help-39

[whole valley]

Yeah

Really shows you how casinos are making their money

[karmic fern]

well no, they make it from the fact that these are negative expectation bets

[whole valley]

Its not small. Its crazy

[karmic fern]

and as soon as you introduce a hard limit on the number of doublings, all of the concentration inequalities come in

[whole valley]

Yeah but the smaller the starting bets with no limits the better their chances get

XaTrillions for a mere 100k profit

[karmic fern]

well, you have seen convergence

a casino doesn't have to worry about whether they get lucky in most games people play

their profits approach a normal distribution with the mean equal to their edge * how many dollars you play

the table limits are to ensure that you can't mess up this distribution with like a billion dollar bet

[whole valley]

Hmm

[whole valley]

Is betting a dollar every time worse then starting at a dollar with martingale?

[karmic fern]

yes, you basically have no chance of going significantly positive

[whole valley]

Lol

[karmic fern]

in general, you want to maximize the variation in your outcomes, so if you want to make money on a negative expectation game, you bet as big as possible

[whole valley]

Yeah makes sense

Just a question, theoretically if you got infinite money you can win infinite money?

100%

Since you can just double down

[karmic fern]

this depends on your strategy

[whole valley]

Martingales

The one i said atleast

[karmic fern]

if you martingale it, you will always go up

if you bet a constant amount of money, you will go down very quickly and never recover

[whole valley]

So infinite yeilds infinite here but almost infinite doesnt yeild almost infinite

[whole valley]

Yeah

[whole valley]

Is that correct?

[karmic fern]

you don't need a martingale to make money with infinite money

just increase your bets exponentially

the whole idea of betting big is the overarching principle

[whole valley]

Same thing but not a stright line

[karmic fern]

yeah it's above the straight line

[whole valley]

Yea

Well what are we missing

[karmic fern]

if you want to actually learn about martingales (generalized) and other things, you should look into stochastic processes

your thing here is technically a supermartingale, in that it goes down

and there's a lot of things already said about them

for one, you can show that, subject to betting limits, i.e. no infinite money, there is no betting strategy / quitting strategy that will make you come out ahead in expectation

[whole valley]

Yeah thats simply the win precentage

So if the odds are the same 50% for win, is martingales the best choice? Or it simply doesnt matter

Or its the same go up if you loose

[karmic fern]

if the game is fair, then the strategy you take doesn't matter, as long as you have zero probability of overshooting your profit target

if the game is unfair against you, then you want to bet big

if the game is good for you, then you want to bet as small as possible to maximize your chances of making a certain profit

[whole valley]

Casino pov

[karmic fern]

well the player and casino are opposite, so just flip everything around

[whole valley]

Actually if you bet every time 50% less then your previous win and double down on a loss you will always loose

Completely useless ✅

[karmic fern]

yeah but your bets will quickly go to zero, and then you'll be sitting on a pile of money

[whole valley]

They wont

How are you sitting on a pile of money that way

[karmic fern]

you're saying that if you win, you bet 50% less, and if you lose, you stay the same? or what is the betting strategy?

[whole valley]

If win bet double

If lose bet 50% less

[karmic fern]

how does that mean I will always lose?

If I win 10 times in a row, don't I end up with a bunch of money?

[whole valley]

Not if you keep doing it

[karmic fern]

but at some point I may hit my profit target

so it's not a sure loser

[whole valley]

Hmm. What if i doible down on winnings until i lose then i bet 50% of the money lost and then 50% of each last bet

Since somewhere along the road to almost recouping your entire money you will lose 1 time your already a sure loser

[karmic fern]

none of those betting variations change the probability that you reach a certain amount, as long as you don't concentrate your probability in the regions above that profit target

[karmic fern]

during the process of doubling down, you might hit your profit target

just make sure to not overshoot on the last bet

[whole valley]

Not if your bet is infinitely small then the profit target, that would mean your winning chance is of a great great win streak, almost guranteed you will lose atleast 1 time

[karmic fern]

you can't start "infinitely small"

you have to start at some finite, but arbitrarily low amount

but if you do that, you'll just be wasting time until your bet size actually gets to the correct size

[whole valley]

Its an expression, i meant like if we need to hit a million i will bet 10 cents

If i use this system my chance of loosing the 10 cents is very big

[karmic fern]

it doesn't matter; basically your expectation is the same at all stages

[whole valley]

I never restart, i just keep betting 50% of the last bet

[karmic fern]

then your bets will go to zero

and you'll have some money left over

[whole valley]

It will never go to zero

Close but never

[karmic fern]

there's a theorem that shows that none of your betting strategies will change the fact

[whole valley]

Even for the worse?

[karmic fern]

no

in a fair game, no betting strategy is going to make your expectation worse

as long as you have a bet limit here, which is your bankroll

[whole valley]

The thing is i never get to use my entire bankroll

[karmic fern]

well yeah, but that doesn't matter, because your bankroll could go positive and then you decay to zero

and then you're sitting on a constant amount of positive bankroll

[whole valley]

It cant

1 lose along the way and its all a lose

[karmic fern]

you can never construct a betting strategy that will surely get you to lose before you hit a given profit target t

basically what happens is that this profit target bounds how far/long you will go

[whole valley]

Ofcourse but you can make one that has a higher chance of loosing

[karmic fern]

now if you don't care about a profit target, then sure betting all in repeatedly will make you go to 0 with 100% probability

[karmic fern]

no you can't, as long as you preserve the condition that none of your probability is concentrated above the profit target

[whole valley]

Lets see. Lets say we got 50% as we said fair game and want to win 10$

Bet 1. Lose bet 1. Win sbet 1 lose bet 1 . Win, eventually we will get 10$ profit

50% of acheiving the profit if we start with 10$

[karmic fern]

how much did you start with?

