help-33

[still temple]

.close

[marsh citrus]

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[marsh citrus]

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[azure trench]

,rotate

[azure trench]

When you take moments about O do you equate those colours

Why can it not be forces going upwards= those downwards when you take moments

[marsh citrus]

@azure trench Has your question been resolved?

[twilit grove]

what is the question asking?

[marsh citrus]

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[fair flame]

Can someone please explain me how the left term is equal to the right term ?

[fair flame]

Yeah

This one

How is that equal

Sooo ?

[marsh citrus]

@fair flame Has your question been resolved?

[marsh citrus]

@fair flame Has your question been resolved?

[marsh citrus]

@fair flame Has your question been resolved?

[fresh oyster]

have you heard about the basel problem?

there is a general solution to the basel problem

this formula is just because both are pi^2/8

both of which are very tricky problems

we can start on the summation first if you do not know about these solutions

we have the mclauren series for sin(x)

$\sin{x}= \sum_{k=1}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!} =x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$

[elfin berry]

Cycadellic

[fresh oyster]

then we can do this

[fair flame]

Yeah that's just the reason why i'm asking this

[fair flame]

yeah ik this and then what can we do ?

[fresh oyster]

so

i had to look up eulers proof

according to wikipdia

we can use this style of proving

to solve the sum

its all based on the roots

basically, we can factor the poloynomials roots

consider the quadratic formula

you how how the quadratic formula can find the roots

d(x-a)(x-b)=0 is all possible parabolas with 0's at x=a and x=b

and we can get to every parabola with these roots because of our d

so, we can do this for any polynomial

3rd degree, 5th degree,... 10000th degree

this will work for all of them

well

this approximation is a polynomial

we use a limit so that we can consider it at finite values, if it works for all finite, we can say it works to infinity

[fair flame]

Yeah

[fresh oyster]

(although there are some slight errors with this logic, and we have to do analysis for full rigor) but this should be convincing enough that we can do this

so, we need to find the factors of the polynomial

in this problem, where does sin(x) pass 0?

at all pis, right?

[fair flame]

in kpi

[fresh oyster]

right

[fair flame]

where k is an integer

[fresh oyster]

so this factorization should make sense

this one is /x

specifically

and they divide the pi off to get it into x, but its the same idea

[fair flame]

yeah

[fresh oyster]

so, if we want to directly solve this instead of just applying the general formula and calling it a day, here is the tricky bit

we need to find the right sin(ax+b) that has the correct roots to get our summation

[fair flame]

yeah i think i understand this

[fresh oyster]

then

[fair flame]

But this is completely different

[fresh oyster]

we take on the massive task of distributing this long product

its the same solution method

instead of sin(x), we need to adjust the roots

then itll work

[fresh oyster]

when we distribute this long product, we will get the power series summation

although

we can avoid this completely by accepting this

[fair flame]

You mean which one

[fresh oyster]

this is the basel problem

[fair flame]

The (1-x²/π²)(1-x²/4π²)+... Or the -(1/π²+1/4π²+...) One ?

[fresh oyster]

i can show you quite easily that this theorem will make the summation easy to solve

then, if you need, i can prove the theorem

[fair flame]

I'm sorry but did u understand my question ?

I may have misspoken

[fresh oyster]

this one?

[fair flame]

Yeah

[fresh oyster]

yeah

[fair flame]

I'm not talking about the Basel problème

[fresh oyster]

i need to show to you that both are pi^2/8

[fair flame]

Problem*

[fresh oyster]

you use the solution in this one

unless you already accept that the summation is pi^2/8, then i just need to show you the integral

[fair flame]

I know how the arcsin integral is equal to π²/8

But not the long sum

[fresh oyster]

okay

[fresh oyster]

then we need to use this theorem to solve the summation

[fair flame]

Oh

I think I get it

Wait

[fair flame]

Yeah it's easy to use this one to solve it

You have 3/4 of π²/6 = π²/8

But without using the Basel problem can you do it ?

[fair flame]

Because I'm trying to demonstrate the Basel problem with this sum

So I cannot use the Basel problem

If I didn't demonstrate it

I'm trying to

Forget it

My question was dumb

[fresh oyster]

there are no dumb questions

[fair flame]

I think you're wasting your time

[fresh oyster]

nah

its good

then you just need to demonstrate the basel problem?

[fair flame]

Yeah I think

With a rigorous demonstration

[fresh oyster]

its quite long, but we can do it

rigorous?

[fair flame]

Not this one

[fresh oyster]

okay, then we need to get into the analysis

[fair flame]

Euler made 2 demonstrations

I tried to do the second one

[fresh oyster]

eulers proofs werent rigorous

[fresh oyster]

remember how i said this?

we have to get into the weierstrass factorization theorem for rigor

[fair flame]

Yh

[fresh oyster]

are we doing this in the real numbers or the complex numbers?

