help-33

[spare idol]

ohh my bad boss

[mossy parrot]

get the quadratic

[wicked hawk]

its fine

[mossy parrot]

make it in vertex form

[wicked hawk]

what would that look like

[mossy parrot]

uh so area is l*w

and u solved for L with the previous function

[wicked hawk]

ya

[mossy parrot]

2L=-5W+120

so plug in for L

[wicked hawk]

ohh

[mossy parrot]

and its in factored form

so the average of the roots

should work

[wicked hawk]

should i factor the 2 out first

so its L = -5/2W + 60

[mossy parrot]

divide by 2 on both sides

yes

then plug in

factor out -5/2

average of roots

is ur width

and plug into the L formula

to get length

[wicked hawk]

okok

[wicked hawk]

I factored out that and i have A= -5/2(w^2 + 24w) now what do i do

[mossy parrot]

dont multiply

wait u know what

its fine

also its -24w

not 24 w

[wicked hawk]

ye typing mistake

sorry

[mossy parrot]

uh so now complete the square

[wicked hawk]

(B/2)^2?

[mossy parrot]

no

(a+b)^2=a^2+2ab+b^2

so u have the a^2 and the 2ab

now add the b^2 inside

[wicked hawk]

im not gna be able to solve it like that on the test cuz my teacher wont give me the mark

[mossy parrot]

and subtract -5/2b^2 outside

[wicked hawk]

she wants me to get the 3rd term inside the brackets

[mossy parrot]

ok

what third term?

[wicked hawk]

like how its aX^2 + bx + c

[mossy parrot]

c

[wicked hawk]

She wants us to get C

[mossy parrot]

the y intercept

but why

[wicked hawk]

Wat

[mossy parrot]

there is no y intercept

the functions c is 0

[wicked hawk]

let me copy paste wat my friend said

you have x^2 + bx + c = 0. solve for x. rewrite it as x^2 + bx = -c. now complete the square, you're gonna add (b/2)^2 to both sides. x^2 + bx + (b/2)^2 = -c + (b/2)^2. the left side is now a complete square, so it's the same as (x + (b/2))^2 = -c + (b/2)^2

[mossy parrot]

ye

imagine u did that all on one side

of the equation

and imagine ur c=0

ill give u a different way

so

[wicked hawk]

I cant use different way

my teacher wont give me mark on test

ik its dumb

[mossy parrot]

oof

what way did they teach u in class

[wicked hawk]

the way i copy pasted

[wicked hawk]

.

[mossy parrot]

ah

ok

[wicked hawk]

ya

[mossy parrot]

so make A zero for now

and complete the square

the way u were taught

[wicked hawk]

kk

[mossy parrot]

so can u ask someone else to help u mate

kinda getting late for me now

just when u find the complete square

the inside part should equal 0

so like it should be w-12 is 0

so w is 12

and L is whatever u solve

i gtg now though so good luck on ur exam

[marsh citrus]

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[gusty lily]

Uh why's my calculator not answering did I input it wrong?

[brave marsh]

You need parentheses

You wrote it down under with the fraction

Your calculator doesn't know the numerator is the whole thing, it just sees -6^2

[gusty lily]

Uh ill see if I got it

[tough igloo]

and u also need another ) at the end bc ur inputting cos(number, and u didnt close it

[gusty lily]

So like this

[tough igloo]

mhmm

[gusty lily]

👌

.close

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[quasi glacier]

Hello I have a question, how do we do arithmetic operations on nonexisting limits? for example liml(x+sinx) as x goes to infinity? because lim sin(x) doesnt exist, but lim(x) does and it is infinity

[devout mauve]

you generally hope to compare the limit to something easier to deal with

eg here x+sinx>=x-1 and x-1 diverges, so x+sinx also has to

[quasi glacier]

yeah i get it from an intuition point of view but if i write that in a homework or exam ill get a very well rounded 0 lmao

what i meant is how do we do it formally

[devout mauve]

that is a legit rigorous argumentation

its the comparison test

otherwise you can just check the epsilon def of divergence

[quasi glacier]

oh you meant bigger or equal

my apologies im blind

yes

thanks man

[marsh citrus]

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[abstract onyx]

ok so i have a silly problem, i can solve this kind of questions with three numbers just for some reason i can't figure out how to do it with 2. sorry for the shaky writing. (is this how you ask for help here?

