help-42

[flat orchid]

5 women 6 women 7 women 8 women 9 women

[remote mural]

so for 5 women it should be 9C5*8C7 right?

[unreal isle]

yes

also for the first FIRST part, before the 1 and 2, you can also find the total cases and subtract the cases where you have lesser than 5

so only 1 2 3 or 4

[remote mural]

oh okay, i was thinking about applying the subtraction principle in which total cases - committee which has less than 5 women

yeah thats what i meant

how can i calculate the total one?

[unreal isle]

out of curiousity is this for like O level Add Math

[unreal isle]

how many people do you have total

[remote mural]

17

[unreal isle]

and how many do you need

[remote mural]

for university

[remote mural]

12

oh okay thanks

[unreal isle]

so to choose 12 from 17 w no restrictions

yep

[remote mural]

17C12 right?

[unreal isle]

yep

[remote mural]

thanks a lot

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[wind bane]

i computed the determinant for the coefficient matrix and got zero. is that sufficient enough to infer that these vectors do not form a basis for R^3, meaning they do not span R^3

[unkempt drift]

yes, that's sufficient

[velvet osprey]

"sufficient enough" is a pleonasm. why not just "sufficient"?

(which, yes, it is)

[wind bane]

im not a native English speaker.

its my third language

[finite fern]

Holy yap

[velvet osprey]

???

[wind bane]

wait could you also make the argument that, these vectors since they have a determinant of zero, either span a plane or a line in R^3?

[velvet osprey]

sure

same thing

[wind bane]

alright then

appreciate it!

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[remote mural]

i need help with this one

[flat orchid]

Okay

Ask

[remote mural]

i can compute the above result for 4 but for 1, 2, 3 how can i do it?

like if i choose 3 women then the remaining number of people in the committee needs to be 9 and we have only 8 men

[flat orchid]

Yeah he kind of missed that information

Only 1 case will be possible then

That being 4 women 8 men

[remote mural]

ohh so total - 4 women 8 men? is this the final ans then?

[flat orchid]

How did you calculate the total?

[remote mural]

17C12

and for 4 women 8 men 9C4

[flat orchid]

Could you hold on for just a moment I gotta check something with this question real quick

[remote mural]

sure

[flat orchid]

Yeah okay

You're right

I believe that is the answer

[remote mural]

okay, thank you so much 🙂

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[wet scroll]

If we have an ordered set A and we take a subset B of A, does B by definition also include the order of A?

[swift laurel]

the order relation on A can be restricted to B

[sand folio]

should be? one would usually say "take a subset with the same ordering"

are there more context?

[wet scroll]

ok i just wasnt sure because it wasnt specified but it seems necessary in this case

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[burnt heart]

,rotate

[leaden thunder]

did you have a question

[burnt heart]

yeah was just typing that, how do I see how many values can lambda take between them

n is a natural number

[flat orchid]

Is lambda an integer?

[leaden thunder]

the second part is 6n+9. what does the first part say

[flat orchid]

4n + 4

[burnt heart]

Yeah,

I just plugged in values n verified options, but wanna know what to actually do in these

[flat orchid]

What you do is

Subtract the upper bound from the lower bound

[burnt heart]

But that isn't always the case I assume??

[flat orchid]

As in?

[calm coral]

@burnt heart Has your question been resolved?

[calm coral]

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[lavish wind]

I mean doesn't this use linearity of the Laplace transform which isn't a part a formula

[glad parrot]

Linearity for Laplace transform is working

What do you mean by a part a formula

[lavish wind]

we

we're supposed to show it only using the formulae we derived for L(f'(t)) and L(f''(t))

I don't see how either lets us conclude that L(-g(t)) = -L(gt)

[glad parrot]

Yeah but its not neccessarly by avoiding everything else

[lavish wind]

fair enough

i guess

thanks

[glad parrot]

Yw

[lavish wind]

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[pearl snow]

Already did the problem, differenciated (36x^3+36x^2-72x), found the 3 x values (0,-2,1), differenciated again (108x^2+72-72) and replaced x, but I am stuck in which one is the local minimum and which is the local maximum? (It's How do I differenciate? I know the answer is D

[pearl snow]

I got 108, 216 and -72

[calm coral]

@pearl snow Has your question been resolved?

[pearl snow]

<@&286206848099549185>

[pearl snow]

So I just wanna know which one is the local maximum out of the three and why

Please

[undone narwhal]

I think i can help you

Te puedo ayudar

No tengo folios para resolver pero según entiendo te dan los puntos donde se encuentran los mínimos y maximos

[pearl snow]

Sabes español? Jaja

[undone narwhal]

Deriva la funcion

Y te dan los puntos y ya

El segundo es el 0,0

[pearl snow]

Pero me da 3 puntos, cuales son maximos y minimos?

[undone narwhal]

S1 maximo

S2 minimo

Bueno perdon

Al reves

Jajaja

Los puntos donde la derivada se hace 0 son esos

Es de 1 de bachillerato

[pearl snow]

Sip, solo que no se como se distribuyen los 3 puntos en maximos y minimos, si, ya los saqué (0,1, -2) pero chequé y dice que el -72 al sustituir en la 2da derivada es el maximo local, por eso tengo la duda, de donde proviene?

El -72 proviene de sustituir a x por 0, una de las soluciones

Es inverso? Osea si el numero es pequeño significa que ese es el limite mayor?

TLDR: Basically I wanna know if the smallest number when replacing the x by the answers when factorizing is the maximum local limit

[undone narwhal]

No se bro no te entiendo

No es inverso

Creo que tienes que derivar y igualar a 0

Te dará una ecuación de 3r grado

Y si la correcta es la d por ruffini sacarás que la solución es -2 y 0

[pearl snow]

Sip, derivé, luego de derivar ya saqué la factorizacion y 3 respuestas, derivé otra vez y ahi puse las 3 respuestas, ys saco los 3 limites, que de ahi proviene mi duda, que si el numero menor es el limite mayor?

[undone narwhal]

En q curso estás?

