help-49

[hexed barn]

why not repost it here?

(for future helpers, that avoids having to jump channels)

[clever sedge]

Next you will have to buy a newspaper to see the question

[midnight plank]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

@vast summit Has your question been resolved?

[vast summit]

🤣🤣🤣

There y'all go

[midnight plank]

@vast summit Has your question been resolved?

[midnight plank]

@vast summit Has your question been resolved?

[marble hedge]

Um

so one cycle

2π/omega

~0.0126

[midnight plank]

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[twilit field]

I would like to solve this using convulsions

we notice$2a<x+y< \infty a$nd thus $2a<z<\infty$

[dreamy lichen]

convolutions?

[twilit field]

Convolutions

[dreamy lichen]

yeah, i was just asking since u said "convulsions "

[grand pond]

wai

[twilit field]

We then have $\int_{2a}^{\infty} \frac{a}{ (z-x)^2} \cdot \frac{a}{x^2}dx$

[grand pond]

wai

[twilit field]

is this right so far?

[dreamy lichen]

seems right

[twilit field]

so I just integrate this to get the density function

[twilit field]

tq

[dreamy lichen]

oh wait

what about the upper bound?

[twilit field]

infty no

[dreamy lichen]

f(x) = a/x^2 for a <= x < inf
f(z-x) = a/(z-x)^2 for a <= z - x < inf

[dreamy lichen]

i think u missed the 2nd inequality

[dreamy lichen]

i too missed it actually, sorry bout that

[twilit field]

so the lower boud is a?

[dreamy lichen]

we should start by solving the inequalitites

so the second one becomes
-a >= x - z > -inf
z-a >= x > -inf
-inf < x <= z-a

[twilit field]

this is the method I'm using

[dreamy lichen]

-inf < x <= z-a
a <= x < inf
yeah, so a <= x <= z-a

a <= x <= z-a
This is the interval in which both f(x) and f(z-x) are non-zero, if they are 0, the whole integrand is 0 anyway

so u should integrate w lower bound a and upper bound z-a

[twilit field]

huh

[dreamy lichen]

Look at the definition of convolution. What you are integrating is f(z-x)f(x)

[twilit field]

yes

[dreamy lichen]

and your f is defined to be 0 for all x other than a <= x < inf

[twilit field]

and it's non-zero when a<x< infty

and a<z-x<\infty

[dreamy lichen]

exactly

upon rearrangement, it becomes a <= x <= z-a

[twilit field]

sure, but adding them

[dreamy lichen]

adding the inequalities? Or adding what?

[twilit field]

the inequalities

[dreamy lichen]

as to get 2a < z < infty?

this is true if both ineqalities are true, but it loses information

it doesnt go both ways

also, u are integrating wrt x, so z is pretty much irrelevant

u need to bound x, not z

[twilit field]

oh right

🤦

It gives me the domain of the PDF

[dreamy lichen]

otherwise it should be correct, just fix the bounds

[twilit field]

so $\int_{a}^{z-a} \frac{a}{ (z-x)^2} \cdot \frac{a}{x^2} dx$

[grand pond]

wai

[twilit field]

cool

I'll just solve one more problem like this

0<x<10

same thing here

.close

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[coral cloak]

hello im trying to reverse a problem

[coral cloak]

this is the problem and im not quit sure if i remember it correctly

and i know that the answer is n=10

but i dont know the problem

[lusty python]

💀

[coral cloak]

furthermore i know that in the differential equation you get from solving this a=3 b=-27 and c=-30

[lusty python]

$C_{x}^{y} = 16(n + 1) + V_{n+1}^2$

Translated version for someone who can't understand his handwritting 💀💀💀

[coral cloak]

the things on the C are question marks

they arent the nimber 2

as thats the part im most unsure about

[lusty python]

What are the question marks for 🙏

[coral cloak]

because i know that there are combinations there but i dont know whats inside

great

this is my problem

[lusty python]

Im guessing the question mark is just some random number

[coral cloak]

one is a number and the opther has n in it

[lusty python]

The C thing obv turns to 1

[coral cloak]

like n+- something

[lusty python]

No need to say

[lusty python]

What's the formula for $V$?

[grand pond]

1 divided by 0 equals Infinity

[coral cloak]

the question marks are 2 different things

[lusty python]

wtf

[coral cloak]

i just marked them the same because

frankly idk why

with the power of hindsight i wouldnt have done that

[grand pond]

1 divided by 0 equals Infinity

[coral cloak]

its variations

2 elements out of n+1 ones

[lusty python]

Where you know what $x$ and $y$ is and you need to find $n$, am I rigjt?

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

Im asking for the formula

[coral cloak]

yes

[coral cloak]

(n+1)*n

[lusty python]

I see

[lusty python]

Substitute that thing you got into the eq

And you got a quadratic of $n$

[grand pond]

1 divided by 0 equals Infinity

[coral cloak]

yes

and i know what the quadratic has to lok like once we get C

and also i know n already n is 10 we are looking for c

it is also possible that the V is wrong

[lusty python]

Restate the question again PLEASE

im down dead if you don't restate the question

[coral cloak]

what are the numbers on C

thats is my question

and i know n already

i got the solution thanks

.close

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[tidal turret]

I have a question somewhat related to this exercise

I am given this Region but I want to find the frontier

aswell as the interior

is there a general method of finding those subsets?

[cerulean oyster]

Lo mas facil es hacer el grafico de las ecuacion y buscar las zonas de interseccion

[tidal turret]

hola buenas todo bien gordo?