[fair flame]

Complex

It's more rigorous I think

[fresh oyster]

its also much more painful

[fair flame]

Ok

[fresh oyster]

how much complex analysis do you know?

[fair flame]

Let's do it in the real numbers

[fresh oyster]

how much real analysis do you know?

[fair flame]

What do u mean by analysis

[fresh oyster]

uh

[fair flame]

Calculus?

[fresh oyster]

do you know mathematical logic and set theory?

yeah, its using that on calculus

[fair flame]

Yeah

I think I know it

[fresh oyster]

like you know the operators $\land\lor\neg\iff\implies$

[elfin berry]

Cycadellic

[fair flame]

Holy

No

[fresh oyster]

i think you just wanted a proof for it

[fair flame]

The 2 last I know it

[fair flame]

Just send me a Wikipedia site

[fresh oyster]

i dont think you want a rigorous proof for it

[fair flame]

And I'm coming back on 5 mind

Mins

[fresh oyster]

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

[fair flame]

Which one is it

[fresh oyster]

https://proofwiki.org/wiki/Basel_Problem

[fair flame]

I mean

Which proof

There are so many

[fresh oyster]

this one has the proof

i need to look for it

[fair flame]

Which number is us it

[fresh oyster]

2

but it uses big O

[fresh oyster]

to account for this

[fair flame]

Ok

How do we do now ?

I explain you what I don't understand?

Or what

[fresh oyster]

go for it

[fair flame]

It seems pretty easy tbh

Short

Efficient

[fresh oyster]

streamline

cause we already have the sin factorization

and the mclaurin series

[fair flame]

Hmm

I don't understand the Pk-Pk-1 thing

On the whole demonstration

That is the only thing that I don't understand

We have a multiplication and then an addition

[fresh oyster]

its just the distributive property

step 1 is given

2 we subtract P_k-1

3 we bring out the kth term

4 we factor

[fair flame]

Oh yeah I get it

well, I still don't understand the O(x⁵)

[fresh oyster]

O tells us the rate of divergence

a constant O

O(2)=O(0)

says we converge

but otherwise its just looking at the leading exponent

O(5x^2+3x+1)=O(5x^2)=O(x^2)

[fair flame]

Oh yeah

[fresh oyster]

we ignor everything but the exponent

[fair flame]

Ok

[fresh oyster]

this is mandatory for the wierstrass factorization theorem

[fair flame]

But how is it equal

[fresh oyster]

O is just looking at what happens eventually

[fair flame]

I'm taking about this

[fresh oyster]

you could think of it like an abuse in notation

its technically proper

but

when i say x^2-1=O(x^2) its true because O looks towards infinity

the way we define O, we can say =

[fair flame]

O seems to be useless here

[fresh oyster]

we just need to show it meets the criteria for converge for the wierstrass theorem

thats what that step is doing

this step says: we can use the factorization theorem

then we use the factorization theorem

[fair flame]

Is the O notation mandatory for the demonstration?

[fresh oyster]

just in the sense that it says we can use a rule in our demonstration

its how we can say the euler formula for sine is correct in this case

[fair flame]

Ok

Let's say this is correct

[fresh oyster]

wait i got it backwards

lol!

sorry

[fair flame]

It's so hard to understand

[fresh oyster]

its for the telescoping series

do you know calc 2?

[fair flame]

the app ?

[fresh oyster]

calculus II

[fair flame]

what are u talking about

[fresh oyster]

i gues

[fair flame]

What's calculus II

[fresh oyster]

calculus of integrals and series

[fair flame]

oh yeah i think ik it

[fresh oyster]

do you know what a telescoping series is?

[fair flame]

Normally yes

[fresh oyster]

we use that here

[fair flame]

How old am I supposed to be to know that ?

just in case

[fresh oyster]

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

[fair flame]

Yeah i know this thing

[fresh oyster]

you know how you have to look at the cancelling terms?

hence "telescoping"

[fair flame]

yeah

[fresh oyster]

the O is just its just saying that after we cancel we have this extra remainder here

[fair flame]

ok so it acts like a constant ?

It's useless

[fresh oyster]

rather, its saying it converges

because x^5 gets big

its saying that x^5 terms are negligable

meaning it converges

does that make sense?