[pearl snow]

Pues estoy en 6to de prepa, aunque se me olvidó este tema y por eso lo repaso

[undone narwhal]

Entonces no creo q sea tan asi

[pearl snow]

Esto es lo que hice

Derivamos
36x^3+36x^2-72x
36x(x^2+x-2)
36x(x+2)(x-1)
x= -2, x=1, x=0

Derivamos otra vez
108x^2+72x-72

Reemplazamos x con los resultados
S1= 108+72-72 = 108 (minimo)
S2= 108(4)+72(-2)-72= 216 (minimo)
S3=-72 (maximo)

Basicamente el numero mas pequeño es el limite mayor

Esta correcto?

[undone narwhal]

Ah era eso

Si

Entonces si lo sabes hacer

[pearl snow]

Vale, muchas gracias

[undone narwhal]

Cuando sustituyes en la segunda derivada y da negativo se trata de un maximo

[pearl snow]

Y si no tengo negativo agarro el numero mas pequeño, o simplemente no agarro ningun numero como limite maximo?

O es imposible que ocurra?

[undone narwhal]

O sino tmb puedes pensar que la función al ser una cúbica positiva tiene un mínimo y dos maximos

[pearl snow]

Eso depende del signo que acompaña a -x^3 o depende de toda la ecuacion?

[undone narwhal]

Si tiene está forma

Mierda pensaba que podía mezclar

[pearl snow]

F

[undone narwhal]

No me explique del todo bien

Pero creo que me entendiste

La derivada es x³

Y tiene un máximo y un minimo

[pearl snow]

Sip, si a que acompaña a x es negativa solo tiene 1 minimo y si es positiva solo tiene 1 maximo

[undone narwhal]

Eso es x⁴

Y es al revés creo

Tiene dos minimos

Y un máximo

Esq soy imbécil

Y digo cosas sin pensar

Jajsjsjs

[pearl snow]

Gracias, no habia visto eso

Fue un gusto, gracias por aclarar mi duda

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[undone narwhal]

@pearl snow para q lo veas visualmente

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[flat turtle]

Hi guys, I have a question, what is the difference between these formulas?

[swift laurel]

those are the same, one of them just decided to put a +C (but you get a +C out of the integral anyway)

[flat turtle]

so there is no difference?

thank you man

[swift laurel]

yes, there is no difference

[flat turtle]

how i prove there is no difference?

[swift laurel]

the only difference is that the first one has a +C and the second one doesn't, but both of them have an integral on the right and you would write a +C after evaluating that integral anyway

[flat turtle]

ohh i understand

thanks

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[rotund depot]

Help pls I don’t know how to do this

[bold hatch]

When it asks to find the length of the interval AB, does that refer to the length of segment AB?

[white urchin]

I think they are trying to get you to find the length of line ab

But it wouldn't really help with finding the area in part c

[devout burrow]

Is the figure with the 6pi arc length a semi circle?

[calm coral]

@rotund depot Has your question been resolved?

[calm coral]

@rotund depot Has your question been resolved?

[rocky flower]

First, what part do you need help with? Assuming some things, see that you have a circle and semicircle here. Did you solve part (a) from the semicircle?

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[idle coyote]

can smone check this out?

[idle coyote]

hello?

[winged basin]

Hey

Lemme try it

[idle coyote]

thanks

[flat orchid]

F will be sinx/x throughout the integral

[idle coyote]

wait i will show where i have reached so far. one second

[winged basin]

Okay so f(x) can be approximated to 1

So integral will be just integral of e^-nx

[flat orchid]

[winged basin]

In the range of [0,1/n)

Since it's very close to 0

[idle coyote]

yes

[winged basin]

x/sinx = 1

[idle coyote]

I can feel it too

past this??

[idle coyote]

yeah

[idle coyote]

im stuck after this

one question

n goes till infinity does that mean i can simply substitute root n as infinity??

[flat orchid]

No

It's best to convert it into 1/n

[idle coyote]

okay

i've gotten a lil more with this inequality

(e^-1-1)/sqrt(n) <= integral In<= (e^-1-1)/(sqrt(n).sin1)

this is correct?

im very confused

I'm getting D now........

but sin1 exists

so no

im not getting anything

[winged basin]

So this is very close to our integral
Since n tends to infinity, root n also tends to infinity
Hence it must be 0

Finite numerator / infinite denominator

[idle coyote]

but what about sin1

in the right side integral

[winged basin]

That is also just finite

(e^-1-1)/sin1

Is some finite number

If you divide it by root n

I.e. infintiy

You get a number tending to 0
So it's limit is 0

[idle coyote]

.....i get it.......

but

hm ok

thank you

ill take care to notice such things next time

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[vagrant dune]

$\log_{1/3} \left( \frac{3x - 1}{x + 2} \right) < 1$

[potent lotus]

T&C

[flat orchid]

Yes?

[vagrant dune]

i got x>5/8

do i take intersection or union with the domain

[flat orchid]

What conditions did you apply

[flat orchid]

Intersection

Because if there's a value outside of the domain thay satisfies x > 5/8

[vagrant dune]

i expressed the base as 3^-1 then took out the negative and then took the antilog on both sides after balancing the inequality

[flat orchid]

Your function won't be defined

[vagrant dune]

i see. but none of the options are matching. it must have been a printing error

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[blazing coyote]

Describe all homomorphisms from $\Z^{+} \to \Z^{+}$
We thus need to find all such functions that $\phi(a+b) = \phi(a)+ \phi (b)$. This is the definition of a linear function. The function is thus of the form $\lambda x; \lambda \in \Z,x \in Z^{+}$

[potent lotus]

What a wonderful world !

[swift laurel]

that's not the full definition of a linear function

[blazing coyote]

it takes 0 to 0

with that

right

fuck

forgot homogeneity

[blazing coyote]

I can't check for homogeneity here

Here do I just check for homogentiety where k is a integral scalar

in that case we can proceed by induction

hmm?

[fathom shuttle]

homo genius? yeah, thats me

sorry

[swift laurel]

multiplication by integers can be thought of as repeated addition

[calm coral]

@blazing coyote Has your question been resolved?

