help-43

[shell horizon]

im guessing you know what to do from now?

[turbid canopy]

Notttt reslly..

Hold on i think

Ill show hownive been working til now

[shell horizon]

For the first step you need to show f(r+1)-f(r) not vice versa

Unless you alr did it on another page

You also missed uhh

All the factorials

😭

[turbid canopy]

No i mean like

This is a diff question

Im just showing u my standard procedure

Which doesnt seem to be working

On this question..

[shell horizon]

It's the same thing but the start of part 2 (hence) requires you to manipulate the first simplification first

[turbid canopy]

God this is hard

The whole changing or substituting thing is confusing

[shell horizon]

Well you'll make the expression look like f(r) thingy

so r=n+1

the expression is just equal for the difference of the functions

But substitute n+1 into r

[turbid canopy]

Wouldnt it mess the whole series up tho as well as the difference

[shell horizon]

f(r)-f(r+1) becomes f(n+1)-f(n+2)

[shell horizon]

no

[turbid canopy]

The difference doesnt have to be rigid?

[shell horizon]

wdym

You're not changing the function itself

Simply the variables

if f(x)=x²

then you plug in y+1

[turbid canopy]

Like it doesnt have to remain as f(r+1) - f(r) ?

[shell horizon]

f(y+1)=(y+1)²

Still holds true

[turbid canopy]

You can change the insides if you wsnt so it matches?

[shell horizon]

yeah I meant why not 😭

[turbid canopy]

Wtf damn

[shell horizon]

ok let's call the function that you simplified $$f(r)=\frac{r}{(r+1)!}$$ \
This implies that \
$$f(n+1)=\frac{n+1}{(n+2)!}$$

Is this not true?

[boreal girder]

AkitoLite

[turbid canopy]

Oh

Wait

I think i got it

[shell horizon]

I would've written it all out but I'm on phone and it takes ages to write out latex 😭

[turbid canopy]

Omfg ok finally got the answer

If i knew you could judt change the difference that would have been so much easier bruh

I thought u couldnt change it and u had to do some weird manipulation to it

[shell horizon]

What'd you write out

[turbid canopy]

Bit of a mess

The thing i wrote Ans next to is what matches the book answer

[shell horizon]

Right yeah

I think you should've made it clear that you multiplied by -1 since it kinda just looks like you flipped the eqn

[turbid canopy]

Yeah i mean as long as I know whats gojng on its ok its just homework

In an exam id have shown it more clearly

[shell horizon]

It looks good

The point is that there's an equality in the f(r) stuff, so if you replace the variable r by anything, the eqn still holds true

For example if you say x²-x=x(x-1), it still holds true for x=n+1, (n+1)²-(n+1)=(n+1)(n+1-1)

[turbid canopy]

Yeahhh i see

Time to struggle on the nexy question

I will.. come back in 5 min if i get stuck yet again

So goodbye…… For now

Thanks for helping

Will close channel

.close

[compact pewter]

Channel closed

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Use .reopen if this was a mistake.

[compact pewter]

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[torpid mist]

Hello I don't understand why the answer is not +- sqrt but just - for d)

previous questions

[torpid mist]

the correct answer

[dusky nymph]

if it was +, then you would have $h^{-1}(x) > 0$

[boreal girder]

Bungo

[dusky nymph]

but $h$ is only defined for negative arguments

[boreal girder]

Bungo

[torpid mist]

ohh it's because it says h(x)<0 and if i have + it gives me positive number ?

[dusky nymph]

yep!

[torpid mist]

okay thank you so much

.close

[compact pewter]

Channel closed

Closed by @torpid mist

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[sudden igloo]

heyyy i need some help on discrete math

its a midterm test and i would want as much confirmation as possib;le tbh

[short lantern]

first off, use an available channel, second we cannot help on tests

[sudden igloo]

its a take home

more so just practice for the main test which is an in class

[sudden igloo]

also where can i find an available channel

[short lantern]

look in the channel list for "Math Help (Avilable)", if you dont have it, look in channels & roles at the top of your channel list

[compact pewter]

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[quartz yoke]

for multivaer limits, how do you choose a path? like e.g. for this how would decided what path to take

[quartz yoke]

nvm i don't need paths, i can just plug it in

.close

[compact pewter]

Channel closed

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[compact pewter]

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[odd lake]

not sure where to start with the proof on tihs one. does anyone haev an idea that could get me started? the most that i've done is define some collection of open sets, but im unsure of where to go from here

[kind crane]

what's your definition of open set

[odd lake]

a set of real numbers A is open if for all x in A there exists an epsilon greater than zero such that (x-ε, x+ε) is a subset of A

[upper shore]

how do we take the union on sets like (a,b) (c,d)

[odd lake]

on open intervals like that the union of them is just (a,d)

[upper shore]

so, start by setting it up, assume a set of open sets, and take the big cup of them

[odd lake]

are open sets always able to be written as open intervals?

i thought that wasnt always the case

[upper shore]

ill be more precise

Let ${A_i}_{i\in I}$ be a set of open sets

[boreal girder]

Cycadellic

[upper shore]

we wish to show that $$\bigcup_{i\in I}A_i$$ is open

[boreal girder]

Cycadellic

[upper shore]

what is the definition of $\bigcup$

[boreal girder]

Cycadellic

[odd lake]

well its just a series of unions between sets right

[upper shore]

nono

not arbitrary unions

[upper shore]

this is a countable union

[odd lake]

oh

[upper shore]

the definition is $$\bigcup_{i\in I}A_i={x|\exists i\in I:x\in A_i}$$

[boreal girder]

Cycadellic

[upper shore]

equivalently, $$x\in\bigcup_{i\in I}A_i\iff\exists i\in I:x\in A_i$$

[boreal girder]

Cycadellic

[upper shore]

i suggest working on this one

but both are identical

does this make it more clear?