[cerulean oyster]

si

[tidal turret]

ok, necesito ayuda con todo eso

[cerulean oyster]

a ver, te vas a tener que acordar/practicar inecuaciones

[tidal turret]

A = { y>= 0, 2y + x^2 <= 9} PONELE

[cerulean oyster]

y >= 0 es basicamente toda la region positiva

[rigid tangle]

Man this sounds so interesting

[tidal turret]

yep basically the from y = 0 and upwards in the y dir

[cerulean oyster]

si, lo voy a hacer yo mismo en desmos

[rigid tangle]

How many plots gonna get deleted today

[cerulean oyster]

Despues tenes 2y + x^2 <= 9

la forma mas sencilla es tratarlo como una ecuacion (=)

Resolver para despejar y

y encontrar la curva

o sea, 2y + x^2 = 9
y despejas la coordenada y

[tidal turret]

tal cual mente

[tidal turret]

dale

9 - x^2 / 2

[cerulean oyster]

con el parentesis, pero si

aca basicamente dependes de tu capacidad de graficar a mano

con < es la parte hacia abajo de la curva

con > es hacia arriba

Tu region "A" es la interseccion entre las dos.

Porque son todos los puntos que cumplen las dos condiciones que te dieron

Para buscar maximos y minimos absolutos, es la misma logica que con una funcion de una sola variable

f'(x) = 0
∄f'(x)
La frontera de la region A

[tidal turret]

ponele, si

[cerulean oyster]

O sea, si lo miras con mucho cariño, la "frontera" de un area de una sola coordenada, son justamente los dos puntos extremos.

Aca tenes que que encontrar una forma de describir los valores de la funcion a lo largo de una curva

[tidal turret]

es dificil graficar esta parabola a mano?

y = (9 - x^2) / 2

jeje

[cerulean oyster]

No, la verdad que no, si sabes como hacerlas no

con x = 0, y=9/2

y sabes que va hacia abajo

[tidal turret]

pero como encontrar el vertice

[cerulean oyster]

Como no tiene termino lineal el vertice siempre esta en x = 0

[tidal turret]

kieee

[cerulean oyster]

todas las parabolas son de forma

ax^2+bx+c

si b = 0, entonces se escribe ax^2 + c

Y se dice que no tiene termino lineal

Porque bx es una linea.

[tidal turret]

b/2a

algo asi es el vertice?

[cerulean oyster]

ni idea

nunca me aprendi la formula del vertice

[tidal turret]

y como sabes que si b = 0 entonces el vertice es 0

no se si me explico

[cerulean oyster]

hace la derivada

te queda 2ax

eso se anula solo en x = 0

que te define un posible max/min

y como las parabolas solo tienen un max o min

[tidal turret]

aya

[cerulean oyster]

entonces esta en 0

[tidal turret]

perfecto

como se deriva esto?

y = (9 - x^2) / 2

[cerulean oyster]

Cuestion con esto, vas a tener que parametrizar una curva para hacerlo de forma analitica

[tidal turret]

9/2 - 1/2(x^2)

[tidal turret]

si

[cerulean oyster]

C1 es una boludez

C2 es una curva que sigue tu parabola original

[tidal turret]

f'(0) = 0 solo en x = 0, la derivada me quedo -x

[cerulean oyster]

si

[tidal turret]

ya sea un maximo o minimo, no nos importa porque sabemos q ese es el vertice

[cerulean oyster]

o sea, igual, mientras sepas dibujar parabolas mas o menos ya deberias poder imaginarte como se ve

[tidal turret]

como describir la parametrizacion o a C1 o a C2, no capte muy bien la idea gordis

[tidal turret]

si

[cerulean oyster]

O sea, si estas haciendo analisis multivariable

casi seguro que ya viste C(t)

una funcion vectorial

[tidal turret]

una parametrizacion de una curva o que

[cerulean oyster]

si

Tipo, C1 es una linea recta sobre el eje x y con y=0

que va desde -3 a 3

[tidal turret]

si

como encontramos la parametrizacion

[cerulean oyster]

O sea que te queda C(t) = (t,0) t=[-3,3]

[cerulean oyster]

tenes que saber vos como hacerlas la verdad

[tidal turret]

ah perfecto

[cerulean oyster]

tipo, en este caso en particular justo podes usar la formula de una parametrica de una funcion

tipo

c(t) = (t,f(t))

[tidal turret]

^

[cerulean oyster]

pero hay veces que no existe una parametrizacion asi

ponele, para circulos, algunas hiperbolas, elipses, figuras cerradas

[tidal turret]

cordenadas polares a veces se usa

[tidal turret]

ok ya tengo una idea de como parametrizar C1

y C2?

[cerulean oyster]

en este caso es lo mismo

[tidal turret]

y = (9 - x^2) / 2

[cerulean oyster]

porque sigue la misma curvatura que nuestra parabola original

si

te quedaria (t,y(t))

[tidal turret]

r(t) = (t, (9-t^2)/2) t ∈ ?

[cerulean oyster]

si

y de 3 a -3 tmb

[tidal turret]

okaaay

[cerulean oyster]

pq son los puntos de interseccion con la linea

[tidal turret]

si

como encontramos los puntos de interseccion entre las dos graficas

[cerulean oyster]

es la interseccion
y=0
y=(9 - x^2) / 2

[tidal turret]

y = 0
2y + x^2 = 9

[cerulean oyster]

tipo, igualas las dos y ya esta

[tidal turret]

x^2 = 9
(x-3)(x+3) = 0

[cerulean oyster]

casualmente aca son las raices de la cuadratica

si mala mia

[pastel mantle]

hi

[cerulean oyster]

!occupied

[midnight plank]

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).

