precalculus

[odd inlet]

is this supposed to be a function or no?

because it doesnt seem to be

you can split 33 to 9 and 24 tho

and get 12x and 24 to the other side

[hoary mountain]

Nvm guys I figured it out and compared it to the lesson pictures @willow bear @odd inlet

[willow bear]

why ping me

[odd inlet]

ok cool

[proven void]

How to solve this limit problem?

Please ping me if possible

[hoary mountain]

My bad

[sharp matrix]

hello , so i kinda need ur help in this im studying polynomial and rational functions so idrk how to start them and im kinda behind my classes so i need to start asap so do u guys know any good channels to study them ?

[winter comet]

i would suggesting starting your homework and if you get stuck on a problem post it here

[near sentinel]

learning by solving sums is better; personal experience: watching yt lectures just made me procrastinate and they often skip a lot of things

do solved examples or smth

[proven void]

,rotate

[proven void]

how can I solve this limit?

25

[odd inlet]

whats sen?

[proven void]

sin

my bad it’s a typo

should be sin

[odd inlet]

alr um for the first part

[odd inlet]

lim n approaches infinity of (n^4+2n^2/n^4+n^2+1) is....you just get the ones with the biggest exponent, so it becomes n^4/n^4 which is 1

raised to the 3n^2 this is still 1

now for the second part with the sin(n^4+2n^2)/3n^2 you can apply the squeeze theorem if you know that and get it to be 0

so the answer is 1 i think

[proven void]

squeeze theorem?

you mean sandwich time

[odd inlet]

lol idk what its called. the one with ...<....< and you find the 2 limits and its equal to the middle one do you know that?

[proven void]

yes can you elaborate how did you used it?

I am still concerned

[odd inlet]

well you can say that |sin(n^4+2n^2)|<1 so |sin(n^4+2n^2)/3n^2|<1/3n^2 so....-1/3n^2< sin(n^4+2n^2)/3n^2< 1/3n^2

lim as x approaches infinity of -1/3n^2 is 0

lim as x approaches infinity of 1/3n^2 is 0

so the middle one is also 0

[mighty orbit]

Okay, so I came up with a solution for the first part of the limit

[odd inlet]

yeah its 1 isnt it

[mighty orbit]

More like e^3

Consider the following limit:
[
I=\lim_{n\to\infty}\left(\frac{n^4+2n^2}{n^4+n^2+1}\right)^{3n^2}.
]
Let $u=n^2$, then $u\to\infty$ as $n\to\infty$ so
[
I=\lim_{u\to\infty}\left(\frac{u^2+2u}{u^2+u+1}\right)^{3u}=\lim_{u\to\infty}\left(1+\frac{u-1}{u^2+u+1}\right)^{3u}.
]
Where the last equation holds because $u^2+2u=u^2+u+u+1-1$. Here's the crux, notice that the quotient alone should behave as $1/u$ (I can't think of an elementary way of proving this 😦 ), so that
[I=e^3.]

[obsidian monolith]

bondalton

[wooden ruin]

what does this mean

what are these hashtags

[surreal crescent]

Looks like matrix’s, but that’s only from my limited understanding of them

[proven void]

how can I evaluate this limit

[mighty orbit]

Can you use L'hopital's rule?

[proven void]

as long as lhopi conditions are met

[mighty orbit]

\begin{align*}
I=\lim_{n\to\infty} \left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}=\lim_{n\to\infty}exp\left(\frac{n^2+2}{2n+1}\log\left(\frac{3n^2+2n+1}{3n^2-5}\right)\right)
\end{align*}
so it suffices to evaluate
\begin{align*}
\lim_{n\to\infty}\frac{n^2+2}{2n+1}\log\left(\frac{3n^2+2n+1}{3n^2-5}\right)=\lim_{n\to\infty}\frac{\frac{n^2+2}{2n+1}}{\frac{1}{\log\left(\frac{3n^2+2n+1}{3n^2-5}\right)}}\to\frac{\infty}{\infty}
\end{align*}
Therefore, applying L'Hopital's rule:
\begin{align*}
\lim_{n\to\infty}\frac{\frac{n^2+2}{2n+1}}{\frac{1}{\log\left(\frac{3n^2+2n+1}{3n^2-5}\right)}}&=\lim_{n\to\infty}\frac{\frac{(2n+1)(2n)-(n^2+2)(2)}{(2n+1)^2}}{\frac{3n^2+2n+1}{3n^2-5}\cdot\frac{(3n^2-5)(6n+2)-(3n^2+2n+1)(6n)}{(3n^2-5)^2}}\
&=\lim_{n\to\infty}\frac{\frac{2n^2+2n-4}{(2n+1)^2}}{\frac{3n^2+2n+1}{3n^2-5}\cdot\frac{-6n^2-36n-10}{(3n^2-5)^2}}\
&=\lim_{n\to\infty}\frac{(3n^2-5)^3(2n^2+2n-4)}{(2n+1)^2(3n^2+2n+1)(-6n^2-36n-10)}\
&=\lim_{n\to\infty}\frac{54n^8+54n^7-378n^6-270n^5+990n^4+450n^3-1150n^2-250n+500}{-72n^6-552n^5-930n^4-776n^3-372n^2-96n-10}\
&=\frac{1}{3}
\end{align*}
Therefore $I=exp(1/3)$.

[obsidian monolith]

bondalton

[viscid thistle]

Hello guys

I got a question

Why if we decrease x in function by some value, the fubction plot is shifted to the right and not to the left?

Is it because the function gains some values "later" than it would do if x wasn't decreased at all?

[willow bear]

yes

[mighty orbit]

Consider $f(x)=x^2$, then the graph is the usual parabola with vertex at $(0,0)$.

Now consider the \emph{vertical shift} given by $f(x-1)=(x-1)^2$, then the graph is still a parabola, but the vertex was \emph{shifted} in the $x$-axis, that is, the new vertex is at $(1,0)$.

In general, a function $f(x)$ is shifted $a$-units to the right when considering the function $f(x-a)$.