[odd lake]

yeah that makes it more clear how i should be treating the question rigorously

seeing it in mathematical notation makes it easier

[upper shore]

to be completely rigorous, you take this for assumption, apply the definition of arbitrary unions to it

suppose x is in some A_i

[upper shore]

conclude that (x-ε, x+ε) is a subset of bigcup A_i

so, if youre still stuck, lets just get everything set up for now

[upper shore]

oh i thought you were asking how to treat it rigorously LOL cause im bad at reading

now for part b, we can abuse induction

all you need to show is that, given any two open sets $A_1,A_2$, that $A_1\cap A_2$ is also open

[boreal girder]

Cycadellic

[upper shore]

if your prof is super nitpicky, you can always easily make this into an inductive argument, but honestly the IH will just be another instance of the BC here, so i dont think its necessary

[odd lake]

oh i dont have a professor ive just been trying to self learn this

[upper shore]

oh ok

[odd lake]

but its very hard cuz this is the first really abstract math subject ive seen

[upper shore]

right

but also really fun, especially since it looks like youre getting into the topology of R

feel free to post your proof here, and i can go over it with you

[odd lake]

its fun when i understand something but its not fun when i sit ifnront of my chromebook for an hour trying to figure out how to do a proof for a theorem

[upper shore]

true

[odd lake]

idk if im wrong but if theres some x in one of these open sets, we already know that (x-ε,x+ε) is a subset of that open set, and since that open set is a subset of the arbitrary union of open sets, then (x-ε,x+ε) is a subset of that union right

[upper shore]

yes

but it does need to come out notationally

[odd lake]

ok ill write it out notationally on paper rq

only for union part

[upper shore]

,rccw

[odd lake]

how do i rotate this

[upper shore]

holds

[odd lake]

ok ill get started on the 2nd part

[upper shore]

yep

[upper shore]

i think you should also add that (x-\epsilon, x+\epsilon)\subseteq A_i, by assumption that A_i is open,
and you should also assume \epsilon>0 after assuming x\in A_i

but regardless, everything works

[odd lake]

ah alright

im thinking for intersection part i can use infinum and supremum and then do like an epsilon is the minimum of a-x and x-b or smth

[upper shore]

you dont need infs or sups

you assume x\in A_1\cap A_2, and A_1, A_2 open

[upper shore]

i mean, you could do this too

to clarify, the key you want here is ||\epsilon = \min(\epsilon_1, \epsilon_2)||

show that such an interval must be in the intersection

||this way, existence of \epsilon follows from existence of both \epsilon_1, \epsilon_2||

[compact pewter]

@odd lake Has your question been resolved?

[static stag]

help whats the best way to learn 11 topics of calc 2 in 4-5 days?

[upper shore]

#❓how-to-get-help

[odd lake]

,rccw

[odd lake]

idk if my justification for the final steps of the proof was good but i think i did it

oh wait i probably shouldve stated x_1 is an element of A_1 and x_2 is an element of A_2

mb mb

[upper shore]

now hold on

you need to show directly from:
\epsilon=\min(\epsilon_1,\epsilon_2) and (x-\epsilon_1, x+\epsilon_1)\subseteq A_1 and (x-\epsilon_2, x+\epsilon_2)\subseteq A_2

prove (x-\epsilon, x+\epsilon)\subseteq A_1\cap A_2

[odd lake]

well because the intersection means that the elemnts must be in both sets, then since this interval is in both sets then wouldnt it be in the intersection

[upper shore]

right

but this needs to come out symbolically

$\epsilon=\min(\epsilon_1,\epsilon_2)$, so consider WLOG $\epsilon_1\le\epsilon_2$

[boreal girder]

Cycadellic

[upper shore]

\epsilon = \epsilon_1

(x-\epsilon, x+\epsilon)\subseteq A_1, by assumption

now you need to prove that $(x-\epsilon, x+\epsilon)\subseteq A_2$

[boreal girder]

Cycadellic

[odd lake]

ok i shall erase some stuff and rewrite it to make it more rigorous also sorry if im not getting this immediately topology is definitely not my strong suit in ra

[upper shore]

its okay

you dont have to apologize for math being hard :P

[lyric sierra]

Well hold on. What's the x in the second part of your proof. You don't define that anywhere and how small your epsilons have to be depends on that. What if I hypothetically choose x1 and x2 at extremities of A1 and A2 but the intersection was some interior part of A1 and A2. What's x in the second part of your proof.

[upper shore]

hes making the assumption that x\in A_1\cap A_2

he just didnt state that

[weak carbon]

I need help with question four.

[upper shore]

#❓how-to-get-help

[odd lake]

,rccw

[upper shore]

you should make it clear the separation between the proofs

[odd lake]

replace arbitrsru witj finite

in the 2nd part

[odd lake]

no its one proof

it has 2 parts tho

[upper shore]

i understand

organization tip to just make a clear division between your assumptions

you put $\epsilon<\epsilon_1$, but we want $\epsilon_1\le\epsilon_2$ and therefore $\epsilon=\epsilon_1$

[boreal girder]

Cycadellic

[upper shore]

also make it clear this is argued WLOG

i would be more comfortable too if you applied the definition of x\in(a,b) to argue via a\le x\le b

otherwise, this works

[upper shore]

although you are right that we can argue that (x-\epsilon, x+\epsilon)\subseteq(x-\epsilon_1, x+\epsilon_1), but this also needs to follow from inequalities

[odd lake]

yeah i am NOT prepared for a real analysis uni course

isnt this one of the more basic proofs

[upper shore]

its likely that im just being unclear

[lyric sierra]

What if I choose x1 and x2 so that they're at the extremities of A1 ∩ A2. And let e1 = e2 so that B(x1, e1) is in A1 but not A2 and vv for B(x2, e2). Then your argument wouldn't work. The point is that x = x1 = x2 for e = min(e1, e2) to work.