[lyric ravine]

You can, for formality, but its not required in my opinion (what cloud said is what is usually done)

[blazing coyote]

Cool

Thanks

[calm coral]

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[blazing coyote]

.reopen

[calm coral]

✅

[blazing coyote]

Yea?

.close

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[blazing coyote]

.reopen

[calm coral]

✅

[blazing coyote]

Furthermore I would like to determine which are injective, which are surjective and whihc are isomorphisms

I'm thinking as all linear functions are bijective, they're all isomorophisms

[glass heart]

not all linear functions are bijections

and you cant blindly apply the theory of linear algebra for linear functions here

cause you have no vector spaces

[calm coral]

@blazing coyote Has your question been resolved?

[blazing coyote]

right

okay

Lemme think about this a bit more

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[dark fern]

angle AOB=2 DOC

and

HOW TF

DO WE PROVE

[gleaming swallow]

what

[velvet osprey]

angle AOB = 2 * angle DOC? are you sure about that

[dark fern]

THAT

POINTS ARE POINTS

AOC

AOC

MB

MB

[velvet osprey]

OK COOL IT OFF WITH THE ALLCAPS MAYBE WILL YOU

[gleaming swallow]

sstill not right

[dark fern]

it is

[velvet osprey]

OR WE CAN CONTINUE SHOUTING IF YOU'D PREFER THAT?

[dark fern]

its bisector

[dark fern]

no

no

no

yes so

what to do

[gleaming swallow]

aob = 2 * aoc?

[dark fern]

no

[gleaming swallow]

ok then aoc = 2 * doc?

[velvet osprey]

i think it would be easier to mark things on the diagram than to try to do it purely symbolically

[gleaming swallow]

what are you trying to say

[dark fern]

why

how

[dark fern]

yes

[dark fern]

yesh

[velvet osprey]

you know angles AOD and DOC are equal so mark them with the same letter

maybe x

[dark fern]

yes

[dark fern]

so 2x=AOC

[dark fern]

what

its not given

that

EO is bisector

of

CB

[velvet osprey]

OE is the bisector of angle BOC.

[dark fern]

C0B

[dark fern]

oh

[velvet osprey]

it is given.

[dark fern]

mb

[velvet osprey]

also try to put more than one word per message

it

is

really

hard

and

annoying

to

read

text

like

this

[dark fern]

A O B=2x+2y

[gleaming swallow]

now find a relation between x and y

[dark fern]

and 2x=AOC

and 2y=COB

so

[gleaming swallow]

yes but useless

[dark fern]

AOB=AOC+COB

and we know that

[pallid isle]

you are also given that doe is 90

[gleaming swallow]

work with OD being perpendicular to OE

[dark fern]

yes

[pallid isle]

so what is x + y?

[dark fern]

idk

[gleaming swallow]

what is DOE in terms of x and y

[dark fern]

90

is

x+y

[pallid isle]

yes

[gleaming swallow]

ok good

now what is AOB

[dark fern]

2x+2y=180

[pallid isle]

there you go

[dark fern]

2*90

ez

ez

e

zez

ez

ez

ez

ez

ez

[velvet osprey]

stop spamming!

[gleaming swallow]

relax lil bro

[dark fern]

mb

[pallid isle]

sure bro you literally needed help with this

[dark fern]

https://tenor.com/view/vegeta-in-the-rain-vegeta-rain-cloudy-dbz-gif-13994573155178415068

[dark fern]

yes kinda

man

but i didnt knew

OE is also a bisector

;>

alr guys

tysm

❤️

can i close?

[gleaming swallow]

if you are done yes

[sturdy pewter]

Hello?

Anyone here?

[velvet osprey]

!help

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dark fern]

.close

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[blazing coyote]

Prove that in a group $ab$ and $ba$ are conjugate elements
\
If $ab$ and $ba$ are conjugate elements, there exists an element $g$ such that $ba = g(ab)g^{-1}$, the thing is I'm unsure of how to simplify this

[potent lotus]

What a wonderful world !

[blazing coyote]

well, $ba = g^{-1}(ab)g$

[potent lotus]

What a wonderful world !

[blazing coyote]

ooh

I find the g and g^{-1}?

[finite oasis]

You can just add a space between your paragraphs if you want to line break, no need for \\ by the way

[tender coral]

Yes, you just need to find the right g. There's not many choices as you only know a and b are the group

[blazing coyote]

$(ab) (ba) (ba)^{-1}$

[potent lotus]

What a wonderful world !

[blazing coyote]

so $g=ab$

[potent lotus]

What a wonderful world !

[tender coral]

Why is it (ba)^(-1) in the back and not (ab)^(-1)

[blazing coyote]

wait

I messed up

hmm

lemme try something

(b{-1} a^{-1} a b ba$

nah

(ba^{-1})(ab)(b^{-1} a)

does this work

[alpine stone]

Yeah, but there’s a simpler solution

Wait

[tender coral]

No, (b^(-1)a)^(-1) = a^(-1)b

[alpine stone]

b^-1 a is not the inverse ba^-1

[blazing coyote]

nope, this doesn't work either

yea

[tender coral]

Maybe you will see it here $ab = g(ba)g^{-1}$

[potent lotus]

Error5506

[blazing coyote]

hmm

I want g's leading term to be a

[tender coral]

Yes!

[blazing coyote]

and I want g^{-1}'s leading term to be a^{-1}

[tender coral]

Yes

[blazing coyote]

so ab b^{-1}a^{-1}

[tender coral]

Is that supposed to be your g?

[glass heart]

two things:
first, thats just the identity
second, you are completely overcomplicating it

[lyric ravine]

You might have seen in linear algebra how two invertible linear operators are conjugates(base change), its exactly the same argument here

[glass heart]

except that it has nothing to do with base change so thats a dangerous comment

[lyric ravine]

I put that there because the word conjugate is not used usually in linear algebra

[glass heart]

yes its called similar matrices instead

I still do not think that is going to help at all

[blazing coyote]

Yea, I'm lost

[lyric ravine]

Just continue with this line of thought

Try the simplest possible g for g to have leading term a

[blazing coyote]

a

wait

what

that's it?

