help-13

[potent fractal]

https://en.wikipedia.org/wiki/Hölder's_inequality#Proof_of_Hölder's_inequality :^)

[lyric narwhal]

😭

because I don't know measure theory

i am familiar with young's inequality though

[potent fractal]

There're multiple proves on wikipedia, and none of them use Holder's inequality, because otherwise it would be circular reference

[lyric narwhal]

we are proving holder's inequality right?

but I don't understand the statement of holder's inequality because I don't know measure theory

I got the basic idea but like..

[potent fractal]

It's this (in a special case) #help-13 message

[lyric narwhal]

isn't my inequality a special case of holder's inequality?

[potent fractal]

It is

But like

It's essentially the statement

[lyric narwhal]

I don't follow

[potent fractal]

Of course the general form requires measure theory and looks like this:

[lyric narwhal]

yep that's the part that scares me lol

[potent fractal]

But if you don't know measure theory then this is enough to grasp the idea:

[lyric narwhal]

okay

so you mean to say that I can in some way adapt the proof they've given to prove this inequality directly?

instead of using the measure theoretic stuff

[potent fractal]

yep

[potent fractal]

This proof can be adapted for your case, I can help with that

[lyric narwhal]

okay then

that's what I wanted, thanks

for our purposes S = R

I assume

so okay the step where they integrated

instead we want to sum

so $\sum_{i=1}^n |x_i y_i|\le\frac1{p}\sum_{i=1}^n |x_i|^p + \frac1{q}\sum_{i=1}^n |y_i|^q$

[wraith dagger]

kheerii

[lyric narwhal]

so if we can prove $\frac1{p}\sum |x_i|^p\le\left(\sum |x_i|^p\right)^{1/p}$ we'd be done

[wraith dagger]

kheerii

[lyric narwhal]

but this does not seem true

[lyric narwhal]

hey I don't know what to do next lol

[potent fractal]

One sec I was adapting the first part

[lyric narwhal]

ahh

thank you so much

for doing all this lol

[potent fractal]

yeah

[potent fractal]

And here we seentially have $\sum_{i=1}^n |x_i y_i| \le \frac{1}{p}\lVert x\rVert_p^p + \frac{1}{p}\lVert y\rVert_q^q = \frac{1}{p} + \frac{1}{q} = 1$

[lyric narwhal]

What??

[wraith dagger]

EQUENOS

[lyric narwhal]

Yeah that's the part that I didn't understand

Are we basically scaling all the values of xi and yi so that the sum of (xi)^p = 1?

[potent fractal]

yes

[lyric narwhal]

Hmm, alright...

[potent fractal]

After scaling both x and y the inequality stays unchanged

[lyric narwhal]

I suppose we can scale the quantities because the inequality is homogeneous

[lyric narwhal]

Yeah

Okay now it makes sense

Turns out I was just intimidated by the measure theory stuff

[potent fractal]

Basiclaly we divided the inquality by $\lVert x\rVert_p \lVert y \rVert_q$

[wraith dagger]

EQUENOS

[lyric narwhal]

Yess

That's awesome

[potent fractal]

indeed

[lyric narwhal]

So it's the young's inequality which is really important here

[potent fractal]

exactly

[lyric narwhal]

That's where the 1/p + 1/q = 1 thing comes from

Because I always found that super weird

I don't understanf thr motivation for this at all though like how and why did people think of this

[potent fractal]

Surprisingly this relation plays a fundamental role in functional analysis

[lyric narwhal]

how so?

[potent fractal]

There're so-called Banach spaces and dual spaces

[lyric narwhal]

oh I've heard of a dual space before

[potent fractal]

Banach spaces constitute a reasonable generalisation of finite-dimensional vector spaces with a norm

[lyric narwhal]

interesting

[potent fractal]

(bascially)

And there's probably the most used class of examples of banach spaces: so-called $l_p$-spaces

[wraith dagger]

EQUENOS

[lyric narwhal]

it is not lol

[lyric narwhal]

hmm

I can't wait to study all this stuff

it sounds really cool

would this fall under abstract algebra?

[potent fractal]

You probably spotted a familiar sum here

And then mathematicians were like "hold on what's the dual of l_p"

[lyric narwhal]

ahhh

and it's just ||x||_q

[potent fractal]

Yeah it's l_q

[lyric narwhal]

yeah sorry

[potent fractal]

Yep the dual space of X is the space of all continuous linear functionals on X

[potent fractal]

Usually that falls under functional analysis
In some universities that might pop up in real analysis

[lyric narwhal]

I don't understand this stuff yet

i haven't studied it at all lol sorry

[potent fractal]

oh that's okay

[lyric narwhal]

ah okay

[potent fractal]

Does jee involve linear algebra btw?

[lyric narwhal]

not much lol

we were taught basic manipulations with matrices and determinants and some vector algebra but that's about it

[potent fractal]

Oh that's enough to understand basic examples of dual spaces

[lyric narwhal]

I have encountered them once before

[potent fractal]

In a finite-dimensional euclidean space any linear function is just f(x) = (p, x) for some vector p

Here ( - , - ) is the dot-product and x is a vector

[lyric narwhal]

yep

so the dual space is just a space consisting of all such f?