[upper shore]

you dont need WLOG, your idea is fine, but you need to show from inequalities that $(x-\epsilon, x+\epsilon)\subseteq(x-\epsilon_1, x+\epsilon_1)$

[upper shore]

hes only using x_1, x_2 as labels to call A_1, A_2 open here

[lyric sierra]

Huh

[upper shore]

everything is reasoned from x

hes saying that A_1 is open and A_2 is open, then he assumes x\in A_1\cap A_2, and only using x

unless i misread something

that 'where' is a 'forall' here

[lyric sierra]

Oh. Oops. It'd be better to just say A1 and A2 are open to start with instead of putting the definition down there twice. You are right

[upper shore]

although i agree that it is definitely more concise to specifically say forall, so it doesnt look like x_1, x_2 are treated free here, but thats a clarity point

[upper shore]

so, does everything i said make sense here? or do i need to explain?

[boreal girder]

Cycadellic

[upper shore]

LOL i messed this one up this is right now

hopefully that didnt confuse you completely

[odd lake]

no everything you said here makes sense
i think the unrigorousness of my proof comes from generally having gpt check my proofs or just from
me not being too used to proofs as i need to be for real analysis

[upper shore]

i wanna see what it spits out

[odd lake]

DUDE THERES LITERALLY NO SOLUTIONS

[lyric sierra]

Ya, use inequalities so you can then use induction in the future. Because it's technically about finite intersections not just A1 and A2. You don't want to split it up by cases even though it's tempting

[odd lake]

FOR THIS BOOK

[upper shore]

this is actually technically a good thing, go through the brutal proof-based pure math with no help, then youll learn the math maturity

cant learn to ride bikes with training wheels

[odd lake]

i cant put this into mathgpt rn i am at a diner rn

[odd lake]

true

[lyric sierra]

this it's for a real analysis class?

[upper shore]

its self study

[odd lake]

yeah self study is rough

[upper shore]

but it is still relevant to show epsilon balls form a topology on R

[lyric sierra]

Whenever I hear self-study, I think meditation or physically studying yourself, I think self-teaching or solo study is clearer, anyway back to the topic

[odd lake]

technically the problem didnt come from the book its from a problem set pdf for real analysis i found online

much better than the book excercises tbh

anyway i am going to eat now

how do i close this

[lyric sierra]

Which book?

[odd lake]

understanding analysis

[lyric sierra]

I'll let ya eat I'm just curious

[odd lake]

.close

[compact pewter]

Channel closed

Closed by @odd lake

Use .reopen if this was a mistake.

[lyric sierra]

Ah okay

[upper shore]

you dont need to close it but

yeah

its that

[cursive harbor]

Why doesn't this channel have OP's name on it

[upper shore]

i think the underscore broke the bot

bot seems to not like special characters

actually maybe thats not it at all

[verbal kiln]

it also broke the vertical bar. I’ve never seen that before

[cursive harbor]

Fair

[compact pewter]

Available help channel!

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[robust robin]

1. Measure the height of a safe step/platform (0.20–0.30 m).
2. One student steps up 10 times while holding the 300 g mass. Record the total time.
3. Calculate Work per step, Total Work, and Power.
4. Compare power outputs among students.

help, one of the tables says "Steps (N)"

so does that mean that all are just 10 N?

• students mass weren't measured so i can't solve for force

the title of the lab is simply " Human Power Output (Step Test)"

[robust robin]

context 🙏

[compact pewter]

@robust robin Has your question been resolved?

[compact pewter]

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[compact pewter]

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[quartz yoke]

yooo

[quartz yoke]

@upper shore you here bro?

[upper shore]

maybe

what do you have

[oak citrus]

What maths are you two doing in undergrad?

[quartz yoke]

youre busy?

[oak citrus]

I have some niche questions

[quartz yoke]

linear algebra

discrete math

[oak citrus]

What is discrete maths?

[upper shore]

im only taking graph theory this semester, but i self study much more

[oak citrus]

Have you had a look at probability?

[quartz yoke]

"Discrete mathematics is a branch of mathematics that studies discrete objects, which are distinct and separate rather than continuous. It is the mathematical language of computer science, dealing with topics like algorithms, set theory, graph theory, number theory, and combinatorics, which focus on counting and enumeration. Discrete math provides the foundations for understanding and solving problems in computer science, engineering, and other fields by working with countable quantities and their relationships. "

[oak citrus]

Oh nice, unfortunately I think that's completely opposite to what I'm trying to do now 😂

[kind viper]

where is your question

[upper shore]

not other than helping people here for now

[oak citrus]

In #probability-statistics and #advanced-probability

[quartz yoke]

does this mean i can choose any matrix or what

heres what they did

[upper shore]

yeah

the coded row matrix is like your secret password here, if im reading correctly

[quartz yoke]

bro this doesnt feel like math 1 bit 😂

its strange

[upper shore]

sure it is, its cryptography

[quartz yoke]

sounds interesting

[quartz yoke]

is it also systems of linear equations with planes and lines

[upper shore]

they can make those, but i think matrix mults are better thought of as weighted sums

[quartz yoke]

wdym by weighted sums

[upper shore]

ax + by + cd + ... is a weighted sum

in fact, matrices generalize this

$$ax+by+cz=\begin{bmatrix}a&b&c\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}$$

[boreal girder]

Cycadellic

[quartz yoke]

whats the difference between this and a system of linear equations

[upper shore]

a linear system is just a system of weighted sums

[quartz yoke]

oh

[upper shore]

the a,b,c here are called the weights

[quartz yoke]

oh

what do weights do

[upper shore]

just multiply by the x,y,z

its just terminology honestly

[compact pewter]

@quartz yoke Has your question been resolved?

[compact pewter]

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[kind viper]

tree diagram time?

[carmine garden]

P(knew answer | correct)= P(correct| knew answer)/[ p(correct| knew answer)+ p( correct didn't know anwer)]

[carmine garden]

I suppose so

[signal willow]

seems like you meant & for some places you typed |

[kind viper]

trying to set up bayes blind like this is going to result in great pain & suffering esp. given that you have 3 scenarios for how a student may treat a question.