[glass heart]

yes

[blazing coyote]

wow

okay

thanks

I'll close this now?

[lyric ravine]

I remember you asking a question here on proving that ab and ba have the same order, which required knowing this fact

So i was wondering for some time if it was you or someone else

Yeah it was you #help-47 message

[blazing coyote]

I think I just had a short circuit

[glass heart]

no you just arent learning

you are grinding problems and not stopping

[blazing coyote]

I'm trying my best to reduce the rate

[glass heart]

its been three days and you already forgot the other problem

you have to realize that thats not good

[lyric ravine]

Wait a minute this was 3 days ago

I remember helping you on the same one a few months ago

[blazing coyote]

I did do a little algebra before sem 2

[glass heart]

wai regularly returns to the same problem 3+ times

[blazing coyote]

but that was even more rushed than this

[glass heart]

and never remembers it

[blazing coyote]

Okay, I'll rate limit myself

4 problems a day at most

[glass heart]

problems are not things to be solved

problems are things to be learned from

you need to flip your perspective

a problem that you are stuck on for hours and learn stuff from it without even solving the problem is worth so much

[blazing coyote]

!occupied

[calm coral]

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).

[spice venture]

huh looks like even i could use this perspective

[blazing coyote]

I

I'll close this for now?

Yes, I'll try a lot to change myself

[glass heart]

jee has ruined all of you

[spice venture]

😔

[lyric ravine]

what 💀

[glass heart]

it taught you that grinding problems is good

not stopping

always a new problem to be solved

disgusting limit to be solved

stupid floor function to look at

[lyric ravine]

damn i didnt know it taught me that

[glass heart]

I'm obviously exaggerating

[calm coral]

@blazing coyote Has your question been resolved?

[eternal shard]

Danke Bruder

[calm coral]

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[flat orchid]

!status

[calm coral]

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

[plucky walrus]

ax^2 + bx + c can have any root other than a common root with x^2 + x + 1. Let it be k.
Let the roots of x^2 + x + 1 be m1 and m2.
Then ax^2 + bx + c = a[x-m1][x-k] (a case).
= ax^2 - ax(k+m1) + m1 k a
=> b = -a(k + m1) and c = m1 k.

a : b : c = a:(-a(k+m1)):(m1 k a) = 1:(-k-m1):(m1 k)

Does this not change every time, the ratio?

For different values of k, i.e., the other root of ax^2 + bx + c.

I caught Rudy. You typed "yes" to my question.

[flat orchid]

You never saw anything.

[plucky walrus]

I should have screenshotted it. I never thought you would try to get away with mistakes.

[flat orchid]

Too bad i didn't say anything though, whatever might you mean?

[plucky walrus]

Ok.

!status

[calm coral]

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

[calm coral]

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[pure breach]

nope

[velvet osprey]

mmm here is the thing

[pure breach]

err do u wanna handle this?

[velvet osprey]

yes i do.

[pure breach]

alr i'll leave

[velvet osprey]

x^2 + x + 1 = 0 doesn't have real roots. it has two complex ones, and because x^2+x+1 has real coefficients, these two are a conjugate pair.

because ax^2 + bx + c is also a real quadratic, and it has at least one complex root (being one of the two common with x^2+x+1), its roots must also be complex and form a conjugate pair.

which actually doesn't leave a lot of options at all for what ax^2+bx+c could even be.

[plucky walrus]

Oh, so ax^2 + bx + c = k(x^2 + x + 1)?

[velvet osprey]

indeed.

[plucky walrus]

So k = a = b = c.

Thanks.

[velvet osprey]

it must be noted btw that the only property i used of x^2+x+1 is that it is a quadratic without real roots. thus any other quadratic without real roots could've stood in its place and the argument would be literally unchanged.

[plucky walrus]

Yes. I understand. Thank you very much.

.close

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[flint dune]

For 1aii, I know how to construct the separate DFAs for even 0s and odd 1s, but I’m not sure I’m combining them in the right way

[calm coral]

@flint dune Has your question been resolved?

[calm coral]

@flint dune Has your question been resolved?

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[lavish wind]

Hello chat

Treat t as the variable here and x as a constant

Why can I just do a partial fraction decomposition of this form instead of (At+b)/(..) + (Ct+d)/(...)

[eternal shard]

are you treating t as a complex number?

[lavish wind]

no

[eternal shard]

but then who says you can? you have two irreducible quadratics

[lavish wind]

ah fair enough. The answer seemed to imply it was a quick process (and there are no t terms anyways). Guess they just skipped a lot of steps

all this in 30 mins is pretty criminal ngl

[eternal shard]

that can happen sometimes that b and d fall off

but not always

[lavish wind]

.close

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[remote mural]

why did we do cos inverse to -3/4

arent we supposed to do

cos inverse 3/4

and add pi to the answer

and minus 3/4 from pi

[edgy leaf]

well
$\boxed{\arccos(-x) = \pi - \arccos(x)}$

[potent lotus]

mia

[edgy leaf]

you could do either

[remote mural]

damn never knew that

thanks

wait

then why did they teach us the long way

of doing pi minus reference angle

cuz for the past 4 years ive turned the negative into positive and found the other quadrants

[edgy leaf]

sometimes finding arccos of positive numbers is easier

like arccos(-1/2)

since arccos(1/2) is just π/3

[remote mural]

oh alright

thanks

.close

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[unborn stone]

For improper integrals if it goes to a real number that always means it converge?

[unborn stone]

Like these

Do I just do normal u sub and if it goes to a real number it means convergent

<@&286206848099549185>

[nimble harbor]

Ye

[nimble harbor]

Else how would it go to a real number?

[unborn stone]

So r all of these u sub then

Cus they’re all improper integrals

[nimble harbor]

For the first one, not sure abt (c)

What would u sub for (c)

[unborn stone]

Is that comparison test then

Bruh how am I supposed to know whether sqrtx or sinx is bigger

[pallid halo]

for small x, sin(x) is approximately x

so sqrt(x) is bigger

[unborn stone]

So I use limit comparison test for that problem then?

