precalculus

[winter comet]

Cuz everywhere you see x you gotta put a 3 instead

f(x) = -72/x + 1
f(3) = -72/3 + 1

Yeah

And then u know how to solve it from there ri?

I mean then you have to simplify -72/3 + 1

Cool 😎

Ok

Yeh

Yep

Yes...

Then subtract 1 ofc

Yeah

No it's just 0 lol

I mean you already did

Like when you plug 4 into x/4 - 1 you get 4/4 - 1 = 0

Right?

So f(4) is 0

XD

Yep

Yeah but it actually divides evenly

Do you know what it's equal to?

Yeah so what do you get

12-52/4

52/4 is 13, what's 12 - 52/4

Right

Cuz that's the entire expression

Yeh

Negative

Yeah

👍

What's ur first step

[whole void]

whats ur logic for this?

[winter comet]

Yeh

Yes

Then what do u get

No

you replace x with 1 right?

So u were multiplying by x

Now ur multiplying by 1

You can think of it like

f(x) = 11-6x
f(1) = 11 - 6(1)

Yea

Ur just replacing x

With the number in the function

Yes

Actually wait

It would be 11 - 6 x 1 right?

not quite

Like uhh

f(x) = 11-6(x) right?

11 minus six times x

So when you have f(something)

Replace x with that something

[white rapids]

Ye I was trying to find the mistake turns out the base 2 wasn’t actually so I read it wrong it was z

[winter comet]

Uhh kinda...

Yea

Yea

XD

Cool 😎

You have to divide 82 by 41 but yeh

And then don't forget

The rest of the expression

Wut

Um

The fraction is, not the answer but yeh

Yea

No

11-2, not 2-11

It's positive

Yeh

💀

☠️

Uh

Maybe?

Lemme see

I think u can as long as it's not derogatory or smthn

Yeh

You know abs value ri?

[cedar juniper]

um

so u multiply 11 by the absolute value of 9 right

and since 9 is positive

9 stays positive

so it becomes 7 + 11 * 9

[proven void]

yeah

[viscid thistle]

i need to find f(x)

[willow bear]

to make everyone else's eyes not hurt:

$f(x) = \lim_{n \to \infty} n^2 \paren{x^{1/n} - x^{1/(n+1)}}$, for $x > 0$

[obsidian monolith]

|Ann⟩

[proven void]

did you rationalized?

[viscid thistle]

it won't help ?

i substituted 1/n as h

where h tends to zero

and this is what i have done

[willow bear]

why does your x look like æ

and your ∞ is impossible to tell apart from 8

anyway did you like

attempt to l'hôpital this or something?

[proven void]

its not zero

also we have x and the limit is from n to infinity?

how is the result log?

[willow bear]

because there's three things you did wrong:

1. you didnt check whether l'hop is even applicable
2. you differentiated wrt x instead of wrt h. x should be a constant
3. you misused thee => symbol. those should all be = here.

[viscid thistle]

i write it like that so i don't get it confused with the multiplication symbol

[viscid thistle]

1. l'hop is applicable

0/0 form

[willow bear]

is it actually 0/0 though?

what is $\lim_{h \to 0} (x^h - x^{h/(h+1)})$?

[obsidian monolith]

|Ann⟩

[willow bear]

$\cdot$ is hard to confuse with $x$

[obsidian monolith]

|Ann⟩

[viscid thistle]

you missed division by h^2

[willow bear]

no i didn't, i am asking you about the numerator specifically

i want you to tell me what the numerator approaches

[viscid thistle]

oh it is 0

[willow bear]

is it?

[viscid thistle]

1-1 ?

x^0 - x^(0/1)

[willow bear]

ah, yes, you're right.

ok

your differentiation was all wrong on the num tho still.

[viscid thistle]

yes

got it

thank you

[viscid thistle]

in most cases i use "x" instead of "."

one more question

for finding f(x)

why can't i use infinity G.P formula ?

[willow bear]

because this is not a GP

[viscid thistle]

it is with a common ratio of 2(x^2)

[willow bear]

no

if that were so, then the last term would be 2^n * x^(2n-1)

but the coefficient increases as an AP

[viscid thistle]

oh

AGP?

[willow bear]

yes (but it's only useful if you have tortured yourself with the memorization of a formula for that.)

[viscid thistle]

no

i know the derivation

for the formula

so i just derive it

thank you

[proven void]

,w D[n^2 (x^(1/n) - x^(1/(n + 1))), n]

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xbz.24

[proven void]

if you factorise n^2 I’m still getting zero

How did you guys calculated this?

[viscid thistle]

convert n as 1/h

[viscid thistle]

you will get a 0/0 form

[viscid thistle]

h tends to 0

[compact spade]

dawg what the fuck does this even mean

[wispy valley]

💀💀💀

is it perms and combs???

i havent done it

[chrome ether]

find the sum of the first term

and the first two terms

and the first three

and the first four

there’s really nothing to it

[wispy valley]

what unit is this???

[chrome ether]

clearly you don’t know what you’re talking about

this is like very basic sequences and series

[wispy valley]

is it a part of precalculus?

[chrome ether]

some US schools teach it in precalc, some teach it in alg2

[wispy valley]

i think i did that in grade 11 i dont remember

i just started trigonometry

am i cooked?

[chrome ether]

you don’t really “need” any series stuff until calc2

[abstract zephyr]

that is the fibonacci series?

[chrome ether]

??????????????????????????????

what the fuck when did I ever say that was the Fibonacci series

do you even know how that sequence is defined?