[carmine garden]

okay , tree diagram it is

I don't think I even get the information

given any random question, the chance they know the answer is 0.5?

[shell horizon]

yes

[carmine garden]

okay

[shell horizon]

question is asking the chance that a correct answer is known and not just guessed

[carmine garden]

I guess this tree diagram is wronfg

[shell horizon]

the diagram looks right if im reading your handwriting correctly

just input the numbers

[kind viper]

it'll be easier if you branch 3 ways from the root

also you need to label each branch with its probability anyway

[carmine garden]

the probability they will be able to eliminate one choice

how do i interpret this

[shell horizon]

well when a student is doing a question, theres a chance that he knows one of the 4 answers is incorrect

[blazing bison]

I must say, beautiful handwriting

[shell horizon]

meaning he wont pick that answer

[carmine garden]

cool

[blazing bison]

Since the teacher set the question paper to measure what the student knows. I think the probability is 0.5

[shell horizon]

no

the question isnt asking what percent of the paper he answered right and knew the answer

its asking, out of the questions that were answered right, what are the odds that the student knew the answer, without guessing

[carmine garden]

so P( knew the answer| the answer is right)

[blazing bison]

Student studied well enough that the prob is 0.5 right

[shell horizon]

i think instead of writing "question" it would be more suitable to write "circled correct answer"

at the top

[blazing bison]

and there is 0.25 chance that the answer is right since there are 4 choice

Let see

[carmine garden]

so 0.5/(0.5+ (0.5) x 0.25 x 0.33)

which is 1/( 1+ 0.25*0.33)

[shell horizon]

woah slow down there

[carmine garden]

well, I suppose I should bayes

[shell horizon]

where did you get that denominator

[blazing bison]

Two cases: Case 1: Student answers without guessing P(A) = 0.5
Case 2: Student answers with guessing P(B) = 0.25

[shell horizon]

you need to find the chance of correct answer on the denom

[blazing bison]

Why can't I simply add P(A) and P(B)?

[carmine garden]

the left branch and the left sub-branch of the right branch

[shell horizon]

"does not know answer" is 0.5

then you divide the 0.5 into 2, ie 0.25

and then you multiply by the odds that the answer is correct

[blazing bison]

(1/2)(1/4) = 1/8 = 0.125?

[carmine garden]

why

[blazing bison]

same question

[carmine garden]

well, I got that bit now. So it's 0.5/(0.25*0.3+ 0.5)?

[blazing bison]

I feel like answer will simply be P(A) + P(B) = 0.75

[shell horizon]

because theres a 25% chance of it being complete guess vs eliminating one answer

[blazing bison]

1/2/1/24 = 1/48

[shell horizon]

why is there only 0.25*0.33

wait

[carmine garden]

the lone 0.5?

probability of knowing an answer and getting it right

[shell horizon]

yeah

the 0.25*0.333 is eliminating 1

what about complete guess?

[carmine garden]

right, I didn't take that into account

just 0.25*0.25?

[shell horizon]

no

theres a 0.25 chance that you guess

and another 0.25 chance that the guess is correct

[carmine garden]

.

[shell horizon]

yea

did you type that out or is that the edit

am i blind

[carmine garden]

edit

got it

0.5/(0.5 + 0.25 * 0.25 + 0.25* 0.33)

[shell horizon]

yeah

thats it]

the denom is the chance of correct answers

and numer is the odds of knowing the answer

[carmine garden]

cool

tq

.closed

.close

[compact pewter]

Channel closed

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[molten grotto]

Represent 5 + sqrt(3) on the number line

[azure vault]

do you know how to represent square roots on the number line?

(and my question is does it have to be "constructed"?)

[proper forum]

You may draw a 1:sqr(3):2 triangle

and the right angle should on (5,0)

then you can sqr3 by 2 and 1 and 30 degree!

[compact pewter]

@molten grotto Has your question been resolved?

[molten grotto]

Not yet

[molten grotto]

I see

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@molten grotto Has your question been resolved?

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[frosty reef]

what i'm thinking of saying is
let z = e^itheta
Re(z^6) = cos(theta)
but even then
idk this question doesnt look so great i havent done enough de moivre practice so what do i do

[kind viper]

let z = e^itheta
Re(z^6) = cos(theta)
correct thus far

recommend abbreviating cos(θ) as c and sin(θ) as s

and then carefully expanding (c+is)^6 and plucking out just the real terms

[frosty reef]

ohhhhh

oh yeah i remember this method

ok

i did the question without much issue

thanks

im now stuck on the next question though

idk how i can use the previous question to answer this

WAIT

yes i do

i know how

because i remember roots of polynomials

i think the last coefficient gives the product of all the solutions

sort of

i think that would mean -1/32 is the product?

but wait there's 2 other solutions multiplied that we dont want

but it's just pi/4 and 3pi/4

i can use that by memory

cos(pi/4) is root(2)/2 and cos(3pi/4) is -root(2)/2 iirc

[frosty reef]

so it's just this /-(1/2) i think

so 1/16?

nvm i can do this question

[elfin finch]

Well i got that using (presumably) another way so ye

[frosty reef]

yeah ok

good

[compact pewter]

@frosty reef Has your question been resolved?

[frosty reef]

.solved

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[brave nebula]

Prove that LCM(a,b,c)/LCM(a,b) = c/LCM(HCF(a,c), HCF(b,c))

[brave nebula]

For any natural numbers a, b and c.

[mortal bay]

lcm(x_i)=prod(x_i) / hcf(x_i)

should help

i.e. lcm(a,b) = a*b / hcf(a,b)

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[carmine garden]

Four coins are flipped. The first two coins are fair, whereas the 3rd and 4th are biased, The latter land on headds with probabilities 0.7 and 0,4. Assume the outcomes are independent
Find the probability that exactly one heads appears

[mild sky]

What did you try?

