help-17

[jaunty coral]

so it would be pi/(3/4)

sine thats where we transform it for tan

[shy willow]

pi/(3/4)pi*

you said you used this? what you said now is correct, this is mistake

what is P?

as in the value

[jaunty coral]

P would be 3pi/4? This is because we are trying to find out the distance between pi and 3pi/4 which is where the next line is

[shy willow]

correct

now substitute that into your equation

what is b?

[jaunty coral]

b = 4/3? Or am I getting confused

[shy willow]

no youre correct

gj!

so ten(4/3*x)

tan*

[jaunty coral]

so it would be tan(1.33x)?

I think the only issue is if I were to put that into desmos it would not align with 3pi/4 unless I put in more decimals like 1.33333x. Want me to do that still anyways?

Also would you be fine checking this.

[shy willow]

if they ask 2 dp then yes

[jaunty coral]

They did, I'll put 1.33x!

[shy willow]

ofc

[shy willow]

this is correct!

ur better

[jaunty coral]

Can you also walk me through these? These are the only questions I have absolutely no clue how to solve

Only if you're able to ofc

[shy willow]

yeah ofc just give me a couple mins

[jaunty coral]

take your time. I think I attempted it around 40 minutes ago so let me see if I have my work

[shy willow]

first one is a cos graph, which starts at a maximum. your graph doesnt start at a maximum. imagine a graph that did (cos(x)) and you where shifting it to the left, how much would you shift it to make them allign?

[jaunty coral]

absolutely unsure. We don't have a point we can specifically use so I would probably find the average of

-3pi/2 and -pi

so it would be (-3pi/2) + (-pi)/2

[shy willow]

correct, it appears to be in the middle

[shy willow]

yes

-5pi/4 i think

idk

[jaunty coral]

what I got was -3.92699081699

[shy willow]

yeah thats about 5pi/4

-5pi/4

so

[jaunty coral]

cool cool

so that would be cos()?

[shy willow]

to shift. graph to the left by 5pi/4, or to have a shift of -5pi/4, what would you do to x?

[jaunty coral]

um. I don't really know?

[shy willow]

so basically, to shift a graph along the x axis, also known as a horizontal shift, we add a constant (the shift) to the x value

so f(x) -> f(x+a)

this denotes a shift of a

it gets shifted to the left

[jaunty coral]

oh. that's why we leave the negative in the equation

[shy willow]

x transformations work opposite to y transformations

[jaunty coral]

so the -(-) both cancels out and becomes a positive

so it moves to the left

[shy willow]

yes

[jaunty coral]

Yeah, yeah. I just got confused by the question. + is how you move to the right

[shy willow]

its better to think of it like this

+a shifts the graph a to the left

if f(x) becomes f(x+a)
for the same f(x) value, x would have to be x-a, so its shifted to the negative (left) side
f(x) = f(x-a+a)

[jaunty coral]

Alright, so basically
y=cos(x--3.93) which translates it to the left.
that leaves y=cos(x- [positive numer] to allow it to translate to the right

but how do you get it for the right shift since there really isn't any points to go off of besides the negatives

[shy willow]

yes

[jaunty coral]

no wait, I'm stupid

[shy willow]

if you had a graph tht starts at a max (cos) and need to shift to the right, whats the minimum shift to overlap the

m

[jaunty coral]

it will be in between the pi/2 and pi again?

[shy willow]

yeah

[jaunty coral]

alrighty, so let me do numbers

[shy willow]

its about 2.4

[jaunty coral]

Yep! 2.356 I think we have to round it up to the second decimal so it should be 2.36

[shy willow]

yeah

[jaunty coral]

Does this seem about right?

[shy willow]

yup

good job!

[jaunty coral]

Yeah, I have an exam on this stuff tomorrow so I'm trying to cram in the review by tonight

so do you know how to solve the other one?

we are supposed to solve y=sin(x-) and y=cos(x-)

[shy willow]

do the same thing and find the shift

the + at the end is just the verticle shift

which is the midline

[jaunty coral]

yeah that's how high it basically goes up from the original point

[shy willow]

yes

what about the shifts?

first lets consider sin

[jaunty coral]

wait, what if it goes up above 2?

[shy willow]

sin starts at the midpoint

[jaunty coral]

only slightly

sin starts at 0

while cos starts at 1

[shy willow]

yes, the midpoint and the maximum

[jaunty coral]

wait we do it for the maximum?

[shy willow]

dont think about 1 and 0 amplitudes can change

[jaunty coral]

so instead of trying to go up 2 we go up 4 which would make sense

[shy willow]

see how much youd shift a sin graph for it to allign, if it starts at the midpoint

[jaunty coral]

OH. 3

correct?

[shy willow]

wait are we takling abt the verticle shift

yes

3

verticle shift = midlie

midline

[jaunty coral]

Yeah. since the new line is between 2-4 we would use 3 for the vertical shift

[shy willow]

yes

now what about horiztonal shifts?

[jaunty coral]

I'm not sure where to start with that

its above the graph

[shy willow]

above the graph?

you mean the x axis>

[jaunty coral]

yes

sorry

[shy willow]

that doesnt matter, thats verticality

we only care about the horizontal shift

so start with part 1

sin graph

starts at its midline

how much and in what direction would you shift the sin graph to overlap

[jaunty coral]

pi/2? I'm getting very confused, I'm sorry I'm pretty tired

[shy willow]

dw

its almost 2 am for me

is this pre calc?