[potent fractal]

All such functions form a space which is called the dual space

[potent fractal]

yes

[lyric narwhal]

okay

so what's the use of it

[potent fractal]

In finite-dimensional spaces this is incredibly boring

Becuase it's just R^n

But in infinite-dimensional spaces it's not so trivial

[lyric narwhal]

I think I came across an example where I created like a dual basis for a set of three vectors

in terms of their scalar triple product and stuff

let me see if I can find it

[potent fractal]

It helps with defining some useful topologies, for example

But that's another story

[lyric narwhal]

yeah I came across something called a reciprocal system of vectors

[potent fractal]

Yeah that looks like a dual basis

[lyric narwhal]

but how is this a basis for the space we defined above?

like how are three vectors in R^3 making up a basis of the dual space

you said the dual space was a space of functionals

[potent fractal]

It's not quite the basis, but there's a natural isomorphism that sends these vectors to the dual basis

Basically we say that f(x) = (a', x), g(x) = (b', x), h(x) = (c', x)

f, g and h act like a basis of the dual space

[lyric narwhal]

ohh okay, so then these three functions make up the basis

alrighty

it also makes sense why we define these vectors like this then

because i believe f g and h turn out to be orthogonal vectors

[potent fractal]

We achieved f(a) = 1, f(b) = 1, f(c) = 1

[potent fractal]

a, b, c were not orthogonal so likely not

[lyric narwhal]

f(a)=1, f(b)=f(c)=0
g(b)=1, g(a)=g(c)=0
h(c)=1, h(a)=h(b)=0

[potent fractal]

yep

[lyric narwhal]

ooh but this means that a vector v in R^3 can be written as f(v)a'+g(v)b'+h(v)c'

right?

[potent fractal]

f(v)a + g(v)b + h(v)c

[lyric narwhal]

[potent fractal]

Let v = x a + y b + z c and use linearity of f, g, h

[lyric narwhal]

f(v)=x, g(v)=y, h(v)=z

oh yeah of course

my bad

okay that's kind of awesome actually

i'm still trying to wrap my mind around this stuff

it's so confusing but so satisfying when you get it

[potent fractal]

Yeah that's the beauty of learning mathematics :)

[lyric narwhal]

if you don't mind can you give me another concrete example of a dual space?

[potent fractal]

Hm we can try a finite-dimensional space with a different norm

[lyric narwhal]

sure :)

[potent fractal]

Let's say $\lVert x\rVert = \sum_{i=1}^n |x_i|$

[wraith dagger]

EQUENOS

[potent fractal]

That's so-called l1-norm

[lyric narwhal]

oh it's the same as the l_p we saw before

[potent fractal]

Yep

p=1

Now we want to find all continuous linear functions on this space

[lyric narwhal]

yes

[lyric narwhal]

are linear functions not continuous by design?

[potent fractal]

Continuous means that f^{-1}(open set) is open

[potent fractal]

Generally they're not

[lyric narwhal]

hmmmm

oh yeah okay that makes sense actually

[potent fractal]

Trying to remember some stuff, 1 sec

[lyric narwhal]

all good

[potent fractal]

One can verify that the space of linear functions on a finite-dimensional space is also finite-dimensional

Because it's enough to define a linear function on basis vectors

[lyric narwhal]

but does it necessarily have the same dimension?

[potent fractal]

The interesting part is what's the norm on the dual space

[potent fractal]

For finite-dimensional spaces, yes

[lyric narwhal]

hmmmmmm

[potent fractal]

Because we can construct the dual basis

Basically, if v1, ..., vn is the basis of X

Then let f_k(vk) = 1, 0 on other basis vectors

[lyric narwhal]

kronecker delta?

f_k(v_i)=delta_{i,k}

[potent fractal]

yep

[lyric narwhal]

is it necessarily that it has to be 1 or is that just some normalisation

[potent fractal]

It's not necessary

But it's convenient for our purposes

[lyric narwhal]

so that something like this holds?

[potent fractal]

yes

[lyric narwhal]

alright that makes sense

[potent fractal]

Next we can prove that f_1, ..., f_n are linearly independent

(assuming you know what that means)

[lyric narwhal]

I do

[potent fractal]

splendid

[lyric narwhal]

I mean, intuitively it makes sense that these are lin. ind. but I do not know how I'd even start trying to prove that

[potent fractal]

Suppose there exists some non-zero tuple of coefficients c_k such that
c1 f1 + ... + cn fn = 0
Substitute v1 to both sides:
c1 = 0
Substitute v2 to both sides, etc.
c1 = c2 = ... = cn = 0

Contradiction

[lyric narwhal]

oh huh yeah

clever

[potent fractal]

So the dimension is at least n

[lyric narwhal]

and assuming the dimension of the original space and the dual space is the same this must be a basis

[potent fractal]

On the other hand, every linear function f can be represented as f = a1 f1 + ... + an fn and therefore the dimension is at most n

[potent fractal]

yes

[potent fractal]

f(v1) = a1, f(v2) = a2, ... and that's how we find all ak

Anyways, currently we once again arrive at a boring dual space. It's of the same dimension

[lyric narwhal]

lmaoooo

[potent fractal]

However we can find the norm

[lyric narwhal]

do we not need the basis first?

[potent fractal]

nah

[lyric narwhal]

oh?

[potent fractal]

Well, we have it anyways

[potent fractal]

.

