or-sqrt(8)

y= +or-sqrt(8) + 2

right?

[tired cedar]

yah

looks right

[timid jacinth]

@tired cedar would it be y= +or- 2sqrt2 + 2?

[tired cedar]

wdym

oh if u want to simplify

yeah. better in this form

[fringe stream]

@timid jacinth Just a note, what you had was valid too

You just didn't realize you had a quadratic in y

But yeah, not expanding the y term is definitely easier

[tired cedar]

yes yes

[autumn verge]

Expand
log2x^-2
please help me with this

[rocky bison]

=tex log\left(2x^{-2}\right)

?

[rocky bison]

@autumn verge

[autumn verge]

the 2 behind the x is below the log

next to, like a small 2

[rocky bison]

Oh

Log base 2?

[autumn verge]

yeah

[rocky bison]

=tex log_2\left(x^{-2}\right)

Ok

[autumn verge]

yep

[rocky bison]

Let's start by just doing

=tex log_2\left(\frac{1}{x^2}\right)

[rocky bison]

Yes?

Lemme know if you don't understand a step anywhere

[autumn verge]

um i think we should use log rules

[rocky bison]

We will

Just wanna break it up a bit first

[autumn verge]

oh ok

[rocky bison]

same stuff happens eitherway

i don't like neg powers

so

Ye

Now we're going to use the rule

=tex log_a\left(\frac{b}{c}\right)=log_a\left(b\right)-log_a\left(c\right)

Yes?

[autumn verge]

yeah

[rocky bison]

Which gives us

=tex log_2\left(\frac{1}{x^2}\right)=log_2\left(1\right)-log_2\left(x^2\right)

[autumn verge]

oh ok

thanks

[rocky bison]

You got it from there?

[autumn verge]

yeah

[rocky bison]

awesome

[viscid thistle]

yo

how come lim as x-> infinity of sinx / e^x = 0

cause sin x to infiinity DNE riht

but sin x osciallytes between -1 and 1

so techincally its -1/0 or 1/0 ?

is that the case

@fringe stream

Why is the bottom 0?

i mean 1/ infinity ooops

which is 0

[thick raptor]

👀

[viscid thistle]

and what is -1/infinity?

0

lol

anything over infinity is 0 right

no, but that's a topic for later

well im in calc

yeah, it's a topic for later

[thick raptor]

∞/∞ = 0?

[viscid thistle]

noo

Fine all constants

[thick raptor]

=tex \infty\in\rm anything?

[viscid thistle]

you happy

lol

[thick raptor]

kek

[viscid thistle]

No

[thick raptor]

=tex \lim_{x\to\infty}\sin(x)\ne\rm constant

[thick raptor]

[viscid thistle]

well thats DNE

lol

[thick raptor]

Wait, why are we in #precalculus

[viscid thistle]

because hes in precalc

im in calc

what I've realized is if you use infinity as a number, it's a topic for later

[thick raptor]

oh ok

[viscid thistle]

its a precalc question

okay so

if Sin infinity DNE

why is DNE/ infinity =0

[thick raptor]

So DNE/∞ = 0?

[viscid thistle]

Yeah

[thick raptor]

[viscid thistle]

like how

[thick raptor]

what about this one?

=tex \lim_{x\to\infty}\sin(x)e^x=\rm DNE

[viscid thistle]

but its / e^x

[thick raptor]

Yeah, but this equals DNE too, right?

[viscid thistle]

You're asking, why does the sequence $$\frac{\sin(x)}{e^x}$$ converge to 0?

[viscid thistle]

im guessing idk

[thick raptor]

So surely we must have $$\lim_{x\to\infty}\frac{\sin(x)e^x}{e^x}=\frac{\rm DNE}\infty=0?$$

[viscid thistle]

bring out the epsilons

[thick raptor]

Yeah squeeze theorem is how you should state these things

[viscid thistle]

yeah but no e^x on top

only bottom

[thick raptor]

@viscid thistle I'm pointing out why your argument is wrong

via counterexamples

[viscid thistle]

Thtats what im trying to say

[viscid thistle]

how can it = 0

yeah I know that

but im assuming they dont know

yeah

[thick raptor]

[viscid thistle]

i neeed this squeeze theorom

$$\forall \epsilon > 0, \exists N$$ such that $$\frac{\sin(x)}{e^x} < \epsilon \ \forall x > N$$

[viscid thistle]

