help-19

[signal yacht]

/nsrs

[brittle plinth]

here they proved e -> f and f -> a

my first actual concern is proving a -> b

because if I suppose x ∈ A
-> x ∈ B, can I go -> x ∈ (A ∩ B) ?

[errant dew]

$(x \in A \implies x \in B ) \implies (x \in A \implies x \in A \cap B)$

[clever fjord]

BBMaths

[odd edge]

@brittle plinth Has your question been resolved?

[brittle plinth]

I mean yeah that's my idea

but is it valid to conclude that

Because x ∈ (A ∩ B) <-> [(x ∈ A) Λ (x ∈ B)]

[odd edge]

@brittle plinth Has your question been resolved?

[brittle plinth]

.close

[odd edge]

Channel closed

Closed by @brittle plinth

Use .reopen if this was a mistake.

[brittle plinth]

will ask this another time

[jagged halo]

Hello

I got question

[errant dew]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[errant dew]

It hasn’t gone back into available yet so please don’t type here

Try #help-14

[brittle plinth]

why isnt this closed yet ffs

[odd edge]

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[timber river]

how do you do c

f'(x) = -3k / (x-2k)^2

[wooden python]

do you know how to tell whether a function is increasing or decreasing

[timber river]

yeah

it's increasing if it's derivative (gradient) > 0

[wooden python]

cool

[timber river]

and vice versa

[wooden python]

can you tell whether the expression $-\frac{3k}{(x-2k)^2}$ is positive or negative

[clever fjord]

Ann

[timber river]

denominator is positive

but we dont know k

that's the problem

[wooden python]

...the question said, putting the key word in bold for increased visibility...

[signal yacht]

or 0.

but yeah

[timber river]

i didn't see that thanks guys

.close

[odd edge]

Channel closed

Closed by @timber river

Use .reopen if this was a mistake.

[devout hedge]

this channel is unfortunately still classified as occupied.

[mystic saffron]

Oh my gos

Gosh

Where am I supposed to ask for help then

[devout hedge]

#help-45 is available right now.

[odd edge]

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[floral cargo]

question 5 pls

[fluid bluff]

又来了

[floral cargo]

是的

又是我

[fluid bluff]

我看看

[distant flax]

积分

[floral cargo]

主要是我连题目都没看懂

[tranquil basin]

would the limit approach infinity?

[floral cargo]

看来是时候转专业了

[fluid bluff]

呵呵

[floral cargo]

actually i Have no clue

[errant dew]

Did you solve (3)

[fluid bluff]

$C_{n}^k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$

[floral cargo]

yes

[errant dew]

Was it 1

[tranquil basin]

given that the top rises faster than the bottom?

[floral cargo]

idk I think so

[clever fjord]

Roy

[fluid bluff]

bruh what how do i take ln of this

[shell haven]

Rip

[fluid bluff]

what about trying to calculate some values

[distant flax]

Stirling approx? @floral cargo

[floral cargo]

Like how?

[fluid bluff]

$x_{1} = 0, x_{2} = ln(2), x_{3} = 2ln(\frac{3}{2})$

[tranquil basin]

isnt x1 = 1

[distant flax]

note that it is symmetric and largest terms are close to the middle

[fluid bluff]

no

[floral cargo]

Why no

[distant flax]

ln(nCk) approx n*H(k/n)

[tranquil basin]

are we still doing problem 5

[floral cargo]

I also thought that x1 is 1 hahahahahahah

Yes

[fluid bluff]

yea

thats what im doing

[floral cargo]

How is there a ln

[tranquil basin]

oh i was looking at 3 (5) not 5

[fluid bluff]

ln(1) = 0

[tranquil basin]

my bad

[fluid bluff]

xD

[floral cargo]

Oh

I notice the ln

Hahahahahahha

[distant flax]

you get the binary entropy function in natural log

-plnp-(1-p)ln(1-p), where p=k/n
integrate from 0 to 1 => 1/2

@floral cargo

[floral cargo]

I have no idea what a binary entropy function is

[fluid bluff]

$x_{1} = 0, x_{2} = ln(2), x_{3} = 2ln(3), x_{4} = 2ln(4) + ln(6)$

[clever fjord]

Roy

[fluid bluff]

oh wait

thats pascal triangle

[floral cargo]

Yep i know it’s pascal triangle

But how to approach and solve hahahahahah

Totally no clue

[distant flax]

from $ln n!\approx n\ln n-n\Rightarrow\ln(nCk)\approx -n(p\ln(p)+(1-p)\ln(1-p)),p=k/n$

there's an error term in O(nln n) but it doesnt matter as you approach inf

[floral cargo]

This is the only way to solve it? Is there another method which does not require this binary entropy function

[distant flax]

i dont know of another way

[clever fjord]

Estelle

[floral cargo]

@fluid bluff Any clue?

[fluid bluff]

Estelle might be right

[distant flax]

maybe try simplifying the sum first using properties of ln

[fluid bluff]

yea

[distant flax]

wait if you move the sum into ln and multiply out n! you can separate the terms

then you have $(n+1)\ln(n!)-2\Sigma_{k=1}^n\ln(k!)$

[clever fjord]

Estelle

[odd edge]

@floral cargo Has your question been resolved?

[dreamy bear]

[
\sum_{k=1}^{n}\ln(k!) ;=; \ln(1) ;+; (\ln(1)+\ln(2)) ;+; (\ln(1)+\ln(2)+\ln(3)) ;+; \cdots ;
]

[clever fjord]

Warg

[fluid bluff]

yea?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[dreamy bear]

.reopen

[fluid bluff]

.reopen

[dreamy bear]

i have trouble typing latex

[fluid bluff]

its not hard

[odd edge]

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[fluid bluff]

We are not finished with the previous question

[dreamy bear]

but this is equal to summation (n+1-j) ln j

all that is left is to approximate summation j ln j

[fluid bluff]

put $ before and after

[dreamy bear]

the rest is stirling approximation

[odd edge]

@fluid bluff Has your question been resolved?

[odd edge]

Channel closed

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[spark shadow]

can someone please explain me inflexion and stationary points?

[spark shadow]

which are when f'(x)=0? and what is the other one

and i googled and if i understand correctly other one is when f''(x)=0?