[wispy valley]

😭😭😭

[abstract zephyr]

ye

that sound like that

[chrome ether]

what

[compact spade]

hey elrichardo just so you know right now you sound like an asshole

not saying you are

but you're definitely sounding condescending and REAL pretentious my guy

[chrome ether]

ok and

i just tend to be impatient when people either

clearly don’t understand what they’re talking about

or

[compact spade]

we can tell

[chrome ether]

are not using sufficiently precise language

[compact spade]

ngl imma just leave you be

[chrome ether]

anyway, i outlined the procedure for finding partial sums above

“nth partial sum” = “sum of the first n terms”

[compact spade]

already solved it

appreciate your time thank you

[abstract zephyr]

Oh, my bad, i dont leard XD

[compact spade]

you don't what

[abstract zephyr]

read*

[compact spade]

ahhh ok

[abstract zephyr]

i read wrong

[chrome ether]

why else is that not the Fibonacci sequence:

the terms alternate in sign
the terms are decreasing in magnitude, while the Fibonacci sequence is increasing

[abstract zephyr]

i read wrong

[wispy valley]

Does anyone know how to solve a

[abstract zephyr]

2x-1=x

i think

no

wait

[compact spade]

yeah not quite lmao

take this advice with enough salt to kill a man, but you might want to start by taking log_8() of both sides

[wispy valley]

and then?

[compact spade]

isolate x?

maybe?

nvm it says it has a link try to follow that

[abstract zephyr]

XD

[compact spade]

im sorry im terrible at math 😭

[willow bear]

better to write both sides as powers of 2, given you can do that with relative ease

[abstract zephyr]

2x-1 = log_8(1/4)·X + log32
2x - log_8(1/4) = log_8(1/4)·X. And I don't know what else

[chrome ether]

no need to take it log 8

take it log 2

in general, try to make everything have the same base

these "textbook" questions will usually make that easy for you by making everything in terms of powers of "friendly" numbers like 2, 3, 5, 7, 10

[abstract zephyr]

1

[true sluice]

help

[abstract zephyr]

in this

2x - log_8(1/4) = log_8(1/4)·X. log_8(1/4 = A)

[willow bear]

well what is b then if it isnt a real number

[abstract zephyr]

i think the same thing

[willow bear]

best answer is "undefined until further info"

@true sluice

[true sluice]

I have to show what happens if b is not real

[willow bear]

??

wait this all seems piecemeal

can you show the full context of your exercise

the whole thing

cause i, for one, am confused as shit

[true sluice]

this

sorry

[abstract zephyr]

si

what is a not real number? i ?

[willow bear]

was there anything before this?

[true sluice]

I need to demonstrate in a hypothetical case what happens to the absolute value of b if b is not real

[abstract zephyr]

I WAS RIGHT, BUT I DONT KNOW HOW XD

[willow bear]

what happens to it is that such a concept makes no sense at all

[abstract zephyr]

i^2 E Real?

[willow bear]

"not a real number" doesn't even imply b is a number of any kind to begin with

so again |b| just isnt a thing

[true sluice]

:v

[willow bear]

you have to be more precise in telling us what b is rather than choosing to only tell us what it isnt

[abstract zephyr]

This is a absolute value or is the definition of Z = sqrt(a^2 + b^2) being b a not real number

??

[peak token]

how to simplify this

15√8x-7√2+6√8x+7√2

[willow bear]

well there are two terms that just cancel out, aren't there

and there's also two like terms

unless you meant $15 \sqrt{8x - 7 \sqrt{2 + 6\sqrt{8x + 7 \sqrt{2}}}}$, which i think doesn't simplify

[obsidian monolith]

|Ann⟩

[unreal lake]

typing math is a skill

[willow bear]

it is

and i have like 6 years of experience with it

[wispy valley]

i already solved it but thank you sm for your time!

[silent oasis]

Can anyone explain how the area is 1/4 alpha?

[willow bear]

does the video not explain it?

can you link the vid itself

[silent oasis]

https://youtu.be/odBiKFAdEXc?si=N25L4HGIrs-q5s93

[silent oasis]

not much.

[willow bear]

ok gimme some time and i'll give this a watch...

[willow bear]

hm

so they just state it without explanation

very good question how to get this thing without resorting to coordinates...

[silent oasis]

yeah. they just said it's simple geometry

is it called Archimedes' quadrature or something?

[willow bear]

seems so

idk what their simple geometry is tho ngl

[silent oasis]

i read few blogs related to this but still couldn't understand.

[willow bear]

can you share some

maybe i will have better luck deciphering what they have to say

[silent oasis]

there's one youtube video i found too.

https://www.youtube.com/watch?v=7bqNw-e2Av4

[silent oasis]

ok

https://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Erbas/emat6690/essay1/essay1.html

https://www.intmath.com/blog/mathematics/archimedes-and-the-area-of-a-parabolic-segment-1652

[viscid thistle]

calculus is math

neet is a bio

[silent oasis]

No idea what neet is

[viscid thistle]

Its an entrance exam for medical uni

[viscid thistle]

Its used in physics bud

[silent oasis]

ohh i see

[viscid thistle]

Neet has physics

What country are u @silent oasis

oh

forgot

[silent oasis]

Nepal

Maybe your neighbour

[viscid thistle]

Yeag

[viscid thistle]

This is hard

[silent oasis]

How's the equation of parabola y=x(k-x)?

[willow bear]

it has x intercepts at x=0 and at x=k

[silent oasis]

so?

[willow bear]

since you are allowing yourself to use coordinates at all, might as well say that the equation of this parabola is quadratic

which you are writing down in factored form based on the known roots

[silent oasis]

By the way, do I need to know what Archimedes' quadrature is for now? I am just beginning to learn calculus.

[willow bear]

it has historical value

but not so much for subject matter

i can tell you i learned calculus without focusing at all on archimedes' quadrature and turned out fine

[silent oasis]

How do I begin?

I see some 12 hours long lectures on youtube. Is that what I need to watch?

[willow bear]

dunno

i'd probably go to khanacademy and do the problems that they offer

and watch their videos according to need

[silent oasis]

alright. I'll do that only.

[mortal vine]

how to determine if smt is a function or not?

[hushed sphinx]

That depends on which information about "smt" you already have.

[willow bear]

,rcw

[willow bear]

1 or 2? if both, which first?

