help-27

[woven radish]

LocalLunatic

[hollow laurel]

Well you plug in- yeah, plug in a and b for the X

[fair juniper]

using this makes sure your signs are correct

third time today?

the scams are relentless

[hollow laurel]

OH MY F&%&*ING

WHO IS SENDING TS

I am trying to do my calculus HW and the next person to send one of these is getting their DM's blown up

[fair juniper]

you know it's most likely a botted hacked account right?

[hollow laurel]

Yeah but still shame on them for clicking on a fake nitro link

nevermind I'll figure it out myself

.close

[devout snow]

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[devout snow]

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[wispy warren]

e^x - 1 rotates around the y-axis ( 0<=x<=1)

[wispy warren]

determine the volume

I'm stuck

[supple knot]

$\int \ln(x) dx = x \ln(x) - x + C$

[woven radish]

riemann

[supple knot]

if you haven't seen that, then either differentiate both sides to confirm or use integration by parts to discover the antiderivative for yourself

oh i'm blind you did use it

[wispy warren]

yes

the integration by parts thingy

I'm stuck at the last step

don't know how to proceed

[supple knot]

what are you stuck on, you just only started the integration by parts

$\int u dv = uv - \int v du$

[woven radish]

riemann

[wispy warren]

don't know how to handle this

[supple knot]

what does that even say

[wispy warren]

y/y+1

primitive of that

[supple knot]

u sub or add and subtract 1 in the numerator

[wispy warren]

how would I u sub this

y+1 = u

du/dy = 1

[supple knot]

yea then use sum integral property

,tex .int rules

[woven radish]

riemann

[wispy warren]

what would I substitute the numerator with

[supple knot]

you don't use a separate substitution. you solve for y here in terms of u

[wispy warren]

duce is confused

[scarlet sequoia]

u = y + 1
du = dy
y = u - 1

and it should be (u-1)/u du

[supple knot]

use $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$

[woven radish]

riemann

[devout snow]

@wispy warren Has your question been resolved?

[devout snow]

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[vagrant crest]

Ahem

[vagrant crest]

According to the limit definition of the derivative, we have: f'(x) = lim [f(x+h) - f(x)] / h or f'(x) = lim [f(x+dx)-f(x)] / dx Now, I'm curious whether this holds true for as well: f(x+dx)-f(x) = f'(x)dx = d(f(x)), as it has allowed for many algebraic manipulations in calculus, including the integral and chain rule.

My strategy closely resembles the concept of infinitesimals. Withou using the limit operator I consider dx to be "0" and I'm not merely stating that it is a "small number". What I'm saying in here is that dx is "0" as quantity we've flagged so we know which "0" we are dealing with

But this approach has led to numerous misconceptions due to the varied syntax used in standard mathematics. What I am really asking is whether there is a rigorous method to work with this concept.

My understanding of calculus is based on certain systematic assumptions and I'm not sure if there are any flaws in them. Could anyone explain with an example where these assumptions may fail? I am currently exploring some examples for multivariable calculus, I would appreciate any guidance.

Also here are the examples I had problems with:

[supple knot]

You should learn existing notation before inventing your own

[vagrant crest]

I mean I always invent the stuff in my head and check if it holds, that's how I learn

[floral heart]

help how can rings be finite if they have to be closed?

[devout snow]

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

@vagrant crest Has your question been resolved?

[vagrant crest]

well, guess I'll trash it out

.close

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[faint gorge]

tuff <@&268886789983436800>

[devout snow]

Channel closed

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[willow helm]

[devout snow]

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[upper cobalt]

for a function that is the sum of 2 functions whos differentiable domains i know, is the differentiable domain of the main function the intersection of these 2 domains?

[upper cobalt]

and same is it for difference, dividing, amplyfying etc

[dark sable]

I think yeah

fundamental theorem of calculus thing

[dark sable]

uhhh, I'm not too sure about dividing and amplifying it'd probably still be the intersection of the 2 domains

[solid perch]

hm, I was thinking something like f(x) = |x| and g(x) = -|x|, then their sum is 0 and differentiable everywhere even if f and g are differentiable everywhere except at x = 0.
correct me if I am wrong though.

[dark sable]

um

er

|x|-|x| is goofy

it is differentiable everywhere except 0 I guess

[solid perch]

it's just a simple example I tried to come up with, but if it is wrong, I apologize.

[dark sable]

it makes sense

[void fox]

the set of points where f+g is differentiable will be a superset of the (intersection of) points where f is differentiable and where g is differentiable

as others have pointed out this inclusion may be strict

[upper cobalt]

then how do i find out its differentiable domain?

[faint gorge]

-# why union?

[void fox]

intersection

my bad

[devout snow]

@upper cobalt Has your question been resolved?

[void fox]

actually i also meant to say superset instead of subset

fucked up doubly

[devout snow]

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[woven radish]

Renato

[rough nova]

To start with, you could let a=5x, b=5y, (x,y)=1 to make things easier

[spring oasis]

sure then what

[rough nova]

Then (a^4+b^4,2ab^2)=...?

[craggy dagger]

hold on this seems familiar

[spring oasis]

5^4(x^4+y^4)

[rough nova]

?

Did you forget about the second part

[spring oasis]

2(5x * 5^2 * y^2)

2.5^3.x.y^2

[spring oasis]

nobody helped

we didnt even solved the Q

after days of the Q being open

[craggy dagger]

well, the first step is to calculate all possible values

[spring oasis]

you remember seeing this question before so in the past you saw this exercise and didnt helped huh

Looks like I ended up .closing because we didnt arrived to anything

[craggy dagger]

start from doing some modified Euclidean algorithm methods

[rough nova]

So you can factor 5^3 out first

[craggy dagger]

Try to find small A, B such that
gcd(a^4+b^4, 2ab^2) divides Aa^B

[spring oasis]

gcd(a^4+b^4,2ab^2) = gcd(2.5^3.x.y^2, 5^4.(x^4 + y^4)) = 5^3 . gcd(2.x.y^2, 5(x^4 + y^4)

with gcd(x,y) = 1

[spring oasis]

this exercise is fucked

I see why nobody helped in like a week this shit was opened

suppose some prime p divides 2xy^2

since gcd(x,y) = 1 then either p | x or p | y but not both

[dapper tiger]

https://tenor.com/blVuB.gif

[spring oasis]

yes!

