help-13

[small dust]

yuh

[plush wharf]

What about these 2

I thought I did 17 right

[small dust]

idr how to do 17 but I can do 18

[plush wharf]

Ok sure

What I do wrong there

[small dust]

hmmm

it looks fine to me actually

idk why 18 is marked off pts

[plush wharf]

Bruh

[glad kestrel]

for 18, your angle is on the wrong side

[hollow trail]

for 17 it seems to be correct other than not being fully simplified (could have canceled out the 13)

[loud elbow]

17 should have just flipped the sign no?

[small dust]

wym

oHHHHH

between ladder and building

[glad kestrel]

the angle he labeled is the angle between the ground and the ladder

[small dust]

not between ladder and floor

[hollow trail]

may also be points off for writing "sin" rather than sin(x)

[small dust]

ur right, tru true

good catch fr

[small dust]

no way........... has that ever happened to you? I'd be livid 😭

[somber lantern]

I need help plz

[small dust]

find ur own channel

this one is occupied rn

[somber lantern]

What do u mean
I am new to this server can u explain

[plush wharf]

So I was suppose to use cosine?

[small dust]

these channels are open for you to post in

once u post there, u "claim" it for some time

[somber lantern]

Ok

[plush wharf]

Wdym i cant simplify it more

[small dust]

this channel is claimed by somebody else rn

[glad kestrel]

technically you could use any trig function as all angles can be discovered, but most simply, yes, relocate your angle to the correct location and use cosine

[hollow trail]

$\frac{-5}{13} \cdot \frac{13}{12}$ you can simplify this without multiplying it out first

[wraith dagger]

cloud

[plush wharf]

Oh right

[cedar kiln]

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[wary estuary]

How come I can somehow cancel the tans?

I am awful in trigonometry, maybe I’m missing something

[untold torrent]

That doesn't seem right

[wary estuary]

Idk, answer is supposed to be sec(x)

[untold torrent]

The last step is wrong

You can't cancel out the sec(x) like that

[wary estuary]

Wait whoops yeah

One sec

,rotate

[wary estuary]

That makes more sense lol, ty

,close

.close

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[ivory finch]

wuts spanning tree and how can i rememberr it? graph theory

[ivory finch]

i think its just duplicating the vertices???

but tbh idk why its called spanning

[odd lagoon]

so given a graph G

with vertices and edges however messy it may be

a spanning tree is a subgraph of it that follows two properties:

1. It has every vertex
2. It is a tree

spanning trees are not unique, so a graph can have multiple

aight check out this ms paint

this is my graph G

[ivory finch]

sure

[odd lagoon]

in blue is some random spanning tree

you can check its a tree by viewing the right vertices as the base

also the blue has no cycle / loop

another spanning tree (there isnt one answer is the point of this)

this is a subgraph but not a spanning tree because it doesnt have all the vertices

that is to say, it doesnt span

this fails in two ways: it doesnt have all the vertics and also isnt even a tree. the red loop goes in a circle

[ivory finch]

so basically spanning in graph theory means it SPANS all the vertices?

[odd lagoon]

yep (and in general, span will mean everything is attained)

[ivory finch]

except edges

[odd lagoon]

so a spanning tree is a subgraph that hits all the vertices and is a tree

yes

spanning all the edges means you probably wont have a tree

so thats implicit in the use of the word tree

[ivory finch]

why they call it tree?

i get that a perfect binary tree looks like a nice xmas tree but other than that i dont think it looks like a

[odd lagoon]

trees look wonky sometimes

a working definition of a tree is that there are no cycles

loops is another word for cycle

trees grow in whatever shape

but they do not connect branches (at least mathematically)

i dont think trees connect branches irl either but

[ivory finch]

is BFS and DFS a spanning tree?

[odd lagoon]

yes!

[odd lagoon]

no clue whats going on here tbh

yellow are both trees

ah

first is BFS

second is DFS

[ivory finch]

how do i find the radius and diameter of this graph (ignoring the yellow subgraphs?

[cedar kiln]

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[wheat atlas]

My question:

Express the quantity without using absolute value:

|2-(sqrt)6|

[cedar kiln]

Please don't occupy multiple help channels.

[digital cliff]

what have you looked for so far

[wheat atlas]

So this question I’ve got wrong multiple times.

I tried 0.44948974278

[digital cliff]

you should be using exact values

[wheat atlas]

So is it a case like the last question where I had to go
-(2-sqrt(6))

[digital cliff]

yup

[wheat atlas]

Okay. I guess that makes sense, the prof. Also said exact values but wouldn’t elaborate on what that meant. Thank you.

[digital cliff]

it just means for example if i had 1/3 id write it like that rather than 0.3333
or if i had sqrt(2) i leave it like that rather than writing 1.414

no rounding basically

[wheat atlas]

Okay. I appreciate the help I think I finally get it.

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[halcyon granite]

Say we do 30 trials of throwing a die. The question is: what's the theoretical median (not mean)?

[halcyon granite]

And how would one go about finding it?

tbh, this is from our «intro to stats» [for pure math students] course, and I have no idea how they got this value:

> qbinom(1/2, 30, 1/6)
5

[muted timber]

i would expect the median to be the same as the mean on this case

[halcyon granite]

by the way, I just tried to simulate it: I've got ~3.5 [not anywhere near close to 5]

[halcyon granite]

hm, still no clue

[muted timber]

i dont know how to go and formally prove it.
However, we're assuming a fair die, so each value has 1/6 probability of appearing, so half the values will be 3 or lower, and half will be 4 or higher.
So the median would be any value between 3 and 4

[halcyon granite]

well, according to the simulation, sounds reasonable

[muted timber]

you should not trust any "simulation" you cant verify yourself by hand given enough time.

because that same simulation gave you a 5 previously as you stated

[halcyon granite]

nope, that was not simulation. that code invoked some math functions

and I didn't write the code that split a 5, so I cannot explain how it works (in fact, I came here exactly to understand why it doesn't work 😆 )

[halcyon granite]

I mean, the same thing I do by hand can be done on a computer

[muted timber]

precisely. Computer should be only used for things that you CAN do, but are annoyingly slow

okay, i know what the thing is doing

[muted timber]

looking at the documentation of qbinom, you're calculating the 50% percentile of a binomial distribution of 30 tries and success probability p=1/6.
since a binomial B(30, 1/6) has a 42% of being below 5, and 61% of being 5 or lower, 5 is the median (also obtained from n*p)

[halcyon granite]

but why does 50% of percentile of bin dist give us the median?
I mean, it tells us that the probabilit of getting 5 or lower is 42%, but I just don't see how it's presicely connected to the notion of median

like, how do we even define the median when we don't have a particular sample?