[errant dew]

Inflexion is when f’’(x) changes sign

[spark shadow]

alright

and stationary?

f'(x)=0?

[errant dew]

It might hit f’’(x)=0 but go back up not inflexion

[spark shadow]

ohhh alright

[errant dew]

That is f’(x)=0

[spark shadow]

but how do i know if it changes the sign

i mean

i put x=0 and f''(0) = 0

[errant dew]

I mean I suppose you could look at 3rd derivative

Or you could see if nearby numbers are different signs

[spark shadow]

so here answer in second question will be b?

cause all -1 0 and 1 when put in f''(x) =0

so it doesnt change the sign

oh or wait

if we do f''(-1) it's 2, f''(0) is 0 and f''(1) is 2

how do we know if these are inflexion points or not

[errant dew]

If they change in sign, so here they are both positive so it isn’t an inflexion point

But I would choose numbers really close to 0 since something about the function can change between -1 and 1

[weary pelican]

oops

[amber schooner]

lol

[spark shadow]

but yeah that still won't change anything

so the answer is b alright

thanks

[errant dew]

It’s bad practice to test nearby numbers though since it might change

[spark shadow]

but how do i check if so

[errant dew]

Stick to derivatives because they are infinitesimal

[spark shadow]

i mean how to prove it

[errant dew]

^

[spark shadow]

second derivative is 5p^4-3p^2

[errant dew]

So when that’s 0 that’s a possible inflexion point

Except neither of those are inflexion i think

[spark shadow]

yeah if we put 0 it gives 0, but if we put 1 or -1 it gives 2

!close

.close

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Channel closed

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fallow geode]

i have a few questions. So first when set A is a subset of set B, every element of A is also B right?

[atomic hornet]

yeah? elements in A are basically the copy of B

[sturdy cape]

That's literally the definition of a subset

$A \subset B \iff \forall x ( x \in A \Rightarrow x \in B)$

[clever fjord]

Waes (Wires)

[fallow geode]

okok and then when set A intersections set B it means that there is atleast one element that is contained in set A and set B that is common right?

[sturdy cape]

"intersection" is an operator; so this doesn't really make much sense

That's like saying "when 7 is multiplied by 8"

[distant flax]

RHS does not imply LHS here—you need subseteq

[sturdy cape]

Depends on the author

[fallow geode]

like if A = { 1, 2, 3,4}
B = {1, ,5, 6, 7}
A intersects B?

[sturdy cape]

This notation isn't standardised everywhere

[fallow geode]

i just want to know the general idea if thats true

[sturdy cape]

Oh, are you referring to e.g. this?

[fallow geode]

yeah

[sturdy cape]

Then yes

Check your course's notes on how it's described, but I'd allow it

[fallow geode]

okay, so when A n B = the empty set, it means that they shared a common element, which is the empty set?

[sturdy cape]

No, you're saying the intersection is empty

[fallow geode]

and all sets have empty sets in it

[distant flax]

if they shared the empty set it would be {emptyset}

[sturdy cape]

A n B is a SET

You can't describe this equal to an element

the objects {} the emptyset and { {} } the set containing only the emptyset are two different objects

[fallow geode]

oh they didnt dont shareanything in cmmon it means?

[sturdy cape]

yes

Because A n B is a set

[distant flax]

this is not true, but all sets have the empty set as a subset

[sturdy cape]

So for instance here, A n B is not equal to 1

It's equal to the set {1}

When we say "x is in A", we mean "x is an element of A", and not "x is a subset of A"

[fallow geode]

im just confused lol doesnt that just mean the empty set is contained in all sets?

[distant flax]

$x\in A\not\Leftrightarrow x\subseteq A$

[clever fjord]

Estelle

[fallow geode]

left is x is inside set A? and right is x is contained in set A?

wait isnt that the same thing

[distant flax]

"in" <-> "is an element of"
"is a subset of" <-> "has all its elements contained in"

[uneven lodge]

{1, 2} ⊆ {1, 2, 3}, but {1, 2} ∉ {1, 2, 3}. You do have 1 ∈ {1, 2, 3} and 2 ∈ {1, 2, 3} though

[distant flax]

$\emptyset\not\in {1}, \emptyset\subseteq{1}$

[clever fjord]

Estelle

[distant flax]

a subset is in the power set but not necessarily the set itself

[sturdy cape]

A strict subset, yes

If you're using $\subseteq$, then it may be the set itself; indeed for any set A, we have $A \subseteq A$

[clever fjord]

Waes (Wires)

[sturdy cape]

If A is a subset of B, then A is in i.e. an element of the powerset of B

$A \subseteq B \iff A \in \mathcal{P} (B)$

[clever fjord]

Waes (Wires)

[fallow geode]

i havent learned powerset im sorry...

[fallow geode]

im still trying to think this lol

[sturdy cape]

Well, this is literally the definition of "powerset" - the set of all subsets

So if A = {1, 2}, then P(A) = { {}, {1}, {2}, {1,2} }, for example

[sturdy cape]

Think of this like sets are boxes

You have a box that contains some apples with the numbers 1, 2, and 3

[fallow geode]

so i understand the first half of the sentence, but not the rest

[sturdy cape]

Each of those numbers is in that box

So the 1-apple is in that box

Same as the 2-apple

However, if I were to talk about the set {1,2}, I'd mean a box with a 1-apple and a 2-apple

[fallow geode]

is just the number 1 a set? or is {1} a set?

[sturdy cape]

There's no extra box inside our {1,2,3}-box that contains that 1- and 2-apple

[outer glacier]

I think it means a set can not belong to another set but an element can belng to a particular set

[sturdy cape]

{1} is a set (it's a "box" with just a single apple, labelled "1")

1 is that apple, i.e. it's not a set

[fallow geode]

1 is just an element of the set?

