help-13

[wraith dagger]

Michael

[carmine sorrel]

Oh shit

[sour tundra]

Oh and yes there are some errors in algebra there

[carmine sorrel]

1s ima do it again 😭🤧

[carmine sorrel]

Wait what

[sour tundra]

You were right, my bad

[carmine sorrel]

Yh it should be only f(x)

Cz differentiation of 1 will be 0

Still getting a different answer

@sour tundra @mild spear

[mild spear]

Should be y/x

[carmine sorrel]

Oh shit😭😭

It should be -d(y/x) right?

[mild spear]

You get y = x(dy/dx) - x^2
x^2 = x(dy/dx) - y
1 = (x(dy/dx) - y)/(x^2)
1 = d(y/x)
Should be positive

Integrate both sides,
Put the value given to find constant and then you'll get your answer

[carmine sorrel]

Nvm

I got it

Thank you so much!

[mild spear]

np

[carmine sorrel]

.close

[cedar kiln]

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[cedar kiln]

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[vale sun]

can someone help me understand this

[floral arrow]

Could you repost it, not cropped?

[vale sun]

is this okay

[opal hinge]

yeah

okay so the solutions can be real or complex but they have the form z=x+yi

[vale sun]

k

[opal hinge]

And you have to find all of them after that you'll have all points that the ellipse passes through and then you find the ellipse

[vale sun]

i get that you have to find solutioons

but how do i find eclipse

[opal hinge]

It has been quite a while since I did a ellipse problem

Im not the best one to help you here

[vale sun]

okay

[cedar kiln]

@vale sun Has your question been resolved?

[cedar kiln]

@vale sun Has your question been resolved?

[frail citrus]

first of all solve the equation $z^{2}+2z+4$ and $z^{2}+4z+6$ to obtain the ordered pairs of (x_k, y_k)

[vale sun]

yeah

[wraith dagger]

Dhairya
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vale sun]

ok

i understand that

z=1

z=-1+ or - sqrt -3

z=-2+- sqrt -2

[frail citrus]

ok

now assume a ellipse centered at (h,k) passing through all these points

[vale sun]

k

[frail citrus]

and use the given data mindfully in braces

[vale sun]

in the complex plane

[vale sun]

wdym

[frail citrus]

i was speaking about this:

like use it to find relations between epsilon and m,n

if you want ill try it myself- i havent done it on a sheet of paper yet

[vale sun]

what is an epsilon

my brother gave me this problem and asked me to solve it but i could not

[floral arrow]

I have a solution but it's lengthy, and judging by the format of the question (competition math) I'm sure there is some kind of trick

[vale sun]

this is the last question on the amc12 2015

last time i took the 12 i got 1 84

[frail citrus]

it is the eccentricity of the specific ellipse we are interested in

[frail citrus]

hmm. !nosols

!nosol

ah

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[floral arrow]

Bruh did I give a solution? As a Helpful I'm pretty sure I understand the rules of the help channels better than you.

[frail citrus]

well well

fine

[floral arrow]

So... @vale sun , do you want the lengthy solution or do you prefer waiting for someone that knows competition math?

[vale sun]

i would prefer waiting

you only get a minute to solve problems like this

and i wanna see if there is a trick to solving

[frail citrus]

waited enough?

no one is appering tbh

my method too is huge but solvable

[jade blaze]

whats the answer?

is it 11

[jade blaze]

whats ur answer mate

[frail citrus]

i didnt try yet

i just blabbered some ideas i had in my head

but never formulated them on paper

seems like this server likes that kind of approach

[jade blaze]

lmao

well idk if my answer is correct

[frail citrus]

ask OP

[jade blaze]

i was waiting for them to be online so i could explain tbh

instead of plainly giving answer since its a guidline here

[frail citrus]

yes

that is a fine line to tread upon

[jade blaze]

yes op if you have made the graph then u will see that u can tell the value of b (minor axis)

so just write the equation of ellipse

get the value of a by putting the initial points in the equation

[cedar kiln]

@vale sun Has your question been resolved?

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[round lava]

here, cardinalities of both of them would be ℵ0​, right? but how do i justify it?

[abstract breach]

uhhh there's a nice way to prove |N^k| = |N| so I guess this would do it

[tropic oxide]

make bijections from N to N^2 and N to N^3.

[wide gorge]

mb

Im dumb fr always jumping:((

[abstract breach]

same lol

[tropic oxide]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

@round lava Has your question been resolved?

[cedar kiln]

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[chrome elk]

Do you know how to show ℕ and ℤ are in bijection

[cedar kiln]

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[last nacelle]

A family consisting of two adults and two children are going to sit in a block of six seats on a
train, arranged as two opposite sets of three seats. Finally, suppose that one of the adults insists on sitting forwards, and the children must not
sit in adjacent seats. How many seating plans are there now?

[opal hinge]

What have you tried?

[last nacelle]

I am trying to make cases, but i don't know how to find the number of arrangements in each

I tried a case where one of the adults is on the front end seat so he can sit in 2 places, then the next adult can sit in 3 places (the middle of the back or either position in the front) and then I think I should multiply by two as the two adults can switch places, but then this double counts the arrangement where they sit on either end of the front row, so I don't know how to accoutn for that

[opal hinge]

This does looks like cases work

Probably need more than 2 cases

[past wave]

try drawing a diagram, that usually helps

[jade blaze]

there are only 2 cases tho?

[opal hinge]

So 2 things have to be satisfied when doing seating, first thing is that an adult has to be in front seat, and second is 2 kids can't sit next each other

[jade blaze]

don't do adult first

try making it on basis of children

[opal hinge]

I would say we satisfy the first requirement first and I think we could subtract the case for when both kids sit next to each other

That would make it less complicated i think

[jade blaze]

nooo

[opal hinge]

xD so what's your idea

[last nacelle]

There is one thing I am struggling with at the moment, say I have already seated the two children in the back with the middle seat empty, how can I now find the number of ways to seat the adults

[jade blaze]

first forward, so adult in between

and 2nd adult has 3 ways

[last nacelle]

Most ways I try to do it end up double counting the ways the adults can sit

[jade blaze]

nah wait sorry

[last nacelle]

the children could also be opposite each other

so that adds a bunch of new ways for them to sit

[jade blaze]

oh no wait

[opal hinge]

Yeah that's why I said we subtract the cases they sit next

[jade blaze]

if one kid goes forward then how many ways does the second child have?