[fair flame]

Yeah

so we don't count it

[fresh oyster]

or rather, that the product is bigger than O(x^5)

right

[fair flame]

But we put it because it exists

[fresh oyster]

so it converges

we put it because we need to demonstrate that it converges in order for the proof to be complete

we can do the method

but we still wouldnt know for sure the telescoping works

[fair flame]

oh ok

i get it

Ok so the value of O(x^5) is really really small

in front of the other thing

[fresh oyster]

right

at least in the context of infinity

or rather, towards infinity

[fair flame]

yeah

[fresh oyster]

also

this is the riemann zeta function

ngl

zeta gets real hard real fast

[fresh oyster]

the summation on the rhs of the equation

[fair flame]

yeah ik

The reason why i'm trying to demonstrate it

[fresh oyster]

i see

thats why you wanted complex?

[fair flame]

is because tomorrow I want to show the demonstration to my math teacher

[fair flame]

Yeah

[fresh oyster]

lol

[fair flame]

i kind of want to flex

to show to my teacher that i'm a

person

that is interested by math

[fresh oyster]

gotta learn the general solution, then you gotta learn the real analytic continuation, then you gotta learn the complex analytic continuation

we love analytic continuation

but this stuff really is a headache in complex

but hey

if it was easy, there wouldnt be a million dollar problem for it

[fair flame]

1 million is too cheap

i think

[fresh oyster]

it is

[fair flame]

i would put more

[fresh oyster]

there are definitely easier ways to get 1 million

[fair flame]

yeah

and a lot

[fresh oyster]

if riemann, dirichlet, and the like couldnt prove it, it def should be more than 1 million

[fair flame]

anyways, too complicated for me

cba to do a rigorous demonstration tbh

i'm going to make a bad one

and it's ok

i'm not even supposed to know that

i was learning quadratic equations at the beginning of the year

[fresh oyster]

you need to make a lot of bad ones to learn to make good ones

[fair flame]

It will take so much time

[fresh oyster]

100% will

ive lost most of my summer breaks from school self-studying

[fair flame]

i don't think i'll do it

How old are u

[fresh oyster]

19

[fair flame]

ok

When you were 15-16

Did u make demonstrations of the basel problem to ur teacher ?

[fresh oyster]

probably not basel specifically

[fair flame]

easier or hardr

[fresh oyster]

but i was like you 15-16

trying to learn it to flex

then i just liked it

[fair flame]

which one did u do for example

if u remember

[fresh oyster]

i really liked stoke's theorem around that time

[fair flame]

lemme take a look at it

[fresh oyster]

basically: the total spin around the outside is the total spin on the inside

[fair flame]

man i'm so nervous right now

[fresh oyster]

still is one of my favorite theorems, honestly

[fair flame]

I still don't understand this sum

[fresh oyster]

amazing result

[fair flame]

it's so annoying when u don't understand something

[fresh oyster]

so, accept the basel problem

i can show you how to solve this

[fair flame]

i know how to solve it with the first method

but it's a bad method

so i'm learning the second one

But with the second one

I'm not supposed to know the answer of the sum

[fresh oyster]

i can show you one really neat trick for this

[fair flame]

Show me

Please

Wait

you were demonstrating stoke's theorem to ur math teacher when you had my age ?

are you crazy ?

[fresh oyster]

lol

[fair flame]

Tbh i don't understand one line

of the theorem

😭

[fresh oyster]

stokes'?

or this summation?

here is the one really neat trick

[fair flame]

you said that u liked stoke's theorem

I'm talking about stoke

[fresh oyster]

on both the summation and the integral, we shorthand the bounds, but actually, they should be x=a, x=b, instead of just a, b

so we can do this

oops

forgot the +1

[fair flame]

btw i have a quick question

[fresh oyster]

whats up

[fair flame]

do u like chess ?

[fresh oyster]

i tried to

[fair flame]

Are u good at chess ?

[fresh oyster]

cant study chess with math

no

absolutely not

[fair flame]

ok nvm

[fresh oyster]

it takes a lot of practice to be good at chess imo

so, stokes theorem

[fair flame]

for me "good" isn't "professional"

anyways

[fair flame]

This is Basel right ?

[fresh oyster]

this is true in general

for anything

with bounds

summation, integral, product, etc.

even derivatives

just because the bounds are =

[fair flame]

Oh

[fresh oyster]

this is actually just called u sub for integrals

[fair flame]

So it's just a formula

Right ?

[fresh oyster]

that one specifically is summations

yeah

but we can apply this thinking on any bounds

is what i meant

[fair flame]

It means f(0)+f(1)+f(2)+...

?

Or not

[fresh oyster]

they both mean f(1)+f(2)+...

cause the top sum starts at 1, whereas the bottom sum starts at 0 and does +1

[fair flame]

Oh yeah

Mb

Didn't see it

Alright

You seem to be a math genius

[fresh oyster]

no

just spent a lot of time learning

im not inherently good at it

[fair flame]

I wasted my time

[fresh oyster]

just got better with practice

how so?