[sturdy cape]

yes

$1 \in {1}$

[clever fjord]

Waes (Wires)

[sturdy cape]

Which should be obvious if we translate to boxes

[outer glacier]

Powerset much simplified is all the possible subsets in a set i think

[distant flax]

is just the number 1 a set?
haha... hahahahahaha.......
in any case 1 is not {1}

[sturdy cape]

"the apple labelled 1 is inside the box with an apple labelled 1"

[fallow geode]

{1, 2} ∉ {1, 2, 3} is because {1, 2} is a set itself not an element of the set {1, 2, 3}

[outer glacier]

Yea and {1,2} is subset of {1,2,3}

[distant flax]

objects behaving like the apples labelled 0, 1, 2, ... can be constructed via n U {n} @fallow geode ignore this

[sturdy cape]

~~Estelle come on ~~

[fallow geode]

im sorry lol im looking on google and then here and then my hw and im still thinking

okay so

[sturdy cape]

Likely because there're two types of "set theory" which might be confusing your search results

The type you need is "naive set theory", if that helps narrow them down properly

[fallow geode]

If A and B are sets,
and A is contained in B,
then A and B are not disjoint
disprove this...

oops

so not disjoint means A n B does not equal to the empty set

it means that theres atleast one element that they share common

[distant flax]

the definition of "contains" seems to vary by author...

[fallow geode]

its the C with the line

xD

[woeful briar]

\subseteq

[fallow geode]

idk how to put it here

yeahh

[distant flax]

it does not imply either is a subset of the other

[fallow geode]

just one direction right?

[distant flax]

consider {1, 2} and {2, 3}

yepp

[fallow geode]

so i need to find a counterexample where they are disjoint or prove its negation entirely

[distant flax]

$A\subseteq B\rightarrow A\cap B\neq\emptyset$

[clever fjord]

Estelle

[distant flax]

where A, B are nonempty

[fallow geode]

the right hand side means that the two sets share at least an element in common

[amber veldt]

it seems to be an ambiguous word in gneeral

whether it means contains as a subset or contains as an element

[distant flax]

empty sets are so annoying when thinking about mathematical statements :(

[fallow geode]

wait this is what i need to prove i just realize

[distant flax]

this is not true in general 🙁
you can have $A=\emptyset,B=\emptyset$

[clever fjord]

Estelle

[amber veldt]

so there is a term

"A intersects B" to mean "A and B have a non empty intersection", but it's only used as a sort of shorthand to make theorems easier to say in the context of a long proof or a long theorem, in my experience

[fallow geode]

WAIT i keep getting things flipped

im so sorry

[amber veldt]

what is the exercise we're doing

[distant flax]

idk

[amber veldt]

iloveanimefivefivefivefive what is the assignment?

[fallow geode]

thats the first question.

..

[distant flax]

oh oops i gave the answer

[fallow geode]

i just want to understand how to do it...

[amber veldt]

so the easiest set to deal with in making counterexamples is the empty set, it should be the first set to consider

[distant flax]

do you understand the definition of subseteq now

[amber veldt]

because it's so often the counterexample

[fallow geode]

mostly i think..

so i need to prove that A and B are disjoint

by giving a counterexample

[outer glacier]

But for A to be a subset of B dont they need to have an “element” in common
I dont think empty set is an element

[amber veldt]

@fallow geode does that make sense?

[fallow geode]

like A is the empty set it self..?

[amber veldt]

try it, maybe thats a counterexample

[distant flax]

when is it true that $\forall x(x\in A\rightarrow x\in B)$ but $A\cap B=\emptyset$

[clever fjord]

Estelle

[amber veldt]

remember for a counter example you just have to find ONE exception, if letting A = {} gives you one exception, you found a counterexample

what sets is the empty set a subset of, do you know?

[fallow geode]

empty set is the subset of all sets right?

[amber veldt]

yes!

[fallow geode]

i mean

[distant flax]

also i disagree pedagogically with ever putting "disprove" in a hw question—"prove or disprove" makes students think more

[amber veldt]

absolutely correct

[fallow geode]

empty set is contained in every set in the universe

[amber veldt]

I agree with you but we cant change that

okay, let's try this

what sets does the empty set have empty intersection with?

[distant flax]

@fallow geode all elements in A must also be in B, but intersection is empty

[amber veldt]

don't give awway the answer

[distant flax]

it's in the qn 😭 😭

[amber veldt]

oh my bad!

🤐

i was wrong to correct you

[fallow geode]

like what what set A which is the empty set have in common with?

[amber veldt]

have NOTHING in common with

the intersection of two sets is the set of everything they have in common

what set has nothing in common with the empty set? or what kinds of set?

[fallow geode]

nothing right xD cause empty set is a subset of all sets

[amber veldt]

mmm I hear what your intuition is saying, but you're wrong, because

for a set to have something in common with the empty set

the empty set has to have elements

which it does not

it's like saying

[distant flax]

all elements in the empty set are flying blue unicorns
^ this is a (vacuously) true statement btw

[amber veldt]

how many pokemon cards are in both my deck and your deck? i dont care about your deck because it doesnt matter, my deck is empty, i dont play pokemon tcg

so my pokemon card deck has nothing in common with any other pokemon card deck

[sturdy cape]

-# Really channelling TomRocksMaths there aren't you

[amber veldt]

who

[sturdy cape]

er... the maths professor from Oxford who has a YT channel

[amber veldt]

oh, sounds like a compliment

[sturdy cape]

He's got a lotta tattoos, and also is a pokemon fan

-# he was also one of my tutors while I was at uni so there's my hall of fame

[fallow geode]

so like what set(s) shares nothing in its elements with the empty set? is it just every set that isnt the empty set..?

[amber veldt]

it's actually every set, including the empty set

now the fact that the empty set has empty intersection with itself is counter intuitive, because

how can it be disjoint with itself!??

the answer is

[distant flax]

convince yourself that the empty set shares nothing with the empty set

[amber veldt]

because it has no elements in common with itself, so, by definition, it has empty intersection

because it has no elements, period

[distant flax]

all elements in the empty set are elements of the set {cats i petted yesterday}

[amber veldt]

:(

[distant flax]

which by the way has a cardinality of 1, so the reverse inclusion does not hold

[amber veldt]

oh you mean "are also"

[fallow geode]

uhmmm so by the definition of empty intersection, and by the definition of the properties of empty sets..
empty set shares nothing in common with itself because empty set has no element and empty intersection wants two sets that has something common with its element

[amber veldt]

"empty set is not an element of empty set" is neither here nor there

[distant flax]

hskshdjsjd i hate set theory in english

[amber veldt]

so you can delete that phrase

[amber veldt]

“It depends on what the meaning of the word ‘is’ is. If the—if he—if ‘is’ means is and never has been, that is not—that is one thing. If it means there is none, that was a completely true statement." - Bill Clinton

[distant flax]

intersection looks at elements in both sets

[fallow geode]

uhmmm okay...
so A = empty set
B = {1, 2, 3, 4}

uhmmm and then

[distant flax]

whats their intersection

[amber veldt]

then A subseteq B, because _____
then A intersection B = _____ because_____

[fallow geode]

nothing... right..? because empty set has no elements

i think??