[muted timber]

No. I'm telling you that the probability of lower than 5 is 42%, and the probability of 5 or lower is 61%.

[halcyon granite]

oh, ok

[muted timber]

In the binomial distribution, the probability of being exactly value k is given by this formula, with n being the number of trials, k the number of successes, p the probability of success

lower than 5 means it's equal to 0, 1, 2, 3 or 4.
5 or lower means it's equal to 0, 1, 2, 3, 4 or 5.

adding the 5 terms is 42%. Adding the 6 is 61%.

[halcyon granite]

ok, good. no questions on that

[halcyon granite]

then this

[muted timber]

the median is such a value where half are below, and half are above

[halcyon granite]

but in a sample, innit? (at least that's what I'm familar with)

[muted timber]

you can obtain the theoretical median

same as you can obtain the theoretical mean

[halcyon granite]

u mean the expected value? (sorry if that's a dumb question; we didn't have a probability class yet)

yes, we had problems involving concepts from probability without knowing probability 😐

[muted timber]

i recall there being a technicality there that prevents me from saying "yes"

i'd recommend you go and look up the exact definition you were given

[halcyon granite]

we weren't

[halcyon granite]

just looked up on wikipidia

so that's what u did?

[halcyon granite]

here

[muted timber]

it's a binomial distribution, computing probabilities on it is pretty easy

there's also a bunch of online calculators that do it for you.
Cuz i know how to do it, but it's annoyingly slow :)

[halcyon granite]

huh? is that a side note or a reply to my last question?

[muted timber]

but yeah, this is the definition of the median. The probability of being at or below the median is higher or equal to 50%, and the probability of being at or above the median is higher or equal to 50%.

which depending on the distribution, means that the median is not necessarily unique

[cedar kiln]

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[halcyon thunder]

Question: Compute the absolute angle and value of z = sin(i).

I've tried moving sine to the left hand side and using Euler's form but when I continue, I get undefined behavior. When I tried to find the angle, I would be dividing by zero so I don't know what to do.

[upper abyss]

How did you use Euler's?

If you take e^(it) and e^(-it), then solve them for sin and cos, you get these:

These are often called "the complex definition of sin(z) and cos(z)"

Weird how I found a picture with sin(i) in it

Google is listening

[halcyon thunder]

I tried converting i = re^theta(i) but the angle became theta = arctan(1/0)

hmmmm

wait that actually is really helpful, let me try and see if I get anywhere

[upper abyss]

Note that sinθ ≠ i

[halcyon thunder]

right okay thank you

.close

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[fervent pike]

hello, is this not right?

[fervent pike]

not sure how you would manipulate it so that 1=tn

[burnt vapor]

You can't go from 1=tn mod sm to 1=tn

[fervent pike]

hmm?

wait Ill send the question

so If m is prime and we say n != 0, and n in {1,....,m-1}

then gcd(m,n)=1

so then by bezouts theorem I get gcd(m,n)= sm+tn

1=sm+tn

and you can get rid of sm to show that 1=tn

so that means n has to be a multiplicative inverse of t, which satisfies one of the field axiom property

but if you dont have m as prime then you cant have 1=tn

[burnt vapor]

Well, if you want to prove it for m and not for sm you should do (mod m) instead of (mod sm)

[burnt vapor]

But you can't write tn=1, it's not true

You can only say tn=1 (mod m)

And that's enough to prove that it is a field

[fervent pike]

should I state that we assume s to be 1 then? so instead of mod sm I write mod m

[burnt vapor]

No

sm = 0 (mod m)

[burnt vapor]

You don't need to

[fervent pike]

because sm is a multiple of m the modulo will be the same

I can write that

[burnt vapor]

Yes

[fervent pike]

how is it enough to show that tn = 1mod m

shows that t is a multiplicative inverse of n

i can see that m divides tn-1

[burnt vapor]

Because you are not considering the product of integer numbers

[fervent pike]

or tn mod m = 1 mod m

[burnt vapor]

But a new product

[burnt vapor]

This says that the product of two numbers will be another number between 0 and m-1

So if the product of two numbers equals 1 for that definition of product, then they are inverses

[fervent pike]

if this holds, why cant I just say that tn=1?

[burnt vapor]

Because you have to specify what tn means

[fervent pike]

well n in {1,...,m-1}

[burnt vapor]

For example, if m=11 and n=8, we can choose s=-5 and t=7, then sm+tn = -55+56 = 1
Then tn = 1 (mod m) means 7×8 = 1 (mod 11). You can't just say that this implies 7×8=1

Because 7×8 is 56, not 1

[fervent pike]

ohhh I can just say m divides tn-1 and 1-1 =0, so you have remainder of 0

so thenn tn=1

[burnt vapor]

But when you write (mod m) at the end, you are specifying that the product you are using is the one that is defined in the question

[fervent pike]

so how can I show that tn = 1mod m satisfies field axiom?

[burnt vapor]

It is the field axiom

The field axiom is: for all n (except for 0), there exists t such that n times t equals 1

[fervent pike]

yea

[burnt vapor]

If tn=1 (mod m), then the product n times t equals 1 because of the definition of the product

But if you just write the equation tn=1 without any more context, you are saying that tn=1 with the usual product

[burnt vapor]

You should still add what nt means at the end

The problem is that there are two products, but they are written the same way

[fervent pike]

what?? i’m so sorry but i’m very confused

can you write it out?

