help-10

[cedar lichen]

Yes

And what's F(a)?

[timid silo]

bruh

what the fuck

is that all he did

[cedar lichen]

Yeah

[timid silo]

i thought he was taking the integral there too

like it's supposed to be equivalent to the prior statement

im like how is c gone here

thanks man

@cedar lichen curious, u a uni student?

[cedar lichen]

Senior in high school

[timid silo]

the fuck

how is ur knowledge of math so superior

[cedar lichen]

Praise me more

[timid silo]

actually ig it was a simple problem but def not hs level

[cedar lichen]

I'm self taught up to calc 3

[timid silo]

american?

[cedar lichen]

Yea

[timid silo]

i decided today i will abandon my 2nd major in maths

i also was self taught to a advanced level

but in linear algebra

@cedar lichen

actually im not sure about G(B)-G(A)

what do next

[cedar lichen]

Use the definition of G(x)

What's G(a)?

[timid silo]

oops

nvm again

[obtuse pebble]

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[fluid meadow]

for a graph transformation f(ax+b)

[fluid meadow]

do u do the stretch first or the translations

[timid silo]

in your opinion is calc 3 easier than calc 2

[wooden cipher]

Hi sr here also self taught to calc 3, most of it is easier yes

But some of it is less intuitive because its harder to visualize

[wooden cipher]

Do the stretch first

[obtuse pebble]

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[obtuse pebble]

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[drowsy girder]

Physics .

[pearl python]

so ive seen the right angle

[drowsy girder]

You know $y = x$
And
$$ y + 60° + x = ???$$

[warm shale]

Pluton

[pearl python]

im already confused

[drowsy girder]

Or just very simply
$$30 + x = \textrm{right angle}$$

[warm shale]

Pluton

[pearl python]

thanks i think i understand

yep thanks alot

i cant even get the first part

i understand 180-31 which gives me 149 but i cant tell the other 2 angles between f and the other one

<@&286206848099549185> can you assist me i dont understand the rest.

[obtuse pebble]

@pearl python Has your question been resolved?

[obtuse pebble]

@pearl python Has your question been resolved?

[obtuse pebble]

@pearl python Has your question been resolved?

[obtuse pebble]

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[civic magnet]

Can someone explain why this works

[twin sapphire]

you want an intuition or a proof?

[civic magnet]

Proof

Like an example

Bc my friend says it’s false

But from what I remember reading it’s true

[twin sapphire]

its true unless im missing something

oh nvm

[civic magnet]

Yea I thought so too, just wanted a second opinion

[twin sapphire]

i'm indeed missing something

[royal basin]

sounds like monotone convergence theorem except somewhat poorly stated

[civic magnet]

Wait so it is false

[royal basin]

er wait no

[twin sapphire]

yeah but the inequalities are not in the right way

[royal basin]

the sequence is increasing and bounded below

[twin sapphire]

yeah

[royal basin]

take a_n = n lmao

[civic magnet]

So it’s wrong?

[solar trellis]

Increasing implies bounded below so that's unnecessary

[solar trellis]

Yes

[royal basin]

the statement is false, yes

[twin sapphire]

its probably a typo though

[civic magnet]

A_n should be less than or equal to M then right?

That’s the only thing I could think of that’s off

[twin sapphire]

a_n verifies every hypothesis

but doesnt converge

[royal basin]

if the inequality said $a_n \leq M$ then yes this would be the monotone convergence thm and it would be true

[warm shale]

Ann

[civic magnet]

So yea it’s that inequality that’s flipped for some reason

Like you guys said prob a typo?

Oh wait

[royal basin]

what were the instructions

[twin sapphire]

this statement isnt even ambiguously false its obviously false so i'm pretty sure it wouldnt be a very interesting exercise

[solar trellis]

Maybe the person who wrote was just wrong? Depends on the context

[civic magnet]

A_n+1 should also have the unequaliyy flipped

[royal basin]

were they "Prove this" or were they "Determine if this is true or false"

[civic magnet]

The true false one

But it seems badly written

[royal basin]

okay then no typos

[twin sapphire]

or maybe it is but you can answer it this way properly

even if theres a typo

so w/e

[civic magnet]

Sorry what’s w/e

[twin sapphire]

what ever

[civic magnet]

Ahh okay

[solar trellis]

Genuinely seems kinda pointless to think about what the guy who wrote this meant

[civic magnet]

So basically if there isn’t a mistake

It is false

[twin sapphire]

this

so answer false by giving a counter example

[civic magnet]

They want me to prove how it’s false

something like m < inf, a_n < M, and a_n+1 < a_n

Correct?

[twin sapphire]

no

if your statement is

A=>B

a counter example should be

A and not(B)

so here its

(an+1>an and an>M)=> (an converges)

[civic magnet]

Wait so an should be less than M?

[twin sapphire]

no you should have an example where A is true but not B

here A = (an+1>an and an>M)

B=(an converges)

[civic magnet]

So A is the original statement and b is the counter

[twin sapphire]

the original statement is

"if A then B"

or equivalently "A=>B"

[civic magnet]

@twin sapphire sorry I’m slow, but can I not just flip the last 2 inequality signs to correct the statement

I think I remember seeing that’s how it worked in the textbook

[twin sapphire]

well its not asking you to correct it does it?

its asking you if its false or true

[civic magnet]

True

And asks for a counter example

But does correcting it not count as a counter example

[twin sapphire]

well no

correcting would be stating that some other statement is true

(even if this true statement is somewhat close to the original one)

[civic magnet]

Ic so what you stated above a -> b is a counter example that works?

