help-33

[marsh citrus]

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[marsh citrus]

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[modern sedge]

f: E to {0,1} is continuous and surjective.
I have to prove that E is a disconnected set.
It's obvious that A := f^-1 (0) and B := f^-1 (1) are 2 disjoint sets.
How do I prove that A and B are open?

[lunar pumice]

549218355765843xt568432t65432?

[whole thorn]

what's the topology on {0,1}

[mighty bramble]

not the place for this

make your own channel

and stop spamming lol

[modern sedge]

Do you mean it's a discrete topology?

[whole thorn]

yes, must be

[modern sedge]

Yes, it is

[whole thorn]

so {0} and {1} are open

and f is continuous

so preimage of open is open

[modern sedge]

Why is {0} open?
For each ε > 0 circle at 0 with radius ε is not in {x}

[whole thorn]

{0} is in the topology of {0,1}

topology literally means open sets

u can't really talk about circles cuz there's no metric

[modern sedge]

Ok, I found it in my lecture notes

ty

.close

[marsh citrus]

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[marsh citrus]

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[lament rose]

From Velleman's book How To Prove It.

a) I would say that the theorem is wrong because:

• m and n are defined in terms of k, without explicitly saying "Let k be an arbitrary integer" prior to setting m and n;
• Nothing implies k is the same for m and n;
• It doesn't always work if n != m + 1, that is, this only applies for consecutive numbers but such restriction wasn't mentioned anywhere in the theorem

b) Counter example could be 9 and 6.

Is this correct? Well, the counter example and the last bullet point are obviously correct but I'd like to know if bullet points 1 and 2 are flaws in the theorem or if I'm just being picky.

[copper raven]

the 2nd bullet point is really the proof killer, the 3rd point is just a consequence of the 2nd

re 1st point, k isn't arbitrary, it's defined in terms of m

here's a rephrasing of the first sentence "since m is even, there exists an integer k such that m=2k"

it's just the definition of being even really

@lament rose

but yeah "we can choose k..." is a bad formulation of that

[lament rose]

Ah, good to know. Thanks mate!

so i wasnt totally wrong regarding k 😄

.close

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[copper raven]

well it's an english weirdness, not a proof destroyer

[lament rose]

oh yeah yeah i meant regarding k being perhaps different for n and m (2nd bullet point). in hindsight the 1st one is probably me being picky

[copper raven]

ah yea fine

[marsh citrus]

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[pale gale]

Topes — Today at 9:18 PM
I can do part a and got 64 + 48k + 12k^2 + k^3
is there a quicker way to do part b?
i went back to 1(4)^3 +3(4)^2(k)^1 +3(4)^1(k)^2 + k^3 and replaced k with [x^2 -x]
but that feels very long as i have to expand (x^2 - x)^3

[marsh citrus]

@pale gale Has your question been resolved?

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[hazy lion]

Trying my best with this problem

[hazy lion]

$y \perp x_n$ and $x_n \to x$, show that $x \perp y$

[elfin berry]

jan Niku

[hazy lion]

i understand i think theres some way to do it with continuity

i dont wanna do that

my thought was that

oh man i dont have ipr macro on here

😭

$<x_n, y> = <x,y> + <x_n-x,y>$

[elfin berry]

jan Niku

[hazy lion]

now $<x-x_n,y> = <x,y>$ because the lhs is 0

[elfin berry]

jan Niku

[spark otter]

why not just use continuity of the scalar product?

[hazy lion]

and $||y|| ||x-x_n|| \geq <x,y>$ and we can take the limit to get $0 \geq <x,y>$

[elfin berry]

jan Niku

[hazy lion]

scalar product? i thought the inner product itself is continuous

[spark otter]

yes inner product

it is continuous

[hazy lion]

because i didnt know that

[spark otter]

so <xn,y> -> <x,y>

[hazy lion]

i think my method works but im not certain about taking the limit

[hazy lion]

here

after applying CS

its kosher right?

obviously you just flip the sign and get the other direction to have <x,y> = 0

but this step im doubting

can you just take the limit of both sides of an inequality

[spark otter]

yes

[hazy lion]

so this method works out then

[glacial hedge]

Yes

[spark otter]

yes, you could have also used

|<xn,y>| <=...

and no need for flipping sign

[hazy lion]

well continuity of the inner product works too i see but

you dont use the assumption that xn is perp to y

do you

[glacial hedge]

It's basically continuous by definition because it defines the topology on the space

[hazy lion]

i dont really understand what continuous operator means tbh

i guess it just means the limit goes through the operator

im not sure ill learn it by the exam so im trying not to use it if i dont think of it

[glacial hedge]

It has the same limit
Said limit is 0

[hazy lion]

it has the same limit?

oh you mean that lim <x_n, y> = lim <x,y>

because of continuity?