[obsidian monolith]

bondalton

[viscid thistle]

Ok, thank you

Also, what do you usually discuss here

Is it trig and functions?

[willow bear]

"precalculus" is a very American class name

almost anything that's too high caliber for algebra, but is not yet calculus, can fall into this

so exponentials and logarithms for example

sometimes complex numbers

sometimes combinatorics

[viscid thistle]

Got it

Okay

[near sentinel]

We were right, ans is -3/4096

That 1/512 was for some other q

@willow bear @proven void

[willow bear]

great

[proven void]

nice ^^

[proven void]

how do I find the limit of this

[wooden ruin]

ok

[willow bear]

try to find the limit of 1/n + 2n/(n+1) first.

[proven void]

1/n tends to zero

2n/(n+1) evaluated as x approaches inf is 2

it doesnt make a difference whether inf +1 or inf apparently its still inf

[willow bear]

n not x

but ok

so the stuff inside the parentheses approaches 2

what do you think (the same stuff)^3 goes to?

[proven void]

8 😛

[willow bear]

i have a feeling you could have gone through the exact same reasoning yourself if you were not so prone to second-guessing yourself

[proven void]

sorry

i’m new to math

[willow bear]

i have seen you ask various kinds of math questions here for at least a year i think

[proven void]

mostly algebra and pre calculus

[willow bear]

analy?

[proven void]

no, pre calculus

here it has a name in spanish idk

analisis matematico

[formal forum]

how can i integrate this
(x^3 - x) / (x^2 + x)

[willow bear]

mathematical analysis

... sounds less pre and more calculus to me lol

[proven void]

it’s one of the courseI have to take to before entering the bachelor program

next year real analysis would be a different thing I pressume

[proven void]

this can be simplified

factor x in numerator denominator

and show me how that looks

whatever, it seems you already finished it

[formal forum]

yes thx anyways

[willow skiff]

#calculus

Precalculus means "before calculus"

For next time

[proven void]

how do i find this limit?

[summer ruin]

divide by n!

[proven void]

,rotate

[proven void]

1/5

how do I find a?

[summer ruin]

what limit solving techniques do you know

[proven void]

not much

lhopi

maybe

?

[summer ruin]

what else

[proven void]

that’s it

[summer ruin]

have you ever solved limits with roots before?

[proven void]

this would be my first time sir

can we rationalize

[summer ruin]

try it

[proven void]

can you elaborate

[summer ruin]

well what was it that you wanted to do when you said "rationalize"

[proven void]

i’m not sure

[summer ruin]

try to make use of a^2 - b^2 = (a-b)(a+b)

[proven void]

how so?

[summer ruin]

wouldn't that be nice if there were no roots?

if you could somehow square them

[proven void]

do i square the whole function

[summer ruin]

this suggests something else

no

[proven void]

how to calculate? a

i need to find a

how to do it

[summer ruin]

look at the numerator

you already have difference of two things

[proven void]

can we rationalize

i mean multiply and divide by the conjugate

of the numerator of the original expression

[summer ruin]

try it

[proven void]

?

can you elaborate

[summer ruin]

you already made a suggestion on what exactly you want to do

what else is there to elaborate on

just do it

[proven void]

,rotate

[summer ruin]

it's -(x^3+ax-1)

[proven void]

sure

sorry

,rotate

[summer ruin]

well work it out

from now on it should be easy to find the limit

[proven void]

how to continue?

[summer ruin]

what can you do in the numerator?

[proven void]

idk

[summer ruin]

why not just compute the difference

[proven void]

,rotate

[proven void]

what now

[summer ruin]

still you can do something with the numerator

[proven void]

this shit is unsolvable

[summer ruin]

the denominator is irrelevant

[proven void]

,rotate

[summer ruin]

1-x = -(x-1)

[proven void]

,rotate

[summer ruin]

see if you can compute the limit now

[cosmic nymph]

you could use l hopital right?

[proven void]

,rotate

[proven void]

depends if the conditions of lhopi are met

e.g. inf/inf form is one I believe

[cosmic nymph]

no like in this question you could just differentiate both

itll be much easier

[proven void]

but this limit is as x approaches 1

[cosmic nymph]

change limit

[summer ruin]

no applying lhopital here would only make it worse

nothing nice will come out from differentiating sqrt(x^3+ax-1)

[cosmic nymph]

why the differentiation wouldnt be hard

[proven void]

how do i finish the exercise

[summer ruin]

you've got that sqrt(a) = 1/4 and you're asked to find a

[proven void]

1/4 = a^1/2

[cosmic nymph]

then you can substitute x=1 after lhopital

itll be easier

[summer ruin]

you should be able to solve that yourself

[cosmic nymph]

it might get be an easier method to use lhopital i think

the differentiation is the only hard part

[near sentinel]

how do I start

[summer ruin]

lhopital

[near sentinel]

We are not taught that yet

I are supposed to do this by trigonometric substitution

[summer ruin]

well I'm not aware of any trigonometric identity for sin(x^2) or pi/(2cos(x))

[near sentinel]

Its supposed to be done by changing converting any trig functions that goes to zero, and the applying Direct Substitution.. but I can't think of a way to convert

[summer ruin]

converting in what way exactly?

[willow bear]

boy this is ugly

sin(sin(x^2)) expands into x^2 + o(x^2) though

[near sentinel]

A friend of mine sent me the solution

I'll send it here

[mighty orbit]

Something doesn't add up with the second equality...

[viscid thistle]

Hello everyone, im a college student studying pre-calc and calc 1 is there anyone that can help me. I recently got 5 out of 100 in the quiz, exam in april 25. I am so bad at math feel like want to die

[fair bear]

Need some help

Question is to write z in z = a+bi form

[fair bear]

5/100 is crazy rip

[viscid thistle]

Yeah, im pretty bad at math

[viscid thistle]

Complex Number Systems?