[jaunty coral]

yep

[jaunty coral]

I have to turn this review in a few minutes.

[shy willow]

midline intersects between 0 and pi/2

midline is y=3

so the midline has been shifted right by that amount

which is pi/4

which is about 0.79

[jaunty coral]

where did you get the pi/4?

[shy willow]

see how the midline of sinx is at the origin

but the midline of your graph is not

see where that middle point of your graph is shifted

[shy willow]

see how the mid point is between 0 and pi/2

average of those is pi/

4

pi/4

does that make sense?

its shifted by pi/4 from the origin

i really gotta sleep tho so if you need answers

[jaunty coral]

Also, do me a favor and check this one

Ah. I mean if you can gimme the answers for the previous one I can talk to you tomorrow about it

[shy willow]

pi/4 + 3
3pi/4 + 3

[shy willow]

max is 1100
min is 900
so A= 100 correct
period = 15 years
B = 2pi/15 = about 0.5 so thats prob right
its maximum at 0, so its shifted by quarter of a period which is 15/4 i think
so it should be -15/4 i think idk my brain is fired rn
midline = 1000 so + 1000 is correct

[shy willow]

if those numbers are right then yeah

how many attempts do you get

[jaunty coral]

Well to recieve the points I will have to get 8/10 during this one attempt

which is due in 6 minutes

[shy willow]

2sin(0.5x) ? for the first one

[shy willow]

yeah with these answers you should get that

[jaunty coral]

We only get one attempt then the questions will change. I believe this may be correct

[shy willow]

cuz thats the only part im not 100% sure and it cant be worth alot

part of 1 questin

rest is right as far as im aware

[jaunty coral]

I can't really blame you if I don't get an 8/10 anyways

my academic advisor tranferred me to this course a week later than everyone else

[shy willow]

just input all the answers i did with you

ohh

thats bad

academic advisors <<<

is there any questions we didnt to together

i can answer them before i sleep

what uni you in btw?

[jaunty coral]

A bunch. I think they are correct tho

[shy willow]

i assume east coast

or like more east

[shy willow]

cool

any other problems u need help w?

[jaunty coral]

I don't think so. I'm just gonna submit and see what I get. Cannot really afford to do anything else

lets see if I get the credit

I LOVE YOU SO MUCH!

[shy willow]

any time

[jaunty coral]

This was the only wrong one

[shy willow]

ah

you forgot the negative

sorry i missed that

-0.79

[jaunty coral]

it doesn't matter since I got a 9 so I got full credit

[shy willow]

yay

[jaunty coral]

also CSU is my institution

only college which accepted a lot of my transfer credits

[shy willow]

ooh nice

[shy willow]

transfer from a diff uni or

[jaunty coral]

AP Exams

[shy willow]

im not sure how credits n shit work tbh i didnt get any

UK syllabus n all

[shy willow]

ah okay

[jaunty coral]

I'm just an undergrad. Yeah, so technically I am a junior from all of my credits

[shy willow]

whats up w that i hear some people get no credits transfered even when they get 5's on their aps

[shy willow]

ah okay

same

or i have no credits

in in umich

[jaunty coral]

Yeah it depends on the institutions. That's why I chose CSU. It's a state school but it works just as well

especially since I'm trying to major in computer science where the insitution does not matter nearly as much

[shy willow]

thats true

the uni itself wont matter that much after a few years employment tbh

unless u went iv or sm then u get a headstart

what courses you doing?

michigan has so many credit req courses only 1 of my courses is my major

[jaunty coral]

Well, I am doing Trig so I can get into Calculus. C++ Fundamentals and Software Development, Computer Systems Foundations

Are the courses I'm doing. I don't really have an electives since I have filled them already out with worthless stuff

[shy willow]

ooh nice. at least ur courses are streamlined

i wanna major math but my courses are math, great books, astronomy, and "video games and learning''

like what

theyre intresting n all but like i wanted to do math yk

[jaunty coral]

that's the dream for every student. Nobody willingly wants to learn about Art History

[shy willow]

fr

[jaunty coral]

Yeah, it's really annoying

[shy willow]

ancient greek texts are not fun

btw if you need help w calc, lmk im pretty good w calc up to like calc 3-4

besides multivariable

[jaunty coral]

I really appreciate that. I really only used calculus for AP Physics 1 back in Highschool but I recall absolutely nothing about it

[shy willow]

no worries i can help you w anything, assuming im free ofc

js dm me or sm cuz i might miss it if you ping here

anyway, this was great but i reallt gtg sleep its like 2 am

and i got lectures tommorow

[jaunty coral]

Alrighty! I really appreciate you helping me. Thank you soooo much ❤️

goodnighty

[shy willow]

goodnight

[shy willow]

ofc <3

[jaunty coral]

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[river temple]

hii

[boreal remnant]

hi! do you have a question!?

[river temple]

yes

<@&286206848099549185>

how do i solve this? 😦

[grizzled halo]

you can't 'solve' a number
you're trying to simplify it right

[river temple]

if it appears that I have to perform the following power exercises to the most simplified form and check results using an alternative method...

<@&286206848099549185>

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[vast shale]

Good day

[vast shale]

I just wanna verify

Nope I got it

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[vast shale]

Where do I begin with proving the division algorithm?