[amber veldt]

well what's "nothing" in the world of sets

[fallow geode]

uhmm wait

[amber veldt]

you're very close to finished!

[fluid bluff]

hi

[amber veldt]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fluid bluff]

ik

im here to help

is this still the original question or is this another one

[fallow geode]

they dont share empty set in common because "they dont share empty set in common" is the same thing as A n B is not equal to empty set

but wait i need to prove that A n B is equal to the empty set..?

but intersection is looking for an element common to both sets, but A has no element because it's the empty set and empty set has no elements... uhmmm

so.. their intersection is nothing.. wut

[amber veldt]

so it may be counterintuitive to say it

the intersection of empty set with any set is the empty set

because it's the set of all elements the two sets have in common, and there are none

so it's the set with no elements in it

which is the empty set

the reason why your intuition doesn't like that, probably, is

because how can the empty set be disjoint from itself? it's the same as itself!

Do you know what axioms are? @fallow geode

[fallow geode]

i think i get it....

[vernal yacht]

This one is still in use, might consider asking at https://discord.com/channels/268882317391429632/905286165316403271

[fallow geode]

its so uhm counterintitive..

its statements thats fundmental

[amber veldt]

yes a lot of things about the empty set are counterintuitive

do you know what axioms are?

[fallow geode]

and mathmatians take it for granted? i think its what my teacherr said

[fluid bluff]

i said im here to help bruhhhhh

[fallow geode]

something with this.. whatever my teacher said

or wrote here

[amber veldt]

mathematicians agree on a set of rules they "take for granted" in the sense they agree they are true but can't be proven because they are too fundamental, they are used to prove other things , these statements are called axioms

[fallow geode]

YES omg the way you said it makes so much sense

or i mean clear

[amber veldt]

this is why mathematicians made rules for set theory, because of paradoxes like this

[distant flax]

naive set theory allows this, so we need axiomatic set theory :(

[amber veldt]

the most important axiom, which will help you deal with the counterintuitiveness of the empty set, is this

Axiom of Extension: Two sets are equal if and only if they have every element in common

So what? this means

If you prove two sets have no elements at ALL, theyre both the empty set

and the empty set is disjoint with itself because

whats their intersection? what property does the intersection set have?

it has no elements in it

any way you create a set with no elements in it

you're creating the empty set, period

does that help?

[fallow geode]

wait wasnt this me trying to figure out why the intersection is the empty set ;-;

cause it didnt make sense xD

[amber veldt]

yes im trying to give you a balm for your burn

you have been scalded by the dangerously rational waters of logic

[fallow geode]

i want to go back to calculus 2 i want to go back to last year.. uhm but this fun..?

omg i need to write this proof

[amber veldt]

by the way

when Is aid "try the empty set", you oinly need one counter example

I expected you to make A and B BOTH empty. turns out you didnt need to

but I thought that was the simplest approach

btw, a tangent

but want to hear something interesting about the history of logic and Russel's paradox?

[fallow geode]

if they are both empty set then the intersection is the empty set not because the common element is the empty set but because "there are none"

sure xD

[distant flax]

whether the empty set is an element is irrelevant

the intersection of {1}, {1,2} is {1}, not 1

[amber veldt]

Bertrand Russell and Alfred Whitehead thought that the main reason Russell's Paradox came up was because set theory was allowed to do self-reference, was allowed to talk about itself in the language of set theory

so Russel and Whitehead developed a "theory of types" where all objects had hierarchies, you start with basic elements called urelements, and you can have sets of urelements, and you can have sets of sets of urelements, and you can have sets of sets of sets or urelements

but you can't just have sets of everything, in the theory of types

around 20 years later a mathematician named Godel proved that any system of math strong enough to prove the laws of arithmetic has to be able to talk about itself in a self-referential way, which means that Russel and Whitehead's approach was an unwinnable contest

the theory of types fell out of use after that as they realized it was pointless

now the usual approach to axiomatic set theory is to make a list of rules of what kinds of sets you can build with other sets, with no distinction between sets of sets vs sets of sets of sets etc

anyway, that's trivia

[amber veldt]

this is important

[fallow geode]

you are better at storytelling than my instructor

this is interesting to listen because it made sense

[amber veldt]

I'm also a game master for tabletop rpgs lol

but thank you :)

[fallow geode]

WHAT

1 is an element and {1} is a set right...

[amber veldt]

intersection is a rule you use to build a set out of two existing sets

[fallow geode]

like i mean 1 is an element just an element, not an element of something..

[amber veldt]

1 is an element of every set that contains 1

[distant flax]

we will not discuss ${\emptyset, {\emptyset}}$

[clever fjord]

Estelle

[amber veldt]

lol, not yet

[fallow geode]

wait the intersection is looking for elements

i thought its just 1 lul

[amber veldt]

$$A \cap B = { x : x \in A \text{ and } x \in B}$$

[clever fjord]

gfauxpas

[fallow geode]

like if it's {1} and {{1}, {2}}, the intersection is {1}

if its like {1} and {1, 2} it's 1

wait im not thinking in words

[amber veldt]

$${1} \cap {1,2} = { x : x \in {1} \text{ and } x \in {1,2}}$$

[clever fjord]

gfauxpas

[distant flax]

it's $\emptyset$ and ${1}$

[clever fjord]

Estelle

[amber veldt]

axiom of extension again iloveanime

{{1}} does not have "1" inside it

the only thing that defines a set is what elements are in it

{{1}} has one element inside it, and that element is {1}

{1} has one element inside it, and that element is 1

[fallow geode]

and {1} is 1 ?