[twin sapphire]

no

its how you should think about counter examples

because you told me something that wasnt a right counter example

so i gave you the rule

on how to do it

you have to give a counter example to a statement

[civic magnet]

Ah ic thanks

[twin sapphire]

of the form A=>B

so you want an example for which

A and not(B)

so a sequence an

such that

an>=m

and an+1>an (A)*

but does not converge (not(B))

[civic magnet]

So you mean an example that proves a and disproves b

Correct

[twin sapphire]

it doesnt prove or disprove a or b

A is true for such an example

and B isnt

A(example)=True

B(example)=False

A and B depend on your sequence

[civic magnet]

Sorry by any chance can you give me a simple example

[twin sapphire]

think about it

its really easy

a_n+1>an so it must increase at each step

and for all n, an>M

with M some number

do you have an idea?

[civic magnet]

Gonna try it in my notebook rq

[twin sapphire]

give me an increasing sequence

[civic magnet]

Sorry I was confused bc I was always under the impression a_n+1 was always less than a_n

Bc that’s what we’ve gone over the most

One sec

[obtuse pebble]

@civic magnet Has your question been resolved?

[civic magnet]

Sorry back, Does (n+1/n)^n work?

[twin sapphire]

whats N?

[civic magnet]

Sorry didn’t mean to capitalize

So now I just plug in n and show how it disproves the statement?

@twin sapphire

Oh I have to put in m somewhere

[twin sapphire]

yeah but you went for something very complicated

[civic magnet]

I did

[twin sapphire]

an= n

is sufficient

[civic magnet]

Oh fr

[twin sapphire]

n>=0 for any n

[civic magnet]

Okay thanks

[twin sapphire]

M=0

and an+1=n+1 > n =an

[civic magnet]

Ty

.close

[obtuse pebble]

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[green flower]

So for BD the gradient is 1/4 right? But for DB would it be -1/-4 or is that just trying to trick me and its always what direction the lines facing for if its negative or positive?

[royal basin]

-1/-4 and 1/4 are one and the same

[green flower]

oh

Ok thanks bye gotta finish that work

[obtuse pebble]

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[waxen condor]

can anyone help me understand this problem? why was A^4p equal to the identity matrix?

[waxen condor]

im havin trouble how to start this

[nocturne minnow]

Because $i^4 = 1$

[warm shale]

dldh06

[waxen condor]

ooohhhhhh

thanks!!

.close

[obtuse pebble]

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[alpine kraken]

is the algebra correct?

Im trying to prove whatever is next to the checkmark

[twin sapphire]

how did you do the first step?

[alpine kraken]

it was given

[twin sapphire]

i mean how do you go from the first line to the second

[alpine kraken]

oh w=dy/dx

and that's why Im here. Idk if I did the integration right

I was just doing whatever to get the proof

[tardy epoch]

You didn't, or at least didn't justify that step Benjamin is asking

Can you just show the original problem

[alpine kraken]

basically transform (6) into whatever it's asking for

@tardy epoch

[twin sapphire]

oh yeah

so thenn you just have to differentiate both sides

with respect to x

start by doing this

you'll replace dy/dx later

[alpine kraken]

with respect to x first?

that's the thing... I'm not sure on how to do that

[tardy epoch]

Use FTC

[alpine kraken]

okay I'll try

so the d/dx cancels out even the integral with the du?

[twin sapphire]

yeah kind of

whats FTC to you?

[alpine kraken]

f(b)-f(a)

Sad thing is I'm starting to forget how to do calculus

[twin sapphire]

is that it? if you dont know just ask

[alpine kraken]

that's what I recall. What is it?

[twin sapphire]

well f(b)-f(a) is noot a theorem it doesnt even state anything

its just a value

FTC makes the link between derivatives and integrals

for f differentiable over [a,b]

2 sec idk latex gotta draw

[tardy epoch]

https://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx

Scroll down. Part 1

[balmy mortar]

learn it, ur green D:

[twin sapphire]

its faster and more convenient for me :p

[obtuse pebble]

@alpine kraken Has your question been resolved?

[alpine kraken]

ngl idk anymore

I've been studying all day for my other classes and my head hurts

[twin sapphire]

okay so the idea is there

[alpine kraken]

for the left part I cant just turn dydx into w and distribute the d/dx into it?

[twin sapphire]

you have to use the product rule

for the left part

[alpine kraken]

yeah I used it but now idk what to do

[twin sapphire]

it doesnt seem like you did

[alpine kraken]

I honestly have no idea

says differentiate both sides in terms of x and thats what I tried to do with the left side

[twin sapphire]

yeah thats what u did but u made a mistake while differentiating

[alpine kraken]

oh I made a mistake

that's helpful

[alpine kraken]

welp

@twin sapphire thank you for your patience and help. Now I understand how it works

the arrangement really confused me and the concept in general

[twin sapphire]

you get how (x-1) * dy/dx differentiates?