[thick raptor]

ew

kek

[viscid thistle]

Actually that one isn't that bad

[thick raptor]

Choose N = ceil(-ln(ε))

[viscid thistle]

so thats the squeeze theorem?

$$\frac{\sin(x)}{e^x} < \frac{1}{e^x} < \epsilon \implies e^x > \frac{1}{\epsilon}$$

[viscid thistle]

So pick N > ln(1 / epsilon)

[thick raptor]

👀

basically copying me

[viscid thistle]

ahhh i see

[thick raptor]

@viscid thistle no

[viscid thistle]

So its cause -1<sinx<1

like sin x is technically bounded

[thick raptor]

=tex \frac{-1}{e^x}\le\frac{\sin(x)}{e^x}\le\frac1{e^x}

[granite stirrup]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[viscid thistle]

but it doesnt convergeeee

yee

[thick raptor]

Noting the upper and lower bounds go to 0

[viscid thistle]

yep

[thick raptor]

then the middle term must go to 0

[viscid thistle]

[thick raptor]

squeeze theorem

[viscid thistle]

sorry im in calc

never took precalc so i skipped this

I just know how to solve limits

Well i know it just converges to 1 or 01

-1

[viscid thistle]

and e to infinity is infinity

and i know 1/infinity = 0 therefore

its 0

So like yeah ig its the same

but the squeeze theorem i never learned

1/infinity = 0

--__--

[thick raptor]

👌

[gilded mesa]

How do I get the inverse function of 2^x?

[viscid thistle]

well it is

lol

Infinity isn't a number

[gilded mesa]

It's a concept

[viscid thistle]

There's no such thing as 1 / infinity

[viscid thistle]

If you use it in class no one cares but if you do it in here our collective math autism will be engaged

1/infinity

nice sotto

[gilded mesa]

How do I get the inverse function of 2^x?

[viscid thistle]

look at your limit

lafom

[gilded mesa]

How?

[thick raptor]

Yeah, pft, 1/∞ = 0

[gilded mesa]

I forgot how to do it using logs

[viscid thistle]

Come on guys, infinity - infinity = 0 r-r-right??

Take the base 2 logarithm

[thick raptor]

If you wanna do it right, you should write st(1/ω) = 0

[viscid thistle]

1/infinity does = 0 wym

Yes i didn't put the lim x-> infinity 1/x =0

but You assuming you plug in infinity for x

therefore its 0

Okay but how come incalc they say its equal to 0

hmm

a limit isn't just plugging the value in

For example, consider a function that is 0 everywhere, but 1 at x = 0

[thick raptor]

What

stop being so stereotypical

[viscid thistle]

The limit at x = 0 is still 0

[thick raptor]

∞ can be a number

so can 1/∞

Shhh

[viscid thistle]

Even though you plug in x = 0 and get 1

it's called REAL ANAL(ysis)

what about 1/ infinity ^2

hmm

just stop lol

well its a real question

[thick raptor]

pft, non-standard analysis

[viscid thistle]

$$\lim_{x \to \infty} \frac{1}{x^2}$$

[thick raptor]

rip

\infty

[viscid thistle]

you mean something like that?

there is no such thing as "1/infinity^2"

yes

[thick raptor]

=tex 0<\frac1{\omega^2}<\frac1\omega

[thick raptor]

👀

[viscid thistle]

That's 0, you can use epsilon-delta to show it

the virgin squeeze theorem

vs

the chad epsilon-delta

[thick raptor]

lmao

[viscid thistle]

ahh i see

i think i messed up

[thick raptor]

who you calling a virgin? 👀

I <3 my squeeze theorem

[viscid thistle]

what i meant to say is lim x-> infinity of 1/ infinity ^2 = 0

Now is that true

LOl

you didn't even use the x

[thick raptor]

"lim x-> infinity of 1/ infinity ^2 = 0"

ikr

kek

[viscid thistle]

well yes it should be X

but chill

lol

[thick raptor]

facepalm

[viscid thistle]

either way 1/ infinity ^2 +1 = 0 + 1

[thick raptor]

👀

[viscid thistle]

and thats through the limit comparison

[thick raptor]

kek

=tex (\omega+\omega)^n=\omega^n+\omega^n

[thick raptor]

👀

[thick raptor]

D:<

=tex (\omega+\omega)^\omega=\omega^\omega

[viscid thistle]

that's not true though

[viscid thistle]

how can you have 2 times infinity though

We are just memeing

infinity isn't a number lol

[thick raptor]