[whole valley]

10$

Said in the end. Im messy i know

[karmic fern]

yeah so it turns out the probability that you make x profit before going broke is given by (10+x)p -10(1-p) = 0

do you want to know the proof for why these betting strategies don't change the situation?

[whole valley]

Now again but now this system. I bet 0.01. 0.02 0.04 0.08 op o lost now i bet 50% of our total.lose. 0.005 win 0.0002 lose 0.000005 win whatever

[karmic fern]

bro there's actually a mathematical proof

you don't have to come up with these random strategies

every single one of them will have a flaw in it

[whole valley]

Is it really flaw if we never get to use our entire money

[karmic fern]

yes it is flawed

if you stop betting at a particular stage (i.e. never get to use all your money), your expectation is also your starting balance

[whole valley]

Only at the starting stage

[karmic fern]

I mean sure, I guess that's "not winning" but that's also not really an interesting expression

[whole valley]

The more you wait the less chance your winning

[karmic fern]

no

you could just as easily restart from where you're going with a diff strategy

[whole valley]

For example if we stop after 100 were.most.likely loosing

[karmic fern]

and end up with the same probability of getting the profit

there's literally a mathemtaical proof

[whole valley]

Yeah but thats diffrent strategy

[karmic fern]

yeah but the point is basically the only way you're not making the profit target involves quitting midway through

[whole valley]

No

[karmic fern]

in that same logic, you could say that "hey, I'm almost at my profit target, let me quit so I have 0% of winning"

or fine

[whole valley]

Look at my example

[karmic fern]

"I have $9 in profit; let me reduce my bets to 0.5, 0.25, 0.125, 0.0625, so I have no chance of winning"

no stop

the point is that none of these strategies get you into a worse spot

you could just as easily change the strategy and you will not have suffered any loss from using your strategy for a number of time steps

there is a mathematical proof for this

[whole valley]

Wait 1 second. You misunderstood me

[karmic fern]

no, you don't get it

[whole valley]

This is not what im saying at all.

[karmic fern]

none of the strategies you proposed can ever hurt a player's chance of winning

stop

I gave you an example of a strategy that purports to reduce the chance of winning but is really degeneraet

if you propose a strategy, run it for 100 or 1000 or 10000 or N time steps

[whole valley]

Its not what i said. Its still profit

[karmic fern]

if I suddenly decide to change to a different strategy, my chance of winning is still the same as if I started with $10

that is the sense in that the strategy that you pick does not harm your chances of winning

[whole valley]

But if you change to a diffrent startegy your not longer using the worse startegy

[karmic fern]

yeah but it's not "worse" because it didn't hurt me at all

[whole valley]

Read this for a second

[karmic fern]

no

stop

you're just parroting a point that I understand and it's not important

[whole valley]

It hurts if you stick to it, as its a system

[karmic fern]

yeah, the same as if you stick to "I won't bet at all"

[whole valley]

No

[karmic fern]

my point is that the class of strategies you describe is basically the same as the examples I've described

[whole valley]

Its not at all

[karmic fern]

yes it is

bro

stop

there are only three outcomes:

1. you go broke
2. you stop in the middle and don't make any more progress
3. you hit your profit target

[whole valley]

Theres no profit target

[karmic fern]

yes there is

bro

you literally just said

"I want to make $10"

stop

[whole valley]

Your right 1s

[karmic fern]

stop

why don't you actually learn the actual probability theory here

instead of just trying to come up with every different scheme possible

this is actually pretty basic shit

it's been studied numerous times

[whole valley]

So the chances of winning 100 times consecutively is lower then 50% I most likely loose somewhere along the way, now i infinitely trying to recover partial of the loss and loss along the way too,

[whole valley]

I agree with it if you change

[karmic fern]

if you have a "what is the probability that I make $10 before I go broke" problem

suppose it's a fair game

[whole valley]

Just point out the logic flaw

[karmic fern]

this is represented as a martingale, where a martingale is defined as E[X_[t+1} | X_t] = X_t

bro I'm not going to dissect every single scheme you come up with

that is a waste of time

you can figure it out yourself

[whole valley]

Everything you said is true

[karmic fern]

I'm telling you that IF YOU PLAY UNTIL YOU GO BROKE OR MAKE $10

i.e. those are the only two outcomes that happen

there is NO strategy that will affect your chances of getting $10

[whole valley]

If you lose once your never winning the 10$!!!! Being constantly losing

[karmic fern]

bro

stop

go read about doob's optional stopping theorem with respect to martingales

in here, you have a 100% bound on what value you stop at

[whole valley]

Im not talking about martingale anymore

We stopped 20 minutes ago

[karmic fern]

I AM talking about the actual martingales in probability theory

it's a shame you don't know what they are

[whole valley]

I know them

[karmic fern]

no you don't

[whole valley]

Their true

[karmic fern]

bro

these are not the same martingale

there is an optional stopping theorem that states that your expectation is the same at all time steps if certain regularity conditions hold

if you play until you either profit $10 or go broke

[whole valley]

.

[karmic fern]

are you trolling

[whole valley]

No

[karmic fern]

then pay attention

it's not just any sequence

[whole valley]

Its a line

[karmic fern]

bro

it's NOT

read the fckin definition

it's a sequence that represents a "fair" game, i.e. the conditional expectation of the next random variable is just the current one

if the game ends in one of two states: you end up broke or you end up with $10

then the time that the game ends is called a stopping time

[whole valley]

Ive read it and its true YES.