So the interesting part here is what's the norm in X*, given that X is endowed with l1-norm

[lyric narwhal]

ooh

hold on I think I can work this out

let me give it a shot

[potent fractal]

By definition, $\lVert f\rVert = \sup{|f(x)| \ : \ \lVert x\rVert \le 1}$

[wraith dagger]

EQUENOS

[potent fractal]

okie

[lyric narwhal]

are we assuming an n dimensional X?

[potent fractal]

Yes

[lyric narwhal]

alrighty

and I have to work out the norm of f in terms of its coefficients?

[potent fractal]

And we already proved that dimX* = dimX

[potent fractal]

yes

[lyric narwhal]

wait it'll just be the largest coefficient

[potent fractal]

exactly

[lyric narwhal]

how is that a norm what?

[potent fractal]

That's so-called $l_\infty$-norm

[wraith dagger]

EQUENOS

[lyric narwhal]

ahh

because it's the max

[potent fractal]

$\lVert f \rVert_\infty = \max|f_i|$

[wraith dagger]

EQUENOS

[lyric narwhal]

yep

[lyric narwhal]

but why choose this norm specifically

[potent fractal]

That's the special case of operator norm

There's a very natural way of endowing the space of operators with a norm

[lyric narwhal]

oh I think I've seen that

[potent fractal]

It looks almost identical to this

[lyric narwhal]

$$\norm{f}=\min{c: \norm{f(x)}\le c\norm{x}\ \forall x\in X}$$

[wraith dagger]

kheerii

[lyric narwhal]

something like this?

[potent fractal]

Yeah it's equivalent

(I assume you meant c>=0)

[lyric narwhal]

well yes

does a norm not have to be non-negative anyways?

[potent fractal]

oh yeah the lower bound for c doesn't matter

mb

[lyric narwhal]

hmm alright

thank you so much for giving me a mini lesson on dual spaces

and for answering the original question lol

[potent fractal]

You're welcome :)

[lyric narwhal]

looking at the wiki it seems like you are right, the infinite dimensional case is much more interesting

since the dual space has higher dimension than the original space

[potent fractal]

They're both infinite-dimensional but behave differently

[lyric narwhal]

the dual space is isomorphic to F^A, where F is the field the original vector space was over, and A is the indexing set of a countable basis of V

[lyric narwhal]

that's cool

[lyric narwhal]

the first thing is what we just proved yes?

[potent fractal]

We proved that in a very special case

They're talking about a more general thing

(Which is amazingly also true)

[lyric narwhal]

yeah under any measure

[potent fractal]

L_p spaces are functional spaces

[lyric narwhal]

i don't get what the second thing is trying to say

[potent fractal]

The main point here is that (L_inf)* \neq L_1 as one might hope

[lyric narwhal]

LOL

the dual space of a dual space giving the original space

that would be funny

almost seems like it should be true

[potent fractal]

(L_2)** = L_2

(yay)

In fact, L_2 is almost too amazing to exist (it's a separable hilbert space)

[lyric narwhal]

wait, is it true that (L_p)*=L_q?

(where 1/p+1/q=1)

[potent fractal]

Yes, if 1 < p < inf

[lyric narwhal]

woahh

[lyric narwhal]

so this means (L_2)*=L_2 itself

that is awesome

[potent fractal]

yes

[lyric narwhal]

is this fact related to the inequality we were talking about initially?

[potent fractal]

Yes

The fact that (L_p)* = L_q directly follows from Holder's inequality

[lyric narwhal]

oh

I thought it would be the other way around

[potent fractal]

Since interchanging p and q doesn't change anything conceptually, (L_p)** = L_p

[lyric narwhal]

very cool

[potent fractal]

yep, that's called being reflexive

[lyric narwhal]

so here considering $X$ to have $l_p$ norm we would have $$\begin{aligned}
\norm{f}&=\sup{|f(x)|:\sum |x_i|^p\le 1}\
&=\sup{|\sum a_ix_i|:\sum |x_i|^p\le 1}\
\end{aligned}$$

[potent fractal]

yep

[lyric narwhal]

clearly the max would just be when norm{x}=1

but what would f look like

hmmmm

[potent fractal]

Using the dual basis we can certainly obtain something

Oh and triangle inequality

[lyric narwhal]

I'm just kind of confused about what f looks like here lol

[potent fractal]

a1 f1(x) + ... + an fn(x)

[lyric narwhal]

I think we need to find the basis

[lyric narwhal]

yes, but what is f1(x)?

[potent fractal]

Where fk(vj) = delta(k,j)

[lyric narwhal]

hmm

[potent fractal]

Where {vj} is the basis of X
(so x = x1 v1 + ... + xn vn)

[lyric narwhal]

this looks right

not sure if it is

okay yeah it should be

[wraith dagger]

kheerii

[lyric narwhal]

okay yeah this just follows from Holder's now

neat

$$\begin{aligned}
|\sum a_ix_i|&\le\sum |a_ix_i|\
&\le\left(\sum |a_i|^q\right)^{1/q}\left(\sum |x_i|^p\right)^{1/p}\
&\le\left(\sum |a_i|^q\right)^{1/q}\end{aligned}$$ so $X^*$ has $l_q$ norm

[wraith dagger]

kheerii

[potent fractal]

Yep

An important remark is that the equality is achievable

[lyric narwhal]

yes, when a_i = x_i^(p/q) for all i, and sum (x_i)^p = 1

right?