=tex \alpha+\beta=\alpha\cup{\alpha+\gamma~|~\gamma\in\beta}

[viscid thistle]

[thick raptor]

=tex \omega={0,1,2,3,\dots}=\bbN

[viscid thistle]

1/infinity ^2 is 0 though

just saying

[thick raptor]

👀

[viscid thistle]

bett

tell thatt to my profesor

[thick raptor]

Its not tho man

[viscid thistle]

infinity is not a number

[thick raptor]

Not even in non-standard analysis

[viscid thistle]

ok

[thick raptor]

where you can have 1/∞

[viscid thistle]

who is your professor

I'll email him

brb im gonna send you my notes

[thick raptor]

oh dear

[viscid thistle]

are you in hs?

or college

probably doesn't want to introduce the formal rigorous ideas

[thick raptor]

yeah, at best its shorthand

@viscid thistle Define what this ∞ thing stands for

[viscid thistle]

understandable if you're in hs

[thick raptor]

And tell us what it means to do arithmetic with this ∞

[thick raptor]

👀 arithmetic over the surreals is pretty hard

oh god ew

[viscid thistle]

[thick raptor]

ordinals = plz no subtraction

[viscid thistle]

[thick raptor]

👀

so all ∞'s are the same?

[viscid thistle]

$$\lim_{x \to \infty} (x - x^2) = 0$$

[viscid thistle]

[thick raptor]

kek

[viscid thistle]

totally makes sense

[patent beacon]

I think you're just drunk

[thick raptor]

I can imagine some kid thinking
"lim (x - x²) = ∞ - ∞ = 0 QED"

[viscid thistle]

implying there is a max int

show that int of 1/ (1+x^2) dx converges using limit comparison. Compare to int from 1 to infinity of 1/x^2 dx . (1/x^2)/1/(1+x^2) mult 1/x^2 top and bottom, = lim x goes to infinity of (1+x^2)/x^2 = lim as x goes to infinity of 1/x^2 + x^2 / x^2 = 1/infinity ^2 + 1 = 0+1=1, 0<1<infinity , so int from 1 to infinity of (1/(1+x^2) dx also converges by limit comparison

haha he still uses computers with finite number of bits

@viscid thistle

still hasn't moved on to infinite-bit processors

what

Can you latex it

its easy to read

lol

but sure i can latex

I can't read

want to tell mehow

im new

no im talking about the fancy code

where it highlights it

kinda like this

the limit DNE

$$show that int of 1/ (1+x^2) dx converges using limit comparison. Compare to int from 1 to infinity of 1/x^2 dx . (1/x^2)/1/(1+x^2) mult 1/x^2 top and bottom, = lim x goes to infinity of (1+x^2)/x^2 = lim as x goes to infinity of 1/x^2 + x^2 / x^2 = 1/infinity ^2 + 1 = 0+1=1, 0<1<infinity , so int from 1 to infinity of (1/(1+x^2) dx also converges by limit comparison(edited)$$

[patent beacon]

Are you trying to evaluate the limit of
x² / (1 + x²)?

[viscid thistle]

no im showing it converges

1/(1+x^2)

[patent beacon]

Through evaluating the limit of x² / (1 + x²) via limit comparison

[viscid thistle]

and i compared it 1/(infiinity ^2)

bruh

[patent beacon]

I am brutally murdered

[viscid thistle]

when 3 people have been explaining this for 1/2 hour and you still do 1 / infinity

[patent beacon]

It happens a lot

[viscid thistle]

also you compared to the integral of "1 / infinity ^2"

infinity squared smh

thats what i said

no?

read smh

you said you just compared to "1 / infinity"

come again

you want to compare to the INTEGRAL

not the sequence

[patent beacon]

You're evaluating the limit of (x² + 1) / x². Notice you can divide top and bottom by x² to get that the limit is 1

[viscid thistle]

[thick raptor]

=tex \frac{x^2+1}{x^2}=\frac{1+\frac1{x^2}}1\to\frac{1+0}1=1

[thick raptor]

:3

[patent beacon]

You instead try to break up the limit which is a huge no-no because you can't be certain the limits you are creating exist (they actually don't)

[thick raptor]

(also why I hate people who write 1/∞ = 0)

(they tend to be the type of people who do that)

(👀)

oh, so that's how you roast someone

[viscid thistle]

1/infinity ^2 + 1 = 0+1=1,

this is what i said.

look at the question

[thick raptor]

yeah and its pretty bad notation if you ask me

[viscid thistle]

but its correct

[thick raptor]

Say, did you get around to telling me what ∞ was?

define what you mean first, then we can judge its correctness

[patent beacon]

OH yeah I understand now.

