help-49

[midnight plank]

@half tangle Has your question been resolved?

[half tangle]

How do I show the square of the norm of p is less than 4 with the right bound?

[midnight plank]

@half tangle Has your question been resolved?

[mortal falcon]

do you have to?

The min enforces both simultaneously

[half tangle]

C = {(x,y): 2< x²+y² < 4}

|p-(x,y)| < min(a,b)

Don't I have to prove that p ∈ C using one bound at a time?
|p-(x,y)| < a
|p-(x,y)| < b

where:
a = 2 - |(x,y)|
b = |(x,y)| - √2

[mortal falcon]

like, the min means both bounds are simultaneously true

So the implications of either bound are also simultaneously true

The min doesn’t generate a union of sets, it generates an intersection

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[regal haven]

can someone help me break this down

[mortal falcon]

Is there any algebraic step you’re confused by

[regal haven]

where they got this

i understand M/2

[mortal falcon]

If you expand the absolute value

-|M|/2 < g(x) - M < |M|/2

add M to all inequalities

[regal haven]

oh wth

what is the intuition behind that

[mortal falcon]

erm

[lavish venture]

the point is that g is bounded in some neighborhood around a

[mortal falcon]

Wait ugh does M have to be in an absolute value

[lavish venture]

because M could be < 0

[mortal falcon]

right 💔 I’m so bad at triangle inequalities there’s definitely a way to just use one here

[regal haven]

so $M - \frac{|M|}{2} \leq g(x)$

[lavish venture]

remove the space

[mortal falcon]

it’s just g(x) on the RHS

[lavish venture]

why is there M still?

[regal haven]

but don't you have to add M to the whole thing

[mortal falcon]

We had g(x) - M in the middle

[lavish venture]

yes and before it was g - M

[regal haven]

oh wait yea

M/2

okay so this is because g(x) around a

so it is the same distance between M/2 and -M/2

[grand pond]

Faduzzle

[regal haven]

i know i was trying to find it but i couldnt

[mortal falcon]

That still isn’t really the right form ugh

[lavish venture]

i mean why did you add the =

[mortal falcon]

knief will save the day

[lavish venture]

,av snowflake0180

[grand pond]

snowflake0180's Avatar

Click here to view the image.

[mortal falcon]

help

my friend drew that for me in 9th grade physics

[lavish venture]

william eyelash

bad boy?

or coy?

[mortal falcon]

bad guy

[lavish venture]

damn his handwriting sucks

[mortal falcon]

Billie eilish 😭

It was a whiteboard 💔

[lavish venture]

oh

makes sense

[regal haven]

from delta epsilon def

[lavish venture]

i mean it doesn’t really matter i was just curious

the epsilon delta definition can use <= or <

same result

[regal haven]

yea, just wanted to be consistent wth the prof

[oak slate]

👍

[lavish venture]

you haven’t been here in ages bruh

[regal haven]

i took a quiz studying from the reccommended textbook and got the question wrong bc i didn't use the =

[lavish venture]

🤔🤔

[regal haven]

and his notes uses the =

[lavish venture]

like on a graded assignment

[oak slate]

forgot math

[regal haven]

yea

[lavish venture]

bro complain

wtf

😭

[mortal falcon]

Help

real analysis is so cursed

I’ve never had a good real analysis experience

[lavish venture]

not too bad

what book did you use

[mortal falcon]

lebl

my class wasn’t cursed bc of that though

it was cursed bc my prof turned it into a writing credit class

[lavish venture]

🤔

[mortal falcon]

And we spent half the class doing writing prompts

🥀

[lavish venture]

lol what

like what

[mortal falcon]

I promise you whatever you’re imagining it was worse

We had to explain analysis concepts and theorems to a 12 year old audience in 3 or less sentences

[lavish venture]

🤣🤣

[mortal falcon]

but if we missed anything important or wrote too much we would lose so many points

[lavish venture]

good lord

[mortal falcon]

a third of our homeworks every week were writing and the TAs were so harsh on the writing

I learned so little in that class

And then we had a pop final

[lavish venture]

dude watched too many of those youtube shorts

[mortal falcon]

with 4 questions that no one knew how to do bc we literally didn’t do any analysis

one of the questions was just “prove l hopital”

[lavish venture]

measure of understating is the ability to explain it to a 5 year old bullshit

[lavish venture]

cauchy MVT

[mortal falcon]

Mind you we spent maybe 1 minute on Cauchy mean value theorem and never used it in any context after that

[lavish venture]

😭

[mortal falcon]

instead I had to explain what Cauchy convergence means to a plumber

[lavish venture]

💀💀💀

[mortal falcon]

I’m pretty sure I answered 2 of the 4 questions on that final

[lavish venture]

yea i don’t blame you tbh

[mortal falcon]

And I still got an A+ so I wonder how everyone else did

[lavish venture]

professors make or break a subject

[mortal falcon]

He didn’t even give us a syllabus 😭

And we had to do these massive portfolios

Again explaining and working through proofs targeted at non math audiences

And then I had to write a paper and give a 20 minute presentation on improper integrals 🥀

And I had to do journal entries every week about my feelings towards real analsysis

[lavish venture]

😭

dawg this guy is evil

[mortal falcon]

They were very negative

[spiral rock]

this is a nightmare

[lavish venture]

is this a joke

[mortal falcon]

it is not 🥀 i hate that class so much and i came out of it with 0 real analysis understanding

[lavish venture]

many people change majors after analysis

at my school at least

my whole class joked about becoming engineering majors

[spiral rock]

In my university real analysis and linear algebra are kind of meant to filter some students

[mortal falcon]

I dropped from honors to regular bc I hated rudin and I honestly regret that decision so much bc rudin would have been better than that mess

[lavish venture]

yea they purposely make it scary

[lavish venture]

rudin isn’t so bad but there are better

[mortal falcon]

yeah I know people like it a lot but I just got overwhelmed very fast by how constants got pulled out of thin air 🥀

but I didn’t realize that’s just what analysis is 💔

I think I’m in a less hateful mindset now

I also like analysis a lot more after taking topology

[lavish venture]

my course focused a lot on metric spaces

[spiral rock]

I learned topology by myself

they didn't let me take the course at the only time I could because it was full

[mortal falcon]

We didn’t even touch topology in my analysis class

I didn’t know what a topology was until I took topology

[scenic wyvern]

um folks, might wanna see if OP has another question

[mortal falcon]

oh true

[lavish venture]

@regal haven

he’s gone

[regal haven]

hi

[spiral rock]

do you have any more questions?