[vast shale]

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[opal obsidian]

Prove $(a_n b_n) \to ab$, given $a_n \to a$, $b_n \to b$\
Proof:
Let $\epsilon > 0$ be arbitrary. We have to produce a stage $N \in \mathbb{N}$ such that $|a_n b_n - ab| < \epsilon$ whenever $n \geq N$.
Since $(a_n) \to a$, we get $N_1 \in \mathbb{N}$ such that $|a_n - a| < \frac{1}{2M \epsilon}$ whenever $n \geq N_1$.

Since $(b_n) \to b$, we get $N_2 \in \mathbb{N}$ such that $\abs{b_n-b} < \frac{1}{|a|\epsilon}$ whenever $n \geq N_2$. \
$|a_n b_n - ab| < \epsilon
= |a_n b_n - ab_n + ab_n - ab| < \epsilon
\leq |a_n b_n - ab_n| + |ab_n - ab| \quad \text{(Triangle Inequality)}
=\abs{b_n}\abs{a_n-a} + \abs{b}\abs{b_n-b}$
$M \frac{\epsilon}{2} +\abs{b} \frac{\epsilon}{2} = \epsilon$

[opal obsidian]

This is what I've attempted to prove the above, can you please check the proof overall writing and also point of stuff I missed

I also used convergence implies boundedness , but i don't how to add it here, please help with that too

[vocal sleet]

@opal obsidian Has your question been resolved?

[worthy citrus]

It looks fine but you need to say what N you are choosing for the original limit

[hard atlas]

the notation at the end is wrong. you are mixing up equalities and inequalities

do you mean => ?

or you could try writing it as one long chain of inequalities

[worthy citrus]

Good catch

[hard atlas]

also what is M

not sure what is happening in the last line

couple typos perhaps?

also eps usually shouldnt be in the denominator

[opal obsidian]

Yes, i mean to write ->

[opal obsidian]

max{N1,N2}?

[opal obsidian]

I’m sorry, I didn’t mention it

[hard atlas]

you cant use |an bn - ab| < eps implies anything. cause you dont yet know that this holds

[worthy citrus]

I think it would be problematic if your N was bigger than both N1 and N2

[opal obsidian]

last line is equally of the previous inequality?

[hard atlas]

is it?

[opal obsidian]

yeah, i can factor |b_n| from |an-a| i suppose

similarly for the other one

that factoring still be same as the above inequality right?

[hard atlas]

you are probably seeing what you want to see. not what you actually wrote

[opal obsidian]

Can you correct it please?

oh wait, i missed a sign

this will work?

[hard atlas]

check again what you originally said about |an-a| and |bn-b|

[opal obsidian]

also, M i should write as worst estimate for |b_n| by convergence implies boundeness

[hard atlas]

also the |b| is wrong, it should be |a|

[opal obsidian]

I’ll do everything on a paper and post again

I messed up a lot typing in TeX

Sorry to close now

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[hushed tree]

When does 3x^a go towards -inf for both x->inf and x-> -inf

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Please don't occupy multiple help channels.

[hushed tree]

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[sweet cargo]

Given for example, the limit: sin(x)/x as x approaches 0, how do we know that the limit is 1 -- since when evaluating for 0 is undefined? How can we prove, or explain, that when calculating for x as x gets closer and closer to 0, that the limit does in fact equal 1?

[potent elm]

You use squeeze theorem

[sweet cargo]

ok thank you!

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[summer bronze]

why is there an N after the (cos mx)

[summer bronze]

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[sinful sigil]

Q1.Imagine that a logician puts four cards on the table in front of you. (Each card has a number on one side and a
letter on the other). On the uppermost faces, you can see E, K, 4, and 7. He claims that if a card has a vowel on one
side, then it has an even number on the other. How many cards do you have to turn over to check this? Explain
why, using reference to propositional logic.

Q2.Imagine that a logician puts four cards on the table in front of you. On the uppermost faces, you can see E, K, 4,
and 7. He claims that if a card has a vowel on one side, then it has an even number on the other. How many cards
do you have to turn over to check this?Explain why, using refera ence to propositional logic

Im confused about how the difference between these 2 statments affect the cards to be turned over.
the difference is in between the brackets in Q1

In Q1 my answer would be p->q / not q -> not p
So i would have to turn over 2 cards: E and 7

While yes I can't assume that every card has a number on one side and a letter on the other side
I can't understand how the answer changes in Q2.
I will be flipping E and 7 regardless to verify the predicate

any thoughts/hints?

[dark kiln]

you need to turn over one more card

the information you don't have anymore let you skip turning K

[paper depot]

yeah an A/K card would render the claim false

[sinful sigil]

is my Q1 answer correct? E and 7

for Q1 only

[dark kiln]

yes

[sinful sigil]

how would K render the claim false?
from my understanding
if vowel then Even
if odd then constant

if p then q
if not q then not p

oh

[paper depot]

consonant*

[sinful sigil]

so if i flip K which causes p to be false and i get a q which is true

it renders it false?