[fair bear]

Yeah ig

[river drift]

try substituting a + bi for z and rearrange a bit

[fair bear]

I did but I cant seems to get the same answer as the answer sheet

[river drift]

show your work, and what the answer sheet says

[tender quest]

Show your work, and if possible, explain where you are stuck.

[fair bear]

ai-b/2+a-bi=b+1 ig this would be the first step

But im not to sure how the equ system would look like

{-b/2+a=1
{a-b=b

Im thinking like this but not sure if the imz = b should be on the imaginary part of the equ system

[river drift]

well you want the numbers on both sides to be equal, meaning they have equal real and imaginary parts

is b + 1 purely real, purely imaginary, or neither?

[fair bear]

Ik the +1 is real

The b im not sure of

[river drift]

remember for a complex number we express it as a + bi

bi is purely imaginary, so is b real or imaginary?

[fair bear]

Yeah ik but for this question idk, the first b is real tho cuz iz = i(a+bi) = ai-b

Apparently the answer is z=4+2i

But idk how they get that

[river drift]

so for example if we have the number 4 + 5i
then 5i is an imaginary number
but is 5 an imaginary number?

[fair bear]

No

[river drift]

so for this particular example a = 4 and b = 5

[fair bear]

Ye ig

[fair bear]

I see then it should be

{-b/2+a=b+1
{a-b=0

[crisp timber]

mod(x,0) = 0?

[fair bear]

So I get a=-2 and b=-2

Apparently the answer sheet is correct tho

[winter comet]

bro took an exam 10 days in advance

XD

[fair bear]

Nvm just forgot some basic algebra rules for some reason but got to the answer now

[fair bear]

For some reason I forgot that both ai and -b was /2 and I forgot the (-) on the bi

[hexed apex]

Does anyone know how the answer for limit of f(f(x) is 1/2?

[mighty orbit]

Note that $f(f(10))=f(2)$

[obsidian monolith]

bondalton

[wild sable]

could anyone help with this optimization problem i have

i know how to solve for b and c but im convinced im not creating the right function for a)

can i ask this in one of the help channels idk how this works

[willow skiff]

Btw this is solved already

[proven void]

,align & \text{knowing that : } \frac{sin(5x)}{x} < f(x) < \frac{\sqrt{x+4} - 2}{x} + \frac{19}{4} & \
& \text{find }\lim_{x\to0} f(x) &

[obsidian monolith]

milanesa de pollo

[proven void]

how do I solve this?

can i use sandwich theorem

[willow skiff]

Yes

[proven void]

how to use it

[willow skiff]

Well the limit as x to 0 on the left is just 5

$\frac{\sin 5x}{5x} \cdot 5$

[obsidian monolith]

south

[willow skiff]

For the RHS ignore the 19/4 for now and multiply by $\frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$

[obsidian monolith]

south

[proven void]

how does that work

$\frac{\cos 6x}{6x} = 1 $?

[willow skiff]

No

Only for sin

lim u to 0 (sin u)/u = 1

[near sentinel]

the cos version is (1- cosx)/x^2 = 1/2

[proven void]

,rotate

[proven void]

what now?

[near sentinel]

cancel x and substitute x->0 in the sq root

1 / [ root(x+4) +2 ]

x -> 0

1/ [root(4) + 2]

answer == 1/4

@proven void

[proven void]

yeah

,rotate

[proven void]

got it , thanks

,, \lim_{x \to 1} \frac{\sin(x\pi)}{\sin(3x\pi)}

how do I calculate this limit?

[obsidian monolith]

milanesa de pollo

[proven void]

should I apply lhopi?

[near sentinel]

i dont know lhopi, i did it using trigo substitution

expand the denominator using sin3theta formula

then simplify

then substitute x ->1

[raven smelt]

might be helpful to sub t = pi(x) to make it simpler

[proven void]

i messed up somewhere

[proven void]

,rotate

[proven void]

something like this you mean

[raven smelt]

yep thats correc

t

now sub x = 1

[proven void]

,w lim x to 1 of (sin xpi )/( sin (3xpi) )

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[proven void]

,w sin pi

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[proven void]

got it, ty lads

[near sentinel]

yupp

help pls, how do i start

[near sentinel]

😦

[compact spade]

uhhhhh

[willow bear]

@compact spade do you still need help with this

[compact spade]

No, I got it :)

Ty tho

[primal girder]

Hii!! is anyone available atm to help me with 2-3 questions (even just one)? I'm going to have my final in 2 days, im trying to 100% it as much as possible 🥹

I guess ima go make some breakie i posted in the help chat if anyone's around!!! thank you sm to anyone who reaches :')

[willow skiff]

Just ask your questions

Don't need to ask to ask a question

[near sentinel]

I still need help

[pine geyser]

Ask your question in https://discord.com/channels/268882317391429632/326138772477575180

they might be able to help

[viscid thistle]

Is the answer 1 ?

[winter comet]

how do you figure?

legit this problem sent me on an l'hop loop ☠️

[viscid thistle]

yes that is ture

[winter comet]

idk if i even did possible things lol

[viscid thistle]

because the question says n tends to infinity and N belongs to Z

which makes no sence

sense

anyways

[winter comet]

it kinda does

it just means infinity but it never hits the fractions

[viscid thistle]

taking n^2 common

[winter comet]

what?

[viscid thistle]

from root over (n^2+n)

[winter comet]

oh

[viscid thistle]

we get n * root over (1+(1/n))

sincen tends to infinity

1+1/n becomes 1

[winter comet]

wait what

[viscid thistle]

?

[winter comet]

you mean sqrt(n^2 (1+(1/n)))?

[viscid thistle]

there is a root

[winter comet]

sry

that?

[viscid thistle]

yes

No!