[potent fractal]

a1=x1=1, ai=xi=0

[potent fractal]

Maybe 😅

Too lazy to check

[lyric narwhal]

lol it's alright

it's just from when equality holds in the normal holder's inequality, so it should be right

thank you so much once again!!

your help is much appreciated

.close

[cedar kiln]

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[tepid wyvern]

Good evening, I've got this thing to resolve when K is -1.

[tepid wyvern]

Also, how do I make this system incompatible by just modifying the third equation?

<@&286206848099549185>

[cedar kiln]

@tepid wyvern Has your question been resolved?

[cedar kiln]

@tepid wyvern Has your question been resolved?

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[rustic storm]

Is this right?

[rustic storm]

Cause twacher didint explained with triple things

And with doble i shouldve shown the space where the both are right

But here idunno

Got this too

<@&286206848099549185>

Its at the start

Blue letters

At the top i mean

[cedar kiln]

@rustic storm Has your question been resolved?

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[static tree]

can someone help me

[cedar kiln]

@static tree Has your question been resolved?

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[olive dirge]

hi

[sweet bay]

hi

[olive dirge]

I need someone to correct me in this

this gotta be false right?

dosent the function need the intervals to have same value

[sweet bay]

Are you familiar with Roulle's Theorem?

[olive dirge]

kind of yeah

[strange hamlet]

Because it must exist, rolle’s theorem

Let me see

[olive dirge]

yeah i am kinda confused..

[strange hamlet]

Let f’(c)=2/2=1

I think it’s mvt

[olive dirge]

if its mvt it has to be flase then

[strange hamlet]

Yes

That’s why I felt a bit weird

[dreamy void]

I think you are right, that this is not always the case e.g. f(x) is strictly decreasing in [-2,0]

[olive dirge]

yeah i think so as well. idk still confused

just to make sure i put it in ai 😂

chat gpt says false gemini says true

im like bro 🤣

[strange hamlet]

F’(x)=0 x in [-2,0]

[surreal pivot]

ye

[dreamy void]

the second condition is not met, so the statement is not necessarily true

[olive dirge]

yeah it is false

thank you guys for the help, I appreciate it!

.close

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[quaint scarab]

so, this is probably stupid, but I'm not sure how to interpret this question:

Lunar astronauts placed a reflector on the Moon's surface, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. (a) To what accuracy in meters can the distance to the Moon be determined, if this time can be measured to 0.100 ns?

[quaint scarab]

could anyone help me with understanding how I give an "accuracy in meters"?

[crimson sedge]

im guessing it has to do with sig figs?

[quaint scarab]

well, even then, that's 10 sig figs and the answer I'm getting, (c*2.5)/2=374740572.5m, is only 10 digits total

[dawn junco]

well ok the laser travels at speed c alright

what distance does the laser travel in 0.100ns then ?

[quaint scarab]

0.0299792458m

[crimson sedge]

how is 0.100 10 sif figs?

[quaint scarab]

ns is 10^-9 s

[crimson sedge]

you would choose the least amount of sig figs in the quesiton

[fair geyser]

i'm just guesssing but it maybe means between 0.01499 and 0.04497 it gets rounded to 0.029979

so you can trust one digit after decimal?

0.1 m

are there options?

[quaint scarab]

there are not options, it's a blank input field

[fair geyser]

well 10 sigdigs is also 0.1 m

[dawn junco]

in 0.100ns the two zeros are to be trusted too

[fair geyser]

ah

[dawn junco]

that's how I understand it at least

[fair geyser]

of course

0.001 m then

idk

well ok, i guess it's what cka said

minimum between inifintite digits and 12

0.000000000001 m?

so confusing

actually it makes no sense that two zeroes count

[dawn junco]

why do you have those huge changes

[dawn junco]

like this answer is correct

[crimson sedge]

is there a unit? m? sig figs? digits?

[dawn junco]

it's just a question of which accuracy you give

[dawn junco]

well the two zeroes are just a way to report accuracy essentially

0.1ns vs 0.100000ns have very different tolerances

[fair geyser]

they don't matter

[crimson sedge]

they do matter. they are significant

[fair geyser]

why not say 0.001 then

[crimson sedge]

bc its not 0.001 its 0.100.

not 0.101, 0.105, 0.111 --> its 0.100

[fair geyser]

0.0299792458m makes sense, i can live with that, but it's ignoring the two zeroes

[crimson sedge]

this makes more sense

no one saw that right

[fair geyser]

i didn't save it

[crimson sedge]

good

ignore it

[fair geyser]

half of what i thought

[crimson sedge]

from my good friend ChatGPT

[fair geyser]

still the zeroes don;t matter

[crimson sedge]

i guess so

i didn't know if it was a sig fig question or what

[dawn junco]

we'll know when OP inputs something ig

[fair geyser]

oh shit

it's divided by 2 because it's roundtrip not because reasons

[dawn junco]

reasons = roundtrip

dang we managed to miss that

[fair geyser]

the problem is it rounded the speed of light

if it didn;t would this answer have around 10 digits?