[opal hinge]

So we don't have to deal with this

[last nacelle]

depends where he sits

if he sits on the end seat

[jade blaze]

no

[last nacelle]

the other kid has 4 ways

if hes in the middle

[jade blaze]

use permutation

[last nacelle]

the other kid only has 3 ways

[jade blaze]

which kid

try to distribute it in forward and backward case

[last nacelle]

yh but those need to be broken down as well

[jade blaze]

1. both kids forward and backward
2. (one kid forward) x 2

[jade blaze]

then break it down

thats what permutation is about

[last nacelle]

but also these cases affect how many ways the adults can sit

ok

i will try to break it down into more cases

[jade blaze]

since we have given that one adult is front then other adult can sit on any available seats so keep it for the end of the case ok

[last nacelle]

ok

[cedar kiln]

@last nacelle Has your question been resolved?

[last nacelle]

Ok, I have decided to try subtracting the invalid options

there are 180 ways they can be arranged with one adult in the front

so my three cases are

both children in the front

both children in the back with 1 adult

and both children in the bac and both adults in the front

case 1: 2* 2 (ways children can sit) * 2 (which adult sits in the front with them) *3 (where the other adult sits) = 24

[opal hinge]

How did you get 180?

[last nacelle]

3 * 5! / 2!

since there 3 places the first adullt can sit

then 5 * 4 *3 for everyone else

[opal hinge]

,calc 3*5!/2!

[wraith dagger]

Result:

180

[opal hinge]

Nope

[last nacelle]

whats wrong with that

🙁

[opal hinge]

Well, an adult has to sit in front so it leaves us with 5 seats

So we have 3 people ( 2 kids and 1 adult)

And 5 seats

[last nacelle]

yeah so we multipy by 543

[opal hinge]

5!/2! is the number of ways we can arrange 3 people to 5 sit

Seats

[last nacelle]

yeah and don't we need to multiply by 3 for where the first guy sat

[opal hinge]

https://tenor.com/view/penguinz0-moist-critical-but-however-gif-15799922108420012050

We have 2 adults right?

Let call them A and B

Case 1: A sits in front seat
Case 2: B sits in front seat

So we actually have to multiply it by 2 which is 2*5!/2!

But....

[last nacelle]

im like 99.99% sure its 180

[opal hinge]

Case 1 and case 2 have intersection

[last nacelle]

inclusion exclusion!!!

[opal hinge]

Which is is when both adult sit in front ( there're two front seats )

[opal hinge]

yeahhhh

[last nacelle]

wait but what was wrong with the way I originally did it

if you just say that one of the adults has to sit in the front

[opal hinge]

I'm not sure where that 3 come from

[last nacelle]

so theres 3 options for the first seat

[last nacelle]

theres three seats in the front

[opal hinge]

Huh

Wait

What?

Okay, two rows right?

[last nacelle]

yeah

six seats

[opal hinge]

So it should be only 2 front seats no?

[last nacelle]

no its a train

theres two opposite rows of three seats

[opal hinge]

Yeah

[last nacelle]

A family consisting of two adults and two children are going to sit in a block of six seats on a
train, arranged as two opposite sets of three seats. Finally, suppose that one of the adults insists on sitting forwards, and the children must not
sit in adjacent seats. How many seating plans are there now?

[opal hinge]

S1 S4
S2 S5
S3 S6

So seat no.1 and no.4 are front seats, no?

[last nacelle]

I thought it was like

s1 s2 s3

s4 s5 s65

maybe i misinterpreted the question

[opal hinge]

Not sure, never take a train so

[last nacelle]

idk I think that would make the most sense

I might just email my teacher

[opal hinge]

Yeah

[cedar kiln]

@last nacelle Has your question been resolved?

[cedar kiln]

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[spiral iris]

guys how can we compute this limit

[spiral iris]

no hopital

no appro

pls help

[last apex]

its not 0/0 or inf/inf

[spiral iris]

its 0/0

[stiff brook]

no

[spiral iris]

wait

its 3+x²

sorry

[stiff brook]

the denominator is 3+x^2?

[spiral iris]

here

the denominator is worng

drop 3x-4 out of the power

i know its messy sorry but ive been stuck for nearly 2h

[stiff brook]

is the denom (x^3) + 3x-4

[spiral iris]

yeah

[stiff brook]

is there any particular reason as to why u cant use lhopital

[spiral iris]

its not accepted in our system

[rare moth]

i am in grade 10th i am new to this stuff

pls teach me

[spiral iris]

its considered "wrong"

[rare moth]

it sounds sick

lhoptical rule and stuff

woww

rn i am solving circles

so easy

[spiral iris]

i think this is a help channel not meant to teach

u can go to yt

and whatever y get stuck on just consult the mighty shcolars here

[rare moth]

cant find a good tutor except organic chem

guy

[rare moth]

ohhh

[stiff brook]

maybe u could try series expansion for sin and sqrt?

around x=1

[cerulean sail]

(first thoughts are that 3 = 1 + 2, and that as 1 is a root of the denominator, you may want to consider manipulating that...)

[spiral iris]

i didnt reach thos yet

[stiff brook]

then im not sure, sorry

[spiral iris]

lemme tr ythat

[spiral iris]

np man lol

[cerulean sail]

Just for a bit of context, what have you reached so far? Just so we know, and we can suggest things that are good ideas
Have you done e.g. differentiation yet?

[spiral iris]

derivative? yeah

we havent done integrale and D E

and exp and Ln

its inf i think

the limit shouldnt exist

[cerulean sail]

Yea, as long as you know how to recognise derivatives from first principles, you should hopefully be able to make some progress with my suggestion

[spiral iris]

wait it hink i got something

btw what are frst principles

[cerulean sail]

Basically the limit definition of the derivative, e.g. do you know that
[
f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}
]
(you may have seen that instead as $\lim_{h\to 0} \frac{f(a + h) - f(a)}h$, but that's equivalent if you replace $x$ with $a + h$)

[wraith dagger]

@cerulean sail

[spiral iris]

im familiar with the first one

why are they even called first principles we call them just the definition?

[cerulean sail]

Well I'm using the term "first principles" because that's the way I've heard it, but also to kind of differentiate (pun not intended) between that and just applying the rules, on reflection I should have said "definition" now

[spiral iris]

btw extra question

a ply of the third degree

has 1 root

so its positive on its right and negeative on its left

right?

even thogh its third degree

this question stupid ik 😭

[cerulean sail]

Well considering the coefficient of x^3 is positive, yea, far enough out right, you're positive, and far enough out left, you're negative

[spiral iris]

well good to know

[cerulean sail]

No no, you're fine, not a bad question, and we all need reminding sometimes

[spiral iris]

i dont have any more qquestions

do i just close it?