[alpine kraken]

yeah derivative product rule

[twin sapphire]

YEAH

[alpine kraken]

but the part where I messed up was failing to recognize that I could get dy/dx when distributing d/dx into y/B

where I previously deleted it in its entirety

[twin sapphire]

its not really distributing

i know it seems like it

but its just that differentiation is linear

like d/dx shouldnt be thougth of as a number

or a value

its more like a function

or an operation

[twin sapphire]

when you think of it as a number you make mistakes and forget about product rule and chain rule

[alpine kraken]

I thought of partial differentiation and deleted y cuz I was thinking of it as a constant with differentiation in terms of x

and that too

[twin sapphire]

w/e

now you know

finish the question

[alpine kraken]

@twin sapphire I think the left side is correct but not sure about the right

[twin sapphire]

well you did nothing on the right side appart from using FTC right?

[alpine kraken]

yeah

[twin sapphire]

so its correct

but something tickles me

why is there a -0?

[alpine kraken]

I watched a video on the FTC and the derivative of the integral is the integrand(b)-integrand(a)

or maybe I didn't understand it correctly

[twin sapphire]

this is it

and now differentiate with respect to b

which is similar to our case

do it

what do you get?

[alpine kraken]

f prime of b?

I don't know

[twin sapphire]

just write it out

d/db (f(b)-f(a)) = d/db(integral)

theres almmost nothing to do

what is d/db(f(a))?

[alpine kraken]

I don't know

[alpine kraken]

is this always the case though? What if f(a) contains the same variable as f(b)? This case would only apply to this problem specifically right?

or do I simply disregard the limits of integration and see them as simply b & a, where d/dx is d/db so f(a) gets eliminated when f(b)-f(a) is the result?

[twin sapphire]

this is always true

here b is the variable

and a is a constant

like the last line

if you replace x with y

and b with x

you might like it better

[alpine kraken]

Thank you so much for putting in so much effort and time into explaining the FTC to me

[obtuse pebble]

@alpine kraken Has your question been resolved?

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[wooden pier]

How to find the anti derivative of this????

[timid silo]

u sub

you know what gives u 1/x when taking the derivative?

[wooden pier]

I think it’s something like this

But not sure with the circled part

[obtuse pebble]

@wooden pier Has your question been resolved?

[obtuse pebble]

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[split carbon]

hello, I've been given this problem and I don't know where to start. I have to solve for a, b, and c. Thanks!

$a+b+c=6,$
$a^2+b^2+c^2=14,$
$a^3+b^3+c^3=36$

[warm shale]

MePushp

[obtuse pebble]

@split carbon Has your question been resolved?

[split carbon]

<@&286206848099549185>

[versed coral]

(a+b+c)²= a²+b²+c²+ 2(ab+bc+ca)

36=14-2(ab+bc+ca)

ab+ bc+ ca= 11

now, a³+ b³+ c³-3abc= (a+b+c)(a²+b²+c²- ab- bc- ca)

abc= 6

[finite matrix]

👍
same way i was gonna say

[versed coral]

lol

then 3 2 1

[obtuse pebble]

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[obtuse pebble]

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[ionic summit]

Hey everyone, Dont really understand this; I asked for the answer and it was just like

"use the squeeze theorm"

and then gave me the answer

[ionic summit]

the only issue is I don't know really how it works at all

oh no

i might have to look at my textbook for an answer 😔 😢

i hate that thing they try to make every possible sentence as confusing and complicated with the least amount of context possible

[short spruce]

the easy thing to do here would be dividing everything by t and using common limit rules

[ionic summit]

omg hi

u helped me with trig in grade 12

[short spruce]

lol

[ionic summit]

sure, but i dont really know what sin/x and cos/x as x approaches inf is

[short spruce]

that's where the squeeze theorem comes in to play

[ionic summit]

right but idk how to

do that

[short spruce]

it's a very useful thing when evaluating some trig limits

you essentially use the fact that sin(x) and cos(x) are always between -1 and 1

so, for example, if you have $$\lim_{x\to\infty}\frac{\sin(x)}{x}$$, you could say that $$-1 \leq \sin(x) \leq 1$$ now you can manipulate every part of this equation to get to what you want... like dividing everything by x
$$-\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$$
what the squeeze theorem says is that if the leftmost side and the rightmost side converge to the same value, then the middle must also converge to that value

[warm shale]

a disappointing son

[short spruce]

i should revise, if the limit of the leftmost side and rightmost side converge to the same value*

[ionic summit]

so wait

hmmm

I see how that would be 0 when we take the limit

but one more question

what if the limit approached 0? not inf?

I know its supposed to be 1 but I dont see it in that demo

[short spruce]

do you know lhopitals yet?