[glacial hedge]

f is continuous iff f(xn) -> f(x) as xn -> x

[hazy lion]

weird

[glacial hedge]

Here it's the continuity of <., y>

[hazy lion]

ah okay so like

yea i guess you just take y fixed

then this is a f(x)

and its continuous

so as xn -> x

then f(xn) -> f(x)

and since f(xn) = 0,0,0,0,0,0,...

then lim f(xn) = 0

and

[glacial hedge]

Well you can also take (xn, y) as your sequence if you want, though that's probably not clearer

[hazy lion]

idk something like that

idk the perp is tripping me up

this is just a sequence of 0's right

and thats the thing you use

okay

i think i get it

thanks folks

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[marsh citrus]

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[glacial hedge]

Do take the time to learn topo at some point
Topo is cool
Though ofc it's not quite your study plan iirc

[hazy lion]

no i have an exam in some few hours then im out of school

next up id like to learn some maybe vector space style optimization methods

career focused though

hope to pivot towards controls or operations

see what happens

[glacial hedge]

What's that

[hazy lion]

.reopen

[marsh citrus]

✅

[hazy lion]

theyre two kinda different things

operations is mainly application of analysis to some real world process

idk i can link stuff

[hazy lion]

controls means maintaining something like

maintaining some function of the parameters describing the state of a system

[glacial hedge]

Physics related stuff

[hazy lion]

its more like industrial math

afaik operations research is one of the highest paying fields related to analysis

but requires a lot of stuff outside analysis

[hazy lion]

easy examples of operations stuff is like scheduling problems

and resolving conflicts in complicated state-machine systems

or you know like if you look at these state machines and this system can you determine if a nonterminating deadlock can be reached

how could you design it away

stuff like that

[glacial hedge]

Alright
Interesting

Sounds a lot more CS related when you put it that way

[hazy lion]

its cs as much as applied math is cs adjacent

but im an idiot you know

im not the expert on this stuff

all i know is from presentations and ppl at my school

theyre trying to start an operations specialty i think its kind of new at least in the usa

anyways

thanks

.close

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[blissful stream]

I need help with all of this

[desert dirge]

too much

select some, g

[blissful stream]

Number 4 then

It makes absolutely no sense at all

[desert dirge]

its essentially asking how many ways you can organise 9 people

[blissful stream]

So do I just do 9!

?

I’ve also tried with nPr and nCr

And none of them appear to be correct

[fathom sun]

If they were in a straight line, how many ways could you do it

@blissful stream

[blissful stream]

362880?

[fathom sun]

keep it in terms of factorials

[blissful stream]

9!

[fathom sun]

but yes

however, in a circle, 12345679

and 234567891

are the same thing

[blissful stream]

But it’s not an option in the mixed answer

[fathom sun]

it was correct for the straight line

the circle is different

[fathom sun]

using this

[blissful stream]

So nCr?

[fathom sun]

no

what would you be choosing

if you did ncr

[blissful stream]

9 nCr 1?

Idk man, probability is so confusing

[fathom sun]

when you are counting, a circle can be thought of a line but with rotations

any ideas with that?

you know that for a line it is 9!

how many rotations would there be for each of those 9! cases?

[blissful stream]

9?

So it’s 9!/9

?

[fathom sun]

nice

which is?

[blissful stream]

40320

[fathom sun]

gj

or 8!

[blissful stream]

So if that’s the answer, then number 6 is wrong

Thanks

Can you also help me with number 6?

[fathom sun]

instead of using an 8-word example, lets go smaller

instead of desserts, we can use "cook"

if all of the letters were distinct, how many ways could you arrange them?

[blissful stream]

4!

[fathom sun]

however, "cook" and "cook" are the same word

[blissful stream]

since o appears twice

I need to do is divide it by 2!?

[fathom sun]

yes

nice intuition

so for desserts?

[blissful stream]

So each time a letter appears more than once, you divide by its factorial

[fathom sun]

yes

[blissful stream]

8!/2!3!

[fathom sun]

correct

,calc 8!/(2! * 3!)

[elfin berry]

Result:

3360

[blissful stream]

Ohhh

[fathom sun]

so the answer is 3360

[blissful stream]

You have to multiply them in parentheses

I thought you just then write out the individual factorial

[fathom sun]

well

depends

you could write it like this

$\frac{8!}{3!2!}$

[elfin berry]

Dork9399

[fathom sun]

but when you put it into a calculator you want the parenthesis

and it probably works wihtout the asterisk

[blissful stream]

When I put that into a calculator, it gives me 13440

[fathom sun]

yea

for calculators you want to add the parenthesis

[blissful stream]

I see

[fathom sun]

when you hand in your work if your teacher wants it as an unsimplified factorial (no clue why they would), then this would work

[blissful stream]

Ya, my teacher says to show it in factorial

[fathom sun]

then you wouldnt need the parenthesis

[blissful stream]

Ok

[fathom sun]

just as a rule of thumb with calculators

when in doubt add parenthesis

[blissful stream]

Yep, worked for number 7

Can you also help me with number 11a

[blissful stream]

On here

[fathom sun]

can you get 9 heads?

[blissful stream]

No

[fathom sun]

can you get 8 heads?

[blissful stream]

Yes

[fathom sun]

how many tails would that be?

[blissful stream]

0

[fathom sun]

can you get 7 heads?

[blissful stream]

Yes

[fathom sun]

how many tails would that be?