[winter comet]

wut

[viscid thistle]

sqrt(n^2 (1+1/n))

[winter comet]

💀

[viscid thistle]

by bad

my bad

i don't know how to use bot commands

[winter comet]

fixed

[winter comet]

same XD

i mean neither do i

anyway

as n tends to infinity, if you plug in you get infinity, no?

infinity^2 * 1

[viscid thistle]

no

cos (pie * n)

cos (infinity)

which means the max value of cos

that is 1

[viscid thistle]

infinity^1 not 2

[winter comet]

oh right

[winter comet]

uh

oh

can you do that ☠️

[viscid thistle]

let me send the solution

give me a minute

[winter comet]

i didnt realize you were TRYING to get infinity, i thought uw ere tryna get an actual number 🤣

i get how the inside is infinity

i didnt know cos(infinity) = 1

nah it shouldnt be

[vapid plaza]

it isnt

[winter comet]

yea

cuz

the domain is getting bigger

doesnt say anything about the y values tho

[viscid thistle]

cos (infinity) mean the max value of cos

[vapid plaza]

it doesnt

[viscid thistle]

no?

[vapid plaza]

no

[winter comet]

no because

the input is max input

but that doesnt mean the output max output

right

like as x gets bigger, y doesnt necesarily get bigger

[vapid plaza]

i don't think you have any idea what you're saying...

[winter comet]

are we sure this problem is solveable ☠️

[winter comet]

i think he does, he was just wrong

💀

[vapid plaza]

i'm quite certain the limit doesn't exist

[viscid thistle]

yes

[winter comet]

interesting

[viscid thistle]

i am sorry

[winter comet]

yeah

i think it doesnt exist

unless its some calculus beyond my knowledge, which is very possible 💀

[viscid thistle]

that is the reason i was asking if the answer was 1

i was not sure

[winter comet]

on second thought

[viscid thistle]

mhm

[winter comet]

couldn't it be 1 or -1

because the domain is z

so it only works when 0pi, 1pi, 2pi, 3pi, ect

so it gotta be 1 or -1

[viscid thistle]

i can't be sure without knowing the final answer

[winter comet]

Oh wait

There can't be two answers to limit

That's like tte definition of not existing ☠️

[viscid thistle]

yes

[near sentinel]

i dont know actually, its a hw question

[near sentinel]

wait, cos infinity aint zero

[near sentinel]

lmao

[willow bear]

cos(+∞) is undefined.

[willow bear]

this is pure bullshit, too.

[viscid thistle]

sorry

can you send the solution

[willow bear]

i'd rather not send the whole thing.

bc like, !nosols and all.

[viscid thistle]

explaination would be fine

[willow bear]

$\cos(\pi\sqrt{n^2 + n}) = \cos(\pi(\sqrt{n^2+n} - n)) + \pi n) \ = (-1)^n \cos(\pi (\sqrt{n^2+n} - n))$

[obsidian monolith]

|Ann⟩

[viscid thistle]

rationalize?

[willow bear]

$\sqrt{n^2+n} - n$ actually approaches a finite limit, which you can work out separately. this does, however, commit the sin of breaking chain-of-equalities continuity --- and that may be unacceptable depending on how strict your teacher is.

[obsidian monolith]

|Ann⟩

[willow bear]

and yes rationalize.

but don't plug n in everywhere yet.

only work out $\lim_{n \to \infty} (\sqrt{n^2+n} - n)$.

[obsidian monolith]

|Ann⟩

[viscid thistle]

1/2?

0?

[near sentinel]

the guys from #calculus told to do with bino expansion

im getting 0

[viscid thistle]

same with Ann solution too

[willow bear]

how many more guesses do you have up your sleeve?

[near sentinel]

reading that rn

[willow bear]

maybe you have countably many, but even that won't save you from the inability to guess every real number

[viscid thistle]

the first one was the answer to this

and the second one was the final answer

[willow bear]

oh yeah and i was obviously supposed to know that

from you just blurting out the numbers

[viscid thistle]

hehe

[willow bear]

i'm not a telepath.

[viscid thistle]

sorry

[near sentinel]

whats chain of equalities continuity?

@willow bear

[willow bear]

it's when you write down your solution as

start
= something
= something else
= something else again
= ...
= answer

a particularly strict teacher may reject outright any solution that doesn't fit this format

[near sentinel]

ohh okkk

we have objective questions so it doesnt matter

lol

got ur explanation too

thankss

[primal girder]

Say, I have found all of my angles using sine law for an SSA triangle, but then using sine law again gives me an incorrect number where c+b>a (that cant happen)

What should I do?

All angles are correct

the two sides too

[willow bear]

can you show what you're looking at

[primal girder]

its an oblique triangle

yes!

[willow bear]

with the actual numbers you got

and the process

to see if there's any screwups in there

[primal girder]

First pic

[willow bear]

and your work?

[primal girder]

I know i have my angles right because, well if one of my angles are right then the other one has to be right too (67% correct = 2/3 of my answers are right)

yes

ill upload it in a sec

my little c should be 24.4

,w 12/sin(24.0483º)=c/sin(55.95º)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by honeyow

[primal girder]

same as b/sinB = c/sinC

these are usually pretty complexe problems, our course supposedly has a reputation for its online homework content (sometimes its hw based on stuff we've never learned)

unless if im overthinking it

[willow bear]

hold on

im trying to retrace your steps here

[primal girder]

okay

take ur time!

ask any q's

[willow bear]

c+b>a (that cant happen)
why can't that happen?

that sounds like a normal instance of triangle ineq to me.

c+b < a would be concerning.

[primal girder]

oh

oh really?

omg

that's what's been really blocking me on this question for 2 hours 😩

i was so convinced 💀 it seemed too easy to remember

thank you so much

yupppppppp

[willow bear]

"detours can't be shortcuts"

[primal girder]

yep, especially not in meth 🤠

[primal girder]

What would you guys do if:
cos(2t) = ? , if cos(t) = -5/7, & π< t <2π/3

we already know its in Q3 obv, and can do pythagorean theorem for sin (my exercise has 4 q's, sin & cos double angles and its half angles - but it gets SUPER messy, especially that it wants exact numbers)

or would anyone know a way i guess to verify this online :')

w/ restrictions and all that

[winter comet]

i would use the identity cos(2t) = 2cos^2(t) - 1

[pulsar cedar]

So I Have two answers for this

this and 650 km/hr

[winter comet]

does ground velocity mean the velocity that an observer may notice from the ground?

which would be the net velocity...