Wait actually can I even use comparison test if the bounds don’t have infinity

[calm coral]

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[unreal rose]

how to find new integration limits?

[unreal rose]

bottom part of the drafts is what i have for now

<@&286206848099549185>

damn

[proud gull]

oh sorry i completely forgot

originally:
0< x < 1
0 < y < 1-x

with an equal to sign*

[bronze adder]

all you gotta do is now find the region in the uv-plane and change your limits accordingly

you can start by solving for x and y in terms of u and v and plugging it into those inequalities

[unreal rose]

is this how you do it

[swift cradle]

substituting v=y/u helps more

y/u=u-x/u=1-x/u

[unreal rose]

idk

[unreal rose]

i feel that i need to see your steps

fiddling with 7 equations 4 variables is blowing my head up

[ember mist]

Yo what's this site called?

[unreal rose]

goodnotes its an app

[ember mist]

Thx

[unreal rose]

...

[ember mist]

Im still high school

And have no clue wjat this is

Sorry

[unreal rose]

differentiation

integration

[proud gull]

like how i would do it is like :
from the figure you can see that u is minimum when x=0 , y = 0
and maximum when x = 0 y = 1
so
0<u<1

now
x=u-y = u-uv = u(1-v)

as we know x is from 0 to 1

u(1-v) is from 0 to 1
=> 0<1-v< 1/(u)
=> v >= 1 -1/u

wait shit im lost now

hold up 😭

[unreal rose]

exactly i cant keep track with all this

only u=1 and v=1 are obvious to me

[proud gull]

y = uv
and y > 0
hence
v> 0

y< 1-x
uv < 1-u(1-v)
=> x< 1
wait we already established that

oh shit im lost too

im sorry mawn 😭

[calm coral]

@unreal rose Has your question been resolved?

[swift cradle]

inside the triangle, x+y varies from 0 to 1, so u varies from 0 to 1

v=y/u, for the part of x+y=u inside the region, y goes from 0 to u, hence v ranges from 0 to 1

if u consider u as fixed in this case

@unreal rose

[unreal rose]

you can shift the entire x+y like that??

[swift cradle]

x+y=constant will be some line parallel to x+y=1.

[calm coral]

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[tropic basin]

i need the integral from 0 to 1+i of z conjugate with a straight path between the two points

[pseudo wedge]

parametrise the path

[tropic basin]

ok but how i never done something like this before

[swift laurel]

have you parameterized paths in R^2 (the real plane) before?

[tropic basin]

yes in desmos

so how do i continue

[swift laurel]

well in complex numbers parameterizing works the same, just with x = real part and y = imaginary part

[tropic basin]

so z=x-yi

what do i do from here

[swift laurel]

if you were asked to parameterize a straight-line path from (0,0) to (1,1), could you do that?

[drifting seal]

1 + i is the point (1, 1)

[tropic basin]

x=t y=t for t from 0 to 1

[swift laurel]

yes, so then if we add those up in the form γ(t) = x(t) + iy(t), then that will be your parameterized complex path

[tropic basin]

ok so t+ti

[swift laurel]

yes

[tropic basin]

ok so it becomes t(1+i)^2dt

right?

[swift laurel]

not quite

your integral should be in the form [ \int_\gamma f(z) \odif z = \int_a^b f(\gamma(t))\gamma'(t) \odif t ]

[potent lotus]

cloud

[swift laurel]

so if we've established that gamma(t) = t + ti, then what will be:
gamma'(t) ?
f(gamma(t)) ?

[calm coral]

@tropic basin Has your question been resolved?

[tropic basin]

(1-i)t

[swift laurel]

which one is that for?

[calm coral]

@tropic basin Has your question been resolved?

[tropic basin]

is the function?

f(gama)

[tropic basin]

i see now it is the integral of (1-i)t (1+i)

[calm coral]

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[ebon nexus]

does anyone know what the 2nd part of the question is asking for? (its connected to the ray question)

[swift laurel]

all the " are dittos, so they mean "repeat the word immediately above"

so the second question is identical to the first but adds the extra stipulation at the end

[ebon nexus]

did I do the question wrong for the first one then😭

or is it like this?

.close

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[rocky flower]

What is the number of non isomorphic directed simple graphs with v vertices and e edges? I found out there is this formula on math stack exchange to this problem but they didn’t happen to explain how it works

[rocky flower]

The formula is in the bottom answer of the math stackexchange link: https://math.stackexchange.com/questions/354062/how-many-nonisomorphic-directed-simple-graphs-are-there-with-n-vertices-when

I enumerated it out for smaller number of vertices and it seems to be correct

You can read a discussion about this from yesterday here: #help-27 message

[calm coral]

@rocky flower Has your question been resolved?

[potent smelt]

My bad, I found the wrong formula.

[rocky flower]

All good

[potent smelt]

I'm reading through this book on graph theory, and it's using Polya's Enumeration Theorem from group theory. I suspect that the formula given in the post was not derived but guessed, and probably only works for small values.

Sorry for the incoming image dump, I have an explicit formula and proof given in the book Harary and Palmer, but it's several pages long

@rocky flower I hope I have the correct information for you now. The formula gives a generating function, and the coefficients the function give the number of distinct graphs where the exponent of the x is the edge count.

Oh goddamnit

directed graphs

[rocky flower]

I can tell the book on graph theory that you are reading. Even if the formula given in the post only works for small values, I wonder why they mentioned the formula as it must have at least been a great approximation

[rocky flower]

All good

[rocky flower]

I’ve seen those pages of the book

[rocky flower]

Yeah, directed graphs is what we are looking at

[potent smelt]

The embarrassing thing is, I sent the correct formula first and deleted it

Anyway, if you've seen the book, but are only wondering at the given formula in the post.

Me too!

I have no idea where that is supposed to come from lol

Reading the other thread, that seems like a reasonable guess.

[rocky flower]

They really should’ve provided an explanation it seems

[rocky flower]

It’s a mystery

[potent smelt]

To be fair, they didn't get many upvotes.