[paper depot]

you misunderstand

[dark kiln]

you don't have to check 4
you have to check 7, it could have a vowel
same logic for K, except you know it can't have a vowel

[paper depot]

you now no longer know for sure that the symbol behind K is a number at all

[dark kiln]

K has the same reason you check 7

[paper depot]

How would K render the claim false?
is a nonsensical question

[sinful sigil]

flipping K i meant

[dark kiln]

same reason you check 7 applies to K

[paper depot]

flipping K doesn't by itself render the claim false

[sinful sigil]

yeah the result

[paper depot]

but flipping K and seeing A on the other side, that's what would falsify

[dark kiln]

both sides of the card are the opposite side of some side

the back side is not special

[paper depot]

that's what i meant by A/K: a card with A on one side and K on the other, which starts out with its K side up

[sinful sigil]

if not p (K) then not q (A)
But that would be end up being true (according to the truth table)

[paper depot]

again you misunderstand

in your notation, p does not necessarily talk about the currently-top side of the card, and q doesn't necessarily talk about the currently-bottom side!

[sinful sigil]

ahhh

[dark kiln]

i don;t understand what causes the confusion, if you can explain why you need to flip 7, this is why you flip K

[paper depot]

you need to flip 7 to ensure you don't hit an A/7, and you need to flip K to ensure you don't hit an A/K

[sinful sigil]

makes sense

i thought that not q would have to be odd

but it can be anything except being even

Thank you guys!

❤️

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[crude flax]

Prove that div(|r|ⁿ r)=(n+3)|r|ⁿ

[crude flax]

r is a vector
r=xî+yj+zk
|r|=√(x²+y²+z²)

I got to this point, but idk if i am supposed to take a dot product of r vector and (x+y+z)

[lusty sentinel]

Two dimensional vectors?

[crude flax]

If i do that the answer will be correct

But it feels like i am winging it without knowing why i took scalar product

[crude flax]

3d

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@crude flax Has your question been resolved?

[crude flax]

<@&286206848099549185>

[crude flax]

Nvm i got it

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[silent geode]

help me find points along a direction with normalized directional vector

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[sage burrow]

i'm not crazy, right-- this is completely wrong?

triple checked but i'm still doubtful of my own work against actual assigned work

[sage burrow]

like it literally doesnt work for any number lol

[paper depot]

can you show a value of n for which it doesnt work

[sage burrow]

n=3

1+2+3 = 6 LHS

3! = 3 * 2 * 1 = 6

(3-2)! = 1! = 1

2! = 2*1 = 2

=> RHS = 6/2 = 3

LHS=/=RHS

i am admittedly very tired please yell at me if i am doing something incoherent

[paper depot]

ah wait of course hold on

[sage burrow]

see when i simplify the RHS i get

n/2 (n-1)

[paper depot]

should be $\sum_{i=1}^{n-1} i$ on the LHS

[twin meteor]

Ann

[sage burrow]

but its a triangular number!

you think it's an error with the question too?

[paper depot]

yes it is an error with the question

[limber wedge]

I think yes

[limber wedge]

The right hand side simplifies to n(n-1)/2

That's not the correct formula for sum i with bounds 1 to n

[sage burrow]

thanks so much! figured i needed a second opinion before i marked it as an error,

.close

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[waxen hawk]

given a random experiment has 2 random variables X and Y, and adding these two variables up can obtain new random variable called X+Y.

then E(X+Y)=E(X)+E(Y)

[waxen hawk]

Above are a session derived from my textbook

[waxen hawk]

Well, that X+Y is obscure tho

So I want to set an example and see what it holds

[hasty lark]

Obscure in what way?

[paper depot]

it may be obscure if you don't have a good concept of what a random variable is at all

[waxen hawk]

What does X+Y stands for?

[paper depot]

your random experiment is defined by a sample space

which, in the simplest form, is a set of outcomes each of which has a probability associated with it

a random variable is then a function from this sample space into the real number line

it's probably easiest to give an example with roulette

have you ever seen a casino roulette? or would you like a reminder

[waxen hawk]

I do see it in gangster movies, but I don’t really know what it is actually

[paper depot]

ok

then let me give you an explanation

here is a roulette wheel

it is a wheel with 37 sectors labeled with numbers from 0 to 36. during a round of roulette, the wheel is spun and the ball is thrown onto it. when the wheel stops and the ball lands in one of the slots, the number the ball landed into is the winning number for that round

now here is a roulette table

there are many ways you can place a bet in a game of roulette

[waxen hawk]

So I place my bet on one of these numbers

[paper depot]

individual number bets exist, yes. but there are also different bets

basically there are different events you can bet on that aren't simply "the ball lands on 17"

different types of bets give different payouts but that is not very important rn

oops im getting a phone call brb

[vocal sleet]

@waxen hawk Has your question been resolved?

[paper depot]

@waxen hawk ok yeah sorry my call took literally 20 whole minutes

ok so

let's say alice and bob are at the roulette table

and alice bets $10 on red, while bob bets $30 on the 2nd dozen (numbers 15--24)

[twin meteor]

Ann

[paper depot]

by the rules of roulette, this means that if the ball lands on a red number, then alice wins $20, and if the ball lands on a number between 15 and 24 inclusive, then bob wins $90. if one of these bets is not satisfied, that person wins nothing ($0).