[mortal vine]

1 first

[willow bear]

for the purpose of functions, all these pairs in the relations should be thought of as (input, output)

and the one thing you CAN'T have in a function is
two pairs with the same input but different outputs

for example the relation {(1, 6), (4, 11), (4, -5)} is NOT a function bc we have two pairs with the input 4 but with different outputs 11 and -5

do you understand this? Y/N

[mortal vine]

ohh yes

[willow bear]

are you able to analyse your X, Y and Z now?

[mortal vine]

wdym

[willow bear]

are you able to look at your relations X, Y and Z, and tell which of them are functions (if any)

[mortal vine]

kinda, i still need help with q2

[willow bear]

in these, it is implied x is the input and y is the output

your goal is to say which formulas can have y isolated in them so that there's never two different values of y for the same x

[mortal vine]

how do i tell

[willow bear]

isolate y in each one

to give you a non-example: y^2 = x would become y = **±**sqrt(x), which is two values for any x greater than zero

[viscid thistle]

how do you find the number of digits in an exponent ?

[willow bear]

do you mean in a power?

[viscid thistle]

yes

[willow bear]

like the digit count of a^b for known a and b?

[viscid thistle]

yes

[willow bear]

the digit count of $x$ is $\floor{\log_{10}(x)} + 1$

[obsidian monolith]

|Ann⟩

[willow bear]

thats the most you can say generally

if you show us what your a and b are then we can tell you what can be done in your case

[viscid thistle]

5^100

[willow bear]

i presume you need to do this calculatorless?

[viscid thistle]

yes

i believe even calculators can't help

[willow bear]

i mean calculating 100 * log_10(5) is very possible on a calculatorr

[viscid thistle]

yes

oh

69.89

i used a calculator

+1

so that would be 70.89

now is it just 70 or 71 ?

[willow bear]

i thought i was clear when i used the floor function

or if you didnt know what i meant you had like 100 opportunities to ask me

[viscid thistle]

what is a floor function ?

Greatest integer function ?

[willow bear]

yes

[viscid thistle]

oh

okay

thank you

[willow bear]

i do not use [] for it because (a) LaTeX has special brackets specifically reserved for it and (b) that way i don't have to type "oh by the way [x] means the greatest integer ≤x just so you know" 10000 times

[viscid thistle]

okay

[tranquil inlet]

$r^{2}=3\sin2\theta$

[obsidian monolith]

Cyberr

[tranquil inlet]

What would be the “range” of r

[-3,3]?

[obsidian monolith]

mrRandomPerson_II

[elfin patrol]

I would like to thank everyone here who helped me with my test review, I got an 86% which is way better than I thought I would've got

Anyways

I have another question I need help with

I need to simplify this and identify the non-permissibles

I got it to the point where the denominators have been factored

[solar olive]

non permissibles ?

u mean x such that the denominators are 0 ?

[elfin patrol]

Like x ≠ 5 for example

[solar olive]

yeah.

thats what isaid

u just have to combine the fraction, not really hard , but tedious

[elfin patrol]

That's the problem

I'm stuck at the common denominator

One sec lemme type it out what I got

[solar olive]

$\frac{(x - 1)(x^2 + 4x + 3) - (x - 2)(x^2 + x - 6)}{(x^2 + x - 6)(x^2 + 4x + 3)}$

[obsidian monolith]

impract1cal

[solar olive]

eitherway it doesnt really matter, u just have to only look at the denominator

[elfin patrol]

Thats not the way my teacher wants us to do it

[solar olive]

what do they want

[elfin patrol]

We have to factor the denominator first

Which I did

I'll take a pic real quick

[solar olive]

then whats the problem?

do ur teacher wants u to factor the denominator and combine?

because if thats so, u arrive at $\frac{4x-5}{(x + 1)(x + 3)(x - 2)}$

[obsidian monolith]

impract1cal

[elfin patrol]

She wants the denominators of both fractions to be the same

[solar olive]

wtf

[elfin patrol]

Because when you add or subtract a fraction, the denominators have to be the same

Common denominator basically

And then you would combine to one fraction

[solar olive]

a/b - c/d = (ad - bc)/(bd) ???

if they want the denominator to be same, then '

a/b - c/d = ad/(bd) - bc/(bd)

[elfin patrol]

Yes

I don't know what to do to make the denominator the same

[solar olive]

this.

literally this

[elfin patrol]

Aside from multiply by each other which seems way more complicated

[solar olive]

a = x - 1
b = x^2 + x - 6

c = x - 2
d = x^2 + 4x + 3

[elfin patrol]

Basically I would have to multiply (x+1)(x+4) • (x+3)(x-2)?

And then vice versa on the other side

[solar olive]

huh? show ur work

[elfin patrol]

K one sec

Idk why it's sideways

That's what I would have to do to get common denominators

But it's confusing me because it just doesn't seem right to me

I swear I'm actually stupid

This shouldn't be difficult

Omg wait

I see where I went wrong

I wrote a 4 instead of a 3 for the numerator

I'm actually dumb

Sometimes I wonder how I have a 90% in math

[elfin patrol]

Thanks for dealing with my stupidity I'm so sorry 😭

I got my homework done

[silver kraken]

when yall got a test tmrw do yall study/review all day even if you understand pretty much everything and did all the reivew

[winter comet]

Not all day, but still practice quite a bit anyway

[silver kraken]

do you just redo the same problems?

[winter comet]

Even if I understand it's still possible to make mistakes yk

[silver kraken]

my main issue is the applications of it so like word problems but i only got like 4 examples

so just been doing through em but not rly helpin

[winter comet]

Oh usually I look in the book or online, r u out of practice problems tho?

[silver kraken]

yeah

[winter comet]

oh dang

do you have any way of looking for more of that type of problem ?