[karmic fern]

bro

I literally have a degree in statistics

you're talking bullshit

Doob's optional stopping theorem says that the expectation of your bankroll at the stopping time is the same as what you started

so P(you end at zero) * 0 + P(you end at $20) * $20 = $10

and this means that P(you end at $20) = $10, no matter what strategy you take

[whole valley]

Your right at everything you said, you just misunderstood that im not switching the way, i can run this on code right now and over 99% of times.it will infinitely run on lose

[karmic fern]

there exists the third possibility that you just stop playing in the middle, but even then, you can show that the expected value of your bankroll in there is $10

and it's not super relevant, because again, those strategies do not harm you at all if you simply decide to switch

[whole valley]

because there is no option to recover

[whole valley]

If you switch its not the same strategy

[karmic fern]

who fcking cares

you weren't harmed by using the strategy

that's the whole point

[whole valley]

Yeah thats true

But if you stick to it forever

[karmic fern]

your saying that "it harms you" is tantamount to saying that just refusing to bet further harms you too

it doesn't matter

yeah if you stick to the strategy of betting nothing forever, you also lose

the point is that those are uninteresting and nobody cares

all of those strategies are the same in that they produce outcome (2)

[whole valley]

Alright well thanks for your help before

[karmic fern]

you can also show that all of these strategies are required to let the bet amount decay to zero

[whole valley]

Now your a bit frustrated or tired i dont know, just admitted its true but unintresting after not listening for 13 minutes

[karmic fern]

what?

I literally told you that there are strategies that involve stopping in the middle and not betting

[whole valley]

You never stop

[karmic fern]

that is basically what all of your strategies amount to

bro

holy shit

you can construct strategies where you bet a nonzero amount of money, but it decays quickly enough so the impact on your balance is negligibly small

[whole valley]

Yeah!

[karmic fern]

I don't care about that technciality; if I had to write that out every time, my fingers would die

[whole valley]

You dont need to, its purely.logical

[karmic fern]

I gave you the example of a strategy where you discover you're +$9 and you decide then to start betting decreasing powers of 2

again, those are basically the same as deciding to stop betting

[whole valley]

The diffrence i said im doubling down on that win

[karmic fern]

bro

I don't care about your particular strategy

all of that class of strategies has to eventually have a betting amount that converges to zero

[whole valley]

I just said if you stick to this strategy your likely to be in the negative when you stop

[karmic fern]

you cannot have a strategy where you bet a certain amount of money infinitely often

[karmic fern]

you didn't learn anything LOL

you have a 50% chance of being negative when you stop

this is the case with every other strategy

[whole valley]

Even after 200 bets?

[karmic fern]

yes, bro that is the whole damn optional stopping theorem

as long as you don't over shoot the $10

[whole valley]

Let me walk you through my mind and you can point out where im being a dumbass okay

Bet 0.1 win

[karmic fern]

if you stop when you're at $20 or at $0, your chance of being at $0 is 50% and your chance at being at $20 is 50%

[whole valley]

50%

[karmic fern]

REGARDLESS of the strategy you take

[whole valley]

Yeah thats true

[karmic fern]

bro

stop wasting time "studying" or "researching" this stuff

just go learn an actual intro to stochastic processes

[whole valley]

I agree both stopping at a lose and not allowing to recover the loss is the same

[karmic fern]

learn what an actual martingale is

you will learn that the things you're talking about here are just trivialities

[whole valley]

Right..

[karmic fern]

if you choose to stop when you hit +$8 or -$3, you can calculate the probabilities the same way

it'll be 3/11 and 8/11, the same way

[whole valley]

Alright i think i got this

[karmic fern]

if you choose to do anything that results in getting stuck in the middle, you will end up betting nothing, and then your strategy will be the same as effectively stopping

the point is that all of these are trivially computed from using optional stopping

and none of them depend on the actual betting strategy you use

[whole valley]

But here your not stuck in the middle your slowly going downhill very slowly for eternity because x=x-50%x will never be 0

You know what fine. Its effectively stopping

[karmic fern]

huh? your bankroll cannot go to zero very slowly

[whole valley]

It doesnt go to zero, it just slowly looses crumbs of the new x

Will never.loose more then 1x

[karmic fern]

yeah but the whole sum of the crumbs doesn't add up to anything

it's just expectation zero

and variation close to zero; it's the same as if you had effectively stopped

[whole valley]

It amount to statistically half of initial bet, no?

Could be more could be less

[karmic fern]

go read the material on stochastic processes

I'm not going to continuously examine every single one of your betting strategies

the general theme is that none of them matter

[whole valley]

Alright i can accept that

[whole valley]

It was only 1 but yeah alright ✅

Thanks for the help

What to google again?

Doobs optional stopping and? @karmic fern

I probably missed more

[karmic fern]

if you look up doob's optional stopping theorem, you can probably follow the links and stuff to find everything else

[whole valley]

✅

[karmic fern]

a useful exercise is to find out why it doesn't apply to the (gambling) martingale

also look up azuma's inequality if you want to know about concentration bounds with respect to martingales

[whole valley]

Could i dm you sometimes questions after i read or no

[karmic fern]

just ask in here; there are people who know about it

[whole valley]

Yeah but it seems cool to keep intact

[karmic fern]

sure then

[whole valley]

Accept friend req. See you

.close

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[flint ibex]

I'm kinda stuck on what to do next here. maybe theres a shorter way i dont know about yet?