Hi wai

[carmine garden]

hi

I chose a coin , that shows heads

and the rest tails

Uh, how do I write this concisely

0.25(0.3 * 0.6 + 0.3 * 0.6 + 0.4 * 0.6 + 0.3 * 0.4)

[mild sky]

Why are you multiplying by 0.25

[carmine garden]

0.5 * 0.5

[mild sky]

Ok you took it a step further

[carmine garden]

The thing is is there any concise way to write this as a summation

[mild sky]

Not really

If there are similar cases you can group them together

Like the two cases where the head is on the first two fair coins

[carmine garden]

tq

Another question if I may

[mild sky]

Go for it

[carmine garden]

One of the numbers , 1 though 10 is randomly chosen. You are to try to guess the number chosen by asking questions with yes-no, answers. Compute the expected number of questions will need to ask in the following two cases

(i) Your ith questionis to be , is it i? i=1,2... 10

[carmine garden]

I'll first try to determine the kind of distrbution this is

This is a geomertic distbution

[mild sky]

Is it?

[carmine garden]

the questions are asked until success

[mild sky]

Your random variable of how many questions you ask has bounded support

i.e. you will stop after 10 questions

[carmine garden]

so it isn't explicitly any type is it

[mild sky]

It is of some simple type

[carmine garden]

uniform

[mild sky]

Yes!

Do you understand why?

[carmine garden]

yea, each number has a probability of 1/10 of being the choeen one

[mild sky]

Yup it has the same distribution as the chosen number

[carmine garden]

so $E[X=i] \sum_{i=1}^{i}= i\frac{1}{10}$?

[boreal girder]

wai

[mild sky]

Yeah by definition

[carmine garden]

okay, cool

thanks!

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[silver canopy]

I have been trying to conceptualise convolution better, and I just want to hear somebody describing what it represents in their own way so I can gain better insight, if anybody would be interested in indulging me?

To preface this, I watched 3B1B's video on it, and while his examples were pretty helpful for me to understand how it works, I guess I am still stumped on what it really is representing. Similar to how you may think of multiplication as repeated addition on a basic level

[compact pewter]

@silver canopy Has your question been resolved?

[hushed magnet]

not everything has a nice basic intuition like that

[compact pewter]

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[flint cliff]

can someone explain this question please

[winged lion]

Do you know the magnitude of a vector?

[flint cliff]

yh ik but why?

[wooden slate]

oh the proof?

[flint cliff]

no

why is it that?

[wooden slate]

alr i need to think abt its proof

[swift sun]

By the Pythagorean Theorem, applied twice

You should try drawing this in a plane

[wooden slate]

yup i thought so

i tried to recreate it in ms whiteboard but erm

kinda failed

i think

[wooden slate]

consider the main vector as the diagonal btw

[flint cliff]

thanks

[swift sun]

It's good to ask yourself why certain formulas school might just give to you without further proof hold

It's kind of unfortunate that sometimes it's done like this, ideally everything should be proven right away

[wooden slate]

yeah we were directly told 😭

thats why i had to think of its proof rn

[compact pewter]

@flint cliff Has your question been resolved?

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[wanton patrol]

$\lim_{n \to +inf} (\frac{n!}{10^{n - 1} + 5})$

[boreal girder]

tetrahedron

[wanton patrol]

how would I go about solving this?

i simplified it to $\lim_{n \to + \infty} (\frac{10n!}{10^n + 50})$

[boreal girder]

tetrahedron

[pine thicket]

$\lim_{n \to + \infty} \left(\frac{10n!}{10^n + 50}\right)$

[boreal girder]

Duck Man

[kind crane]

you just compare the largest terms in each numerator and denominator

[wanton patrol]

10n! is the largest one in this case, right?

Not sure what's the next step

[wanton patrol]

10^n is the larger one actually

[kind crane]

yes for the numerator

[kind crane]

yes for the denominator

then you can just compare 10n! with 10^n

[wanton patrol]

In this case, the denominator is larger, right?

what would I do after that

$\frac{10}{\frac{10^n + 50}{n!}}$

[boreal girder]

tetrahedron

[wanton patrol]

can one say that the limit of this is zero?

[thorny kestrel]

n! is larger than 10^n

[wanton patrol]

What would be the next step after identifying the largest term?

Since the numerator > denominator, you just say that the answer is infinity?

[thorny kestrel]

you need to show (if you haven't already) that n!/10^n goes to infinity

[wanton patrol]

How would I show that? pretty new to this

Is it via this rule?

Although the numerator/denominator are flipped

[thorny kestrel]

yes

think about why it's true by writing a few terms, then you can formulate a lower bound the goes to infinity when n is large enough

[wanton patrol]

The thing is that

The rule says that the limit of it is zero

Not sure what's so differnet in that expression

Except for the fact that the numerator/denominator are flipped over

I'm confused on why the answer is infinity

[kind crane]

instead of this, you can divide by 10^n. the denominator goes to 1 and you just figure out the numerator

[kind crane]

by using this

[wanton patrol]

Makes sense

Thanks 👍

.close

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[sudden igloo]

hiiii

[south sage]

!da2a

[compact pewter]

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/

[sudden igloo]

so i found the pattern

but i have no clue how to state is

cuz its moving along fibonaccis triangle

<@&286206848099549185>

[onyx agate]

i mean i guess you coul call it

[sudden igloo]

i mean not the fibonacci but like ykwim 😭

[onyx agate]

but that would make it a spcl ccase

of the camtor triangle

[sudden igloo]

no clue what that is ngl 😭

[compact pewter]

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.

[onyx agate]

Let s=m+n. Then k=s(s+1)2+n, k=2s(s+1)+n,so k sits on the s-th “diagonal” of the Cantor pairing grid.
fr positive m, we stay on the same diagonal and move one step:
(m,n)-->(m−1,n+1)
(m,n)-->(m−1,n+1).

If m=0, we finish the diagonal and jump to the next:
(0,n-->(n+1,0)
(0,n-->(n+1,0).