[burnt vapor]

Just write
nt = 1, where nt denotes the multiplication modulo m

[burnt vapor]

Look at this example

nt = 56 with the usual product
nt = 1 with the product mod m

[fervent pike]

ok! sry i have to leave quickly for smth

but i’ll be back

[cedar kiln]

@fervent pike Has your question been resolved?

[fervent pike]

ohhh i see now

yup ok thank you!

.close

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[violet trout]

I know that this can be written as (w-1)(1 + w + w²... + w⁶)

Not sure what to do after

Looked at the answer, confused by this part

Yeah I am confused overall with this

Can someone explain:
a) How we can say w⁷ - 1 = 0 implies (1+w + w² ... w⁶)(w-1)

b) How it is factorized
c) The remaining steps essentially

<@&286206848099549185>

I just dont understand how any of this works

despite spending some time on it

Help will be appreciated

[warm crescent]

w = 1 is an obvious solution

then divide w^7-1 by w-1

nevermind

I didn't read everything

[violet trout]

ok but what about the rest

and how do I solve this

[warm crescent]

rewrite 1 in exp complex form

[violet trout]

1 is e^(i π)

wait no

that's -1

[cedar kiln]

@violet trout Has your question been resolved?

[violet trout]

No

[violet trout]

,close

.close

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[north smelt]

Why they put “ +0x “ ?

[slate lintel]

becuase they're about to do long division

[north smelt]

❓

[slate lintel]

polynomial long division

[north smelt]

Yeah that’s the lesson….

Doesn’t explain 0x

[slate lintel]

well you should watch what they do

[north smelt]

Ok

[north smelt]

Ok I did and it’s to line everything up correctly but where did 0x come from

[slate lintel]

it's to line everything up correctly

[north smelt]

Or can I put it there whenever I need to line up correctly

Free to use ?

[slate lintel]

since +0x is just +0

you can add 0 to anything and it stays the same value

[north smelt]

Ohh

Wow nice

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[willow haven]

The base function 𝑓 ( 𝑥 ) = √ 𝑥 is transformed by a vertical reflection in the 𝑥 - axis, followed by a vertical stretch by a factor of 2, a horizontal compression by a factor of , 1/3 , a horizontal translation of 2 units to the left, and a vertical translation of 1 unit down.

can someome see if my answer is right
i got -2/3(√x+2)^2-1
am i correct

[slate lintel]

i see √x unmodified in the center there which makes me think no

also that ^2 which seems to come from nowhere

[willow haven]

in vertex form isnt it supposed to be like that

[willow haven]

wdym unmodified?

[slate lintel]

i mean it's not like √(17x + 3)

[slate lintel]

vertex form
that applies to quadratic functions, which you don't have

[willow haven]

oh ur right

[willow haven]

mb im in grade 10 trying some grade 11 questions 😭

in that case i dont think i need to worry about this

thanks tho

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[shadow carbon]

Please help me get a. I need help

[shadow carbon]

Im really lost.

I think it’s
y = mx + 35.4 but idk what the other numbers r

(0, 35.4)
(10, 33.6)
(20, 28.3)
(30, 27)
(40, 22.4)
(50, 20)

<@&286206848099549185>

Sorry I’m in a huge rush

[clear umbra]

just find the slope

you have a bunch of points

[shadow carbon]

Each slope is different though

[clear umbra]

use the first and last point

[shadow carbon]

y =-0.3x + 35

Is that right

[clear umbra]

yeah

[shadow carbon]

Thanks

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[vale musk]

this should be a rlly easy question to answer for some reaon it's showing incorrect

[vale musk]

am i just stupid

[supple flume]

(1/5)^-1 = 5

(-11/23)^0 = 1

(1/2)^2 = 1/4

5 - 1 - (1/4)/-2

4 + 1/8

[vale musk]

fuck pemdas

[supple flume]

everything right except you needed to divide first lol

[vale musk]

yeah ik forgot abt that

thanks

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[maiden harness]

can someone help me solve these type of questions

[crimson sedge]

have you tried anything ?

[cedar kiln]

@maiden harness Has your question been resolved?

[maiden harness]

i was able to start off question a by useing trig identities but dont know how to solve it

[crimson sedge]

okay then share your progress here

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@maiden harness Has your question been resolved?

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[dreamy fern]

how can i prove this

[dreamy fern]

?

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@dreamy fern Has your question been resolved?

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[native elm]

Hey I got a nitpicky question about something that seems skipped over in Folland's real analysis book

[native elm]

Basically I'm just asking about prop 2.16, because it makes sense to me but after it is proven the book basically seems to refer to this prop as if the statement is f=g a.e. iff their integrals are equal, and it seems to m that makes sense but I'm trying to figure out the simple explanation of how to get from 2.16 to that

(thats why I included 2.17 because it kind of uses it that way)

It also refers to 2.16 as giving us that we can modify a function on a null set and the integral stays the same which seems to be what I stated earlier.

L+ is measureable fcns to [0,inf]

Guess where I am mostly confused is, it seems like we can get this result pretty obviously by replacing f with f-g in the theorem statement but that implies f-g would need to be in L+ so g<f

[crimson delta]

well, g<=f

[native elm]

Right

[crimson delta]

2.17 doesnt even seem to use 2.16

[native elm]

The first and last equalities in the last part of it does, right?

which is why they show that the difference is 0 a.e. for those two functions

[crimson delta]

oh right

ok thats what they mean

[mental trail]

yeah they apply 2.16 to f - fchiE

[native elm]

Which this one makes sense I guess because fn increases to f so f-fn would be nonegative

[mental trail]

but that's about it

[crimson delta]

if they wanted to be precise they would need to show that the differences are in L^+

[native elm]

Yeah and I was reading ahead, they straight up prove what I was looking for in the next section so I think they were just using that temporarily in L^+

[crimson delta]

I mean int f=int g iff f=g a.e. is completely false

[mental trail]

they kinda skipped the fact that f - fchiE >= 0 everywhere but that's because f >= fn on E for all n

[crimson delta]

well, one direction is

[native elm]

Oh is it not true?

