help-36

[lapis wedge]

.close

[final saddle]

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[opaque ember]

looks good

[final saddle]

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[earnest shuttle]

i found the derivaitve of the first equation and i got 2x-4, now i dont know what to do

[final saddle]

Please don't occupy multiple help channels.

[earnest shuttle]

also i rearranged the second equation for clarity to y = 2x - 1

and have determined that the slope is 2

[barren hound]

well, thats the slope of the tangent line. whats the slope of the line you were given?

ok cool

[earnest shuttle]

yuh

[barren hound]

so where is the slope of the tangent equal t othat?

[earnest shuttle]

the slope of the tangent is 2

[barren hound]

yes, at what x value is the slope of the tangent of x^2 - 4x - 5 equal to 2?

[earnest shuttle]

x value

is

i set y = 2

so x value is

[barren hound]

the slope of the tangent is the derivative (you calculated this) -- where is that equal to 2?

[earnest shuttle]

oh

3

because i set 2x -4 = 2 and got x = 3

[barren hound]

yea

[earnest shuttle]

wait so i just had to set the derivative equal to the slope and i get a tangent that is parrallel to the second equation?

[spare summit]

well the first derivative of a function at a point is the slope of the tangent to the function at that point. here though you want the tangent to have a particular slope, so you know the value of the first derivative, just want to know which x gets you that value.

do note though that you're not quite done yet

[earnest shuttle]

that makes sense

i am not done because i have not yet found tthe y value of hte point?

or is it something else

[spare summit]

correct!

oops, slow internet

[earnest shuttle]

ok thanks

you can ask for help in the math help channels

just find one that is not occupied

you can use this one

yes

i think i have to close it first though

[barren hound]

ummm its better if you grab your own

[earnest shuttle]

.close

[final saddle]

Channel closed

Closed by @earnest shuttle

Use .reopen if this was a mistake.

[barren hound]

so the bot knows its yours

use one of these

[earnest shuttle]

does that work now kitten?:

[barren hound]

its still yours for about 3 minutes in case you change your mind

[earnest shuttle]

oh

never knew that

.clsoe

.close

i think ask

then people will come help

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[lapis wedge]

So im not really to sure how to approach this

[lapis wedge]

I can take some element in sigma(F) and show its in some arbitrary sigma algebra containing G and vice versa?

[tiny gorge]

it would suffice to show that $G \subseteq \sigma(F)$ and $F \subseteq \sigma(G)$

[soft zealot]

Bungo

[fast smelt]

Show $\sigma(F) \subseteq \sigma(G)$ and $\sigma(G) \subseteq \sigma(F)$.

[soft zealot]

Pfhrohug

[lapis wedge]

I remember that also but I was wondering if that would be harder to show

we did that method when he had a concrete set

hmm ill give it a try tho

[tiny gorge]

why would it be harder?

[lapis wedge]

I dont know I just thought it was really only used with concrete sets

[tiny gorge]

you know exactly what the elements of F and G are, whereas sigma(F) and sigma(G) have much more complicated stuff in them

[lapis wedge]

I was thinking sigma(F) has less complicated stuff though

or I was thinking it would be the same

[tiny gorge]

not less complicated than F itself, surely

[lapis wedge]

because taking an element if F wouldnt be much different from taking an element of sigma(F) right

they would give me the same info to work with no?

[tiny gorge]

how would you even describe an arbitrary element of sigma(F), to show that it's in sigma(G)?

[lapis wedge]

oh wait

ah

I was thinking backwards

your right

let me try again thanks

[tiny gorge]

yw

[final saddle]

@lapis wedge Has your question been resolved?

[lapis wedge]

Ok I think I was stuck for a while because I was confused on what T1 and T2 was a sigma algebra on

but I think we just assume they are both a sigma algebra on some set X

After that it wasnt too bad unless Im underestimating the problem and bypassing some logic

but shouldnt the inclusion follow from

$X \setminus (A_1 \cap A_2) = X \setminus A_1 \cup X \setminus A_2$

[soft zealot]

Branshi (Hints only plz)

[lapis wedge]

wgere both of those sets on the union is in T1 and T2

so $X \setminus (A_1 \cap A_2) \in \sigma(G)$

[soft zealot]

Branshi (Hints only plz)

[lapis wedge]

thus $X \setminus X \setminus (A_1 \cap A_2) = (A_1 \cap A_2) \in \sigma(G)$

[soft zealot]

Branshi (Hints only plz)

[lapis wedge]

.solved

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[jade fable]

Hello

[jade fable]

Hello

[candid pulsar]

i can hear you

[radiant igloo]

They're messing around

[tender pollen]

what do you need help with?

[jade fable]

When I deal with a physic

Phenomenon

That is unfamiliar to me

[tender pollen]

can you be specific?

[jagged wave]

(just ask the question!)

(help comes faster if you do!)

[fathom meteor]

!da2a

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[jade fable]

I feel like I have to depend on the god, instead of using my logic to solve it

[candid pulsar]

pcmg basically a celebrity atp

[jagged wave]

dude just ask it can't be that bad

[jade fable]

There are plenty of them

[jagged wave]

ok ask one of them

[tender pollen]

please, just ask one at a time 🙏

[radiant igloo]

Since so many ppl online I'm gonna open my own help thing

[jade fable]

I want some advice

[tender pollen]

could you translate?

sry i dont know chinese

[final saddle]

@jade fable Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

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[rocky lantern]

a

[soft zealot]

a random guy

[final saddle]

@rocky lantern Has your question been resolved?

[rocky lantern]

<@&286206848099549185>

[final saddle]

@rocky lantern Has your question been resolved?

[final saddle]

@rocky lantern Has your question been resolved?