[silver kraken]

i was thinking about using chat gpt

putting my used examples and asking for similar new ones

[winter comet]

to generate questions? 😂

I mean what kind of problems r u doing

[silver kraken]

stuff on inequalities

[winter comet]

U might be able to like look it up online or smthn if u don't have any book

[silver kraken]

il try that

[winter comet]

Hmm ok

[torn atlas]

Would this be considered the proper way of solving for sinø over the domain 0<=Ø<2pi

If I’m missing any small step or the tiniest detail please let me know, cause my teacher won’t let it slide

[elfin patrol]

Yes. I find that doing the same questions over and over again helps it imprint in your memory

[echo cove]

Sorry if this is the wrong channel but can anyone help me with the answers for these questions i have my own but trying to see if they are right or wrong

[willow bear]

,rccw

[willow bear]

@echo cove you have to send your own answers here first

also this is #calculus

[echo cove]

Ok should i send it in calculus then

Oh yeah I can’t read

[willow bear]

-> #calculus

[willow bear]

-> #calculus

[echo cove]

Sorry I wasn’t able to see the channel before now

[whole oyster]

I am sorry but that is political incorrect.

[echo cove]

?

[willow bear]

troll?

oh, looks like they left

[midnight pawn]

Just wondering what are the main mathematicians that created precalculus as it is today?

[summer ruin]

archimedes, pythagoras, euclid and descartes

[marble rose]

Can someone help

[fallow quiver]

Why do you no longer have x in the denominator for the second term?

[marble rose]

ive solved it already but yeah that was one of the issues

[fallow quiver]

The other is 1/9 thingy

[marble rose]

was going too fast and looked over a ton of things

yes

[fallow quiver]

But you solved it already so whatever.

[marble rose]

dw ty either way

[proven void]

how do I solve this limit?

[summer ruin]

factoring 5^n out

[proven void]

how?

[summer ruin]

a+b = a(1+b/a)

[proven void]

I know but I meant how do I take the 5^n out of the radical

[summer ruin]

do this

then use the fact that root(xy) = root(x) * root(y)

[proven void]

what to do doe

[summer ruin]

👆

[proven void]

wait

ignore e^ln

is just the below

what now?

[lone maple]

the thing at the left is how much ?

n-root of (5^n) is ....

[proven void]

damn, 5

thanks

[lone maple]

u okay for the right part?

[proven void]

ye

is 1

[lone maple]

👍

[plucky totem]

I had a question about some math terms not necessarily about calculations per se. When talking about Der Moirve's therom. What is a modulus and argument? Specifically my notes say r is the modulus of Z and theta is the argument of Z. I understand the mechanics of the algebra and arithmetic. I'm just curious what the verbiage actually means.

I.e what is a modulus and argument.

[hushed sphinx]

The modulus of a complex number z is the same as "absolute value", i.e. the straight-line distance between 0 and z.
Argument is the direction from 0 towards z, measured counterclockwise from the positive real axis.

[plucky totem]

THANK YOU. I was just mindlessly doing calculations but had no idea what these two words meant. Makes tons of sense.

[winter comet]

is this kinda like if you were to graph it on a polar plane, the modulus would be r and the argument would be theta? and the point Z would be (r,theta)?

[hushed sphinx]

Yes.

[midnight pawn]

Just wondering is there any other important mathematicians that contributed to precal?

[proven void]

how to do without differentiation

[river drift]

power series, if you know that (although i suppose that's really just advanced differentiation)

[proven void]

thats differentiation though

also i’m bad at taylor expansions

is there any other way ?

[arctic sundial]

Can someone pls help me

[willow bear]

no not really

youd have to have an a-priori definition either for ln(x) or for e^x

[proven void]

,, \lim_{x \to \infty} \frac{1}{x} ln(1+x) = ln\left(\lim_{x \to \infty} (1+x)^{\frac{1}{x}}\right)

[obsidian monolith]

milanesa de pollo

[proven void]

mmm

,, ln\left(\lim_{x \to \infty} (1+x)^{\frac{1}{x}}\right) = ln\left(\lim_{x \to \infty} (x(1 + \frac{1}{x}))^{\frac{1}{x}}\right)

[obsidian monolith]

milanesa de pollo

[proven void]

,, ln\left(\lim_{x \to \infty} (x(1 + \frac{1}{x}))^{\frac{1}{x}}\right) = ln\left(\lim_{x \to \infty} (x)^{1/x} (1 + \frac{1}{x}))^{\frac{1}{x}}\right)

[obsidian monolith]

milanesa de pollo

[proven void]

i messed up somewhere

whatever

[willow bear]

this step is sus

[river drift]

tbh the question seems like it's set up for using series expansions, since ln(1+x) has a well-known maclaurin series

[proven void]

is this true?¡

[proven void]

if so, this step is fixable

1/inf is similar to 1 + x when x approaching 0
1/0 == ^infty

Idk if I am making myself clear

whatever

this exercise is fucked

[winter comet]

where did the extra x come from?

oh i see u factored out nvm

[winter comet]

yea thats true

[viscid thistle]

you can realize this by taylor expansion of (1+x)^1/x

which is just e-ex/2 + 11e/24x^3 +......

and so on

you can use this for any limit of form (1)^infinity

[viscid thistle]

L'hospital's rule

lol

how do i solve

for

f(x) = sgn(sin^x - sinx - 1)

where i want the minimum value of n for x belongs to (0,npi)

for which function has exactly 4 points of discontinuity

<@&286206848099549185>

sgn is sign function

1 for positive -1 for negative and 0 for 0

[silent oasis]

How do I find limit of composite function?

[willow bear]

question too general to give a conclusive and useful answer

however, if f and g are the two functions you are composing (as f(g(x)) specifically), and by a miracle of fate the following are satisfied:

• lim[x -> a] g(x) = b, which may be a finite number or +∞ or -∞ (but it must exist)
• lim[t -> b] f(t) exists
  then lim[x -> a] g(f(x)) = lim[t -> b] f(t)

passing from lim[x -> a] g(f(x)) to lim[t -> b] f(t) is often called substitution

it would be better if you posted the limit you're looking at

@silent oasis

[willow bear]

ok, and the limit?

limit of what as x goes to what

[silent oasis]

lim of g(h(x) as x approaches 1

[willow bear]

ok

so does $\lim_{x \to 1} h(x)$ exist, and if it does, what is it equal to?