[dapper tiger]

I like your style, but you should have chosen that p is not 5 to be able to use Maribel's

or not 2

so you can use maribel's on 2xy²

2 and 5 will clearly be a bit more special than other prime factors, from the fact they are in your gcd

[spring oasis]

its always the same pattern, I think is something I can add to my toolbox

[spring oasis]

suppose some p != 2 divides 2xy^2

[dapper tiger]

well it divides x or y²

but not both because they are coprime

[spring oasis]

yes and since p is prime if p | y^2 => p | y

[dapper tiger]

now your problem is the 5, because you won't be able to say that p | 5(x^4+y^4) => p | x^4+y^4

that's the reason why the smart trick is to say p is not 5

and do the case p = 5 separately

[spring oasis]

this exercise is so nasty

is worse than the previous one

[dapper tiger]

it's the same thing we did in disguise tbh

[spring oasis]

ok

suppose some prime p != 5 divides 2xy^2

then either

1. p = 2,
2. p | x and gcd(p,y) = 1
3. p | y and gcd(p,x) = 1

[dapper tiger]

can p divide 5(x^4+y^4)?

[spring oasis]

1. no
2. no
3. no
   rigurous justifications: no

well I mean, suppose p = 2 then we know gcd(x,y) = 1 worst case scenario p | x or p | y but not both

[dapper tiger]

you should be careful with p = 2

he is a tricky one

when x and y are odd, x^4+y^4 is even

that's why p = 2 is also a special case

you can care about it later

but 2) no and 3) no are right

[spring oasis]

suppose p != 5 and p != 2 but p is prime such that p | 2xy^2 => p | x or p | y but not both

from here we get that gcd(p, x^4 + y^4) = 1

[dapper tiger]

yep

so you know gcd(2xy², 5(x^4 + y^4)) isn't divisible by p

d can only be divisible by 2 and 5 as primes

so is of the form 2^k * 5^q

[spring oasis]

how to do the special cases

[dapper tiger]

well, now you know the form of gcd(2xy², 5(x^4 + y^4)) as 2^k * 5^q, and that's what was important

like the exercise we just did, we'll try to find the possible values for k and q

we know that x and y are coprime, so 2 can divide x or y but not both

same for 5

let's list them all like pokemon

case 1: 5 divides x but not y

case 2: 5 divides y but not x

case 3: 5 divides none

[spring oasis]

why

[dapper tiger]

because I went too fast and I'm wrong

the case is possible, corrected

[spring oasis]

ok

what about 2

[dapper tiger]

we'll do it after

we'll do case 1, 2, 3 on it too

for 5;
case 1: 5 divides x so 5 divides 2xy² and 5(x^4+y^4), but not x^4+y^4, since x and y are coprime, only x^4 is divisible by 5, not y^4, so x^4+y^4 isn't
case 2: 5 divides y so 5 divides 2xy²... same stuff but x and y inverted

[spring oasis]

d = gcd(2xy^2, 5(x^4 + y^4))

[dapper tiger]

case 3: 5 divides none, nothing to say, it just won't be in the gcd

so in both cases 1 and 2, q is at most 1

and can be 0 when 5 divides nor x nor y

now, let's do the cases about 2

2 divides 2xy²

and we have several cases

if 2 divides y (so not x)

it means 8 = 2^3 divides 2xy²

if 2 divides x (so not y), 4 = 2^2 divides 2xy²

if 2 divides none, 2 divides 2xy² and it's no more division by 2

so now we have to check what cases can really divide our gcd

ie how many times 5(x^4+y^4) will be divisible by 2

and you were right, it's actually a bit harder than the other exercise because of 2

but we'll manage it

let's look at x^4+y^4 mod 8, so we'll check at the same time if it can be divisible by 4 or 8

[spring oasis]

case 2 | y and gcd(x,2) = 1
5(x^4 + y^2) = (x^4 + y^2) = x^4 (mod 2)

[dapper tiger]

yes I treated this case

if 2 divides y (so not x)
it means 8 = 2^3 divides 2xy²

that's why we have to check if 8 can divide 5(x^4+y^4) too

it will tell if our gcd is divisible by 8 or not

5 doesn't matter mod 8, it's invertible

the question is can x^4+y^4 be 4 or 0 mod 8

a power of 4 mod 8 can only be 0 or 1

so the possible values of x^4+y^4 mod 8 can only be 1 or 2 (not 0 since x and y are coprime, 2 can't divide both)

so the max power of 2 that divides our gcd is 2

and k = 0 or 1 too

so gcd(2xy^2, 5(x^4 + y^4)) = 2^k 5^q with 0 <= k, q <= 1

ie it can be 1, 2, 5 or 10

gcd(a^4+b^4,2ab^2) = 5^3 gcd(2xy^2, 5(x^4 + y^4))

= 2^k 5^(3+q) with 0 <= k, q <= 1

the gcd we were looking for can only be 125 = 5^3, 250 = 2*5^3, 625 = 5^4, or 1250 = 2*5^4

[spring oasis]

wtff

[dapper tiger]

it's the same stuff we did in the exercise before, the only difficulty added here was to count how many times 2 divides x^4+y^4

which is at most 1

[spring oasis]

how

[dapper tiger]

if you're divisible by 4 or 8

you're equal to 4 or 0 mod 8

the question is can x^4+y^4 be 4 or 0 mod 8
a^4 mod 8 can only be 0 or 1 (you can list all values mod 8)

and x^4 and y^4 are coprime

so x^4+y^4 can only be 1 or 2 mod 8

so the max power of 2 that divides it is 2

since it's never divisible by 4

[spring oasis]

gcd(a^4+b^4,2ab^2) = gcd(2.5^3.x.y^2, 5^4.(x^4 + y^4)) = 5^3 . gcd(2.x.y^2, 5(x^4 + y^4)
with gcd(x,y) = 1

[dapper tiger]

yep

and gcd(2xy^2, 5(x^4 + y^4)) is 1, 2, 5 or 10

[spring oasis]

how is gcd(2.x.y^2, 5(x^4 + y^4) = 2^k * 5^q

[dapper tiger]

because we eliminated the other possible prime factors

in fact, you did

[spring oasis]

when

[dapper tiger]

^

[spring oasis]

ahh I remembber now, suppose p is prime != 5,2 then it divides one argument of the gcd but not the other

[dapper tiger]

yep

[spring oasis]

can we go through the cases p = 5 = 2

dont get it

[dapper tiger]

p = 2 is a little more hard here so let's start with p = 5

the goal is to count how many times 5 can divide each argument of gcd

because the min will be the number of times it divides the gcd

like if 5 | a and 5 | b, 5 | gcd(a, b)

and same with higher power of 5

and it's very clear that 5 | 5(x^4+y^4)

[spring oasis]

what

?