[radiant igloo]

Hmmm

[radiant igloo]

@rocky lantern okay wait in this image does 'i' belong to the set of natural numbers or is it just 1,2

As shown in image

[brave lake]

@rocky lantern what is 3345535+5556655645655

[radiant igloo]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[rocky lantern]

oh, i is 1 or 2 :)

[radiant igloo]

Okay thanks that makes much more sense now

[rocky lantern]

wow, you’re so funny, man ._.

[brave lake]

im not trying to be funny

[rocky lantern]

yeah, you weren’t trying hard enough :)

[brave lake]

you neither

[rocky lantern]

i’m not though .-.

i’m serious

[brave lake]

nice

[radiant igloo]

The question is kinda too vague to give a specific numerical domain of theta honestly...

[rocky lantern]

yeah… you have to divide it into cases :). that makes it convoluted, which is why i’m struggling :(

[radiant igloo]

Yeah I'm kinda too early to math to be doing stuff this complex 🥲

'I can comprehend it but I can't explain it at the moment'

[rocky lantern]

you know what… i’m just gonna spill the beans. it’s related to that sphere problem i had :)

[radiant igloo]

Oh

[true urchin]

Hi

[final saddle]

@rocky lantern Has your question been resolved?

[rocky lantern]

.close

[final saddle]

Channel closed

Closed by @rocky lantern

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[glacial breach]

Apologies to all those who feel like victims of my poor communication skills and maths vocabulary

I am trying to understand dependent events, I usually go about doing this by observing examples and forming a hypothesis.
Here is what I have surmised:
-# Examples used:
-# drawing a particular colour marble from a bag containing marbles of different colours without replacement.
-# the chances of rain on the next day given that it also rained the previous day
What I understood from these was that I could determine dependent event by considering that if the event has already occurred once, then upon a subsequent experiment, would the probability be different from the first.

This was working until I encountered a different archetype of examples
-# someone being arrested, given that they have committed a crime
-# someone getting a parking ticket, given that they parked illegally
this time there isn't a reiteration of the experiment, but instead, a premise is defined using which, probability is inferred.
For these examples, I surmised a different definition.
Dependent event would be the event wherein, the sample space is redefined, such that, the chances of another event occurring changes. That is to say, using the arrest example, the chances of getting arrested for a person is low, because the sample space in that case would be the entire population, however, if we define the premise that the person has committed a crime, we have now redefined the sample space to 'only criminals', which significantly increases the chances of arrest.

The visualization of two venn diagrams intersecting always comes to my mind when I imagine dependent events, but this diagram appears flawed to me.
Because I also found cases wherein, intersection did not directly mean dependence. This fact also seems to have invalidated my last defintion that dependent events are when the premise changes the sample space of the experiment.

[glacial breach]

Such can be illustrated by the example: chances of drawing a face card given that the drawn card is a spade

I am sorry for whatever nonsense I have just spewed

[glacial breach]

here spade redefines the sample space from all 52 cards to just the 13 spade cards
despite that, the probability of drawing a face card does not change

I understand the mathematical definition P(A intersection B) != P(A) * P(B)

but I was just trying to understand this from a more non mathematical way

[opal plinth]

Events being dependent just means the outcome of one affects the outcome of the other

[glacial breach]

the sample space is redefined, such that, the chances of another event occurring changes.
i seem to have made progress on the answer on my own, midway, and then become oblivious to it

so proceeding now, there would be two conditions that must be met for events to be dependent:

• that there is a change in sample space
• and that that change transpires a change in probability as well

[tender pollen]

its kinda in the name

dependent events depend on each other

[opal plinth]

In the bag of marbles example, the first event is drawing the first marble of a particular color, and the second event is drawing the second marble of that same color

[glacial breach]

problem is, that definition relies too much on common sense which I seem to lack while looking at questions 🥲

[tender pollen]

thats kinda the most intuitive way to understand it tho

dependent events depend

[glacial breach]

if we are being absolutely mechanical with this, then in that case, would this be a good way to go about understanding dependent events

[tender pollen]

but is it always a change in sample space?

[glacial breach]

when is it not?

[opal plinth]

Drawing marbles with replacement

[tender pollen]

when there’s not really a sample space

[glacial breach]

yes but that case is not dependent is it?

[glacial breach]

could you give an example?

[opal plinth]

Ah sorry I misunderstood what you meant

[tender pollen]

maybe flipping coins infinitely?

would that count

[glacial breach]

that is again not dependent

[tender pollen]

you win after 10 heads

[void valley]

Imagine, for example, rolling a die twice (or two dice). Does what you roll on the first roll (on the first die) have any effect on what you roll on the second?

[opal plinth]

There is always a sample space, otherwise you can't have events

[glacial breach]

it does not, they are independent

[tender pollen]

oops youre right

sorry

[glacial breach]

i only said a change in sample space, not none at all

[opal plinth]

I was responding to this

[void valley]

Another example:
We roll a die and then choose one of two urns to draw. Let's assume that if we roll an even number, we'll draw from urn number 1, and if we roll an odd number, we'll draw from urn number 2. Here you can see that these events are dependent, because the number of dice determines the choice of urn

This is basically how it works, one event affects another

[tender pollen]

yeah theres not really any other way to phrase this

thats just how dependent events work

[opal plinth]

I think you might achieve a better understanding by considering what independent events are

[glacial breach]

for some examples, it is quite obvious, but sometimes it could be obscure

[opal plinth]

Event A might restrict the sample space, but if event B is orthogonal, meaning its probability is the same over the original space or over the restricted space, then A and B are independent

[glacial breach]

let's see, if I encounter any such example in the future, would you mind if I contact you guys?