[dawn junco]

but 0.100ns only has 3 digits precision as we argued before

so assuming the question gives a shit about sig figs, the answer should have 3 digits precision then

[fair geyser]

no i don't buy that

i can't explain why they say 0.100 instead of 0.1

but it's the increment, it doesn't have an additional precision to it

[dawn junco]

yeah that could make sense also

the only question is whether OP's answer grading program is a sig fig hardass or not

[cedar kiln]

@quaint scarab Has your question been resolved?

[fair geyser]

@crimson sedgeask it to use the exact speed of light

[crimson sedge]

chatgpt is not the most reliable

so i'd take it with a grain of salt

[fair geyser]

thanks

no of course not

[quaint scarab]

I've now tried .149 and it's counted as wrong

[fair geyser]

.149 is wrong in any case

you would round to 0.15

(i don't know if it's 0.150)

i would try 0.15

[quaint scarab]

sorry, input -8 by mistake and didn't notice, not -9, answer is .015m

[fair geyser]

ok

[quaint scarab]

now to figure out part b....

[fair geyser]

i just noticed chatgpt messed it up the second time

[quaint scarab]

ehh, might need help with this part too...

(b)
What percent accuracy is this, given the average distance to the Moon is 3.84*10^8 m?

[fair geyser]

uh

[quaint scarab]

initial thought was to treat .015 as the top difference value and perform (0.015/3.84*10^8)

[fair geyser]

yeah

but you maybe divide by 2?

[quaint scarab]

bottom value? it's already the average distance

[fair geyser]

not becuase of the roundtrip

[quaint scarab]

ah

hm

[fair geyser]

because 5% is ±2.5%

but i don't know

it would be strange

[fair geyser]

i would do this

wait wait

"top difference" is half of that

so yeah i guessed right

or at least your instinct was the same

or i misunderstood

[quaint scarab]

so we agree it should be (0.015/3.84*10^8)?

[fair geyser]

yeah

no complications first

[quaint scarab]

awesome

it was correct, value of 3.9e-9

[fair geyser]

cool

[quaint scarab]

appreciate all your help everyone, very weirdly phrased question I hadn't had to deal with yet, so I was a bit lost haha

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[amber elk]

derivatives of inverse, what am I even doing?

[lime tartan]

First try doing what it's asking below

The function compositions

And it'll become immediately clear

[amber elk]

oh i see

thanks 🔥

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[languid cypress]

This is what I initially got stuck on

[languid cypress]

Trying to figure out Cot (22.5) without a calculator. I’m trying to do 1/tan(45/2) but I’m stuck in how to get the answer of Root square 2 +1

Perhaps something like this could work?

[stable oasis]

I think you're right , keep going

[languid cypress]

Wouldn’t it be Rad2/2? since it’s 45 degrees

[stable oasis]

Do you want the final answer?

[languid cypress]

I have the final answer.

[stable oasis]

A/2 is 22.5° so A is 45°

[languid cypress]

That’s the final answer

Oh I have an earlier image, there’s two images.

[stable oasis]

Oh i think i did it wrong lmao , i got

.

[languid cypress]

Yeah I am confused

Ok my calculator does show that this makes the final answer. How would I solve it algebraically?

[stable oasis]

Strike out the 4 then multiply with conjugate of √(2-√2) which is √(2+√2)

[stable oasis]

I confirmed with a friend that my answer is correct

[languid cypress]

I have no idea. I’m just trying to get to the key answer lmao

[stable oasis]

Okay , so what are you confused with?

[languid cypress]

The algebra mainly

[stable oasis]

Did you try this?

[languid cypress]

Got stuck here since operating would just cancel it with the -Rad2 downstairs

[stable oasis]

To solve this

[cedar kiln]

@languid cypress Has your question been resolved?

[languid cypress]

Where did this 4-2 come from? When you moved the 2 from beneath the square root it counted as a power of 2 to multiply the 2 -> 4 and remove the square root on the other 2?

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[wraith rapids]

Why would this not be even?

if you plug in -x

you get -x^2sin(-x)

which cancels out to get back to x^2sinx

[humble karma]

$(-x)^2 = x^2$

[wraith dagger]

Azyrashacorki

[wraith rapids]

When I do x^4 it wants me to take it as -x^4

Wouldn't -x^4 also equal x^4

[dull oxide]

what?

[empty bobcat]

what?

[dull oxide]

A function is even if $f(x)=f(-x)$ for every $x$ in the domain. $f(1)=\sin 1$ but $f(-1)=-\sin 1$, so it cannot be even.

[wraith dagger]

SWR

[wraith rapids]

Yeah you are right lol

[wraith rapids]

thanks

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[shell agate]

idk if I did tis right but it converges for certain a and p but not all so is that divergent or did I do it wrong

[dire geode]

Show

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[slim thicket]

I have a study guide that I really need help running through even if its like one question per section please and thank you

[slim thicket]

This is the study guide

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[vernal sentinel]

'Surprise' boxes of chocolates each contain 15 chocolates: 3 lemon, 5 are orange and 7 are strawberry.
Petra has a lot of Surprise chocolates. She chooses 3 chocolates at random from the box. She eats the chocolate before choosing the next one.

"Find the probability that none of Petra's 3 chocolates has orange flavour."

Guys, i found the answer by 10/15 * 9/14 * 8/13 = 24/91

will i get the same answer if i multiply one by one like LLL, LLS, LSL, LSS, SLL, ... (etc)?