[twin meteor]

Ann

[paper depot]

(the reason why those payouts are like that is beyond the scope of this explanation)

so then

the sample space for the roulette "experiment" is {0, 1, ..., 36}, with each of the 37 numbers assigned probability 1/37 (we are assuming the roulette is fair)

Alice's payout is a random variable which assigns the value 20 to red numbers, and the value 0 to black numbers and zero (which is neither black nor red)

Bob's payout is a random variable which assigns 90 to numbers between 15 and 24, and 0 to everything else

we can call them A and B

then A+B is simply the total amount of money that Alice and Bob win this round

[vocal sleet]

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[paper depot]

.reopen

[vocal sleet]

✅

[paper depot]

the value of A+B is:
110 if the ball lands on 14, 16, 18, 19, 21, or 23 (probability 6/37);
90 if the ball lands on 13, 15, 17, 20, 22, or 24 (probability 6/37);
20 if the ball lands on 1, 3, 5, 7, 9, 12, 25, 27, 30, 32, 34, or 36 (probability 12/37);
0 if the ball lands on 0, 2, 4, 6, 8, 10, 11, 26, 28, 29, 31, 33 or 35 (probability 13/37)

but in principle random variables are just real-valued functions on a sample space

so you can do with them anything you can do to functions

[vocal sleet]

@waxen hawk Has your question been resolved?

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[outer flower]

Why is the exponent for the numerator and denominator placed different?

[paper depot]

that's just how it be

it is just notation, nothing else to it really

[outer flower]

That is difficult for me to deal with when it comes to math lol

Whelp okay then

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[fallow ivy]

I’ve been trying to solve the problem I put the brackets next to by creating two separate cases but honestly I’m not sure how to get to the desired answer from here and also don’t think this is the most efficient method

[delicate stirrup]

so $x\le1$ here?

[twin meteor]

TimK

[fallow ivy]

I think so but everything past the ] at the top is just me trying to solve it

[delicate stirrup]

ok, why there is no $x\ge0$?

[twin meteor]

TimK

[fallow ivy]

I think there should be I just don’t know how to get there mathematically

Is it enough to just say it has to be greater than or equal to zero bc of the square root thing

[delicate stirrup]

yes

[fallow ivy]

Ok got it

But also

1 >= x^3 seems wrong

[delicate stirrup]

for second, don't you just reverse signs?

[delicate stirrup]

why?

[fallow ivy]

I take it back it makes sense again if x has to be greater than or equal to 0

Ok I’m understanding now

Why is it and and not or

If the sign is >=

[delicate stirrup]

,w -x^2+sqrt(x)>=0

[twin meteor]

Results provided by WolframAlpha

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Requested by tim_oder

[delicate stirrup]

right

[delicate stirrup]

wdym?

[fallow ivy]

Why do both the cases need to be true if the original problem is suggesting value and value

Wait

It’s not in standard form

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[rancid roost]

this is the answer of a homework (it's optional so it isn't too important but still)

[rancid roost]

but this are the options

in the book, ik it's pretty basic but I just keep getting mixed up as to which is the right one

[delicate stirrup]

i guess, the closing [ here is )

and vice versa

so 1st

[rancid roost]

yes, that's the one I noted, but wasn't sure

thanks for that

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[unborn vapor]

Is this right ?

[delicate stirrup]

,rotate

[loud pivot]

Oops

[delicate stirrup]

lol

[rocky igloo]

ok just saying for the future but write multiplication as a dot

[unborn vapor]

Ok sure

[rocky igloo]

it gets complicated if your teachers cant tell if its x or x

[delicate stirrup]

asterisk?

[unborn vapor]

t’es thx for the tips

[silk comet]

Should be -4 instead of +4

[rocky igloo]

it would be minus 4

[silk comet]

The constant term

[unborn vapor]

So is it wrong ?

[rocky igloo]

im pretty sure it should also be 2x

yes because the 2 on the second part is negative

[unborn vapor]

I founded 2 I was just messing with thoses - and +

So its

,rotate

[rocky igloo]

the constant should also be negative becuase positive*negative is negative

[unborn vapor]

The constant ?

[rocky igloo]

4

[unborn vapor]

What 4

The first one ?

[rocky igloo]

in your answer, the last number shouldnt be positive

[unborn vapor]

Oh t’es I Forget thx

Yea *

[rocky igloo]

what i do when j dont want to do mental math is write out the box

[unborn vapor]

So 20*x^2+2x+-4

[unborn vapor]

What type of method is that

[rocky igloo]

idk what its called but it helps with large equations

so you basically put each different number on the outside to make a box

[unborn vapor]

But wait So the final résultat is What

[rocky igloo]

https://youtu.be/szGj9C0Djaw?si=2JU5QEYvmAl8Guaw

[rocky igloo]

this is correct

[unborn vapor]

Okay perfect thx ill Watch that vidéo right away.

[delicate stirrup]

$(4x+2)(5x-2)=4x5x+10x-24x-2*2$

[twin meteor]

TimK

[unborn vapor]

.close

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[rapid shore]

might be a dumb question but can somone help me understand if say f(x,y) = x^2y + y^2 why does the y dissapear if i take a partial derivative of x?

[hard atlas]

its constant

its like differentiating 2x^2 + 2^2

[rapid shore]

how is it simillar

[patent pasture]

If you are deriving something with respect to x, anything other than x is a constant

So y^2 would be treated the same as any number such as 4

[rapid shore]

ahhhhhhhh

[patent pasture]

And if you derive 4 it's 0

[rapid shore]

but the product of xy^2 is dependenant on the value of x hence the y stays right?

[patent pasture]

Yes because it's linked with x it can continue to exist

[rapid shore]

right

makes sense now

ty

[patent pasture]

Np

[rapid shore]

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[mellow rampart]

Is this correct?

[mellow rampart]

If so, the solution is then just a line?

[vocal sleet]

@mellow rampart Has your question been resolved?