[dapper tiger]

for example, if 5 divides c twice, it means 5² | c right ?

and if 5 divides c 3 times, 5^3 | d

so when you'll look at gcd(c, d)

it will have a 5²

because you will be able to factor it out of both c and d!

what we're trying to do is separating cases, and looking how many times we can factor 5 out in each

[spring oasis]

gcd(a,b) = 5

[dapper tiger]

I meant a and b as a general truth

not our exercise

I should have called them c and d

to not be confusing

[spring oasis]

sure

[dapper tiger]

5 can be factored out of 5(x^4+y^4), the 5 is already there
now, the question is about 2xy²

and how many times can we factor 5 out of 5(x^4+y^4)

[spring oasis]

1

[dapper tiger]

well, the answer is only once, since 5 can't divide x^4+y^4

indeed

it will be 0 or 1, depending on when 2xy² is divisible by 5

(ie if x or y is divisible by 5)

[spring oasis]

5 | 2xy^2 and 5^2 | 2xy^2 if 5 | y

[dapper tiger]

that's why q = 0 or 1

[dapper tiger]

or if 5 | x

works too

it gives the same thing

[spring oasis]

?

[dapper tiger]

if 5 | x and not y

it still divides 2xy²

and 5(x^4+y^4)

that's a one 5 too

from the gcd

[spring oasis]

yes but not 5 ^2 | 2xy^2 just 5 | 2xy^2

[dapper tiger]

I agree

[spring oasis]

ok

[dapper tiger]

however it doesn't matter

since you can't factor a 5 out of x^4+y^4

[spring oasis]

so 5^k with k in Z s.t k in [0,2]

[dapper tiger]

no, only [0, 1]

because for the 2

[spring oasis]

i see

[dapper tiger]

we would need a case where 5² | 5(x^4+y^4), the other argument too

which doesn't happen bc 5 doesn't divide x^4+y^4

[spring oasis]

ye

[dapper tiger]

proof: x^4+y^4 = 1+1 = 2 mod 5 by little fermat

or 0+1

[spring oasis]

def?

we can just do a residue table

[dapper tiger]

fermat little theorem: either x is 0 mod p, or x^(p-1) is 1 mod p

that's true but fermat make the table much shorter

only 2 cases

doing the table does work tho

[spring oasis]

ok

[spring oasis]

ok

[dapper tiger]

(or 0+1 = 1 but not 0+0 since x and y coprime)

[spring oasis]

yes

[dapper tiger]

so it proves our gcd(stuff with x, y) which is of the form 2^k 5^q

has a q which is 0 or 1 at most

[spring oasis]

ye

now help with 2

[dapper tiger]

now we want to prove that k can only be 0 or 1 too

[spring oasis]

why

[dapper tiger]

I'll prove it wait a sec I'm looking for the expression of the gcd

[spring oasis]

.

[dapper tiger]

we're looking at gcd(2xy^2, 5(x^4 + y^4))
and 2 can't divide both x and y (they are coprime)
so we have 3 cases:
case 1) 2 divides x (so not y)
case 2) 2 divides y (so not x)
case 3) 2 divides none

[spring oasis]

1. 4 | 2xy^2

[dapper tiger]

in case 1 and 2, 5(x^4+y^4) is odd

so 2 won't divide your gcd

[spring oasis]

why

odd + even = odd?

[dapper tiger]

because the sum of an odd and even is odd

yes

and 5 times odd is odd

[spring oasis]

but we have x^4

[dapper tiger]

x^4 has the same parity as x

if x is even x^4 too

[spring oasis]

why

why

odd ^n = odd

[dapper tiger]

and even^n = even

[spring oasis]

odd = 2k + 1

[dapper tiger]

it's just about mod 2

if you're one mod 2

1^n = 1

if you're 0 mod 2

0^n = 0

either you have 2 in your prime decomposition or you don't

and it's true for your powers too

[spring oasis]

so hard

can you imagine doing this in exam

[dapper tiger]

case 1) 2 divides x (so not y)
case 2) 2 divides y (so not x)
case 3) 2 divides none
so in case 1 and 2, 2 doesn't divide 5(x^4+y^4) so doesn't divide the gcd, k = 0

case 3) is where the k = 1 comes from

if x and y are both odd

x^4 and y^4 are both odd

so their sum is even

and a 2 comes from here

so k = 0 or 1

why not 2 or more ?

because x^4+y^4 is never divisible by 4

you can make a table accouting for gcd(x, y) = 1

to check it

or look mod 8 like I did, same thing

can x^4+y^4 be 4 or 0 mod 8
a^4 mod 8 can only be 0 or 1 (you can list all values mod 8)
and x^4 and y^4 are coprime
so x^4+y^4 can only be 1 or 2 mod 8

[spring oasis]

why

[dapper tiger]

the proof of it I gave earlier is mod 8

being divisible by 4 means being 4 or 0 mod 8

but mod 8, x^4+y^4 can only be 1 or 2

since they are coprime there is no 0+0 case

(they can't both be divisible by 2)

so you can make the table

[spring oasis]

wut

[dapper tiger]

yes, that's because something to the power of 4 can only be 0 or 1 mod 8

you can make the table of residues to check it

that's why k can't be more than 1

and we have 0 <= k, q <= 1

[spring oasis]

is too hard

[dapper tiger]

this one was a bit harder than the previous one

because of this thing with the 2

but again, the more number theory you do

the more you'll know stuff like a^4 mod 8

it's about habits and doing a lot of different exercises

also you're not very comfortable with stuff about prime decomposition or p | x^n => p | x

it needs to be more automatic

[spring oasis]

i am

[dapper tiger]

you asked for a proof that even^n = even

and odd^n = odd

it should be more automatic

[spring oasis]

well since the residues of even under mod 2 are 0 or 1 is evident if you say it like that

in general x^n = x (mod 2)

[dapper tiger]

I agree, it should be obvious from p | x^n => p | x too tho

a multiple of 2 to the power of 2 is still a multiple of 2...

the 2 didn't go anywhere

and for odds it's the contraposite

[spring oasis]

this can be argued by meribeth or prime factorization

[dapper tiger]

indeed

[spring oasis]

exercises are nasty

[dapper tiger]

that's why you could say the same thing with prime facto instead of mod 2

[dapper tiger]

this one was a bit meaner yeah

[spring oasis]

i will continue tomorrow, try to pick up this two and re do

[dapper tiger]

gn

[spring oasis]

because there is a lot going on with this 2

[dapper tiger]

yeah

also ideas like x^4 = 0 or 1 mod 8

come from experience

you also need to do number theory exercises that are not just gcd

but also other things

[spring oasis]

care to elaborate

[dapper tiger]

like even exercises like "do the table of residues of n²-n+7 mod 9" is not necessarily a bad exercise

sometimes it can help you recognize patterns later

and helps you see the tables faster

even this can be useful

[spring oasis]

i am taking like 4 courses simultaneously, I barely have time to keep up with everything

but i will try to keep it in mind

[dapper tiger]

for example when I see n²-n+7 mod 9, which was chosen at random
I see immediately n²-n+7 = n²-n-2 = (n+1)(n-2) mod 9

so I know the trivial 0 are 2 and 8

that's the kind of automatisms you can only have by doing a lot of basic manipulations