[amber birch]

this is Cramer's rule right?

[abstract onyx]

it isn't difficult i just got to figure out how to multiplay it then i can find x,y,z,w

[abstract onyx]

pretty sure yeah

[amber birch]

yeah then you just forgot how to calculate 2 by 2 determinants right

the determinant will be just ad - bc

[abstract onyx]

so like

-1 - 0

5-12

[amber birch]

no, the determinant is a number

[abstract onyx]

right so for this spesificly

wouldn't that be right?

ok yeah it is the right answer thank you a lot :)

[amber birch]

no worries!

I'll just close this channel then

.close

[marsh citrus]

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[keen patio]

Why did we use sin in Ax 😭

This is how i imagine itd look like

[amber birch]

you want to be able to use the distance formula

that requires knowing Ax, Ay, Bx, By right?

[keen patio]

what the hell is that

[amber birch]

but then to find those you just apply right-angled trig

[amber birch]

the same thing as Pythagoras

[keen patio]

yeah but we dont have that in physics

The distance thingy

the question just wants the C vector

[amber birch]

YOU DO!!!!

[keen patio]

i dont see it in the solution tho

[amber birch]

otherwise use the cosine rule, that's another option I guess

[amber birch]

they've done this calculation

but they didn't mention the distance formula explicitly

I'm telling you that's what they used

[keen patio]

Okay yeah... But why is Ax the mag multiplied by sin

[amber birch]

you can calculate it yourself from the solution

[amber birch]

right-angled trig

[amber birch]

your 60 degrees is labelled wrong

that angle should be 90 - 60 = 30 degrees

[keen patio]

okay this might sound stupid asf... But what is right angled trig and when do i use it

[amber birch]

so $A_x = -26 \cos 30, A_y = 26 \sin 30$

[elfin berry]

south, just south

[amber birch]

https://www.youtube.com/watch?v=PUB0TaZ7bhA

[keen patio]

awh hell nah 😭

[amber birch]

are you self-studying?

[keen patio]

this is my first year of uni but in highschool they didnt teach us well

[amber birch]

yeah then unfortunately you're playing catch-up

[keen patio]

Do i just have to watch that video

[amber birch]

cause these are like STEM fundamentals, trigonometry, algebra-precalculus, and I guess stats as well
this is probably like physics I or something at your uni I guess

[amber birch]

and you need to do a ton of practice

to succeed

[keen patio]

im actually gonna cry rn

.close

[marsh citrus]

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[marsh citrus]

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[hazy verge]

Ik its in french but i just need to prove this inequality 😦 could anyone help

[limber hearth]

c'est xi * xj en bas ?

[hazy verge]

oui

[unborn lake]

▶•၊၊||၊|။||||။၊|။

▶•၊၊|||||။၊|။

[hazy verge]

[limber hearth]

essaie de développer et de voir ce que cela peut faire

[unborn lake]

732-76=
732-70-6=
662-6=656

[amber birch]

stop spamming

[limber hearth]

not here then

<@&268886789983436800>

[proud arch]

<@&268886789983436800>

[limber hearth]

they here

[hazy verge]

wow

that was badass

[proud arch]

Literally called the purge lol

[remote mountain]

Im so bad at math

[hazy verge]

ong

[limber hearth]

ducoup tu as essayé de developper ?

[hazy verge]

oui, rien

[proud arch]

Could we just take arctan on both sides

And proceed from there?

[limber hearth]

arctan(quotient) < pi/6

[proud arch]

And it becomes

0 < arctan(xj) - arctan(xi) < pi/6

[hazy verge]

u mean pi/3?

[proud arch]

No

[limber hearth]

arctan(sqrt(3)/3) = pi/6

[proud arch]

That would be for arctan( (3)^(1/2))

[hazy verge]

arctan squar three is pi/3

[proud arch]

.-.