[obsidian monolith]

|Ann⟩

[silent oasis]

it's equal to 2

[willow bear]

ok

so then does $\lim_{t \to 2} g(t)$ exist?

[obsidian monolith]

|Ann⟩

[silent oasis]

no

does it mean limit of g(h(x)) doesn't exist?

??

[willow bear]

yes, also your second closing parenthesis is missing

g(h(x)) not g(h(x)

[silent oasis]

but

[willow bear]

but what

if you have something to say, then say it in full. don't cut yourself short after one word.

[silent oasis]

he says it's -2

i don't really underestand the above and below thing he says

[willow bear]

ah

oh my bad

yeah ok i overlooked it

there's a caveat here in that h(x) approaches 2 from below

so in fact we only need to look at the left-hand limit of g at 2

which exists, even though the two-sided one does not

[silent oasis]

is there any definite rule for this?

[silent oasis]

like you mentioned here.

[willow bear]

i mean hopefully you should be developing some graphical intuition for limits instead of memorizing 100 different rules and 35000 different exceptions and caveats

[silent oasis]

is limit all about intuition?

[willow bear]

no

[drowsy loom]

How do i solve the one i circled in white

[willow bear]

do you know the range of the raw absolute value function, ie y=|x|?

[drowsy loom]

No

[willow bear]

do you know what its graph looks like?

[drowsy loom]

No it doesnt give an example question

[willow bear]

do you know what the absolute value function is?

[drowsy loom]

No idea

[willow bear]

bruh ok well we found the problem

you don't know what |these things| mean

one moment...

[drowsy loom]

|this| is modulus right?

[willow bear]

modulus yes

[drowsy loom]

I underatand how to sketch a graph but i dont know how to

[willow bear]

it's what i have been asking you about all along

https://youtu.be/ld4UD98yHio

watch this ig

[drowsy loom]

Without the domain

Will watch later cuz i have aome other business suddenly

[narrow briar]

Hi! Does someone know how to solve this limit? I would appreciate it a lot, thanks.

When I try to solve it i get into an indetermination, but when i evaluate the function on large numbers it approximates to pi!

[summer ruin]

the limit doesn't exist

[hushed sphinx]

I think a limit of pi sounds reasonable, assuming the cosine works in degrees.

[narrow briar]

yes, i have worked on degrees

[narrow briar]

so, why when i evaluate the function with larger and larger numbers, its still equal to 3,14159265...

[hushed sphinx]

Note that "indeterminate" just means "the approach you're trying right now is not strong enough to find the answer" -- it doesn't mean there is no answer.

[narrow briar]

so, when working on limits, an indetermination just says that you are not trying it on the right way, but there is a way with which you can solve it?

every function has a determinate limit on infinite?

[winter comet]

no, indeterminate means that you can just get a more conclusive answer. either it doesn't exist or it does.

but indeterminate doesn't tell u if it exists or not

[narrow briar]

interesting, thank you

[hushed sphinx]

There are functions that don't have limits. "Indeterminate" tells you something about the method you're trying, not anything about the function.

[narrow briar]

so when i get into an indetermination, the limit may exist or not, i dont know

[winter comet]

yeh

[hushed sphinx]

Right.

[winter comet]

but when you get indeterminate don't go like ok this is the answer xD

[narrow briar]

and how do i know if it exists or not?

trying ona different way?

[winter comet]

yeh

[narrow briar]

is there a limited amount of different ways to solve a limit?

[hushed sphinx]

In this case, good first steps would be to note that $$\cos\bigl( \frac{90x-180}x \bigr) = \cos(90-\tfrac{180}x) = \sin(\tfrac{180}x)$$

[obsidian monolith]

Troposphere

[hushed sphinx]

But to get onwards from there, I think you need to switch to radians.

[narrow briar]

i can try it although im not very used to radians

[summer ruin]

probably because you assumed writing 90 inside a cosine would mean 90 degrees, but it actually means 90 radians, it doesn't even make sense to divide degrees by real numbers

[narrow briar]

ok ok, im trying to pass it to radians

[hushed sphinx]

They specified that they're working in degrees, in which case pi is a correct limit.

[proven void]

he never said he is working in degrees

[hushed sphinx]

(Why wouldn't it make sense do divide a number of degrees by a real number?)

[hushed sphinx]

Here.

[proven void]

mb

[winter comet]

technically wouldn't he be dividing by infinity which technically wouldnt be a real number XD

its irrelevant tho 💀

[hushed sphinx]

No, he wouldn't. The limit is about what happens when you insert ever larger, but always finite numbers in place of x.

[narrow briar]

thats the limit on radians

[winter comet]

oh ig

[hushed sphinx]

Do you know that $\lim_{t\to 0} \sin(t)/t = 1$ when the sine works in radians? (Otherwise I'm not sure how you would get through).

[obsidian monolith]

Troposphere

[winter comet]

you may also wanna use this identity if u havent to do yeh

[narrow briar]

how would you solve this limit?

[winter comet]

using the trig identity

[winter comet]

and then using this

[proven void]

you were asked a question, do you know this special limit?

[winter comet]

basically everything troposphere said XD

[narrow briar]

sorry for asking too much, but i dont know how to continue without the infinite*0 indetemination...

[hushed sphinx]

You haven't answered this question yet.

[narrow briar]

i dont understand it

limit to cero of sin(x) isnt 0?

sin(0) = 0, so if x=0, it makes sense that the limit is = 0, where am i wrong?

[hushed sphinx]

I said $\lim_{t\to 0} \frac{\sin(t)}{t}$, not $\lim_{t\to 0}\sin(t)$.

[obsidian monolith]

Troposphere

[narrow briar]

ohh sorry

i understood wrong

well, now i know it xd

but my limit is tending to infinite, not 0, how im supposed to use that indentity on my limit?

[hushed sphinx]

Note that the argument to the sine actually goes to 0, so if you switch variables to t=pi/x, you get a limit for t->0 instead of x->infty.