[spring oasis]

wat

[dapper tiger]

(n+1)(n-2) = 0 mod 9

has 2 trivial 0

which are 8 and 2

[spring oasis]

right

[dapper tiger]

it also has 5

[spring oasis]

6x3 = 18

[dapper tiger]

= 0 mod 9

yep

the more you see a lot of basic number theory, the more you know how to use your tools quickly

[spring oasis]

yes

i appreciate the help

.solved

[devout snow]

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[pine spear]

Usak

[pine spear]

Ask

[solid perch]

hm?

[crystal dawn]

got a question?

if you have, send it and I'll pin it

[pine spear]

I m just checking

I will use it later

[crystal dawn]

!done in that case

[devout snow]

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

[pine spear]

.close

[devout snow]

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[devout snow]

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[arctic mountain]

need help solving this limit: $\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$ idk where to start

[woven radish]

solomoncyj

[silk panther]

u can try multiplying its common factor

[silk panther]

the common factor of the numerator

[solid perch]

I was thinking the conjugate of the numerator instead.

[unborn yew]

Try Rationalization?

[silk panther]

rip english

[vital edge]

Just use the binomial approximation?

(1+x)^p = 1+px for x very close to zero

[devout snow]

@arctic mountain Has your question been resolved?

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[ocean haven]

Limit definition of derivative tells us the slope for infinitesimal change in function on infinitesimal change in x, so what do we mean by derivative at a point

[gloomy aurora]

first off the concept of infinitesimal is not a very good way to start off calculus

it's not rigorous and it confuses beginners

As for what is a derivative at a point, its the instantaenious rate of change at that point

[ocean haven]

I see

But what is instantaneous rate of change

[gloomy aurora]

Instantaneous rate of change at particular point $c$\
= $\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}$

[woven radish]

Annie Maqionde

[gloomy aurora]

Its the rate of change at a particular point

[ocean haven]

hmm But what do you mean by "at"

[gloomy aurora]

hm i'd say we'll take the geometrical approach then

Do you know what is a tangent AT a point?

of the function obvi

[ocean haven]

Yes

[gloomy aurora]

the rate of change at a point is the slope of the tangent at that point

[gloomy aurora]

i'd use very vague language here: but in evaluating the limit, the more and more we decrease h, the more and more the points $(c,f(c))$ and $(c+h, f(c+h))$ come nearer. So we could imagine a hypothetical scenario where $h$ is so miniscule that the points are together. That's what we mean by that limit and by the concept 'at' a point.

[woven radish]

Annie Maqionde

[ocean haven]

I see but we are still using the fact how the function is in an area with very small range of input

[misty crest]

what

you don’t seem to understand limits

you keep mentioning this infinitesimal nonsense

[gloomy aurora]

I mean thats how they were defined...............

Do you know the epsilon-delta definition of a limit?

[misty crest]

nowhere in the limit does it say for super duper close points

[gloomy aurora]

If not I'd suggest you read that up first.

[misty crest]

this notion of closeness isn’t even defined

i’d suggest you review limits

[ocean haven]

But isnt that what limits is

[gloomy aurora]

NO

[misty crest]

🤔🤔

no

stop watching pop math videos

[ocean haven]

Lmao

[void fox]

the start to differentiation is the observation that for a nice function its graph looks like a straight line if you zoom in enough

[misty crest]

that’s just an introductory way to explain it to someone who can’t be bothered to learn the formalities

[ocean haven]

Then what is it

[misty crest]

before we go on

what class is this for?

calculus 1?

[ocean haven]

Hs

Uhm we don't have that system

[misty crest]

first year/semester calculus

[gloomy aurora]

Have you heard of something called the epsilon-delta definition of limits?

[ocean haven]

I have heard but not taught

[gloomy aurora]

that provides a very rigorous pic

please read it up

[void fox]

does op want a technical explanation or an informal one

[gloomy aurora]

Well since I've already provided an informal one and OP doesn't seem very satisfied in it and also because of the fact that OP's use of the word 'infinitesimal' leads to very wrong conclusions, I'd guess a technical explanation would be better.

[ocean haven]

Not university like ppl aged 17-18

[misty crest]

same shit

[ocean haven]

idk

[void fox]

you don't know?

[misty crest]

💀

[gloomy aurora]

So to be clear, you're self-studying calculus? And you've a doubt in differentiation

[ocean haven]

I am confused

[void fox]

are you studying for a "calculus" class or an "analysis" class

[ocean haven]

No, but they didn't teach eps delta

[misty crest]

high schools usually don’t

[ocean haven]

Uhm no it doesn't I don't have that here

[void fox]

have what?

[misty crest]

analysis

i assume

[ocean haven]

Different class like "calculus" "analysis "

[gloomy aurora]

Well how is this explained in your textbook?

We'll follow that.

[misty crest]

do people in high school even use textbooks though

i think i was the only one who bothered to actually read it

[ocean haven]

Ok I'll get it in some time, i need lunch

@devout snow don't close this

[gloomy aurora]

idts it works like that......

[mystic scarab]

This doesn't do nothing 😅, the bot can't read the messages

[ocean haven]

Bruh

[misty crest]

what’s the ZH mean

in your bio

[gloomy aurora]

Chinese

[misty crest]

thought that was CN

[gloomy aurora]

<@&268886789983436800>

[gloomy aurora]

hm ZH-CN for putongua, ZH-TW for taiwanese

[misty crest]

so mandarin

[devout snow]

@ocean haven Has your question been resolved?