[ionic summit]

I do but its not allowed

:(

[short spruce]

that's alright, lhopitals is just the proof of the lim -> 0 sin(x)/x = 1

[ionic summit]

di u mind using the squeeze theorm to demo

[short spruce]

honestly that's just a limit you need to memorize, no explanation is really needed whenever you use it

[short spruce]

what do you mean by that exactly

[ionic summit]

that as x approaches 0 sinx/x is 1

i really liked the example u made last time

[ionic summit]

this

but in here if x approached 0 wouldn't it break everything

[short spruce]

in that way, yes

proving that limit using the squeeze theorem requires a lot more steps, as far as i'm aware

uses geometry

[ionic summit]

yeah i remembe

my prof doing something with a unit circle and drawing a whole bunch of stuff on it

lol it was pretty fast so I dont exactly remember the explination

I just hope he doesen't ask prove it

then I would be sad :(

anyways what about

cosx/x as x approaches 0

[short spruce]

that limit doesn't exist

[ionic summit]

oh

well thats simple

[short spruce]

it does have one sided limits though if you have learned about those

[ionic summit]

i have

[short spruce]

but since the one sided limits aren't equal, the limit does not exist

[ionic summit]

indede

but

what about tanx/x

[short spruce]

you can break the tan(x) into sin(x)/cos(x) and use limit rules to manipulate your limit

[ionic summit]

ooooh cheeky cheeky

[short spruce]

note that you can break it into $\frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}$

[warm shale]

a disappointing son

[ionic summit]

how

wut

[short spruce]

$\tan(x)=\frac{\sin(x)}{\cos(x)}$, therefore $\frac{\tan(x)}{x} = \frac{\sin(x)}{x\cos(x)}$

[warm shale]

a disappointing son

[short spruce]

which is equivalent to this

[ionic summit]

ah right

makes sense

thank you!

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

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[broken lichen]

i dont really get why is it wrong

[short spruce]

"enter vector answers in terms of i and j"

speed sure is a vector hoo boy 💀

[broken lichen]

but for a

isnt it the magnetude

[gilded needle]

speed is a scalar

unless they misused the term

how did you get sqrt(45)?

find the x and y components of the velocity

[broken lichen]

as

[gilded needle]

yes but you don't want the magnitude of the r (position) vector

you want the magnitude of its derivative

[broken lichen]

yepyep but isnt it

-3i+(t^3)/2 j?

[broken lichen]

yep yep thats what i did

[gilded needle]

what's the derivative of t^2 - 3t + 4?

[fossil crag]

velocity can be expressed as a vector, it is just the derivative of position wrt time

[gilded needle]

i don't think they mean that you should enter speed as a vector, they mean that for those answers that are vectors (like the acceleration), that's the format to use

[short spruce]

yeah i barely read the question before i said that

[broken lichen]

2t-3?

[gilded needle]

right, and what does that give you when you plug in t=2?

[broken lichen]

1

[gilded needle]

right

now do the same thing with the j component

[broken lichen]

and the second part is 3t^2 /2

so it will be 6

[obtuse pebble]

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[humble dust]

Hey! I have the following problem: Present the following by not using any set-theoretic operations. Prove the equality of these sets.

[humble dust]

I have no idea where I should start from

[balmy mortar]

have an idea what the answer should be?

[humble dust]

not really

[balmy mortar]

well that thing there, you have some idea what it is maybe?

like descriptively

whats in that set

[humble dust]

well the x should be a real number

[balmy mortar]

yes

so an example element on the set?

[humble dust]

well that is the thing that confuses me. Can I just take any real number?

and if it is any R then how do i find the specific set

[balmy mortar]

in words

uh

you have a union over all possible x in R

ie. for all real numbers, your set contains that algebraic expression

thats what that union means

,,\bigcup_{x=\bZ}{3x}

let me give this example

[balmy mortar]

#❓how-to-get-help can u not see this is occupied. delete pls

[timid silo]

Oh sorry :/

[humble dust]

i understand the example yes

[balmy mortar]

,,\bigcup_{x=\bZ}{3x}

[balmy mortar]

sorry back

[balmy mortar]

yeah so whats in this set

[royal basin]

do you by any chance meant x in Z?

[balmy mortar]

lmao oops

,,\bigcup_{x\in\bZ}{3x}

[humble dust]

union for x in Z, so every integer times 3?

[balmy mortar]

yes every integer times 3

ie the multiples of 3

,,\bigcup_{x\in\bR}{3x}

what if i gave u this

[humble dust]

every R times 3?

[balmy mortar]

yes which is?

[humble dust]

the whole set of R?

[balmy mortar]

yes

you can think of it like that

including the original question

maybe thay will help?

,,\bigcup_{x\in\bR}{|x|}

[balmy mortar]

you can do it step by step starting with this say

[humble dust]

the absolute value of all R

the set itself is only gonna be positive right

[balmy mortar]

not quite

its not just the positives.

[humble dust]

hmm

[balmy mortar]

what if x = 0

[humble dust]

wont it be (0, inf)

[balmy mortar]

no

[humble dust]

im not sure then

[balmy mortar]

what if x = 0?

[humble dust]

still gonna be 0 no?

[balmy mortar]

here

yh |0| = 0

[balmy mortar]

but u havent included 0

[humble dust]

oh

of course

[balmy mortar]

and 0 is neither positive or negative

[humble dust]

right

so would the solution to the original problem be (-inf, 0]

[balmy mortar]

yes

do u know how to prove equality of sets?