[pulsar cedar]

im not sure

[winter comet]

i would assume it does...

[winter comet]

how did you get 650km/hr?

[pulsar cedar]

500 + 150 to find net

resultant ^

[winter comet]

ah

maybe it would help if you could draw a picture with vectors/arrows?

[pulsar cedar]

if i knew woujld but this question is super confusing

im just trying to kno wwhich answer it would be

[winter comet]

actually i'm pretty sure it isnt that hard

[pulsar cedar]

im overthinking it rn

i personally believe 650 is right

[winter comet]

wouldn't it look like this?

because the wind is moving south at 150 km/hr

and the plane is moving north at 500km/hr

[pulsar cedar]

yeahh pretty much

[winter comet]

why?

[pulsar cedar]

i think that should be right, cause its not saying they are against the wind and theyre saying the wind came from the south

[winter comet]

if the plane is going north, and the wind is going south, that means they are going in opposite directions

[pulsar cedar]

so your saying we would subtract?

[winter comet]

yes

[pulsar cedar]

so then it would be 350?

[pulsar cedar]

this

[winter comet]

yes

but do you see why?

like the forces are acting against each other

so they neutralize each other a bit

[pulsar cedar]

yeah i understand

so 350 is 100 percent right

[winter comet]

assuming we understood the problem correctly, yes

[pulsar cedar]

but i think it should be 650 cause its worded that the wind is coming from the south, so blowing north

[winter comet]

that is a good point

[pulsar cedar]

yeah so im not sure

[winter comet]

☠️

[pulsar cedar]

the question got to you

[winter comet]

i would still ASSUME its 350 but honestly its a poorly worded question

[winter comet]

😂

[pulsar cedar]

should i ask my teachr

[winter comet]

yes

[pulsar cedar]

let me ask

@winter comet

can you tell me what to say

cuz idk how to aask him

he said its correct

[winter comet]

he said 350 is correct?

you can ask him, "does the wind from the south mean that the wind is going south or coming from the south"

[pulsar cedar]

no no

he said the question is worded fine

he didnt give an answer

he said look at the word coming from visaulize which way your heading when your coming from somewhere

@winter comet

[winter comet]

it doesn't say coming from though

it just says from

[pulsar cedar]

yeah

so from

[winter comet]

hmm

i guess it would be 650 then lol

you know what it probably is 650 because i think "from the south" doesnt mean going south anyway

💀

[pulsar cedar]

bro im doomed

so 650

[winter comet]

yeah 💀

i would say thats more of an english/geography question than a math question 💀💀

[pulsar cedar]

ok do this

reword it

and then lets solve

[winter comet]

An airplane is heading north with an airspeed of 500 km/h. The plane encounters a wind coming from the south at 150km/h. What is the magnitude of the resultant ground velocity? Round to the nearest km/h

💀

[pulsar cedar]

gauth math is giving 350

chatgpt is giving 350

[winter comet]

they're wrong

its 650

💀

don't rely on gpt or gauth math

or like

bots in general

they unreliable

[pulsar cedar]

bro idk wat do to

[winter comet]

i guess its 650

the bot is wrong

i was wrong

u were right

cuz the wind is from the south

which heavily implies it is going north

in which case you add the velocities

[pulsar cedar]

i wrote 350

💀

[viscid thistle]

!nogpt

[tender quest]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[winter comet]

it is possible to say this without using a robot

💀

[viscid thistle]

The bot’s response is much nicer than mine

Also it’s easier

[winter comet]

you know what else is easier

using chat gpt to solve problems

☠️

[viscid thistle]

This isn’t correct

What’s your point

[winter comet]

its funny 💀

[winter comet]

cat :D

[weak glen]

'~'

cute

[fickle stag]

verifying identies? anyone

[fluid stone]

guys weird question when did u know about integrations ?

[winter comet]

that is a weird question

i'm not sure what you mean

but if i do

you learn integration in calc 1

as in people in general

but idk if that what ur asking

[fluid stone]

ok look i am not american im actually Egyptian so i am asking + calc 1 is in what stage high school or uni ?

[winter comet]

late high school or early uni

somewhere around there

[fluid stone]

k thx alot

[winter comet]

👍

[primal girder]

So I never fully understood radicals (when the +/- applies), only that you take their positives in distance/radius/hypotenuse, is what this person on Reddit wrote true? Did I apply it correctly to my exponential EQ, where I had to find the exact solutions (x)?

125^(x+1) = 5 x 25^(x/2) is my eq btw

[summer ruin]

you made the same mistake twice

5 * 5^x is not 25^x

and why do you bring up root if it has nothing to do with the equation you're solving?

[shadow summit]

You don’t use roots.

,,b\times b^{n}=b^{n+1}

[obsidian monolith]

TheLord26

[near sentinel]

School

12th grade for mathematics, 11th for physics

[fluid stone]

thx alot

[viscid thistle]

Can anyone help me to solve the equation?

[willow bear]

how did sin(pi/3) happen?

[viscid thistle]

I managed to simplify it a little

[willow bear]

0.375 is 3/8

[viscid thistle]

Yeas

You are right

[willow bear]

also you tried to square-root both sides

but

[tribal ferry]

how do i solve lim z-->0 (1+x)^n - 1/x

[viscid thistle]

Yes

[willow bear]

i would be careful with that (bc you can lose solutions)

also there is a way to not have to do that here

first off why not just write $\frac{3}{8} \sin^2(-\pi/4) = \frac{3}{8} \cdot \frac{1}{2} = \frac{3}{16}$

[obsidian monolith]

|Ann⟩

[viscid thistle]

Okay

[willow bear]

second, write sin^2(x/4 - pi/4) as cos^2(x/4 + pi/4)

[viscid thistle]

I dont understand this part

[willow bear]

sin(t) = -cos(t + pi/2)

that's assuming you're confused about the what rather than the why

[viscid thistle]

But shouldn't be there cos^2(x/4 + 3pi/4)?