.close

[cedar kiln]

Channel closed

Closed by @spiral iris

Use .reopen if this was a mistake.

[cerulean sail]

Wait one monent

.reopen

[cedar kiln]

✅

[spiral iris]

okk

[cerulean sail]

Just to check, did you find the limit to be nonexistent for this one?

[spiral iris]

yeah

its +inf on the right and -inf on the left

why?

[cerulean sail]

I disagree

[spiral iris]

or the opposite ?

wiat its the opp

+inf on the left and vice versa

[cerulean sail]

The limit exists and it's finite, if I'm not wrong

[spiral iris]

prove me wrong 😭 pls

[spiral iris]

how

[cerulean sail]

Well, how about we work together and we see where we get

[spiral iris]

first i seperated -3 to -2and -1

and sqrt(3-x²)-2 was taken to q different side

wait can i wrire it on paper??

is that okay?

[cerulean sail]

You can, but a quick check, did you mean sqrt{3 - x^2} or sqrt{3 + x^2} there?

[spiral iris]

+X

i alwys mix them

[cerulean sail]

It's easy to do, mistakes just love sneaking their way in sometimes

I otherwise like the way you seem to be going though

[abstract breach]

what I would do first is to substitute 1+ t =x to have a nicer limit

then you'll see what you can cancel

[spiral iris]

SORRY FOR THE TARDINESS

@cerulean sail

1-==> deno<0 and vice versa

[abstract breach]

what did you do between 2nd and 3rd line?

[spiral iris]

It just turns into a mess tbh moving the limit to 0 doesn't help at all cuz we're not using the trigo identifies

[cerulean sail]

,rccw

[spiral iris]

cpnjugate 3+x²-4=x²-1=(x-1)(x+1)

[abstract breach]

are you sure?

[spiral iris]

sorry for skipping lines

and sloppy handwriting

tried to make it look presentable

[abstract breach]

you did not multiply the bottom by the conjugate

[spiral iris]

wait mb

[cerulean sail]

(it's worth noting that the idea I had was that on factoring the denominator, you could then notice that you have some definitions of derivatives that you're dealing with for both terms, after factoring out 1/(x^2 + x + 4) )

[spiral iris]

but its still a real number regarless

[spiral iris]

i dont get it

but its correct no? i dont see where coulve messed up

[abstract breach]

could you clarify what you meant on the left?

[cerulean sail]

Let me take that second term you have, from the second line, and show you what I mean, as you're most of the way there: you could, after factoring, write that as
[
\lim_{x\to 1} \frac1{x^2 + x + 4} {\color{green} \frac{\sqrt{3 + x^2} - 2}{x - 1} }
]
then of course split that into the product of the limits, the green being the derivative of $\sqrt{3 + x^2}$ at 1

[wraith dagger]

@cerulean sail

[spiral iris]

i get it now

[spiral iris]

when x app. from the left the denominator is negative?

is that what ur asking

[abstract breach]

numerator is negative as well

anyways, both expressions have a limit as n->1 so the entire expression as well

also it's supposed to be n^3 + 3n - 4 instead of n^3 + 3n + 4

[spiral iris]

how does the left term have a limit

the num is real and the den is 0

pls help

i need to understand

[cerulean sail]

(the numerator is zero as well, and you can e.g. do a similar trick as to before)

[abstract breach]

as I said, it's easier to reason with 1+t = x

[spiral iris]

whast the answer

i got 5/3 now

andits 100 perc wrong

[spiral iris]

yeah sorry man

[cedar kiln]

@spiral iris Has your question been resolved?

Show your work, and if possible, explain where you are stuck.

[spiral iris]

i think its 1/4

final answer

. close

.close

[cedar kiln]

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[cedar kiln]

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[naive hollow]

can someone explain me how to prove the well ordering principle using mathematical induction

[fast osprey]

What do you mean by ‘well ordering principles’

[ocean sentinel]

Imo the easiest way is to first prove it for finite sets

Then you can generalise it without much trouble

[old perch]

what class is this for?

[naive hollow]

the principle that states that there exists a smallest element say L which would always be </= all other of its elements

[ocean sentinel]

*for a set of natural numbers

[naive hollow]

a topic in my integers and divisibility module

part of discrete math afaik

[naive hollow]

oh yeah

[naive hollow]

could u elaborate

i just dont get the textbook proof

[ocean sentinel]

As in step 1 would be "Let S be a finite nonempty subset of N. Prove that S contains a minimum element"

Then step 2 would be to generalise it to infinite sets

Do you need help with step 1 or can you do it?

[ocean sentinel]

I was wrong earlier, step 2 is the interesting step

[naive hollow]

ah pls elaborate, i would to hear the entire thought from u lol

[ocean sentinel]

Okay

If S has size 1, then S = {n} for some n

so n is the minimum

[fast osprey]

Well if we dont know whether there exist such an L in N or not we cannot use induction because the whole foundation of induction is that axiom

[ocean sentinel]

If $S$ has size $k + 1$, then $S = T\cup{x}$ for some $n$ and $T$, with $|T|=k$

[wraith dagger]

depression

[ocean sentinel]

Not really. Well ordering and induction are equivalent. If you assume one of them you can prove the other

[naive hollow]

ah wait

can we think of S as a union of {n} and a set T with k elements?

[ocean sentinel]

This means that the minimum element of S is just min (min(T), n)

[naive hollow]

cuz that way S is k+1

[ocean sentinel]

That's what I said isn't it?

But yes, if you take an element out of a finite set, its size goes down by 1

[naive hollow]

oh shoot right, its just u used {x}

[ocean sentinel]

Oh shit yeah sorry

[naive hollow]

yea

[fast osprey]

Are you sure its not an implication not equivalence

[ocean sentinel]

Yes

The implication goes both ways

It's much easier to prove induction from well ordering than the other way around, but both can be done

[ocean sentinel]

So does that make sense? That's the induction step

[naive hollow]

yep it does

[ocean sentinel]

Nice

[fast osprey]

Alr but still absurd reasoning is better in this case

[ocean sentinel]

Wdym?

So the next step is to take an infinite set S, and somehow get a finite subset T that has the same minimum element

Do you have any ideas?