[glacial breach]

yeah, that seems to be what the drawing face card from only spades is an example of

[opal plinth]

Yes

Because a deck is composed of 4 sets (colors) of cards, each with the same number of cards and the same number of face cards, the choice of set does not affect drawing a face card

Color and face-or-number are orthogonal attributes

Not sure what the word is for whether a card is a face or not

[glacial breach]

hm, very well

[opal plinth]

More generally, rank, I guess

Suit and rank are orthogonal in a standard deck

[glacial breach]

yeah

I guess, I will close this here

thank you for your time everyone @opal plinth @void valley @tender pollen

.close

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[rain sentinel]

i dont understand how they've got the area

[rain sentinel]

as in why they have 2* and the limits are 3 and 1/2

[loud sundial]

,w graph y=1/x^2, y=4, x=3, x=-3

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[tender pollen]

2 is for symmetry

[loud sundial]

oh sniped

rip

[tender pollen]

the bounds are symmetry too

since the area from [1/2, 3] is equal to the area from [-3, -1/2]

just some ways to be more efficient 🤷‍♂️

[rain sentinel]

and how would be know they're symmertic?

without drawing the graph

[tender pollen]

since 1/x^2 is even

and 4 is even

1/x^2 - 4 is even

and thus the area from [a,b] is the same as the area from [-b,-a]

you know about even and odd functions, no?

[rain sentinel]

yes

[tender pollen]

so my argument should make sense

[rain sentinel]

thanks

[tender pollen]

type ".close" if you're done

🙂

[rain sentinel]

.close

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[latent pelican]

Let $P = \frac{2\sqrt{x} - 1}{\sqrt{x} - 2}$. Find all values of $x$ satisfying $\sqrt{1 - 2P^2} = \sqrt{1 - 2P}$

[soft zealot]

An

[tender pollen]

square both sides is what i would do first, no?

(of course there's probably extraneous solutions but let's ignore those)

[latent pelican]

Ok

[tender pollen]

and then move the terms to one side

what solutions are there for P?

[latent pelican]

I don't understand

[tender pollen]

ok

what do you get when squaring both sides?

[latent pelican]

Of P?

[tender pollen]

the equation

$\sqrt{1-2P^2}=\sqrt{1-2P}$

[soft zealot]

nadat12

[tender pollen]

what do you get when squaring both sides?

[latent pelican]

$|1 - 2P^2| = |1 - 2P|$

[soft zealot]

An

[tender pollen]

ok

remove the absolute value bars for now

(it doesn't make sense anyway, they can't be negative)

[winter lava]

absolute value is for sqrt(x^2), not sqrt(x)^2

[latent pelican]

Like this: $1 - 2P^2 = 1 - 2P$

[soft zealot]

An

[tender pollen]

yes

now move all terms to one side

say, the left-hand side

[latent pelican]

$1 - 2P^2 - 1 + 2P = 0$

[soft zealot]

An

[tender pollen]

nice

can you convert this into a quadratic in P?

[latent pelican]

Replace P with $\frac{2\sqrt{x} - 1}{\sqrt{x} - 2}$?

[soft zealot]

An

[tender pollen]

not yet

we're just solving for P first

it's easier on you, i promise, if we solve for P first

[latent pelican]

Ok

[tender pollen]

turn it into the form $aP^2+bP+c=0$

[soft zealot]

nadat12

[latent pelican]

$- 2P^2 + 2P = 0$

[soft zealot]

An

[tender pollen]

nice

what's an obvious root of the quadratic?

[latent pelican]

idk

[tender pollen]

ok

what does the quadratic factor to?

[latent pelican]

idk

[tender pollen]

uh

try dividing everything by ||-2||

[latent pelican]

$P^2 - P = 0$?

[soft zealot]

An

[tender pollen]

yep

do you know the roots of this quadratic?

[latent pelican]

Idk

[tender pollen]

ok

ummm

the roots are ||0|| and ||1||

so either ||P = 0|| or ||P = 1||

[latent pelican]

…

[tender pollen]

out of the roots, which actually work?

as in, which are solutions to

$\sqrt{1-2P}=\sqrt{1-2P^2}$

[soft zealot]

nadat12

[tender pollen]

which ones are actually solutions to this?

[latent pelican]

idk

[tender pollen]

uh

[tender pollen]

plug these in

[latent pelican]

$\frac{2\sqrt{x} - 1}{\sqrt{x} - 2} = 0$ and $\frac{2\sqrt{x} - 1}{\sqrt{x} - 2} = 1$?

[soft zealot]

An

[tender pollen]

no

[tender pollen]

to this equation

because P = 0 and/or P = 1 may not actually work

(spoiler alert: ||one of them doesn't work||)

...@latent pelican?

[latent pelican]

I’m here

[tender pollen]

ok

[tender pollen]

which one works and which one doesn't work for the equation here

which one of P = 0, P = 1

it shouldn't be that hard

just plug and chug

[latent pelican]

P = 0

[tender pollen]

yep

so what equation do we have now?

[tender pollen]

(hint)

[latent pelican]

$\frac{2\sqrt{x} - 1}{\sqrt{x} - 2} = 0$

[soft zealot]

An

[tender pollen]

yep

[split cipher]

set the numerator to 0

[tender pollen]

what can we multiply by to clear denominators?

[latent pelican]

…

[tender pollen]

we just need to cancel the bottom of this expression

[latent pelican]

I don’t understand

[tender pollen]

ok

if i have a/b = 0

what do i multiply to get rid of b?

[latent pelican]

By 0?

[tender pollen]

no

if you multiply by 0

then you get 0 = 0

that doesn't solve anything

i still want an equation

but i want b out of the picture

[split cipher]

I think he has a conceptual problem

[latent pelican]

Move b to the right side?

[split cipher]

0 is 0 as long as it’s on the numerator

whatever the fraction on the left is, you can sub in the bottom on the right

[tender pollen]

uh

the answer was

[split cipher]

for example x/y = 0/y

[tender pollen]

multiply by b

[split cipher]

so just set it as x = 0 since the bottom is the same

[latent pelican]

a = 0?

[tender pollen]

yep

[tender pollen]

now do the same for this

[latent pelican]

$2\sqrt{x} - 1 = 0$

[soft zealot]

An

[tender pollen]

yep

can you solve for sqrt(x)?