We must add pi/2, right?

So t = x/4 + pi/4. And t + pi/2 = x/4 + 3pi/4

[primal girder]

I’d never seen it that way (n+1) thank you!!

[willow bear]

no actually i am taking x/4 - pi/4 as my t

[viscid thistle]

Got it

I would never come up with it

But what's next

[willow bear]

well,

write down the equation after the rewrites that i suggested

since you don't know what to do next, do nothing else, only that

[viscid thistle]

I almost solved

[willow bear]

then why ask "what's next"?

[viscid thistle]

It was after i asked

But

What is sine of 3/8

Oh, i forgot about squares

[willow bear]

why would you even want to calculate sin(0.375 rad)?

also yeah you forgot about squares

lhs should have been 1/4 sin^2(x/2 + pi/2) = 3/16

[viscid thistle]

I dont want to

Or i misunderstood something

[willow bear]

"What is sine of 3/8"

exact quote from you

[viscid thistle]

Okay

Got it

[willow bear]

you either misspoke or misunderstood

[viscid thistle]

But, when i do square

It will be +-sqrt(3)/4

Okay

[willow bear]

??

you either made an arithmetic mistake or let one of the things i said in one ear and out the other.

[primal girder]

,,b\times b^{n}=b^{n+1}

[obsidian monolith]

Honey

[primal girder]

I’m feeling off about the way I’m going about this

[willow bear]

is 125^(x+1) = 5 * 25^(x/2) an equation that you are solving?

[primal girder]

Yes!!

[willow bear]

ok

what does 2(x/2) simplify to?

[primal girder]

I need to find the exact x solution to it

2x/4, right?

Also, did I do that okay?

Reducing it to 5 and multiplying it with my parentheses up top?

[willow bear]

no, 2(x/2) does not simplify to 2x/4

[primal girder]

Oh right

Because we don’t have a plus

In there

So nominator time nominator

Simplifies to x?

[willow bear]

the top part of a fraction is the numerator.

but yes, 2x/2 simplifies to just x

[primal girder]

Nice!

[willow bear]

what does your equation look like now, after you do this simplification but nothing else?

[primal girder]

5 . 5^x

I could easily use structure here

I suppose that would probably be the way to go, but my x on on the other side isn’t going to be a quadratic

[willow bear]

i said to write the whole equation not only half of it

[primal girder]

Oh sorry !!

125^(1+x)= 5 . 5^x

[willow bear]

breaking the right hand side back up was not a good idea

[primal girder]

I’m on my iPad so it’s not the easiest to type on xd I’ll do my best to be quick

[willow bear]

125^(x+1) = 5^(x+1)

this is your equation

[primal girder]

Ahhhh

Thank you!,

[willow bear]

two options:
a) you continue yourself and tell me what you end up with
b) you ask me what to do next, and then do that and only that and nothing more n

[primal girder]

I cancel one of my logs (or can just do any log and bring my x down and factor it)

I’ll do it!

I’ll be right back gimme a min!

Tysm for the help :))

Omg is it supposed to look that crazy or am I doing this wrong

@willow bear

Idk if I should have combined them before hand too, on my left hand side at the bottom

Maybe it would’ve switched them up? Or no because of the parenthesis?

[willow bear]

you are making at least one mistake

[primal girder]

With my combining of logs right?

[willow bear]

i am in class right now so can't concentrate on retracing your steps

[primal girder]

Oh my bad!

I can always ask in the help chat, dw about it :) u helped plenty

[shadow summit]

Do you still need help?

[primal girder]

I tried following what u told me. Did I do it correctly this time? @shadow summit

I realized that I was confusing myself too much w the change of base formula and division before

[uncut mulch]

you messed up at the VERY last step
doing what i told you NOT to do yesterday

[primal girder]

Combine the two 😳

[uncut mulch]

notation is also suboptimal in the last two lines

[primal girder]

I’m looking at our prob from yesterday on my phone

[uncut mulch]

you should be writing stuff like \
$\begin{aligned} x \ln(7) &= \ln\br{\frac{7}{21}} \
&= \ln\br{\frac 13} \
&= -\ln(3) \
x &= -\frac{\ln(3)}{\ln(7)} \end{aligned}$ \ \
and that's where you stop

[obsidian monolith]

ℝαμΩℕωⅤ

[primal girder]

Found it!

Also yeah it gives me a syntaxe error which is why i combined them

[uncut mulch]

well you could apply change of base law to that if you want

which will get you
$$x = -\log_7(3)$$

[obsidian monolith]

ℝαμΩℕωⅤ

[primal girder]

Damn really!

I need a refresher on the change of base law honestly

But

Ok no syntax error!

That’s it, that’s my solution?

Would both -log7(3) and -ln(3)/ln(7) be a correct answer for an exact solution?

[uncut mulch]

yes

[primal girder]

Omg wait I must’ve screwed up earlier, no syntax error now if I do the fraction version

Ok thank you again :’)

[uncut mulch]

np

[queen shoal]

how do i derive trig functions

i know how to derive functions with powers and root but not trig functions

[willow bear]

when you say derive do you mean take the derivative?

if yes:

sin'(x) = cos(x) and cos'(x) = -sin(x). everything else can be calculated from those. but it's convenient to remember tan'(x) = 1/cos^2(x).

@queen shoal

[queen shoal]

thanks

what about taking the derivative of functions with natural logarithms

ln(sin(2x)) this function

[willow bear]

ln'(x) = 1/x

you will need the chain rule for ln(sin(2x)) of course.

[queen shoal]

ok ill watch a video about chain rule now

[proven void]

How do I do this?