[humble dust]

no

[balmy mortar]

u show they are a subset of each other

,,A \subseteq B

[balmy mortar]

and subset, do u know how to prove?

so u need to show $A \subseteq B$ and $B \subseteq A$ to show $A = B$

[humble dust]

huh

but im a bit confused, what would the A and B be in my context

[balmy mortar]

the q wants you to show equality of 2 sets

the 2 sets are A and B

[balmy mortar]

ie. this set and the union

[humble dust]

so (-inf, 0] subseteq {-IxI**2}

and the otherway around

[balmy mortar]

yh do u know how to show a set is a subset of another

[humble dust]

well i probably would normally but the union thing is messing with my head

[balmy mortar]

shouldnt do

to show

A subseteq B

you need to show all elements in A are also in B

So your proof should start 'Let x be in A'

and eventually end with something like 'Therefore x is in B'

[humble dust]

yeah im not really sure what to do. Let x be in A, but what next? Do I bring an example of x and show that it is equal in both the union and the set?

[balmy mortar]

no example

you need to show a general x

is also in tbe other set

to show $x \in \bigcup X_i$

[balmy mortar]

in general

you need to show x is in one of the X_i

to show this statement

[humble dust]

ive probably done this a little differently in the past so its hard for me to understand. I really have no idea how im supposed to show it

[balmy mortar]

,,\bigcup_{x\in\bR}{-|x|^2} = (-\infty, 0]

[balmy mortar]

is what u need right

[humble dust]

yeah

[balmy mortar]

,,\bigcup_{x\in\bZ}{2x+2} = evens

[balmy mortar]

ok try this example

,,evens \subseteq \bigcup_{x\in\bZ}{2x+2}

[balmy mortar]

so lets say u try this as first step ok?

So we first take n an even number

so n = 2k for some integer k

you then need to show 2k is in that union

with me?

[humble dust]

yes

[balmy mortar]

ok, so how do u think u can do that

[humble dust]

let 2k be in said union?

[balmy mortar]

no

u need to show 2k is in it

ok lets be concrete

take 4

an even number

how can u show 4 is in that union

[humble dust]

umm

substitute it in there?

[balmy mortar]

,, \bigcup_{x\in\bZ}{2x+2}

[balmy mortar]

4 is in here because

2*1+2 = 4

so letting x = 1, you see 4 is in the union

[humble dust]

huh

[balmy mortar]

2(1) + 2 = 4

{2(1) + 2} = {4}

?

,,4\in{4}={2(1)+2}\subseteq\bigcup_{x\in\bZ}{2x+2}

[balmy mortar]

3 steps of working here, see if they all make sense to you

,,4\in{4}\
{4}={2(1)+2}\
{2(1)+2}\subseteq\bigcup_{x\in\bZ}{2x+2}

[humble dust]

yes i get that

[balmy mortar]

yh so ive managed to show 4 is in that union

how about in general

if you needed to show 2k is in it

,,2k\in{2k}={???}\subseteq\bigcup_{x\in\bZ}{2x+2}

[balmy mortar]

you need to think about that middle step

you need to write 2k in the form {2x+2} for some x (that is in terms of k)

[humble dust]

2(k-1)+2

[balmy mortar]

yh.

,,2k\in{2k}={2(k-1)+2}\subseteq\bigcup_{x\in\bZ}{2x+2}

[balmy mortar]

therefore the evens are a subset of the right thing

Let $y\in\bigcup_{x\in\bZ}{2x+2}$

[balmy mortar]

Then for the other direction, you need to start like this

well letting y be in there is the same as choosing some x

in this particular union

choosing some x in Z so that 2x+2

and then u can proceed

The original question is done similarly

[humble dust]

okay

[obtuse pebble]

@humble dust Has your question been resolved?

[obtuse pebble]

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[long briar]

f(11) = 11
f(x+3) = (f(x)-1)/(f(x)+1), f(2006) = ?

[long briar]

things ive tried = noticing that f(20) = -12/10 and f(26) = -12/10, which means the next occurance of -12/10 is 6 ahead, so f(32), f(36), ...

other than that, im stuck

Actually im wrong

[obtuse pebble]

@long briar Has your question been resolved?

[sage geode]

f(23) = 11

So there is a cycle

[high lily]

isn't it f(23) = 11

[long briar]

yeah i mad a mistake but there is a cycle

annd idk how to abuse it

f(14) = 5/6
f(17) = -1/11
f(20) = -6/5
f(23) = 11

f(26) = 5/6
f(29) = -1/11
f(32) = -6/5
f(35) = 11

[royal basin]

you end up knowing that f(x+12) = f(x), no?

,calc (2006-11)/12

[warm shale]

Result:

166.25

[sage geode]

And consequently f(x + 12n) = f(x) for any integer n

[royal basin]

so 166 cycles later you will end up at f(2003) = 11

[long briar]

so the answer is 5/6 ?

what i did was basically 2006 = 14 + (n-1)3 and n = 665, and i did 665 mod 4 which is 1 so i assumed that 1 means the first position which is like 5/6

[obtuse pebble]

@long briar Has your question been resolved?

[long briar]

So is it 5/6?

[obtuse pebble]

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[dull hornet]

can anyone help me in proving trigo identities

[obtuse pebble]

@dull hornet Has your question been resolved?

[dull hornet]

<@&286206848099549185>

[obtuse pebble]

@dull hornet Has your question been resolved?

[timid silo]

which part do you want help with

and the send a clearer image

[dull hornet]

19

okay

[timid silo]

so you have to show them as equal ?