[mellow rampart]

<@&286206848099549185>

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@mellow rampart Has your question been resolved?

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@mellow rampart Has your question been resolved?

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[brisk dragon]

How do you do negotiate numbers tot he power of a negotiate

[brisk dragon]

For example -4 to the power of -2

Is it just -4x-4?

[sly sierra]

$$(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$$

[twin meteor]

Bungo

[brisk dragon]

Oh oh I see

Thx

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[karmic abyss]

im wrong?

[gaunt sparrow]

Maybe they just want the general form y=2-x

[karmic abyss]

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[noble solstice]

bit stuck on the first show that, feel like i'm missing something obvious

i can show that g^r = h^s = 1 from that

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[woven rivet]

4(k+3)+2=4.5(k+1)

[left jay]

I want to use algebra to determine the largest possible delta that satisfies this solution

[left jay]

oh sorry, i took the channel, i can help you with this in another channel though

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[mild epoch]

the answer shows e^9 but where tf does e come from

[glossy maple]

$\lim_{n \to \infty} \parens{1 + \frac1n}^n = e$

[twin meteor]

RedstonePlayz09

[vast shale]

damn

[mild epoch]

how

[vast shale]

binomial expansion

[mild epoch]

telling me the answer when i already had the answer does not help me

9/inf is 0 right

so (1+9/inf)^inf = (1 + 0)^inf

[vast shale]

redstone lit gave the definition of e

[mild epoch]

still

[vast shale]

https://math.stackexchange.com/questions/2642478/proof-of-e-as-a-limit

[mild epoch]

is my logic wrong?

i really dont care about e

is my logic wrong?

[vast shale]

its wrong

indeterminant form

infinity is not a number

[mild epoch]

bro

if u dont know the answer u can just say so

[vast shale]

[mild epoch]

9/inf is 0 right

[glossy maple]

manipulate ur limit to make it have the form

(1 + 1/something)^something

[mild epoch]

it already has that form

[glossy maple]

where that something is the same in both places

no it doesnt

u have 9/n

not 1/something

[mild epoch]

no

does 9/inf approach 0?

[glossy maple]

it does

so?

[glossy maple]

.

[glossy maple]

.

(1 + 1/f(n))^f(n)

where f(n) approaches infinity as n does

[mild epoch]

if 9/inf approaches infinity

lim n -> inf (1-9/inf)^inf = (1-0)^inf

[glossy maple]

again, read what I was asking

EXACTLY of the form (1 + 1/f)^f

that 9 on the numerator MATTERS

Basically, get it to something like [(1 + 1/f)^f]^g

where f approaches infinity and g approaches something else

in this case it will just be 9

[mild epoch]

i do not care about form

i am trrying to evaluate a limit

you are only able to apply the form because you already know the answer

[vocal sleet]

@mild epoch Has your question been resolved?

[glossy maple]

how else are you gonna evaluate it

what's your definition of e

are you allowed to use the (1 + 1/n)^n limit?

[vocal sleet]

@mild epoch Has your question been resolved?

[mild epoch]

i dont see e in the equation

[mild epoch]

no, we havent ever even heard of that

[glossy maple]

but you're not answering my question

the solution to the limit has e in it

so obviously it involves some usage of it

[mild epoch]

i dont see e anywhere in the question i posted

so where would the e come from

[vocal sleet]

@mild epoch Has your question been resolved?

[glossy maple]

I don't know how to give you an answer that will satisfy you. this limit has the form 1^inf and it just turns out to be equal to some positive irrational number we call e

some people define e using this limit, and some define it in other ways and then use those definition to evaluate this limit. it's a well known limit

so we can't continue UNLESS you tell me what you already know about e, mainly the definition

and you should probably double check whether or not you've encountered the limit of (1 + 1/n)^n in your class. seems unreasonable to give your question unless there's been some introduction to it

that's all I can say

[twin citrus]

end of discussion

[glossy maple]

@twin citrus You aren't supposed to give away solutions

[twin citrus]

when there is no other way than i am

[glossy maple]

??

"No other way"?

Please delete it

@twin citrus

[twin citrus]

you wanna torture him?

[glossy maple]

I told them the correct approach, but they haven't even tried applying it

[vocal sleet]

@mild epoch Has your question been resolved?

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[blissful prairie]

hi, anyone can help me?

[rugged orchid]

Have you tried anything

[maiden iron]

maybe switch every a for 1-2b 🤔

[blissful prairie]

I tried, but I don't know how to do next

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[astral apex]

correct?

[astral apex]

(b)

[rugged orchid]

That just says there are non-multiplicative identity elements in S

Let me pick x = 2

There exists a y let’s say 3

Such that xy ≠ y

Well, 2 * 3 = 6 ≠ 3

So yes the statement is true, but it doesn’t have to do with the multicaptive identity

Since I could’ve picked S = ℝ for this case and clearly ℝ has a multiplicative identity

[astral apex]

so im right?

[rugged orchid]

No you’re saying your subset has elements that are not the multiplicative identity

Which I mean, is true for ℝ

You didn’t show the right thing

[astral apex]

what should it be

for all
for all
does not equal to

?

[rugged orchid]

Can you write what P is first?

[astral apex]

P = there is a multiplicative identity in S?

[rugged orchid]

Oh

You are right

I’m sorry

[astral apex]

oh

[rugged orchid]

Do you know why you’re right?