[tidal turret]

en este caso ponele que cambiamos las inecuaciones por ecuaciones

eso fue un poco raro

[cerulean oyster]

las cambiamos porque las ecuaciones son siempre el borde de la inecuacion

sabiendo el borde podes definir toda el area

[tidal turret]

tal cual

[cerulean oyster]

tipo, si lo pensas
a >= b
es la composicion de
a > b y a = b

[tidal turret]

sii

[cerulean oyster]

y en su defecto, si solo tenes >, excluis los bordes

que igual en la busqueda de maximos y minimos los usas igual

si te queda que un maximo esta en el borde cuando no esta incluido, decis que se aproxima a un valor maximo alrededor de ese punto de frontera

[tidal turret]

kieee

[tidal turret]

es posible en una expression matematica parametrizar la region que incluye la frontera de A y el interior de A?

osea el area ese que pintaste con los bordesitos incluidos

sera posible?

[cerulean oyster]

si, pero no con una curva

necesitas una parametrizacion de superficie,
en vez de C(t), tenes S(u,v)

[tidal turret]

ahh me acabo de dar cuenta que eso es justo lo que es A

[tidal turret]

ah mira vos, igualmente, A es una region de dos dimensiones ponele que las superficies son 3 o mas dim

[cerulean oyster]

lo que el "parametro" te dice es cuantas dimensiones hay

los puntos son vectores exactos, las curvas son de 1 dimension, las superficies tienen 2, etc...

hay parametrizaciones de volumen tambien

pasa que generalmente uno se confunde con la cantidad de elementos de la pametrizacion
tipo, C(t) = (x(t),y(t),z(t)) es una curva de 1 dimension que recide en un espacio de 3

[tidal turret]

ya veo, la matematica es complicxdx

[tidal turret]

tal cual

[cerulean oyster]

bue, yo me tengo q ir

bai

[tidal turret]

nos vemos gordi, gracias por la ayuda

.solved

[midnight plank]

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[tidal basalt]

how would i start solving this

i got this done

what about the other one?

can i just rewrite it like this

[small jasper]

Yes as long as $x \neq 2$ and you put the $\geq 0$

[tidal basalt]

does this seem correct

i think i did it

[grand pond]

Civil Service Pigeon

[small jasper]

,w x^2-4 \leq 0, (x-3)^2 (x+4)/(2-x) \geq 0

[grand pond]

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[small jasper]

Looks good

You can check like this in #bots

[tidal basalt]

thank you very much

[small jasper]

[tidal basalt]

.close

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[tidal basalt]

i have to find the area of this green shape

does this seem correct for the left one

[small jasper]

if you only want the part to the right of x=1, then yeah

[tidal basalt]

okay thanks

but im just confused since

the teacher marked the correct one as

1.5

which you get if you do

not sure what those r called sorry

[sharp coral]

that is an integral

[tidal basalt]

i meant the numbers on top and bottom

[sharp coral]

they are called bounds or limits

[tidal basalt]

ah okay

well the teacher marked the bounds as 4 on top and 1 on bottom

not 3 on top, 1 on bottom

which is odd cause i dont know if the teacher did a mistake or

[sharp coral]

did your teacher use absolute value inside the integral?

[dusty portal]

It might be because they did int from 1 to 4 of |f(x)-g(x)|? not entirely sure but that would seem like it

[tidal basalt]

she didnt, she just tried it the usual way but 4 on top instead

i'll ask her tomorrow

[tidal basalt]

the other one seem correct?

[small jasper]

,w area between y=x^2-4x+5 and y=5-x for 3<x<4

[grand pond]

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[small jasper]

good

[tidal basalt]

appreciate it thank you

texit thing seems like rocket science at first glance

.close

[midnight plank]

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[twilit field]

here <k> is supposed to be the group generated by k, right

[hard shard]

yes

[twilit field]

so the elements look like a_n+<k>

[runic hamlet]

depends on what you mean by a_n

[twilit field]

elements of Z_n

I'm trying to figure out what the mapping looks like atm

[runic hamlet]

how about you do an example

[twilit field]

Z_4, with <2>

<2>; 1+<2>, 2+<2>, 3+<2>

[hard shard]

are any of those equivalent?

[twilit field]

<2>+2=<2>

and <2>+1=<2>+3

[midnight plank]

@twilit field Has your question been resolved?

[hard shard]

wai quick question: do you know the first isomorphism theorem

[twilit field]

oops, I thought I'd do that later

I'll do that now

tq

.close

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[hoary drift]

When is k! + 1 a perfect square?

[hoary drift]

I tried 0 to 9 and got 4, 5, and 7

[modern sapphire]

https://en.wikipedia.org/wiki/Brocard's_problem

[midnight plank]

@hoary drift Has your question been resolved?

[hoary drift]

Thank you

.close

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[buoyant linden]

daily integral a few days ago

$\int_0^\pi \frac{x^2\cos^2x-x\sin x - \cos x - 1}{(1+x\sin x)^2} , dx$

[grand pond]

Flatus

[buoyant linden]

im lost

[obsidian bay]

substitute?

[buoyant linden]

is the strat
cos^2x = 1-sin^2x

so the numerator becomes

x^2 - x^2 sin^2x - xsinx - cosx - 1
= -xsinx(1+xsinx) + x^2 - cosx - 1

[buoyant linden]

sub what

u = xsinx?

[obsidian bay]

just spit balling

[buoyant linden]

oh

[buoyant linden]

i heavily suspect it's reverse quotient rule but i don't see the vision

[obsidian bay]

nah that would be really confusing to get du/dx trust me

u want a substitution to get you easier to work with numbers

maybe substitution isnt the right method here

[buoyant linden]

i don't think it's gna be a basic sub

ye

[obsidian bay]

I would expand the denominator

see what u get

[buoyant linden]

oh?

[obsidian bay]

I think this is the correct method

[buoyant linden]

i never considered expanding

because i thought since denominator squared,

reminds me of quotient rule

[lethal path]

yep!