[hazy verge]

yes but thats whats in the problem no?

[proud arch]

It's given 1/ (3)^1/2

[proud arch]

@hazy verge see it now?

[hazy verge]

help im blind

sorry

[proud arch]

Issok

Happens

[hazy verge]

i think we will use rolle for this

[proud arch]

The one with differentials?

I am unsure of its application here

[hazy verge]

or mean value theorem

[hazy verge]

ngl me too but i found this in the chapter

<@&286206848099549185>

[timber mantle]

sorry I don't speak french, is the goal to find just a pair x_i, x_j with that inequality holding? or that for any i, j

[proud arch]

Obviously xj > xi

[hazy verge]

to prove that it exists yes

[proud arch]

In order for inequality of exist

Got any ideas

??

[hazy verge]

im so lost ngl

[timber mantle]

sorry trying to come up with something what math course is this? just to know what concepts are relevant

[proud arch]

Could just use induction

[timber mantle]

but its finite n no?

there isn't a general case to show

[proud arch]

We could just set different range of values for x1 and x7 to prove its existence

Like x7 could be infinity

And x1 be sqrt(3)

This a sequence that would be true for this inequality

[hazy verge]

im in hs and this is the mean value chapter tho idk how i can use it here🥲

[timber mantle]

for calc

hmm this actually does look like some MVT could work here, let me ponder

hmm j doesn't even need to be bigger than i I don't think, since if x_i *x_j is sufficiently negative the sign flips right?

[hazy verge]

xi*xj could be inferior to one tho

[timber mantle]

right but we can't assume there isn't a pair with i > j such that the lower bound of 0 works

[hazy verge]

hmm true

either way ill ask my prof tmrw if anybody is still interested leave a dm

.close

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[dapper gazelle]

im a 14 yr old home sick really struggling with this question on a test thats due today. not asking for exact answers, just a simple explanation on how to solve it
question is 'write an equation for each graphed line' which im unsure how to do

[faint narwhal]

what are they asking for? equation of the line?

[dapper gazelle]

yes

i dont know where to start

[faint narwhal]

so linear funtion could be writtern in mx+c

where m is the slope

do you get a hint now?

[dapper gazelle]

yes man. thank you

absolute lifesaver

all i needed was a refresh on those specific letters

.close

[marsh citrus]

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[potent temple]

I tried proving that 2^k . j choose 2^k+1 will be odd but failed, and i tried using binomial theorem but failed, and induction but failed

Its obviously true for powers of 2

But if they are not idk how to prove that atleast one will be odd

[main idol]

did you try contradiction

"Suppose n isn't a power of 2..."

[potent temple]

Yeah with 2^k . j choose 2^(k+1)

But i did not have any idea other than that

[marsh citrus]

@potent temple Has your question been resolved?

[potent temple]

<@&286206848099549185>

[cedar stone]

(a C b)

= a!/b!(a-b)!

try (nCr) where r is the largest power of 2 smaller than n

for n not a power of 2

this should work

[potent temple]

This no?

[potent temple]

But i did not manage to prove it

[cedar stone]

do you know Vp ?

[potent temple]

Greatest power of p that divides smth?

[cedar stone]

yes

[potent temple]

I mean i dont know its properties but the lte lemma

[cedar stone]

actually there is a method without vp n

yk that nCr = n-1Cr-1 + n-1Cr

try using this

[potent temple]

No i dont

I mean its hard to see writting like this

[cedar stone]

oh

[potent temple]

Oh yeah i do

[cedar stone]

good

[potent temple]

Ok i will try

Thx

[cedar stone]

if you still dont get it, this image might give you an idea.

[potent temple]

Okie

[potent temple]

No i dont get it

[marsh citrus]

@potent temple Has your question been resolved?

[vague zinc]

For n=2^k you can easily prove by induction: If P(x)=(1+x)^n=1+x^n+2T(x) (where T(x) is some polynomial with integer coeffs) then P(x)^2=(1+x^n+ 2T(x))^2=1+x^(2n)+2S(x) where S(x) also has int. coeffs.
And if n=2^k * d, where d>1 is odd, then you have (1+x^(2^k)+2T(x))^d=1+d*x^(2^k)+2S(x), so we see that coefficient at x^(2^k) is always odd.