[blissful stream]

1

Oh

[fathom sun]

so you can have anywhere from 0 to 8 heads

[blissful stream]

So there are 9 outcomes

[fathom sun]

yes

[blissful stream]

Why isn’t it in the mixed answers

[fathom sun]

i do not know

[blissful stream]

31 answer and 31 mixed answers

Could it be that I have to write it in fraction form or something

[fathom sun]

i see a lot less than 31 problems

[blissful stream]

There is a back side

[fathom sun]

oh lol

[blissful stream]

Ya

[fathom sun]

ohh

i think I see what they meant

they should have said that order matters

[blissful stream]

Oh

[fathom sun]

any ideas on how we would count that?

[blissful stream]

2^8

[fathom sun]

nice

so 256

which is in the answers

[blissful stream]

Oh

And then for 11c, would I do (1/2^8) * (1/2^8)?

Just put it in the calculator, that’s is not what I need to do

[fathom sun]

yea

its not what you should do

the easiest way to do this

is to count the number of ways that you don't get at least one of each

and then subtract it from 256 (the total count)

and then for probability you just divide by 256

[blissful stream]

So 7/8 + 7/8 - 256?

[fathom sun]

count the number of cases, not the probability

or you could sum the probabilities and subtract them from 1

either way

[fathom sun]

7/8s is not the correct probability

[blissful stream]

Hmm

1/2 * 1/2?

Im so sorry man, probability and statistics is very confusing for me

[fathom sun]

np

thats what im here for

what did you get for 11b?

[blissful stream]

1/128

[fathom sun]

nice

in words

what would it mean for you not to get at least one of heads or tails

[blissful stream]

255/256?

[fathom sun]

.

[blissful stream]

Oh

If you flip it 8 times, 7 times you could not get heads or tails?

[fathom sun]

so what must you get?

[blissful stream]

1 heads or tails?

[fathom sun]

no

for the complement

if you have 1 head or 1 tail it still satisfies

where will it not satisfy

[blissful stream]

If you get 8 heads or 8 tails in a row?

[fathom sun]

yes

what is the probablity of that

(you already found it)

[blissful stream]

1/128

[fathom sun]

so what do you get when you subtract that from 1?

[blissful stream]

127/128

[fathom sun]

and there we go!

[blissful stream]

Finally

But it still kinda dont get it

[fathom sun]

lol

sure

which part?

[blissful stream]

Like where did the 1 come from

[fathom sun]

when you add up all of the probabilities, you will always get 1

because something always must happen

if they added up to like 7/8 or something like that

what would happen that 1/8 of the time?

does the world just stop ending?

so it must always add to 1

[blissful stream]

So nothing happening is still something happening in probability

[fathom sun]

yes

[blissful stream]

Ok

Ohhh

[fathom sun]

a better explanation is that we will track the variable's data no matter what

so it will always add to 1

[blissful stream]

So i take the chance of something happening and subtract it from the probability of getting all heads and tails

[fathom sun]

yes

and you find the probability of everything else happening

its a method called complementary counting

[blissful stream]

Ok, that makes more since

Sence*

[fathom sun]

because you are counting the complement

[blissful stream]

Alright

Thank you so much

[fathom sun]

anything else?

[blissful stream]

But i think ill need help with the back side so should I keep this open

[fathom sun]

just post the paper

keep it open

[blissful stream]

Kk

[fathom sun]

any specific or just all?

[blissful stream]

Im already stuck on 15😭

[fathom sun]

just count the number of cases

in how many cases are the numbers equal or add up to 8?

[blissful stream]

There are 6 cases of them being equal and 5 cases of them having he sum of 8

Minus the 1 inclusive (4 and 4) (4+4)

[fathom sun]

for a total of?

[blissful stream]

10/36 or 5/18

[fathom sun]

and theres the answer

anything else?

[blissful stream]

Number16

Do i do 16/30 + 15/30 - 9/30?

Ooh wait, its - 7/30

[fathom sun]

yes

[blissful stream]

So 4/5

[fathom sun]

yes

[blissful stream]

And means multiply in probability right

[fathom sun]

yes

[blissful stream]

Ima take a little break

[marsh citrus]

@blissful stream Has your question been resolved?

[blissful stream]

Number 19

And 20

21

[blissful stream]

<@&286206848099549185>

[blissful stream]

On here

[marsh citrus]

@blissful stream Has your question been resolved?

[blissful stream]

.close

[marsh citrus]

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[fathom sun]

@blissful stream

srry i had to go somewhere

im back now

if ur ready

just .reopen

[blissful stream]

I would, but I have to leave somewhere

Thanks for the help

@fathom sun

[fathom sun]

np

[marsh citrus]

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[thin island]

why do you need to multiply through by xy?

is it possible to continue without doing so?

[obsidian trail]

oh because they tried to eliminate denominators because the other side is zero anyway

[thin island]

so its kind of just so its visually simpler?

[obsidian trail]

Yeah, I mean I wouldn't want to deal with denominators either so heck yeah multiply both sides with xy

There's no need to write that step tho tbf, just try to proceed like you solve a normal algebraic equation.