[mental trail]

oh yes we need f <= g or g <= f as a presupposition

[native elm]

Oh at least in L^+ it is

[mental trail]

that's why f in L+ is important

[crimson delta]

otherwise its very easy to come up with counterexamples

[native elm]

Alright yall I think this helped me out, I'll go ahead and close it

.close

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[twin panther]

Find the remainder when 3737373737373737373 is divided into 36

[slate lintel]

!status

[cedar kiln]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@twin panther Has your question been resolved?

[twin panther]

i think we should do it with mod but i am not sure @slate lintel

[slate lintel]

just do the long division

[cedar kiln]

@twin panther Has your question been resolved?

[raven shard]

what's the remainder when 3737373737373737373 is divided by 9?

what about 4?

you can uniquely determine the remainder (mod 36) with these two values

[cedar kiln]

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[wind nexus]

hi

[wind nexus]

https://i.imgur.com/0n7pLY6.png

I'm confused about the last line

its clear how we got to (nb/n)p(b/n)

I don't see how that is equal to b^(p+1)/n

holy shit nevermind its not times p

its to the power of p

freaking font

.close

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[rotund dawn]

Am i right in saying that if one value is directly proportional to another, then whatever happens to one value then must happen to the other?

[ancient lodge]

I think you have at least kind of the right idea, but your statement needs to be more precise

can you give an example of what you're talking about

[rotund dawn]

so if y=10 and x=-2
when the value of x=15 then y=-75?

[ancient lodge]

so y=-5x

so do you mean to say "If I multiply one quantity by some amount, the other quantity must also be multiplied by the same amount"?

[rotund dawn]

Thats what im assuming

[ancient lodge]

yeah that's right then

let's say you multiply by k, then you have x become kx

and substituting that in, y=-5(kx) = (-5x)(k) in this case

See if you can justify it in general (the logic is very similar)

[rotund dawn]

okay thank you, i'll work with that for now

.close

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[latent rover]

How would you solve a following problem:\
Suppose there's 5 ports numbered from 1 to 5 and three days. Ships that travel between ports in these three days can spread the epidemic. Rules of contracting epidemic: \

1. the epidemic emerges by itself only once during the 3 days in one of the ports with equal probability \
2. an adjecent port contracted epidemic day before. in this case the probability of contracting epidemic is $\frac{1}{2}$\
3. port stays infected after epidemic was contracted previous day\

On the third day epidemic was detected in 1st port. On day 1 and 2 there ware no checks, so we don't know where epidemic emerged and when.\
a) what is the probability of epidemic emerging in port 2 on the first day\
b) what is the probability of port 4 being infected with epidemic

[wraith dagger]

ta

[slate lintel]

make a tree diagram ig

[latent rover]

what would that diagram look like

The answer is supposed to be 1/6 and no matter what I try I can't reach it

[slate lintel]

basically every level is a day

[cedar kiln]

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[latent rover]

.close

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[quasi imp]

what does forward and back slash mean in set theory? smthng like
A/B , A\B

[crimson sedge]

hey

it means division

[quasi imp]

Hi

[quasi imp]

How r we supposed to divide set A from set B

[crimson sedge]

(A)/(B)

[quasi imp]

i thought its smthng to do with addition subtraction of sets

[crimson sedge]

Hmmmmmmmmmmmmmm, n o.

/ means divide

[quasi imp]

and what does backslash A\B mean

[crimson sedge]

Ummmm

It means:

B(A-B)^B(-A)

Since thats the opposite of forward slash

[wheat ocean]

A\B means difference of sets probably

[quasi imp]

A-B right?

[quasi imp]

pardon?

r u messing with me or smthng

[wheat ocean]

not exactly

A\B is notation for set that contains all elements from A that isn't in B

[crimson sedge]

???no

this is how to do math

It’s very simple

[wheat ocean]

if you talk about notation, then difference of sets can also be written as A-B, yes

[quasi imp]

hmm and what about A/B is it B-A ?

[cedar kiln]

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[spring gulch]

Hi i have a problem determing if this sentence is true : if corr(X,Y) >0 then : D2(X+Y) >D2(X-Y)

[spring gulch]

<@&286206848099549185>

[cedar kiln]

@spring gulch Has your question been resolved?

[spring gulch]

<@&286206848099549185>

[crimson sedge]

What do you mean by corr and D2?

And is your bday by any chance on October 18th?

[spring gulch]

no its not

[spring gulch]

corr is corelation and D2X is D^2X and its variation

i think the other symbol for it is σ2

[crimson sedge]

Sad

[crimson sedge]

Okay it went over my head, mb

[spring gulch]

why tho? its such a specyfic question

[crimson sedge]

Idk i saw 1810 and asked you so

[spring gulch]

ooooo nah it was my old discord number b4 they changed the way the nicknames worked, i got used to the numbers so i kept it like that

[crimson sedge]

Aye aye

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@spring gulch Has your question been resolved?

[cedar kiln]

@spring gulch Has your question been resolved?

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@spring gulch Has your question been resolved?

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[coral jewel]

Given the equation $f(x)=x^4-4x^2+m$. There exists a value of m such that f(x) intersects with the x-axis at 4 separate points and the total areas formed by the x-axis and the graph above the x-axis is equal to the area below the x-axis. The m value is a simplified fraction under the form $m=\frac{a}b$. Find a + 2b

[wraith dagger]

FungusDesu

[coral jewel]

the original question is in vietnamese, so this is my best translation

i find that 0 < m < 4 in order for the equation to have 4 real roots

from there im stumped

[cedar kiln]

@coral jewel Has your question been resolved?