[latent pelican]

Yes

$2\sqrt{x} = 1$

[soft zealot]

An

[tender pollen]

so what's sqrt(x)?

[latent pelican]

$\sqrt{x} = \frac{1}{2}$

[soft zealot]

An

[tender pollen]

nice

square both sides

what do you get?

[latent pelican]

$x = \frac{1}{4}$

[soft zealot]

An

[tender pollen]

nice

i would check that this works. does it?

if it does, you have your answer!

[latent pelican]

Ok

[tender pollen]

when you're done type ".close"

[latent pelican]

Ok

.close

[final saddle]

Channel closed

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[latent pelican]

Thanks

[final saddle]

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[lime crest]

Let $k \ge 2 $ be a positive integer. Suppose that $n$ is a natural number such that $k^2 \le n <(k+1)^2$. Prove that [ \lfloor (k+\sqrt{n}+1)^{5^m} \rfloor \equiv 2k+1 \pmod{10} ] For all natural numbers $m$.

[soft zealot]

Copter

Copter

[lime crest]

oh wait

should there be a -1

and then the result follows by chinese remainder?

<@&286206848099549185>

[thorn zenith]

Hello! I'm not very good at number theory but this problem looks interesting, might try give it a go xD

I wonder if trying to prove the expression mod 2 && the expression mod 5 is a reasonable approach?

Let $k \ge 2$ be a positive integer. Suppose that $n \in \mathbb{N}$ satisfies $k^2 \le n < (k+1)^2$. Prove that for all $m \in \mathbb{N}$, $\left\lfloor (k+\sqrt{n}+1)^{5^m} \right\rfloor \equiv 2k+1 \pmod{2}$ and $\left\lfloor (k+\sqrt{n}+1)^{5^m} \right\rfloor \equiv 2k+1 \pmod{10}$.

[soft zealot]

Happy Meal Enjoyer

[lime crest]

yup, thats basically what i did tbh

[lime crest]

^ with a -1

expanding follows

[thorn zenith]

Oh I c, OK

Yeah you're right

[lime crest]

im just checking if this is valid

[tiny kraken]

How did you use chinese remainder? I can see the equation is right, with a -1 ofc

[lime crest]

[…] ≡ 1 mod 2 and ≡ 2k + 1 mod 5

i think that follows, right

[tiny kraken]

If you have already been able to show both then yeah

[tiny kraken]

not sure about mod 5 lol

[lime crest]

its 2(k+1)^5^m - 1 mod 5

[tiny kraken]

hm

[lime crest]

FLT (k+1)^5^m = k+1 by induction

[tiny kraken]

ig that works

[lime crest]

yippe :3

[tiny kraken]

Tho I don't know how you get those but I do trust olym people

[lime crest]

a^5 = a mod 5 for any a

you get a^5^m = a^5^{m-1} = ...

i wonder if theres any other way to do this

[tiny kraken]

Oh wait, damn my dumbass, I forgot that FLT exists

[lime crest]

oki thanks guys :3

.close

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[tiny kraken]

[glass oriole]

Hello, I've switched to home education. Can someone explain the topic of rational fractions?

[final saddle]

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[brazen stump]

hello where is the mistake in this

[brazen stump]

wdm

[final tangle]

a - b - c isn't a - (b-c)

messing up order of operations at the very start

[brazen stump]

where

[lucid nymph]

You took the x+2/x-3 to be positive in the abstraction

When it should be negative

[brazen stump]

i did the 2 fractions first then i moved onto 2/1

[final tangle]

given something like
$$5 - \red{4 - 1}$$
you're not allowed to compute $\red{4-1}$ first, doing so would reach the incorrect value of $5-3 = 2$ instead of 0

[soft zealot]

ραμOmeganato5

[brazen stump]

oh

so i need to do 2/1 first

move from left to right?

[lucid nymph]

Well the point is that the second term is just negative

[brazen stump]

which second term

[final tangle]

if you want, you could combine the fractions first, but you'd need to be careful,
its a good idea to write everything out in each line

[lucid nymph]

Imagine it as 2+(-[x+2/x-3])+(-[x-6]/[x+3])

[final tangle]

e.g.
$$= \frac 21 - \br{\frac{x+2}{x-3}\blue{+}\frac{x-6}{x+3}}$$

[soft zealot]

ραμOmeganato5

[brazen stump]

the negative turns positive?

• and -= +?

[lucid nymph]

Yes its as if you factored -1

[brazen stump]

so i do inside brakcter first then i can move on to 2/1

with this way

[lucid nymph]

The same way you could factorise 3x+6 into 3(x+2)

[brazen stump]

my way you cant because the - did not become +

I think the safest way is to move from 2/1 from left to right

[lucid nymph]

Correct but if you can see the way it works then you can indeed just as well do the last two parts first

[brazen stump]

i have a question

if it was all + signs it wouldnt matter right

wht i do

bc i dont need to be careful with the signs

[final tangle]

addition is commutative and associative
so you can add in in order you want

[brazen stump]

this case i do the + first then move onto 2/1

or i can do 2/1 minus the fraction then add it do x-6/x+3

[final tangle]

you'd do whats inside the () first
or equivalent

[lucid nymph]

The thing is its technically all addition anyway, just the fractions are negative

[brazen stump]

what did i do wrong now bru

i redid it in red

i did 2/1 first

[final tangle]

when subtracting (x+2)/(x-3)
you only applied the - to the first term of (x+2) on the numerator

[brazen stump]

omddddd

[brazen stump]

aaa

ok

yooo i got it

[brazen stump]

thanks

[brazen stump]

thanks

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[rain sentinel]

I don't understand what s is

[rain sentinel]

And how they've subbed in infinity and 0 for t to get those values

Any explanations are appreciated

[void valley]

e^(-st) = 1/e^(st), therefore when t approaches inf 1/e^(st) tends to 0

because 1 divided by a really huge number is close to zero

I hope the next part i.e. plugging the lower bound is clear

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[rain sentinel]

can someone help me integrate the bit in blue

[lucid nymph]

u sub

what could you substitute to make it doable

[final saddle]

@rain sentinel Has your question been resolved?