[marsh citrus]

@potent temple Has your question been resolved?

[potent temple]

I will read it thx

[marsh citrus]

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[mystic gust]

Let $f(x) = \frac{-xlnx}{1+x^2}$ if $x \ne 0$ and $f(0) = 0 \ \$ 1) prove that f is continuous in $]0, +\infty[ \$ 2)a) prove that $\forall x \in ]0, +\infty[ , f(\frac{1}{x}) = -f(x) \$ 2)b) prove that $\exists \alpha \in ]0,1[ , f'(\alpha) = f(\frac{1}{\alpha}) \ \$ 3)a prove that $\forall x \in ]0, +\infty[, -xln(x) \le \frac{1}{e} \$ 3)b) deduct that $\forall x \in ]0, +\infty[, f(x) \le \frac{1}{e} \$ 4) We admit that $\alpha$ and $\frac{1}{\alpha}$ are the only solutions to the equation $f'(x) = 0 \$ a) prove that $f(\alpha) > 0$ and $f(\frac{1}{\alpha}) < 0 \ \$ b) Prove that $f'$ is continuous in $]0,+\infty[\$ c) Prove that $f'$ doesn't change its sign in every one of the following intervals: $[0, \alpha]$, $]\alpha, \frac{1}{\alpha}]$ and $[\frac{1}{\alpha}, +\infty[$

[mystic gust]

i am not able to do anything past 3)b)

[elfin berry]

Froggy

[mystic gust]

how do we prove that $f(\alpha) > 0$ just from the fact that $f'(\alpha) = 0$

[elfin berry]

Froggy

[mystic gust]

$\text{and the same for } f(\frac{1}{\alpha}) < 0$

[elfin berry]

Froggy

[mystic gust]

.close

[marsh citrus]

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[timber mantle]

did you figure it out?

[marsh citrus]

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[silent marten]

if PACF gives us the direct correlation between an observation Yt , and Yt+k , why do we care about ACF ?

[marsh citrus]

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@silent marten Has your question been resolved?

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[shy shard]

Can someone check if I did problem 50 correct

[shy shard]

Find volume using integral

To find volume of solid

[marsh citrus]

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[marsh citrus]

@shy shard Has your question been resolved?

[outer lodge]

No, the region revolves about the y axis so you should use the washer method with dy integral

[marsh citrus]

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[still temple]

0

[marsh citrus]

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[undone arch]

What is pi x x.?

[undone arch]

helllo

What is x=32

[tender lily]

you gotta be a bit more specific

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[undone arch]

oh ok ty

[marsh citrus]

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[hollow spear]

How do I find the inverse laplace of this?

[marsh citrus]

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@hollow spear Has your question been resolved?

[marsh citrus]

@hollow spear Has your question been resolved?

[marsh citrus]

@hollow spear Has your question been resolved?

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@hollow spear Has your question been resolved?

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[vital juniper]

You have the opportunity to purchase a pair of x-ray goggles that will allow you to see through exactly 1 door before the battery runs out. How much are you willing to pay for these goggles?

[vital juniper]

How is this different from monty hall

here you can directly just say that the probability becomes 1/4

[unborn condor]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@vital juniper Has your question been resolved?

[vital juniper]

original problem:

You're on a gameshow and are tasked to choose the lucky door. Behind 1 door is $10,000. The remaining four have nothing. How much are you expected to win?

part 2:

You have the opportunity to purchase a pair of x-ray goggles that will allow you to see through exactly 1 door before the battery runs out. How much are you willing to pay for these goggles?

[unborn condor]

!show

[marsh citrus]

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

[unborn condor]

also, considering it has absolutely nothing to do with the Monty Hall problem, i'd say it's quite different

[vital juniper]

what makes monty hall diff

that's all i'm curoius about

[unborn condor]

On monty hall you have two choices, and on the second you already have knowledge.
On this one you only make one single choice

[vital juniper]

gotcha guess that makes sense

.close

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[finite wigeon]

why cant i find the area under the curve by integrating with respect to x axis? and the video wanted to integrate with respect to y axis?