[sudden igloo]

that makes sense

but idt thats what my prof is looking for per se

well ok hows this sound

it says by induction

so lets say we use the base case

where k = 0

it satisfies of m and n are 0 yes

and uhhh inductive step

so we use k+1

[onyx agate]

ya

[sudden igloo]

so then 2(k+1) = ....

(m' + n' )(m' + n' = 1) + 2(n')

but what do i do with that 😭

[onyx agate]

We construct (m′,n′) by k+1
Se m′=m−1,n′=n+1.

[sudden igloo]

how do yk that

[onyx agate]

proceed

because we set s =m +n

and summin m'+n'

gives us m +n

[sudden igloo]

so what we get

s(s+1) + 2n?

[onyx agate]

yes

+2

s(s+1) + 2n +2

[sudden igloo]

so how do we get n' = n+1

[onyx agate]

by def s(s+1) + 2n +2 = 2(k+1)

since k k=s(s+1)2+n

[sudden igloo]

im lost

[onyx agate]

how about you tell me what you have rn

[sudden igloo]

well i have

s(s+1) + 2n

= 2(k+1)

[onyx agate]

so then you're done

now this fr m>0

the edge case is m = 0

you need to prv fr that and et voila you're dn

you're missing a +2

[sudden igloo]

from where would the +2 come

[onyx agate]

gimme a sec, i'll just typeset the whole thng

[sudden igloo]

is it cuz we added 2 to the k side?

that we have to add 2 to the right side as well?

and hten u can group that with n

making

2(n+1)

?

[onyx agate]

ya

[sudden igloo]

so

2(k+1) = s(s+1) + 2(n+1)

[onyx agate]

yup

[sudden igloo]

so now

we do a couple of cases?

[onyx agate]

this was if m>0

if m=0 its trvial

anf qed

[sudden igloo]

now can we set n =s?

[onyx agate]

yes

[sudden igloo]

in whcih case

m = 0?

so

n(n+1) + 2(n+1)

[onyx agate]

yes

and done

bt induction true $\forall m,n \in \mathbb{N}$

[boreal girder]

ben

[onyx agate]

hope i was helpful

[sudden igloo]

thank u

can we do the next problem as well

ur very helpful 😭

@onyx agate

[onyx agate]

sure

[sudden igloo]

so basically // is being distributed

[onyx agate]

i dont know what // as distributed means ?

i havent this notation in combinatorics before

[sudden igloo]

whats // do btw

[onyx agate]

i don't know check yr reference, i am guess floor div but i might be wrong

[sudden igloo]

whats floor div do

[onyx agate]

basically lower mod

[sudden igloo]

ok so like 5 mod 2 would be 1

what would 5//2 be

[onyx agate]

i don't know you gotta tell me 🙂

im just guessing since its the only time i see //

in python mostly

i think its flr div

i am 98 percent sure

but the fact you said it distributes threw me off

cuz div doesnt dist

[sudden igloo]

sorry sorry

and it isnt true right

we can counter with an ordered pair like

(2,1,2)

[onyx agate]

ya

i mean there are mny counters

[sudden igloo]

ofc

i just gave an example

[onyx agate]

btw before i ask is this an exam or smthn

[sudden igloo]

no its not

just a problem set my prof cooked up

its was kinda hard ngl

a review guide for the midterm next week

[onyx agate]

ok

so you found a counter

so done

we cn also prv it rigorously that it's nt true

usn the euclid thm

[sudden igloo]

well it says a give a counterexample if its false

so i think an ordered pair would suffice

im going in for office hours on tuesday

so i was tryna complete the problem set over the weekend

[onyx agate]

so done

any other q

oh you're teaching ?

or not ? i dnno been a while since ive been in the grad system

[sudden igloo]

wdym

[onyx agate]

by oh i thought ya meant you're teaching the oh hrs

[sudden igloo]

nah nah

im a student

first yr of my undergrad

[onyx agate]

oh i see

gl on yr math journey

its gonna be fun

[sudden igloo]

yea hopefully

i love math

this class is just funky imo 😭

i can do the definitions on my own...

[onyx agate]

i think maybe someone else can come in and help, i gotta go, sprry

[sudden igloo]

ur good

tysm

[onyx agate]

okay i'll do one more

[sudden igloo]

oh ummm do yk how to close this

i posted again in help-31 😭

[onyx agate]

my ocd doesnt allow me to leave a problem if i havent done it

[sudden igloo]

can we leave this channel tho 😭

[onyx agate]

you can close by . close and then reopen by . reopen

[sudden igloo]

.close

[compact pewter]

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[torpid mist]

Hello I'm not sure how to do b) for this question any tips?

[alpine canyon]

you know r
so what would happen if you sub in A and B to what you have there

[timber lodge]

i am korean and can you guys help me? (three following conditions are 가,나,다 in korean) i used gemini to translate the problem to english.

[verbal kiln]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[torpid mist]

Like this?

[alpine canyon]

the B went a bit awry

[torpid mist]

oh wait yeah

[torpid mist]

Fixed it I think

should I make the two equations equal to eachother ?

[alpine canyon]

i would expand them first

[torpid mist]

I got a final equation h-2k=8, how can I find h and k ?

[alpine canyon]

nicely done, you can now make h or k the subject and sub back into your initial equation (either)

[torpid mist]

Is this okay? And then I just find K and plug it back in without x and y right?

[alpine canyon]

thats alright yeah, you find k, then you get h from that and youre done pretty much

[torpid mist]

okay thank you so much

[alpine canyon]

you did it all, i just gave a hint

[torpid mist]

not too sure if I did this right since I got 2 values for k but there is only 1 centre

[alpine canyon]

thats fine, reread the question

[torpid mist]

ah your right

ok thank you for the help

.close

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[alpine canyon]

nw

[compact pewter]

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[novel narwhal]

Does anyone have a hint for this problem (where to start)

[novel narwhal]

I did get this

[compact pewter]

@novel narwhal Has your question been resolved?

[novel narwhal]

<@&286206848099549185>

[grizzled geyser]

After I multiply the constraints together and plugging in your result, I found two answers for xyz

I don't know how to determine which one is correct, though.