[queen shoal]

derivative of sin(2x) is cos(2x)2 i think. ln'(cos(2x)2) = (1/cos(2x)2)(sin(2x)2)

is this correct

[willow bear]

ln'(cos(2x)*2)

this notation is messed up in at least 2 ways i think

[ln(sin(2x))]' is what you wanted, right

the derivative of sin(2x) is indeed 2 cos(2x)

[queen shoal]

so why is it not ln'(cos(2x)*2)

[willow bear]

the only reason i was able to give you the derivative values as sin'(x) = cos(x) etc is because i was talking about the sine, cosine etc. functions themselves -- in their pure form with nothing mixed in. if you tried writing "the derivative of sin(2x)" as sin'(2x) you would not be understood.

the derivative of $\ln(\sin(2x))$ is $\frac{2\cos(2x)}{\sin(2x)}$.

[obsidian monolith]

|Ann⟩

[queen shoal]

i dont think i understand the derivative of ln yet

[willow bear]

maybe you also misunderstand the chain rule.

i can't tell where your gaps lie.

[queen shoal]

it says you first work out derivative of sin(x) which is cos(x) the you replace x with x^2 so you get cos(x^2) times by the derivative of x^2 so you get cos(x^2)*2x

is the derivative of sin(2x) = cos(2x)*2

[willow bear]

.

[queen shoal]

oh yeah

derivate of ln(x) is 1/x and i replace x with cos(2x)*2 so i get 1/cos(2x)2 then times by the derivative of cos(2x)2 which is (sin(2x)2)2 so i get (1/cos(2x)2)(sin(2x)4)

[willow bear]

and i replace x with cos(2x)*2
no

you replace x with sin(2x)

unless you've now suddenly decided the original function is now ln(2 cos(2x)) and not ln(sin(2x)) as i thought

[queen shoal]

i thought because sin(2x) also needs chain rule you first find derivative of sin(2x) then replace x with that

[grave meteor]

formally you do use the chain rule

sin(2x)' * (2x)'

[willow bear]

you first find derivative of sin(2x) then replace x with that
you misunderstand

[grave meteor]

which is just cos(2x) * 2

[willow bear]

$[\ln(\sin(2x))]' = \frac{1}{\sin(2x)} \cdot [\sin(2x)]'$

[obsidian monolith]

|Ann⟩

[willow bear]

and THEN you apply chain rule to get [sin(2x)]'

[grave meteor]

(when you have a composition you pretend that the argument of the outermost function is just x, and multiply by the derivative of the argument, which is another function)

like peeling an onion

[queen shoal]

is the derivative of ln(sin(2x)) just 1/sin(2x) * sin(2x)

dont you have to find derivative of sin(2x)

[grave meteor]

yes you do

[queen shoal]

derivative of sin(2x) = cos(2x)*2 right?

[grave meteor]

yes

[willow bear]

no and i did not say that it was

[grave meteor]

therefore

[queen shoal]

so do i replace sin(2x) in the thing the bot sent with cos(2x)*2 for the derivative?

[queen shoal]

ok

[grave meteor]

here?

[queen shoal]

yes

[grave meteor]

yes the derivative is 2cos(2x) replace it accordingly

it's just evaluating the expression

[queen shoal]

so the derivative of the function ln(sin(2x)) = (1/cos(2x)*2)(cos(2x)2)

[grave meteor]

nope

[queen shoal]

oh ok

[grave meteor]

first

you take the derivative of ln

assuming what's inside to be x

thats 1/x, substituting back

1/(sin(2x))

and you're done with this

multiply this

by the derivative of what you assumed to be x

i.e. the derivative of sin(2x)

again a chain rule

assume 2x to be x,
the derivative is cos(2x)

and multiply by the derivative of 2x which is 2

[queen shoal]

yes

derivative of sin(2x) is cos(2x)*2

[grave meteor]

yes it us

but the original ln derivative being 1/sin(2x) doesn't change

[grave meteor]

its just multiplied by this

[queen shoal]

so is it 1/sin(2x) * cos(2)2

[grave meteor]

yes

[queen shoal]

thanks so much for help @willow bear and @grave meteor

[tawdry viper]

can anyone explain summation notation to me simply?

[grave meteor]

just ask the question

[proven void]

this notation?

[chrome ether]

there is nothing to explain, it's literally just "do you know what each part of the notation means"

[primal girder]

anyone got a sec to reply to this https://discord.com/channels/268882317391429632/1230615624611266570 ?

[fathom night]

The Symbol: The Greek letter sigma (Σ) represents summation. It tells us that we are adding something up.

Basic Example:

Suppose we have the expression:

∑n=13(2n−1)
n=1∑3​(2n−1)
Here’s what each part means:
    The index variable is (n), which starts at 1 and goes up to 3 (inclusive).
    The expression inside the summation is (2n - 1).
Let’s evaluate it step by step:
    For (n = 1): (2(1) - 1 = 1)
    For (n = 2): (2(2) - 1 = 3)
    For (n = 3): (2(3) - 1 = 5)
Add these results: (1 + 3 + 5 = 9)

Problem 1:

Evaluate:

∑n=14n2
n=1∑4​n2
    The answer is (1 + 4 + 9 + 16 = 30).

Using Different Indices:

You can use any letter for the index. For example:

    ∑i=02(3i−5)
    i=0∑2​(3i−5)
    Evaluate similarly: ((-5) + (-2) + 1 = -6)

[warped cipher]

!nogpt

[tender quest]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[vestal fossil]

cAN I ASK DIFFERENTIAL EQUATION DOUBT HERE?

[willow bear]

yOU MIGHT WANNA TURN OFF CAPSLOCK AND THEN ASK IN #calculus

[kind glade]

Does anybody know the book Differential and Integral calculus by feliciano and uy reddit

[queen shoal]

i uhh forgot what derivative of tan is pls can someone tell me thankss

[uncut mulch]

easily searchable

[willow bear]

go to http://wolframalpha.com/ and type in "derivative of tan(x)"

[queen shoal]

is this natural log or what

[winter comet]

it should be yeah 💀

[queen shoal]

ok

why skull

[summer ruin]

it says what it is

[queen shoal]

oh ok sorry

[winter comet]

wdym

[winter comet]

cuz its wrong? it doesnt show the base or ln

[winter comet]

oh 💀

[uncut mulch]

wolfram uses log to represent the natural log

[queen shoal]

ok

[proven void]

how to calculate

is $-1^{\infty} = 1$ or $-1$

[obsidian monolith]

milanesa de pollo

[willow bear]

neither

but also, if you want to talk about raising -1 to any power (aside from ∞, which makes no sense) then you have to use parentheses

$(-1)^n$ and not $-1^n$

[obsidian monolith]

|Ann⟩

[willow bear]

consider: where does sqrt(n+2) - sqrt(n) approach?