[sweet pawn]

you can

but it'd be two integrals

because y is not a function of x

[finite wigeon]

what do u mean by this?

cant i just sqrt both side?

Y = SQRT x

[sweet pawn]

sqrt only gives you half of the parabola

there are two solutions to x = y^2

[trail hamlet]

you can also integrate the function of difference of the the values of y for a given x
so basically intergral of 2 * sqrt(x)

[finite wigeon]

so x is a function of y but y is not a function of x?

[limpid pond]

y=+-sqrt(x) 1 input value yields 2 outputs not a function

[sweet pawn]

yes

[limpid pond]

Yes

[finite wigeon]

ok than i have a qn if i just have a x=y^2 graph

can i integrate it to find area?

from maybe 3 to 0

wrt to x axis

[fresh nymph]

That's what I thought

Idk tho

[finite wigeon]

can i integrate like this to find area? @limpid pond @sweet pawn @trail hamlet ?

seperate qn

[sweet pawn]

no

that only gives you half of the parabola

you would need two integrals

[limpid pond]

If you want you would need to integrals
one for the sqrt(x) and the other for -sqrt(x) and it’ll yield zero

[finite wigeon]

ok so i add them together?

[trail hamlet]

wait why though. if you integrate the difference, aren't you ultimately calculating the are

[sweet pawn]

the original problem had a line. that would make the region non symmetric

[limpid pond]

This will yield the area inside the curve rather then area under the curve tho

[trail hamlet]

then can we subtract the y coord of line with the sqrt of x? isnt that the same thing as taking two integrals

[finite wigeon]

yeah but to find area under the curve just divide answer by 2?

[trail hamlet]

Oh nvm

The line ends before the origin

[finite wigeon]

im so confused rn

[finite wigeon]

this is a new seperate qn

jus this graph alone

[limpid pond]

The area under the curve of this function of y (integrating with respect to y) doesn’t equal the area inside the function (integrating with respect to x)

[finite wigeon]

is it possible to find area inside the function with respect to x?

from x = 0 to x = 3

[limpid pond]

If you bound it to the x axis then yea
but the question you showed above states the area between 2 functions

[limpid pond]

Referring to this

[finite wigeon]

i already said its a seperate qn T.T

anyways thanks for help

.close

[marsh citrus]

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[thorn sage]

this might sound like a dumb question

but can we expand sqrt(a+b)

[marsh citrus]

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[thorn sage]

stop snitching dawg

[night mica]

no

[thorn sage]

i said "hello"

lame ahh message

stop exposing me in front of the fine shyts

[thorn sage]

😡

what am i supposed to do now

[night mica]

why are you angry ?

[thorn sage]

because idk how to solve the cubic discriminant

like derive it

from the form ax^3 + bx^2 + cx + d = 0

[limpid pond]

Cubic discriminant damn

[night mica]

maybe you can find something useful searching the cardano formula

[thorn sage]

the regular one is easy i even derived it myself but this one hard

[thorn sage]

NO

i have to derive it myself

no cheating

[wheat rover]

first try with depressed cubic before doing with the original cubic equations

that helps

[thorn sage]

alr fine

ax^3 + bx + d = 0

this right?

[wheat rover]

yeah

[thorn sage]

okayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

x^3 + x(b/a) + d/a = 0

ok i will play around with this equation

thanks gng

[wheat rover]

yw

[marsh citrus]

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[tight ice]

In triangle ABC angles A = C45°. Side a is 20 m. Calculate height hb and sides b and c.

[main idol]

,rccw

[wooden shore]

What have you tried?

[tight ice]

Nothing

Cause I don’t know what to begin with

I can’t do Pitagoras

Bc I only know C

[fossil field]

hit: sum of angles in triangle is 180

[tight ice]

Oh I know!