[regal haven]

i too am comtemplating my math major

where did they get the epsilons

[spiral rock]

yeah it seems like they pulled these out of nowhere

[regal haven]

great

[spiral rock]

but usually we guess $\varepsilon_{1}=C * \varepsilon$ and then when we continue we understand what C needs to be

[grand pond]

ExpertSqueeSQUEE

[regal haven]

wait i think i got it

is it to control it to be less than epsilon

[lavish venture]

do you have a second major

[regal haven]

yea econ

[lavish venture]

stick with it then

math + econ is good

[regal haven]

yea like 6 classes overlap

wait more

but it is an extra two semesters

but anyways

to control it to be less than epsilon

make it less than epsilon/2

first term $< \epsilon/2$

[grand pond]

Faduzzle

[regal haven]

and $\epsilon/2 < \epsilon$ so the inequality holds and the def is true

[grand pond]

Faduzzle

[spiral rock]

well you want both terms here to be less then epsilon/2 so their sum is less then epsilon

sry man but I need to go

cya

[midnight plank]

@regal haven Has your question been resolved?

Channel closed

Closed by @regal haven

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[real minnow]

is this like actually useful

ive never used it but it's in my textbook

[scenic wyvern]

this is completing the square

normally you learn this when you have not learnt the quadratic formula

but when you have, you normally use that instead since it's much faster

still, learning this is good, because then you can see how the quadratic formula came about

[real minnow]

ok

.close

[midnight plank]

Channel closed

Closed by @real minnow

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[scenic wyvern]

what's the question? can't read spanish

[signal wadi]

Let A be a set. Describe all relations on A which are simultaneously:
i) symmetric and antisymmetric
ii) equivalence relations and orderings

(i did not get whether they mean total or partial ask Renato)

[scenic wyvern]

the first one sounds like you can force your way to the relation

by considering what symmetry means, then what antisymmetry means

actually, where are you stuck at?

[tidal turret]

partial

poset

[scenic wyvern]

have you done i)

[tidal turret]

i am trying

symmetric relations are such that x R y => y R x

[scenic wyvern]

use this definition. then, apply the antisymmetry definition on top of this

what does antisymmetry say if xRy and yRx are present?

[tidal turret]

x R y, y R x => x = y

[scenic wyvern]

so that means x must be equal to y, always

there's only one kind of relation this can be

[tidal turret]

?

[tidal turret]

reflexive

[scenic wyvern]

well, reflexiveness can come later

actually

you can describe the relation using reflexiveness, yes

but the relation itself will be a subset of all the reflexive pairs in A

[tidal turret]

so?

whats your point?

[scenic wyvern]

my point is that you've already arrived at the relation definition

at least for i)

[tidal turret]

how?

[scenic wyvern]

if R is both symmetric and antisymmetric and x != y (x is not equal to y), then the presence of xRy forces the presence of yRx (symmetry), which would then force x = y by antisymmetry (so the x != y becomes a contradiction)
so the only kind of relation this can be is a relation that maps an element to itself

[tidal turret]

wdym?

[scenic wyvern]

which part do you not get?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

how to formalize this argument

[scenic wyvern]

let A be a set and $R \subseteq A \times A$ be a relation

my god

[grand pond]

Hanako

[tidal turret]

then?

[scenic wyvern]

then if R is both symmetric and antisymmetric, and $x \neq y$

[grand pond]

Hanako

[scenic wyvern]

continue from here

based on what i said

[tidal turret]

how?

[scenic wyvern]

right here

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

how do I join all the pieces together for the proof?

[prime hornet]

there's not much else one can say without giving away the entire solution at this point

[prime hornet]

hm, I can give you an outline and ask you to fill in the blanks, maybe

on second thought, it'd probably be more effective if you just went and strung the argument together yourself

you'd definitely learn more that way!

do give it a shot!

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[graceful drum]

Five men and nine women stand equally spaced around a circle in random
order. The probability that every man stands diametrically opposite a
woman is m/n , where m and n are relatively prime positive integers. Find
m + n.

[graceful drum]

is my reasoning correct:

if we assume all men and women are distinct

then there are a total of 13! possible circular permutations

now there are 5 pairs of diametrically opposite men & women

4 women left

10 spaces for them in between the pairs

so 10^4 arrangements

therefore the probability is 10^4/13!

[quartz heart]

The number of rearrangement is going to be = no.of men + no of women|| (i e. 14) ||

Since it's circle so first person will have only 1 chance to sit there and rest will choose to sit on the basis of how much seat have left like :

||1*13*12*11*10*... = 13! ||

Here each value is how many seat left to choose from

So total no of arrange ment will be || 13! ||

I don't know what to do after this

[graceful drum]

first of all no of men + no of women is 14

and for the seccond part the arrangements matter

[quartz heart]

Oh mb

[graceful drum]

can you please check my argument?

[quartz heart]

1st argument is correct and I am not sure about the second one

[graceful drum]

<@&286206848099549185>

these kinds of things are always tricky

[worldly trellis]

WHY ARE YOU SO SWEET WHAT

Anyways where is the question

[graceful drum]

Five men and nine women stand equally spaced around a circle in random
order. The probability that every man stands diametrically opposite a
woman is m/n , where m and n are relatively prime positive integers. Find
m + n.

[graceful drum]

also the question is always in the pinned messages

[worldly trellis]

That sounds like one of the question i would have lmao but yeah

You can do this easily

Just draw it out and

See whats its asking for

[worldly trellis]

Yeah alr! Thamks

[graceful drum]

i had a solution but its probably wrong

[worldly trellis]

Mind showing me?