[fair flame]

I prefer algebra

[fresh oyster]

get very good at algebra

get like

very good at that

very important in math

[fair flame]

Played Fortnite instead of learning maths

[fresh oyster]

lol

never too late to just start right now/very soon

[fair flame]

Math becomes boring sometimes

[fresh oyster]

yeah

you gotta find your pace

hey im blue now

[fair flame]

??

[fresh oyster]

apparently, im active

[fair flame]

Oh yeah

Congrats

I didn't even see it

[fresh oyster]

anyways, keep the balance between the tedious workload and the brilliant proofs

if you can manage that, you can learn whatever math you want

just need time

[fair flame]

Yeah

I get you

Everything takes time

[fresh oyster]

yeah

also

revisit

if you dont use it, you lose it

[fair flame]

I want to solve Riemann's problem

Imagine if I solve it

Omg

[fresh oyster]

thats good spirit

[fair flame]

I believe in you

I think you can solve it

[fresh oyster]

[fair flame]

You're too clever

[fresh oyster]

i dont think i could

i dont know the number theory

its crazy

ive looked at zeta a little but i cant

wayy too much

[fair flame]

You, talking to me is like Elon musk talking to a homeless kid

[fresh oyster]

i mean, it is true anyways, it works for the first 10^80 zeroes

[fair flame]

It seems to be true

Like Syracuse conjecture

Why am I talking about it

Idk man

[fresh oyster]

if you have no other questions, lets move to #serious-discussion

free up this channel, yk?

[fair flame]

Yeah I think I'll stop it here

Too much math for today

.close

[marsh citrus]

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[marsh citrus]

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[velvet violet]

can I get some help please?

[still temple]

use remainder theorem

so p(a) = -3

where p(x) = x^2 + ax^2 + 3x -5

[still temple]

I feel like you're conflating the a in the question with the a in the theorem you gave

If the binomial in question is x-a then the a we're talking about is the 2 from x-2

[velvet violet]

ok thanks I thought it was more than that

[still temple]

So wouldn't it be p(2)=-3 where p(x)=x^3+ax^2+3x-5?

[still temple]

i wasn't though

i should've used another variable but yeah

[velvet violet]

x^3 but doesn’t matter

[still temple]

yeah that too lol

anyway the overarching point is to use that principle

[marsh citrus]

@velvet violet Has your question been resolved?

[velvet violet]

I assume I use the same principal of the remainder theorem right?

[marsh citrus]

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[velvet violet]

.reopen

[marsh citrus]

✅

[marsh citrus]

@velvet violet Has your question been resolved?

[modest pulsar]

factor theorem is your friend

[modest pulsar]

yes

[marsh citrus]

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[lost hound]

Help

[marsh citrus]

@lost hound Has your question been resolved?

[lost hound]

<@&286206848099549185>

.close

[marsh citrus]

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[solemn plover]

having a bit of trouble over here

[solemn plover]

its about logarithmic equations

[hardy girder]

Show the question

[solemn plover]

i dont have he question wih me b

[random palm]

nice

how do we help you then?

[solemn plover]

suppose i have (ln(a))/c

how can i write this expression in terms of only 1 ln function

in other words simpliify it

[late geode]

its already simplified

[solemn plover]

nothing more yes?

thanks

[cobalt sentinel]

another way of writing it:

Doesn’t necessarily simplify it tho

[elfin berry]

Stephen

[marsh citrus]

@solemn plover Has your question been resolved?

[marsh citrus]

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[still temple]

j

[still temple]

whats that thing called where you put the limit inside the function e^lim...?

i feel like thats what i need to do here right?

[devout mauve]

continuity of e^x?

[marsh citrus]

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[marsh citrus]

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[vale yacht]

Just looked at it.
In the 6 >= 2t^2 - t
He moved it so it was
0 >= 2t^2-t-6
I moved it so it was
6-t-2t^2>=0

[vale yacht]

Aka the difference is >= or <=

now based of what does it change? or what can I do

I didn't multiply by - so why is that error occuring?

[marsh citrus]

@vale yacht Has your question been resolved?

[vale yacht]

<@&286206848099549185>

[tardy bridge]

@vale yacht I'm confused by your inequality manipulation, you should have -2t^2+t+6>=0

then multiply by -1 so 2t^2-t-6<=0
solve by factoring or quadratic formula

[vale yacht]

I did solve with quadratic

but why is the multply by -1 a must in what you said?

is it because the a is negative and I need to account for it?