[dull hornet]

yes

prove that they’re equal

[timid silo]

the tan to cot part is easy

right ?

just write tan@ = 1/cot@ , simplify and you're done

[dull hornet]

how about the first part?

lhs

[timid silo]

expand using the identities

then divide numerator and denominator by cos(theta)cos(phi)

[timid silo]

...

[obtuse pebble]

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[dull hornet]

.reopen

[obtuse pebble]

✅

[dull hornet]

is it like this? hahaha

[timid silo]

you're multiplying, i asked to divide

[dull hornet]

im gonna multiply the reciprocal of cos(tetha)cos(phi)?

[timid silo]

yes

[dull hornet]

like this?

[timid silo]

no

and how can you do that

you can't multiply by any number like that

if there is
a+b=c

then you have to multiply as well as divide, add and subtract the same value, so that the expression remains the same

2(a+b)/2 = c

a+b + 3 -3 = c

[timid silo]

YES FINALLY

[dull hornet]

i thought that when dividing the nr and dr with cos(tetha)cos(phi), i would have a /1 on the dr hahaha lol

[obtuse pebble]

@dull hornet Has your question been resolved?

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[robust drum]

How do I calculate the curved surface are (M) and the surface area (O)?

[robust drum]

I am not really sure how to calculate this

its a pyramid

[halcyon plover]

Using trigono

[robust drum]

haven't that yet in the school

[halcyon plover]

Can you give all the information then? Or there is only this picture?

[robust drum]

one pic

[halcyon plover]

Where is the question?

[robust drum]

there are some formulas, but they will be of no use since we are calculating with angles and nothing is said about this in the formulas

[halcyon plover]

There is no height information or anything?

[robust drum]

the task only says "calculate O and M of the pyramid" more not

[halcyon plover]

Where is the O and M

[robust drum]

of height, ofc there are some ^^

[halcyon plover]

And look at the 60°

[robust drum]

M and O are areas

yes

[halcyon plover]

So you're asked the area of the triangle?

[robust drum]

bro no, of a pyramid

[halcyon plover]

Yep you can do this

Notice the 60°

What kind of triangle is it?

[robust drum]

equilateral

all sides are the same long

[halcyon plover]

Yep now you can you get the height

[robust drum]

and the angles are the same

[halcyon plover]

Wait the height doesnt matter

[robust drum]

how. show me how to put the values in a formula or something

[halcyon plover]

You know phytaghoras formula right?

[robust drum]

ofc

so n=60^2 + 60^2?

[halcyon plover]

But the height doesnt matter to find the surface area

[robust drum]

ok

[halcyon plover]

I want you to draw the net of the shape

[robust drum]

you mean this?

[halcyon plover]

Yes

Put the information you have right there

[robust drum]

ehm.. how do you mean😅

show me to understand

or a tip

[halcyon plover]

look at the a

What is the a equivalent in your problem

[robust drum]

tell me

[obtuse pebble]

@robust drum Has your question been resolved?

[robust drum]

<@&286206848099549185>

[worldly nacelle]

You can calculate surface area by multiplying base length with height

So O=a*h in this case

[robust drum]

what is h?

a is 90m

but what is h

where are the helpers?

[worldly nacelle]

Sorry had to do smth

Do you know side S?

Ok so when there is an angle of 60° and one of 90° this rule applies: longside= 2x, side on the side of the 60 percent is x and the one on the side of 30° angle is x√3

Maybe this can help https://study.com/learn/lesson/30-60-90-triangle-rules-ratio.html#:~:text=A 30-60-90 triangle is a specific type of,right angle is the hypotenuse.

This rule is valid on every triangle with these angles so you only need 1 value to calculate all sides

[obtuse pebble]

@robust drum Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[stone raptor]

ok guys let me guess if i get a bound of 0 <= y <= 1/0, which is undefined, can i interpret as infinity, and then do lim b-> inf integral b to 0 ....

[obtuse pebble]

@stone raptor Has your question been resolved?

[simple marsh]

Context?

[obtuse pebble]

@stone raptor Has your question been resolved?

[stone raptor]

context is i gotta set up equation

for surface area around y axis

so its integral 2pix * arclength function

and i was given 0 <= x <= pi/2

function is: y=tanx

however i need to set up both interms of x and y

right so when i get bounds for y its 0 <= y <= 1/0

i assume 1/0 approach infinity

so did limit as b-> infinity for that integral

would that be correct to say?

@simple marsh

[obtuse pebble]

@stone raptor Has your question been resolved?

[obtuse pebble]

@stone raptor Has your question been resolved?

[timid silo]

Help anyone? Either one

[obtuse pebble]

@stone raptor Has your question been resolved?

[forest sinew]

you should use a linear system solver for these

if you have no idea how to solve it

[forest sinew]

#❓how-to-get-help what the hell is wrong with u ppl lol

[lavish quail]

My bad gang

[obtuse pebble]

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✅

[obtuse pebble]

@stone raptor Has your question been resolved?

[obtuse pebble]

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[warped fulcrum]

Bit confused on the definition of a neighbourhood of an element. Say $X$ is a topological space. Is the neighboorhoud of $x \in X$ an open set $U$ that contains $x$? Or is it a set $U$ that contains an openset $V$ that contains $x$? My book states the first one

[warm shale]

Yeetus

[royal basin]

this might vary from author to author

[warped fulcrum]

Okay! I will assume the first one then. Thank you! Quick follow up question: I assume that given a finite set. If I take a finite subset then the completement is finite too since well infinite subset is simply not possible?