[graceful drum]

this ^

[worldly trellis]

Il tell you where you went wrong

[graceful drum]

@worldly trellis this message

10^4/13!

[worldly trellis]

OH ALR

Are you familiar with basic permutation and combination?

[graceful drum]

yeah.

[worldly trellis]

Represent this in the form of a combination

[graceful drum]

5! ?

[worldly trellis]

NOPE

Ok

Hear me out

Since each man must be opposite to a women

We would pair 7 people

Out of which 5 pairs

Are men women pair

And rest are women women pairs

Now since we have 7 pairs

To chose 5 pairs from those 7 pairs

What we would do is

[graceful drum]

wait wait 7 men?

[worldly trellis]

5c_7

Pair*

[last gate]

theres 7 pairs of people

[graceful drum]

7 c5 right?

[worldly trellis]

Yes

[last gate]

u can choose 5 male women pairs

then arrange them in 5! ways

so 7C5 * 5!

[worldly trellis]

Yup purmute them

Now

[last gate]

yea basically then i think times 2^5 as well because each side could be a male or a women so 2 options for each pair

[worldly trellis]

There is this trick

To sove circular permutation

Arrange all 14 people in a circle and

And fix one person to remove rotation symmetry

So 13!

[graceful drum]

one moment

[worldly trellis]

Ways to arrange

[graceful drum]

the women pairs dont need to be diametric right

[last gate]

then remaining 4 women fill the remaining 2 pairs

[graceful drum]

ach this is confusing

[worldly trellis]

The answer is most certainly 509

[last gate]

so overeall the woman arrangements would be 9!/(5!*4!) but then 4! remianing positions for the other 4 woman so 9!/5!

[worldly trellis]

Is kinda is

[last gate]

then the total is what evix said 13!

[graceful drum]

can someone correct my reasoning:

first theres 7 c 5 pairs

[last gate]

or u can fix a male and women pair

[graceful drum]

and we can arrange these pairs in 6! ways

in a circle

[worldly trellis]

Piney takeover please i have classes

[graceful drum]

theres 10 gaps

[worldly trellis]

Sorry cherry man he got this thoe

Cya

And you got this too!

[graceful drum]

👍

[last gate]

yea

[graceful drum]

actually @last gate can you explain your whole procedure again

[last gate]

hm ok

i think a better way fo thinking about this

is fixing a male and woman pair

to break the symmetry

so then there are 4 males left

and 8 women left

so we can pace the 8 women in 8p4 ways (since order matters since they different)

[graceful drum]

bruh theres 5 men and 9 women

[last gate]

and then for each of those placmeents theres gonna be one man and we can arrange the 4 men there in 4! ways

no im fixing one pair

so one men and one woman is gone

[graceful drum]

oh

[last gate]

do u understand so far? or which part confusing

[graceful drum]

i can understand

[last gate]

ok

then after placing all the remaining men

we already picked 4 women from the 8p4 expression

so therers 4 women left to arrange

in 4 places

so 4!

so it should be like (8p4 * 4! * 4!)/(13!)

[graceful drum]

hm

what if we tried to find the number of pairs first

so 9c5*5!

no?

or

[last gate]

uh

there would be 5 pair

[graceful drum]

oh i mean possible sets of 5 pairs

[last gate]

so ig choose the woman for the pair in 7c5 ways

since u fix one

[graceful drum]

no im not fixing

[last gate]

why

[graceful drum]

i want to try this way

[last gate]

bruh

u have to fix

cause for circular arrangements that will overcount otherwise

[graceful drum]

9*5 ways to fix one pair then

[last gate]

no when u fix u can choose it arbitrarily

so no need to consider order or anything

just place one man and one women and consider whats left and how many ways can i arrange

[graceful drum]

uhhh

[last gate]

where u stuck

[graceful drum]

ok can you just outline the general method you are using

like not the calculations

[last gate]

yes

[graceful drum]

just the steps you are taking

[last gate]

1. fix one pair of men and women

2. consider how many spaces are left and then choose 4 women to place there

3. arrange the 4 men diametrically opposite to those into those spaces

4. arrange the remaining women

[graceful drum]

ok this makes sense

[last gate]

i choose 4 women and not 8 cause i wanna get rid of all the men

[graceful drum]

ok now how to calculate

[last gate]

ok

so after fixing

theres gonna be 8 women left

and 4 men left

[graceful drum]

yes

[last gate]

there will be 12 spaces left on the table

since 2 are already seated

so how many ways can i pick 4 women to sit (order matters)

[graceful drum]

8p4*4!

i mean 8p4

for the first part

[last gate]

yes

[graceful drum]

then *4! for the men

[last gate]

eys

then for the remaining women?

(their positions are fixed so its just the order)

[graceful drum]

it depends on the spaces right?

like the gaps between the men and women

[last gate]

like there will only be 4 spaces left

so u can just arrange the remaining women

[graceful drum]

but i think that since we have placed 5 men and 5 women, theres 10 spaces between them

so 10 * 10 * 10 * 10 ways to arrange the 4 women in the spaces

whats wrong with this

[last gate]

wait what

sorry i dont undderrstand

[graceful drum]

like we have placed 5 women and 5 men diametrically opposite to each other in a circle\

that means that the total number of gaps between men and men/ men and women / women and women = 10

so when we want to place the next women, she has 10 spaces

the next also has 10 possible spaces

and the next

[last gate]

wouldlnt there be 4 places left?

[graceful drum]

and the last has 10 possible spaces

[last gate]

theres 14 spots

10 are taken up by the 5 men and women

theres 4 women left in 4 spots

4!