[amber veldt]

not "is", but "has"

[fallow geode]

oh yeah xD

omg thank you so much you two

i think i finally understand something in this 3 weeks

other than truth tables

uhm also..

gfauxpas can i message you directly? 😭

i do not think i can find a tutor thats this helpful uhm yeah

[amber veldt]

yes

[fallow geode]

okok

.close

[odd edge]

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[fallow geode]

help i cant.. its not accept wgwe'w

accepting*

[odd edge]

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[tired obsidian]

i need help with point b)

i have to verify first if x(t+T) = x(t) ?

can i write $x(t+T) = A\cos(\omega_0 (t+T) + \theta)$

[clever fjord]

alee

alee

[tired obsidian]

so $A\cos(\omega_0m + \theta) = x(t)?$

[clever fjord]

alee

[tired obsidian]

so its periodic

i have to verify if its energy content per period is finite , yes ?

if yes its a power signal

$E_T = \int^{t_0 + T}_{t_0} |x(t)|^2 \mathrm dt$

[clever fjord]

alee

[tired obsidian]

then $P = \frac{E_T}{T}$

[clever fjord]

alee

[tired obsidian]

and $T = \frac{2\pi}{\omega_0}$

[clever fjord]

alee

[tired obsidian]

<@&286206848099549185>

[clever fjord]

alee

[errant dew]

Isn’t it given as a variable

[tired obsidian]

ok

[tired obsidian]

here

can i choose t0 = 0 ?

$E_T = \int^{\frac{2\pi}{\omega_0}}_0 |A\cos(\omega_0t+\theta)|^2 \mathrm dt$

right ?

[clever fjord]

alee

[tired obsidian]

$\frac{A^2}{\omega_0} \int^{2\pi + \theta}_0 \cos^2(u) \mathrm du$

[clever fjord]

alee

[tired obsidian]

right ?

then $\cos^2(u) = \frac{1}{2} (1+ \cos(2u))$

[clever fjord]

alee

[tired obsidian]

$\frac{A^2}{2\omega_0} \int^{2\pi + \theta}_0 1 + \cos(2u) \mathrm du$

[clever fjord]

alee

[tired obsidian]

$\frac{A^2}{2\omega_0} \Big[2\pi + \theta + \frac{1}{2} \sin(4\pi + 2 \theta)\Big]$

[clever fjord]

alee

[tired obsidian]

and

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

[clever fjord]

alee

[tired obsidian]

$= \frac{A^2}{2\omega_0} \Big[2\pi + \theta + \sin\theta\cos\theta\Big]$

[clever fjord]

alee

[tidal matrix]

this only gives you x(m), and m doesnt equal t :( i think you would need to expand the addition identities out if you wanted to prove algebraically.
or you could say that cos is a periodic function, so x(t) will be too because its just a bunch of scalar multiplications and additions applied to cos (phase shifts & stretches)

[tired obsidian]

i read this

[tidal matrix]

yeah but $x(t+mT)$ would look like $$A\cos(\omega_0(t+mT) + \theta),$$ whereas you just had $$A\cos(\omega_0m + \theta)$$

[clever fjord]

χασιβ ♥

[tired obsidian]

oh wait

[tidal matrix]

i could do the same thing with a straight (not horizontal) line, even though it isn't periodic

[tired obsidian]

$A\cos(\omega_0 t+\theta+\omega_0 T)=A\cos(\omega_0 t+\theta)$

[clever fjord]

alee

[tired obsidian]

i have to find when this is true

$\alpha = \omega_0 + \theta$

[tidal matrix]

ye :)

[clever fjord]

alee

[tired obsidian]

$\cos(\alpha + \omega_0 T) = \cos(\alpha)$

[clever fjord]

alee

[tidal matrix]

$\alpha = \omega_0 t + \theta$ i think, but that should be enough given properties of cos

[clever fjord]

χασιβ ♥

[tidal matrix]

this integration checks out, and its finite (given a finite phase angle and nonzero frequency ofc)

[tired obsidian]

actually

i can just do cos(u) = cos(v)

$u = 2\pi k \pm v$

[clever fjord]

alee

[tidal matrix]

oh yeah you can

[tired obsidian]

$\omega_0 T = 2\pi k$

[clever fjord]

alee

[tidal matrix]

nvm im stupid, thats good

[tired obsidian]

$T = \frac{2\pi k}{\omega_0}$

[clever fjord]

alee

[tired obsidian]

so $x(t) = x(t+T)$ when $T = \frac{2\pi k}{\omega_0}$ ?

[clever fjord]

alee

[tidal matrix]

yeah

sorry its been a while since ive done SHM

[tired obsidian]

but so

[errant dew]

That looks like what a SHM answer should look like

[tired obsidian]

$P = \frac{A^2(2\pi + \theta + \sin \theta \cos \theta)}{2\omega_0 T}$

[clever fjord]

alee

[tired obsidian]

do i have to put $T = \frac{2\pi k}{\omega_0}$ ?

[clever fjord]

alee

[tidal matrix]

i think anyone familiar with SHM will assume that to be the case, but no harm in the additional info

up to you :)

[tired obsidian]

ive read that the fundamental period is the smallest positive

so for k = 1

T = 2pi/w0

thx

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[tired obsidian]

$Y(e^{j\omega}) = e^{-j4\omega}!\left(\frac{\sin(5\omega)}{\sin(\omega)}\right)^{!2}.$

[clever fjord]

alee

[tired obsidian]

how can i find

$\angle Y(e^{j\omega})= ?$

[clever fjord]

alee

[faint knot]

omega is real, right?

[tired obsidian]

yes

[faint knot]

this is non-negative

my god that image is massive

[forest sky]

,, \angle(z_1 z_2) = \angle z_1 + \angle z_2

[clever fjord]

κλαοδ ☁ (cloud)

[faint knot]

so really you have non-negative * e^(j4w)

so you can read its angle off as 4w

[tired obsidian]

so $\angle e^{-j4\omega} + \angle \left(\frac{\sin(5\omega)}{\sin(\omega)}\right)^{2} ?$

[clever fjord]

alee

[forest sky]

yes

[tired obsidian]

ok the first is $-4\omega$

[clever fjord]

alee

[tired obsidian]

wait

isnt $\angle \left(\frac{\sin(5\omega)}{\sin(\omega)}\right)^{2} = 0$ ?