I thought you broke the limit no you did the right thing

[thick raptor]

👀

[viscid thistle]

I compared that limit to 1/(infinity^2)

then i found that it also converges through limit comparison

[thick raptor]

🙄

[viscid thistle]

look maybe my professor is wrong

but i learned it this way

be nice )):

[thick raptor]

Well it's just terrible notation is all we're saying

lacks rigor and can be ambiguous

[viscid thistle]

it's ok, we all did that at some point

[thick raptor]

👀 you sure?

[viscid thistle]

yes

[thick raptor]

👀 frankly I'll be taking that as rude stereotyping

[patent beacon]

I mean there never is a 1/infinity² step.
lim 1/x² + 1 = 1

To write infinity in there implies 1/infinity² is a real number, to which it is not. It also suggests breaking a limit, which didn't happen here, but can happen in other cases.

[thick raptor]

all I ever wanted was a definition of this infinity

[patent beacon]

Like lim (1 + 1/x)^x
You'd likely know not to write infinity into this, but if you did you'd get the limit is 1

[thick raptor]

?

[patent beacon]

Where the limit is actually e

[thick raptor]

@pine kindle are you talking about my college admission essays?

oh

no, I already have definitions of "infinity"

[patent beacon]

Little did he know, he would never get there

[thick raptor]

kek

[viscid thistle]

PhD in large numbers

[thick raptor]

[viscid thistle]

how can math majors even compete?

[thick raptor]

Founder of PhD in large numbers

kek

can't compete when it comes to large numbers I'll give 'em that

[viscid thistle]

you are talking about notation on discord tho lol

if i had enough time in the day to have good notation on discord i would, but as a busy physics major i dont got time

physics

[thick raptor]

[viscid thistle]

[thick raptor]

So is this professor/teacher who writes ∞ like this

is (s)he (or other non-binary gender) a math prof.... or a phys prof?

[viscid thistle]

what o you mean infinity like that

do*

[thick raptor]

👀 cuz in rigorous math no-one writes 1/∞ = 0 AFAIK

especially without any definition of ∞

[viscid thistle]

but this is when take the lim as x-> infinity

but she did write that

1/infinity^2 +1 = 0+1

thats what she said

[thick raptor]

you missed the question tho

[viscid thistle]

well we dont write 1/infinity = 0

[thick raptor]

👀

then what are you saying

[viscid thistle]

i already said i was wrong with the infinity

[thick raptor]

well anyways ur still ignoring my question

[viscid thistle]

let me try and write this better so you can see my notes

``Ex: show int from 1 to infinity where f(x) = 1/(1+x^2)dx converges using limit comparison, compare to integral from 1 to infinity where f(x) = 1/x^2 dx.

We know it will converge because p= 2>1.