[tardy bridge]

yes

[vale yacht]

like I get what you mean

but not 100% sure why it's needed

can you explain?

[tardy bridge]

if you use quadratic formula it's not needed
but you said you got the inequality 6-t-2t^2>=0

this is impossible, a and b always have opposite signs

[vale yacht]

Nevermind, forgot to change the - of the t to a positive when switching sides

I thought I missed something and looked but didn't quite catch that

Yeah no I get it now. thanks

.close

[tardy bridge]

okay great!

[marsh citrus]

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[serene kestrel]

May I ask how to get the corresponding angles and corresponding side??

[serene kestrel]

im stuck in in number 3 and 5

<@&286206848099549185>

[paper raptor]

For number 3, imagine flipping the top over the bottom then find which angles correlate, as for number 5 imagine if one triangle rotated 180º then find which angles correlate

As for sides, repeat the same exercise but find which sides correlate

[marsh citrus]

@serene kestrel Has your question been resolved?

[marsh citrus]

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[gritty sequoia]

the limit as n goes to infinity of 3^n+2^n/3^n+1 + 2^n+1 = 1/3 , how?

[gritty sequoia]

i know that 3^n/3^n+1 = 1/3 and 2^n/2^n+1=1/2 but why is the final answer 1/3

[static quarry]

please use brackets to disambiguate, do you mean: $$\frac{3^n + 2^n}{3^{n+1} + 2^{n+1}}$$

[elfin berry]

Bungo

[gritty sequoia]

yes

thats correct

[static quarry]

cool

nonrigorous intuitive explanation: 3^n grows much faster than 2^n, so for very large n, the num is dominated by 3^n and the denom is dominated by 3^(n+1)

so the fraction looks like 3^n / 3^(n+1) = 1/3

so that gives you some feel for why it should be true, now you need to make it more rigorous..

for that, try dividing the num and denom by 3^n

[gritty sequoia]

hm alr

ill work on more practice problems

ty

[static quarry]

sure, gl

[marsh citrus]

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[still temple]

Hey

[still temple]

The question asks state the principal angle

How would I find it

-42/15

[sleek lake]

you add 2pi to it until it's positive

[jovial star]

it also needs to be less then pi/2

[sleek lake]

unless it's in degrees?

wait really?

i think that would be called reference angle

[still temple]

@sleek lake Wait, but it’s only in radians we don’t use degrees

Generally it’s radians only

[sleek lake]

sure

[still temple]

Why do you minus 2 pi tho

[sleek lake]

you plus

[still temple]

How come?

Can you explain why?

Is it because the principal angle can only be positive?

Or am I incorrect

[sleek lake]

no sorry

yes it's because it can only be positive

[jovial star]

Idk
But i have seen the term base angle or principal angle used for the angle between 0 and pi/2

[sleek lake]

that's valuable info

[still temple]

What would the final answer be

I assume it cannot be negative ever

[marsh citrus]

@still temple Has your question been resolved?

[sleek lake]

the final answer is 2pi − 42/15

[still temple]

That’s not the final answer

@sleek lake

[sleek lake]

it would be mine

idk what's wrong

[still temple]

I don’t think that’s wrong, but you have to solve for it

@sleek lake what do you get when solving

[sleek lake]

i can't solve it

it would be like solving pi − 1

you can't

[still temple]

It says the answer is 18 pi over 35

[sleek lake]

oh you;re right i didn't notice it's not simplified

[still temple]

How do we even get this

[sleek lake]

42/15 = 14/5

[still temple]

How?

[sleek lake]

dude what

[still temple]

How did they get the answer as that

Can you explain

[sleek lake]

not yet

[still temple]

Huh

[sleek lake]

18pi over 35 is a different number even

i give up

[still temple]

Show your work…. @sleek lake

[sleek lake]

,calc (2pi - 42/15)

[elfin berry]

Result:

3.4831853071796

[sleek lake]

,calc (18pi /35)

[elfin berry]

Result:

1.6156762218462

[sleek lake]

my work

[still temple]

2pi-42/15 is 18pi/35? @sleek lake

[sleek lake]

no

[still temple]

Oh okay

So what do we get when we minus 2pi-42pi/15?

[sleek lake]

why do that?

[still temple]

You did this

[sleek lake]

you can do it too it should work for you

,calc

[elfin berry]

Please give me something to evaluate.
See ,help calc for usage details.

Apr 20, 2023, 4:58 AM

[still temple]

,calc (2𝞹-42𝞹/15)

[elfin berry]

The following error occured while calculating:
Error: Undefined symbol 𝞹

[still temple]

,calc (2pi-42pi/15)

[elfin berry]

Result:

-2.5132741228718

[still temple]

^

[sleek lake]

what's interesting about it?

what do you mean lol

[still temple]

You asked me to do this calculation !!!