[clever fjord]

alee

[forest sky]

yes, it is

[tired obsidian]

because it is > 0

right ?

[forest sky]

yes, because it is a positive real number

[tired obsidian]

alright thx

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[hardy yew]

yo

QQ

huh

bot doesnt work?

[faint knot]

damn

.close

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[lavish jackal]

hi

[lavish jackal]

how do i do the 2nd line

bc there’s no a’s

do i just need to add 0?

[late dust]

To make a matrix? Yeah, it's 0

[errant dew]

u’s

[wanton bison]

-# 0u laughing

[late dust]

au+bv+cw

[lavish jackal]

so i assume this is correct?

ok i think i got it

[late dust]

Sure, now it depends what you do with it

[lavish jackal]

yeah im using the cramer's rule 🙂

i can continue the rest, thanks for confirming!

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[wanton bison]

.pin

perfect timing

[lavish jackal]

.unpin

[wanton bison]

.deletethis

[odd edge]

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[crystal radish]

let r_L & h_L be radius/height of larger cone, same but _s for the smaller one

$\frac{r_L \sqrt{r_L^2+h_L^2}}{r_s \sqrt{r_s^2+h_s^2}} = \frac{20}{9}$

[clever fjord]

UCYT5040

[crystal radish]

$\frac{r_L^2 h_L}{r_s^2 h_s} = k$

[clever fjord]

UCYT5040

[crystal radish]

$\frac{r_L}{r_s}=\frac{h_L}{h_s}$

[clever fjord]

UCYT5040

[crystal radish]

Stuck here

tried using CAS, it must have gotten stuck in a loop or something cuz it took forever processing, i had to stop it

never takes that long ever

[odd edge]

@crystal radish Has your question been resolved?

[crystal radish]

<@&286206848099549185>

[orchid torrent]

https://www.onlinemathlearning.com/scale-factor.html#google_vignette

[crystal radish]

does it hold true with this being lateral surface area and not full surface area?

[orchid torrent]

yes

you should be able to justify this to yourself using the exact same logic used to establish the facts in the image

[crystal radish]

ok, makes sense. thank you!

.close

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[plucky cobalt]

How does one go about solving this and finding A (which also solves for finding B) and the angle between A and B?

[orchid torrent]

law of cosines

[https://math.libretexts.org/Courses/Cosumnes_River_College/Math_384%3A_Foundations_for_Calculus/11%3A_Non-Right_Triangle_Trigonometry/11.02%3A_The_Law_of_Cosines]

[plucky cobalt]

so with this, I solve for a and b and get 10.02 and 20.05

[orchid torrent]

,w a^2 + 26.5^2 - 2a26.5*cos(41 deg) = 4a^2, a>0

[clever fjord]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

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[orchid torrent]

,calc 10.0225*2

[clever fjord]

Result:

20.045

[orchid torrent]

both are fine to 2 decimal places

[plucky cobalt]

So I get the angle opposite of A to be 19.149 degrees @orchid torrent

with respect to that, can I then just do 180 - 41 - 19.149 to solve for the last one?

[orchid torrent]

,w arcsin(1/2 sin(41 deg)) * 180/pi

[clever fjord]

Results provided by WolframAlpha

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[orchid torrent]

yes

[orchid torrent]

yes

[odd edge]

@plucky cobalt Has your question been resolved?

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[misty drum]

sooo i am clearly doing something wrong here.. i graphed the equation on desmos to find its around 1.5 , but my calculations are giving me 5x10^-11… 😗

can someone please explain where i’m going wrong here.. i’m so confused… thank you 🙏

[misty drum]

sorry the picture kinda sucks when u click on it 😔

[errant dew]

What does DS mean

[vivid wyvern]

x²+x-2

It was factorized wrong

[vivid wyvern]

Same question

[misty drum]

uhh direct substitution

[errant dew]

I’ve seen it in a few previous help requests from other people, never heard of that before

[misty drum]

omg

i see it now

😭😭😭😭

i have to go to sleep or something holy

[errant dew]

When you get so really smaller number it might be because it’s actually 0 and your calculator is having float precision issues

[misty drum]

yeah i factored that wrong and i could not for the life of me see that

[errant dew]

The limit in the thing you calculated is 0

[misty drum]

but it’s okay cuz we’re rockin now

[misty drum]

oh okay

don’t worry guys i’m gonna get it now

thank you for your help!!!!!!!!

.close

[vernal yacht]

!bnuuy

[odd edge]

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[errant dew]

!bnuuy

[odd edge]

[vivid wyvern]

!bnuuy

[odd edge]

[misty drum]

!bnuuy

[odd edge]

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[spice veldt]

Mainly need help with the diagram

[errant dew]

,rccw

[mystic saffron]

,calc 989898989898989898898898989898 * 99

[clever fjord]

Result:

9.8e+31

[spice veldt]

<@&286206848099549185>

[wicked viper]

by equilibrium does the question mean roots or points where slope shifts?

[odd edge]

@spice veldt Has your question been resolved?

[spice veldt]

i believe it is roots

[wicked viper]

sweet

plot the graph of y = rx, and y = sinx, taking values of r=0, 0<r<1, r=1, r>1 etc

and find how many points both the graphs intersect

[orchid torrent]

https://math.stackexchange.com/questions/1991510/how-to-figure-out-what-the-graph-of-frac-sin-xx?noredirect=1

[spice veldt]

how do i find this

[wicked viper]

okay lets start with an example

take r = 0

it will yield y = 0

which is nothing but the x axis

y = sinx intersects x-axis at infinitely many points

now take some value of "r" between 0 and 1

which is nothing but tilting the line y=0 in the anticlockwise sense with respect to the coordinate axes

[spice veldt]

are you able to draw it out for me? i think its better for me to learn visually

[wicked viper]

go to https://www.desmos.com/calculator

first box type in y = sinx

second box type in y = rx

a slider will appear which will allow you to choose the values of r

play around with the slider and notice what happens at the values i just mentioned

[spice veldt]

alright

yes, the line is rotating anti-clockwise as the r goes up

[wicked viper]

yes

notice what happens to the number of intersection points

[spice veldt]

more intersections closer to -1 and 1, and less at 0

nvm

the other way around

more at 0

and less at -1 and 1

[wicked viper]

infinitely many at zero

one at r = 1

[spice veldt]

oh yea,

one at -1 aswell

[wicked viper]

yes

so the number of intersections between 0 and 1 is finitely many

we know for sure that number of equilibria change at r=0, and +/- 1

but for 0 to 1 its a bit tricky

hmm

[spice veldt]

what type of bifurcation it will be

[odd edge]

@spice veldt Has your question been resolved?