[plush bramble]

There isn't any shortcut, just keep computing

[pearl pond]

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[flint ibex]

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[rocky flume]

Can somebody walk me through this question

[pearl pond]

@rocky flume Has your question been resolved?

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[formal swift]

Question 20

It’s for a math contest

I’m lost in every part

[cinder flower]

what matters is how many 5s and 2s are in the prime factorization of n!

because that tells how many 10s there are

which is how many trailing zeros the number will have

[formal swift]

Mmm

I still don’t get it

[cinder flower]

and a little side note it turns out the only thing that matters is how many 5s there are because there will always be at least that many 2s

ok take 5! for example

[formal swift]

Ok

[cinder flower]

5*4*3*2*1

this is 12*10, so just one trailing 0

a number like 12*10^5 would have 5 trailing 0s

[formal swift]

Okay but what is the variable m for

What is it defining

Oh I’m pretty sure m is the amount of trailing zeros

Or I’m assuming

[cinder flower]

yea that's what it says

so when m = 1, yes, there is a value of n such that n! has m trailing 0s

5 is an example

what about 2? 3? etc

that's what it's asking

[formal swift]

Mm is it like a trail and error sort of questions

Where I pretty much have to go through a set of numbers

To find what I’m looking for?

Wow I reread everything and analysed it and now I’m more lost than ever why did m equal 5,11,17,23,29,30

[cinder flower]

legendre's formula could streamline it a bit maybe

[formal swift]

What the legendre formula

[cinder flower]

a formula for the highest exponent of e.g. 5 that divides n!

works for any prime but only 5 matters here

[formal swift]

So basically this formula is another way I get into the answer?

I’m sorry if I seem a bit slow since this isn’t a concept I’ve been taught

It’s a math competition from last year that has been given out so I’m assuming similar concepts will be in the next one

[cinder flower]

i don't think it would be that much different than the solution they have

[formal swift]

Oh alright

This sort of question I should probably take up with my teacher I’m assuming since it probably requires a bit of understanding

I’m not Tryna memorize anything because obviously I’m gonna get different questions so it’s probably better to learn why something happens than how something happens

[cinder flower]

it's not that complicated but yea i've seen this kind of stuff in competition math

[pearl pond]

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[midnight haven]

sorry this is a really simple question but uhdoes anyone see the error in this??

[midnight haven]

on the part d

[west sapphire]

your expressions use just t, which implies that the particle leaves the helix at t=0

probably you want (t - t0), where t0 is the time it actually leaves the helix

[midnight haven]

oh so instead of just cos(8) - tsin(8) do like

cos(8) - (t-8)sin(8)?

ah yup that fixed it-.

ty~

.close

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[dry quail]

hey I dont really understand what I’m supposed to do or how to do it, cab anyone help out thanks

[hardy kindle]

Substitute the values in the ordered pairs

[north sierra]

input x and y into each equation and if they make both equations true then it's a solution

[hardy kindle]

Wait no it's graphing

Nevermind uh

Dkw

I'll leave it to the professionals

[north sierra]

so for example for that first solution the 0 would represent the x and the -13 would represent the y

[dry quail]

ohh ok thanks

this is what happens when i miss one lecture lmao 💀

im trying out the problem ill lyk if i run into any issues

[pearl pond]

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[dry quail]

wait how do i do this now 💀

.reopen

[pearl pond]

✅

[light helm]

identify the intersection point

[dry quail]

is that all

[light helm]

yes

[dry quail]

okk tysm

.close

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[manic sparrow]

for the second problem

when ur evaluating the inner sum u would just assume 2^(-i) is constant right

[west sapphire]

yes

[manic sparrow]

but then you have $\sum_{i=0}^n\left(\frac{2^{-i}(1-3^{n+1})}{1-3}\right)$

and then u have two variables

[west sapphire]

should be 3^(n+1) but otherwise yes

[manic sparrow]

is it because it starts at j=0

[west sapphire]

it's becaues it ends at n

if it ended at n-1 then your expression would be right

[jolly parrot]

Jash

[manic sparrow]

yea i was wondering about that when i was typing it

[west sapphire]

yea that's correct now

you can move everything except the 2^-i outside the sum because nothing else depends on i

[manic sparrow]

$\frac{1-3^{n+1}}{-2}\sum_{i=0}^n 2^{-i}$

[manic sparrow]

but isnt the upper bound n

[west sapphire]

sure

wait

why are you suddenly summing over n instead of i

[manic sparrow]

oop

[jolly parrot]

Jash

[west sapphire]

yea that's good now

[manic sparrow]

doesnt that mean u cant assume its constant

[west sapphire]

the summation variable is i

anything that does not depend on i is a constant with respect to the sum

[manic sparrow]

oh

so $\frac{1-3^{n+1}}{-2}\left(\frac{1-\left(\frac{1}{2}\right)^{n+1}}{1-\frac{1}{2}}\right)$

[jolly parrot]

Jash

[west sapphire]

yes that's correct

[manic sparrow]

ty

.close

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[swift pilot]

Hey, could someone help me to solve a differential equation, here is my equation: y'(x) = sinh(x)cos²(y) with y(0)=pi/2

[pearl pond]

@swift pilot Has your question been resolved?

[pearl pond]

@swift pilot Has your question been resolved?

[pearl pond]

@swift pilot Has your question been resolved?

[pearl pond]

@swift pilot Has your question been resolved?

[plush bramble]

Write y' = dy/dx then use separation of variables

[swift pilot]

Ok thank you !