[novel narwhal]

The one in the problem statement?

[grizzled geyser]

Yes

[novel narwhal]

Ok

[novel narwhal]

.close

[compact pewter]

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[split lance]

I am currently working on logarithmic functions for precalculus. can anyone tell me the steps to evaluating equations because the lesson didn't explain it all that well.

[sullen merlin]

Okay so first up do you understand what log_x(y) means?

$\log_x(y)$

[boreal girder]

lance lance

[split lance]

uhhhh not really the way that they teach us is they just tell us that it is what it is they don't explain it

[pure orchid]

just take the square root really

[sullen merlin]

So what $log_x(y)$ means is its the answer to the question "What power of x equals y". So as an equation $x^{\log_x(y)}=y$

[pure orchid]

$text_bf{hello}

[boreal girder]

lance lance

[sullen merlin]

if you want to shitpost go to #chill

[pure orchid]

[pure orchid]

no i am trying to latex bot

am not sure how it works

[sullen merlin]

go to #latex-testing

[pure orchid]

thx

[sullen merlin]

So for an example, $\log_2(8)$ is the number that makes $2^x=8$

[boreal girder]

lance lance

[sullen merlin]

can you think of a number that does that

[split lance]

i think i kind of get it?

is it basically flipping around the question until its true?

[pure orchid]

,tex \int \log\left(\frac{1}{2}\right) , dx = x \log\left(\frac{1}{2}\right) + C

[sullen merlin]

idk if until its true is correct

[pure orchid]

here you go

huh its gone

[sullen merlin]

mate this isn't helpful, leave this channel before you get timed out

[pure orchid]

how is that not helpful its a log question

[sullen merlin]

trolling in help channels is against the rules

[pure orchid]

man whatever ill help somewhere else

bye maxwell

[sullen merlin]

It's more "If I want to evaluate log_x(y) I should answer the question x to the power of what equals y, the answer to that question is the log's value"

[split lance]

ohhh okay

in that case what would i do with this?

[sullen merlin]

So can you solve log_2(8) for me?

[sullen merlin]

Well ln is log base e

so the question here is, e to the power of what equals e to the negative fourteen

[split lance]

OHHHHHHH OKAY

that makes wayyyy more sense than the way my teachers explained it

okay i think maybe i've got it now :)) thank you!!!

[sullen merlin]

Then for some other info you need for these questions \
log of a negative number is undefined \
$a\log_x(y)=\log_x(y^a)$ is a useful identity \
If the thing inside the log isn't a perfect power of the base there's not much you can do there \
$\frac{\log_a(b)}{\log_a(c)}=\log_c(b)$ is also true

[boreal girder]

lance lance

[sullen merlin]

Nice! type .close then to close this channel c:

[split lance]

.close

[compact pewter]

Channel closed

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[tawdry loom]

I'm trying to figure out expected exp game from a game, and I'd really like to optimize/simplify my function a little, primarily to make it easier and faster to compute.
This is my complete function, but most of it is irrelevant, since it's already computationally easy:
s>0 and 1>t>0

[tawdry loom]

This is the relevant computationally expensive part:

I'm halfway sure there's no way of doing this better, but I'd like input from somebody who's smarter than me :3

[mild sky]

if a computer needs to calculate this once, why even optimize it?

its hard to believe this takes any meaningful time

[tawdry loom]

I mean, I'd like to have a reasonably detailed 3d chart

mmmmmmm

the end goal was having it on a website

should proabably just precompute it and put it into a CSV?

[mild sky]

if its static definetly

heck precompute the image

ofc see what works best for you

[tawdry loom]

lol, being able to move around the chart would be fun
but idk how good 3d chart libraries are

maybe I'll have to settle for a picture

[mild sky]

if you are developing a game, this shouldn't really be something you focus on too much

because you still need a functioning and enjoyable game

[tawdry loom]

well it's an informational site for a game

[mild sky]

oh

[tawdry loom]

but not a game itself

similar to a wiki

[mild sky]

I see

maybe a more useful feature would to list the formulas, and then key features of them

like specific values, and general behavhiour

[hushed magnet]

I mean, fun idea but if its only for exp it doesnt sound that important?

[mild sky]

and maybe include tips on how to maximize it

[hushed magnet]

not something to dedicate too much time to

[tawdry loom]

I like nichés

and it's really fun looking into math like this

so even if it's not super important, I really like crafting stuff like this

thank you both for your input!

[mild sky]

a bit to answer your original question

[tawdry loom]

oh?

[mild sky]

try to look into how many iterations of the sum you really need

if you sum a lot of terms that barely effect the result you can skip them

and get a good enough approximation

[tawdry loom]

oh yeah, totally, I was already planning to do that

the sum converges onto a specific value as you approach infinity

going as high as s is a reasonably good accuracy

wait, well

a value depending on s

lol

I'll stick with my current function for now, I do hope it's good enough
see you around!

<3

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[hazy obsidian]

Let ${a_n}$ be an infinite sequence of integers, and let $p$ be a prime not equal to 2 nor 5. Prove that the sequence $$b_n = \frac{a_n}{p^n}$$, when written as decimals, for sufficiently large $n, b_{n+1}$ has a repeating decimal whose period is $p$ times that of $b_n$

[hazy obsidian]

so how do i even start this💔

[hushed magnet]

do you know modulo

and what a multiplicative order is

[hazy obsidian]

i know modulo but not that

oh wait i translated the problem wrong

[boreal girder]

Copter

[hazy obsidian]

i dont know multiplicative order much

[hushed magnet]

take a number like 1/23 and compute it by hand

think about how what you are doing has to do with the number 10 mod 23

the obvious problem with the question is that if a_n=p^n, then b_n=1 always and the period doesnt get bigger

so you mistranslated somewhere

or the problem is wrong

[hazy obsidian]

multiplying by 10 until it repeats?