[marsh citrus]

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[worldly dome]

Is this a good induction proof

[lilac siren]

nop

To be proven: n<2n for all n e Z, n≥2 (or Z≥2 if that's a notation you were taught)

Base Case: the statement is true for n=2: 2<2*2=4 ✓

Induction Hypothesis: the statement is true for an arbitrary fixed n: (repeat statement usually)

Induction Step: to prove: the statement is true for n+1: ...

And then you'd perform the main proof

we want to show n+1 < 2(n+1)

and are allowed to use n < 2n

so we can start with n+1 and use the hypothesis:

n + 1 < 2n + 1 (using the hypothesis)

< 2n + 2

< 2(n+1)

✓

@worldly dome

[worldly dome]

Oh ok thanks

.close

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[hasty comet]

I am I’m Pre calc, and we’re doing polar curves, and I need help solving this part of the problem to find the zeros

[clever spade]

Find values where cos(2theta) will equal 0

Try and think about where cosine might equal 0 using the unit circle

[hasty comet]

Omg, it’s that easy!!

Bruh

[mint root]

can someone help with my math hw?

[hasty comet]

What math?

[mint root]

so im in sixth grade so this might be rly easy for u

[hasty comet]

Ok, what’s the question?

[mint root]

um hold on i'll send a screenshot

[clever spade]

@mint root you can just send in #help-12

[hasty comet]

.close

[marsh citrus]

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[spring bridge]

I’m very confused with this question how do I do this

[spring bridge]

I don’t understand anything can someone please explain

I have no clue

I don’t even know what that is

Teacher assigned

I have been absent for like a week so not really

Not really

Wdym

How much an item is discounted

And cumulative frequency

So how do you work it out

Okay

Nah

Lemme go watch rq

How do you do the interquartile range

@still temple

Okkk

Thanks

[marsh citrus]

@spring bridge Has your question been resolved?

[marsh citrus]

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[near meteor]

How do you find the values of a and r for a geometric sequence?

[near meteor]

$\left{2-\left(-\frac{1}{2}\right)^n\right}$

[elfin berry]

Luca M

[sand vigil]

value of a depends on value of n

does n ranges from 0 to inf?

[near meteor]

it doesnt say. It just says to determine if it converges or not. And then if it does find the limit

im guessing its from 0 to inf

[sand vigil]

if n's starting value is 0 then just substitute n=0 and you will find a

substitue n=1 and just divide 1st term with 0th term you will get r

[near meteor]

so a is 1?

i tried plugging that into the formula for the limit but i get 2/3

2-2/3 is 4/3

but when i test it on desmos it seems the limit is actually 2 of the entire sequence

[sand vigil]

5/2 is the second term

[near meteor]

howd you get that

[sand vigil]

see 2-(-1/2)^1

2+1/2

= 5/2

so r is 5/2

[near meteor]

ohhhh

i was treating the 2 and the other term differently

[sand vigil]

not its your part of GP

[near meteor]

so the limit is -2/3?

[sand vigil]

i dont know what limit means here in GP, could you explain what limit here means?

[near meteor]

The limit of the sequence

like if you plug in n=100000

what is the value of the terms approaching

in this case it seems to be 2

but i dont know how to work it out algebraically

[sand vigil]

okok

when n is infinity right?

[near meteor]

yes

[sand vigil]

see what is our r?

[near meteor]

5/2

[sand vigil]

so is it converging or diverging?

[near meteor]

well its >1 so diverging but how

the 100000th term is 2

[sand vigil]

do you have the answer of this?

[near meteor]

nah

[sand vigil]

okay i see the graph now

[sand vigil]

see the negative term after 2 is making it converging to 2

[near meteor]

yeah

Like it makes sense that it converges to 2 but how do i show it algebraically

[sand vigil]

that i think i need some time to think

how to prove it

[near meteor]

wait

could i use p-series?

[sand vigil]

its for 1/n^p

[near meteor]

oh yeah

[sand vigil]

we can use limits

limit n->inf 2 - lim n->inf (-1/2)^n

the second term approaches 0

[near meteor]

yeah i know

[sand vigil]

hence it converges to 2

[near meteor]

is that sufficient?

[sand vigil]

yes ye s

[near meteor]

oh

[sand vigil]

its more than sufficient

[near meteor]

i thought i had to somehow show the second ones limit is 0

[sand vigil]

teacher will not even give marks but will kiss you for this solution also

[near meteor]

great

[sand vigil]

keep doing maths and if your doubt has been resolved free this channel

[near meteor]

just wanna check i did the next one right then

{sin(1/n)}. I know the limit as n approaches infty of the 1/n is 0. That implies the limit of the sequence is sin(0)=0

would that be sufficient?

[sand vigil]

no this is wrong

see lim of this is undefined

[near meteor]

why?

[sand vigil]

https://www.youtube.com/watch?v=y8C9eZG_CEk

see this short video

[near meteor]

but its not a series right?

sin(1/1),sin(1/2), sin(1/3), sin(1/4), ....

im not adding anything

[sand vigil]

sin(1/n) is itself quiet interesing graph

[near meteor]

so its not really harmonic series

[sand vigil]

yes not in your case

allright my bas

bad*

i thought you were talking about series

[near meteor]

all good

[sand vigil]

this is the video you might find your answer in

[near meteor]

isnt that n*sin(1/n)

[sand vigil]

https://www.youtube.com/watch?v=npT9VEEl4rs

[near meteor]

im pretty sure the answer i got {sin(1/n)}=0 is right

i follow math sorcerers courses lol

[sand vigil]

just its proof is something different

[near meteor]

yeah the epsilon definition of a limit

.close

[marsh citrus]

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[marsh citrus]

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[warm elk]

I dont know what to do with this

[rough fox]

Do you know about the sine cosine formulas?