[coral jewel]

<@&286206848099549185>

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@coral jewel Has your question been resolved?

[dire geode]

Use discriminant

And quadratic formula on u=x^2

[cedar kiln]

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[static sorrel]

what is x 😭

[peak minnow]

the figures are similar so that means the ratios of the sides should be same

[static sorrel]

ik

im so confused on how

[peak minnow]

x : x+3 = 18 : 20
solve the rest

[static sorrel]

um

how tho

is it 27?

[peak minnow]

u can just change them into fractions and solve
$$\frac{x}{x+3} = \frac{18}{20}$$

[wraith dagger]

JustToPro

[static sorrel]

?

[peak minnow]

no

[static sorrel]

oh

[peak minnow]

actually yes

[static sorrel]

beettt

thx

how do i close this thing

i js got here

[peak minnow]

".close"

[static sorrel]

.close

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[static sorrel]

ita not closing

wtv

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[digital star]

Sorry for two questions today, I'm not really sure how to start with this question?

[flint plinth]

what does the denominator look like when n is large

[digital star]

Would the best way to go forward multiplying by the comp?

Infinity?

[flint plinth]

not that large

large enough that the 1 is irrelevant

but still finite

[digital star]

I'm not sure I really follow

[slate lintel]

if n is really big then it's wayyyyy bigger than 1

and n^4 is definitely way bigger than 1

so we can just pretend it's not there when we're thinking about big n

[flint plinth]

like say n is 100
then n^4 is 100000000
and n^4 + 1 is 100000001

the n^4 part is dominating

[digital star]

So I could say n = 0 in this case?

[flint plinth]

wha

n = 0 is not large

you're taking the limit as n->infinity, that's why we're considering what things look like for large n

what happens at n=0 is irrelevant for this purpose

[digital star]

Right, would you be able to give me a hint on how to apply that to the problem

Or if there's a video on Khan that would explain the concept more thoroughly?

[flint plinth]

i tried to hint above... if n is very large, what does the denominator look like

let me elaborate on that

if n is much bigger than 1, then n^4 + 1 is basically the same as n^4

the 1 only makes a tiny difference

like imagine this example but with a million zeros instead of 8
if n = 100
then n^4 is 100000000
and n^4 + 1 is 100000001

[digital star]

Does it not break any rules to discard the 1?

[flint plinth]

we can make it more formal

this is just for getting some intuition about what the limit should be

i guarantee most people think about it this way before proceeding with a more formal argument

but if you don't like this way, i can suggest another way

try dividing the numerator and denominator by n^2

[digital star]

No I would like to actually understand the concept behind this

[flint plinth]

ok

[digital star]

Knowing how to solve this equation is one thing but if I can't apply it I'm stuck in a circle

[flint plinth]

let me just proceed with that line of thinking then

we'll justify it later

i claim that for large n, the difference between n^4 +1 and n^4 is so slight that we can just pretend the denominator is sqrt(n^4), which is n^2

and then your sequence (for large n) looks like (13n^2 + 2n) / n^2

which is the same as 13 + 2/n

now what's the limit of that as n -> infinity?

[digital star]

15/n or 15/infinity?

[flint plinth]

13 + 2/n

not (13 + 2)/n

13 + 2/n means 13 + (2/n)

so it's the number 13 plus the fraction 2/n

the fraction 2/n becomes very small as n grows large

in fact it goes to zero in the limit

so what are we left with?

[digital star]

13?

[flint plinth]

yes

[digital star]

So more formally had I divided everything by n^2 it would make the 1 even less significant because it would be 1/n^2 and thus discardable?

[flint plinth]

yea if you divide everything by n^2, you will have:

$$\frac{13 + 2/n}{\sqrt{1 + 1/n^4}}$$

[wraith dagger]

Bungo

[flint plinth]

which makes it clear that when n is large, the 2/n in the numerator goes to zero and the 1/n^4 in the denominator go to zero, leaving you with 13/sqrt(1) = 13

[digital star]

Okay I think I got it

Thank you sm for taking the time to explain things, this is all very new to me rn so I'm sure my grasp on it is flimsy at best

.close

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[grave halo]

can someone help me with this problem

[grave halo]

<@&286206848099549185>

[wraith dagger]

futureneverlands

[grave halo]

you’re a life saver thank you 🙏🙏

[wraith dagger]

futureneverlands

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@grave halo Has your question been resolved?

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[dim elbow]

Is 1/1+cosx equal to 1/1-cosx?

[crimson sedge]

no

[dim elbow]

But if you had 1/1+cosx and you multiple top and bottom by 1-cox

You get 1-cosx/1-cosx^2

Then cancel and get 1/1-cosx

[crimson sedge]

so?

,,\f1{1-\m\cos x} = \f1{1-\m\cos x} \c b{\cd \f{1+\m\cos x}{1+\m\cos x}} = \f{1+\m\cos x}{1-\m{\cos^2}x}

[dim elbow]

Where is the error here?

[crimson sedge]

1- cos²x doesnt decompose like that

[clear umbra]

1 - cos^2(x) = (1+cosx)(1-cosx)

[crimson sedge]

yeah

[clear umbra]

1-cos^2(x) is not (1-cosx)^2

[dim elbow]

Ohhh I see now

Thank you

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@dim elbow Has your question been resolved?

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[waxen tartan]

What is difference between geometry and planimetry?

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[crimson sedge]

can someone help me with this differential equation?

[royal loom]

What're the points for?

[crimson sedge]

from an old exam

[cedar kiln]

@crimson sedge Has your question been resolved?

[brittle stag]

what have you tried

did you get the homogenous solution

[crimson sedge]

no

that's the point I have looked at countless examples on how to do it and I still don't understand how I do it

[brittle stag]

get the homogenous solution, the HDE is the form of euler's equation, you take y=x^r

[crimson sedge]

and this is where u lost me lol

[brittle stag]

what's the problem with what i said

[crimson sedge]

"euler's equation, you take y=x^r"

How do I do that?