[sleek lake]

you suggest we do −42pi/15 instead?
okay

you asked me first

[still temple]

Bro we cannot do a negative

The principal angle cannot be negative

[sleek lake]

my angles is positive

[still temple]

Yeah, but what is that in radians

Obviously you minus 360

[sleek lake]

you're not making sense

[still temple]

Bro 2 pi is 360 in degrees

You have to minus the 2 in the same symbol

You can’t minus degrees and radians together

[sleek lake]

you're talking about something completely new

[still temple]

No it’s the same topic dude

[sleek lake]

well i can't keep up

[still temple]

2 pi is 360 degrees

It’s one full rotation

[sleek lake]

i gave up
then you asked what's 2pi-42pi/15

[still temple]

Yeah

[sleek lake]

that went nowhere now you're saying we're in degrees

i can't keep up

[still temple]

It’s basically the same thing as 360-42pi/15

[sleek lake]

why do anything with 42pi/15 at all

what is it, how ddid you get it

[still temple]

Apparently you add 30pi over 15

[sleek lake]

that would be my answer

that's good, but you already said it's wrong

[still temple]

…..

🤦‍♂️

[sleek lake]

showing my work, i liked that part

,calc (30pi/15 - 42/15)

[elfin berry]

Result:

3.4831853071796

[sleek lake]

,calc (18pi /35)

[elfin berry]

Result:

1.6156762218462

[still temple]

You add it

[sleek lake]

i do

[still temple]

yeah

[sleek lake]

yeah

[still temple]

Why do we do that

[sleek lake]

to force it to be positive?

[still temple]

Why do we add 30pi over 15

[sleek lake]

i can't explain the why

you can add 30pi /15 to angles

they stay the same

[still temple]

Or is it because you don’t know ……?

What……

That doesn’t even make any sense bro….

[sleek lake]

i indeed don't know why angles stay the same when you add a full turn

[still temple]

Okay then don’t help

[sleek lake]

okay

[still temple]

We should wait for someone that knows how to do the question

To show us how to do it

Ok @surreal hull

[sleek lake]

go ahead

[still temple]

Bro wtf

Is wrong with you guys

You guys join to waste time

I’m going to bed

You guys are wasting my time

Not you since you tried @sleek lake you’re good

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[pale tiger]

oof

[pale tiger]

30pi/15 simplifies to 2pi, so it's adding a full rotation

imagine if you walked 360 degrees around a pole, you'd end up at the same spot. that's what frownyfrog was trying to say

[marsh citrus]

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[sage gale]

,,\lim_{x\to0}\frac{sin x}{x} = 1

[sage gale]

can someone help me show this without fancy methods

as in, geometrically

wait ill just look it up

.close

[marsh citrus]

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[marsh citrus]

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[errant lance]

need help , Im stuck

[bitter dawn]

Sin can't be bigger than one

[errant lance]

oh. thanks 😭

[bitter dawn]

And you can use the formula for sine

[marsh citrus]

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[opaque wolf]

bye

@still temple your channel is going to close please open a new one

[still temple]

k

[marsh citrus]

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[cunning tartan]

x^2+y^2-6x-14y+42=0 have a quick question is the discriminant -4?

[marsh citrus]

@cunning tartan Has your question been resolved?

[oblique karma]

Not sure if circle functions are meant to have discriminant?

It should only apply to quadratic expressions

Ahh yeah I wouldn't know

[tulip shale]

I believe a discriminant is only related to quadratic equations

The discriminant determines how many real zeros the quadratic equation has

Oh I am wrong the discriminant determines the type of conic section

So what you do is find b^2 - 4ac and if it is < 0 and a = c then it is a circle

Rewrite the equation in general form:
Ax^2 + Bxy + Cy^2 + Dx + Ey + F
In this case there is no xy so B = 0 and A = 1 and C = 1

[cunning tartan]

hm yea the homework asked for a discriminant

So Idk

[tulip shale]

0^2 - 4(1)(1) = -4 so yes the discriminant is -4

[cunning tartan]

yea that’s what I calculated aswell so I’m right thanks

[tulip shale]

So then since the discriminant is < 0 and B = 0 it is a circle

[cunning tartan]

mm yea

thank u

👍

.close

[marsh citrus]

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[left plaza]

Vectors are getting me confused, I need someone to lighten it up for me.

[left plaza]

Translation for #3:

Make an equation for the smallest vector using the difference or the addition of other vectors

For A it's just u - v = w

But b can it go like.
CA - CB = AB

[marsh citrus]

@left plaza Has your question been resolved?