(cause 4p4)

[graceful drum]

is there a way to draw a diagram

1 min

[last gate]

yea its getting kinda late for me so i prob have to go

[graceful drum]

aw ok

also m+n is a two digit integer

so i think its too high in your case

[last gate]

actually i think its 12p4 * 8c4 cause u have 12 spaces and pick 4 with order matters and then theres 8 women choosing 4

yea

instead of 8p4

yea well hopefully someone can take over bye

[graceful drum]

k thanks so much

.close

[midnight plank]

Channel closed

Closed by @graceful drum

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[inland patio]

I'm a bit confused. I'm reading a proof concerning the fact that $C^\infty(\mathbb{R})$ is a Fréchet space, with respect to the topology such that $f_n\to f$ iff $f_n^{(k)}\to f^{(k)}$ uniformly on compact sets for all $k\geq0$ (the family of seminorms that generate this topology are $p_{n,k}(f)=|f^{(k)}|_{K_n}|_u$, where $K_n=[-n,n]$).\

At one point, we are considering completeness, i.e. a Cauchy sequence $(f_j){j=1}^\infty$ in $C^\infty(\mathbb{R})$. Then we have that $(f_j^{(k)}|{K_n})_{n=1}^\infty$ is uniformly Cauchy and hence uniformly convergent for each $j,k\in\mathbb{N}\cup{0}$. Then it is claimed that by taking pointwise limits it is clear that the limits of these sequences agree. What is meant by this last sentence?

[grand pond]

psie

[midnight plank]

@inland patio Has your question been resolved?

[ripe forge]

What was question ❓

[midnight plank]

@inland patio Has your question been resolved?

Channel closed

Closed by @inland patio

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[past jay]

<@&268886789983436800>

[midnight plank]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[lethal path]

<@&268886789983436800>

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[viral dagger]

Let $S$ be the set of all positive real triples $(a,b,c)$ such that $a+b+c=ab+ac+bc \$
a. prove that $$\text{min}(a+b,a+c,b+c)>1$$ for any triple $(a,b,c)\in S \$
b. does there exists a triple $(a,b,c)\in S$ such that
$$\text{min}(a+b,a+c,b+c)<1+\frac{1}{20^{25}}$$

[grand pond]

ihave<skissue>

[viral dagger]

done a, stuck on b

my solution on a was wlog a>=b>=c then essentially making the top equality into a^2-a+1=(b+(a-1))(c+(a-1)), then am gm to get b+c>=2(sqrt(a^2-a+1)-a+1)>1

[midnight plank]

@viral dagger Has your question been resolved?

[chilly cobalt]

cant you pick the minimum abitrarily close to 1?

let s = b + c = 1 + epsilon or something like that

[viral dagger]

theres the a+b+c=ab+ac+bc thing ig

[chilly cobalt]

you can just solve for a in that eq

[viral dagger]

ehh

really

a(1-b-c)=bc-b-c
a=(bc-b-c)/(1-b-c)

aight vro

well we still need the a>=b>=c dont we

[chilly cobalt]

fix very small c maybe

[viral dagger]

wait

if 1-b-c is really small then a shoots to infinity which woud mean a>=b>=c

._.

[chilly cobalt]

:3

[viral dagger]

thank you

.solved

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[chilly cobalt]

skibidi slicers

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[rain kayak]

hi guys, this should be an easy question but im a bit unsure

[worldly canopy]

Stretch it from the x axis by a factor of 2

Like a deeper V shape

[austere blade]

isn't it y?

[past jay]

Taller V shape

[worldly canopy]

Along the y axis

[austere blade]

ah okay

got confused reading it a bit

[worldly canopy]

By from the x axis I mean like using the x axis as a sort of centre point

If that makes sense

[austere blade]

yeah makes sense now

[rain kayak]

so the 2g just means multiply the y value by 2?

[sudden yacht]

Sure

That's called vertical dilation (or stretching)

[misty mirage]

It literally means “for every value of g(x), multiply the result by 2”

[rain kayak]

yeah the answer is (3, -2)

basically for every x value from -1 to 7 the value is doubled

thanks for the help

[midnight plank]

@rain kayak Has your question been resolved?

Channel closed

Closed by @rain kayak

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[molten bay]

Hello everyone can anyone tell me a list of maths books which an American or UK students read at high school level and graduation

[past jay]

#books #books-old #book-recommendations

[molten bay]

I wanted to compare the educational system and books level

I can't search for exact answers

[austere blade]

I guess in the UK they don't really use books for gcse or a levels?

there's stuff like physics and maths tutor, tlmaths, bicen maths, etc in a levels

there's the CGP revision books though for gcse

[midnight plank]

@molten bay Has your question been resolved?

Channel closed

Closed by @molten bay

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[rain kayak]

guys pls help

[rain kayak]

how am i supposed to know whether meghan chose a blue or green or red counter

[lyric charm]

you don't, you're supposed to calculate all 3 cases

although... think about it

if meghan draws a red counter, what would hector have to draw to make the "3 counters all of different colors" thing happen?

@rain kayak

[rain kayak]

he couldnt

[lyric charm]

right.

and if meghan drew a green counter?

[rain kayak]

he still couldnt

[lyric charm]

right.

you've just worked it out haven't you

[rain kayak]

soo

[rain kayak]

so all 15 are blue?

[lyric charm]

no

it means that the event can only happen if meghan draws a blue marble from her jar

(and hector draws two different colors from his bag)

you're asked to find how many blue marbles meghan's jar contains

maybe call it a variable, something like b

work out the probability that the 3 drawn counters have different colors in terms of b

then set it equal to 7/24 and solve for b

bog-standard

[rain kayak]

the probability of hector frawing two different colours is 35/72

so 35/72 + b/15 = 7/24?

[lyric charm]

why plus

[rain kayak]

youre adding the probability of getting b

[lyric charm]

"you add because you add"

[rain kayak]

so what do i do

[lyric charm]

the event says that meghan draws blue AND hector draws two different colors

and these are independent

so you multiply

[rain kayak]

oh

thanks

so 35/72 * b/15 = 7/24?

the answer is 9

thanks for the help

.close

[midnight plank]

Channel closed

Closed by @rain kayak

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hushed fox]

Am I tripping or for x=-1, the sum to infinity wouldn’t be convergent as lim of (-1)^n as n—> ∞ DNE as it oscillates between -1 and 1?

[midnight plank]

@hushed fox Has your question been resolved?

[hushed fox]

<@&286206848099549185>

[mystic condor]

$W_r$ is really just $\frac{(r+9)!}{(r-1)!}$ and any factorial above 1 is even, so actually -1 wouldnt matter each term in the sequence still becomes positive

[grand pond]

ImOakley

[hushed fox]

Oh so even though Wr would tend to infinity, we know for sure it would be even(as weird as that sounds)?