[devout snow]

Channel closed

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[ocean haven]

.reopen

[devout snow]

✅ Original question: #help-27 message

[dark sable]

a derivative is change of y or f(x)/change of x
it is just slope but at a point, so to speak, tangent line at a point

the limit

tells us how much the function essentially rises over run at a point rather than over two points

since h approaches 0, f(c+h) is approximately equal to f(c) or relatively close

it's like how we do f(x1)-f(x2) / x1-x2 for slopes except in this case it's just f(c+h)-f(c) / c+h-c = f(c+h)-f(c) / h

[ocean haven]

so they defined to delta to be a small interval , an epsilon such that for every every x, |x -x0| the function may deviate from L within a boundary of L+epsilon ad L-epsilon( 2 eps )

or am i wrong

[ocean haven]

yes but that is what i am confused about

very small h

infintesimal

[dark sable]

h approaches 0
h very very small but not exactly 0

[ocean haven]

yes but then we are still speaking of the slope of the function in a boundary where delta x is small

[dark sable]

yes

that's literally a derivative

dy = change in y, dx=change in x, dy/dx = slope

it's just that the change is over an EXTREMELY SMALL period such that it's essentially a point

[ocean haven]

so what does at a point mean

[mystic scarab]

You'd already been given an explanation earlier iirc

[ocean haven]

is "essentially" rigorous

[mystic scarab]

@ocean haven

[mystic scarab]

@ocean haven

[ocean haven]

yes but

is this rigorous

[devout snow]

@ocean haven Has your question been resolved?

[ocean haven]

<@&286206848099549185>

https://tenor.com/view/mr-bean-esperando-mr-bean-waiting-gif-4513135960021631723

[hardy trail]

What do you mean by "rigorous"? The argument presented above is the usual interpretation of a derivative, which works for pretty well as an introductory concept for calculus.

The level of rigour required will depend on what you are going to use the definition for.

[ocean haven]

uhm i dont understand , at a point. limit feels like approximating

[hardy trail]

In order to approximate, we'd need to stop at a specific value, right?

Like for example, if we wanted to approximate the digits of "pi", you'd have to stop somewhere:
pi = 3.14159

The limit argument isn't an approximation - it's giving an argument that is representing infinity in some way. That is, no matter how far you nudge a point in a particular direction, you can find that the slope of that particular nudge is bounded (and converges to a particular number - the instantaneous rate of change).

Because the nudge is arbitrary (and in particular, arbitrarily small), the natural conclusion is that whatever this slope approaches, that must be what the slope at that point is equal to.

[ocean haven]

oh i see this makes sense

[hardy trail]

I.e if the derivative exists, then as h approaches 0, the slope between these two points approach this value (let's say L), which leads us to conclude that the slope at the point x = c must be this "value", which is L.

Or in other words, since h -> 0 implies rate of change -> L, then we have that: f'(c) = L

[ocean haven]

but this does not say at instant, truly instantaneous at that point

[hardy trail]

No it does

Because as I said, it's true regardless of how big or small your nudge is

Like let me put it this way

[ocean haven]

but we never are at point , it always approach ?

[hardy trail]

Yeah but it doesn't matter that we are at the point or not.

[hardy trail]

Out of curiosity, have you ever used a compass before?

If you have, you might know that the compass needle always points north. So, in order to find where north is, you need to align yourself such that the compass needle is point forward (and that tells you where north is).

[ocean haven]

not an actual, one in phone

[hardy trail]

Okay yeah good enough for me lol

[hardy trail]

Here's the thing though:

Can you ever point it exactly at the north pole?

Like, can you ever line up the needle so that it's 100% facing directly towards north?

[ocean haven]

idk

[hardy trail]

Good! So in fact, you actually can't - it's quite literally impossible to get that accurate on a compass tool (since it's a measuring device, it will always have some inaccuracies).

However, the point is,

Assuming the compass is working properly,

You can always orient the needle in a way such that you know where north is.

Like, you won't ever get the spot that's exactly facing north. However, if you literally keep trying to orient the compass, you will always always get to a spot that is facing in some direction.

[ocean haven]

oh

so but this is like minimizing

the error

[hardy trail]

Yeah but the thing with derivatives is that we have actually proven that there is no error.

As in, we have proven that with absolute certainty, that if you were to draw a line tangent to some particular point on a graph, then you can find the slope of that tangent line exactly.

And the argument is basically the compass argument but you have almost an infinite amount of needles:

All of these needles will be pointing in some general direction - that general direction will be the exact position of north.

Of course, the way that you prove these slopes are exact require a little bit of work and understanding (hence the discussion on epsilon delta proofs, which is the most common form of proof needed to prove these sort of things), but the general idea remains. You are trying to sort of show that all of these needles will point somewhere in the direction of true north, and that they will all eventually hit that point.

[ocean haven]

ohh

this makes sense

🤯

but this was intuitive type

arguement

if i want to read a technical type proof what would i search for

[hardy trail]

👍

As you look into the mathematical arguments, you'll see that the definition of limits and derivatives will closely match with these intuitive arguments. We define what it means to be "close" to something, we define what it means to be "arbitrarily small", etc etc.

[hardy trail]

Um a technical type proof? You'd probably need to look at an analysis or calculus book or something

I would say asking on #book-recommendations is probably your better bet to get more informed recommendations. Though for an introductory level, I think any undergrad analysis or calculus book will do the trick.

I mean you do seem to have a textbook already - look at that in depth first before you try other sources.

[ocean haven]

it doesnt have eps delta 🥀

and they say it is naive

[hardy trail]

Probably too intro level then. But if you're not confident in calculus, you should probably still stick to this book.

[gleaming dragon]

meow

[hardy trail]

I mean, there's nothing stopping you from reading other literature

[ocean haven]

but tysm

[gleaming dragon]

Yw

[ocean haven]

https://tenor.com/view/bite-cat-gotcha-playing-catsdoingthings-gif-16767804

[hardy trail]

The way people learn calculus is that they are first introduced to a sort of benign form of calculus (i.e 'intuitive arguments', just use the power rule, integration spam, etc). Then after mastering all those concepts, that's when you pick up an undergrad book and really start to break down the logic behind everything that you learned in regular calculus.

So unless you are prepared to sit down and have a lot of patience, you may want to avoid picking up a technical textbook. Though, if you're ambitious, then yeah go for it.

[ocean haven]

👍

[ocean haven]

ic yeah we been through that

tysm

everyone

.close

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[prime hemlock]

If m,n be any two distinct odd primes then \ $\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^\frac{(m-1)(n-1)}{4}$

[woven radish]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[hasty cargo]

is that a fraction

[lyric hornet]

and you need to prove this?

[lyric hornet]

Legendre symbol

@prime hemlock are you familiar with Gauss' Lemma?

[prime hemlock]

This ?

[lyric hornet]

yep!

[candid maple]

interrupting but I recall this being related to quadratic reciprocity?

(will leave after this question, sorry.)

[lyric hornet]

this is quadratic reciprocity!

[candid maple]

ah, I see! alright, sorry for interrupting, please proceed.

[lyric hornet]

so yeah you're pretty much halfway there equipped with this lemma

do you have an idea on how to proceed?