[obsidian bay]

expand trust me

[buoyant linden]

oh wait

it is expand

i see a vision with expand

[lethal path]

it's indeed in the form f(x) / (1 + x sin x)^2 + c

[buoyant linden]

but im sure there's an rqr

[obsidian bay]

huh

[lethal path]

what you can do is to add $\frac{(1 + x \sin x)^2}{(1 + x \sin x)^2}$ then subtract back $1$

[buoyant linden]

i actually see it

i think it splits into 2 parts, where one part is expanded and one part needs the factored form

[buoyant linden]

how did u see this vision btw

i saw it after babba mentioned expansion

[obsidian bay]

u need identities

i think

[buoyant linden]

ok lemme try it rq

brb

[lethal path]

WolframAlpha

[obsidian bay]

hacks

[buoyant linden]

😭

[obsidian bay]

not a certified OG using only brain power

[grand pond]

south

[buoyant linden]

uhh

got up to this

[obsidian bay]

alien language

[buoyant linden]

say ong my handwriting is ineligible 😭

[obsidian bay]

ON GOD

[buoyant linden]

AW HELL NAW

😭

[obsidian bay]

i thought mine was bad

nvm

[buoyant linden]

yea a lil lost rn

dunno if im on the right track

[obsidian bay]

i cant read it

[buoyant linden]

SAY ONG

SURELY THE 2ND LAST LINE IS READABLE 😭

(ping me if u respond)

[obsidian bay]

ues

[lethal path]

might seem like magic but rewrite the numerator as $x^2 \sin^2 x + x^2 \cos^2 x + x \sin x - \cos x$

[grand pond]

south

[lethal path]

also note that $\frac{d}{dx} (x \sin x +1) = \sin x + x \cos x$

[grand pond]

south

[lethal path]

(rewriting it so babba can read)

[buoyant linden]

oh my lawd

[obsidian bay]

huh

now im lost

[lethal path]

IDK how to explain it

[buoyant linden]

i dont see the vision

like

what happens after

[obsidian bay]

give final answer and log off

[lethal path]

,w integrate \frac{x^2\cos^2x-x\sin x - \cos x - 1}{(1+x\sin x)^2}

[grand pond]

Results provided by WolframAlpha

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Requested by sooouth.

[obsidian bay]

thanks boss

[buoyant linden]

do i rqr this?

[obsidian bay]

im trying a different appriach

[obsidian bay]

are u sure the 2nd last line is correct

[lethal path]

it's correct I checked

[obsidian bay]

k

idk man this is pretty hard

[buoyant linden]

yuh

[buoyant linden]

ts wasn't even the hardest difficulty

it was on medium

normally it isn't this fried

[obsidian bay]

I might have someting

[buoyant linden]

like yesterday it was
$\int_1^\infty \frac{1}{x(x^{10}+1)^2} , dx$

which was chillin

[grand pond]

Flatus

[obsidian bay]

yeh easy

is this like an question generator

where r u getting this

[buoyant linden]

https://dailyintegral.com/

[buoyant linden]

handmade questions

[obsidian bay]

ok

cool

copyrighted?

[buoyant linden]

wdym

[obsidian bay]

nvm dont worry

[buoyant linden]

oh wait ur wace?

it'll prolly be interesting for u to know that the daily integral site was made by HSC students

damn and im pre sure south is from vce

[obsidian bay]

good stuff

[buoyant linden]

@lethal path

[obsidian bay]

I might have a solution bro

[buoyant linden]

yes pls 🙏

[obsidian bay]

i gotta try and make sure it works first hang on

[buoyant linden]

bet

oh my days

wait

[lethal path]

damn I thought this was some Indian curriculum shit

no way

[obsidian bay]

what is that test called

[buoyant linden]

nvm i didnt find it

[obsidian bay]

you know the one I mean

[buoyant linden]

nah

[buoyant linden]

JEE

[obsidian bay]

JEE lol

[buoyant linden]

putnam type shi

[obsidian bay]

cooked stuff bro

[buoyant linden]

interesting that everyone in this thread is from aus but diff parts

now we need a qld

[obsidian bay]

QCE?

[buoyant linden]

right

QCE

[obsidian bay]

I was trying to do the log law stuff

like natural logs

not right I dont think

not log laws but integrating then turning into logs

[modest heart]

no, south is from a made up state

[obsidian bay]

tasmania

NT

not a state but still made up

im not gonna continue on this question

[buoyant linden]

😭 🙏

[obsidian bay]

are u yr12

[buoyant linden]

would u like another integral

[buoyant linden]

ye

like freshly turned y12 this term

[obsidian bay]

like graduated or class of 26

oh

what maths u taking?

I swear HSC uses some random names like extesnion or something

[buoyant linden]

ye

[buoyant linden]

4u maths

[obsidian bay]

wtf is this

[buoyant linden]

yeah uh

everywhere else says

applications methods specialist

[obsidian bay]

the f is 4u?

[buoyant linden]

but we say
standard, standard 2, advanced (your methods i think), extension 1, extension 2

[buoyant linden]

it's slang for extension 2 maths

[obsidian bay]

are you doing the top two?