@mystic condor that makes sense, thanks

.close

[midnight plank]

Channel closed

Closed by @hushed fox

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[umbral scroll]

my results are different than my teacher's

[umbral scroll]

i found

z = 11/4 +2i

z' = 3/4 - i

[lyric charm]

let's check these

,calc 3 * (11/4 + 2i) + 5 * (3/4 - i)

[grand pond]

Result:

12 + i

[umbral scroll]

he finds dis

[lyric charm]

,calc (11/4 + 2i) - (3/4 - i)

[grand pond]

Result:

2 + 3i

[lyric charm]

looks like you miscopied the RHS of the second equation and forgor the minus

so -2+3i became +2+3i by your mistake

[umbral scroll]

fuck

sigma

thanks

i'm on 3hours sleep

.close

[midnight plank]

Channel closed

Closed by @umbral scroll

Use .reopen if this was a mistake.

[midnight plover]

8z' - 6 +9i = 12 + i so z' = 9/4 - I

[lyric charm]

are you also on a deadline?

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[jade magnet]

can i do this by induction?

or can i do this without induction too

for the base cases 0, 1, 2 it's true

so i'm going to assume n^5 - n^3 = 24a

[midnight plover]

i dont think induction is a good idea

[jade magnet]

but then if you try (n+1)^5 - (n+1)^3

[jade magnet]

yeah

[jade magnet]

this isn't super nice

[austere blade]

which question are you doing?

[midnight plover]

you are working in N

[jade magnet]

iii

but i struggle with iv too

[midnight plover]

you have in N a nice axium saying that all natural integer are either even or odd

[jade magnet]

oh yeah i did this with the previous questions before this

[austere blade]

factor n^5 - n^3

[jade magnet]

i just wasnt sure because it was a multiple of 24

[midnight plover]

do it with this one too

[jade magnet]

before when it was even or multiples of 3 it was easy

[midnight plover]

jst try

[jade magnet]

but i felt since it was a multiple of 24

i'd have to test 24a, 24a+1, 24a+2, 24a+3 etc

and it wouldn't be feasible

[midnight plover]

i dont think you get the idea

just set n = 2i where i in N so n is even

and prove it by replacing in the values

[austere blade]

I think factorising it is a better idea

[midnight plover]

ye factorise first

[jade magnet]

ok so even if i just test even and odd it'd work?

let me try

[austere blade]

Here's what I'd do. factor n^5 - n^3 to n^3(n^2-1)

[midnight plover]

factoris first as MxRgD said

[austere blade]

then factor that further down to n^3(n-1)(n+1)

[compact copper]

then to n^3(n + 1)(n - 1)

[compact copper]

Yeah, maybe we can do for cases now, n is even or odd

[midnight plover]

king

[jade magnet]

[8a^3][2a+1][2a-1]?

[midnight plover]

this is a etter idea

[jade magnet]

if n = 2a

[austere blade]

well for starters n, n+1 and n-1 are 3 consecutive integers, so it must be a multiple of 3

Now check for the case for n being even and n being odd @jade magnet

[jade magnet]

oh yeah and 2a+1 and 2a-1 is always a multiple of 3

actually no idk how to explain it

basically you can either have a multiple of 3, multiple of 3 + 1 or a multiple of 3 + 2

[compact copper]

yeah, and when n is even then maybe n^2 should provide the remaining factors

[jade magnet]

and if a is a multiple of 3 then the entire thing would be a multiple 8(3^3) so it would be a multiple of 24 anyways

and if it's a multiple of 3 + 1 then 2a-1 is a multiple of 3

[compact copper]

Well maybe not for n = 2 though, but it works for all other even numbers

[jade magnet]

we can substitute n = 2 easily

so like

that part's okay

[compact copper]

Yeah, so we've proved for when n is even, just n^2*n(n - 1)(n + 1)

n(n - 1)(n + 1) is divisible by 3, and for all even natural numbers aside from 2 (which as you said we'll just show), it gives the remaining 8 we need (with a 2 from n(n - 1)(n + 2))

Wait and also, if n is odd, then either of (n + 1)(n - 1) gives one even number with 2 once as a factor, and the other gives an even number with 2 twice as a factor

[mortal mirage]

@jade magnet from this, can you see that if n is even, the expression is automatically a multiple of 8?

if you set n=2k, then n^3=8(k^3)

[jade magnet]

yeah

and (n-1)(n+1) is a multiple of 3 when n^3 isn't

so yeah

i get how to do this now

in fact

i've done iv and v too

[mortal mirage]

ok good

[vivid yoke]

is the problem solved?

[austere blade]

n(n-1)(n+1) is a multiple of 3. What I thought of was just change the expression of n^3(n+1)(n-1) to get n(n+1)(n-1)(n^2)

[vivid yoke]

(n-1).n.(n+1) is divisible by 6

anyway Idk if we have proved for odd n yet so I will share my way here

[austere blade]

that too

[compact copper]

what about this?

[austere blade]

Yeah that would mean one will be a multiple of 4

[vivid yoke]

n^2(n-1)n(n+1) , so (n-1)n(n+1) is divisible by 3, now if n is odd that mean n-1 and n+1 are even so we can rewrite it as
2k(2k+1)(2k+2)=4k(2k+1)(k+1) but k(k+1) is divisible by 2
=>4k(2k+1)(k+1) is divisible by 8

[compact copper]

Also one of them must be divisible by 3

[compact copper]

Yeah, and the other of 2

[vivid yoke]

but 2k(2k+1)(2k+2) is also divisible by 3 that makes it divisible by 3.8=24

[austere blade]

essentially just 3 * 2 * 4 = 24

[compact copper]

Yeah nice, that sums it up

[compact copper]

Exactly

[jade magnet]

yeah ok

[compact copper]

IDK how to write this, I just thought of the logic, but Alexis has shown how to write this

[jade magnet]

oh yeah it is

i usually leave it open since like

i might come back anyways

[vivid yoke]

n^3-1=( n - 1 )( n^2 + n + 1 )=(n - 1)(n^2 - 2n + 1 + 3n)=(n - 1)[(n-1)^2+3n)

the last one is divisible by 9, I wonder how the substitution way look like

[midnight plank]

@jade magnet Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[languid mica]

doesnt this have to be >0 only?