[radiant scaffold]

lemon and strawberry perhaps

[vernal sentinel]

yes

bcs i didn't realize we can do 10/15 * 9/ 14 * 8/13 previously

and i counted it one by one, but i got it wrong

anyways thank you!!

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[cold dust]

If Desmos(?) help isn't welcome here, please let me know.
Anyway, I want to fill the insides of the areas bounded by multiple equations (image 1). Here's the link: https://www.desmos.com/calculator/7rpk3fyz9f?lang=en

[cold dust]

Yes, but I haven't graphed it yet

I'm currently doing the waves

That's exactly what I want to do

Shade an area

I want to shade the red area with red and blue area with blue

No I want to shade them

I already did the calcs

Yeah

I have the bounds

I don't know how to shade them

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@cold dust Has your question been resolved?

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[gray isle]

Suppose that m and p are real numbers for which the polynomial
f(x) = x^2 + mx + p
has two distinct positive real roots. Prove that the polynomial
g(x) = x^2 - (m^2 - 2p)x + p^2
has two distinct positive real roots

Hi, I ended up using vieta's formula to set m and p equal to the roots, then I subbed in those roots into the B and C terms for g(x), and found the discriminant which I got R_1^4 - 2r_1^2r_2^2 + r_2^4

[gray isle]

can someone confirm

[pastel vault]

don't you just want to show that $(m^2 - 2p)^2 - 4p^2 > 0$ given that $m^2 - 4p > 0$

[wraith dagger]

south, just south

[pastel vault]

yeah you've massively overcomplicated the problem

[gray isle]

could you show me a simpler process?

[pastel vault]

this is the simplest

just use the fact the discriminants have to be > 0

[gray isle]

I used the method from gpt lol

thats why it got so complicated

[pastel vault]

DO NOT TRUST GPT

[gray isle]

is the method that GPT uses still correct?

[pastel vault]

I have no interest in reading through GPT anyways to see if it's correct

GPT suggested a horribly inefficient method

you do not need to find the roots of f(x) explicitly to do this problem

[gray isle]

Thank you

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[crimson sedge]

There are only one pair of values.

Use the factor theorem.

i am getting -23

i dont know whether it is correct

i am not sure

If (x-a) | f(x), then f(a) = 0

[crimson sedge]

of what?

a or b?

?

a = -23

well

a = 23

and b = -14

1 sec

wdym

yup

[wraith dagger]

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[crimson sedge]

,w subsitute x = 1 in x^3 - 10x^2 + ax + b

[wraith dagger]

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[crimson sedge]

,w substitute x = 2 in x^3 - 10x^2 + ax + b

[wraith dagger]

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[crimson sedge]

,w solve {2a+b-32=0} and {a+b-9=0}

[crimson sedge]

what is the mistake in this

[wraith dagger]

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[crimson sedge]

well

you have made a mistake

where?

Ok bro

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[crimson sedge]

thank you bro

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[crisp halo]

Hello!

I am trying to learn how to see what kind of infinite series I am analyzing

[crisp halo]

I think 2 is a geometric series

11-14 is telescoping like the instructions say

but i dont know what the ones to the power of n are called

[dusk finch]

geometric

[crisp halo]

and is 2 also geometric?

[crimson sedge]

yes

[crisp halo]

okay thanks

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[dusk finch]

everything else seems to be either geometric or a sum of 2 geometric series

if you do some appropriate manipulations

[crisp halo]

I will try!!

.reopen

[cedar kiln]

✅

[crisp halo]

Hello again!

I am stuck on 3.

I don’t understand how to see what r or a is

also i dont understand how to get it into the form so that the power says n-1

[vivid wadi]

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

[wraith dagger]

kaue

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[grave nexus]

black is f(x) and red is f'(x)

is the f'(x) a correct sketch?

[dull oxide]

Acceptably close enough

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[sullen wind]

I have to solve it, dont have any certain idea

[edgy sun]

you could start by saying 2^2x instead of 4^x

idk might not help much

[sullen wind]

yea, but what next...

[cerulean sail]

noticing 10 is 5 * 2 might help a bit too!

[sullen wind]

okey, maybe it's stupid question but after deviding like that what I should do with a "x"

it should be like (5*2)^x

or what

[cerulean sail]

You can then write that as 2^x * 5^x, and if you spot it, you can turn this into a quadratic (in)equation if you e.g. divide by something "nice"

[merry solstice]

(a^x)*(b^x)=(ab)^x

[merry solstice]

Chartbit is talking about substitution

so he will let (a^x) be another variable so that this equation is easier to solve

[cerulean sail]

(you can and may find it easier to substitute, but what you'd make your "substitution" for probably isn't integer!)

[merry solstice]

a is a variable

[merry solstice]

so what do you think should be this a term

and how should we manipulate the equation so that it is easier to substitute this variable

[sullen wind]

I think I should do something with 2^x

but I'm not sure

ah

omg I'm so stupid with this example

it kills me

[merry solstice]

Think of it as you are learning, not that you are stupid

[merry solstice]

look at chartbit's hint, the "substitution isn't an integer

or you could use two variables

I think

[sullen wind]

yea, it`s a good way to think about it and I'm trying to

[cerulean sail]

Stealing the previous hint, you effectively have as your inequality
[
2^{2x} - 2\cdot 5^{2x} < 2^x \cdot 5^x
]
if that helps any more...