``

[thick raptor]

``` ... ```

Use that format

 
We know it will converge because p= 2>1. ```

[viscid thistle]

alright

[thick raptor]

It just looks better imho

[viscid thistle]

ight

would you prefer teaching me how to write fractions

or is this fine for you

im talking about in the way you guys write code in here with the $$

[thick raptor]

#bot-testing

[viscid thistle]

Its gonna take me a while to type this out but i will

also, sorry if i sound dumb guys, this is the way my professor taught me

and if im wrong

please blame my proff

[thick raptor]

k lol

[viscid thistle]

yeah i always trust them

[thick raptor]

:P

[viscid thistle]

$$ \frac{1}{1+x^2}dx \text {converges using limit comparison, compare to} \int_{1}^{\infty} \frac{1} {x^2}dx { \frac{1}{x^2} * \frac{1}{x^2}} \frac{ \frac{1}{x^2} * \frac{1}{x^2}} {1}^{\infty^2} \frac{1} {x^2}dx + {1} = {0} +{1} = {1} \text {so} \int_{1}^{\infty} \frac{1} {x^2}dx \text also converges by limit comparison. $$

[thick raptor]

Um

dollar signs in the wrong places

\text needs to be followed by {...}

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[thick raptor]

Problem? \frac{ \frac{1}{x^2} * \frac{1}{x^2}} {1}^{\infty^2}

[viscid thistle]

im multiplying a fraction by a fraction

but that frac multiplied by afraction all together

[thick raptor]

But the first \frac in the part I mentioned is all alone it seems

Does the last {1} go to the fraction or the ^{\infty^2}?

[viscid thistle]

it goes to the integral i believe

wait where at

nvm ill try something

$${so} \int_{1}^{\infty} \frac{1} {x^2}dx \text {also converges by limit comparison. }$$

so this works

[grizzled hull]

I can't help but notice that the integral is in precalculus. Might #calculus be a better place?

[viscid thistle]

yeah I guess

i was talking about precalc earlier my bad

[wind igloo]

@thick lichen I got it.

It's pretty straightforward once you have the recurrence relation. The thing that's weird is that which one is the P and which one is the P+1 flips every iteration.

[thick lichen]

@wind igloo I'll post the solution here so you can check just in case

tell me if you can't read something

that should round the edges of your proof if you miss one or two formal steps

also tagging @zealous shell for this when he wakes up ^ 😄

[wind igloo]

Hm.

That does a lot less algebra than I did...

[thick lichen]

maybe you went a hard way 😛

[wind igloo]

Maybe.

[thick lichen]

maybe you did a mistake in reasoning?

not sure

[wind igloo]

I don't think so...

[thick lichen]

there can be multiple ways to a solution

I don't feel too bad that I didn't get it under 20 minutes then

gotta work though because some people in my class 4 years ago would have

[wind igloo]

Well, I haven't done formal proofs like this for a class in a very long time.

[thick lichen]

I can get more coming when I have interesting problems, I'm reviewing my courses 😃

[wind igloo]

I explicitly found the recurrence relation, and then ground through the algebra to show that multiplying by root 2 preserves the relationship.

[thick lichen]

I see

so it is a constructive proof?

[wind igloo]

It's ultimately inductive.

[thick lichen]

No I mean do you have a closed expression for p

[wind igloo]

No.

[thick lichen]

ok so it's just the existence

I wonder how hard it would be to have the value of p for all n

[wind igloo]

I have explicit formulae for a and b in terms of a_n-1, b_n-1

[thick lichen]

maybe we can try and solve it

what's the relation you have?

[wind igloo]

a_n = a_n-1 + 2b_n-1
b_n = a_n-1 + b_n-1

[thick lichen]

will try to look at it tomorrow

no reason it can't get solved

[wind igloo]

IIRC, most recurrence relations can be transformed into closed form solutions...

[thick lichen]

maybe using some kind of linear algebra magic

[wind igloo]

But the process requires a differential equation.

[thick lichen]

can't it be expressed as some kind of linear endomorphism?

A= B*C

where B = (1 2; 1 1)

sorry

A = B*A-1

shit

A_n = B*A_n-1

[wind igloo]

There's a matrix representation, yes.

[thick lichen]

Well can't we just calculate B^n then?

[wind igloo]

Hm.

Point.

[thick lichen]

shit I gotta sleep

[wind igloo]

Go to sleep. We can discuss this more tomorrow.

[thick lichen]

I'll try it tomorrow when I'm in the train 😛

yeah I'll get back to you if I find something

[viscid thistle]

instead of using the law of tangents cant i just solve the missing angle and use the law of sines to solve the missing sides?

[viscid thistle]

oof

[rocky bison]

@viscid thistle ???

[thick lichen]

I'm gonna post all the exercises I have trouble with while reviewing all my undergrad courses 😄

$$\text{arcsin}(2x) - \text{arcsin}(\sqrt{3}x) = \text{arcsin}(x)$$