[narrow briar]

so, the next step would be this?

did i do the variable change right?

[hushed sphinx]

Looks right (except x->0 should be t->0, of course)

[narrow briar]

oh yes yes

but i havent understood something of before

limit to infinite of sin(pi/x) isnt 0?

[hushed sphinx]

☝️

[narrow briar]

lim to infinite of sin(pi/x) = lim to 0 of sin(t)

[hushed sphinx]

☝️

[narrow briar]

thats the reason because i can do the varible change

lim to 0 of sin(t) / t = 1

but lim to infinite of sin(pi/x) = 0

so these limits arent equal

[hushed sphinx]

Why would they be?

[narrow briar]

im so sorry if the answer is very easy, im not seeing it

[narrow briar]

if they arent equal, why would the variable change be right?

[hushed sphinx]

You've got the division by t right here.

[narrow briar]

:0

wait im trying to understand it 😆

[hushed sphinx]

$$\frac{\pi}{t}\sin(t) = \pi \frac{\sin(t)}{t}$$

[obsidian monolith]

Troposphere

[narrow briar]

the thing i dont understand is why the variable change from pi/x to t that changes the limit from infinite to 0, is right

sorry, maybe its too easy

i dont want to annoy with silly questions

[hushed sphinx]

When x is very large, pi/x is very small, and vice versa.
To be completely pedantic we should perhaps have said t -> 0+ instead of just t -> 0.

[narrow briar]

i think i get it

yes, i get it

thank you

so much

[hushed sphinx]

You're welcome.

[narrow briar]

😄

[proven void]

how do I find all $x \in \mathbb{R}$ such that this sequence $\frac{x^{2n +1}}{n^3 4^{n+1}}$ converges

[obsidian monolith]

milanesa de pollo

[willow bear]

do you actually want something with that sequence, or are you actually looking at a series (infinite summation)?

("yes" isn't a valid answer, we need to know which)

[proven void]

I need to find all x in R for which this sequence converges. then for those x values I should calculate the limit of the sequence

sorry, my english is dogshit

[willow bear]

ok so limit and NOT sum, yes?

like there is no $\sigma$ anywhere, yes?

[obsidian monolith]

|Ann⟩

[willow bear]

augh

[proven void]

no sigma no sum

[willow bear]

well ok whatever

[proven void]

😁

[willow bear]

you can write this as the product of a constant, a geometric sequence, and 1/n^3

[proven void]

constant is $x^{2n + 1}$ probably

[obsidian monolith]

milanesa de pollo

[hushed sphinx]

That's not a constant in any way ...

In particular, note: 4^(n+1) = 2·2^(2n+1)

[proven void]

okay

$\frac{1}{n^3} \cdot \left(\frac{x}{2}\right)^{2n+1}$ mayhaps

[obsidian monolith]

milanesa de pollo

[hushed sphinx]

Yes, except you forgot the last remaining factor of 2 in the denominator (not that it actually matters in the end).

[proven void]

I dont get it

what did I missed in the denominator?

on a side note, how can I write this as the product of a constant, a geometric sequence, and 1/n^3

[hushed sphinx]

Do you know what those terms mean? Above you have a product of 1/n^3 and a geometric sequence, and I just told you you've misplaced the constant.

[proven void]

$\frac{x^{2n +1}}{n^3 4^{n+1}} = \frac{1}{n^3} \cdot \frac{1}{2} \cdot \left(\frac{x}{2 }\right)^{2n+1}$ mayhaps

like this?

😭

4^(n+1) = 2·2^(2n+1)

[obsidian monolith]

milanesa de pollo

[proven void]

1/2 is the constant

(x/2)^2n+1, is the geometric sequence

[hushed sphinx]

Yes.

[proven void]

what do I do now?

[hushed sphinx]

You need to know, hand-wavingly speaking, that a geometric sequence always wins over a polynomial unless it is constant.

[proven void]

okay

how do I find x doe?

[viscid thistle]

Simplify and compare @proven void

Oh it's a sequence problem

[compact spade]

the fuck is this

[thin thorn]

It's some binomial expansion shit

That I haven't learnt about till now

[compact spade]

I know

I thought I knew how to do it but I checked and it's wrong

[thin thorn]

I'm sorry man I just know the name

This uses some permutations and all as far as I've seen it, vaguely

[compact spade]

Nah you're good dw

[proven void]

how to find the limit of this sequence? $\2 - \frac{3}{2^n} < 5 - 2a_n < 1 + (n)^{\frac{1}{n}}$

[obsidian monolith]

milanesa de pollo

[proven void]

this is my progress so far: $\\frac{2 - \frac{3}{2^n} -5}{-2} < a_n < \frac{1 + (n)^{\frac{1}{n}} - 5}{-2}$

[obsidian monolith]

milanesa de pollo

[proven void]

oh, 3/2 due to squeeze theorem

I wanted to ask, to calculate the limit of the $a_n$ in this case $\\frac{1}{a_n} > \left(1 + \frac{1}{n}\right)^{n^2}\\$ would be something like: $\\frac{1}{a_n} > e \implies 1 > e \cdot a_n \implies ln(1) > ln(e \cdot a_n) \implies 0 > ln(e \cdot a_n)\\$ which is only true when $\lim_{n \to \infty} a_n = 0$?

[obsidian monolith]

milanesa de pollo

[summer ruin]

limit of (1+1/n)^(n^2) is not equal to e

moreover you have to justify why that inequality for some n implies 1/a_n > e (if it were equal to e in the first place)

[proven void]

aaaa

how to solve this then?

[summer ruin]

notice that a_n > 0

transform the inequality and compute the limit of a_n instead of 1/a_n

[proven void]

this maybe?

[summer ruin]

the inequality sign flips

and you have to write limit on both sides

and make inequality not strict

[proven void]

why?

[summer ruin]

because a > b <-> 1/a < 1/b

[proven void]

ok

makes sense

1/4 vs 1/2 is 0.25 vs 0.5

so?

also how am I going to apply squeeze theorem here if I only have RHS

[summer ruin]

.