[prime hemlock]

Till here is what I did

[lyric hornet]

,rccw

[lyric hornet]

fantastic place to start!

actually @candid maple do you wanna take over? I just got summoned to go somehwere

the rest of hte proof is just counting lattice points

[candid maple]

no actually, I... do not know how to work this.

I've just watched one video about this, I cannot in good conscience claim to be able to help. I'm very sorry.

[lyric hornet]

okay well transparent shadow, imagine a rectangle of integer lattice points ranging from (i\in{1,2,\dots,\frac{q-1}{2}}) on the horizontal axis and (j\in{1,2,\dots,\frac{p-1}{2}}) on the verticle axis

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[lyric hornet]

you can count such points as conclude that there are (\frac{q-1}{2}\cdot\frac{p-1}{2}) many

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[lyric hornet]

likewise, you can also count them by constructing the line (y=\frac{p}{q}i) and showing that there are no lattice points on the line (since p,q prime)

[woven radish]

ΠαϳαμαΜαμαΛλαμα

[lyric hornet]

then count the ones above and below the line

oh you used m,n not p,q

just replace throughout the proof is teh same

[green crypt]

looks like it requires mathematics...

[prime hemlock]

Umm ok lemme try

[devout snow]

@prime hemlock Has your question been resolved?

[prime hemlock]

@lyric hornet correct?

[lyric hornet]

beautifully done!!

[prime hemlock]

Tysm

[prime hemlock]

.close

[devout snow]

Channel closed

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[devout snow]

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[woven radish]

Renato

[spring oasis]

d = gcd(7 . 3^n - 5.5^n, 3.3^n + 7.5^n)
d | 7 . 3^n - 5.5^n
d | 3.3^n + 7.5^n
d | x(7 . 3^n - 5.5^n) + y(3.3^n + 7.5^n)

,w simplify x(7 * 3^n - 55^n) + y(33^n + 7*5^n) with x = 7 and y = 5

[woven radish]

Results provided by WolframAlpha

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Requested by renatoxbz

[pure cedar]

reduce the ratio in mod 4

[devout snow]

@spring oasis Has your question been resolved?

[dapper tiger]

1. what are the possible prime factors of your gcd?
2. use Mr Smith hint to say the max power of 2 that divides your gcd

Conclude

[pure cedar]

idk if im mistaken but , the question asks that p:q can have two values right , and not (p,q)

[spring oasis]

the gcd dude has 2 possible values

[pure cedar]

right, then this cant be used cuz 5,3 and 1,3 dont have same gcd

[dapper tiger]

You took x = 7 and y = 5 and found your gcd divides 64 3^n
With x = -3 and y = 7 you could conclude it divides 64 5^n too
So it divides gcd(64*3^n, 64*5^n) = 64

[spring oasis]

yes!

[dapper tiger]

So you can immediately say your gcd is a divisor of 64 = 2^6

[spring oasis]

yes because gcd(3^n, 5^n) = gcd(3,5)^n

[dapper tiger]

The question is what are the possible 2^k, the ones that actually happen

[spring oasis]

yes

[dapper tiger]

So it depends on how many times you can factor a 2 out of the arguments of your gcd

That's why Mr Smith tells you to look at them mod 4

Then mod 8 if needed

Etc

[spring oasis]

mod 2 no?

[dapper tiger]

You already know mod 2

They are the sums of odds

So they are even

You can check it mod 2 it's quick

[spring oasis]

wut

[spring oasis]

how

[dapper tiger]

Odd are 1 mod 2

Even are 0 mod 2

That's kinda their définition

It's not supposed to be hard to check 2 divides the arguments of your gcd

[spring oasis]

1.1 - 1

= 0

1 + 1 . 1 = 2

gcd(0,2) = 2

@dapper tiger

[dapper tiger]

You can conclude that 2 divides the gcd

But not that It's 2

It could be possible that 4 divides your 2 arguments too

You can read what we said about that earlier

[spring oasis]

why

[spring oasis]

under mod 2 both arguments then the gcd is 2

[dapper tiger]

No

You can't mistake the numbers with their values mod 2

[dapper tiger]

This doesn't have much meaning relatively to your original gcd

Besides the fact that 2 divides it

[spring oasis]

can we start from scratch

under mod 2 we get that both arguments are 0

@dapper tiger

2 | arg1 and 2 | arg2 => 2 | gcd(arg1, arg2)

so d = 2^k with k = 1 holds

[dapper tiger]

yes

[spring oasis]

what about k = 0?

it always holds?

[dapper tiger]

impossible

[spring oasis]

care to elaborate

[dapper tiger]

2 | arg1 and 2 | arg2

so 2 | gcd(arg1, arg2)

2 can't divide 1

gcd(arg1, arg2) won't be 1

[spring oasis]

ok

what about 4 | gcd

@dapper tiger

[dapper tiger]

well, there are cases where it happens

when you say 2 | arg1 and 2 | arg2
so 2 | gcd(arg1, arg2)
you proved k >= 1
when we said d | 64 = 2^6, we proved k <= 6

k = 2, 3, etc could be valid

however, if you look at it mod 8

you'll find k < 3

so k = 1 or k = 2, and you have to find examples for both

[spring oasis]

i would like if we check mod 4 and mod 8

becuase seems out of the blue mod 8 doesnt work

[dapper tiger]

mod 4 you'll just find that it sometimes work

which is why you'll have to check for 8 after

to see if k >= 3 is possible

[spring oasis]

why

[dapper tiger]

check it

4 can divide both the args for some n

[spring oasis]

7 . 3^n - 5^(n+1) = 3^(n+1) - 3^(n+1) = 0 (mod 4)

[dapper tiger]

that's false

-5 is not -3 mod 4

[spring oasis]

oops

its + 3

[dapper tiger]

yep

[spring oasis]

7 . 3^n - 5^(n+1) = 3^(n+1) + 3^(n+1) = 2.3^(n+1) (mod 4)

[dapper tiger]

= 2*3^(n+1) = 0

[spring oasis]

how

[dapper tiger]

for some n

[spring oasis]

wut

[dapper tiger]

wait

[spring oasis]

we need forall n in N

[dapper tiger]

no

we're listing all cases

[spring oasis]

care to elaborate

[dapper tiger]

well even if a case exists only for one n

it's still a case you'll have to list and find an example for

[spring oasis]

why

[dapper tiger]

because you have to find all possible values for the gcd

with examples

the exercise asks for it

you have to give what the exercise asks for

if exercise asks for all cases with examples, you give all cases with examples

[spring oasis]

forall n in N tho

[dapper tiger]

doesn't mean anything

the value of the gcd depends on n

the exercise asks what can it be

and examples that it can be it

[spring oasis]

you are right

[dapper tiger]

https://cdn.discordapp.com/attachments/903486480075354182/1497257348824039555/image0-6.jpg?ex=69ecdd11&is=69eb8b91&hm=544ae735f62f6cedf0c651cc31db423a6cd05cfd4e852d285f2c3a0f3d632d17&

[spring oasis]

yes

[spring oasis]

how

[dapper tiger]

because it's wrong because you did a sign error

and I assumed you didn't

[spring oasis]

where

[dapper tiger]

here

-5^(n+1) is not 3^(n+1)

[spring oasis]

-5 = 8 - 5 = 3

[dapper tiger]

yes but the ^(n+1) isn't on the -

-5^n is not the same thing as (-5)^n...