[rough fox]

For example, what's Sin(theta) here

[dusky violet]

SOHCAHTOA 🙌🏼

[warm elk]

yes

[warm elk]

ive done other stuff with this I just dont understand this phrasing

[rough fox]

Basically you need to make AB the subject
Cos(theta)= something something
Rearrange and you'll get your answer

[sand vigil]

just use simple trignometry

PBP

---

HHB

SCT

sin cos tan

[warm elk]

what would ab be represented by though

[dusky violet]

AB is a segment

[rough fox]

Cos(theta) = adjacent/hypotenuse
Adjacent = cos(theta) x hypotenuse
AB = cos (theta) x 1
AB = cos(theta)

[warm elk]

oh I see I handt associated AB iwth adjacent

thanks

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[marsh citrus]

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[sly fox]

Ive been struggling with this for awhile now and I couldnt come up with a way to start this. Im asking for a guide or something to at least get me started or put me in the right direction.

For context the highlighted portion is the number on question (number 4a)

[marsh citrus]

Please don't occupy multiple help channels.

[sly fox]

Sorry for this, didnt notice my previous channel was still open

<@&286206848099549185>

[marsh citrus]

@sly fox Has your question been resolved?

[leaden monolith]

Can’t you just pick a cyclic group with only 5 things and list those as your generators

Like say ℤ₅

[sly fox]

That has 3 right?

[leaden monolith]

Wdym

[sly fox]

Oh i meant 4 generatora

[leaden monolith]

<0, 1, 2, 3, 4> are things in ℤ₅ and they indeed generate the group

[sly fox]

Does 0 generate Z5?

[leaden monolith]

All 5 of those things together generates Z5 no?

If you need them individually just pick ℤ⁷ or something

[sly fox]

Wait what binary operation are we using?

Is it addition?

[leaden monolith]

Ya

[sly fox]

Im so sorry im totally new to this but isnt a generator an element with the same order as the group?

*cyclic group

But the order of 0 is 1 and the order of Z5 is 5

[leaden monolith]

That’s just from Wikipedia

I’ve not really taken a class on group theory but from what I know, the things in 1, 2, 3, 4 generates Z5 individually

And if you needed more you can always pick some Z_prime

[sly fox]

Oh okay, i can sort of see where this is going. Im just going to ruminate on it for awhile and see what i get from it. Thanks!

Im quite new to these abstract algebra stuff

[leaden monolith]

Me too haha

[marsh citrus]

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[wintry yacht]

how to do part a

this is the answer key

[marsh citrus]

@wintry yacht Has your question been resolved?

[wintry yacht]

<@&286206848099549185>

[still temple]

whats confusing you about the answer key

[wintry yacht]

where do we get 2pix

times 41 over pix squared

[eager hemlock]

.rotate

[still temple]

so the surface area of the curved portion of a hemisphere is 2*pi*x^2

you have two hemispheres in total so the surface area of both is 2*pi*x^2 + 2*pi*x^2 = 4*pi*x^2

[wintry yacht]

wait dont u have to divide this by 2

then subtract the area of the cirlce

underneath

[still temple]

no but like that's the thing

you have two hemispheres right?

[wintry yacht]

ye

[still temple]

if you put them together you get a full sphere

[wintry yacht]

yea

but then u have to subtract the 2 areas of the circles right

[still temple]

no actually because

those are not considered on the surface

[wintry yacht]

OHHHH

[still temple]

they are inside coated up by the cylinder

so you dont care about that

[wintry yacht]

ophhhhhh

kk

then where do we get the 41 / pi r squared

[still temple]

for the cylinder part

[wintry yacht]

yeah

wait is that just height

because u divide by area of circle

[still temple]

yeah

[wintry yacht]

k then were do we get

the 2 pi r

[still temple]

yk how the surface area of a cylinder is [
2\pi r h +2\pi r^2
]
right

[still temple]

but like

[still temple]

like with the hemispheres the second term is describing the top and bottom of the cylinder

but thats on the inside now

so we don't care about it and we just have 2pirh

r is x in ur case

but we dont know the height h

[wintry yacht]

so the final thing is canceldd

[still temple]

yeah

the surface area of the curved part of a cylinder is 2pirh

[wintry yacht]

k i get it now

ty

can u help with part c now

ii

[still temple]

so what they do is use the fact that we know the volume of the cylinder and solve for $h$: [
\pi x^2 h = 41 \Implies h = \4{41}{\pi x^2}
]
so we put it in the surface area formula [
SA = 2\pi x h = 2\pi x\8{\4{41}{\pi x^2}}
]

[wintry yacht]

i get i

[still temple]

oh ok

[still temple]

ok uh

do you know the second derivative test

[wintry yacht]

i kinda forgot it

[still temple]

aight so here is the idea

[wintry yacht]

i remember learning abt it tho

[still temple]

[0.6\textwidth]if the solution of $f'(x) = 0$ is something like $x = c$, then if $f''(c) > 0$ the point $c$ is a minimum, and if $f''(c) < 0$ then the point is a maximum

[still temple]

so first step is to find what solves f'(x) = 0

so uh, what did u get for f'(x)?