[fast osprey]

Proof by contradiction

[ocean sentinel]

I don't think so but I'm sure you can do it like that

In general, proofs without contradiction are better unless they're a lot more complicated

[naive hollow]

hmm i cannot realy think of any

[ocean sentinel]

It keeps the constructivists happy at least

I'll give you a clue

It has to be an argument that doesn't work with the integers Z

[naive hollow]

ohh

[ocean sentinel]

Also proof by contradiction doesn't exempt you from proving the implication both ways

[naive hollow]

i cannot get the solution

[ocean sentinel]

Okay

It relies on the fact that for any natural number, there is a finite number of smaller natural numbers

[ocean sentinel]

RAA just allows you to prove that ¬B implies ¬A instead. You still have to prove two implications

So if you let $n\in S$ be any element, then you can let $T={x\in S\mid x\le n}$

[wraith dagger]

depression

[ocean sentinel]

Of course you still need to prove that it works, but that's the idea

[naive hollow]

hmmmmmmm

i think i would want to try it out once again by myself with this

[ocean sentinel]

Here's an example

Let S be the set of even numbers

Name me any even number

[naive hollow]

18

[ocean sentinel]

Great

Now let T be the set of even numbers at most 18

That's a finite nonempty set, so it has a minimum element

[naive hollow]

yeah

[ocean sentinel]

And so that minimum element has to be the minimum even number

[naive hollow]

hmmm

.close

[cedar kiln]

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[cedar kiln]

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[obtuse coral]

Could someone plz explain #6 to me?

[split ice]

"throws a dart up at 60 degrees" means he throws it diagonally, and the angle that that diagonal makes with the ground is 60 degrees.
"travels 11 metrers" simply means it went 11 meters, horizontally speaking.

[cedar kiln]

@obtuse coral Has your question been resolved?

[ornate hearth]

where are you stuck

[cedar kiln]

@obtuse coral Has your question been resolved?

[cedar kiln]

@obtuse coral Has your question been resolved?

[frail citrus]

.

[frail citrus]

after this tho it is manageable only- you have the range of a projectile's formula- use it to find v_0. now plug v_0 into the height attained by a projectile's formula

you will then obtain your answer

@obtuse coral

[obtuse coral]

Yeah, but I’ve never learned that formula

[frail citrus]

learn it

otherwise this problem will be full of derivations and stuff

[obtuse coral]

Just to be clear, what are the formulas I should know?

[frail citrus]

wait sir

ill send em

the red ones are imp for these type of ques

!done?

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[obtuse coral]

Thank you

Yes

.close

[cedar kiln]

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[uneven dock]

can someone explain this pls

[whole nexus]

Take sqrt(at) common maybe

[uneven dock]

what does that mean

[long swan]

You're name is misspelled

[whole nexus]

Oh my bad, I meant as x

[long swan]

It should be "Avocado"

[whole nexus]

I meant to say substitute

[steady anchor]

Uhh one question... wat grade question u guys solving..

[uneven dock]

my name is avanya ppl call me ava so my username is avacado

[uneven dock]

this is undergraduate calc 2

[steady anchor]

Oww alr

[uneven dock]

im a chemistry major im forced to deal with this

[steady anchor]

Thank You🙏

[coral jewel]

try integrating $\int\sin(\sqrt x)\dd x$ first

[coral jewel]

if you can do this, then you can also do the original problem

[uneven dock]

would i do IBP

[coral jewel]

IBP is involved, yes

[whole nexus]

Just substitute sqrt(at) as x

[uneven dock]

ok im still lost

[whole nexus]

You did this first yeah?

[uneven dock]

wait

like turn sqrt(at) using u sub?

[whole nexus]

Yes, then IBP

Not sure though (Well usually I get an idea of how I'd reach the answer with a good level of clarity, here I just haven't thought that far, so still should be fine IG)

Alright I'll think

Okay I reached somewhere, still not sure though, you try it on paper, maybe you'll cross this threshold I'm getting stuck at

Oh wait I'm an idiot, yeah it's pretty basic, IDK why I kept on sending x at the denominator after substitution. it's fairly simple now

You still here @uneven dock ?

[uneven dock]

YES SIR

help..

[whole nexus]

Where are you stuck?

[uneven dock]

where do i go after usub

[whole nexus]

Well you get 2xsinx dx yeah?

[uneven dock]

ok

im confused hold on

[whole nexus]

(I just thought for sin(sqrt(x)) for this)

[uneven dock]

when you usub youre doing u = sin(sqrt(at)) right

but what does that accomplish

[whole nexus]

No no

[uneven dock]

oh ok

[whole nexus]

Take u = sqrt(at)

The one you did would be very complicated

[uneven dock]

ok

wait so i get du = sqrta * (1/2)(t^(-1/2))

i still

dont see where this is going

[whole nexus]

It'd be easier if you take u^2 = at and then differentiate

[uneven dock]

huh

[whole nexus]

Well u = sqrt(at)

So u^2 = at

[uneven dock]

yes

[whole nexus]

Now differentiate u^2 = at

[uneven dock]

2u = a

[whole nexus]

Whoops, missing the du and dt

[uneven dock]

2u*du = dt

[whole nexus]

Alright, so 2u*du = adt

[uneven dock]

but if a is a constant wouldnt it just disappear

[whole nexus]

Now dt = (2u/a)du

[uneven dock]

wait nvm

im dumb

thats wrong

[whole nexus]

No you're Avacado

[uneven dock]

im wrong i should clarift

wait so

[whole nexus]

Okay, now substitute dt as (2u/a)du, sin(sqrt(at)) as sin(u)

[uneven dock]

we get 28/a * integral of (sin(u) * u du) right

[whole nexus]

Yes (assuming the 8 was accidently pressed)

[uneven dock]

2*14

28

cuuz the 14 was pulled out

[whole nexus]

Ahhh yes, apologies

Okay, so focusing our attention to u*sinu du

Apply IBP

Where integrate sinu and differentiate u

[uneven dock]

28/a * (ucos(u) - sin(u))

right

[whole nexus]

One second

Can you check if it would be +sinu instead of -

Also -ucosu

You still here @uneven dock ?

[valid thorn]

what is the remainder when X6 is divided by (x2 +1)?

[whole nexus]

Sorry but get your own channel mate

[valid thorn]

okay

[cedar kiln]

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[wild plaza]

I don't understand what this shape is supposed to be

[chrome elk]

The shape S?

It's an ellipse. Like a stretched out circle

Try plotting it in desmos to get a better look

[wild plaza]

the shape is a circle with an isosceles triangle? the base of the circle is the hypotenuse of the triangle?

[chrome elk]

Oh nvm I'm blind

The base is an ellipse

[flint plinth]

"cross sections perpendicular to the x axis"
in other words, cross sections parallel to the yz plane, if that helps

it's a bit hard to visualize mentally but the description seems clear enough that you could just work with it

like draw a typical 2d slice parallel to the yz plane

[cedar kiln]

@wild plaza Has your question been resolved?