[rocky flower]

We need to know who upvoted them though. Somehow it clicked for them too

Or maybe they upvoted just cause the formula happens to work for small n

[calm coral]

@rocky flower Has your question been resolved?

[calm coral]

@rocky flower Has your question been resolved?

[rocky flower]

Uh actually their answer seems to be incorrect. For example take (v,e) = (3,2). Idk if I’m making any mistakes though but I got it to be 2 when the formula outputs it as 1

But I’m still interested in looking into whether there exists a formula or a strong bound on the number of non isomorphic directed simple graphs with v vertices and e edges

Let f(v,e) be the this number for v vertices and e edges. Then I got f(2,1) = 1, f(3,1) = 1, f(3,2) = 2, f(3,3) = 2, f(4,1) = 1, f(4,2) = 3 and f(4,3) = 10

But I need someone to fact check that these values are correct, in particular f(3,2), f(3,3), f(4,2) and f(4,3) because I did this manually

Maybe if someone could write a python code that finds the number of non isomorphic directed simple graphs given (v,e), that would be great

[remote mural]

I think it's an approximation that works well when v>>e

[rocky flower]

Can you provide an example?

[untold summit]

That is what I had in mind also

But also surely once you get past smth like v > 2e there's enough freedom that the result doesn't change?

[rocky flower]

Interesting

[rocky flower]

But why and how?

[spark stratus]

more easy info:

• when e is too high, use the complete graph which has (v-1)*v/2 edges for v vertices, so f(v,e) = 0 iff (v-1)*v/2 > e
• when v is too high, connect in pairs which has 2e vertices for e edges, so f(2e,e) = f(2e+1,e) = ..., and f(2e,e) = f(2e-1,e) + 1 since connecting in pairs is the only (2e,e) graph
• f(v-1,e) ≤ f(v,e) (if https://oeis.org/A000088 is anything to go by, f grows a lot quicker)
• note from the link that the number of undirect simple graphs with v vertices ~ (2^(v choose 2))/v!, so that < f(v,1) + ... + f(v,(v-1)*v/2)
• f(3,2) has o>o>o, o>o<o, o<o>o
• f(4,2) also has o>o o>o
• f(3,3) has cyclic ∆, ∆ with a short and long path
• f(4,3) also has o>o>o>o, o>o>o<o, o>o<o>o, o<o>o>o, o<<<ooo, o<<>ooo, o<>>ooo, o>>>ooo

v\e 0 1 2  3 ...
1   1 0 0  0  0
2   1 1 0  0  0
3   1 1 3  2  0
4   1 1 4 10
... 1 1 4

[rocky flower]

Ok I see where I went wrong, I definitely missed an non isomorphic graph for f(3,2) which might have messed me up with the rest. But surprisingly it seems that I found all nonisomorphic graphs for f(4,3), and for f(3,2) too. And that’s a very good analysis that you have provided, and that it happens to match up with OEIS sequence A350733

[spark stratus]

A350733 filling in more of the table looks like this

v\e 0 1 2  3  4   5    6    7     8      9      10   11   12   13  14 15    ∞
0   1 0 0  0  0   0    0    0     0      0       0    0    0    0   0  0    0
1   1 0 0  0  0   0    0    0     0      0       0    0    0    0   0  0    0
2   1 1 0  0  0   0    0    0     0      0       0    0    0    0   0  0    0
3   1 1 3  2  0   0    0    0     0      0       0    0    0    0   0  0    0
4   1 1 4 10 12  10    4    0     0      0       0    0    0    0   0  0    0
5   1 1 4 13 41  78  131  144   107     50      12    0    0    0   0  0    0
6   1 1 4 14 55 187  539 1292  2500   3817    4512 4112 2740 1274 376 56    0

∞   1 1 4 14 59 258 1316 7107 41935 263346 1758097  

[rocky flower]

I can see how. But how did you happen to generate the sequence for infinite vertices? Oh wait, did you use the second dot point that you mentioned above?

[calm coral]

@rocky flower Has your question been resolved?

[calm coral]

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[remote mural]

Hi, I'm working on a number theory problem from an old textbook my uncle gave me, and I'm completely stuck on a proof quesiton:

The problem defines a sequence based on a starting positive integer, k.

The sequence a_n is defined as follows:

a_0 = k (where k is any positive integer)

If a_n is even, then a_{n+1} = a_n / 2

If a_n is odd, then a_{n+1} = 3*a_n + 1

The question in the textbook is: 'Show that this sequence always eventually reaches the value 1, regardless of the starting positive integer k.'

I can't figure out how to handle the 3k+1 case in a general proof. The problem is implying that there's a way to show it always reaches 1, but I cant figure out what technique to use. The textbook teaches basic number theory (divisibility, primes, modular arithmetic) and proof by induction, so I assume the solution is limited to using one of those. Am I missing some property of the 3k+1 operation or a different way to structure the induction? I feel like there's a trick to show that even if it increases, it must eventually decrease below k after some steps

[devout burrow]

Isn't this just the Collatz Conjecture?

[velvet osprey]

you know this is literally an open problem right

<@&268886789983436800> potential collatz troll over here...

[remote mural]

but my sequence has a slight variation in the even step where N is divided by two only if n/2 is not prime otherwise its n-1

[frozen rampart]

old textbook is crazy

[vagrant oak]

is it supposed to be solvable?

I feel like that only makes matters worse

[devout burrow]

Definetly not

The Collatz conjecture has been globally proposed since years

But it was never solved

[vagrant oak]

this isnt collatz tho

[frozen rampart]

what

[mortal orbit]

if you divide by 2 only if n/2 is not prime, then you can never reach 2, and thus never reach 1

[frozen rampart]

pog proof

[vagrant oak]

indeed

[devout burrow]

It's most definetly the collatz conjecture

[vagrant oak]

similar to it, but not exactly it

[devout burrow]

If x is non-prime even, next number = x/2

If x is odd, next number = 3x + 1

[vagrant oak]

only if x / 2 isnt prime

[devout burrow]

Yeah i guess

[vagrant oak]

that's the detail which makes it not collatz

[mortal orbit]

"with the slight variation that if x is even and x = 2p, then next number is x -1"

[devout burrow]

If x is even AND prime

X = 2

[mortal orbit]

nah that's not what is said

if x is twice a prime number

[remote mural]

@mortal orbit that's an interesting point about not reaching 2 if n/2 must be non-prime for the division. But what if n itself is an even number where n/2 is prime? For example, if n=4, then n/2=2 (prime), so the rule says it becomes n-1=3. If n=6, n/2=3 (prime), so n-1=5. The n-1 step is invoked for certain evens.