[summer ruin]

.

[proven void]

I see

but why inequality not strict

because its equal to zero

[summer ruin]

because limits don't preserve strict inequalities

1/n > 0, the limit of 1/n as n->infinity is not > 0

[proven void]

yeah

damn, really hard, thanks

[proven void]

how do I clauclate this limit?

,, \lim_{n \to \infty} (3^n \cdot n^2 + n)^{\frac{1}{n}}

[obsidian monolith]

milanesa de pollo

[proven void]

now what do i do?

,, \lim_{n \to \infty} \left(3^n + \frac{1}{n} \right)^{\frac{1}{n}}

[obsidian monolith]

milanesa de pollo

[proven void]

i can factor again my 3^n and apply multiplication of roots to get the 3 out and make it a 1

maybe

[summer ruin]

factor out 3^n

[proven void]

limit is equal to 3

[summer ruin]

except I don't think you factored out n correctly

[proven void]

where?

[summer ruin]

$3^n n^2 + n = 3^n n\left( n + \frac{1}{3^n}\right)$

[obsidian monolith]

Transparent Elemental

[proven void]

then?

[summer ruin]

n + 1/3^n isn't the same as 3^n + 1/n

[proven void]

how do i calculate the limit tho

now that I look closer

n^{1/n} is not 1

[summer ruin]

it is 1 in the limit

[proven void]

why

isnt inf^0 indeterminate form

[summer ruin]

it is

that doesn't mean we don't know what the limit is equal to

[proven void]

?

[summer ruin]

do you think indeterminate form implies we can't ever find out the limit?

[proven void]

no

but you said n^1/n is 1

why?

i dont get it

[summer ruin]

I said the limit is equal to 1

you can consider it an exercise if you want to prove that

you're calculating limits far more involved that n^(1/n) anyway

[proven void]

,w lim n to infinity n^(1/n)

[obsidian monolith]

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[proven void]

so its 3?

i will re do the algebra

[summer ruin]

why did you assume that

[proven void]

,w lim n to infinity (n^2)^(1/n)

[obsidian monolith]

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[proven void]

,w lim n to infinity (3^n)^(1/n)

[obsidian monolith]

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[proven void]

im not sure about the last one tbh

is that 1 aswell?

,w lim n to infinity (1 + 1/(3^n * n))^(1/n)

[obsidian monolith]

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[proven void]

so strange tbh

but ill take it

[river drift]

it's in the form (1 + 0)^0, so i'd say that's expected

[proven void]

mmm

i guess yeah

[plucky totem]

One nit pick. My professor today said that it's measured counter clockwise not clock wise. Which makes sense to me. We read the unit circle counter clockwise from 0,90,180,270,360. Unless you were eluding to something else I just wanted to make sure this is correct.

[hushed sphinx]

Whoops, horrible typo on my part, fixed now.

[plucky totem]

No worries. I was just comparing my notes and noticed it. Then asked my prof and was like. So which one is it?

[river drift]

just kind of unfortunate that "counterclockwise" is the default direction chosen by mathematics whereas "clockwise" is the default direction picked by the english language

[solemn parcel]

I'm assuming you have graphs that give possible answers?

you probably need to decide what form this equation would take depending on unknowns

try graphing it out with constants and you might see some similarities

[plucky totem]

Just like cartesian coordinates lol. Who ever made the compass system wasn't in touch or just wanted to be a jerk. Ya lets tilt this 90 degrees lol. Like WHY. just make them the same. So annoying @river drift

Like why did we have to do this lol.

[winter comet]

i mean they were probably made completely seperate from one another

its like why do ppl speak different languages, why doesnt everyone just speak 1 language lol

[river drift]

it makes sense for north (or south) to be the reference point based on the way a compass works

[plucky totem]

I get you but I think you missed my point slightly. I'm not saying everyone should. More along the lines of it would have been nice if. Hell the Greeks were so close to figuring out basic calculus they just didn't have the Arabic numbers and letters systems we use today. If only they had a faster way to communicate.

[winter comet]

yea

world issues

lol

[plucky totem]

Oh I totally agree. If they were made at the same time with people who were in touch maybe they'd be the same. But like Vicious viper said. World issues

[white rapids]

Can someone tell me how I can get this f(x) to f(-x)?

f(x) = x * ^3√x-1
f(-x) = -x * ^3√-x-1
Do I just leave it like that or is there anything i have to do

[winter comet]

is that cubed root?

if so, you can factor a negative out of the cube root to cancel with the one outside

if not, i have no idea what that says 💀

[chrome ether]

that looks fine

[remote ivy]

why is this statement incorrect?

[summer ruin]

because the left hand side doesn't depend on x

[proven void]

help

can i use ratio test here?

first off

how do I show that (n^2)^(1/n) is 1?

if possible without proofs

just algebra

[hushed sphinx]

(n^2)^(1/n) is not 1.

[proven void]

i mean

the limit of that

sorry for not being precise with my wording

,w lim n to infinity n^(2/n)

[obsidian monolith]

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[junior willow]

When x approaches infinity we have 1/ very large number so the limit is 0

[remote ivy]

so does that mean they are equal?

since the are nothing in-between 1/x and 0

[summer ruin]

1/x is not equal to 0 for any x

[remote ivy]

that must mean it equlas zero

isnt this just the same argument for 0.99... =1

[summer ruin]

no

[remote ivy]

why not?

[junior willow]

But it approachrs

Since it's a limit

[summer ruin]

then one should write the limit of 1/x as x->infinity equals to the limit of -1/x as x->infinity, this is not what was written however

[summer ruin]

tbh I don't even see where you're getting this from

[junior willow]

The limit is zero

If u mean

Nvm can't display latex

Oops

[summer ruin]

it's \lim

[junior willow]

Oh

$\lim_{x \to \infty} \frac{1}{x}$ = $\lim_{x \to \infty} - \frac{1}{x}$

[obsidian monolith]

w

[junior willow]

It only took 3 tries

[summer ruin]

you can edit the message to recompile

[junior willow]

Oh

[junior willow]

@remote ivy if that's what you meant, it's correct, both limits approach 0

[willow bear]

both limits are 0

[agile sparrow]

+0 and -0

🙂

[junior willow]

Ye that's what I meant

[willow bear]

and the point of this correction is?