[pastel vault]

this is the side view

hopefully it makes more sense now

the flat shape that these triangles trace on the xy-plane is an ellipse

[cedar kiln]

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[dark crater]

Question 4.e)
Along with my attempt, I wanna know how far off I am and where I could have gone wrong please and thanks.

[cedar kiln]

@dark crater Has your question been resolved?

[dark crater]

<@&286206848099549185>

.close

[cedar kiln]

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[lone parrot]

Given a natural number n, knowing that when we add another 3 digits to the end of n + 26, we get the sum of the natural numbers from 1 to n. Find n and the 3 digits.

[lone parrot]

I'll call n after the change as t.

so we have: $t = \frac{n(n+1)}{2}$

[wraith dagger]

Thomas

[cedar kiln]

@lone parrot Has your question been resolved?

[lone parrot]

<@&286206848099549185>

anyone here?

[north wyvern]

i dont quite understand the question

so, we let n be a natural number, consider n+26 and attach any three digits at the end
then n+26 with those digits is the sum of naturals 1 to n
find n and its digits

is this correct

regardless, its key to note that (assuming base 10) we have done $1000(n+26) + k$, where $0\le k<1000$

[wraith dagger]

Cycadellic

[lone parrot]

yes

[zealous pawn]

So basically what you can do is Equate 1000 * (n + 26)+ x to this n(n+1)/2......eqn 1
Now X belongs to [0 , 999)
Obtain a expression in n It will be a quadratic expression Then apply the quadratic formula to find the solutions of n In terms of x
Then basically put x= 0 one time and x = 999 second time
I have done a rough work And basically the range of n Coming out like this is gonna give you a range in decimal places So there will be only one integer between them.....most prolly, And that integer will be your "n"
Now once the N is figured You can put that in equation one And find X

[north wyvern]

so, whats your question exactly?

[zealous pawn]

This is quite lengthy But I don't I am not able to think of any other approach

and the numbers are gonna be huge

[lone parrot]

find x and the three digits

[zealous pawn]

so brace it

[zealous pawn]

yesss

[zealous pawn]

x is gonna be a 3 digit number if u did everything right
and u can find those 3 digits like that

[north wyvern]

we arent allowed to do problems for you, but you have the keys necessary to solve it

if you need help, we need you to show us your process from here

[lone parrot]

So the quadratic is $x^2-1999x-52000$

[wraith dagger]

Thomas

[north wyvern]

wait wait

which quadratic? we need a full statement to understand if youre doing this correctly

so, show work, or set it equal to something

[zealous pawn]

uhm well there should be -(52000 +2 (that three digit number)) in the end
ur missing that 2(bla bla) term

[zealous pawn]

i think he made quadratic in "n"

[lone parrot]

yeah

i'm just familiar with writing x

than n

[north wyvern]

so, we need an equation to work with, what do you have to start

[lone parrot]

so if we have $\frac{n(n+1)}{2}=(n+26)*1000$ then we have the above equation

[wraith dagger]

Thomas

[lone parrot]

and when calculated, it gives n of about 2024.68

[north wyvern]

you forgot the three digits

[zealous pawn]

nooo
u have to add those 3 digits as well in the RHS to give u a range
ukw nvm n=2024.68

[north wyvern]

right now it just ends in 000

[zealous pawn]

with those 3 digits

iw ould have been from 2024.68 to 2025.something somethinh cuz u wouold have added 999 as one extrema
hence n=2025

[zealous pawn]

yess

[lone parrot]

so we replace n with 2025 in the new equation: $\frac{2025(2025+1)}{2} = (2025+26) * 1000 + x$

[wraith dagger]

Thomas

[zealous pawn]

yess sir
now u can find x

[lone parrot]

so we have x = 325

which should be it

[zealous pawn]

yes idk the value of x but it should be a 3 digit number and now u have ur 3 digits

[lone parrot]

testing seems correct

btw, any tips for solving competitive math?

[north wyvern]

study study study

you need to solve all the problemsets from the past that you can

cause they are sadistic

[lone parrot]

like how sadistic actually?

[north wyvern]

depends

turns out even basic algebra and geometry can be incredibly hard lol

[lone parrot]

like my question?

[north wyvern]

oh this question is nothing comparatively

[lone parrot]

oh god

i'm scared now

[north wyvern]

imo has some real wonders on it

[zealous pawn]

imo? good luck buddy

[north wyvern]

im just using comparison, not that hes necessarily doing the imo

[north wyvern]

do you have any other questions?

close this channel if not, ask if so

[lone parrot]

oh i forgot

thanks for the help

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[restive bison]

find all polynomials P(x) and Q(x) such that
Q(P(x)+y)=P(P(x))+P(y)
for all real x and y

[restive bison]

the solutions i found are
Q(x)=k, P(x)=k/2 (for any real k)
Q(x)=ax+b, P(x)=ax+b/2 (for any real a and b)

[north wyvern]

show your work

also, what is your question, concretely

[cedar kiln]

@restive bison Has your question been resolved?

[feral sapphire]

sure, these are solutions, but any progress in showing whether there may/may not be other solutions?

[restive bison]

i'm not really sure where to start in finding other solutions tbh

trying a quadratic seems impossible

[feral sapphire]

typical ideas include:

1. the degrees on both side should be equal, maybe that will restrict the choices for P and Q
2. what term might appear on one side but not the other? Since they are equal, it might give some info about P and Q.
3. Treat one of the variables as constants

[restive bison]

if i make y=0
Q(P(x))=P(P(x))+P(0)
and if theyre quadratic, then Q(x)=ax^2+bx+c and P(x)=ax^2+bx+2c?

because P(0)=c

but i'm not sure if it holds for all y

[feral sapphire]

thats a big claim right there, need justification

[restive bison]

i'll try expanding it then

[feral sapphire]

expanding the whole thing does work, but it might be easier if you focus on specific terms, that is, 2) mentioned above

[restive bison]

how? like just focus on the P^2(x) terms?

oh wait i think i get it

[feral sapphire]

no, for example, you can look at whether there are "xy" term on either side

or x^2 term, or y^2 term, etc

[restive bison]

oh yeah, there's a missing P(x)y term on the RHS if it's a quadratic

[feral sapphire]

P(x)y is not a term but you are on the right track

After you expand P(x)y, it might cancel with other term

[restive bison]

so basically there's no like x^2y and xy terms on the RHS?