[spring oasis]

i see then -5 = -1

[dapper tiger]

mod 4, 5^(n+1) = 1

so -5^(n+1) = -1

[spring oasis]

depends on n + 1

[dapper tiger]

nope

[spring oasis]

?

[dapper tiger]

^

you're making the same sign error again

[spring oasis]

dammit

[dapper tiger]

so 7*3^n - 5^(n+1) is 3^(n+1) - 1 mod 4

so is 0 when n is odd

now for the other arg 3^(n+1) + 7*5^n
= 3^(n+1)-1 mod 4

same thing as the other arg

it's 0 when n is odd

so when n is odd, the gcd is at least 4

because 4 divides both args

[spring oasis]

ye

[spring oasis]

how

[dapper tiger]

solve 3^(n+1) - 1 = 0 mod 4
<=> 3^(n+1) = 1 mod 4
<=> 3^n = 3 mod 4

true when n is odd

[spring oasis]

how?

[dapper tiger]

how what?

[spring oasis]

<=> 3^(n+1) = 1 mod 4

<=> 3^n = 3 mod 4 (

[dapper tiger]

3 is its own inverse mod 4

3² = 9 = 1

multiply 3^(n+1) = 1 by 3
you get 3² * 3^n = 3

[spring oasis]

ah so smart

[dapper tiger]

so 3^n = 3

true when n is odd so there are cases for 4 | d

since 4 can divide both args

when n is odd

[spring oasis]

3^(n+1) = 3.3^n
3.3^n = 1 mod 4
3^n = 3 mod 4

[spring oasis]

how

[dapper tiger]

you wrote how

brother

wym how, you wrote it

[spring oasis]

ok

under mod 4 you mean

[dapper tiger]

yes obviously, otherwise the only sol would be n = 1

[spring oasis]

?

[dapper tiger]

when n is odd, 3^n = 3 mod 4

when n is even, 3^n = 1 mod 4

[spring oasis]

it does not depend on n

[dapper tiger]

it does

3² = 1 mod 4

not 3

[spring oasis]

but i never used n is odd or even when proving 3^n = 3 (mod 4)

[dapper tiger]

that's not what you proved

you proved
3^(n+1) - 1 = 0 mod 4
<=> 3^(n+1) = 1 mod 4
<=> 3^n = 3 mod 4

to solve the first line, you have to solve the last

[spring oasis]

ah

so arg1 is divisible by 4 iff 3^n = 3 mod 4

[dapper tiger]

arg2 = arg1 mod 4

we calculated it

so you can say the same thing for arg2

btw

[spring oasis]

why

[dapper tiger]

arg1 = 7*3^n - 5^(n+1) = 3^(n+1) - 1 mod 4
arg2 = 3^(n+1) + 7*5^n = 3^(n+1) - 1 mod 4

both are divisible by 4 iff 3^n = 3 mod 4

so iff n is odd

[spring oasis]

no

[dapper tiger]

you're wrong

probably because you make the same sign error as earlier again

[spring oasis]

how is 7.5^n = -1

[dapper tiger]

5^n is 1

7 is -1

their product is -1

[spring oasis]

fair

[spring oasis]

why

[dapper tiger]

if n is even, n = 2k
3^n = 3^(2k) = 3²^k = 9^k = 1

if n is odd, n = 2k+1

3^n = 3*3^(2k) = 3*9^k = 3

3^n is either 3 or 1 mod 4 depending on if n is even or odd

[spring oasis]

i see

[dapper tiger]

so to sum up what we proved:

1. d | 64 = 2^6
2. 2 always divides d
3. 4 divides d iff n is odd

if you prove 8 never divides d, you'll have proven that d can only be 2 or 4

you're one last step away from finishing the exercise

[spring oasis]

mod 8 is prolly hard

[dapper tiger]

let's look at the eq arg1 = 0 mod 8
7*3^n - 5^(n+1) = 0 mod 8
<=> -3^n = 5^(n+1) mod 8
<=> 3^n = -5 * 5^n
<=> 9^n = -5 * 15^n mod 8 (I multiplied by 3^n on both sides)
<=> 1 = -5 * (-1)^n mod 8
<=> -5 = (-1)^n mod 8

it never happens, since (-1)^n mod 8 can only be 1 or -1

so 8 never divides arg1, so never divides d

[dapper tiger]

so 4) 8 never divides d
conclusion, d can only be 2 or 4, depending on if n is odd or even
d is 4 iff n is odd so d is 2 iff n is even

if you take n = 1, you'll find d = 4, if you take n = 2, you'll find d = 2

you have all cases with examples

you need to train solving equations mod things

[spring oasis]

7*3^n - 5^(n+1) = 0 mod 8
<=> -3^n = 5^(n+1) mod 8
how

[dapper tiger]

7 is -1

[spring oasis]

oh right

sorry

for the ping

[spring oasis]

<=> -3^n = 5^(n+1) mod 8
<=> 3^n = -5 * 5^n

is multiplying both sides by -1 legal in a congruence

[dapper tiger]

yes, -1 is in the class of 7 mod 8

[spring oasis]

ok

[dapper tiger]

it's just a multiplication by 7

[spring oasis]

yes I see

[spring oasis]

<=> 1 = -5 * (-1)^n mod 8
<=> -5 = (-1)^n mod 8

what

[dapper tiger]

multiplication by (-1)^n on both sides

[spring oasis]

this is only true if -5 is the inverse of (-1)^n on mod 8

[dapper tiger]

you're wrong

(-1)^n is always the inverse of (-1)^n

since ((-1)^n)² = 1

[spring oasis]

i shee

this is why you multiply by (-1)^n both sides

similarly when you multiplied by 3^n both sides, 3^n is the inverse of 3^n

@dapper tiger 💡

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

i see

so gcd = 2^k with 1 <= k <= 2

thats the two cases d = 2 and d = 4

[spring oasis]