[wraith dagger]

@cerulean sail

[merry solstice]

I need to learn how to code TeXit lol

anyways

[merry solstice]

so what are your variables that you will substitute?

[sullen wind]

em 2^x and 5^x?

[merry solstice]

good

we'll denote 2^x as a and 5^x as y

so

what would this make our equation

I'll help you out

we can denote 2^2x as (2^x)^2

because of the identity (a^bc)=(a^b)^c

[merry solstice]

so what would we write our equation as?

[sullen wind]

so it's gonna be a^2 - y^2 -(a*y)>0?

[merry solstice]

oops I meant 5^x as b

remember, there is a -2*(5^x)^2

[edgy sun]

I have a follow up question after he’s done

[merry solstice]

So it would be a^2-2b^2-ab=0

@cerulean sail

how would we solve this equation

[merry solstice]

what is the first step

in doing this

[sullen wind]

I don;t get this one, it is from where?

[merry solstice]

sorry, I forgot the negative

the term is (-2*5^2x)

which we can denote as (-2*5^x^2)

We get a^2-ab-2b^2<0

[sullen wind]

oh okey I get it now

[merry solstice]

sorry I had a brain fart

[merry solstice]

what is the first step in solving this?

[sullen wind]

reduicng it to (a-b)^2?

oh no

no

we cant

fck\

[merry solstice]

good

[sullen wind]

we can XD?

[merry solstice]

another thing we can try is factoring

[merry solstice]

what's that

[sullen wind]

I'm suprised that we can simplify it to this form (a-b)^2

with 2b^2

[merry solstice]

no

we can factor the equation

not as (a-b)^2

but as (a-2b)(a+b)

do you get how I got here?

[sullen wind]

yea you just modify it right?

with

2b

bneacuse

we have it here

[sullen wind]

.

so it need to be in our

how we can say that in english

omg

[merry solstice]

good

[sullen wind]

formula?

[merry solstice]

quadratic formula?

well I find factoring easier if possible

[cerulean sail]

My original suggestion was more that e.g. if you divide through the $2^{2x} - 2\cdot 5^{2x} < 2^x \cdot 5^x$ by, say, $5^{2x}$, you then get $\qty(\frac25)^{2x} - 2 < \qty(\frac25)^x$, from where you could, say, substitute $u = \qty(\frac25)^x$ and solve the resulting quadratic inequality, making note that you must therefore have $u > 0$

[wraith dagger]

@cerulean sail

[merry solstice]

ok so now (a-2b)(a+b)<0

[merry solstice]

oh

Since you know how to do it chartbit's way

I"ll just show you my way

If (a-2b)(a+b)<0, then -b<a<2b

or

-5^x<2^x<2*5^x

If you divide by 5^x

you get

-1<(2/5)^x<2

and if you use a log, then 0<x

Hope that helps!

[edgy sun]

may i ask a question rq?

[merry solstice]

sure

[sullen wind]

the answer should be something like that, sorry for not showing it before, I didn't see it

[edgy sun]

how is (a*b)^x equal to

[merry solstice]

yes

[edgy sun]

a^x *b^x?

[merry solstice]

um

it's something like x^a*x^b=x^ab

but with the bases

I was just taught the formula and not how to derive ti

[edgy sun]

oh i might just be dyslexic

[merry solstice]

oh lol

it's ok

[edgy sun]

for some reason i thought you were saying (a+b)^2=a^2+b^2

[merry solstice]

do you understand how to get the answer nwo?

[merry solstice]

lol you're good

[edgy sun]

so do you distinguish ab^2 and a^2*b^2 with parentheses?

like ab^2≠(ab)^2

[merry solstice]

wdym?

[edgy sun]

like how do you know when it’s just b squared versus ab squared

[sullen wind]

tbh? no...

[merry solstice]

ok

so we go back to the part where (a+2b)(a-b)<0

do you understand how we get -b<a<2b?

[sullen wind]

yes, it's simple

why is that

[merry solstice]

It's (a^2)(b^2)

[merry solstice]

ok

so when we substitute out the variables, we get

-5^x<2^x<(2^x)*5^x

dividing by 5^x, we get that -1<(2/5)^x<2^x

sorry I mis-substituted the variable

to isolate the x, we put a log(2/5) on all terms

However, logs can't be negative, so that means that 0<x

also, we know the limit of log(2/5)of 2^x goes to infinity eventuall

so it's like 0<x<infinity

[sullen wind]

ok. I agree and then?

[merry solstice]

Hmm

I don't know how you get the log(2/5)2

as the lower limit

[sullen wind]

okey, I undestand it, don't worry

[cedar kiln]

@sullen wind Has your question been resolved?

[cedar kiln]

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[whole parrot]

how am i wrong bro

[lime tartan]

You got the height, what’s the width?

[whole parrot]

what

[lime tartan]

e^4 + e^4.5 will tell you the sum of the heights, but we want the sum of the areas

[whole parrot]

i dont get it

[lime tartan]

We go from e^4 to e^4.5 when we split it in half, then e^4.5 to e^5 like you correctly identified. But this is only half of the work. We’re finding the areas of the rectangles, not just the heights

So, each rectangle is that tall, but what’s the width of the rectangle?