[feral sapphire]

nice observation

[restive bison]

and in that case, the RHS of this equation doesn't have any x multiplied with y, but on the LHS for a power above 2, there's always an xy, x^2y and other terms, so there's no other solutions for a power above 1 right?

[feral sapphire]

"there's always an xy, x^2y and other terms" is a big claim, how do you know if they would cancel or not? what if after you expand there is a xy and there is a -xy so they cancel?

again, right idea, but need to be more careful

[restive bison]

i think we take the highest power of the (P(x)+y) term on the left, and then the 2nd highest term of y, so if it's a power 3 polynomial, we take (P(x)+y)^3 then the part where there's P(x)y^2, and since P needs to be a power 3 polynomial too, the highest term is x^3y^2, which doesn't exist on anywhere else

and for any power n polynomial, the term that can't be cancelled out is the x^ny^(n-1) term

[feral sapphire]

ok, looks ok, but justify the claim: "P(x) need to be a power 3 polynomial too". More generally, justify that P(x) and Q(x) must have the same degree

technically your claim "x^3y^2 doesn't exists on anywhere else" also need justification, but this is pretty much just computation, so I won't bore you with it

[restive bison]

oh wait, sorry for ignoring what you said
that's basically the whole question, and i just translated it

[restive bison]

oh ok, thanks

how about this one
f(xy+f(f(x))=yf(x)+x

my first thought is substituting y=0 and that gives f(f(f(x)=x

[feral sapphire]

wait we aren't even done with the first one

[restive bison]

do i just generalize it to n instead of 3?

or how do i do the computation?

[feral sapphire]

yeah, and you need some pretty long case-work to first conclude P and Q must have the same degree

[restive bison]

how would that case-work look like?

[feral sapphire]

degree argument + constant cases are annoying, you should try it yourself

[restive bison]

are degree arguments like 2 polynomials with a different degree can't always be equal to each other for all values of x?

[feral sapphire]

Let the degree of P be n, and degree of Q be m, then P(Q(x)) has degree nm

[restive bison]

but it's good enough to just say that the degree of LHS and RHS is equal right?

so LHS has degree mn, and RHS has degree n^2
mn=n^2
so n=m or n=0

but n has to be equal to m because if n=0 and m is not equal to 0 then degree of LHS and RHS is different

[feral sapphire]

there is a chance n^2=n and in which case the degree of P(P(x))+P(x) might be 0

ok nvm its P(y)

[feral sapphire]

yeah that should work

[restive bison]

and the constant cases are basically putting P and Q to k like i did?

[feral sapphire]

yeah let P(x)=c and Q(x)=d and you just solve

[restive bison]

so basically
just prove P(x) and Q(x) have the same degree
check the constant case
prove there's no solution for degrees >=2
find the linear case

[feral sapphire]

yeah pretty sure that works

[restive bison]

how do i solve this then?

all i can think of is just that f(f(f(x)))=x and now i'm not sure how to proceed

[feral sapphire]

idk what else to say other than plug in some nice values

plug in x=0 for example

[restive bison]

so f(f(f(0)))=yf(0)

yeah i don't really know how to continue this though

[feral sapphire]

^

[restive bison]

oh, f(0)=0 then

[feral sapphire]

and you can show that $f(x)\neq 0$ when $x\neq 0$

[wraith dagger]

qwertytrewq

[restive bison]

i guess i can try to prove this by contradiction
if f(x)=0 for x not equal to 0
f(xy+f(0))=x
f(xy)=x
but f(x) can't be 1/y because y isn't constant

[feral sapphire]

wdym f(x) can't be 1/y

you showed f(xy)=x

you already fixed your x to be a value where f(x)=0

[restive bison]

oh i thought it was impossible because y is a variable

[feral sapphire]

need more rigour

[restive bison]

more rigour? like testing in the initial condition?

[feral sapphire]

because what you said didn't make sense to me

once you picked your x, to be a value such that f(x)=0, you should treat x as a constant

so your f(xy)=x implies f(x)=1/y doesn't make sense (also it should be x/y)

[restive bison]

oh wait yeah i messed up there

so f(x)=x/y, then i should try it in the original function?

[feral sapphire]

no f(x) is not x/y (you have not shown that), you've only shown that f(xy)=x (because x is a fixed value you picked)

its like saying f(1y)=1 thus f(x)=x/y

[restive bison]

so something like f(y)=1?

wait or does it show that f(x) is linear?

[feral sapphire]

think about your argument: "suppose that f(1)=0, we want to reach a contradiction, and bluh bluh bluh you deduce that f(1y)=1" it doesnt imply from here that f(x)=x/y for all x,y,

[feral sapphire]

no, i just did the case where you picked x=1, and in that case yes f(y)=1 which is impossible

[restive bison]

wait so it just shows that f is linear right?

[feral sapphire]

again, only in the case where you picked x=1

[restive bison]

oh is it just f(x)=x? it makes sense for the question

[feral sapphire]

our deduction is nowhere close to that conclusion yet

I think you might be a bit lost in this

we have f(0)=0, f(f(f(x)))=x

[restive bison]

yeah i haven't had much experience in solving functional equations

[feral sapphire]

and we are trying to show f(x) is nonzero for nonzero x

[restive bison]

oh wait i think i got it
f(xy)=x
f(f(f(xy)=f(f(x)
xy=0 which is a contradiction

because x is not 0, and y in this case is treated as a variable

[feral sapphire]

why does f(f(f(xy)))=f(f(x)) gives you xy=0?

[restive bison]

because in this case f(x)=0 and f(0)=0 so f(f(x)=f(0)=0
and since f(f(f(x)=x, f(f(f(xy)=xy

[feral sapphire]

ok so you deduced that for all y, xy=0, which is a clear contradiction

the argument can be made much simpler: take y=0, then x=f(xy)=f(x0)=f(0)=0

An alternative solution: if f(x)=0, then x=f(f(f(x)))=f(f(0))=f(0)=0.

[restive bison]

oh yeah those are simpler

[feral sapphire]

either way, f(x) are non-zero when x is non-zero (useful for later

[restive bison]

oh yeah can i also set x to like -a and y=0
so f(f(f(-a)=-a and it shows f(x) is odd?