[buoyant linden]

becuase if u take it, in total you're doing 4 units of math

[buoyant linden]

yes

[obsidian bay]

Basically methods and specialist are the "hard" maths and applications is "easy"

[buoyant linden]

specialist is my ext1 + ext2 iirc

[obsidian bay]

thats all subjects

[buoyant linden]

but easier

[obsidian bay]

nice bro

hang on DM me

[buoyant linden]

ok so if u take advanced, it's 2 units
extension 1 is 1 unit
extension 2 is 1 unit

[obsidian bay]

I honestly wont be able to solve this integral any time soon i cant lie

[buoyant linden]

if u take extension 2, u need to complete the advanced course and extension 1 course with it

[buoyant linden]

it's alr

ill let it marinate here

$\int_0^\frac{\pi}{4} \frac{(1+x)\tan x}{(1+\tan x)^2} , dx$

[grand pond]

Flatus

[buoyant linden]

@obsidian bay try this one

[midnight plank]

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[fathom onyx]

https://tenor.com/view/a-carai-smile-blink-gif-14907522

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[royal dirge]

initially i just try to make a giant system and use CAS to solve it but

i cant really find any relations here

i think it probably has something to do with angles, but im not sure what

[latent wadi]

I think you were supposed to use menelaus' theorem

[royal dirge]

i need to add this theorem to my calc

let me look it up

[hallow basin]

Is the problem supposed to be difficult... it would be way too easy using menelaus'...

[royal dirge]

these tests are super easy

its more memorization/application than it is true math

which is kinda irritating, because i love math

[latent wadi]

If you're looking for harder problems, I recommend trying Olympiad math

[hallow basin]

I don't

nah i'm just kidding

[royal dirge]

i think im going to take AMC 12 next year

[hallow basin]

But might take the fun out of math if you going for competition

consider that

[royal dirge]

this is a competition question here

i won first place at state last year in the algebra 1 tests

now i am taking geometry

this year

[latent wadi]

Good luck then!

[hallow basin]

I didn't know AMC has these kind of geometry problems

[royal dirge]

no no this is not AMC

this is just IL state competition

im going to try AMC seperately from this

[hallow basin]

I mean uhh you won't encounter menelaus' theorem or similar thing that often even in AMC

[royal dirge]

wait so how do i do this one lol 😭

i found e:f = 9:4

i think im overthinking this

[hallow basin]

bruh why do you even need those for

[royal dirge]

what triangle should i be looking at here

[hallow basin]

Apply Menelaus' for triangle BCE with C,J and F collinear

then that same triangle with C,H and G collinear

[royal dirge]

oh wait i misunderstood menelaus

i was thinking of

ceva's

[hallow basin]

well ceva's is kinda a twin of menelaus' lol

[royal dirge]

i still dont see it

looking at this example

oh wait its it

nah

i dont see it

i think one of the ratios is BJ/BE

or no BE/BJ

but then what

[delicate helm]

maybe to BDE triangle instead?

[hallow basin]

line FJC passes inside triangle BDE

[hallow basin]

yeah lol

[royal dirge]

ok yeah i see it now

DF/FE * EJ/JB * BC/CD = 1

so 5/3 * EJ/JB * 2/3 = 1
EJ+JB=494

[delicate helm]

oh this is two eqs and 2 unknowns

[hallow basin]

CHG passes inside BDE next

Then you'll have both JE and BH

Subtract those 2 to get JH

Menelaus' is just too overpower man

[royal dirge]

EJ = 234, JB=260

[delicate helm]

why does menelauses theorem work?

[hallow basin]

A bunch of similar triangles

the proof for it is quite lenghty tho

or not i dunno it has been years since I learnt it

[royal dirge]

DG/DE * EH/HB * BC/CD = 1

9/2 * EH/HB * 2/3 = 1

EH/HB=1/3, EH+HB=494

EH=247/2, HB=741/2

finally HJ = 260-741/2

uh oh

negative number

what did i do wrong

BJ should be greater than HB

[hallow basin]

,w a/b=1/3 and a+b=494

[grand pond]

Results provided by WolframAlpha

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Requested by 5tailfox

[hallow basin]

wait

HB looks shorter than EJ

So it might be the other way around

[royal dirge]

oh wait thats true

[hallow basin]

so 247/2 is HB

[royal dirge]

so final answer is 273/2 then

let me check it

nope

[hallow basin]

lmao

hmm, check your calculation ig

[delicate helm]

Yo this is cool https://en.wikipedia.org/wiki/Menelaus's_theorem#A_proof_using_homotheties

I can finally remember the order as well

[royal dirge]

ive just double checked, all my calcs seem right

i used a calculator to find them as well, so they should be correct

[hallow basin]

sorry dude my head's hurt so badly after I did ~50 SAT reading questions today, I can barely do any calculation in my mind rn

[royal dirge]

haha no worries, if you are unable to help im sure someone else can

in the meantime i will just write a summary of all my calculations here

$\frac{DF}{FE} \cdot \frac{EJ}{JB} \cdot \frac{BC}{CD} = 1$

[grand pond]

UCYT5040

[royal dirge]

Since $\frac{DF}{FE} = \frac{2}{3}$, then $\frac{DF}{FE}=\frac{2+3}{3}=\frac{5}{3}$

[grand pond]

UCYT5040

[royal dirge]

It is given that $\frac{BC}{CD}=\frac{2}{3}$

[grand pond]

UCYT5040

[royal dirge]

Then $\frac{EJ}{JB}=\frac{1}{\frac{5}{3} \cdot \frac{2}{3}}=\frac{9}{10}$

[grand pond]

UCYT5040

[royal dirge]

,w x/y=9/10 and x+y=494

[grand pond]

Results provided by WolframAlpha

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

Requested by ucyt5040

[royal dirge]

$EJ=234, JB=260$

[grand pond]

UCYT5040

[royal dirge]

$\frac{DG}{DE} \cdot \frac{EH}{HB} \cdot \frac{BC}{CD} = 1$

[grand pond]

UCYT5040

[delicate helm]

I got 32

[royal dirge]

Since $DE:EF:FG = 2:3:4$ then $\frac{DG}{DE} = \frac{2+3+4}{2}$

[grand pond]

UCYT5040

[royal dirge]

Do you spot my mistake then?