Asking because I assume that the cofinite topology has all sets open?

[obtuse pebble]

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[vernal fern]

Hello! Not looking for the answer, more just a nudge in the right direction - "Prove or give a counterexample: If p is a prime, then 6p + 1 is a prime"

[kind hawk]

just as a general note, primes are not easy. claims of the form "this and that is always prime" are rarely true

[vernal fern]

Fair enough; I started going down the list of primes, plugging and chugging, (I got to 6*17 + 1 which is prime); should I continue that route?

[kind hawk]

yes

it surprises me that it even holds for this long tbh. then again for small numbers anything can happen

[vernal fern]

facepalm 6*19+1 115.

[kind hawk]

stopped 1 step before the finish line

[vernal fern]

...would have saved myself a lot of time had I gone one further about 2 hrs ago

🤷‍♂️

Thanks though!

.close

[obtuse pebble]

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[obtuse pebble]

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[twin folio]

anyone know how to solve i?

VW vector

[obtuse pebble]

@twin folio Has your question been resolved?

[obtuse pebble]

@twin folio Has your question been resolved?

[hazy vine]

what did u try?

[obtuse pebble]

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[slender escarp]

hey i wanted to know is there was a way to express k with n given these values of both

[slender escarp]

when n=2 k =1

when n=3 k=4

when n=4 k = 18

when n=5 k=96

[novel knoll]

There is always a way

[kind hawk]

for any finite list of these pairs there is always some polynomial that fits them

that said, here we probably don't have a polynomial relationship

think factorials

[slender escarp]

i tried (n!-n+1) works for the first 2 but not the rest apparently

[final thunder]

96 = 244 18=6 * 3 4=22

[slender escarp]

huh?

[kind hawk]

\* for *

96=24*4
18=6*3
4=2*2

[covert crag]

You kinda have the right idea
Try to modify it
Hint: try to involve more factorials

[slender escarp]

oh

n!-(n-1)!

it works

[covert crag]

Yea

[slender escarp]

thanks man

.close

[obtuse pebble]

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[teal osprey]

Hello! I'm trying to write the functions 5cos(3t) - 12sin(3t) on the form

[teal osprey]

C*cos(ω(t−t0))

I have figured that it should be 13cos(3(t- ?))

but i can't figure out what t0 is. I know that it is arctan(-12/5). But I only get -1,18 which is not in the interval (π,3π/2)

Anyone?

[obtuse pebble]

@teal osprey Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[viscid wolf]

.close

[obtuse pebble]

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[mild gorge]

how do i get rid of the powers like ^5 ^6\

want to make it quadratic so i can factorise

[floral canopy]

lim m--> 1?

but the variable there is n

[mild gorge]

x->1

thats my x

sorry haha

[floral canopy]

wait I am confused which ones are x's which one's are n's and which are m's

[mild gorge]

all of them are x

just the way i write

[floral canopy]

ok

then why do you want to get rid of the higher powers?

[mild gorge]

i just want to know if there is a way to factorise them

[floral canopy]

you don't need to though

[mild gorge]

i remember my teacher said smth about dividing them with smallest x power

[mild gorge]

i knoww cause my denominator nto gonna be 0

just curiouss

[floral canopy]

yea in questions where x---> inf you can divide both sides by the highest x power, then everything cancels besides the highest powers

[mild gorge]

oh yea the inf

thank you

.close

[obtuse pebble]

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[glad inlet]

I know i have to subtract cos(4pi/5 - 3pi/5)

[tranquil arch]

so you want to calculate cos(pi/5)

you can use cos(2 *pi/5)=-cos(3 *pi/5) , to solve cos(pi/5)

[glad inlet]

where did you get 2 from

[tranquil arch]

I fixed

cos(pi-x)=-cos(x)

[glad inlet]

im a little confused

[tranquil arch]

you can refer to https://math.stackexchange.com/questions/113466/show-4-cos2-frac-pi5-2-cos-frac-pi5-1-0?noredirect=1&lq=1

[glad inlet]

that still dosent explain

how im supposed to rewrite the expression

[tranquil arch]

ok, I can explain my method

first, can you understand
$\cos \frac{2 \pi}{5} = -\cos \frac{3 \pi}{5}$

[warm shale]

秋水

[glad inlet]

yes

[tranquil arch]

then use double angel formula and triple angle formula

[glad inlet]

so 2sin x cos x

[tranquil arch]

set $\cos \frac{\pi}{5} = m$

[warm shale]

秋水

[tranquil arch]

$\cos\frac{2 \pi}{5} = 2m^2-1$

[warm shale]

秋水

[flint tartan]

bro u dont need to find the value of cos(π/5)
The question only asks u to rewrite the expression as a single trig function

cos(π/5) is the answer

[tranquil arch]

$\cos\frac{3 \pi}{5} = 4m^3-3m$

[warm shale]

秋水

[tranquil arch]

you can solve
$$2m^2-1=3m-4m^3$$
to get m

[warm shale]

秋水

[tranquil arch]

oh, you just need to write cos(pi/5)

[tranquil arch]

if you do after this,
$$2 m^2 - 1 - 3 m + 4 m^3=(1 + m) (-1 - 2 m + 4 m^2)=0$$

[warm shale]

秋水

[tranquil arch]

you can get
$$m = \cos \frac{\pi}{5} = \frac{1+\sqrt{5}}{4}$$

[warm shale]

秋水

[glad inlet]

.close

[obtuse pebble]

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[crimson hill]

hi! can someone explain this to me?