[still temple]

well this goes back to ur part b of the problem. What did u find the first derivative to be, and a?

[still temple]

ok, and a?

[still temple]

no i meant the value of a 😅 part ii of question b

[wintry yacht]

oop

[still temple]

nice

ok so uh

just evaluate $f''(a)$

[still temple]

thats it

put a inside the second derivative

if it gives u a positive value then its a local minimum (it will)

[wintry yacht]

wioth justy

second deriv?

no need to use first?

[still temple]

well did u understand whats happening in part b or did u just yoink the solutions from the manual

[wintry yacht]

mannual

i think i know how to get them tho

[still temple]

well uh yea so the process is this

the first half is in part b

the second half is in part c

they found a from f'(x) = 0, which got them x = a

then, in part c, you are meant to substitute a in the second derivative f''(a)

[wintry yacht]

so f(a) = o?

0

[still temple]

no

f'(a) = 0

ok u seem to understand Leibniz notation better so imma use that

[wintry yacht]

why is negative second derivative maximum

and postivie second derivative minimum

it should be the other way around 😭

[still temple]

its because like

the second derivative gives u the "eventual feel" of how the first derivatives will be like

like

if u have a minimum

you know how the derivatives will be sloped down from the left

then equal to 0 at the minimum

[wintry yacht]

ye

[still temple]

and finally, sloped up onwards?

right so f''(c) finds the eventual behaviour of teh first derivatives to be positive

after the minimum

like f'(x) > 0 for x > a

so f''(c) > 0

[wintry yacht]

wat

eventual feel

[still temple]

You want to find the minima of the function $y = f(x)$. What you do is find the first derivative $\dv[y]x$ and the second derivative $\dnv2[y]x$ of it. Then the following process follows:
\e{enumerate}{
\ii Find the solutions to the equation $\dv[y]x = 0$. Those solutions are called \textbf{the critical points}. $x=a$ in your case.
\ii Plug those solutions into the equation $\vertr{\dnv 2[y]x}{x = a}$ and if $\vertr{\dnv 2[y]x}{x = a} > 0$ the point $x=a$ is a local minimum and if $\vertr{\dnv 2[y]x}_{x = a} < 0$ the point $x = a$ is a local maximum
}

[still temple]

this is the general id ai made it a bit more concise

[wintry yacht]

ill jyst remember

its opposite

[still temple]

https://upload.wikimedia.org/wikipedia/commons/7/78/Animated_illustration_of_inflection_point.gif

like look at this for example

the line is the tangent line at every point

[wintry yacht]

oh wait that makes sense

because maximum u go down after

[still temple]

after the maximum on the left, the tangent line's slope is negative (first derivative is negative) and after the minimum on the right, the tangent line's slope is positive (first derivative is positive)

[still temple]

yeahhh exactly

[wintry yacht]

k now iii

[still temple]

uh

just evaluate f(a) lmao

[wintry yacht]

do u put the a as ur x value

in the original surface area

formula

[still temple]

yes

[still temple]

the a u got here (how did they fuck up this formatting actually )

[wintry yacht]

ib 😭

exam in in 5 hours

[still temple]

lmao rip

[wintry yacht]

u prolly know what it is right

since ur so smart

[still temple]

heard of it yea

i think its probably the best thing for highschoolers to take out there

probs

[wintry yacht]

why do they say to sub it into graph

when u said i could just put it in the formuka

they gave us

[still temple]

yeah i mean

idk if yall have calculators during exams

[wintry yacht]

we have calculators

for paper 2

[still temple]

but evaluating f(a) is kinda bruh

[wintry yacht]

so this version

[still temple]

but yeah just sub it in and u r gucci

maybe if u were in a hurry or smth idk

[wintry yacht]

ib exams were leaked this year

[still temple]

woops

[wintry yacht]

and the grade they give u is based of averages

and im not cheating so my grade isnt gonnna be as good

sadge

[still temple]

[wintry yacht]

ok well ty

im prolly gonna open so many more tickets in the next 5 hours

lolololo

.close

[marsh citrus]

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[still temple]

Lmaoo alrighty good luck

dont die

[wintry yacht]

tyty ur prolly gonna see me again

[marsh citrus]

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[small vector]

what have you tried?