[agile sparrow]

I made it up

-1 * 10^inf and +1 * 10^inf

[willow bear]

??

[agile sparrow]

I'm sorry

I'm not funny

[junior willow]

[remote ivy]

yeah thanks, i did write it incorrectly

[clear scarab]

How to integrate the sin t part

[viscid thistle]

Let f and g be such that g(x) = 3x − 2 and f(g(x)) = 9x^2 − 3x + 1

Calculate f(4)

[willow bear]

!status @viscid thistle

[tender quest]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[viscid thistle]

4

[willow bear]

show what you've got then

[viscid thistle]

a) g(4)=3.4-2=12-2=10
f(10)=9.10^2-3.10+1=9.100-30+1=871
f(4)=81.4^2-117.4+43=f(4)=871

f(x)=9(3x−2)^2−3(3x−2)+1
=9(9x^2-12x+4)-(9x-6)+1
=81x^2-108x+36-9x+6+1
=81x^2-117x+43
f:R -> R is f(x)=81x^2-117𝑥+43

[willow bear]

are you using . for multiplication?

[viscid thistle]

yes

[willow bear]

how did you get that formula for f(x)...?

that sounds like you calculated f(g(g(x)))...

[viscid thistle]

i exchanged

[willow bear]

you... what?

f(x)=9(3x−2)^2−3(3x−2)+1
this is wrong

[viscid thistle]

how do I do it?

[willow bear]

find x such that 3x-2, a.k.a. g(x), equals 4.

then evaluate f(g(x)) with that value of x.

[viscid thistle]

I will try...

[willow bear]

well it is not like there is a minefield on that route

[viscid thistle]

kind of. I'm still learning

[willow bear]

what grade are you in?

[viscid thistle]

finished high school, but didn't have math because of covid and also education problems

didn't learn pre-calculus at school, nor calculus, now I'm learning at my first year of college

I don't have a good background in math so sometimes I also get stuck with the basics

[proven void]

you got this brother

[viscid thistle]

I'm not that confident loll

[proven void]

solve 3x-2 = 4 then for the x you got plug it in the f(g(x)) to get f(4) I think

[willow bear]

congratulations you have successfully repeated my instructions almost to the letter

[proven void]

sorry

[viscid thistle]

3x-2=4
3x-2+2=4+2
x=6/3=2

[viscid thistle]

should I keep going?

[proven void]

read

[viscid thistle]

but is it correct?

[proven void]

,w solve 3x -2 = 4

[obsidian monolith]

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[viscid thistle]

.

[proven void]

read

[willow bear]

then evaluate f(g(x)) with that value of x.

the formula for f(g(x)) is given directly in the problem, and no work needs to be done to find it.

[viscid thistle]

I'm brazilian, I'm having trouble trying to comprehend you guys

[willow bear]

f(g(x)) = 9x^2 - 3x + 1

and x=2

[proven void]

you got this

[viscid thistle]

I need help with the 5th one

[indigo dew]

How do I prove this

[viscid thistle]

binomial expansion

of (7+1)^n

[supple sky]

Does anyone know the best online website to teach myself Pre calc for free?

[uncut mulch]

khan

[supple sky]

Does it give you like the full course

Because like the videos are only like 5 minutes long

[proven void]

there is prolly not best website, khan academy is good, openstax is good, pauls online notes is good, why not take a little of everything?

[supple sky]

Just picked this up at the library is this book good for learning the subject matter?

[river drift]

try it out and see if you like it

[proven void]

best book is the one you stick with, imo

[near sentinel]

Try Cengage Calculus for JEE Advanced.. It's pretty good

[supple sky]

Found this YouTube channel I dunno if this guy is legit or not

https://youtube.com/@ProfessorLeonard?si=7Nmtxos0222Y3yQF

[indigo dew]

I can’t do it with that

Any other way?

[supple sky]

Every time I look through Pre calc I’m just getting what seems to be a copy of algebra 2 and trig

Doesn’t seem legit

[uncut mulch]

precalc isn't clearly defined

[summer ruin]

if it walks like a duck and quacks like a duck...

[uncut mulch]

it's more or less knowledge required for calculus

which is mostly stuff related to algebra, trig, exponents, logs, functions

[supple sky]

So yeah I’m looking at the right stuff

I thought I was just looking at stuff that seemed to be a copy of the trig and algebra I was learning in 10th grade

[uncut mulch]

if you've already learned the stuff, it can serve as a review

[supple sky]

That’s good then

Because I was going over functions and it seemed a little sus to me

It was like as almost the exact thing I already learned

Just a little bit more complex

But now it makes more sense since it’s just like a class where you put different stuff you learned in the past and just add a tiny bit more

[winter comet]

yea thats basically all precalc is

[supple sky]

Right now there’s a chapter on evaluating a function which seems to be Pre calc

[supple sky]

Oh I thought I was just wasting hours finding stuff that seemed to be the same thing

[summer ruin]

if you feel like you already know this it's better to skip to calculus to not waste time

[supple sky]

Smart

But I think I might continue to watch this videos made by this person named Professor Leonard he seems pretty legit.

To prepare me for my college pre test

It seemed pretty confusing at first because when I was looking up Pre calc it was a lot different then the other classes

Like with algebra 2 most of them started with factoring and then went into the rest of the stuff

But with pre calc you have some of them starting with functions some starting with slopes and some starting with trig

[winter comet]

it goes over a lot of the stuff you see in algebra 1 and 2 and geometry, but then it goes into introductory concepts from calc 1 or calc 2 or linear algebra

ish

[supple sky]

Which makes it super confusing on where to start