I see

I wouldn't say this was easy, all this exercises are nasty

[spring oasis]

multiplicative inverses still trip me up

[dapper tiger]

I liked some of them tbh

[spring oasis]

this method of characterizing the gcd is so hard but is pretty much was it is expected, i have seen solutions from other people and even though they dont use modular multiplicative inverse so much they make a lot of residue tables

[dapper tiger]

yeah they check every possible residue instead of solving for finding them

same thing, longer

[spring oasis]

me 2 when we manage to solve, when I get stuck it sucks

😂

skill issue on my part

[spring oasis]

anyways. I appreciate the help. As I said, this exercises are a lot of work, thank you for helping me with them until the end, the mod 8 was the worst part

since both 5 and 3 were very less than 8, the mod 8 case was specially difficult for me, but you handled it gracefully

using modular multiplicative inverse and what not

[dapper tiger]

I mean even for me it's not instantaneous, I think about it before writing it to you

I see an eq mod 8, I try to solve it

there is no secret, you have to solve more eq in modular arithmetic

[spring oasis]

yes but for you it should be like this

https://tenor.com/view/hardpretzel-spongebob-patrick-roller-coaster-screaming-on-roller-coaster-gif-15822836723246643102

jaja

[spring oasis]

yeah definitely, I will try doing more exercises over the weekend

[dapper tiger]

do you still have exercises about gcd?

I think they are fun

but you should do other types of exercises too

[spring oasis]

i have more than I can think of, I have like gazillion of them, I have more than I can do in a lifetime

[dapper tiger]

lol

[spring oasis]

i still have a little less than 8 weeks b4 the exam

[final valve]

Hy

[dapper tiger]

hello, if you need help you have to take an unoccupied channel

[spring oasis]

certainly they are, if you manage to make them then you are good on divisibility

[dapper tiger]

see #❓how-to-get-help

[spring oasis]

I appreciate the help

.solved

[devout snow]

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[devout snow]

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[tawny wolf]

Having trouble with this grade 11 trigonometry question. I believe I have all I need to find theta using cosine law, but I always get an error using my calculator. I believe it’s because I’m using inverse cosine on a number greater than one but I don’t know any other methods to find Theta.

[winter patrol]

firstly, what did you get for cos(theta)

actually 1 sec

yeh, cos(theta) > 1
the question is fked

side would actually round to 6.48,
regardless that fails triangle ineq, 6.48 + 6.48 < 13.5

[covert mango]

method looks perfectly correct

probs just typed it wrong into calc

[devout snow]

@tawny wolf Has your question been resolved?

[devout snow]

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[undone plaza]

what is r?

[sacred sedge]

what is the context to the question?

[undone plaza]

find what vector r is

[grand warren]

do you have more context lmao

[undone plaza]

like if r=w-v

finding what its a combination of from the blue lines

[little sonnet]

Isnt it just lineair combination of u, v and w

[undone plaza]

ok its r=u+v+w?

[midnight echo]

Send the complete question, not just a photo and "what is r?"

[sacred sedge]

wait can you take cross product of r and u?

[midnight echo]

<@&268886789983436800>

[sacred sedge]

because they are different lengths

[little sonnet]

r=k.u+l.v+z.w => k,l,z are scalars

[midnight echo]

r is some linear combination of basis vector u v and w. so is any other vector in R3. we can't help until OP provides more context on the problem; ideally a screenshot of the problem itself where it was posed.

[sacred sedge]

is that a quiz/

[undone plaza]

r=u+v+z ?

[undone plaza]

no

[sacred sedge]

assignment?

or what?

[midnight echo]

its Webwork, assignment

[undone plaza]

how do i do it

[midnight echo]

You see, now we have numbers. We didnt have numbers when you said "what is r?"

r is a vector from one corner of the cube to the furthermost corner (think diagonally). The origin (where r starts) is <0,0,0>. What would the coordinates of the furthest corner of this cube be? Given that is is of side length 3

[undone plaza]

3,3,3?

[midnight echo]

Yes

so what is vector r?

[undone plaza]

r=3,3,3

[undone plaza]

is s also 3,3,3

[midnight echo]

how could two vectors originating from the same point and ending up in different points have the same value?

[undone plaza]

how do i find its value then

[midnight echo]

Label each corner

[undone plaza]

isnt it 3

<@&286206848099549185> how do i find s

[sharp laurel]

u + w no?

[undone plaza]

yeah thanks

[sharp laurel]

bosh

[undone plaza]

.close

[devout snow]

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Closed by @undone plaza

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[devout snow]

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[bold dock]

when I was solving this question in two different ways, I got different probabilities, which I don't think should happen.
so the first way I considered to use counting with repetition but without order and the second way I use counting with repetition and with ordering.

[bold dock]

so did something went wrong with my calculations or is considering with and without ordering changing the probabilities?

so I used na/N for both ways.
my calculations for with ordering gave me the answer:
24 /81outcome 3-1-0
18 /81 outcome 2-2-0

without ordering it gave me
6/15 outcome 3-1-0
3/15outcome 2-2-0

[polar chasm]

<@&268886789983436800>

in the channel below and above too

[bold dock]

can u by any chance help me?

[polar chasm]

i dont understand what you did

wdym by without ordering

[bold dock]

oh let me explain

[bold dock]

so u know these formulas right?

its like counting but then u consider if order matter or not

and when I considered counting where order matters and also counting where order did not matter I did nto get the same probabilities

so thats what I found weird I thought both approaches should be valid

[polar chasm]

i still have no idea what you actually did

like how you got those numbers

[bold dock]

oh that one, okay so u know the na/N right?

the amount of combinations of ur specific sitaution the 3-1-0 outcome / total outcomes

so for with ordering total outcomes should be 3^4 right?
since 4 apples each has 3 choice

that is the 81 I had

then to get the 3-1-0 outcome u have 6 possbilities to order the 3-1-0, 1-3-0 etc etc

but u have to multiple it with 4 since u have 4 different choices for the apples that goes in that single basket and the other 3 apples will together go to the same basket

so therefore 24 since we used with ordering

[polar chasm]

yeah, right, thats 24

[bold dock]

then the 18

[polar chasm]

yeah, 18 makes sense too

i dont really get the without ordering tho

[bold dock]

I see

so I thought u can count multiple ways with and without ordering so now I consider where apples order does not matter

I thought that should be a valid approach because now N becomes n+k-1 choose k

andd that was 15

and then for na u can say its 3! for the outcome 3-1-0 and 3 for the 2-2-0

[polar chasm]

oh so 15 is the number of ways I can put apples into the baskets, where each apple is identical?