[thick lichen]

Solve this in $$\mathbb{R}$$

[thick lichen]

@wind igloo 😄

because you liked the previous one

this one is easier and reminded me of a trick I had completely forgotten about

[fringe stream]

$$\arcsin(x) + \arcsin(y) = \arcsin(x\sqrt{1 - y^2} + y\sqrt{1 - x^2})$$

[thick lichen]

you guys know everything by heart 😛 when we solved it we didn't know about this formula we had to have the idea of appying sin to both sides, and finding necessary conditions on x etc

I wish I knew all these formulas lol

can't even properly memorize tan(a+b), cos(p) + cos(q) etc

equation is not solved yet just with that though

[thick raptor]

Or just take sin() to both sides

[thick lichen]

yes that's what I did

not equivalent though

but then there is still this $$x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1$$ and that's where I was stuck

[thick raptor]

it ought to simplify

[thick lichen]

it doesn't there are two intermediary steps to get to the potential solutions

do tell me if you don't want me sharing these little exercises, I just feel less alone revising this way 😄

[viscid thistle]

yo

i've a question about sketching a function

[dense zealot]

@viscid thistle

[viscid thistle]

o

well

how can i find a range of a function fast & graph an asymptote fast

[dense zealot]

U just

For the range that depends on the graph

And that's biggest y minus smallest y

for asymptote

Look at the line where the function is appearing to become

then look at the end and graph the line back

[viscid thistle]

@dense zealot how do i know which quadrant they are in

[dense zealot]

@viscid thistle hi

wdym

[viscid thistle]

uh

lemme show

https://gyazo.com/9e1c6e1238c1f443fd97867102ca6130

[dense zealot]

What about it?

[viscid thistle]

how do i find where the lines are going

[dense zealot]

on both sides of the function, they're closing onto a point on the top and bottom , you just connect these two points make a line and go forever

[viscid thistle]

i know but i'm saying how do i know where they will be

[hexed ermine]

You talking about the asymptotes?

If so, you can determine where they will be by doing simple transformations, a vertical asymptote is made when the denominator of the function is =0, since dividing by 0 is undefined

To shift it vertically, you just add/subtract a constant after the function

[sacred eagle]

https://gyazo.com/2ccc607b1f46fb8c8a469a302edf576f

this is so weird how do i solve these

[dull hinge]

Use the multiple angle formulae and sin^2+cos^2

[lean owl]

compound angle formula

cos(a+b) = cosa cosb - sin a sin b

sin(a+b) = sina cosb + sin b cos a

tan(a+b) = sin(a+b)/cos(a+b)

[viscid thistle]

Can anyone help me with this?

is it like domain of function?

[lean owl]

for what values of x is f(x) = y greater than 0, and less than or equal to 0

in other words when is it strictly above the x axis

and when is it below or lieing on the x axis

you will have 3 intervals if you can see it just from looking at it

[dense zealot]

Hai

[ancient hatch]

Hi is there a way to prove two equations using general solutions

*prove two equations are sometimes true, always true, never true

[dull hinge]

are those one equation?

[ancient hatch]

Two different equations but when plugged into a graphing software, they are sometimes true

However I need to be able to prove using general.solutions somehow...

[stray mica]

ohoh

yes

i have experience in the field of teaching

what is it you need help with @ancient hatch @unreal plaza

i see im not wanted

thanks for wasting my time

dohyeong kim brother of dohwan kim from nangok seoul in korea

[unreal plaza]

why is there something unrelated to maths in a math channel??

[stray mica]

son of younmin yoo

[unreal plaza]

and why are you talking to yourself

[stray mica]

iYGEW H&WG

I WILL END YOU

[unreal plaza]

someone kick this dog1

[stray mica]

@grand shale

dohyeong kim brother of dohwan kim from nangok seoul in korea son of younmin yoo lover of mark minaj

[unreal plaza]

btw how do you differentiate f(x) = x^2+3x ?

[fringe stream]

Power rule

[sleek mesa]

f’(x)=2x+3

[unreal plaza]

oh

im really bad lol ty guys

[sleek mesa]

Was it the fact that it had 2 terms that tripped you up?

[grand shale]

i didnt do nothing

[sleek mesa]

If so just think about everything that’s in terms of x as another function of x

[unreal plaza]

oh

[sleek mesa]

And f(x) as a composition of those

[unreal plaza]

oh yeah

is f(x) and dy/dx the same thing kind of

[sleek mesa]

So f(x) is just a function f of x. dy/dx is the derivative of y with respect to x

It’s usually implied that y = f(x)

[stray mica]

f'(x) is dy/dx righ

[unreal plaza]

oh yeah

[sleek mesa]