$\frac{DG}{DE}=\frac{9}{2}$

[grand pond]

UCYT5040

[royal dirge]

It is given that $\frac{BC}{CD}=\frac{2}{3}$

[grand pond]

UCYT5040

[royal dirge]

$\frac{EH}{HB}=\frac{1}{3} \ EH+HB=494$

[grand pond]

UCYT5040

[royal dirge]

,w a/b=1/3 and a+b=494

[grand pond]

Results provided by WolframAlpha

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

Requested by ucyt5040

[royal dirge]

$EH=\frac{247}{2}, HB=\frac{741}{2}$

[grand pond]

UCYT5040

[royal dirge]

There must be some mistake so far, because it is imposible for HB > JB

so I did something wrong here

either in finding HB or in finding JB

[delicate helm]

This is wrong I think?

I have GD/GE

[royal dirge]

ohhh

$\frac{DG}{EG} \cdot \frac{EH}{HB} \cdot \frac{BC}{CD} = 1$

yes

[grand pond]

UCYT5040

[royal dirge]

you're right

[delicate helm]

yep, now it's correct

[royal dirge]

finally

i hate menelaus 😭

but i will still take note of this

in case it ever curses me again

[delicate helm]

But seriously look at https://en.wikipedia.org/wiki/Menelaus's_theorem#A_proof_using_homotheties

[royal dirge]

will do

i dont understand that

[delicate helm]

And follow link for homothety if you don't understand

[royal dirge]

oh its like dialation?

[delicate helm]

It's scaling by a factor, about a point

basically imagine that point is origin, draw vectors to everything, and multiply vectors by some scalar

positive or negative

But importantly the three transformations have 3 centers E, D, F on the line.

And they move the triangle vertices in a circle

So in this diagram it would be A -> C -> B -> A

[royal dirge]

i dont really understand what you mean 😭

but i watched this video

I like this explanation: https://www.youtube.com/watch?v=IHn2OAO1i6Y

it uses similar triangles

[delicate helm]

I find that less helpful

but ok

[royal dirge]

lol

[delicate helm]

to each their own

[royal dirge]

im def going to look more into this later but

my brain is not working right now

anyway

thank you so much for your help

.close

[midnight plank]

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[midnight plank]

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[edgy crater]

hi i was wondering whether my solution would be valid? the part im a bit dubious about is that when you split x into more than 2 terms, for example 3 terms, would it still hold that we get the highest product if we split x into 3 equal numbers, i wasn't too sure how to show this, but i was able to show this fairly easily in the case of splitting into 2 terms. thanks🙂

[delicate helm]

calculus yes or no?

[edgy crater]

wdym

[delicate helm]

with calculus we can show that partial derivatives at the solution are all zero...

[edgy crater]

idk what partial derivates r sorry

[delicate helm]

nvm that then

but yes it does hold, I'm just trying to figure out how to prove tha

[mellow sand]

i think i agree, i came to the same conclusion using am-gm inequality

[edgy crater]

okok ty🙂

[delicate helm]

You know what you can do I reckon?

you can prove it by saying consider picking out 1 term, and then the rest separately

but then consider changing the 1 term

clearly we would need to split that (additive) excess between the rest somehow

but the "best" way to split it is the one that maximizes their produc

And so it transitions into n-1 case, and you can prove by induction

[mellow sand]

did u ever find a specific way to say what the best y to use?

[edgy crater]

no i dont think theres a markscheme / answer for this

[mellow sand]

i was playing around with it, and it seemed to me that the best y is the number closest to x/e

i was thinking about the limit definition of e and how this looked close to it

this isn’t rigorous or anything but interesting to mess around with

[edgy crater]

does that mean my last part is wrong

can i see how u did it with am-gm

[mellow sand]

yes

you know what am gm is right

[edgy crater]

yh

acc i will try it out

[delicate helm]

maximize ln(n^x x^(-x)) = x ln(n) - x ln(x)?

[mellow sand]

ok, it isn’t very difficult just imagine some sum of terms and play from there

[delicate helm]

we can actually do that, right?

Are you absolutely sure you don't know any calculus? y = n/e

specifically max(floor(n/e), ceil(n/e))

[mellow sand]

i agree

more rigorous solution

[graceful drum]

not necessarily

we have to find the closest integer to n/e

[delicate helm]

if (n/floor(n/e))^floor(n/e) > (n/ceil(n/e))^ceil(n/e) then y = floor(n/e) else y = ceil(n/e)

is the real (corrected) answer

for the number of terms

[graceful drum]

you wrote the same two things on either side of the inequality

[delicate helm]

thx

[edgy crater]

i just differentiated (x/y)^y with respect to y and got that y = x/e

[graceful drum]

yeah thats right for the critical point

[mellow sand]

did u figure out the am gm version or nah?

[delicate helm]

I actually would like to know how to use am/gm to find the number of terms?

[edgy crater]

well i know that the am = gm if all the terms are the same, so all the terms are meant to the same since if they weren't the gm would be less than am? and the am is the constant?