[astral apex]

i do not

[rugged orchid]

Ok here’s how I would do it

First we want to write what P is saying

Then we just want to negate P

Which is usually pretty easy

So for P

[astral apex]

how do u negate that sentence

[rugged orchid]

We want xy = y, where x is 1 and y is anything

Right?

[astral apex]

yes

[rugged orchid]

So we want to say there exists an x (in particular the 1)

Such that for all y (y can be anything)

xy = 1*y (is equal to) = y

[astral apex]

yea but i did the opposite

[rugged orchid]

Yes

So when you have $\exists x\forall y : xy=y$

[twin meteor]

Frosst

[rugged orchid]

The negation of this statement is really easy

[rugged orchid]

Remember this is P

We want not P

[astral apex]

ohh this is what P is saying

[rugged orchid]

Yeah

[astral apex]

and we have to negate it

exists becomes for all and for all necomes exists

equal becomes opposite

ohhh

[rugged orchid]

$\not(\exists x\forall y : xy=y)$

[twin meteor]

Frosst

[rugged orchid]

Exactly!!

What the heck

[astral apex]

i get it nkw

tysm

[rugged orchid]

Ok ignore that you get what I mean

[astral apex]

yes i understand it now

ty 🙏

[rugged orchid]

Sorry for the confusion at the beginning

[astral apex]

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[vast shale]

QUESTION IN COMBINATORICS

In how any ways can you devide 71 identical balls into 5 drawers such that:
in the top drawer there are exactly 3 balls. in the middle drawer there are at least 4 balls and in the bottom drawer there are at most 5 balls?

[paper depot]

well put 3 balls in the top one and forget about it

[vast shale]

i know i need to devide 68 balls into the 4 bottom drawers

[paper depot]

now you have 68 balls going into 4 drawers

and put 4 in the third (middle?) drawer

[vast shale]

okay that is as far as i got

[paper depot]

so now you have 64 balls going into drawers #2-#5, but with at most 5 balls in drawer #5

feels like a little bit of case work

[vast shale]

so inclussive and exclussive theorem?

if that is the translation idk

Inclusion–exclusion principle

נוא ים' גם ן ודק ןא

but how do i use it

[paper depot]

no you don't need that i don't think

like

[vast shale]

ok

then what

[paper depot]

figure out the number of ways to put k balls into 3 drawers

and then sum that for k from 59 to 64

[vast shale]

k+3-1 C 3

so that will be $\sum^{64}_{k=59}{\binom{k+2}{3}}$

[twin meteor]

HadarS

[vast shale]

Right?

@paper depot

[paper depot]

yes

[vast shale]

well i hope its the answer

thanks!

[grand frigate]

What did you get?

[vast shale]

.

.close

[vocal sleet]

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[jolly obsidian]

hello, could someone work me through this roots of polynomials question? or show me what the answer would look like

[crimson jetty]

I think vieta's formula

(or figure it from first principles)

Multiply this out
k(x - a)(x - b)(x - c)

Then multiply this out
k(x - 1/a^3)(x - 1/b^3)(x - 1/c^3)

[jolly obsidian]

sorry what is k and x in this?

[crimson jetty]

x is the same x

[jolly obsidian]

oh okay

[crimson jetty]

k is the coefficient

Are you aware this is the general factored form for a cubic
k(x - a)(x - b)(x - c)

[jolly obsidian]

ah yeah okay i remember now

do i need to find the individual roots or do i just need to like relate the formulas to the questions?

[crimson jetty]

no.

[crimson jetty]

expand this out

[crimson jetty]

expand that out

and that should likely help

[jolly obsidian]

okay i'll have a go at that one sec

[crimson jetty]

uh actually, I don't think so. Think about the 1 drawer case

[vocal sleet]

@jolly obsidian Has your question been resolved?

[jolly obsidian]

man i think i need to go through factorising cubics first and come back to this

thank you

[vocal sleet]

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[vast shale]

.reopen

[vast shale]

ddd

[crimson jetty]

right so yeah this

isnt quite right

[vast shale]

i choose drawers to the balls for each case

[crimson jetty]

figure out the number of ways to put k balls into 3 drawers

Focus on this question

[vast shale]

and if im not mistaken the cases are seperate

ok

[crimson jetty]

The answer is not k + 3 - 1 C 3

Whats the answer if its:

figure out the number of ways to put k balls into 1 drawer

[vast shale]

1?

[crimson jetty]

yes

but your pattern would be

k + 1 - 1 C 1

which is generally not 1

[vast shale]

but the k represents balls

not drawers

the drawers stay 3

number of balls change according to the last drawer

[crimson jetty]

I get that

But just take this example

figure out the number of ways to put 20 balls into 3 drawers

The answer isnt 20 + 3 - 1 C 3

[vast shale]

why is that?

[crimson jetty]

number of ways to put 20 balls into 1 drawers

Because this isnt

20 + 1 - 1 C 1 = 20 C 1 = 20

like you've got the wrong formula is what im tryna tell u

[vast shale]

we never put 1 on the right side of C tho

[crimson jetty]

Do you know how to convert this into a stars and bars problem

[vast shale]

it stays 3

[crimson jetty]

yes but if your formula isnt correct for 1 drawer

it wont be correct for 3 drawers

[vast shale]

you need to give me a k such that it isnt the answer

[crimson jetty]

number of ways to put 3 balls into 3 drawers

[vast shale]

well go ahead and prove it

[vast shale]

that would be 5C3

[crimson jetty]

3 + 3 - 1 C 3 = 5 C 3 = 10

god 3 into 3 is annoyingly large to verify by hand

number of ways to put 1 balls into 3 drawers

1 + 3 - 1 C 3 = 3 C 3 = 1

Ok there we go, its clear this is wrong

[vast shale]

right

[crimson jetty]

2 balls into 3 drawers would be 4 C 3 = 4 according to you, but also nope.