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[cursive portal]

help

[stoic path]

Have you drawn a rough diagram?

[pearl pond]

@cursive portal Has your question been resolved?

[cursive portal]

yeah

[cursive portal]

how to do it without graphing

[pine light]

you need to calculate the angles knowing $\tan(\frac{\theta}{2}) = \frac{a}{|a|/\sqrt(3)}$

[jolly parrot]

The Mad Pirate

[pine light]

@cursive portal give me a second to revise it

$$ L \cdot \cos(\frac{\theta}{2}) = a $$
$$ L \cdot \sin(\frac{\theta}{2}) =\frac{|a|}{\sqrt{3}} $$

[jolly parrot]

The Mad Pirate

[pine light]

@cursive portal now
$$ \tan{\frac{\theta}{2}} = \frac{1}{\sqrt{3}} $$

[jolly parrot]

The Mad Pirate

[pine light]

so, that gives you $\theta$ at the bottom

[jolly parrot]

The Mad Pirate

[pine light]

the remaining angle is just $$ \beta = \frac{\pi - \theta}{2} $$

[jolly parrot]

The Mad Pirate

[pine light]

now if all thre angle are the same, it is an isoceles triangle

if theta is $\pi/2$ it is a right angled triangle

[jolly parrot]

The Mad Pirate

[pine light]

Sorry, I got confused with the definition of an isoceles triangle

you will most certainly get an isoceles triangle if $\beta \neq \theta $

$$\beta \neq \theta \neq \frac{\pi}{2}$$

and an equilateral for

$$\beta = \theta = \frac{\pi}{3} $$

[jolly parrot]

The Mad Pirate

The Mad Pirate

[pine light]

@cursive portal I think that should do it

[cursive portal]

hmm alright

.close

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[grim swift]

hello, trying to find the derivative of this using definition of derivative and am getting stuck.

[grim swift]

i think i need to make a common denominator to combine them ?

[native gull]

yes

simplify

[fathom dragon]

You need to use (f(x+h) - f(x))/h?

[grim swift]

yep, can't use quotient rule yet

[fathom dragon]

so as I understand, using (f(x)'*g(x) - f(x)*g(x)')/g(x)^2 is forbiden

[grim swift]

yeah they r makin me do it with the definition

[fathom dragon]

so Ig you have to simplify as other guy said, but this aint help for you

[grim swift]

how can i simplify it ? thats what im stukc on

[fathom dragon]

I could try help you with that

or just gave you link to site

which would show you

how to do it

[grim swift]

if u have one thatd be great

when i looked at symbolab they only use quotient rule

[fathom dragon]

wolfram alpha

https://www.wolframalpha.com/input?i=dervitave+of+(3%2Bx)%2F(1-3x)

oh hold on

sadly

it will use

always

qoutient rule

[grim swift]

yep lol

[fathom dragon]

dang

thats

a lot

of

math

I think you could simplify the (3 + x + h)/(1 - 3x - 3h) using definithon of function limit

[pearl pond]

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[cloud eagle]

Hi

[cloud eagle]

How do I figure out if it's an overestimate or underestimate

It doesn't tell me on what intervals it is increasing/decreasing or where it's conc up/down

It doesn't even guarantee that it changes on the interval we are integrating on

So I'm confused

If anyone could help it would be much appreciated 👍

One of the potential answers could be not enough information, but the equation for the sum looks funky to me

It's 1/2(f(1.25^2+1.25)+f(1.75^2+1.75)+f(2.25^2+2.25)+f(2.75^2+2.75))

Wait

I think it's the not enough info one

Because it's a midpoint sum, the x values are 1/2(x_1+x_2)

Not 1/2(f(x_1)+f(x_2))

Because that is y value

Now I'm just looking for confirmation

<@&286206848099549185>

[dusky swallow]

Shut the fuck up

[pearl pond]

@cloud eagle Has your question been resolved?

[cloud eagle]

<@&286206848099549185>

Sorry if I'm being disrespectful in any way, I'm just looking for help and trying to follow the rules to the best of my ability

<@&286206848099549185>

<@&286206848099549185>

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[cloud eagle]

.reopen

[pearl pond]

✅

[cloud eagle]

My question was not answered

[plush bramble]

If you don't want to be pinged, remove your helper role and don't be an asshole about it

[plush bramble]

You did nothing wrong

[pearl pond]

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[trim solstice]

let's look at a polynomial of degree n which is a Taylor expansion around the point (0,0) of the function $g(x,y) = (-3xy-3x-4y+6)sin(x)ln(1+y)$ so the coefficient of $x^3y^ 2$ is?

[jolly parrot]

mtr123

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[fleet idol]

.close

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[lyric helm]

how do I solve this without any angels

angles

like can I do like (n-2)*180 or something?

cuz I’m finding the sum of the angles anyways

[autumn narwhal]

Look at the sum S + Q + U and R + T + P separately

And consider the circle that passes through all of those six points.

Well I gotta think why that circle exists tho

If it does then this works

Oh

[brisk scarab]

I don't think the circle does exist

[autumn narwhal]

Look at the circle around S, Q, U

and a different circle around R, T, P

Then it still works

[brisk scarab]

(those do)

[autumn narwhal]

Yeah you are right

👍

@lyric helm

[pearl pond]

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[magic bluff]

I know its impossible for the statement and its negation to both be true. Just wondering what im doing wrong here cuz somehow im getting both of them true.

[worldly glacier]

,rrcw

[magic bluff]

Whats that mean?