[mellow sand]

|| let x be your final sum number
x = a + b + c + …
by am gm
a + b + c + … / n >= (abc…)^1/n)
(a + b + c + … / n)^n >= abc….
(x/n)^n >= abc…
so max is when the two are equal. Thus max is (x/n)^n then solve in integers ||

[graceful drum]

basically am gm is for the first step when you prove that any maximal condition must have all terms equal

[mellow sand]

yea pretty much

[edgy crater]

ahh i see

[mellow sand]

just an easier way to get to that form

[edgy crater]

tysm everyone🙂

.close

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[feral heron]

hi guys, since i'm on vacations, i would like to ask you a simple question and that is: what are the most useful math formulas that i should learn, formulas that could help me throughout the last years of highschool or even my first years of college

[dreamy lichen]

simple question, difficult answer

[simple olive]

if by learn you mean how to derive them then volume formulas, quadratic formula and uh maybe some other stuff idk

[dreamy lichen]

this heavily depends on your country, college, what degree you wanna pursue etc...

if u have some extra time and wanna spend it learning math, go on khanacademy, look at every topic, take the "unit test" and see what you're weak at, then focus on that

[simple olive]

ye thats a good idea

[upbeat monolith]

Take this with a grain of salt, I'm redoing calc this year because my school didn't have a higher level math, so that's the furthest I've gone, but I'd look into
trig identities,
calculus rules (both differentiation and integration),
quadratic formula,
basic volume and area formulas,
polynomial interpolation,
the basics of Taylor/McClaurin series,
e^(i*theta),
and maybe vector/matrix multiplication and determinants depending on how far you want to go.

[feral heron]

this year im moving to a school with a high focus on things such as physics and math and since i've always felt kinda like im losing my time on vacations and that i should use it to learn, i wanted to see if i could go any further. im currently in argentina and to give you an idea, the last thing they taught us is the bhaskara formula and quadratic ecuations in general

[dreamy lichen]

https://www.khanacademy.org

Yeah, this would definitely be a good resource

you can switch it to spanish if im not mistaken

yeah, u can

[upbeat monolith]

I'm not that familiar with the Argentinian (sorry if that's wrong) curriculum, so this is based more on the American curriculum.
Calculus rules, Taylor series and vector/matrix stuff might be a little far, but i'd look into the rest of what I said

[delicate helm]

polynomial interpolation is actually crazy tbh (as in, weird that's a suggestion)

[feral heron]

i dont have any problem with language since i found learning languages kinda easy (i'm currently preparing for a c1 fce exam and i'm learning russian too)

[feral heron]

imma take it

[upbeat monolith]

It's not usually taught, but it's a really nice skill to have

It's surprisingly useful

[delicate helm]

But you can always just derive it if you really need to and can't look it up

[upbeat monolith]

*at least, it's not usually taught in the US

[feral heron]

anything that could make my brain melt from how challenging it is is sort of fun to me

[upbeat monolith]

I'd watch 3blue1brown's essence of calculus series on youtube, and then look into the rest of the things I talked about in that case

[delicate helm]

I'd say
e^(ix)
(a+b)^n (binomial, or multinomial formulae)
ad - bc (determinant of a 2x2 matrix)
quadratic formula
common trig identities
common taylor series

oh and the pythagorean theorem

[feral heron]

thank you all guys

ill definitely spend time on learning from what you guys suggested

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[graceful ferry]

For the least squares problem, in the normal equation, is the residue vector multiplied by A^T always zero? even if the columns of A are linearly dependent?

[graceful ferry]

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[rich yew]

[plucky lark]

???

.close

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[frank wolf]

Perhaps you can use the classification of finitely generated abelian groups

[twilit field]

I am

I'm just unsure of how to go ablout it

3,2^3,5 are the factors I care about

[frank wolf]

I think the prime factorisation of 120 is a good place to start

[twilit field]

.

[twilit field]

yup

so $Z_{3} \oplus Z_{8} \oplus Z_5$ is one possiblity

[grand pond]

wai

[frank wolf]

I think you’re forced to have $Z_3 \oplus Z_5 \oplus H$ where $H$ is an abelian group of order 8

[grand pond]

Pseudo (Cat theory #1 Fan)

[twilit field]

hmm, why

oh

right

yes

[frank wolf]

And there are only 3 options for H

[twilit field]

ye

[frank wolf]

So you can probably just go through each case and see what works

[twilit field]

But not all have 3 elements of order 2.,do they

[frank wolf]

Yeah

[twilit field]

so I have to examine all cases?

[frank wolf]

Yeah

[twilit field]

yup

It's $Z_3 \oplus Z_5 \oplus Z_4 \oplus Z_2$

[grand pond]

wai

[runic hamlet]

we have done this question before

why are you always forgetting how to do things

grinding problems is useless if you keep forgetting

do less problems and make sure you actually remember them

[frank wolf]

yep!

[twilit field]

Tq

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[narrow edge]

so i was trying to derive the volume of a cone by myself what is the error in my reasoning:
drop a perpendicular to the center of the base circle. we can see that 2pir triangles have been formed with base length r and height h which means that they have an area of rh/2. 2pir * rh/2 is pi*r^2 * h which is the volume of a cylinder not that of a cone

[bold peak]

What do you mean by "2pir triangles are formed"

[narrow edge]

there are 2pir points on that circle which means that there are 2pir hypotenuses connected to the top of the cone

[bold peak]

Again, what do you mean by there are "2pir points on that circle"

[narrow edge]

2pir is the circumference of a circle

[bold peak]

A circle has an uncountably infinite number of points

[bold peak]

Circumference ≠ Number of points

[narrow edge]

wait what

what is the circumference supposed to be

[exotic stratus]

length of the boundary of the circle

[bold peak]

The length of the boundary

[narrow edge]

but like

[exotic stratus]

I mean just think of it this, way let's say you have a circle with radius 1u, it's circumference will be 6.28u, what do you think 0.28u points are?

[bold peak]

They're concepts of a point

[narrow edge]

while deriving say the volume of a cube we note that there are x squares with an area of x^2 and get the result

[bold peak]

What when

What derivation have you been shown

[narrow edge]

no i derived it myslef i trusted it cuz it made sense 😭

[bold peak]

...