[marsh citrus]

@left plaza Has your question been resolved?

[marsh citrus]

@left plaza Has your question been resolved?

[left plaza]

.close

[marsh citrus]

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[marsh citrus]

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[hard schooner]

hi
we know sinx ~ x (and the rest of the maclaurin series)

1. This is correct at x=0 or any x?
2. Can we say sinax~ ax?

[woeful phoenix]

1.if x approaches 0
2. If x approaches 0

[hard schooner]

ty

.close

[marsh citrus]

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[marsh citrus]

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[severe briar]

idk how to do part B or if there's a better way to do part A

[shrewd grotto]

Can someone help me with this pls

[marsh citrus]

@severe briar Has your question been resolved?

[severe briar]

😭

[marsh citrus]

@severe briar Has your question been resolved?

[severe briar]

.close

[marsh citrus]

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[severe briar]

.reopen

[marsh citrus]

✅

[blazing gazelle]

@severe briar about the part B you follow the same steps, you try to find θ

[severe briar]

the same steps?

im not entirely sure

[blazing gazelle]

@severe briar oh sorry i saw wrong

you use the half angle formula

[severe briar]

how

[blazing gazelle]

sin(θ/2) =± √((1-cosθ)/2)

[severe briar]

like ik the formula

but what would cos be

[blazing gazelle]

(cosθ)^2+(sinθ)^2=1

you find cos from here

[severe briar]

ommmgg

i did not think of that in all this time

yes

[blazing gazelle]

dont worry my friend

[severe briar]

i got around .72

,w sqrt(1-(2sqrt(3)/5)^2)

[elfin berry]

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[severe briar]

is that right

@blazing gazelle

[blazing gazelle]

is that the sin(θ/2) value?

wait

[severe briar]

thats the cos

[severe briar]

from this

cos = sqrt(1-sin^2)

sin = 2*(sqrt(3)/5)

[blazing gazelle]

its right

i calculate

that θ is 0.765

so the cos is something like 0.72

[severe briar]

yeah it confuses me a little bit though

why is it sqrt(13)/5 instead of sqrt(12)/5

[blazing gazelle]

listen you can do the other thing

go to a calculator

[severe briar]

and?

[blazing gazelle]

calcute θ from the sin

and then calculate the cos

just to see if its right nothing else

[severe briar]

,w cos(arcsin(2sqrt(3)/5))

[elfin berry]

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[blazing gazelle]

you found the cos thats right

its 0.7211

actually is

what is the problem? i cant see it

[severe briar]

its probably right

[blazing gazelle]

yes it is

[severe briar]

w, sqrt((1-sqrt(13)/5)/2)

,w w, sqrt((1-sqrt(13)/5)/2)

[elfin berry]

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[severe briar]

rip

,w sqrt((1-sqrt(13)/5)/2)

[elfin berry]

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[severe briar]

is that accurate

[blazing gazelle]

cos is 0.37?

[severe briar]

this is sin x/2

i think its right

i need to do this next

[blazing gazelle]

its right

cause

θ=43.85377861

[severe briar]

im not sure what to do here

[blazing gazelle]

are you studying engineering?

[severe briar]

im not its just a math class with applications

but id love to know more

[blazing gazelle]

for school?

[severe briar]

yep

but yea idk how to do that problem

[severe briar]

do you know any way to do this

[marsh peak]

f(theta) is defined as the difference in length of the spring from its non-compressed length and in part A they are asking you to find all theta such that the spring's length is equal to its non-compressed length

So you simply need to solve f(theta) = 0

Part B is the same except you solve f(2theta) = 0

[severe briar]

oh the way they worded it made no sense

[marsh peak]

And part C is just solving f(theta) = g(theta) (for that you may want to rewrite 1 - cos^2 as sin^2)

[severe briar]

so theta = pi/4 or 3pi/4

for a

[marsh peak]

Isn't it sin(theta) = -sqrt(2)/2 though?

[severe briar]

oops

so 5 and 7 pi over 4

[marsh peak]

Yes

[severe briar]

wait f(2theta)?

i made sin(2theta) that explains what i was doing wrong lol

[marsh peak]

Yes

Yea you are supposed to set 2sin(2theta) + sqrt(2) equal to zero

[severe briar]

yeah that makes sense i still have the same issue then

like i get

sinx*cosx = -sqrt(2)/4

how can i simplify this

[marsh peak]

Why would you expand though, just solve for 2theta in sin(2theta) equal to -sqrt(2)/2

-> 2theta is either 5pi/4 or 7pi/4

[severe briar]

oh rightr

5pi/8 or 7pi/8

[marsh peak]