[eternal pawn]

Constant is decreasing

[languid mica]

isnt it an > an+1

[keen widget]

yeah

[runic hamlet]

decreasing generally means >=

strictly decreasing means >

[keen widget]

2 successive terms of this series cannot be equal

[languid mica]

oh

alr yea so it is correct

since it just says decreasing and not strictly decreasing

.close

[midnight plank]

Channel closed

Closed by @languid mica

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[molten bay]

A takes 7 hours B takes 5 hours

[lyric charm]

(we have to assume the trains move at constant speed and nobody is delayed or stops anywhere intermediate. otherwise the question is unanswerable)

[molten bay]

Of course

That should be obviously

What graph did you make?

[lyric charm]

you can think of it as a position-time graph

[molten bay]

Yes right. It starts at

[lyric charm]

time axis runs horizontal, position axis runs vertical (and is aligned with the track between M and N)

[molten bay]

But but ma'am.... how can I find the exact timing?

Any idea? I meant answer is given that 1:35PM

I am thinking of relative speed formulas

2 hour distance of A will be 2D/7

So remains 5D/7

Sane for t_b which is D

Ohh wait I can do it least common multiple and find the sppeds

.close

[midnight plank]

Channel closed

Closed by @molten bay

Use .reopen if this was a mistake.

[molten bay]

LCM of time 35km distance

S_a---35/7=5km/h

S_b---35/7=7km/h

2 hour m kia A n 10km

Remaining h 25km is time se relative speed shuru

25km/12h = 2h 5m

11:30+2h5'

@lyric charm

[lyric charm]

that works out to 13:35 and matches me

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[gaunt nimbus]

progress?

[tidal turret]

if the plane contains both lines, then both direction vectors are contained in the plane, and thus, the direction vectors are orthogonal to the normal

[tidal turret]

the normal of the plane is (1,-1,2)×(-1,1,0)

[gaunt nimbus]

cool

[tidal turret]

,w (1,-1,2)x(-1,1,0)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[tidal turret]

do you follow?

[gaunt nimbus]

yes

[tidal turret]

-2x -2y + 0z = d

[gaunt nimbus]

👍

[tidal turret]

it needs to contain the points aswell

[gaunt nimbus]

yes, to find d

[tidal turret]

(1,1,0), (2,0,2)

d = -4

-2x -2y + 0z = -4

(t+1,-t+1,2t) = (-λ+2,λ,2)

t = 1

λ = 0

no? they don't intersect

@gaunt nimbus

oh they do

they intersect at (2,0,2)

[gaunt nimbus]

these two.. lie on the plane?

,calc -2(2) - 2(0) + 0(2)

[grand pond]

Result:

-4

[tidal turret]

ye dude

that's how we found d

oh no

we found d by plugging the points sorry

[tidal turret]

if they are orthogonal to the normal they are contained in the plane

because the normal of the plane is orthogonal to all the vectors contained in the plane

@gaunt nimbus

[gaunt nimbus]

ye

[gaunt nimbus]

ye

[tidal turret]

.solved

[midnight plank]

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sand flume]

however I realized my work cannot be right because the base we found was for the right triangle but then the answer we get doesn't solve the question

[lyric charm]

would've been nice to see an image somewhere in your work ngl

[sand flume]

this is represents the problem. The plan was to find the area of one triangle and multiply it by 12. I did that and got the answer they were looking for. But when looking it over I realized I was calculating the area of a right trianglel inside one of the green triangles not the green triangle itself

[lyric charm]

hmmm

heres something i cooked up just now

[sand flume]

sorry I sent the wrong photo this is the right one:

[lyric charm]

well you're gonna need both.

[sand flume]

but for the 3 < pi part I just need the most recent one

[lyric charm]

ok right

zoom in on this one slice

24 of these, by mirroring and rotation, recreate the entire picture

[sand flume]

oh I used the wrong angle

[lyric charm]

you've got a piece of the dodecagon (green) enclosed inside the same-angled circular sector (red)

the area of the green triangular slice is 1/2 * sin(15°) * cos(15°) while that of the circular sector is pi/24

[sand flume]

ok I see. if we were to ammend what I wrote thought it is still wrong

@lyric charm

[lyric charm]

your work is lacking in diagrams and i honestly cbf to decipher what means what here.

The a, b, and c here are not related to the ones above.
if you have to write this you know you're doing communication wrong for sure 💀

[sand flume]

ok how is this we first find the value of c

next step is to find d

[lyric charm]

i mean a=b=1

[sand flume]

is this clearer

wait

c/2

here

[midnight plank]

@sand flume Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hasty flume]

yo guys, for part C why is the answer 14.4/14.8?

[hasty flume]

Its on he MS and it says 2*7.2 but why?

mechanics btw

[midnight plank]

@hasty flume Has your question been resolved?

[hasty flume]

<@&286206848099549185>

[rough birch]

let me see

part (a) looks fine

part (b) looks right

your calculation for T looks right

[rough birch]

they are now calculating the average speed

read the problem - it asks for T, and then for the average speed

total distance traveled = (36/5) in first 10 seconds, and another 36/5 after that. that gets us the 14.4 of total distance traveled

then, T=14.8, meaning 14.8 was the total time

hence: the average speed over everything was 14.4/14.8

does that make sense? @hasty flume

[hasty flume]

Sorry, wait ill read it rn

ah wait im dumbb

[rough birch]

i need to go now btw

[hasty flume]

I get it yeah

thanks 😂😅

distance/time

.close

[midnight plank]

Channel closed

Closed by @hasty flume

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[civic radish]

My question will be two parts due to character limit

[civic radish]

I would like to prove algebraically that the sum of two discrete random variables with Pascal distributions is itself a Pascal distribution. Emphasis on algebraically, I'm aware of logical arguments for this fact.\\