[narrow edge]

they dont teach derivations in hs they just make you rotelearn the formulas

[bold peak]

Your idea works not because it's correct, but because it's a restatement of a correct idea

If a shape has uniform cross section, then it's volume is the area of the cross section * height perpendicular to that cross section

Unfortunately this idea stops working the moment your cross section is not uniform

Like a cone. Or a sphere

And unfortunately proving the volumes for these shapes requires integration. Which I imagine is beyond your syllabus for now

[narrow edge]

but the cross sections can be uniform cuz all of these right triangles are congruent right?

[bold peak]

In which direction are you taking the height

[narrow edge]

perpendicular to teh base

[bold peak]

For the cross sections to stay uniform?

[bold peak]

What's the cross section there

It's circles

Circles that reduce in radius the higher you go

[exotic stratus]

If you want to keep going with your idea, I recommend you use circles which get progressively smaller rather than drawing lines from points on circle to the top because infinity isn't a good way to work on things. Take really thin circles and progressively make them thinner and eventually you'll see the pattern

[bold peak]

(at that point you've basically reinvented integration no?)

[exotic stratus]

Yea but that's kinda the idea

I don't think OP knows integration

[narrow edge]

yea

[bold peak]

I'm just wondering if that's a feasible path to take

[exotic stratus]

If they did they would've figured out the error they made

[narrow edge]

ik differentiation tho if thats useful for this

[exotic stratus]

Integration is anti-differentiation

[bold peak]

Well integration is the opposite of differentiation

So you might as well look up some integration and come back to this

You shouldn't need more than the very basics

Godspeed

I'd say do three things

See basic integration

Try to derive the volumes of a cube and a cylinder through integration

Then try to derive the volume of a cube cone

[exotic stratus]

I think you meant cone

but yea

[bold peak]

Right yes

[midnight plank]

@narrow edge Has your question been resolved?

Channel closed

Closed by @narrow edge

Use .reopen if this was a mistake.

[midnight plank]

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[wispy mortar]

yo

[wispy mortar]

how do i show increasing

i only have doubt in showing monotone sequence

i know that we need to get xn+1 > xn

but idk how to start

[rose trout]

If you show it's bounded above by 2 first I think it's a bit easier to show it's increasing.

[graceful drum]

try using induction

[wispy mortar]

thats what im not getting

how to start that

[graceful drum]

ok lets start with the base case

how do we show x2 > x1

[wispy mortar]

we can do that by putting in value of x1

[graceful drum]

right, so what are the values of x2 and x1

[wispy mortar]

root3 and 1

[graceful drum]

then

[wispy mortar]

root3 > 1

[graceful drum]

yeah

[wispy mortar]

we can take a xk > 1?

[graceful drum]

yes

but it wont help much

[wispy mortar]

ohhh

[graceful drum]

now for our inductive ste

p

[wispy mortar]

yess

[graceful drum]

we have xk > x(k-1)

how do we prove x(k+1) > xk

[wispy mortar]

how do we have this?

are we assuming

[graceful drum]

yes

if we assume xk>x(k-1)

and prove this implies x(k+1)>xk

that combined with our base caves proves that xn+1>xn for all n

*case

[wispy mortar]

OH

[wispy mortar]

i didnt know what to assume to start this proof

thanks man

[graceful drum]

a big step is still left

proving the inductive step\

[wispy mortar]

yess

[wispy mortar]

which is?

[graceful drum]

this

[subtle blaze]

Essentially you want to show the sequence is bounded from above by 2

Then you want to show that the sequence is increasing

Then you use the theorem that a bounded monotonic sequence converges

[wispy mortar]

from the equation

[graceful drum]

yes

go on

[wispy mortar]

xn-1 is just xn^2 - 2

oh wair

[wispy mortar]

add 2 on both sides and then root

[graceful drum]

correct

$x_{k} > x_{k-1} \implies \sqrt{x_{k}+2} > \sqrt{x_{k-1}+2} \implies x_{k+1} > x_k$

[grand pond]

CherryMan

[graceful drum]

so you proved that x_n is monotone

[wispy mortar]

damn thats it?

[graceful drum]

can you prove that it is bounded similarly

[wispy mortar]

yesss

now we can claim that

since the limit is 2

xk < 2

[graceful drum]

your claiming all this right

go on, how would you prove it

[wispy mortar]

im assuming that there is xk < 2

base case x1 approves

1 < 2

[graceful drum]

yup

[vague seal]

In induction we have 2 steps we want to do:
step 1) base case, proving that for some specific n P(n) is true, in your case it is showing that P(2), which is x_2 > x_1 is true.
step 2) induction step, this is the step where you prove the implication P(n) ⇒ P(n + 1) where in your case P(n) is x_n > x_(n - 1).

[wispy mortar]

and then we +2 and root on both sides

[graceful drum]

nice

[wispy mortar]

it gives xk+1 < 2

[graceful drum]

$x_k < 2 \implies \sqrt{x_k+2} < \sqrt{2+2} \implies x_{k+1} < 2$

[grand pond]

CherryMan

[graceful drum]

nice

[wispy mortar]

yesss

[graceful drum]

so you proved that xn is bounded and monotone

so by mct, you proved that it converges

[wispy mortar]

lessso thats it?

[graceful drum]

well you still have to find the limit

[wispy mortar]

we can find that by putting both L

and solving the quad

[graceful drum]

yes

[wispy mortar]

L = root 2 + L

thank u so muchh

[graceful drum]

welcome

[wispy mortar]

.close

[midnight plank]

Channel closed

Closed by @wispy mortar

Use .reopen if this was a mistake.

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sullen bolt]

just wanted to check my thought process

this just involves fraction decomposition, then using the identity \int(1/(x+blah blah blah)=ln(x+blah blah blah) right

[sullen bolt]

please dont smite me for forgetting the dx in thought process

[merry shale]

...

yeh gl bro that is cooked... 😭

[sullen bolt]

its not that bad I think