Yes

Hm there may be a third solution though

I think 13pi/8 also works

[severe briar]

wait im confused

,w sin(5pi/8)

[elfin berry]

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[severe briar]

,w sin(7pi/8)

[elfin berry]

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[marsh peak]

Double it

,w sin(2*5pi/8)

[elfin berry]

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[severe briar]

confusion 100

[marsh peak]

5pi/8 makes sin(2theta) turn into -sqrt(2)/2

Not sin(theta)

[severe briar]

oh right

[marsh peak]

,w solve sin(2x) = -sqrt(2)/2 for x in [0, 2pi)

4 sols

Yup, guessed it

So, basically, since 0 < theta < 2pi

[severe briar]

i never knew wolfram alpha could do that omg ;-;

[marsh peak]

It means that 0 < 2theta < 4pi

So we have to solve for 2theta in sin(2theta) = -sqrt(2)/2 in [0, 4pi)

Which yields four solution

[marsh peak]

It can do many things

[elfin berry]

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[marsh peak]

Okay clicking "more" didn't do much in this case

[severe briar]

XD

do you have pro version?

[marsh peak]

No

You can use it on the server as well

[severe briar]

oh right

wait can you make it show steps thru the server?

[marsh peak]

Not sure

,w solve t^2 + t - 6 = 0

[elfin berry]

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[marsh peak]

Yeah no

[severe briar]

lol i ended up with this for part c

could i like complete the square or something trigonometrically

is that possible

should i do something else

[marsh peak]

Factor to get sin(x)(sin(x) - 2) = 0

Meaning sin(x) = 0 or sin(x) - 2 = 0

But note that sin(x) - 2 can never be 0

[severe briar]

oh yeah thats better

so x=0 + 2pi

or x=pi +2pi

[marsh peak]

I think they want x to be in [0, 2pi) per usual

So x = 0, pi or 2pi

[severe briar]

they actually didnt specifiy this time

in part a they said determine all values

here it says determine what times

ok i think ill turn it in

everything seems right

[marsh peak]

Yeah

[severe briar]

ok now i have a bunch of vector questions

[severe briar]

do you like vectors?

[marsh peak]

Can you tell the components of v?

[severe briar]

<-4-4,-1-3>

<-8,-4>

right?

sqrt(64+16)?

[marsh peak]

Yeah

[severe briar]

and the most obvious angle is in the third quadrant

so theta = 206

i dont think i would be able to calculate the angle right?

idk how i would do that

[marsh peak]

You don't need to calculate but it would be 180 + arctan(4/8)

[severe briar]

oh ok

ok cool that gets me the angle

is it always plus 180 or does it depend

[marsh peak]

Depends

If v was, let's say, <3, 4>, then the angle would be just arctan(4/3)

[severe briar]

oh so like on the quadrant?

so which quadrants require you to add and how much

[marsh peak]

In 2nd you add 90
In 3rd you add 180
In 4th you add 270

[severe briar]

i think this is accurate

[marsh peak]

And btw it switches between y/x and x/y from quadrant to quadrant

[severe briar]

ah ok that makes sense

wait isnt adding 360 full circle?

oh right ok

[marsh peak]

Yeah I meant 270 there

[severe briar]

oh so quadrant one y/x then quadrant 2 x/y?

[marsh peak]

,w sqrt(9^2 + 19^2)

[elfin berry]

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[marsh peak]

Yeah

,w pi + arctan(19/9) radians into degrees

[elfin berry]

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[marsh peak]

[severe briar]

rip

XD

[marsh peak]

,w (pi + arctan(19/9))* (180/pi)

[elfin berry]

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[severe briar]

i think all these are right just making sure

my best guess just by looking here is o

[marsh peak]

,w sqrt((-16)^2 + 6^2)

[elfin berry]

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[marsh peak]

,w (pi/2 + arctan(16/6)) * (180/pi)

Weird

Oh did I type in wrong order

[elfin berry]

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[marsh peak]

Yeah

[marsh peak]

It's correct

[marsh peak]

o is <-4, -3>

[severe briar]

this was wrong i think cuz of the direction slightly

[severe briar]

ohh

[severe briar]

i confused the numbers

thats alr its only one question

now over to forms

which goes first im guessing x then y

so b?

[marsh peak]

Yes

<a, b> = ai + bj

[severe briar]

so based on that this should be right as well?

[marsh peak]

Yes

[severe briar]

trigonometric form the cos is x and sin is y right

how do you find the values though just find the angles?

and whats the number in front

theres probably a formula actually lemme find it

[marsh peak]

The number in front is the magnitude

Let's see

,w (pi/2 + arctan(15/8)) * (180/pi)

[elfin berry]

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[marsh peak]

Yeah it's a