[rain sentinel]

i did (x+1)^-2

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[dull compass]

yea thats right

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[rain sentinel]

is my method for integration correct?

thanks

[storm haven]

seems all good to me

[rain sentinel]

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[earnest shuttle]

why is the restriction 3/5x ??

[earnest shuttle]

how did he even determine that

[proud igloo]

well u know the denom cant equal to 0

[earnest shuttle]

yeah

[proud igloo]

thats why you wrote $10y-6x=0$

[soft zealot]

Nyxzore

[proud igloo]

but this isnt to useful to read

so if we solve for y we get

...

[earnest shuttle]

oh yeah solving for y we get that

but like how do we know this would the denom 0?

[proud igloo]

because $10y-6x$ \neq 0$ is the same thing as $y\neq \frac{3}{5}x$ just with terms shifted around. I.e the 2 equations are equivalent

thats the <=> symbol

obvs u should be using $\ne$ symbol at not =

[soft zealot]

Nyxzore

[earnest shuttle]

o wait i see

rearranging for either

and plugging it in

both gets u 0

[soft zealot]

Nyxzore
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[earnest shuttle]

yaeh?

[proud igloo]

in the denominator

yes

[trail mango]

[crude rampart]

$10y - 6x = 0$ so $10y = 6x$ so $y = \frac{6}{10}x$ and there is a factor of 2 in the numerator and denominator so it simplifies to $y = \frac{3}{5} x$. as you can see in the work, on the bottom line, he divides $10y-6x$ so there is a domain restriction to make sure it cannot equal zero.

[earnest shuttle]

ahh yes

[soft zealot]

9k

[earnest shuttle]

ok thanks guys!

[earnest shuttle]

isaac newton appreciates the support deeply

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[earnest shuttle]

.reopn

.reopen

[final saddle]

✅ Original question: #help-36 message

[earnest shuttle]

wait can the restriction also be x does not equal to 5/3y?

instead of y cant be 3/5x?

[lilac bison]

oh my god its issac newton

do you mean 5y/3?

[spare summit]

bracketing!

[earnest shuttle]

yes

thats exactly what i mean

[earnest shuttle]

alr next time

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[vast iris]

How do I check 4th

Quickly no need to prove

[spark swan]

consider bases for both M and N

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@vast iris Has your question been resolved?

[vast iris]

Dim is same

So they have same number of vectors in basis

[opaque ember]

if M!=N then N contains a vector not in M. it should lead to contradiction

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[abstract nacelle]

is there an easier way to solve this then to simplifying it as an exponent of seven?

[void valley]

hint: notice that 32 is 4 mod 7

and use Fermat’s Little Theorem

[abstract nacelle]

great idea except for the fact that i have no idea what that is

[void valley]

This is helpful here because 7 is prime

[abstract nacelle]

woaw i dont think our grade teaches on this level

what is mod p?

[void valley]

a mod p means the remainder when a is divided by p

Example:
12 mod 5 is 2 because 12 = 5 * 2 + 2

[abstract nacelle]

and then apply binomial?

[void valley]

So, you aren't familiar with modular arithmetic, yeah?

[abstract nacelle]

i think in my country they teach that at college level

or senior high school

im still a junior in regards to american standards

[void valley]

The binomial theorem itself probably won't help here (it can help to simplify it slightly I guess)

[abstract nacelle]

the way it was taught to us was that

after applying the binomial theorem

every element would be divisible by 7 except the 1st one

and then we would have to solve it from there

[gritty willow]

good point

[abstract nacelle]

oh wait

what do you mean?

you mean straight up dividing 32^32^32 /7??

[void valley]

ok, I see, the binomial theorem + recognizing cycles of powers can help

[abstract nacelle]

yea honestly that method takes too much time

you know it wouldnt be a problem if the exam im preparing for didnt allow calculators

there is a fine for upto

1 lakh rupees

or about 1000 dollars

i will try with the mod method once thx a lot guuyss

.closed

[soft zealot]

Modus

[void valley]

Something like this?

[abstract nacelle]

yess

exactly like that

so then i would have to simplify 4^32^32

[void valley]

Yes

[abstract nacelle]

in terms like 16-14)^16^32

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[obtuse tangle]

guys i want someone to help me in this bac i can't solve it

hey?

is anyone here?

[void valley]

Open your own channel, this one has been already closed

I'll help you

[obtuse tangle]

ok ty

[obtuse tangle]

what is the chanel i want ?

idk what u r talking about

[abstract nacelle]

go into help 14 and reupload the question there

[vital crag]

#❓how-to-get-help read

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[obtuse tangle]

ok

[abstract nacelle]

uhh just use this thread now

its already urs

[obtuse tangle]

ok

[void valley]

Can you find the height of parallelogram given it's area and base?

[abstract nacelle]

hint: area of paralellogram = base into height

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[loud phoenix]

<@&268886789983436800>

[loud sundial]

.close

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[spice tangle]

$\int{\frac{1}{\sqrt{-x^{2}+6x-8}}dx}$ I don't even have an idea on how to start this

[spice tangle]

And yup it's me again... still

[soft zealot]

Flappy the Turd/Frost

[spice tangle]

Fixed the compile I suppose

[lilac bison]

Do you know trigonometric substitutions?

[spice tangle]

Not really

[olive lantern]

yo

find the domain of the function first

u can also write the thing under the root at 1-(x-3)^2

[spice tangle]

How would the domain of the function help?