[whole parrot]

do they want me to multiply by deltax too

im lost

[lime tartan]

Yep

[whole parrot]

i hate math

thanks

[lime tartan]

Because that will give you the area of the rectangle

No problem

[whole parrot]

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[cedar kiln]

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[sacred socket]

What am I doing wrong in calculating the time the ball is in the air for this problem? Sorry its messy I was using Word.

[cedar kiln]

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@sacred socket Has your question been resolved?

[still ledge]

hello

i’m here to help

This is just a kinematics questions, so we can neglect the mass here

(for A)

@sacred socket so what i’m looking at is that you were only calculating the time it was taking to only go up

not up and down

Also, are they counting for sig figs?

[sacred socket]

I don't think significant figures matters.

[still ledge]

I got a similar answer

but I think maybe just use more digits just like you did with the 2nd one

18.32390191 is a good approximation

[sacred socket]

Thanks for the help.

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[simple torrent]

Hey I'm trying to prove that |zw|=|z||w| with the exponential form of complex numbers. Here's the supposed solution:

[simple torrent]

Isn't this self-referential? (step 4 separating one magnitude into two products)

[cedar kiln]

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[dire geode]

Did your class already prove the absolute value for polar forms of complex numbers

[cedar kiln]

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[simple torrent]

Think so, it's just the r right

sqrt(r^2(sin^2+cos^2))=sqrt(r^2)=r

If r>=0

[grand stone]

they use the property |az| = |a||z| for real a, complex z, which can be shown independently

[dire geode]

Yea so just use this for rs and exp(...)

[cedar kiln]

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[sharp veldt]

how is is 5,10

[sharp veldt]

for context, u have to derivate these functions and equal each to 0 or 10

so for second equation, dy/dx=10 and I have to solve for t

so wouldnt it be
10t - 50 = 10
10t = 10 + 50
10t = 60
t=6

[vivid furnace]

Why r u setting it equal to 10 if your looking for critical points

[sharp veldt]

heres the full answer

[plucky merlin]

I think those points are (x, y), following the parameterized functions
5t^2 - 50t + 83 -> 10t - 50 = 0 => t = 5, given x = t, y = 0 therefore (5, 0)
483 + 5t^2 - 50t -> 10t - 50 = 0 => t = 5 given x = t, y = 10 therefore (5, 10)
10t^2 - 60t + 83 -> 20t - 60 = 0 => t = 3 given x = 0, y = t therefore (0, 3)
83 + 10t^2 - 60t -> 20t - 60 = 0 => t = 3 given x = 10, y = t therefore (10, 3)

This is what I noticed, the (t, f) {as I am calling the output f here} is going to be (t, 0) but then consider the parameters. Is this not what is going on?

[sharp veldt]

no

actually u know what

u might be right

maybe my prof was explaining it wrong

[plucky merlin]

I was gonna say the next step has f(x, y) and uses those points meaning they have to be (x, y), not (t, whateva)

[sharp veldt]

yeahh ur right

thanks a lot

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[bleak marsh]

I need help with a triple integral using polar coordinates, cause I am getting stumped on the bounds

:(

tried reworking the bounds and is still not given an answer

<@&286206848099549185>

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[tender cipher]

how would you determine the answer between B and C?

[coral jewel]

great solution, except you should keep it to yourself and let op do the work

give hints

[cedar kiln]

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[tender cipher]

aight i get that thanks

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[coarse junco]

I can't see the connection between the cone's slant and the radius of the sphere (10cm).

Which topic should I revisit?

[cedar kiln]

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[coarse junco]

. close

.close

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[crimson sedge]

Help me understand 2 math subjects.

Why?
I do not understand and Youtube doesn't help me (My opinion but I will try again if you suggest). I also have a quiz tomorrow with these 2 subjects.

What subjects?
General Chemistry - Alkanes & Cycloalkanes (Naming ang Drawing Line and Structural Formula)
Mathematics - geometry and data sufficiency

[crimson sedge]

Which do you prefer to start first?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

Come onnn

YES

[crimson sedge]

.

I just don't... understand the topic...

Uhh

Want me to grab the powerpoint...?

okeh

I can send you the photos instead

As soon as I saw Page 11

I didn't understand it

Wha

WHAT?

okay-

SORRY

BOI

I got into an argument in the gc

Anyways

Wait

NO

I UNDERSTAND THAT

I just.. don't know anything about the properties of triangle

GIRL

Focus

I need to know like

I don't understand the properties of triangle

[lean lotus]

basic properties?

or like trig properties

[crimson sedge]

DON'T JUDGE ME I HAVE A WEIRD PROBLEM WITH MY BRAIN OKAY

HOLY

Im 11.........

SHUT

IM STILL A MINOR GUYS

[dusk goblet]

you’re 11 years old!?

[crimson sedge]

GRADE

11

Now let's go back to the topic

:3

What the skibidi are they saying

What

How

yeah

OHHHH

Okay

What the 1 sum

I don't get it

So

A is 180

B is 180

C is 180?

Now I don't get it

What

Oh

90

90

Correct?

WHAT?!

Oh

oh.

Okeh

Yeah

Broading

Soo..

is that all the explanation?

[crimson sedge]

Really????

[cedar kiln]

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[cedar kiln]

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[cinder jay]

"The solutions to the equation sin kv = a that are in the interval 0° ≤ y ≤ 360° are marked in the figure.
Determine the constants k and a."

[cinder jay]

I dont even know where to start tbh

<@&286206848099549185>