[still temple]

I'm stuck

[small vector]

what method do you usually use to solve these problems

sohcahtoa?

ok that's good

we have the opposite and adjacent though, not the hypotenuse

[still temple]

Wait I'm solving wrong questions lmfao I sent a different one oops

this is where I am ^

[small vector]

are you familiar with $\tan^{-1}$

[elfin berry]

b

[still temple]

yeah it just threw me off today as my calculator is very unique

i assume i just gotta do that with all my answers from there

[small vector]

for this question yeah

same with the one you sent before

[still temple]

so tan -1 (3) 71.57 degrees and that's it

[small vector]

that's it yeah

[small vector]

if you look you'll see that angle looks to be about 71 degrees

[still temple]

tan -1 w this one too right

[small vector]

-1 yes, tan no

[still temple]

ah

ty

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[fallen rain]

Two tanks were filled with water with 9 identical pipes. At first all the pipes filled the first tank, when it was 1/5 full the two pipes were switched to fill the second tank. At the moment when the first tank was completely full, the second tank was only 3/10 full. What is the ratio of the volume of the second tank to the volume of the first tank?

[fallen rain]

<@&286206848099549185>

.close

[marsh citrus]

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[topaz hull]

Is this right? For question 3 a ?

[marsh citrus]

@topaz hull Has your question been resolved?

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[exotic depot]

cna anyyone help with ungrouped data

[exotic depot]

12, 15, 18, 24, 26, 31, 34, 37, 40, 42, 42, 44, 44, 44, 45, 45, 46, 46, 46, 48, 48, 49, 49, 49, 49, 49, 49, 50, 50, 50, 51, 51, 52, 52, 52, 53, 53, 54, 54, 54, 55, 55, 55, 55

i arranged it already from lowest to highest but i dont know what step to do next

[marsh citrus]

@exotic depot Has your question been resolved?

[marsh citrus]

@exotic depot Has your question been resolved?

[marsh citrus]

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[flint jolt]

Range of sin^-1(x)+cos^-1(x)+tan^-1(x)

[flint jolt]

<@&286206848099549185>

Is it (0, π)

[eager hemlock]

pi/2 + arctan x just add pi/2 to the range of artan which yields 0 to pi

it should be [0, pi]

[flint jolt]

Not square

Bracket

But wolfram alpha showing the range
[π/4,3π/4]

[eager hemlock]

my mistake sorry yes at infinity so (0, pi)

[eager hemlock]

does it provide a reason?

[flint jolt]

Nope

[eager hemlock]

i don't think it is correct stick to 0 to pi

[still temple]

look for x

between [ -1,1]

sin^-x + cos^-x is pi/2

and for [-1,1] tan^-x

is

between -pi/4 and pi/4

so we add pi/2 to those

so we get [pi/4,3pi/4]

[still temple]

its correct

[marsh citrus]

@flint jolt Has your question been resolved?

[marsh citrus]

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[still temple]

can someone explain how the radius is 6?

[late geode]

do you know the formula for arclength and area of a sector?

[still temple]

l = r x theta
a = 1/2r^2theta ?

[late geode]

yeh, apply those here

[still temple]

I'm not sure where to start

like i have 5pi/2 and 15pi/2 as arc length and area respectively

not sure where to work it to get 6 for the radius though

[late geode]

use those values in your formulae

sub r * theta from the first equation into the second

[still temple]

1/2(5pi/2theta)^2 x theta?

[still temple]

yeah?

[late geode]

um

you should have an equation

[still temple]

well a = 1/2(5pi/2theta)^2 x theta?

[late geode]

no

not what i want from you

starting from the formula
use the given values for area and arclength ONLY
and tell me the two new equations you now have

[still temple]

theta = 5pi/2r
a = 5pi/4
?

[late geode]

no

[still temple]

Sorry i have no clue then

[late geode]

you're doing extra stuff i'm not asking for

don't overthink

you have the arc length

replace what represents the arctength with the given arclength ONLY
do nothing else

same for area

[still temple]

l = 5pi/2
a = 15pi/2
?

[late geode]

no

l = r x theta
literally just replace the l with the value you're given

a = 1/2r^2theta
literally just replace a with the given value

[still temple]

Oh okay sorry

5pi/2 = r x theta
15pi/2 = 1/2r^2theta
?

[late geode]

yes.

now note that r^2 = r * r

15pi/2 = 1/2r^2theta = 1/2 * r * r * theta
and you know the value of r * theta from the first equation

$\blue{\frac{5\pi}{2} = r\theta} \ \
\frac{15\pi}{2} = \frac12 r^2 \theta = \frac12 r \cdot \blue{r\theta}$

[elfin berry]

ℝαμΩℕωⅤ

[still temple]

Oh i see so
15pi/2 = 1/2 * r * (5pi/2)?

[late geode]

yes

[still temple]

So now we can just solve for r?

[late geode]

yes

[still temple]

one second

back sorry my dad was calling me

15 pi = 5pir/2
r = (15pi*2)/5pi
r = 30pi/5pi
r = 6?

[late geode]

that works

[still temple]

Is it the most efficient method? What'd you use?

oh i see online that someone multiplied both sides by 2, got 5pir = 30pi
then divided both sides by 5pi

and got 6

anyway thanks i'll close the ticket

.close

[marsh citrus]

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[hazy lion]

Hello I have another analysis question

[hazy lion]

I wonder if I am missing some piece of information that relates the norm of functionals in the dual and the norm of elements in X

they are bounded though

I figure you must take some nonzero x in X then $||x|| > 0$

[elfin berry]

jan Niku

[hazy lion]

but the given information doesnt directly conflict with this, at least as far as i can tell to use it

because then clearly $0(x) \leq ||x||$ is satisfied just by $||\cdot||$ being a norm

[elfin berry]

jan Niku

[glass silo]

The definition of the dual norm?