[frail oyster]

^

[frail oyster]

this is the easiest method, follow this

[olive lantern]

ur right u dont need to

[spice tangle]

Got it I suppose.
Oh wait.
u-substitution always snakes into these things 😭

[olive lantern]

yep

but it would be helpful to learn trigo substitutions

u should defo explore that area

[frail oyster]

in general yes, not for this problem

[spice tangle]

I'll look into it if it's not taught in class

[olive lantern]

hm atb

[spice tangle]

Oh wait, this is completing the square too

[barren cairn]

would you use partial fractions?

[spice tangle]

I already have an idea going

[lone talon]

nah partial fractions for when the denominator has polynomals

[barren cairn]

wdym polynomials

[spice tangle]

I think I got it

[barren cairn]

doesn't it have a degree that is larger than the top?

[lone talon]

Yeah but the bottom is a root function

If the root wasn’t there you could factor it and use partial fractions

Imma find you

[barren cairn]

wdym by root

[spice tangle]

Well I managed to reach arcsin(x-3)+c

[barren cairn]

sorry I just haven't done this type of math since last year lol

[lone talon]

Square root

[olive lantern]

yes

[barren cairn]

oh alr

[spice tangle]

Cool

[olive lantern]

good

[spice tangle]

Ill close this now

.close

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[barren cairn]

bye

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[rustic wedge]

potd today

[rustic wedge]

reduced it to $$19 \cdot 502 - \sum_{r=0}^{501} \sum_{m=0}^{18} \left( \floor{\frac{502m + r}{503}} + \floor{ \frac{10000 - 502m - r}{503} } \right)$$

bruh ugly floor

nvm i aint fixing

[soft zealot]

rak³en

[merry trench]

whats the problem exactly?

cuz its just the series you can see the pattern

then it simplies to somwhat

(1+w^k)^502

somewhat

[rustic wedge]

tried that too suprisingly difficult

[merry trench]

what was ur answer

[rustic wedge]

i got $\sum_{r=0}^{501} \sum_{k=0}^{501} \omega^{k(502-r)} (1+\omega^k)^{10000}$

and i gave up on simplifying this

[soft zealot]

rak³en

[rustic wedge]

i cant compute either series

[fickle vortex]

what math are you in

[rustic wedge]

idk

math i guess

[fickle vortex]

?

what course

[rustic wedge]

well i am a jee aspirant (soon to be in grade 12)

but i like to poke stuff around when i have time

[merry trench]

we could try binomial approximations

[rustic wedge]

on complex numbers?

[merry trench]

wait I have heard of it

jee?

[rustic wedge]

can we focus on the problem

my 'course' is not really important

[merry trench]

we can k≡r

then series would simplify to 462 C r

[rustic wedge]

??

[fickle vortex]

just curious on when id reach this kind of math

[rustic wedge]

who knows depends on your country and its curricula

[merry trench]

the mod zeta functions?

[rustic wedge]

the what?

[merry trench]

ok then lets try a different approach

what topics have u learnt?

[rustic wedge]

no wait wtf are mod zeta functions

i know multiple and hurwitz

eta beta

riemann zeta but wth is mod zeta??

[merry trench]

like u can simplify the sum to a different transcript

[rustic wedge]

?? uh?

i dont understand sorry

[merry trench]

roots of unity and all the stuff

[rustic wedge]

this is an olympiad problem its definitely uses only hs math

[rustic wedge]

thats not called mod zeta and those roots of unity is how i got the omega series

[rustic wedge]

this right here

[merry trench]

(1+ζj)502≡1 would be one technically

[rustic wedge]

💀 i call them Roots of unity

[merry trench]

ok

[rustic wedge]

omega is 502th root of 1

[merry trench]

yes ok thenw hats the problem

jst simplify it

[rustic wedge]

??

how am i supposed to simplify that

[merry trench]

ζj=−1 then u proceed with r part

[rustic wedge]

i dont understand

[merry trench]

im getting the answer as 39 which is probably wrong

[final saddle]

@rustic wedge Has your question been resolved?

[ebon agate]

Take a smaller case for ex. 4m+r or 2m+r

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[rustic wedge]

reopen pls

[rustic wedge]

does that help with computing the sum or the problem in general

[ebon agate]

Simplified expression for Sr

[rustic wedge]

i see ok

[ebon agate]

Why is there a summation on r?

also you can simplify that to w^(-kr)

Also missing dividing by k

[rustic wedge]

ik that

[ebon agate]

502

[rustic wedge]

wait good question

right

it shouldnt be sum on r

[ebon agate]

Yes

[rustic wedge]

it should be solve for all r such that > 0

err listen not free rn

will take a look at this later

ty for the hint!

.close

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[proud igloo]

,tex
Given a circle with centre (0,0) and radius 2. Find the equation of the tangent line at $(\sqrt3 , 1)$

Now this is pretty trivial stuff just using implicit differentiation.
$$x^2+y^2=4$$
$$2x+2y\cdot\frac{dy}{dx} =0$$
$$\frac{dy}{dx}= -\frac{x}{y}$$
$$(y-1)=-\sqrt{3}(x-\sqrt{3})$$
$$y=-\sqrt{3}x+4$$

Now my lecturer suggested writing the circle as
$$f(x,y)=\begin{bmatrix}
x^2 & y^2
\end{bmatrix}$$
Then by our definition of matrix differentiation I have
$$f'(x,y)=\begin{bmatrix}
\frac{d x^2}{dx} & \frac{dy^2}{dy}
\end{bmatrix}$$

$$f'(x,y)=\begin{bmatrix}
2x & 2y
\end{bmatrix}$$

But I can't tell whats exactly useful about this result i.e. where to go from here.

[soft zealot]

Nyxzore

[digital birch]

f(x, y) = [x^2, y^2] doesn't seem right, maybe you mean f(x, y) = x^2 + y^2 - 4?