[brittle tinsel]

,rccw

[brittle tinsel]

lol

[magic bluff]

Ohh my bad

[brittle tinsel]

well the first statement does not appear to be true

it says there exists a rational number x such that for ANY integer y, xy=1

(or y = 0)

so choose some rational number, say 1/3

it is obviously false that (1/3)y = 1 for all integers y

[magic bluff]

Is 1/y not rational?

[brittle tinsel]

you wrote y is an integer in your problem

[magic bluff]

y is an integer

[brittle tinsel]

maybe you wrote the problem wrong?

[magic bluff]

I let x=1/y which is rational right?

Cuz y is an integer

Or is that wrong

[brittle tinsel]

i think it's worded incorrectly then. what you're saying is that for all rational numbers x, there exists an integer y such that xy = 1 or y = 0

and i mean that's still false

are you sure the problem is as you wrote down?

[magic bluff]

Yea. there exists a rational number x such that for all integers y, y=0 or xy=1

[brittle tinsel]

i would think a little harder about what this means, because it is false

[magic bluff]

And i used x=1/y as a rational number

[brittle tinsel]

the issue is that x is assumed to be fixed

if it depends on y then it varies

you need there to exist a single fixed x

such that that statement is true

meaning, take xy = 1. y is allowed to be anything in the world since it says for all. y could be 17, 100, 4, etc. is there a single x value such that x(17) = 1 and x(100) = 1 and x(4) = 1?

[magic bluff]

I understand the question i just thought you could use x=1/y

[brittle tinsel]

i just explained why you can't do that

you need a single x

x can't be multiple things

these kinds of problems are always so hard to explain lol

[magic bluff]

You’re explaining is fine it’s just my brain is just not comprehending it lol. Just exhausted from midterms. I appreciate the help tho thank you.

[pearl pond]

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[safe oasis]

y=-2x+3

[safe oasis]

equation of the line i think it's called

i have to get the perpendicular equation

<@&286206848099549185>

[solid coyote]

The perpendicular line's slope is the opposite reciprocal

[pearl pond]

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[pearl pond]

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[still plover]

How would I convert y = x^2 + 3x + 2 into slope intercept form?

[frank goblet]

like y=mx+b?

[still plover]

Yes

[frank goblet]

we have a polynomial of degree 2, that can't just be reduced to a polynomial of degree 1
the function you have given is not a straight line, therefore there is no constant slope
rather the slope depends on the x value

[still plover]

So it cannot?

[frank goblet]

yep, it can't be expressed as y=mx+b

[still plover]

Alright thank you

.close

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[smoky saffron]

what ami doing wrong ehre?

[feral leaf]

Why do you have x^2 - 4x + 16 in the numerator?

[midnight haven]

before you start algebraic manipulations try and see if your limit is even indeterminate

[smoky saffron]

because when i do limits the teacher makes us put the difference of cubes on both sides i dont know why tho

[smoky saffron]

oh ok ill try right now

[smoky saffron]

what does it mean if its 0/128

[midnight haven]

then it's just 0

[smoky saffron]

oh

[pearl pond]

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[lunar condor]

stuck on this part don't know how to find fx and fy given what I have

[lunar condor]

<@&286206848099549185>

[runic veldt]

where(chaos)

[lunar condor]

uh

like

fx

and fy

I'm not sure

how to calculate for them

I'll show u my math

idk I'm not very good at this unit 💀

and it's eating me alive

[pearl pond]

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[lapis ledge]

I need help factoring Polynomials and trinomials

[lapis ledge]

stressing that I might not pass math already bro

<@&286206848099549185>

[plush bramble]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[plush bramble]

And

[lapis ledge]

?

[prime bramble]

!da2a

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[prime bramble]

if you dont post a question, we cant help you

[lapis ledge]

oh sorry

I need help understands how to solve questions like these until I'm used to them

I'm on number 5

teacher said only odd numbered questions

[plush bramble]

What even is the question

[lapis ledge]

factoring

[prime bramble]

ok so

you probably want to rewrite 5) as y^2 + yz - xy - xz

because then you can group the terms like this

(y^2 + yz) - (xy + xz)

[lapis ledge]

ok

is there a step after doing common terms?

[prime bramble]

examine each of these grouped terms

can you pull out a GCF from each?

oh wait

did you already do that?

[lapis ledge]

yeah

itd be 1

no?

[prime bramble]

...no?

what's the GCF of y^2 + y?

[lapis ledge]

oh

its y

[prime bramble]

okay

and whats the GCF of xy + xz?

[lapis ledge]

x

[prime bramble]

okay good

so (y^2 + yz) - (xy + xz) = y(y+z) - x(y+z)

[lapis ledge]

okay

[prime bramble]

now we have y(y+z) - x(y+z)

can you factor out something from this?

[lapis ledge]

what even happened

[prime bramble]

we pulled out GCF from both terms

[lapis ledge]

and did what after

[prime bramble]

nothing

[lapis ledge]

what?

[prime bramble]

y^2 + yz = y(y+z) we agree?

[lapis ledge]

ye

[prime bramble]

and (xy + xz) = x(y+z)?

[lapis ledge]

oh now I see

yes yes

[prime bramble]

ok good

so we have y(y+z) - x(y+z)

[lapis ledge]

whys x out

[prime bramble]

hm?

wdym

[lapis ledge]

whyd it get pulled out from brackets

[prime bramble]

do we agree with this?

[lapis ledge]

sure but if it factored all the x's why does it go outside brackets

[prime bramble]

because that's how factoring works?

ab + ac = a(b+c)

[lapis ledge]

why didn't we put y outside brackets?