[feral sapphire]

how does that show f(x) is odd?

f(x)=-f(-x) is what you need to show to show that f(x) is odd

[restive bison]

-f(f(f(a)))=-a=f(f(f(a)))
f(f(f(a)))=-f(f(f(-a)))
so i guess it only shows that f(f(f(x))) is odd, but not f(x)

but if f is even then f(-a)=f(a) so f(f(f(-a)))=f(f(f(a)))
that becomes f(f(f(a)))=-f(f(f(a))) which is a contradiction, so f(x) is not even

[feral sapphire]

yup, but f(f(f(x))) being odd is obvious, since it is just equal to x

[restive bison]

oh ok

[feral sapphire]

true

[restive bison]

ok so f(x) cant be even, but that doesn't guarantee it's odd

[feral sapphire]

just doing f(f(f(x)))=x is pretty futile, this equation has LOTS of solutions

you need to look back at the original equation f(xy+f(f(x)))=yf(x)+x. there is this annoying f(f(x)) on the LHS, but you know f(f(f(x)))=x, so, what do you think we should plug in for x?

[restive bison]

like f^-3(x)?

so the f(f(x) turns to f^-1(x)

[feral sapphire]

you don't know if f have inverse

for example f^{-1} is illdefined when f(x)=x^2

[restive bison]

so do i just plug in f(f(f(x))) to every other x?

[feral sapphire]

why not just plug in f(u) into x

f(f(x))) turns into f(f(f(u))) which is just u

[restive bison]

wait so it's kind of like inverse then?

but the inverse doesn't really have to exist?

f(yf(u)+u)=yf(f(u))+f(u)

[feral sapphire]

remember for u non-zero, f(u) is nonzero. so you can plug in c/f(u) for y

which is natural since you want to get rid of the yf(u) on the left hand side

[restive bison]

so f(c+u)=cf(f(u))/f(u)+f(u)

then do i just plug in something for c+u?

[feral sapphire]

as a function of c, you notice that the ffunction is linear right?

for example fix u=1

then f(c+1)=cf(f(1))/f(1)+f(1)

this is just a linear function

and this holds for all c

[restive bison]

oh yeah since all those f(1)s are constant

so to match th variables
f(c+1)=f(f(1)/f(1)*(c+1)
=cf(f(1))/f(1)+f(f(1))/f(1)
so f(f(1))/f(1)=f(1)

since it's linear and f(0)=0, then f(x) has no constant term

[feral sapphire]

um.....

you can just say f(x)=mx+b for some m and b and plug into f(f(f(x)))=x

[restive bison]

oh

[feral sapphire]

why does f(c+1)=(c+1)f(f(1))/f(1)?

[feral sapphire]

u r making the claim that f(f(1))/f(1)=f(1) in order for that to be true

[restive bison]

f(f(f(x)))=f(f(mx+b))=f(m^2x+bm+b)=m^3x+bm^2+bm+b=x
m^3=x
b(m^2+m+1)=0
b=0 because m^2+m+1 has no real roots
so f(x)=x

[feral sapphire]

did you confuse f(f(x)) with (f(x))^2?

[restive bison]

my reasoning was that f(0)=0 and since it's linear, then f(x)=mx and b=0

[feral sapphire]

but i didn't get the first line

how did you even get f(c+1)=(c+1)f(f(1))/f(1)

[restive bison]

like to match the c variable i guess

[feral sapphire]

we had f(c+1)=cf(f(1))/f(1)+f(1)

I don't see how it follows that f(c+1)=(c+1)f(f(1))/f(1)

[restive bison]

like my reasoning was f(x) has no constant term
so f(c+1)=m(c+1) and to match to c variable then i set the m=f(f(1))/f(1)

[feral sapphire]

again ur idea is fine i just didn't get how f(c+1)=(c+1)f(f(1))/f(1)

It should've been f(c+1)=(c+1)f(f(1))/f(1) -f(f(1))/f(1)+f(1)

[restive bison]

oh ok

[feral sapphire]

this is f(c+1)=cf(f(1))/f(1)+f(f(1))/f(1)
what we had was f(c+1)=cf(f(1))/f(1)+f(1)

u see the discrepancy right?

[restive bison]

yeah i think i just skipped a step that wasnt supposed to be skipped

so the only solution is f(x)=x right?

[feral sapphire]

minor typo: m^3=1 not m^3=x

[restive bison]

oh yeah mb

well, thanks for helping

.close

[cedar kiln]

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[feral sapphire]

np

[cedar kiln]

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[obsidian pelican]

how can i calculate the determinant of this?

[tropic oxide]

are you familiar with elementary row/column operations

[obsidian pelican]

i think so yes

[tropic oxide]

ok

consider looking at the two columns that don't contain any scary shit

consider what col op you can perform to make things suddenly so much nicer

[obsidian pelican]

well

c4 - c3 it becomes in the c4:
0
-1
0
0

[tropic oxide]

rather "subtract c3 from c4" to be more precise but yes thats the idea

do you see now how to proceed

[obsidian pelican]

yep sorry, english isnt my firzt language so i suck a lil with the terms

[tropic oxide]

what is your first language

[mental trail]

also it's c4+c3 you meant right?

[obsidian pelican]

yep i think so, i can quit the third column and the seconf row, because the determinants would be 0 so its canceled out right?

[tropic oxide]

i mean why not just expand along the new C4

then you just get a single 3x3 determinant with no scary shit

[obsidian pelican]

yeah you cna do it in that way too i think

[obsidian pelican]

what do u mean?

[tropic oxide]

laplace expansion

[obsidian pelican]

oh yeah okay, yes thats what i meant i didnt know the name

this rifght?

[cedar kiln]

@obsidian pelican Has your question been resolved?

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[torn marsh]

why is b wrong for the second derivative

[night wedge]

You have to use the product rule the second time

What’s the other way?

[north wyvern]

my silly misreading is the other way

sorry lol

[night wedge]

That sounds like a great method

[north wyvern]

well, i think you have to do the product rule for this one

[torn marsh]

how will I know which one is u and which one is v

[north wyvern]

do the product rule first

[torn marsh]

I mean I know e^-2x^2 is u and -2x is v but whats the rule for determing which one is u and v

[north wyvern]

then we will use chain rule

$$f'(x)=-2xe^{-x^2}$$ is right, start from here

[wraith dagger]

Cycadellic

[nova snow]

there is no rule you pick which one to make u and v

[torn marsh]

cuz I was wondering what if I do -2 as u and e^-x^2 as v but I left x out

[north wyvern]

we generally dont sub the coefficients

we only use chain rule on functions like f(g(x))

we know how to find derivatives of exponents, and of squares, chain rule tells us how to find the derivative of a squared exponent

[north wyvern]

so, you should start by applying the product rule here

[buoyant latch]

you can do this even with constants but you didn't apply it correctly

if in 2xe^x you let u = 2 and v = xe^x you should still get the same answer

[torn marsh]

got the answer now

thanks

.close

[cedar kiln]

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[sinful wyvern]

I apologize for the language barrier: << Show that, using case disjunction reasoning, that:>>

[slender atlas]

hi, but this channel is not freed up yet.

please post your problem in an available channel, like #help-48.