[last oriole]

Is this a good place I ask a question?

I have this graph and its asking for its Interval(s) of decrease (enter DNE if none exist)
my answer was (-infinity, -4) U [1,5) U [5,+infinity) and it said that is wrong. so I entered (-infinity, -4) U [1,+infinity) which also was tagged as wrong answer so Im just confused

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[odd edge]

@spice veldt Has your question been resolved?

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[odd edge]

Please don't occupy multiple help channels.

[mystic saffron]

How do do this?

[opal path]

First, do you understand what delta x and delta y mean?

[unkempt lichen]

might wanna close your previous help channel first

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[misty drum]

hello can someone verify that my calculations in pink are correct?
i am not sure if thats how im supposed to handle absolute value limits or not

[warped glacier]

$|x - 2| = x - 2$ if $x - 2 \ge 0$

and $|x - 2| = -(x - 2)$ if $x - 2 < 0$

[clever fjord]

south

[warped glacier]

start from the piecewise definition of |u|

then replace u = x - 2 everywhere; this is what you get

[misty drum]

sorry what is u

where is that coming from

[warped glacier]

ah I see, you wrote this part wrong

but everything else is correct

[warped glacier]

since we used the variable x already, we can't use x again

so it comes from this

is what I mean

[warped glacier]

so the correction is x - 2 < 0

same for the other one, it's x - 2 >= 0

[misty drum]

okay

[warped glacier]

but then after that you did the problem correctly

[misty drum]

my friend was telling me that you had to make it 2 for some reason but i dont understand why she said that

[warped glacier]

she's wrong

or she meant x < 2 and x >= 2 directly

[misty drum]

oh okay

and so then is the work for this one right as well?

i also asked for my friends help here but she confused me more undortinately

[warped glacier]

ah, so $\frac{|x| - 1}{x - 1} = \frac{x - 1}{x - 1}$ (when $x \ge 0$) $=1$ is correct

[misty drum]

x =/= 0 right'

[warped glacier]

the issue is that around $x \to 1$

[clever fjord]

south

[warped glacier]

you don't have x < 0

[clever fjord]

south

[misty drum]

what does this mean??

[warped glacier]

the limit is around x = 1 right?

so what happens when x < 0 is irrelevant

[misty drum]

umm am i missing x =/=0

or am i confused

[warped glacier]

I feel you don't understand what it means to take a limit around a certain value, where x is not 0

[misty drum]

probably not

[warped glacier]

have you seen the table of values method before?

[misty drum]

yes

[warped glacier]

you sub in values of x that get closer and closer

yeah, so you could sub in x = 0.5, 0.9, 0.99, 0.9999... from the left

or x = 1.5, 1.1, 1.01, 1.0001.... from the right

in all of these cases, x will be positive

[misty drum]

okayyyyy

[warped glacier]

the point of the limit is figuring out what happens really really close to x = 1

we usually can't sub in x = 1 directly in these problems

[misty drum]

yea

[warped glacier]

so hopefully you can now understand how x < 0 is irrelevant

we can choose some neighbourhood around x = 1

say, x being in (0.9, 1.1), that completely excludes x < 0 altogether

we started from x = 0.5 and x = 1.5 before, so that would be the neighbourhood (0.5, 1.5)

we can have as tiny a neighbourhood as we want, as long as the neighbourhood includes x = 1 and both sides (less than, more than x = 1)

[warped glacier]

okay I added a bit more on why

[misty drum]

i think ive kind of lost the plot

im not really sure what we're talking about anymore

[warped glacier]

why x < 0 is irrelevant

so you can just assume that |x| = x for the limit

[misty drum]

sorry could you maybe circle what you mean from the original screenshot? im not very good with just words it seems

i still don

't really know what you

are referring to

[warped glacier]

the limit as $x \to 1^-$ is also $1$

[clever fjord]

south

[misty drum]

okay but for the last question (pink one) i turnned the absolute value negative to do the other side... so why is this one different

[warped glacier]

cause $x < 0$ is irrelevant to $x \to 1$

[clever fjord]

south

[misty drum]

i just dont understand

[warped glacier]

which part don't you understand?

[misty drum]

this i think

and why the pink version turns the absolute value negative but the green one doesnt

[warped glacier]

I mean, for the green one, you don't even need to consider the negative case

|x| = -x when x < 0 isn't wrong, but it's useless to consider it

how will considering the case when x is negative tell us anything about what happens as x gets very close to 1?

[misty drum]

but how did it tell me anything when i go to 2

[warped glacier]

cause |x - 2| changes around x = 2

you're dealing with two different functions that split at x = 2 exactly

x - 2 for x >= 2, and -(x - 2) for x < 2

[odd edge]

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[robust olive]

for b)

[robust olive]

this is the solution

but what if n is zero?

[forest sky]

well we would need n to be greater than 1 for this sequence

if you require sequences to have a 0th term, just map it to 0 or 1 or something, it doesn't affect the limit

[robust olive]

oh okay

also i have another question

if we were to apply squeeze therom to eval this

is it nessessary for us to start off with -0.5 <= sin(n)cos(n) <= 0.5

because technically u can also do -1 <= sin(n)cos(n) <= 1

and thats also true

basically is it nessessary to have the lowest/highest bound

[forest sky]

it doesn't really matter how tight the bound is if both the upper and lower bounds end up converging to the same thing anyway

in this particular limit i suspect either bounds will work, in other limits it might end up being important

[robust olive]

i see

[robust olive]

do u have an example where it would be important to have the "tightest" bound

[forest sky]

let's take [ \frac{n}{n^2 + n^2 \sin n \cos n + 1} ]
then the worse bounds would be
[ \frac {n}{2n^2 + 1} \le \frac{n}{n^2 + n^2 \sin n \cos n + 1} \le n ]
where the upper bound is useless, but the better bounds are
[ \frac {n}{3n^2/2 + 1} \le \frac{n}{n^2 + n^2 \sin n \cos n + 1} \le \frac {n}{n^2/2 + 1} ]

kind of a contrived example but oh well

[clever fjord]

κλαοδ ☁ (cloud)

[odd edge]

@robust olive Has your question been resolved?