[brittle stag]

you substitute y with x^r?

[crimson sedge]

xx^r = x^r + x^2sin(x)

like that?

[brittle stag]

not exactly... you have a first derivative, you have to differentiate f(x)=x^r

[crimson sedge]

so rxx^(r-1) = ..

[brittle stag]

right

also we are solving the homogenous differential equation

so r x^r - x^r = 0

[crimson sedge]

so we will set x^2*sin(x) to 0?

okay

[brittle stag]

right

[crimson sedge]

and now solve for x?

[brittle stag]

we solve for r, (r-1) x^r = 0, x^r can never be zero

[crimson sedge]

r = 1?

[brittle stag]

right so we can say y=x is a particular solution

[crimson sedge]

alr

[brittle stag]

y_h = c x

we have our homogenous solution now

we want to finish it with a particular solution, to get the general solution

you know method of variable coefficients?

[crimson sedge]

"knowing" yea

again, seen many examples

still don't know how to do it myself lol

[brittle stag]

aight, in method of variation of constants, we take c = z(x), the constants in our homogenous solution are now functions of x, we substitute y=zx into our DE and solve for z

[crimson sedge]

so x * z'(x) = z(x) + x^2*sin(x)

or wym?

[brittle stag]

you misunderstood, ig y = z(x) * x is more accurate

[crimson sedge]

ah

x * z'(x)*x + z(x) = z(x) * x + x^2 * sin(x)?

[brittle stag]

add necessary parentheses

[crimson sedge]

ah yea ofc

x(z'(x)*x + z(x)) = z(x) x + x^2 sin(x)

[brittle stag]

mhm, continue

[crimson sedge]

and now we can solve for z'(x)

which is z'(x) = sin(x)

integrating that yields -cos(x) + C

[brittle stag]

correct

[crimson sedge]

means the general solution is y = C_1 * x - cos(x) + C_2

[brittle stag]

no

remember we took f(x) = z(x) * x

[crimson sedge]

but isn't the general solution y_h + y_p

[brittle stag]

right but z isn't the particular solution

[crimson sedge]

ah yea

so

y = C_1 * x + (-x * cos(x) + xC_2)

something like that?

and then we have y = x(C_1 - cos(x) + C_2)

[brittle stag]

f(x) = (C - cos x) * x

[crimson sedge]

and simplify to just x(C - cos(x))

yes

[brittle stag]

yah but idk what you did

[crimson sedge]

had constants from hom. and part. solution

and then we merge them together

but yeah I know what i did and got the same result so everything is fine

[brittle stag]

just replace z(x) with the expression you got

[brittle stag]

don't do whatever this is

[crimson sedge]

yes alright

and now

we would need to find out what C is s.t. f(pi) = 0;

[brittle stag]

right

[crimson sedge]

which is cos(pi)

yea and then we can compute f(2pi)

thank you very much

[brittle stag]

🫡

[crimson sedge]

.close

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[wanton pecan]

Hi, I am having trouble calculating this with logs, is there a simpler way?

[candid mason]

So what have you tried

[wanton pecan]

I tried graphing it (I can't do logs)

I put the 7.6bn years, in red, then one half life, should be 1,25 (about 1 1/4 of the red) (for the first 50% of the Potasium to decay.

Then there is 50 percent (which I thought would be 10^10

Half of 1.25 bn years should be 0.62 rem 5 bn years...(going to an 1/8 of the original...

[candid mason]

I don't think logs is the issue. It is interpreting the problem

[candid mason]

This is not correct review exponent rules

[wanton pecan]

So, 10^20 / 2 = was wrong, it is :

[candid mason]

Refer to this. Answer is basicallly given in the article. no logs required. https://en.wikipedia.org/wiki/Half-life

So we have 10^20 * 2 ^-(t/t_(1/2)) where t_(1/2) = 1.25 and you let t= 7.6

[wanton pecan]

Ah, I am going to study this article, and I will show my workings when I am done. Thank you so much. I thought I was never going to get it!

[cedar kiln]

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[wanton pecan]

Yes, I got good advice and I need to go and apply it for further study.

.close

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[kindred hornet]

how would i do this? im confused on where to start

[quiet plover]

that's interesting

[dusty hazel]

Consider \
$S = \sum x_i^2 = \sum (x_i + 1)^2$
Then expand.

[wraith dagger]

! What the hell am I doing here?

[kindred hornet]

to 100 terms?

@dusty hazel

[dusty hazel]

Well you only have yo expand (a+b)^2 here and keep the summation.

And the sum splits.

[wild ocean]

I’m pretty sure this means you add 1 to the numbers before squaring them again, (A+1)^2, not afterward, (A^2)+1. Consider the case if there are only just two numbers written on the board.

For two numbers A and B, you get the equation A^2 + B^2 = A^2 + 1 + 2A + B^2 +1 + 2B, which can just apply to every number on the board arbitrarily.

[cedar kiln]

@kindred hornet Has your question been resolved?

[kindred hornet]

What do you mean by this?

[dusty hazel]

$\sum (a+b) = \sum a + \sum b$

[wraith dagger]

! What the hell am I doing here?

[wild ocean]

The equation I mentioned would show that for any two numbers A and B, the sum of their squares wouldn’t be changed if you added 1 to both of them, if B is the negative of (A+1). Adding yet another 1 to each will change the sum of their squares though, where the difference between the old sum and the new sum is simple to find by plugging in (A+2) ^2 and (-A-1+2)^2 and comparing that to (A)^2 and (-A-1)^2.

[cedar kiln]

@kindred hornet Has your question been resolved?

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@kindred hornet Has your question been resolved?

[cedar kiln]

@kindred hornet Has your question been resolved?