[hazy lion]

dual norm

lemme look it up

[hazy lion]

im not sure how this helps

so the norm of an element in the dual is the smallest C such that $|f(x)| \leq C ||x||$ for all $x$

[elfin berry]

jan Niku

[hazy lion]

but then just take f to be 0

and its fine

norm of all of the functionals (say its only 0) is 0 and theyre bounded

well i mean

theyre bounded anyways

[glass silo]

There's an equivalent formulation that $\norm{f} = \sup_{x\neq 0} \frac{\abs{f(x)}}{\norm{x}}$

[elfin berry]

@glass silo

[hazy lion]

i dont see the problem with this either

what is this supposed to conflict with?

why would this force us to pick up additional functionals

[glass silo]

Because the dual space is the set of all possible functionals

[hazy lion]

all bounded linear functionals yea

[glass silo]

There's the zero functional, but if you have a nonzero element in the space you can cook up a new functional

[hazy lion]

? how

we dont know anything about the space other than that its normed

so it doesnt seem possible to just create some functional and ensure that its in the dual still

[glass silo]

Well, if x is in X, any scalar multiple of x is as well

[hazy lion]

i dont think thats true

[glass silo]

You can define one by fixing f(x) to be nonzero and demanding linearity

[hazy lion]

we dont know that X is linear

only that its normed

[glass silo]

aren't normed spaces defined over vector spaces though?

[hazy lion]

idk lemme see

okay so sure

normed space is closed by definition

wait thats not true

well bad use of the word closed

anyways

[hazy lion]

not sure what you mean here

[glass silo]

You know e.g. how you can define a linear map by instead showing how it acts on basis elements

[hazy lion]

hrm

[glass silo]

Then saying it's linear tells you how it acts on all other elements

[hazy lion]

hmm can we just use the norm

no

because the norm isnt linear

[hazy lion]

im struggling with this so we have to create a basis?

[glass silo]

Well, you don't have to create a "full" basis (you only know that there's a nonzero element in x, but you don't know if there are other ones that aren't scalar multiples of said x)

[hazy lion]

i dont follow

[glass silo]

hmmm

[hazy lion]

im not sure what to ask im just not following

this is in the section for hahn banach

but i couldnt figure out how to work the theorem in

because it seems hard to create a subspace of X

well, you arent sure a subspace of X exists that isnt only the 0 vector

so its hard to apply the theorem

[glass silo]

Take the span of {x}

Anyways, the idea was that if I told you that f is linear and told you that f(x) = 1, for example, you should hopefully know that something like f(5x) = 5f(x) = 5

[hazy lion]

oh because its a vector space

so create a subspace span {x} where x!=0

then

we can define $f(x) = ||x||$

[elfin berry]

jan Niku

[hazy lion]

and its necessary to show that f is bounded and linear

which, its not

oh wait well

maybe you define it on the span of (x) then extend it?

its linear on the span of x

since any ax+bx = (a+b)x

[glass silo]

Sure though with that said, the OG problem seems like you don't really need that much?

[hazy lion]

im just trying to answer it at all lol

idk what im doing really

actually now im not sure this works

well idk id have to track it out im worried about the extension

[glass silo]

is it like part of a bigger problem?

[hazy lion]

no

thats the whole problem

which i guess i get what youre saying and i had the thought too, just find some nonzero functional, and show its linear and bounded on X

finding it is hard though

[glass silo]

Defining it to be linear

[hazy lion]

you know actually the norm thing doesnt work

because its not linear

not even in just the span of a vector

[hazy lion]

but we have to define it explicitly right

or least in some way that demonstrates its not equivalently zero

if not explicitly

[glass silo]

Saying it's linear and fixing some nonzero output $f(x_0)$ should be enough, no?

[elfin berry]

@glass silo

[hazy lion]

dont you have to show its linear?

i mean we have to demonstrate that its in the dual

and that its defined on all of X

and we have to show its bounded

how do you do that without a definition

[glass silo]

Well, you're defining it to be linear, you could e.g. set it to map to zero on all other vectors that aren't multiples of x0

[hazy lion]

there is a theorem in the book that tries what i tried but it appears they made the same mistake and just didnt catch it

equation 10 isnt correct

[glass silo]

It is correct

[hazy lion]

$f(x) = f(ax_0) = |a|||x_0||$

[elfin berry]

jan Niku

[hazy lion]

norm isnt linear

[glass silo]

"we define a linear functional"

You're not defining f(x) = norm{x}

[hazy lion]

but then this extension doesnt agree with the original functional on the span of the vector

or do we not require that

[glass silo]

You're using Hahn-Banach to extend it, no? So it's forced to agree, you're extending f to the whole space

[hazy lion]

but maybe HB isnt necessary because we can just define it and not worry about it existing?

so we dont need to satisfy the assumptions of HB

because we can just find it directly?

it exists by virtue of we built it

is that right

[glass silo]

I mean I would think that for the OG problem, that's fine, as we only need to show there's a nonzero functional, we don't need to show it has any particular norm