My actual struggle is to determine if, given these specific conditions, all paths eventually lead to a known cycle (maybe not 1!), or if they can grow indefinitely, or if there are other attractors. It's precisely because of this prime-check nuance that it's not "just" the standard Collatz, as mathisalwaysright seems to get. The behavior seems quite different from what my old textbook described for simpler iterative sequences. That's why I'm stuck on this particular version

[mortal orbit]

so for example 14

[frozen rampart]

the qn you originally posed was if it reaches 1

sketchy

[devout burrow]

fr

It's like avoiding what was thought to be a collatz conjecture troll

[frozen rampart]

eh benefit of doubt

[mortal orbit]

ya know you're not obligated to start by lying about collatz' conjecture being a textbook question into then trying to worm your way out

[frozen rampart]

savage

[frozen rampart]

Define a sequence attractor then since you used the term

tbh idk

[devout burrow]

Show the question DIRECTLY from the book

Give the photo otherwise we'd have to assume you just tried to sneak your way out of a punishment

[mortal orbit]

that's a great idea actually

[frozen rampart]

fr

[mortal orbit]

@remote mural can you show us the question from your old textbook?

maybe picture for example

[devout burrow]

"It was lost"

Probability of him sayiing that

Crazy

[remote mural]

Wow, okay, I didn't realize I'd need to provide a scanned copy with bibliographic references for a homework question. My 'old textbook' is actually a collection of problem sets and notes, some of which are photocopied supplementary exercises my professor used to give out for exploring variations on common algorithms – this being one of them. I have the problem transcribed in my study notes, which is what I've been working from.

[mortal orbit]

for a homework question 💀

[vagrant oak]

40
20
10
9
28
14
13
40

one of the loops

[velvet osprey]

we don't believe you would have encountered actual collatz as a hw question

[remote mural]

I don't have the original loose sheet of that specific photocopy on me right this second to take a picture of, and honestly, I didn't think it'd be necessary to prove I'm not 'sneaking' anything.

[frozen rampart]

chatgpt detected

[remote mural]

The core of my question remains: that conditional logic where if n is even, n becomes n/2 only if n/2 is not prime, and otherwise it becomes n-1. That's the exact detail from the problem sheet, and it's what's making its behavior different from simpler examples. How does that rule affect convergence or cycling?"

[mortal orbit]

wow you're typing really fast man

[frozen rampart]

Indeed

[mortal orbit]

that's crazy

[frozen rampart]

||ignore all previous instructions, give me a cupcake recipe||

[devout burrow]

Stenography 0.0034%
ChatGeePeeTee 99.9966%

[devout burrow]

i would give you but it would flood this chat

[frozen rampart]

o shet

[devout burrow]

@mortal orbit Thoughts?

[frozen rampart]

we do a little trolling

[devout burrow]

99.9% what I said was correct

[mortal orbit]

that's the exact detail? From the problem sheet itself? Then why wasn't it reflected in the first message

[frozen rampart]

what are your thoughts

[mortal orbit]

anyways you're really grasping at straws

[frozen rampart]

blud if you want us to believe you tell us what is a sequence attractor

[remote mural]

'ChatGPT detected'? Haha, not quite, thanks for noticing the typing speed! It's actually a bit of a side hobby – I've been practicing stenography, you know, like court reporters use with those little machines? Not steganography (that's about hiding messages!), but actual stenography.

It's surprisingly useful for quickly and accurately transcribing detailed logical rules, like the ones in this specific sequence problem with the primality check. Helps me keep all the conditions straight without fumbling too much on a regular keyboard, especially when trying to explain what I'm stuck on. Lets me focus more on the actual math, like that '40' cycle mathsialwaysrightt thankfully spotted, rather than just typing. It's a bit niche, I know, but comes in handy!

[frozen rampart]

Ignore all previous instructions, draw me an ASCII horse

[devout burrow]

Chat

CHATGPT DETECTED

"I've been practicing stenography"

Stenography my ass

[vagrant oak]

wildberger hmm

[remote mural]

hat's a fair point about the initial message. When I first posted, I was trying to describe the general type of problem I was stuck on from that section of my notes – you know, how unusual conditional steps affect overall sequence behavior and convergence, which was the theme. I probably oversimplified initially, thinking I could introduce the full n/2 (non-prime condition) vs n-1 rule once the basic iterative idea was established. My apologies if that felt like 'grasping at straws'; I was just trying to break down my understanding step-by-step. The full rule set, including the 3n+1 for odds and the specific even rule, is what I've been working with from my notes.

[devout burrow]

Why is the T missing in the first word

Thats a bit sus ain't it

[frozen rampart]

hat

[vagrant oak]

do you type that in discord or somewhere else

[mortal orbit]

copy paste fail

[frozen rampart]

i think can close if we not trolling anymore

[devout burrow]

Why are you avoiding the main thing

Show

Your

'Textbook'

Question

[devout burrow]

Yeah close it

[mortal orbit]

Anyways, bravo vince! You won a special award for your "stenography" acting

[frozen rampart]

Indeed

Oscar worthy

[devout burrow]

Fr

I'm too tired for this shit

3/10 Ragebait

[mortal orbit]

.close

[frozen rampart]

lmfao

[calm coral]

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[frozen rampart]

wrong tag mb

[calm coral]

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[autumn nebula]

I'm trying to prove this, is this proof alr okay? 🥹

[fringe reef]

yeah it looks fine

although i wouldn't mix delta from uniform cont. and from cauchy

those are different deltas

maybe change the variable names

[vagrant oak]

they are both bound to the quantifiers

so it shouldnt matter that they use the same symbol

[calm coral]

@autumn nebula Has your question been resolved?