[crimson hill]

specifically, why doesn't A infinity+1 count as a part of the set?

[haughty coyote]

no element is in all the An

by definition of the infinite intersection

[crimson hill]

but that wouldn't be the definition of the set, right? like it would be incorrect to equate any set as an empty set even if it contains some elements

[civic zealot]

if the set is empty, then it is the empty set

[haughty coyote]

How did you define the infinite intersection ?

[crimson hill]

as i see it, the issue is that the thing above says there is no m+1 that is true for every set An

e.g. you can't say, idk, 4 is a part of it bc A5 wouldn't include 4

but surely all of them include Ainifnity+1

[haughty coyote]

A_inf probably doesn't exist

and if it does, it isn't part of the question anyways

you're talking about things that don't exist

[crimson hill]

huh?
why does it not exist?
this isn't part of the question...it's just a statement from a book, i'm asking a question

[haughty coyote]

what's A_inf ?

[crimson hill]

{inf, inf+1, inf+2, inf+3...} so on

[haughty coyote]

doesn't that sound wrong ?

since inf + k = inf for any finite k ?

[crimson hill]

no

i'm not sure why it would be wrong

btw n is a natural number, forgot to mention
but still so is a countable infinity
so

[haughty coyote]

first time seeing an infinite intersection/union ?

[crimson hill]

yes actually :(

i'm not sure why it would be wrong, i don't think "it seems wrong" really answers things for me

[haughty coyote]

$\bigcap_{n=1}^\infty A_n = \bigcap_{n \in \mathbb{N}} A_n = {a, \forall n \in \mathbb{N}, a \in A_n}$

[warm shale]

themateo713

[crimson hill]

well...yes, that's what the set is, but i still don't know why this wouldn't include infinity or inf+1, etc etc

sorry if i'm being dumb, i just genuinely don't get it

[haughty coyote]

as an index ?

infinity is always just a notation for a limit

remember that

[crimson hill]

not necessarily

[crimson hill]

why can it not just mean an actual countable infinity?

[haughty coyote]

it does

but it's like in a series: the sum up to infinity means the limit as n goes to infinity of the sum up to n

here it's the same thing, but you can't use the notion of limits

but it's the same thing of it having to hold true for all n as you let n go to infinity

[crimson hill]

so infinity+n would be considered...not an option?

[haughty coyote]

the 2nd equality actually defines the intersection of a family indexed by another set

what would inf+n be ?

inf+n = inf

[crimson hill]

i get what you're saying now partially, but i still don't understand why the limit has to be infinity and not inifnity+n
which...i guess would approach inifnity

ohhh

i kind of

get it now i think

so inf+n (inf as n inclusive) can only be used like
conceptually? but not as elements to a set

[haughty coyote]

for instance, what's $\bigcap_{n=1}^\infty [-\frac1n, 1+\frac1n]$ ?

[warm shale]

themateo713

[crimson hill]

....

give me a second

....0?

[crimson hill]

maybe 0 ? but
even though i understnad the question i asked
i still dont fully get why this would be 0 intuitively

[haughty coyote]

it isn't

[crimson hill]

oh

[haughty coyote]

if I said [-1/n, 1/n] it would be

[crimson hill]

wait

[haughty coyote]

but I said [-1/n, 1+1/n]

[crimson hill]

everything
under 1/n

positive

OHHH

wait 3.0

wait maybe i was right
[0, 1] ?

[crimson hill]

is that right ?

[haughty coyote]

yes it's [0, 1]

does that make the notation a bit clearer ?

[crimson hill]

okay
again, i get it but i dont intuitively get it
would you mind explaining why it's [0,1] ?
(so sorry)

[crimson hill]

it does :)

[haughty coyote]

all the intervals contain [0, 1] so this interval is included in the intersection

if I take x < 0: there exists n big enough (here n = -floor(1/x) - 1) such that x < -1/n, so it isn't included in the intersection because it isn't in the nth interval.
If I take x > 1: same argument: I find n such that x > 1+1/n, meaning x isn't in the intersection because it isn't in the nth interval

therefore [0, 1] contains the intersection, so by double inclusion the intersection = [0, 1]

[crimson hill]

(i promise ill get it soon-ish) but any n larger than 0 would make the first term (-(1/n) smaller than 0 and the latter larger than 1

ill read your explanation, maybe it'll clear it up

[haughty coyote]

yes

therefore [0, 1] is in the nth interval, for all n. So it's in all the intervals, so it's in their intersection

[crimson hill]

what does floor mean?

[haughty coyote]

the only integer n such that n <= x < n + 1 is denoted floor(x)

$\floor{x}$

[warm shale]

themateo713

[crimson hill]

so x would always necessarily be n ?

[haughty coyote]

the floor function is useful to talk about things that aren't integers

so no, in general x isn't equal to n

actually, you have the equivalence x is an integer <=> x = floor(x)

[crimson hill]

thats so cool

still trying to comprehend your explanation, sorry for the wait