[kindred hornet]

what do you mean by this?

the problem is each x_i is not the same

im kind of confused

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[kindred hornet]

Anyone

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[magic solar]

i can't seem to do 3

[magic solar]

specifically, i can't find the number of arrangements where 2 second years are together and the other one is not

i tried 2 different ways but they both seem to be wrong

<@&286206848099549185>

[lucid stump]

Take.them as 1 term and use permutations

[magic solar]

i tried doing that, but it didn't work because you need to consider the cases when the 3rd second year is the next to the 2 others

which would invalidate it as an arrangement

[raven shard]

there is a general solution to this problem that one can find by starting with this line of reasoning (or something similar):

consider a line with 5 first year students. in how many ways can we place 3 second year students in this line such that no two second year students are next to each other?

[magic solar]

i dont understand, is that not just the question

[raven shard]

yes but imo it sort of reframes it in a way that's easier to understand

[magic solar]

well initially, 6 places to place the first person

then it gets more complicated, because the next person's spots depend on where the first person is: if the first person was on the very edge then the second person would have more spots to be

compared to if the first person was somewhere in the middle, then there would be less spots for the second person to be which i find hard to do

[raven shard]

the trick is that this is not actually true

[raven shard]

nice yes

[magic solar]

could you explain?

[raven shard]

with our 6 initial slots, we can see that each slot is independent of the other slots

if we have our first second-year on the very edge, that leaves us with 5 slots to put our next second-year

but if we have our first second-year somewhere in the middle, that also leaves us with 5 slots to put our next second-year

[magic solar]

ah yeah counting again, i think i just counted wrong

does the next person simply have 4 slots?

[raven shard]

yeah it goes 6 5 4

[magic solar]

ah yep i got the answer

tysm, on second thought, ur rewording is super helpful

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[kindred adder]

Yeah I don't even know where to start

[dire geode]

Well did you learn contour integration

[kindred adder]

yeah but im not sure what the best way to start is

like i found the poles

i went from there

i am doing hte limits

but im not sure if its right

this is what i did

[vagrant elbow]

probably better to use $f(z) = \frac{ze^{iz}}{9z^2 + 4}$

[wraith dagger]

NEON

[vagrant elbow]

semicircular contour

take the imaginary part of the result

because that will give you the sine part

[cedar kiln]

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[kindred adder]

hmm

okay ill try thgat

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[dusty hazel]

<@&268886789983436800>

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[lapis kiln]

I just need help with this im a bit dumb and ik its probably easy for you guys

[dusty hazel]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[lapis kiln]

1

ion know where to start

[dusty hazel]

Break it into multiple rectangles.

Then individually calculate their areas.

And add them.

[lapis kiln]

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[dry swallow]

Probability - basics

Given a bag with 2 black balls and 2 white balls

1. Person 1 picks a ball at random and throws it in the trash
2. Person 2 takes a ball at random and puts it back in
3. Person 3 takes a ball at random

Question: What is the probability that person 2 and person 3 both took black balls?

So I got 5/18. Supposedly, that’s correct. But I don’t understand why it isn’t 1/4 intuitively. Any help?

[dry swallow]

Basically I don’t understand why step 2 and step 3 are suddenly dependent when you add step 1, whereas if you remove step 1, step 2 and step 3 are independent

[nova glacier]

Because the outcome of step 2 reveals which color is more likely to be picked

Without step 1, the outcome of step 2 reveals nothing

With step one it is unkown whether you are in the case of 2 black balls and 1 white ball or 2 white balls and 1 black ball

And, for instance, given that step two is black, it is more likely that you are in the case - 2 blacks, 1 white

@dry swallow

That's my intuition.

[dry swallow]

That’s really hard to grasp, just a moment

[dry swallow]

But step two isn’t really “given” that it’s black, the question asks what’s the chance that both person 2 and person 3 got black balls in their pick

What’s given is the fact someone threw a ball out no?

[nova glacier]

You asked why step 2 and 3 aren't independent

So they aren't independent given step 1

Intuitively, one way to see that two event's aren't independent, is to see that given the first event, the probability of the second changes...

And here we are looking at the events of step 2 and 3

Now given that step 2 is black

the probability of step 3 to be black gets larger

So step 2 and 3 aren't independent

[dry swallow]

When you say it like that it makes sense

And still, the answer to the original question still feels like it should just be 1/4. Is this just really unintuitive or am I lacking some fundamental knowledge?

[nova glacier]

I don't see why it feels like it should be 1/4

Here, let me show you why I think intuitively it shouldn't be 1/4

without throwing anything out you have equal probabilities for (b,b),(b,w),(w,b),(w,w)

If you throw a ball out

You are more likely to have two same colored balls

so the outcomes of (b,b),(w,w) are more likely than (b,w),(w,b)

It skews the probabilities

[dry swallow]

Huh. Okay that’s a new perspective I haven’t seen

I’ll contribute it some more thought, thank you very much. I’ve been bashing my head on this for about an hour now lol

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[nova glacier]

You are welcome

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[digital umbra]

<@&286206848099549185>

[livid hound]

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[crimson sedge]

I need help. (Arithmetic Progression Series)

1. Given that the first and second terms of an arithmetic sequence are 6 and 12 respectively and the last term is 162, find the sum of the whole series.

[crimson sedge]

I know the formula Sn = n/2(a+l)

i have a(first term) and l(last term) but im struggling to figure out how i could find the total number of terms(n) given this information

[livid hound]

do you know the general expression for the nth term of an AP

[crimson sedge]

yes a+(n-1)d

[livid hound]

are you able to determine d?