Let Pascal(n, p) denote a random variable containing the number k of indpendent Bernoulli trials with probability p performed before n successes are observed.\\

Let X: Pascal(n, p) and Y: Pascal(m, p)\\

So, what I want to prove algebraically is that X + Y: Pascal(n + m, p).\\

For reference: $P(X = k) = \binom{k - 1}{n - 1} p^n (1 - p)^{k - n}$\\

Here's what I have so far:

$P(X + Y = k) = \sum_{i=n}^{k - m} P(X + Y = k|X = i)P(X = i)$ by the law of total probability

$P(X + Y = k) = \sum_{i=n}^{k - m} P(Y = k - i|X = i)P(X = i)$

$P(X + Y = k) = \sum_{i=n}^{k - m} P(Y = k - i)P(X = i)$ by independence of X and Y

$P(X + Y = k) = \sum_{i=n}^{k - m} \binom{k - i - 1}{m - 1} p^m (1 - p)^{k - i - m} \binom{i - 1}{n - 1} p^n (1 - p)^{i - n}$

$P(X + Y = k) = \sum_{i=n}^{k - m} \binom{k - i - 1}{m - 1} \binom{i - 1}{n - 1} p^{n + m} (1 - p)^{k - n - m}$

$P(X + Y = k) = \lparen\sum_{i=n}^{k - m} \binom{k - i - 1}{m - 1} \binom{i - 1}{n - 1}\rparen \space p^{n + m} (1 - p)^{k - n - m}$\\

[grand pond]

master jonsie

[civic radish]

What I'm ultimately trying to prove is:

$P(X + Y = k) = \binom{k - 1}{m + n - 1} p^{n + m} (1 - p)^{k - n - m}$\\

So I must show the following to be true:

$\sum_{i=n}^{k - m} \binom{k - i - 1}{m - 1} \binom{i - 1}{n - 1} = \binom{k - 1}{m + n - 1}$\\

And finally we've reached my actual problem. I have no idea how to prove this, or if I've made a mistake somewhere. If I haven't made a mistake, what I'm lacking here is binomial coefficient identities where the top argument is being summed over. I know Vandermonde's identity and have solid intuition for why it works, but I struggle to even conceptualize summing over the top argument, and I don't know any algebraic tricks for manipulating the top argument besides the recursive formula.\\

What I'm looking for here is insight, a nudge in the right direction, maybe some keywords to research. Thank you!\\

I will pre-emptively answer how I found my summation bounds in the first step since they look odd out of context. X and Y must be equal to at least n and m respectively since you can't observe more successes than total trials, so I chose summation bounds that ensure those conditions.

[grand pond]

master jonsie

[eternal pawn]

https://en.m.wikipedia.org/wiki/Vandermonde's_identity

It looks like this

Maybe with some change in indices?

Oop i didn't read carefully

[midnight plank]

@civic radish Has your question been resolved?

[civic radish]

<@&286206848099549185>

[languid glade]

Have you tried finding a story proof for that?

[civic radish]

Can you please clarify what you mean by "that"?

[languid glade]

for the thing that you are "ultimately trying to prove"

[civic radish]

Yes, my question is about proving it algebraically

My real goal here is to prove that combinatorial identity

Or understand where I went wrong if it's not valid

[languid glade]

maybe the right side counts the number of ways you can choose from the first k-1 integers (n+m+1) numbers, and you can prove the left side does the same

i did not see any miscalculation until there so it may be worth a try

[civic radish]

My question is where specifically to start with this

Or where vaguely to start I suppose, I think I just need a nudge. Like I said above, what's throwing me off is summing over the top argument, I'm not sure how to think about what that means from a counting perspective.

[languid glade]

i think if you define j := i-n, and you write everything in terms of j instead of i you may be able to prove it strictly algebraically. Also I think summing over the top arguments (if i understand what you mean) is related to choosing k-1 from m+n-1 naturals by focusing on one number and then counting the number of ways to have all other numbers "fit in". That is what i meant by story proof

but i can't quite formalize yet what i mean

hope it helps at least a bit

[eternal pawn]

Expanding on DJ's idea, maybe you can count the number of ways to pick n+m+1 integers from 1 to k-1, depending on the value of the m-th integer? (Imagine they are sorted)

[civic radish]

I did start down this path earlier and got:

$\sum_{j=0}^{k - m - n} \binom{k - j - n - 1}{m - 1} \binom{n + j - 1}{n - 1}$\\

which can further be manipulated with the identity:

$\binom{n}{k} = \binom{n}{n - k}$\\

to get:

$\sum_{j=0}^{k - m - n} \binom{k - j - n - 1}{k - n - m - j} \binom{n + j - 1}{j}$\\

which is nearly Vandermonde's identity, but not quite because of the j terms in the top argument of both binomial coefficients

[grand pond]

master jonsie

[tender dew]

which is perhaps more neatly written when r = k-m-n as

$\sum_{i=0}^{r} \binom{r-i + m-1}{r-i} \binom{i + n-1}{i} = \binom{r+n+m-1}{n+m-1}$

[grand pond]

Peter

[civic radish]

If you're asserting this as an identity, I'm not familiar with it. Does it have a name?

[tender dew]

No I'm not asserting it as an identity, just a rewrite of what you ended up with

A double-counting proof of this seems to be here

though, I understand this is not what you asked for, so I'll keep trying the algebraic route and see what gives

nvm, this seems to be it

[civic radish]

If I'm not mistaken, applying these to what we have would give the wrong answer since the bottom of our righthand side is (n + m - 1) and should be (k - m - n)

Just to be clear, this is all in terms of our chosen variable names

Just to be clear: part 2, when I say "applying these" I mean the theorem that these posts are proving

[tender dew]

$\binom{r+n+m-1}{n+m-1} = \binom{r+n+m-1}{(r+n+m-1) - (n+m-1)} = \binom{r+n+m-1}{r} = \binom{r+n+m-1}{k-m-n}$

[grand pond]

Peter

[civic radish]

Oh duh

Could you share the search terms you used to find these posts?