[proven void]

0

sqrt(inf + 2) isnt it similar to sqrt(inf)?

[proven void]

im not sure tbh

I just threw it since I dont know what else to answer

[willow bear]

don't guess

guessing will get you nowhere

plugging in n=+∞ gets you ∞ - ∞, which means direct substitution by itself won't work and you need to do something else.

before i tell you what that "something else" will be, do you understand that idea

bad notation.

also doesnt answer my question.

[proven void]

mmmm

why are you saying that n = +∞ gets you ∞ - ∞

[willow bear]

$\lim \sqrt{n} = +\infty$ and $\lim \sqrt{n+2}$ is also $+\infty$.

[obsidian monolith]

|Ann⟩

[willow bear]

(n -> ∞ in both cases, dropped for brevity)

[proven void]

in case thats true, I think I understand ∞ - ∞ then

what were you about to say?

[willow bear]

in case what's true?

[proven void]

.

[willow bear]

i mean

on the one hand, it's good that you are not taking my words as gospel.

on the other hand, i can tell you for certain that both of my statements there are correct.

ok anyway

direct substitution doesn't work as is, so we need to do something else.

what we can do here is multiply and divide by the conjugate, like so:

$\sqrt{n} - \sqrt{n+2} = \frac{(\sqrt{n} - \sqrt{n+2})(\sqrt{n} + \sqrt{n+2})}{\sqrt{n}+ \sqrt{n+2}} = \frac{n - (n+2)}{\sqrt{n} + \sqrt{n+2}}$

[obsidian monolith]

|Ann⟩

[willow bear]

do you understand what i just did?

[proven void]

I do

[willow bear]

ok

do you see how to continue, or should i keep explaining?

[proven void]

you rationalized I think

you simplify the n - n to 0

and the denominator is infty so the limit evaluated to zero

it doesnt matter whether $-1^{\infty}$ is $-1$ or $1$

[willow bear]

but also, if you want to talk about raising $-1$ to any power (aside from $\infty$, which makes no sense) then you have to use parentheses \
$(-1)^n$ and not $-1^n$

[obsidian monolith]

|Ann⟩

[willow bear]

$\lim (-1)^n$ does not exist.

[obsidian monolith]

|Ann⟩

[willow bear]

but $(-1)^n$ is a \textbf{bounded} sequence, and bounded times zero-limit is zero-limit.

[obsidian monolith]

|Ann⟩

[proven void]

?

what does zero-limit mean . . .?

[willow bear]

a sequence or function whose limit is 0.

i made that term up, hoping its meaning would be clear.

it was not.

[proven void]

😛

[willow bear]

and i also didn't want to introduce any more notation unnecessarily.

[proven void]

I think the skull is because if the log is without base its prolly natural log, since I heard somewhere that log base 10 or log base 2 does not have practical use cases other than cookbook exercises

prolly I am wrong

[willow bear]

log_10 is usually the one used for logarithmic scales, and log_2 shows up in computer science.

[proven void]

why does wolfram use log base e as default base

[willow bear]

convenient for higher math purposes

in part bc the derivative of ln(x) is 1/x

[queen shoal]

Oh ok

[queen shoal]

why is the derivative of f(x)=f(x-1)*x not 0

f(0)=f(-1)*0=0

so f(1)=0*1=0

so f(2)=0

[willow bear]

f(x)=f(x-1)*x
this does not unambiguously define a function

@queen shoal where did you see this function?

[queen shoal]

i asked my friend to give me a function to find derivative of and he gave me this

[willow bear]

verbatim?

[queen shoal]

?

[willow bear]

was f(x) = f(x-1) * x the exact formula that he sent you

[queen shoal]

oh yeah he also said something about f(0)=1

but i asked him why he said because he said it was

[vapid plaza]

Is there any restriction of what x can be (positive integers?)

[queen shoal]

he didnt say anything about positive integers

[willow bear]

can you send a ss of the exchange

because this is looking suspicious a

[queen shoal]

he told me at school

[willow bear]

and your friendship should be presumed to be in danger

[queen shoal]

he did send this

[willow bear]

before that?

[willow bear]

like here is the issue: i dont think he actually gave you anything from which a differentiable function could be extracted

[proven void]

whatsapp

[willow bear]

...

no wait

ok

show me where he gave you the function

[vapid plaza]

What is your friend smoking

[willow bear]

bruh

[queen shoal]

ok

[willow bear]

4 screenshots and you can't find the original problem stmt

how on earth

[queen shoal]

he didnt text it he just told me at school

[willow bear]

eh

[proven void]

😛

[willow bear]

so he didnt give you a formula

only a recursive equation sort of thing

and ordered you to take the derivative under threat of 10000 years of torture

[queen shoal]

he wrote f(x)=f(x-1)*x on a sheet of paper

[willow bear]

f(x) = f(x-1) * x, f(0) = 1 at best only tells you the value of f at positive integers.

[queen shoal]

i asked him for a function so i could practise finding derivatives and he gave this

[willow bear]

this is not nearly enough to even establish that f is differentiable, let alone speak of finding its derivative

[willow bear]

is it a legally binding contract

i think not

[queen shoal]

ok

[vapid plaza]

[willow bear]

i think your friend is bullshitting you

[queen shoal]

oh ok ill tell him

[proven void]

for f to be differentiable we need to check if its continuous?

[willow bear]

i mean

its domain needs to be made of intervals and not just isolated points

for a start

[proven void]

ok

[vapid plaza]

Gamma function is not easy to describe without the integral thing