[prime bramble]

we did

[prime bramble]

see here

[lapis ledge]

we did?

hmmm

okok

[prime bramble]

so we do both at once for (y^2 + yz) - (xy + xz)

to get y(y+z) - x(y+z)

does this make sense to you?

[lapis ledge]

sure

[prime bramble]

okay

so now

we have y(y+z) - x(y+z)

do you see anything we can factor out?

[lapis ledge]

yeah

y and x

I mean z

[prime bramble]

y+z you mean?

or just y and z

[lapis ledge]

yeah

[prime bramble]

right

so what is the final answer?

try factoring it out yourself

[lapis ledge]

-x

bruh i did it wrong probably

[prime bramble]

how did you get this?

[lapis ledge]

or y-x

[prime bramble]

this is part of the answer

but not quite the whole thing

[lapis ledge]

I wouldn't know what to do

how would you factor this?

[prime bramble]

here

we have y(y+z) - x(y+z)

[lapis ledge]

yeah

[prime bramble]

let's call y+z k shall we?

[lapis ledge]

ok

[prime bramble]

so y(y+z) - x(y+z) = yk - xk

all we did was let y+z = k

[lapis ledge]

hm

ok sure

[prime bramble]

now how would you factor yk - xk?

[lapis ledge]

factor k

[prime bramble]

right

so we have yk - xk = k(y - x)

[lapis ledge]

mhm

[prime bramble]

now let's substitute k = y + z back

to get (y+z)(y-x)

this is our final answer

[lapis ledge]

so we just break down the question as much as possible?

[prime bramble]

essentially yes

in general, math becomes much easier when you break it down

[lapis ledge]

but when will I know if I broke it down enough or if I even got the answer right

[prime bramble]

i cant really answer your first question, but as for the second: you know you have the answer right (especially in high school math) if you can examine your process and find no logical mistakes or leaps of faith

if you've made no logical mistakes or leaps of faith, then math is actually quite forgiving for a subject

because you basically cannot get the question wrong

[lapis ledge]

k, can we do a different question

[prime bramble]

sure

[lapis ledge]

x^2 - x - 6

[prime bramble]

right

do you know how to factor quadratics?

[lapis ledge]

is that like

4 or more terms?

[prime bramble]

quadratics are polynomials with a maximum power of 2

they can have a maximum of 3 terms

[lapis ledge]

I got a idea but not the best

I know that its a + b + c

and you have to find the numbers for those that have variables

so rewinding time pretty much to find the number that got the original number in the question

[prime bramble]

hmm, i dont quite know what you mean

[lapis ledge]

ok let's just do it and see

[prime bramble]

so we need to find two numbers which multiply to make the last coefficient (-6) and add to make the middle coefficient (-1)

[lapis ledge]

mhm that I know

on that fact

is there a simple easy fast method to find those numbers incase it difficult for me?

[prime bramble]

ehh

if there are, i dont know of any

[lapis ledge]

alright continue

[prime bramble]

if the numbers are difficult, just use the quadratic formula

[prime bramble]

right

[lapis ledge]

?

[prime bramble]

we're here

[prime bramble]

you can factor any quadratic with the quadratic formula, but it's slowing than factoring simple quadratics by hand

it is more powerful though

because it works in more situations

[lapis ledge]

when won't it work?

[prime bramble]

it always works, but if you are considering real numbers only (think of these as the collection of possible decimal numbers), then it doesnt always work

because some quadratics have no real roots

and cant be factored into factors that are only real numbers

[lapis ledge]

okay continue with the question

already confused

[prime bramble]

we're still here

[prime bramble]

dont worry about roots of quadratics too much yet

i suspect you'll learn about when quadratics fail to have real roots soon enough

(it's when the discriminant is less than 0)

but again, dont worry too much about it

[lapis ledge]

I'll try

[prime bramble]

currently, we must find two numbers which multiply to make -6 and add to make -1

[lapis ledge]

-1 and 6

or backwards

-6 and 1

[prime bramble]

do those add to make -1?

[lapis ledge]

oh

[prime bramble]

remember, the two numbers have multiply to make -6 and add to make -1 at the same time

[lapis ledge]

ok

seems so easy but I genuinely can't find the number

[prime bramble]

try something related to the numbers 3 and 2

[lapis ledge]

yeah

3 -2

backwards srrt

-3 and 2

[prime bramble]

yes

-3 and 2

this is actually enough to just get straight to the answer

x^2 - x - 6 = (x-3)(x+2)

i could show you the long way, if you wish

it's more tedious, but it shows you the details of what just happened

what i just did was a shortcut of sorts

[lapis ledge]

I want to learn a simple way so whatever one it is

[prime bramble]

okay, then writing the answer straight from finding the numbers is always quicker

but this quickness only works when the coefficient of x^2 is 1

[lapis ledge]

I need to show my work though

[prime bramble]

if its not 1, then you need to go the long way

[prime bramble]

okay

let's do the "long" way then

[lapis ledge]

ok

[prime bramble]

we found -3 and 2 to be the numbers we needed

so now we break apart the central -x term into -3x + 2x

[lapis ledge]

ok

[prime bramble]

x^2 - 3x + 2x - 6

now this is like the problem i helped you with earlier

let's group

(x^2 - 3x) + (2x + 6)

and factor out GCF from each term

[lapis ledge]

x

[prime bramble]

can you do that for me?

[lapis ledge]

ok

(x-3)+(2+6)

[prime bramble]

not quite

what's the GCF of x^2 - 3x?