And now? @fossil field

[wooden shore]

What's the angle of B then

[tight ice]

90°

[sleek topaz]

this might help

[wooden shore]

Use the Pythagorean theorem now

[fossil field]

Now when you have all angles you can calculate sides. You have square triangle.. so it’s pretty easy

[fossil field]

That’s like using cannon on fly

[tight ice]

400+b^2=400

It’s : gonna be 0

[tight kite]

b is the hypotenuse

[tight ice]

Ok I get it

[crystal lintel]

hi swerriee ily

[marsh citrus]

@tight ice Has your question been resolved?

[marsh citrus]

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[vocal cloud]

can someone give me some suggestion on where to start

[vocal cloud]

this is fourier transformation question though

[sturdy flower]

what have you tried?

[main idol]

What's eqn (*)

[marsh citrus]

@vocal cloud Has your question been resolved?

[vocal cloud]

the fourier transform equation

[vocal cloud]

nothing yet. i was hoping someone could give me some kind of hint so that i can start. i dont need the answer, just some kind of guide

[sturdy flower]

first thing you can do is expand the first integral using the definition of F(k)

[vocal cloud]

how do i expand it? use by parts?

[marsh citrus]

@vocal cloud Has your question been resolved?

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[wide osprey]

Hello guys, how are you doing? I need help with this complex numbers problem; it's honestly really simple and I can't believe I am yet to figure it out:

I need to find a polynomial p(x) ∈ R of minimum degree such that the solutions of the equation z^3=-8i are roots of p(x) and p(i)=-3

Here's where I got to, I now believe I should multiply that P(x) with their respective conjugates... but I have no clue how to multiply them all

[proud ice]

If the solutions to $z^3=8i$ are roots of $p$, then you can start with the polynomial $q(z)=z^3-8i$. Then just let $p(z)=\alpha q(z)$ and solve for $\alpha$ to satisfy $p(i)=-3$

[elfin berry]

SWR

[wide osprey]

So how would I solve q(z)=0 with complex numbers? Don't I need the trigonometrical ecuation for complex numbers?

[proud ice]

Why do you need to solve q(z)=0?

[wide osprey]

I may or may not have thought of something

I may not have a brain, but I've had an idea

[proud ice]

Hm?

[wide osprey]

Give me a min I'm trying stuff in my sheet here

Thanks for the help, I'll be back in 5

[wide osprey]

Well, I'm back... I didn't get much further

alpha HAS to be a real number according to the image right?

[proud ice]

Why did you do +8i?

[proud ice]

Oh i missed that the polynomial needs real coefficients

Okay that adds some wrinkles

[wide osprey]

mb

[wide osprey]

No wait, it is -8i

[proud ice]

Oh your problem showed $z^3=-8i$ i see

[elfin berry]

SWR

[wide osprey]

Yeah... Already asked 5 AIs about it and they all answer something different

[craggy veldt]

ok so

he wants p(i) = 3 and p(z) = 0

well for starters you can solve the equation by converting -8i into is exponential form

once you get that, establish a polynomial P(X) = Q(X).(X-z1).(X-z2).(X-z3)

he wants P(i) = -3

Q(X) would be degree zero in this case and you would solve for it

[wide osprey]

Brother this was the first topic we saw as review

I'm preparing the obligatory final

[wide osprey]

At the bottom of the pic there's P(x)

I would need to do their conjugate

HOLD ON

I do see a pattern here

[craggy veldt]

buddy he said minimal degree

you don't have to do conjugates

just the 3 roots would do

plus the constant that verifies P(i) = -3

[wide osprey]

But they're complex

If we're talking about a polynomial p(x) ∈ R

[craggy veldt]

oh you're doing R[X]?

ok

[wide osprey]

ye

Guys....

@proud ice I'm sorry for pinging you, but what do you think

[wide osprey]

Look

[marsh citrus]

@wide osprey Has your question been resolved?