[tender dew]

another useful post which has one of my fav books

[tender dew]

Use https://approach0.xyz with latex

or just search for sum of negative binomials

[civic radish]

Wow, this is amazing

[tender dew]

indeed :D

[civic radish]

Thank you for all this! Any last words before I close?

[tender dew]

not really, thanks for bringing to my attention this problem, the double counting was fun

[civic radish]

.close

[midnight plank]

Channel closed

Closed by @civic radish

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[safe notch]

guys

[safe notch]

i cant get abt the mean positions in shm

they be taking position anywhere

when x=0 and y=0

are they both mean positions ??

[compact copper]

yeah technically, depends on your axis

[safe notch]

yoo

u came

i guess its easier now

[compact copper]

I hope🫡

[safe notch]

abt displacement

if a particle starts at A
say theres a circle u make axis by two lines
on the furthest east or the right is A and B A' and B' respectively in anticlockwise direction

u start at A

[compact copper]

okay

[safe notch]

after time t u reach a point p

such that theta = wt

[compact copper]

Yeah

[safe notch]

w= omega

the book considers OA to be the amplitude

[compact copper]

yeah yeah, it's similar to traditional kinematics, where s = ut when a = 0

here theta is s and u is w, rotational dynamics formulae are very similar to kinematics formulae, so are the concepts

[safe notch]

wait im making the fig

[compact copper]

Yes, the higher the peak the more the amplitude

[safe notch]

its something this

[compact copper]

Yes

[safe notch]

welll make this our base figure of understanding

oh i forgot to mention origin

[compact copper]

Alright, traditional figure

[compact copper]

I get it, you don't have to add it

[safe notch]

k

here

why is OA teh amplitude

and where is teh mean position

at A or B

??

i dont get it

[compact copper]

You see, this is a graphical representation, and if you could see the sine wave (which is just the simpler form of this), you'd be able to grasp it better

Basically A is the mean position, and B is the extreme position

[scenic wyvern]

basic sine wave if you guys need it

[compact copper]

Thanks mate, welcome back

Yeah so you see the circle

When you graph it with t>0, then you get a sine wave like the one above

[safe notch]

what

if A is mean position what does it make of A'

[compact copper]

A' is just to simplify the figure, you don't have to count it in

The figure moves ahead in a sine wave

[safe notch]

like what

man we neeed the figure as a whole dont we ?

how does the wave complete

[compact copper]

Hanako has given that for reference

[safe notch]

without it

[compact copper]

Alright see

How good's your visualisation?

[safe notch]

u can try

i might get it

[compact copper]

Alright

Imagine I have a pendulum under ideal conditions

[safe notch]

k

[compact copper]

and I get it on a cart which moves with a constant velocity of 1 units in +x direction

[safe notch]

k

[compact copper]

And we have a magical pendulum on our hands

which leaves a line in space over whichever region its bob moves

[safe notch]

k

[compact copper]

So I pull it back, and release it the moment the cart starts moving

See it from above

[safe notch]

pull the cart

??

[compact copper]

No no, the bob of the pendulum

to the extreme position

[scenic wyvern]

i think

[safe notch]

ahhh

[scenic wyvern]

OP needs to learn how to look at his circle from the side

[safe notch]

its jus a line

no

?

[scenic wyvern]

no

it will be a point moving back and forth

[safe notch]

yea

thats what i meant

[safe notch]

yeah

[compact copper]

Thanks for taking over, I'm back

[safe notch]

so u pull the bob to one extreme position( say left) the very moment teh cart starts to move

[scenic wyvern]

https://www.youtube.com/watch?v=snHKEpCv0Hk
maybe watch this video, starting from 2:40

[compact copper]

yes

[scenic wyvern]

this might help you visualize exactly what kind of motion is going on if you look at it from the sides

[compact copper]

and see it from the sky

[safe notch]

what

[compact copper]

doesn't it leave curves as the cart moves

[compact copper]

from above

Well those curves, as it turns out, are in fact sine waves

[safe notch]

why see it from sky lmao we looking at the curves the pendulum left no

[compact copper]

Okay, whatever floats your boat

It leaves certain curves that oscillate yeah?

along the x axis

[safe notch]

hmmm u are right

it has to be above

if its horizontal it dosent add up

[compact copper]

Yeah so as it turns out, those oscillating curves are sine waves

[safe notch]

okayyyyy....!?

[compact copper]

You got it yeah?

[safe notch]

yeah

up to this point yeah

[compact copper]

Okay, now just imagine a stationary pendulum

[safe notch]

ok

[compact copper]

At time t=0, it's at mean position, at time t = t1, it's at one extreme position, at time t2, it's back at mean position, at time t3, it's at the other extreme position and at time t4, it's back at mean position

Like it's oscillating like this

[safe notch]

yeah

[compact copper]

Don't spam

[safe notch]

what

[compact copper]

NP

[safe notch]

lmao

[safe notch]

fr

[compact copper]

OMG thank you dude

[safe notch]

i get it

[compact copper]

And we represent it like this, with time as x-axis

[safe notch]

bro the frquency both be working with

😭

[compact copper]

through a sine wave

[safe notch]

k k

[compact copper]

Thank Hanako for saving our times

[safe notch]

thanks @scenic wyvern hanako

[compact copper]

Yeah you get it now?

[safe notch]

ahmm yeahhh

buttttt

i got what u taught me

but i dont get what book is saying

[scenic wyvern]

take a pic of the book maybe?

[compact copper]

yup

[safe notch]

the fact i dont have a mobile

😭

i work on laptop

[scenic wyvern]

windows has a camera app

[safe notch]

click from there

???

[compact copper]

This or something else?

[safe notch]

its theory

all texts

[scenic wyvern]

or tell me the title of the book, its edition, author and page number

[compact copper]

Alright try to get a pic then

[safe notch]

what

[safe notch]

ull find the pdf

no wat

way

[compact copper]

or that

[safe notch]

the heart of physics grade xii

[scenic wyvern]

author?

[safe notch]

binil aryal phd
mn singh

[scenic wyvern]

page?

[compact copper]

Share a pic if you find it Hanako

[safe notch]

34