[crimson sedge]

yes its 6 because the first term is 6 and second term is 12

[livid hound]

set the general expression equal to 162
plug in your a,d
solve for n

[crimson sedge]

ah thank you

i got it

when my teacher done it in class, they done it in a different way however

he done

162 - 6 = 156

156 / 6 = 26 gaps

n = 27 (26+1)

s27 = 27/2[6+162]

[twin ivy]

that is faster way to get answer :)

[livid hound]

that's literally the same

[crimson sedge]

ok

[livid hound]

just more broken up

[crimson sedge]

but when they divided 156 by 6

[livid hound]

so its actually harder to read/follow

[crimson sedge]

is that 6 the first term or common difference

[livid hound]

which demonstrates my point

[crimson sedge]

lol

[livid hound]

highlighting their steps in bold
starting with
a_n = a+(n-1)d
162 = 6 + 6(n-1)
subtracting 6 from both sides:
162 - 6 = 6(n-1)
156 = 6(n-1)
dividing both sides by 6
156/6 = n - 1
26 = n - 1
adding 1 to both sides
26 + 1 = n
27 = n

most of that don't really need to be written for efficiency

[crimson sedge]

thank you

[livid hound]

162 = 6 + 6(n-1)
156 = 6(n-1)
26 = n- 1
n = 27
starting witht he full equation, instead of unnecessarily splitting that into smaller sections
its easier to keep track of where n is, what's happening to the equation

[crimson sedge]

yeah

[livid hound]

something like: \
$y = 3(1 + 2 + 4)( 7 + 3 \times 2) \
1 + 2 +4 = 7 \
3 \times 2 = 6 \
7 + 6 = 13 \
3 \times 7 = 21 \
21 \times 13 = 273 \
\therefore y=273$ \
is extremely hard to follow

[wraith dagger]

ℝαμΩℕωⅤ

[crimson sedge]

yeah you're right

i think he used that example because he wanted to show us the number of "gaps" between each term

like methodically

[livid hound]

its essentially what the general expression does

structuring work like that is more suited if there's a diagram showing what the calculations are intended to do

[crimson sedge]

right

ok thank you, understood i made my own question with smaller numbers to help me understand it and i got it

!done

[cedar kiln]

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[crimson sedge]

.close

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[abstract pollen]

Prove $ln(x)$ is uniform continuous on $[1,\infty)$

[abstract pollen]

But the caveat is I have to do it without any results that come from differentiation

$e^x$ is defined as $$\lim_{n \to \infty} (1+\frac{x}{n})^n$$

[wraith dagger]

Xetrov

[abstract pollen]

and $\ln$ is the inverse of $\exp$

[wraith dagger]

Xetrov

[mental trail]

uniform continuous?? on which domain?

[wraith dagger]

Xetrov

[abstract pollen]

i'm trying to first prove lipchitz continuity

[abstract pollen]

sorry, fixed

But I also hit a roadblock when I'm trying to prove that $|\ln x - \ln y| \leq |x-y|$

[wraith dagger]

Xetrov

[mental trail]

start from $e^x \geq ...$

[wraith dagger]

rafilou2003

[abstract pollen]

i'm assuming that's 1+x

[mental trail]

correct

[abstract pollen]

I don't know how to prove that without differentiation

is the issue

so no taylor series

[mental trail]

(1+x/n)^n >=...

do you need more hints?

[abstract pollen]

yes please

[mental trail]

expand

and suppose x>=0

[abstract pollen]

binomial theorem

[mental trail]

yes

is it clear enough?

[abstract pollen]

oh yeah

thank you so much

i was just writing everything down

[mental trail]

k

[abstract pollen]

that was the roadblock i'd hit

cheers

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[paper cloud]

This is a small question but how come -1^2 is -1 even though -1 times -1 is 1?

[paper cloud]

Oh ok, thank you

Got it 🙏🏿

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[hollow trail]

let's say i have a coin, and i want to test whether it's "fair" (if you flip it it has a 50% chance of heads), or it's somehow unfair (it has some different chance). You can flip it a whole bunch of times, and record the result. Note that if it was a fair coin, you can calculate that it's extremely likely that if you flipped this coin 100 times, it would get somewhere around 50 heads. so if you were to flip it 100 times and got 97 heads, then it's technically possible for that to happen with a fair coin, but so unlikely that it would be more reasonable to conclude that it is actually unfair.

[hollow trail]

(note that we would usually consider both the chance of getting 97 heads and the "more extreme" cases of 98 heads, 99 heads, and 100 heads with a "fair coin" at the same time)

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[lapis arrow]

compute the integral of f(x.y) = (ln y)^-1 over the region D:
D bounded by the cures y=e^x, y=e^square root(x), y=1 and y=2

[lapis arrow]

<@&286206848099549185>

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[lost osprey]

How do I do this?

[flint plinth]

do you know what (f + g)(-1) means?

[lost osprey]

I have a small understanding of this, I just wanted to make sure I'm doing it right.

[lost osprey]

So, since it says (F+G)(-1), that means I take that (-1) which is a X coordinate. That would be 1, and 4 on the Y-Axis. So, I would add 1 and 4 and then get 5?

[flint plinth]

yes

(f+g)(1) means f(1) + g(1), which means add the value of f at x=1 to the value of g at x=1

[lost osprey]

Alright, thank you!

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[crimson sedge]

@tacit spade yer killin me man

[crimson sedge]

😭

@flint plinth 𝑃(𝐴∩𝐵)= 5/36 * 5/36 since they are not mutually exclusive

[flint plinth]

wait what happened to the other channel

[crimson sedge]

some guy and i opened it at the same time. he typed in it again and it just shut it randomly

[flint plinth]

oh, haha

ok

[flint plinth]

and because they are independent

that's the condition for P(A ∩ B) = P(A)P(B)

ok, how about the union

(d)

[crimson sedge]

ok so P(𝐴∪𝐵)= P(A)+P(B)-𝑃(𝐴∩𝐵) = 0.1389*2 - 0.0193 = 0.2585

[flint plinth]

yep

[crimson sedge]

letsgoo theyre all right

https://tenor.com/view/lets-go-ready-baby-charge-up-revved-up-gif-14820405

[flint plinth]

nice!

[crimson sedge]

thank you sir

[flint plinth]

yw, cheers!

[crimson sedge]

you are a real one

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[queen jackal]

What have you tried so far?

[cedar kiln]

@crimson sedge Has your question been resolved?

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@crimson sedge Has your question been resolved?