So dy/dx = d/dx(f(x)) = f’(x)

[unreal plaza]

and e=mc2

ok i guess im understanding now

thanks for the help guys

[sleek mesa]

any time

[hexed ermine]

Use double angle formula

Okay give me a bit im helping with someone

hmmm i've never dealt with double angle in this sort of way

but I can try to help

cos(2a)=-7/8

we can use the 2cos^2x-1 to find cosx

So -7/8=2cos^2x-1; add one; 1/8=2cos^2x; divide by 2; 1/16=cos^2x; take square root; cos(x)=+-1/4

And since it says quadrant ii, we can say that the cosine is indeed negative

so -1/4

and then make a triangle out of that and you can find the sin

np

[lone bloom]

So, I’ve got a test tomorrow and I don’t remember how to solve this. (It’s also been years since high school math so I’m very rusty)

[hexed ermine]

very tiny lol

[lone bloom]

Okay what

[hexed ermine]

micropixels man

I cant see that

[lone bloom]

Glitch with the phone, my apologies.

[hexed ermine]

@lone bloom do you by any chance play gd?

[lone bloom]

I'm unaware of what that is.

[hexed ermine]

Oh nvm, just from the name I thought you play gd

anyways i've never dealt with those problems so I am unable to help 😦

<@&286206848099549185>

[lone bloom]

It's from something silly a long time ago. I appreciate it though.

There’s also this, which I got one answer to but wolfram alpha got a different answer to.

I'd say C, wolfram says E.

[hexed ermine]

I believe it would be E

Because I believe when finding the composition you just plug the g(x) function for x in f(x)?

[lone bloom]

Is there something that works between the 2 and the root to cancel them, which would leave E?

Yes.

[hexed ermine]

If so it would be e^2ln(1/sqrt(-x)+3-3)

the -3 and 3 would cancel

[lone bloom]

Correct

[hexed ermine]

so e^2ln(1/sqrt(-x)

and you should know that 2ln(x) is ln(x^2)

so it's essentially e^ln((1/sqrt(-x)^2)

and the e and ln cancel

so it's just (1/sqrt(-x))^2

Or 1/-x; -1/x

[lone bloom]

That makes sense now.

I did the wrong thing to start.

Now I'm feeling that if I just brush up on interest problems I can make a nice C on this test.

[void patio]

C eh

damn ur school must be vigorous

[lone bloom]

Me and Math aren't friends.

[void patio]

oh ic

same here i suppose.

i'm pretty bad as well

i had to review geometry and algebra 2 and 1 bymyself

[lone bloom]

Everything I'm learning is 100% useless to me. My dream job only involves calculating dosage.

[void patio]

i'm preparing for calculas atm

ic

the stuff i'm learning rn is just to prepare me to get an A on calculas 1

and hopefully i can get A's after calculas 1 as well

[lone bloom]

The university I'm going to doesn't require vet students to take Calc 1.

[void patio]

ah ic thats good

[lone bloom]

That's a bomb deal.

[void patio]

calculas isn't worth the effort unless you're doing science or engineering courses

that's pretty good man

[lone bloom]

This class has been frustrating to the point I've considered just dropping out and being poor for the rest of my life.

[void patio]

damn really

don't stress out too much about it, you won't need to take calculas

at least

[lone bloom]

It literally causes my brain pain.

[void patio]

well it use to for me as well.

i had to go back and master algebra 1

before it became much easier for me

[lone bloom]

Ow.

[void patio]

indeed

it wasen't a nice time to be alive

but i managed it. i average an 80 on tests atm

i'm planning on bringing that up to 90

it's possible to get through it without much trouble. you just have to put effort into mastering the basics.

divide sqrt3 on both sides

then you take the csc of it.

[hexed ermine]

And For problem 4 im pretty sure it's B. @lone bloom

[void patio]

just change it then, if you simplify it, it'll be -(2*sqrt3/3)

all you have to do now is go to the unit circle

[hexed ermine]

Are you trying to find degree measures that satisfy?

[void patio]

and figure out which quadrant it's in

yeah ik.

you find the csc of that value on the quadrant

then you find a way to write out all the possible radians

[hexed ermine]

I would divide by sqrt3 first

then take the reciprocal of that giving you the sin(x)=-sqrt(3)/2

Yes exactlyu

[void patio]

ok so the question is asking for csc right? sin is negative on two quadrants. 3rd quadrant and 4th quadrant

you can do that

it's actually the easiest way to see it

imo

[hexed ermine]

Yeah

[void patio]

anyways

yes those are the values you use

you go to the unit circle

and look at it

lol

sin=-sqrt3/2 right

for that value it's negative on 3rd and 4th quadrant

yeah

it asks for all values

so you then add 2pin

which means whenever you add 2pin continiously

it'll always be the same value

because that's where sin is -sqrt3/2

lol