[proud igloo]

well i'll quote this cuz I just wrote this; I was hinted to "write the circle as a function of x and y defined by a matrix"

the rationale behind the matrix setup is strange

[digital birch]

The resulting derivative f'(x, y) = [2x, 2y] should be what you have already written at the end. Now, to get the tangent line, you consider the set of all points (x, y) such that there is no change in f locally at (x_0, y_0) = (sqrt(3), 1), that is {(x, y) | f'(x_0, y_0) [x - x_0; y - y_0] = 0}.

Plugging in the values yields 2 sqrt(3) (x - sqrt(3)) + 2 (y - 1) = 0 and thus y = -sqrt(3) x + 4, which is the same as what you have already computed.

[proud igloo]

sorry i dont quite get this ${(x, y) | f'(x_0, y_0) [x - x_0; y - y_0] = 0}$ why is the multiplication of these 2 things 0?

[soft zealot]

Nyxzore

[digital birch]

For all (x, y) lying on the circle, you have that f(x, y) = 0. Taking the derivative on both sides at (x_0, y_0) yields f'(x_0, y_0) [h; k] = 0 for all vectors [h; k] tangent to the circle at (x_0, y_0).

Setting x = x_0 + h and y = y_0 + k to get the set of all points on the tangent line yields the condition f'(x_0, y_0) [x - x_0; y - y_0] = 0.

This is essentially implicit differentiation.

[proud igloo]

thank u zan

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[minor sandal]

let X:=x_n be a sequence of real numbers.

x_n = (1+ 1/n )^n

We have to find the limit of this sequence.
In my real Analysis book, they have shown that the limit lies between 2 and 3 .
But what it is precisely is not discussed.
What should I do

[open geode]

Do you know what e is?

[minor sandal]

Yes

[spare summit]

this is the limit/asymptomatic definition of e, if that is what you are asking.

[minor sandal]

Can't we prove it ?

[open geode]

What def. of e do you know?

[minor sandal]

I only know e=2.718 ...

[spare summit]

if you don't already have a definition of e, proving this defn becomes circular reasoning

[minor sandal]

So for now , i should stick to the book?

[spare summit]

yes.

[minor sandal]

Not go for specific value of e

[spare summit]

unless you have another definition to work with

[minor sandal]

And only 2<e<3

[spare summit]

then we can prove that the limit defn of e matches that other defn

but without another defn, we are proving that the limit defn is... well, there's no reference, so

[minor sandal]

As such there no discussion about definition of e , for now

[spare summit]

not yet I guess

[minor sandal]

In book, they have only written it's Euler number and

Very important

Thank u btw

.close

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[harsh quail]

solve this using bpt, converse bpt or similarity please cause thats whats in our portion and answer is expected that way we have coord geometry others too idk how to solve this

[spark swan]

first, draw a line from D to AC, such that it is parallel to BY

you can label the point on AC as say E

[harsh quail]

ok then u will use bpt on it ig then ce/ey=cd/bd

[spark swan]

now, what ratios can you derive using BPT?

[harsh quail]

CE/EY=1/1 then

this paper was in board exam today

many ppl were not able to solve

[spark swan]

also what ratios so you get in ∆ADE and ∆BYC

[spark swan]

howd your boards go

[harsh quail]

decent

time managament went bad

[spark swan]

seems like many people messed up a probability question

[spark swan]

anyway how's this going

[harsh quail]

idk

i left that q cause i wasnt able to think anything

like how ur getting bx=4y when u cant see anyth

😭

and ur knowledge here limited to similarity and bpt

[spark swan]

well you have DE||XY

[harsh quail]

ax/xd=ay/yc then

[spark swan]

have a friend who also wrote boards today said the toppers in his class were crying

[harsh quail]

ay/yc=2/3

[harsh quail]

i am going to give boards again they this time introduced 2 exam so if u dont get well enough u want in board 1 u can choose to appear for 2nd

[spark swan]

=XY/DE as well

you can set XY as 2k and DE as 3k for some k

[spark swan]

gl

[harsh quail]

can u give me a answer which ahs everything given to prove and proof for this q

if u are able to solve

[spark swan]

I'm not supposed to give you the whole solution

i mean just give it straight away

[final saddle]

@harsh quail Has your question been resolved?

[final saddle]

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[wheat delta]

Idk if this would count as math but whats the differnece between q=cv and e = 1/2 * cv^2. Also why is E in Joules?

[strange sparrow]

what do the variables stand for

maybe it's better to ask in discord.gg/physics

[wheat delta]

oh Ok ty

[loud sundial]

.close

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[little python]

q = cv tells you how much charge (q) is on the plates of a capacitor when a voltage is applied

e is in joules because energy is measured in joules

[wheat delta]

.reopen

[final saddle]

✅ Original question: #help-36 message

[little python]

and i think e = 1/2 cv^2 is just the energy stored in the capacitor

[wheat delta]

ok thank you. SO they basically tell how much of Coloumbs and Joules are stored in the capacitor at the give moment?

So it has no relation to how much max energy or charge it can hold

[split ocean]

Hello

[wheat delta]

hey

[split ocean]

Nice to meet u

[wheat delta]

You too

[split ocean]

Thanks

Have fun great day man

[wheat delta]

You too

[split ocean]

Thanks mate

[little python]

lovely conversation

[little python]

it does not have a fixed "max charge" or "max energy"

they tell the instantaneous stored values when voltage is V

[wheat delta]

Oh okk

[little python]

you can just keep increasing V and the charge will keep increasing forever

[wheat delta]

Last q why isnt time related to these formulas cuz as time passes, charge / energy would increase right?