[hazy obsidian]

uhhh ill find it wait

,rccw

[hazy obsidian]

the wording is so goofy

oh shoot an is coprime to p

my bad😭

[hushed magnet]

the amount of times you do that is the multiplicative order of 10 mod 23

but I really dont know how to continue from here without more knowledge about orders

[hazy obsidian]

so like, 10^k = 1 (mod 23) then k is the multiplicative order?

[hushed magnet]

for the smallest k, yes

[hazy obsidian]

the book hint says

"Use the fact that the number of digits in the period of each of the fractions $\frac{a_n}{p^n}, \frac{a_{n+1}}{p^{n+1}}$ is equal to the smallest of natural numbers $k,l$ such that $10^k -1$ is divisible by $p^n$ and $10^l -1$is divisible by $p^{n+1}$"

[boreal girder]

Copter

[hazy obsidian]

this is just multiplicative order right?

[hushed magnet]

yes

[hazy obsidian]

what am i supposed to derive from this😭😭

[hushed magnet]

hmm what other results has your book covered

I'm actually not quite sure how to finish either, this was easier in my head

do you have LTE?

[hazy obsidian]

its just problems (also the book is like 30+ years old so modern number theory things probably dont exist)

[hazy obsidian]

probably not

[hushed magnet]

30 years is not that old

well look in the book in that section

the problem is written for one of the results in that section

[hushed magnet]

but you know what it is?

[hazy obsidian]

yep

[hushed magnet]

I think it finishes the problem

but not quite sure

[hazy obsidian]

most of the book uses only gcd and euclidean algorithm arguments so

[hazy obsidian]

whats your idea?

[hushed magnet]

rough idea, p^n divides 10^k-1 so p adic valuation of 10^k-1 is >=n

similar for 10^ell-1

we want to show ell=pk

which would make 10^pk-1=(10^k)^p-1^p a situation for LTE

then some fiddling around

maybe works, maybe doesnt

[hazy obsidian]

ohh

ill try it out, thanks!

ill be back if i havent figured it out

.close

[compact pewter]

Channel closed

Closed by @hazy obsidian

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[compact pewter]

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[valid mist]

i need help with calculus

[mild sky]

with what

what in calculus

[valid mist]

so school js started kinda but our teacher hasn’t made the recheck but she just did. I failed the first exam because i had lower than a 50 percent. i corrected myself and i jjst want to see if i’m actually correct or no

[mild sky]

ok

send?

[valid mist]

yes this is the first page

[mild sky]

you shouldn't put your name in the picture

[valid mist]

oh yes my bad

[mild sky]

ok so its ok for a limit to be infinite

[mild sky]

so it doesn't mean it doesn't exist

[mild sky]

b here

[valid mist]

oh yes thats true oh no i have to fix it

[mild sky]

for c, the limit does exist

[valid mist]

really? i thought horizontal asymptote

[mild sky]

it doesn't exclude the limit from existing

the constant function is a horizontal assymptote of itself

and it has a limit everywhere

[valid mist]

oh yesss

so would it be 2 then? bevause it’s approaching 2

[mild sky]

ye

now d and e are also wrong

[valid mist]

d is 1?

i dont know whag e is though bevause there’s 2

[mild sky]

for d, the limit does exist, because limit don't care about what happens at the point they care what happens around the point

[mild sky]

yeah so the explanation is innacurate

you would explain it has the one-sided limits being different

[mild sky]

not quite

[valid mist]

so instead of saying there’s a hole i say because there are 2 different limits at 3?

[valid mist]

i meant -1*

[mild sky]

yes

[valid mist]

oo okay okay

[mild sky]

the one-sided limits are different

it approches one thing from the right and another from the left

[valid mist]

mhm so at x>3- it goes to 1.5 but at x>3+ it goes to 3 so the actual limit doesnt exist

right?

[mild sky]

yes

now question 2, a and c are correct

[valid mist]

okay i made sense of the first page

[mild sky]

b is not correct

d I can't read your explanation

[valid mist]

for d i said for c you would have to redifine the function f(5) so there’s a point at 5

basically

wait so for (a) there is a limit right? just not defined

[mild sky]

yeah

continuity requires the function to be defined at the point

[valid mist]

and for b thats an asymptote so thats not considered a discontinuity?

then it would be 3 right?

[mild sky]

2 would not be a jump discontinuity

3 is right yes

[valid mist]

okay oaky and then is d correct?

[mild sky]

yeah

question 3 is also right

[valid mist]

yeah at first i got it wrong but we just plug in 3 at first then put it equal to each other right?

[mild sky]

can't read 4b

[valid mist]

because it wouldn’t be defined and we wouldn’t be able to find anything if there’s a whole. (hole=empty=no number)

[mild sky]

it might be defined

[valid mist]

ooooo

[mild sky]

but we still can't relate the value of the function at the point to the value of the limit

and definetly we can't subsitute x=2 like you did in part a

[valid mist]

ooo okay let me rewrite this

wai kt

so 4a is right?

and what should i say for 4b to explain why it must be continuous in order for me to find an answer

[mild sky]

this pretty much

don't copy 1-1

put it in your own words

[valid mist]

oohhhh bevause it’s still f(x)?

[valid mist]

i understand but i also dont understand whag we said for that

[mild sky]

continuity tells you how f(a) is related to lim x-->a f(x)

without it you can't just substitute x=2

[valid mist]

Oooooooo wowwww yess

soooo am i wrong on anything on 5?

[mild sky]

no

[valid mist]

okayy

i think 6 is pretty easy but 7 i dont know if i messed up or not

[mild sky]

7a is fine

I think the rest is fine

[valid mist]

okay

looking at my work

is there any way how i could solve any of these easier?

i’m sorry for the late reply im in class right now loll but i need to study for my quiz after class

[mild sky]

nothing specific comes to mind

[valid mist]

okayyy thats all thank youu!! bye byee

[mild sky]

.clsoe

.close

[compact pewter]

Channel closed

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Remember:
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• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.