[robust olive]

i see, alright thanks

also sorry, one more question

so according to the highlighted definition, the sequence {5, 5, 5, 5, 5....} would be increasing?

[forest sky]

yes, constant sequences would be included

[robust olive]

but it also mathces with the decreasing definition, so constant sequecnes are both increasing and decreasing?

[forest sky]

yes

in fact that's the only type of sequence that's both increasing and decreasing

[robust olive]

i see, alright

thank u so much for ur help

i rlly appreciate it

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[forest sky]

provable in one line if you want to try it

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[tawny thorn]

Hi, I'm trying to solve this integral (lambda and r is real). First, I change sin(kr) into Im(e^{ikr}). Then, I rearrange it into the following expression(below this text). After rearranging it into this expression, I am not sure if I can directly change the variable k to k + i/lambda.

[tawny thorn]

$\Im\int_{0}^{\infty}\frac{1}{2}\left(\frac{1}{k+\frac{i}{\lambda}}+\frac{1}{k-\frac{i}{\lambda}}\right)e^{ikr}dk$

[clever fjord]

EleMc_115

[odd edge]

@tawny thorn Has your question been resolved?

[tawny thorn]

<@&286206848099549185>

[odd edge]

@tawny thorn Has your question been resolved?

[odd edge]

@tawny thorn Has your question been resolved?

[errant dew]

<@&268886789983436800>

[stoic cloud]

Damn cod and gambling? Typa person to censor the word shower 💔

[dreamy bear]

Use Residue Theorem

[dreamy bear]

it's standard if you know contour integration

[amber veldt]

,w fourier transform of (x²+1)^-1

[clever fjord]

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[amber veldt]

if you are willing to accept this, you can reduce your integral into something related to this using feynman's trick

but the easiest way to find the Fourier transform of (k²+lambda‐²)‐¹ is contour integration so maybe that doesn't help

[tawny thorn]

ohoh, i get it, nevermind

thx for your helps

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[karmic lotus]

this implies that absolute AC is always greater than B^2 for extremas, how?

[odd edge]

@karmic lotus Has your question been resolved?

[quasi sparrow]

what is "this"

[karmic lotus]

this paragraph

i edited my question, im gonna ask about cases of max and min only

[quasi sparrow]

your question is vague/incomplete

what is "this". "always" and for "max and min only" are contradictory

[karmic lotus]

ok
i read these 6 rows, and realised that for points 1. and 2. to be correct, AC has to be greater than B^2. but there are cases where both a and b are positive/negative, and i thought this would automatically make the critical point a max/min, yet AC turns out to be less than B^2

i think i might now know why tho

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[mystic saffron]

,rccw

[mystic saffron]

I have no idea on how to approach 2nd question

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

<@&286206848099549185>

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[lyric dust]

this question is pretty straight forward if i use power series but i was wondering if there are alternative ways

[lyric dust]

current tutorer for the course told us that we should avoid going to power series for these question as our default approach if there are other ways possible so i thought i should try other methods at least

[odd edge]

@lyric dust Has your question been resolved?

[short terrace]

I see a possibility of triangle inequality

Actually no I'm blind

@lyric dust have you done complex differentiability

[lyric dust]

yeah

[short terrace]

Ah cool then you have ways

e^z -1 is the integral of d/dt(e^tz) from 0 to 1

[lyric dust]

ah ok i think i get wat u mean instantly

[short terrace]

That's a hint, see how far you can take it

Good luck

[lyric dust]

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[honest ore]

looking to get help with the algebra/calculus aspect of my ap physics c homework
the answer key was provided with the assignment, but im struggling with understanding how to reach it and id like to understand that before the unit test

[honest ore]

answer key

oh theyre sideways one second

current progress on each question

[odd edge]

@honest ore Has your question been resolved?

[turbid comet]

@honest ore how did you go from step 4 to 5

[honest ore]

which question

[turbid comet]

8 part d

Sorry c

[honest ore]

theyre being multiplied so i separated them so i could turn it into the equation that comes after that

[turbid comet]

But the denominator won't multiply like that

[honest ore]

oh right yeah

i think i was treating it like addition

[turbid comet]

Also this term cannot be taken out of the integral

Since it contains a variable

[honest ore]

it stays constant though

[turbid comet]

-bv

And we integrating wrt v

[honest ore]

oh right yeah ill leave it under the dv

i thought u mean m for some reason

sorry its 1 in the morning rn

[turbid comet]

We live in completely different time zones

It is 12pm for me

[honest ore]

gg

at least one of us is coherent rn lol

[turbid comet]

In the second question what do they mean by radius of curvature

[honest ore]

i assumed that meant what u would use for r in stuff like a = mv^2/r

ok now im here

[turbid comet]

Is it this distance

[honest ore]

i think so

in any case it worked to get the right answer for 14b

[turbid comet]

How did you get this

[honest ore]

^ from my ap physics c notes video

[turbid comet]

Yes but if we substitute mgsin@ - bv = u

[honest ore]

(the teacher was absent and posted this and i think thats why im strugglign to understand it)

[turbid comet]

We get a different answer

[honest ore]

yeah we havent properly learned u sub for integrals

thats what u mean right

[turbid comet]

Yes

[honest ore]

anyways we learned u sub for derivatives in ap calc bc

since they said u could take the latter with the former at the same time, i figured we would learn what we needed before it came up

which sucks cuz now its screwing me

anyways back to the hw

[turbid comet]

Are you aware of the basics of u sub

[honest ore]

separate it into multiple parts so each part is solveable yeah

[turbid comet]

That is not u sub

[honest ore]

fuck

[turbid comet]

That was something else which I cannot remember

[honest ore]

wait no it was like

for derivatives its like g[f(x)] = g'[f(x)] * f'(x)

so whats the rule for u sub in integrals

[turbid comet]

Yes similar to the chain rule but for integrals