[little python]

because those formulae give the charge and energy at a given voltage

as time passes, charge / energy would increase right?
yes

[little python]

but ^

[wheat delta]

But the charge changes as time goes on so what would be the use of the equation? You would get the charge but theirs nothing you can do with becuase its now how much actual charge the capacitor is actually holding

also nice pfp

[little python]

i don't get it

it does tell you how much charge the capacitor is actually holding

because the value of V used in the formula is the voltage at that instant

if the voltage changes with time, you use the new value of V to know how much charge it's holding at that moment

[harsh quail]

no

[little python]

thank you

,av

[soft zealot]

kowaru373's Avatar

Click here to view the image.

[wheat delta]

no im talking about when the V is the same but since capacitors dont charge instantly so they would increase charge over a period of time.

but q = cv tells charge right now without relation to Time, so u wouldnt know if 1 second(less charge) or 10 seconds(more charge) passed so how would be q = cv be accurate and usefull

[final saddle]

@wheat delta Has your question been resolved?

[autumn oracle]

youre asking for an expression for q as a function of t?

i think theres a formula for that for an rlc circuit

[wheat delta]

yea i just learnt that their is func for that. What would be the use of the q = cv

Ik it gives the charge at the moment but

[autumn oracle]

to find out the charge at a specific voltage

[wheat delta]

it wouldnt be accurate since you didn't know how much time has passed. Also when mean use im talking like uses in real life problems

[autumn oracle]

its a simplification / approximation

[wheat delta]

ohh ok ty

[severe canyon]

To find the charge if you know how the voltage changes, for instance

[autumn oracle]

the formula for charge in an rc circuit is Q0(1 - exp(-t/RC))

here Q0 would be the Q in Q = CV

usually when we use Q = CV its implied that the circuit has been connected for a long time

so we say the exponential term is 0

[wheat delta]

kk now i under stand whats the use

[autumn oracle]

yeah

Q = CV is for an ideal capacitor, which would have 0 charge time

[wheat delta]

would these ideas also apply to energy

[autumn oracle]

a real circuit ofc has resistance due to the capacitor, wires, and source voltage, so we model that as a resistance in series

[wheat delta]

kk

[autumn oracle]

wdym

[wheat delta]

Like E = 1/2 C * V^2

this formula

[autumn oracle]

yeah so for an rc circuit theres also an expression for V across the capacitor

V = V0(1 - exp(-t/RC))

so at any time, E can be expressed in terms of time since V can be expressed in terms of time

usually we say the circuit has been closed a long time, so the exponential term is 0, hence just E = 1/2 CV^2

[wheat delta]

Ahhh, thank you so much

[autumn oracle]

np

[wheat delta]

One last q

[autumn oracle]

yeah

[wheat delta]

Can capaicators be charged forever or are their maximums. The guy above told me their were no maxes

[autumn oracle]

the theoretical maximum is the Q0, also sometimes called Qmax

thats the Q in Q = CV

youd need an infinite amount of time for that to happen

since its multiplied by 1 - exp(-t/RC), which can never be zero, a capacitor can never be fully charged

[wheat delta]

Kk now everything makes sense

Tysm

[autumn oracle]

np

[wheat delta]

.close

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[void valley]

Have you already found the position of Q3?

[mint mulch]

yea

but i dont know if its right

[void valley]

What is it then?

[mint mulch]

there are multiple ways to find the position using different formulas

and they all give varying answers

[void valley]

it's just 3/4 of n

and here n is 32, right

[mint mulch]

yea so .75 * 32

[void valley]

and because the result is even

we're supposed to take 24th and 25th result

[mint mulch]

.75 * 32 = 24

oh wait is it supposed to be .75 * n + 1?

so .75 * 33?

[void valley]

No

[mint mulch]

= 24.75?

[void valley]

(24th + 25th)/2

[mint mulch]

yea so 667

[void valley]

it's like the median, if the number of data is even, there is no "middle" and you take the arithmetic mean, the same here

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[mint mulch]

so this is it?

[mint mulch]

= .75 * n 
= .75 * 32 = 24th position
= (n24 + n25) / 2
= (661 + 673) / 2 
= 667```

if n = 33 and we want to find the Q3

we do

= .75 * 33
= 24.75th position

so if its a decimal value, do we round up?

or round the closest nmber?

[heavy stump]

yeah if it's a decimal you usually just round up to the next integer position

[mint mulch]

You always round up?

Even if value is like 20.1?

[muted prairie]

Well it's always going to be .25 .5 .75 or .0

[mint mulch]

But that's only if question asks for a quarterly percentile

What if they ask for 12% for example

[muted prairie]

Idk the formal way to round then

[mint mulch]

Ok

[muted prairie]

But with quartiles alone you don't always round up

Sometimes you round up, sometimes you take the median, probably sometimes you round down

General percentiles probably work differently

Since you probably don't take medians there usually

But idk

[mint mulch]

Hmm ok

Thabks

.done

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@mint mulch Has your question been resolved?

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[hollow dove]

Uhhh how do I do the integral for 2c

[loud sundial]

Integrate $P(x)$. (why?)

[soft zealot]

Civil Service Pigeon

[hollow dove]

wait but how do i set up g(x)

[loud sundial]

I just told you

[hollow dove]

.close

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[jagged root]

what grade math is that?

[spare summit]

uh, you opened a channel btw

got a question to ask?

like a math question of your own

[jagged root]

yes

Well

[spare summit]

send it

[jagged root]

Its a concept question

[abstract bramble]

math isnt really associated with grade level you know
it varies a lot place to place and even then there's a lot of honors/advanced sections

[spare summit]

it's fine. go ahead!

[jagged root]

so for this type of problem right

for the range

how do ik when theres a restriction

or for domain

for parabolas ik x = all real numbers

for the domain

[spare summit]

when the graph does not cover the entire y or x axis (for range and domain respectively)

[jagged root]

fair it just looked hard to me thats why

[abstract bramble]

ok well i will tell you it's likely 11/12th grade or entry college level in the states

[jagged root]

Ohh

[spare summit]

here's a thing though

the domain of all polynomials is always all real x

[jagged root]

is all real numbers

right