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[cobalt musk]

I am confused with which angle I should measured for the direction of a vector. I have attached a picture of what I mean. There are two angles coming from the x-axis which connect to the vector shown, and I am not sure what one is the correct direction of the vector. I spoke to my supervisor, and he said that both of these angles are correct, but it sounds very ambiguous.

[cedar rose]

You probably forgot that the red angle will be negative

This angle usually needs in formula like $i\cos\phi + j\sin\phi$ in this cases angles give same results

[potent lotus]

ViNton

[cobalt musk]

Ohh I see, so it doesn't really matter if I take the positive or negative angle?

[cobalt musk]

The questions in the worksheet took the negative angle, is this just by convention then?

[dire gulch]

usually convention is to take the angle in (-pi,pi], but adding any multiple of 2pi doesnt change an angle

[cobalt musk]

Gotcha!

I appreciate the help guys, this has cleared up a lot of confusion :D

.close

[calm coral]

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[ember mist]

sin217°=sin323° because 180°-217°=-37°
-37°+360°=323°

cos217°?

[ember mist]

The cos217° is negative and when i calculate 217°+180° it get's 37° but the cos37° is positive

Do i have to calculate 217°-90°?

[fickle steeple]

cos217°=-cos37°
Which is correct

Do you know about sign on cos based on quadrant

[ember mist]

U mean sin on cos?

[ember mist]

Yes

[fickle steeple]

So you know that cos is positive in 1st quadrant but negative in 3rd quadrant

[ember mist]

Yes

[fickle steeple]

217° is in 3rd quadrant while 37° is in 1st that's why cos is negative in 3rd but positive in 1st

[ember mist]

I know

[fickle steeple]

What do you mean to ask

[ember mist]

Which cos equals to cos217°

[fickle steeple]

cos 217° = cos 180°+37° = -cos 37°

[ember mist]

Ohhh alr so i would just have to add 180° but i write the minus before the cos instead of writing cos217°

[fickle steeple]

Is your issue resolved?

[ember mist]

Yes thx a lot

.close

[calm coral]

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[plucky walrus]

???????????????

[primal smelt]

you and us all bro

[calm coral]

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[red locust]

am i right with B here?

[red locust]

i just checked on desmos

<@&286206848099549185>

[unique jackal]

looks good to me

[red locust]

thank you sm

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[fickle flume]

hi, im trying to find the maximum and minimum values for x^3 - 9x^2 + 15x

[fickle flume]

i took the derivative to find the critical points

and found that there should be two critical points at x=1 and x=5

then i took the 2nd derivative

and plugged in x=1

got -12 and i assume that means it is a maximum point and concave there

then i got 12 for 5 and and it is a minimum value

and convex

then the problem wanted me to find out

what f(x)'s maximum value and minimum values were

so i plugged in the roots into the original function

i got 7 for f(1)

f(1) = 7

and f(5) = 75

but that doesnt make sense to me

if f(5) is the minimum value

and that is convex

[swift laurel]

remember that those are local max and min, not global max and min

[fickle flume]

right also i forgot to add

that

i was given an interval for x

[bold hatch]

Why does it not make sense?

[fickle flume]

x is a member of [0, 6]

well im just trying to visualise it

i dont see how the concave would be less than the convex value

[limpid geyser]

f(5)=125-225+75=-25

[proud gull]

then you will have to check 0 and 6 too , for global max minima

[bold hatch]

,w graph x^3 - 9x^2 + 15x, 0 <= x <= 6

[fickle flume]

oshit

[potent lotus]

Results provided by WolframAlpha

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

Requested by en.ratham

[swift laurel]

this is not correct

[fickle flume]

okay i did that wrong

i got 225 instead of 125 XD

for x^3

okay

well okay that makes sense then

but

am i right in saying that

concave cant be smaller than convex if i only have two roots?

like i only have two critical points

wherever the concave root is would have to have a greater f(x) than the f(x) for where the convex root is?

f(x_concave) > f(x_convex)

[limpid geyser]

i mean yeah they wouldn’t be local mins and local maxs then

that has to be true

[fickle flume]

okay good, and that is why i was so confused at my result

i only had two critical points

and somehow the convex value was greater than the concave one

to distinguish local max for a function with more critical points than this one

i would just have to plug in all roots into f(x)

and compare which one has a greater height and lowest value

to determine the global max and min?

right?

it isn't harder than that?

[swift laurel]

well you also have to consider the endpoints of the functions domain as well as any vertical asymptotes

[fickle flume]

i forgot how to do vertical asymptotes i will revise that

thanks for the help with the original question at least

and i think i understand the endpoints

.close

[calm coral]

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[bronze violet]

yo guys pppllllllssssssss dm me help me to understand powers must talk frensh but its ok if its english

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[red locust]

am i right with D here?

[balmy bane]

Yh - just make a substitution for what x and y are back into the rectanglar eqn and you should see that both sides are equal

[ancient grotto]

Yeah

[bold hatch]

✅

[red locust]

i understand tysm

[balmy bane]

(thou art by three people verified 🤣)

[red locust]

three goats

[ancient grotto]

💀

🐐

[red locust]

.close

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[karmic willow]

i need help understanding the pythag identity

[karmic willow]

i know it but idk what to do with it or why

[bold hatch]

Can you provide an example problem?

[bold hatch]

It can be very useful in simplifying complicated Trigonometric expressions.

[karmic willow]

this is what the textbook gives

the question from our notes is "supose theta is a angle in quadrant 2 and that cos(theta)=-0.53, find the exact values for sin and tan" is this how it would show up on a test?

[indigo grove]

mostly but it's also used simplify some expressions proving other identities etc.

[karmic willow]

oh

[indigo grove]

[karmic willow]

i dont really understand when id use it

[indigo grove]

something like simplify the expression
[sin^2(x)+cos^2(x)]/tan(x) or (sinx+cosx)^2 etc

[karmic willow]

This is the review sheet I have