[proud ice]

If you need the real coefficients, then do $P(z)=\alpha (z^3-8i)(z^3+8i)$

[elfin berry]

SWR

[proud ice]

@wide osprey

Multiplying by the conjugate polynomial will give you real coefficients

[marsh citrus]

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[marsh citrus]

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[uncut forum]

how would i approach this? (This is discrete math)

we have to manipulate this expression in a way that allows to take the limit to give us e.

i think we can represent everything but the 2 as a sum rule of some binomial coefficient, just not sure how.

i cant progress other than that i can represent that as c(n,k) times 1/n^k

[uncut forum]

i believe you can approach this algebraicly or using the sum rule

[sweet pawn]

rewrite the binomial coeffs using the factorial formula

[uncut forum]

so itd be

lim as n ap -> inf of 2 + n!/(k!(n-k)!) * 1/n^k

[sweet pawn]

,w n!/(k! (n-k)! n^k)

[elfin berry]

Results provided by WolframAlpha

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Requested by the.mr.gamer

[uncut forum]

yeah

we care about the input here correct? not the others

i mean we could put it ontop but i dont see how that helps

i write it out and like,if we get k at 0 = 1 , wouldnt that go above the num of the expoenential? since u have 2 and then k at 0 is 1, it becomes over??

[marsh citrus]

@uncut forum Has your question been resolved?

[uncut forum]

<@&286206848099549185>
Ok i figured out the first 2 terms give a 1, when k = 0 and 1

so i see why the +2 exists now,
not sure what to do after

[uncut forum]

ok <@&286206848099549185> i got an extra thing,

now we can represent that as n times n-1 times n-2 ... times n-k +1, since we can divide it by n-k, which leaves us with k! on the denominator with n^k, not sure what to do with it though.

[frozen cape]

I got an idea of how to do it

If you expand every combination( i.e) nC1=n!/(n-1).1!
nC2=n!/(n-2).2!

[uncut forum]

yup

[frozen cape]

I will take what i did photo

[uncut forum]

i will show you what i ended up with so far aswell

[sharp vessel]

Try to use the binomal formula

Expanding the choose term is not the way I think you're supposed to do it

[uncut forum]

like the sum rule of x^k-j*y^j ?

[sharp vessel]

Yeah

[uncut forum]

how would that work?

like howd you able to evaluate it

[sharp vessel]

$\parens{a+b}^n = \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$

[elfin berry]

RedstonePlayz09

[sharp vessel]

Notice how the right side is similar to what you have

[uncut forum]

yeah but here we have 1 term instead of 2 no?

the 1/n^k

with n choose k

[sharp vessel]

Well not really

They are just hidden

What did you choose for a and b for this to fit into your case?

If you make a good guess you'll see that it fits for the first terms aswell

But focus on the general term first

[uncut forum]

like you mean i try to represent this from the start with the bionomial coefficient?

wait

the other term is a 1?

[sharp vessel]

Try to take your sum, and express it as (a + b)^n for some a, b

There are specific a, b that you can choose, for which if you take (a + b)^n, and expand, you get exactly the sum in your limit.

[uncut forum]

so (1/(n^k) +1) ^n?

[sharp vessel]

Close

What is k?

k is only a variable that appears in your summation. it's the index variable.

[uncut forum]

this bounded by (1/(n^k) +1) ^n?

since we are looking at every index?

[sharp vessel]

What do you mean bounded?

[uncut forum]

like that OF what i said

the sum of

[sharp vessel]

No

Your original expression contains only n

(1/n^k + 1)^n has k in it

And if you put that in the summation, although it now has only n, it's just wrong

Maybe it'll be easier if you first take your sum

2 + (n choose 2) * 1/n^2 + ...

And write it in summation notation

THEN use the binomial formula

[uncut forum]

i dont get what you mean, it has k in it but it doesnt ?

[sharp vessel]

Forget that

Just write your sum in summation notation

With sigma

[uncut forum]

you mean i write the same exact thing in the question in summation form?

[sharp vessel]

Yes

[sharp vessel]

Refer to this if you are confused

[tight furnace]

Turn the sum there into summation form

[uncut forum]

so the sum up to n starting at k = 0 of n choose k times 1/n^k ?

[sharp vessel]

Yes

At what k does the sum end?