[sinful wyvern]

Thank you

[cedar kiln]

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[wild belfry]

Discrete maths question, above is the question and below is the solution

I dont understand the proof they give

I understand x^2 - y^2 = (x-y)(x+y)

[chrome elk]

Which step do you not understand

[wild belfry]

And I understand the part about if x=y then its equal to 0

But what is the 'otherwise' part?

[chrome elk]

Basically they're proving that x² - y² is never 1

[pastel vault]

because $x \ge y$, $x - y \ge 0$

[chrome elk]

And falsity implies everything

[uneven dome]

hint: vacuous truth

[wraith dagger]

south

[chrome elk]

Also I'm assuming your definition of ℕ does not include 0?

[wild belfry]

Not mine but the courses lol

[chrome elk]

Yeye

[wild belfry]

Ok this makes sense

[pastel vault]

yeah that's how you can have $x \ge 1, y \ge 1 \implies x + y \ge 2$

[wraith dagger]

south

[wild belfry]

Okkkk I think I see

[pastel vault]

so essentially the whole point was to show that x^2 - y^2 is never equal to 1

so since the premise is false, it's vacuously true

anything follows from a false statement

[wild belfry]

Yeyeye I think I get it now

Thanks everyone you explained it well!

[chrome elk]

It was mostly south

[wild belfry]

Haha you all helped

❤️

.close

[cedar kiln]

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[pastel vault]

yeah it's the small steps that always get people to trip up

[nova snow]

hello odie

[wild belfry]

Hi water beaaaamm 🤩 ❤️

[cedar kiln]

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[crimson sedge]

Should I be using continuity corrections for part g to 19, or is it not necessary here? I should complete it in R

[north wyvern]

i meant its not strictly necessary, but that really depends on what youre doing to solve for it

[crimson sedge]

I should write a R code. I am not sure, it just asks these d) Find the probability that X is equal to 17.
(e) Find the probability that X is at most 13
(f) Find the probability that X is bigger than 11.
(g) Find the probability that X is at least 15.

[north wyvern]

if you wanted faithful code, id actually run random trials, and find the distribution for a huge number of random trials

that way you are showing the math does do what we want it to

[crimson sedge]

Ok, thank you

(d) Probability that X = 17 (exact Binomial)

dbinom(17, size = 31, prob = 0.447)

Parameters

n <- 31
p <- 0.447
mu <- n * p
sigma <- sqrt(n * p * (1 - p))

(d) Probability that X = 17 (Normal approx with continuity correction)

pnorm(17.5, mean = mu, sd = sigma) - pnorm(16.5, mean = mu, sd = sigma)

I meant this, in terms of which method to use

It worths like 1 mark

I want to know like at what instance should I do the normal approximation with continuity correction versus the exact binomial

Is it that I should use the continuity correction if I am doing it manually?

[flat mica]

you use normal approximation if exact calculation is tol computationally intensive

if it would take your computer hours or days to run for example

[crimson sedge]

Ah I see

In this case, would this be ok dbinom(17, size = 31, prob = 0.447) and no need to use the continuity correction

[flat mica]

right

[crimson sedge]

Awesome

[flat mica]

also note that the larger n is, and the smaller p is, the less the continuity correction matters

[crimson sedge]

Thank you

If I should do at most 13, would it be like this pbinom(13, size = 31, prob = 0.447)

[flat mica]

if n=2000 and p=0.0005, i wouldn't even bother

[crimson sedge]

It's the first time I am coding in R

[flat mica]

no thats a case where you would

um, how do say it

at most 13 means 0 or 1 or 2 or 3 or 4 or ... or 12 or 13

in which case its easier to find the complement of the event

[crimson sedge]

oHH

[flat mica]

or

normal.approximation works well there

if you dont want to do 13 different calculations

but complement tricks often avoid it

[crimson sedge]

Ok, so I can either do complement or normal approximation?

Sorry, I accidentally resent the image

[flat mica]

or you can use

in R

[crimson sedge]

Oh, well, the entire question should be done in R

[flat mica]

do you know what vectorization is in r

[crimson sedge]

Is it like assigning a varaible?

x -> 9

or something like that

[flat mica]

assigning a tuple of variables

experiment wuth this:

pbinom(1:13,...)

it will.do it 13 times

with 1,2,3,4,5

actually you needvto start with 0

[crimson sedge]

yes

[flat mica]

so if you put in 0:13 it will run it for 0 successes, 1 success, 2, 3, ..., 13

[crimson sedge]

I did this part before pbinom(0:31, size = 31, prob = 0.447) # cdf values

[flat mica]

this is another way to do it

[crimson sedge]

for the question before

[flat mica]

ah okay

[crimson sedge]

Yeah

[flat mica]

anyway when to use normal distribution? when it becomes impractical or slow to do exact

[crimson sedge]

Alright, so for clarity, to do this in R (e) Find the probability that X is at most 13

I should use complement?

[flat mica]

for at most 13 with n=31 id say to approximate

because the complement is also annoying

[crimson sedge]

Ah okay

(f) Find the probability that X is bigger than 11: 1 - pbinom(11, size = 31, prob = 0.447)

and the last one like that

(g) Find the probability that X is at least 15. Is it same as bigger than so, 1 - pbinom(14, size = 31, prob = 0.447)

[flat mica]

i have to go

gl

[crimson sedge]

Thank you

[flat mica]

tag helpers if you need more help

[crimson sedge]

Sure

[flat mica]

came back to say that

this was dumb, its not annoying @crimson sedge

the complememt is simple

[crimson sedge]

Alright

[flat mica]

anyway the answer should be roughly the same if n is large enough, p is small. enough, and np is medium enough

though many books only focus on n being large enough, its actually a balance between n and p, though

secret math knowledge

[crimson sedge]

Ah, thank you

.close

[cedar kiln]

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[median sparrow]

hello

[sudden cairn]

Dude

Post your question first

[median sparrow]

can anyone help me with exponential

oh mb

[sudden cairn]

Question please

[gray hamlet]

lol

[median sparrow]

wait

[sudden cairn]

Plants vs Brainrot server is crazy

[median sparrow]

chilll

[sudden cairn]

Sorry

[median sparrow]

log2 x^2/y