[lapis ledge]

I guess 1x?

[prime bramble]

right

whats the GCF of 2x - 6?

[lapis ledge]

2

[prime bramble]

okay

so x^2 - 3x = x(x-3) and 2x-6 = 2(x-3), do we agree with these?

[lapis ledge]

if 1x/x^2 = x

does that mean a number never affects the invisible one?

[prime bramble]

i... dont know what you mean?

[lapis ledge]

brahh moving on i agree yes

[prime bramble]

so we combine these to see that (x^2 - 3x) + (2x - 6) = x(x-3) + 2(x-3)

i made a mistake earlier, i wrote 2x + 6 when its actually 2x - 6, as given in the question

[prime bramble]

but we're here now

do you understand what's happened thus far?

[lapis ledge]

hey I honestly appreciate you for helping, its not you but I just gotta find a way for me to understand where I can remember and not struggle

.close

[pearl pond]

Channel closed

Closed by @lapis ledge

Use .reopen if this was a mistake.

[prime bramble]

hmm, my apologies ig

i probably could have done a better job explaining

but i'm still practicing my pedagogy so

do forgive me if i was not being clear

[lapis ledge]

no it really isn't you I just suck at math

[prime bramble]

i dont think anybody "sucks" at math

[lapis ledge]

Its a huge challenge and I really need to do something for me to pass this semester

[prime bramble]

with enough time and practice, virtually anyone can get "good" at math

[lapis ledge]

I'm trying brahh I swear I'm trying

[prime bramble]

i dont doubt you are

but you arent "bad" at math

you just need practice

[lapis ledge]

thank you again hope you have a goodnight

[prime bramble]

you too

[lapis ledge]

bless

🙏

[pearl pond]

Available help channel!

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[midnight haven]

I'm not sure what do to after this

I've just started calculus 2 a few weeks ago and I'm confused

[placid badge]

use partial fractions**

you can rewrite the denominator-- nvm lol just go to partial fraction

[summer juniper]

Factor the denominator

And use partial fractions

As he said

[midnight haven]

would integration by parts also work

Or splitting up the integrand

[placid badge]

ive never done integraration by parts with those, but ye split up the integrand

[wild fable]

You should try completing the square

[midnight haven]

not sure how

(x + ) (x - )

i cant think of anything that makes -6x and 25

[wild fable]

Its completing the square so you aren’t doing that regular factorising stuff

[midnight haven]

oh

[hollow rose]

To see if its factorable do b^2 -4ac

i got -64 soo

[midnight haven]

(-6)^2 - 4(1)(25)

[wild fable]

You want to get it into the form (x+a)^2 + b

[midnight haven]

36 - 100

ok

[wild fable]

I can show you how to complete the square

[midnight haven]

yes please im still confused

[wild fable]

first set x^2 - 6x + 25 = 0 right

[midnight haven]

yes

[wild fable]

then the first thing you do is get the constant on the right hand side so in this case we get

x^2 - 6x = -25

i purposely left a space there by the way

Then you have to take half of the middle term (-6x) and then square it, then add it to both sides

So in this case we get

x^2 - 6x + (-3)^2 = -25 + 3^2

$x^{2}-6x+\left(3\right)^{2}=-25+3^{2}$

[jolly parrot]

water beam

[wild fable]

this is what you should have

[midnight haven]

yes

[wild fable]

makes sense so far?

[midnight haven]

yes

9 on each side

[wild fable]

yes but heres the trick

hold on

we can group the left hand side easily if we take the first term, x^2 and the third term, the 3^2 which we just added

note that when we bring this into the form of (x+a)^2 + b

we use this sign

now if it was $x^{2}+6x+3^{2}=-16$

[jolly parrot]

water beam

[wild fable]

with a plus instead of a minus

we use that sign

so $\left(x+3\right)^{2}+16$

[jolly parrot]

water beam

[wild fable]

but we have a minus here so we do this

[midnight haven]

ohh

[wild fable]

do u have any questions so far

[midnight haven]

(x+a)^2 + b

what makes 3 the a and the -16 the b

[wild fable]

look

if we expand (x-3)^2 + 16

what do we get

[midnight haven]

x^2 - 6x + 25

[wild fable]

exactly

so do you see we have it factored now

[midnight haven]

yes

Ohh

[wild fable]

lets try one more example to make sure u know how to complete the square

[midnight haven]

Ok

[wild fable]

$x^{2}-4x+3$

[jolly parrot]

water beam

[wild fable]

complete the square on this

[wild fable]

Nice

You got it

[midnight haven]

Awesome

[wild fable]

Now we can proceed with our next step

By the way have you learnt the derivatives of inverse trig functions?

Because we’ll be needing that here

[midnight haven]

Yes my teacher gave me a formula sheet I'm still having trouble memorizing the cosecants and cotangents

[wild fable]

Okay great,

We won’t be needing csc or cot here so down worry

Dont*

[midnight haven]

ok

[wild fable]

are you happy with the fact that we can rewrite the integral as $7\int_{ }^{ }\frac{1}{\left(x-3\right)^{2}+16}dx$?

[jolly parrot]

water beam

[midnight haven]

Yes

[wild fable]

okay good

[midnight haven]

u substitution?

[wild fable]

do you have your sheet with u?

[midnight haven]

Yes looks like we're using 17 I think

[wild fable]

give me a sec

[midnight haven]

ok

[wild fable]

alright im back

so actually we'll be using something else here

we can use that but i reckon this is an easier way