[vast shale]

yea

[crimson jetty]

Do you know how to convert this into a stars and bars problem

[vast shale]

so whats your idea

[crimson jetty]

So yeah, lets go back to this

[vast shale]

bits?

[crimson jetty]

Well really - i wanna ask you where u got k + 3 - 1 C 3 from

[vast shale]

0001000100?

[crimson jetty]

no

[crimson jetty]

the idea is almost correct

[vast shale]

where the 1s are the divisors?

[crimson jetty]

it just needs some fixing

[crimson jetty]

yes ok, thats stars and bars

but how many divisors are there for 3 drawers

[vast shale]

we got 4 drawers tho

so 3 bars

[crimson jetty]

no, not with ann's idea

Ann's idea is to fix how many balls are in the last drawer

and then you do stars and bars on the middle 3 drawers

The last drawer can be 0, 1, 2, 3, 4, 5

We fix that number. And that gives us 6 cases

[vast shale]

no ,i got the idea of let it change and count according to the current number of balls in the last one

[crimson jetty]

Yes exactly

Then in that case, the last drawer is a constant

so there are only 3 drawers to split into

[vast shale]

ohhhhh

right

so k+2-1C2

right?

[crimson jetty]

why is it k + 2 - 1

(nope)

[vast shale]

cuz 2 divisors

oh

nvm

it comes from the -1

discard it

[crimson jetty]

yh, so 2 divisors and k balls

k + 2 C 2

or k + 3 - 1 C 3 - 1

if u wanna write it like that

[vast shale]

can you explain it?

[crimson jetty]

so im gonna pick a number for k

k = 6

and 3 drawers

000000 <-- balls

|| <-- divisors

right?

then we need the number of ways to order these 8 things

6 + 2 C 2

divisors = slots - 1

so

(things + slots - 1) C (slots - 1)

OR

(things + divisors) C divisors

[vast shale]

so things are k

[crimson jetty]

yes

[vast shale]

divisors are k-1

[crimson jetty]

not k-1

[vast shale]

3-1

[crimson jetty]

0|000|00

^example of valid way to order the 8 things

[vast shale]

yea

divisors are 3-1

[crimson jetty]

So the usual formula

lets n be the things

k be the sets or slots

,,{n+k-1}\choose{k-1}

[crimson jetty]

and you get this.

n balls

k-1 divisors

[vast shale]

and that is for identical balls?

[crimson jetty]

yeah identical balls

[vast shale]

where we dont care about the order

[crimson jetty]

distinguishable sets

[vast shale]

i have no idea what this word means xD^

[crimson jetty]

distinguishable = they're different

the sets have a label on them

eg. A B C

A: 00
B: 0
C: 000
is different from
A: 0
B: 00
C: 000

[vast shale]

so their intersection is empty?

[crimson jetty]

well yeah sure, sets is a bad word, I should use another word

just say groupings

I dont mean a mathematical set, no.

[vast shale]

oh i got a private lesson

thank you so much

[crimson jetty]

👌 cya

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

@dim hatch Has your question been resolved?

[vocal sleet]

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[shell laurel]

how can I prove that this is monotone?

I dont know what equation to set for the denominator

I know that it is decreasing because I tried the first few values and the denominator grows faster than the numerator

<@&286206848099549185>

[modest coral]

look at a_{n+1}/a_n

[shell laurel]

yeah but what would the denominator be?

of a_n+1

3n*3n+1?

[modest coral]

what?

What is denom for n=1, n=2, n=3, n=4?

[shell laurel]

3, 18, 162 ...

isnt it?

[modest coral]

what?

[shell laurel]

aren't the first terms of the sequence: 2/3, 4/18, 8/162

[modest coral]

no they are not

[shell laurel]

how are they not?

[modest coral]

3 * 6 * 9 * ... * (3n) means continue that pattern until you reach 3n

so n=1 its 3 then stop because 3 * 1=3

n=2 its 3 * 6 then stop

[shell laurel]

yeah

for n=3 its 3 * 6 * 9

which is 162

[modest coral]

oh sure

[shell laurel]

so isnt this right?

[modest coral]

it is

[shell laurel]

yeah so what equation can i set for the denominator

[modest coral]

just like this but continue until 3(n+1)?

[shell laurel]

but that wouldnt work for this

[modest coral]

?

[shell laurel]

cause the denom of second term isnt 3(n+1)

its 3n*3n+1

[modest coral]

its still not that

plug in n=2, what you have isn't 6 * 9 in denom

its 3 * 6 * 9

[shell laurel]

oh yeah

<@&286206848099549185>

[modest coral]

?

just do what I say

compute a_{n+1}/a_n

[shell laurel]

HOW

literally what will you put for a_{n+1}

4/18?

[modest coral]

no?

it depends on n

just like a_n does

[shell laurel]

what

[shell laurel]

answer this

how would you even compute it

[modest coral]

simplify it

[shell laurel]

WHAT

how would you simplify a subscript

[modest coral]

what is a_n?