[uncut forum]

when its n

[sharp vessel]

$\sum_{k=0}^n \binom{n}{k} \frac{1}{n^k}$

Here you go

[elfin berry]

RedstonePlayz09

[uncut forum]

when k = n it stops

[sharp vessel]

Now it's just a special case of this

So what a, b give you the sum you want?

[uncut forum]

wait for if it goes for every single number, that means, the second element b is k ?

[sharp vessel]

No

[sharp vessel]

Just try to see what you can plug in to a, b so that you get this sum

[uncut forum]

uh idk if i understood what you said wrong, but i tried to plug in if n was 2 and i tried to do the sum at 0 1 and 2, i got 1 for both 0 and 1 but i got 1/4 for 2

because 2! cancels with 2! and 0! is 1

[sharp vessel]

We already said the sum you have is this

Are you trying to check it?

For k = 0 you get:

$\binom{n}{0} \frac{1}{n^0} = 1 \cdot \frac{1}{1} = 1$

[elfin berry]

RedstonePlayz09

[sharp vessel]

For k = 1 you get:

[uncut forum]

i thought by observing some pattern we'd get b

[sharp vessel]

$\binom{n}{1} \frac{1}{n^1} = n \cdot \frac{1}{n} = 1$

[elfin berry]

RedstonePlayz09

[sharp vessel]

So for k = 0 and k = 1 you get 1

Their sum is 2

[uncut forum]

yes

[sharp vessel]

That's the first term in your sum

For the rest, it's exactly as written

For example, the (n choose 2) * 1/n^2 is just k = 2

So this sum is correct

All you have to do it use the binomial formula now

$2 + \binom{n}{2} \frac{1}{n^2} + \binom{n}{3} \frac{1}{n^3} + ... + \frac{1}{n^n} = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} = ?$

[elfin berry]

RedstonePlayz09

[sharp vessel]

Left side, what you have

Then we got this sum

Now using the binomial formula, this should be equal to...?

[uncut forum]

the exponential?

[sharp vessel]

?

[uncut forum]

oh wait

[sharp vessel]

Use the binomial formula

[uncut forum]

express it in (a+b)^n ? To be able to do n choose k times that?

[sharp vessel]

You just write it as (a + b)^n

Because that's not a summation anymore

It's a very simple expression which you'll be able to take the limit of

Find a, b for which:

$\sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^n \binom{n}{k} a^kb^{n-k}$

[elfin berry]

RedstonePlayz09

[sharp vessel]

If you find such a, b

Then your sum is just equal to (a + b)^n

[uncut forum]

i have no idea other than a being 1/n^k and b being a 1

am i missing something

wait

[sharp vessel]

Plug in a = 1/n^k to the right side

You wont get back the left side

b is 1 yes

a is not 1/n^k

[uncut forum]

umm, does it have to do with a being 1/n^k which can be expressed as n^-k and that ^k gives 1?

so its one?

[sharp vessel]

No

Ok first I told you that b = 1 is correct

Plug that in

And look what you have left

The a^k term needs to become 1/n^k

What is a for this to happen?

[uncut forum]

does a have to be negative then?

[sharp vessel]

No

a^k = 1/n^k

Solve for a

Hint: ||1/n^k is equal to (1/n)^k||

[uncut forum]

so if they have same exponent they have the same original num at the non ^ of k so they are the same so a = 1/n?

[sharp vessel]

I don't know what you said but a = 1/n

yes

[uncut forum]

or you just take the root for both sides of k

[sharp vessel]

You just notice that when a = 1/n, you get the 1/n^k term

Now take the sum with a^k * b^(n-k)

And plug in a = 1/n, b = 1

And make sure you understand how you get the original sum

I have to go now, but the final step is to plug in into (a + b)^n

That will be equal your sum

And then just evaluate the limit

Good luck

[uncut forum]

ok i will try , thank you man

[sharp vessel]

Np

[sleek topaz]

ily tooo <3

[uncut forum]

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[marsh citrus]

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[sweet glade]

Uhh, I can’t use logarithms

And I don’t know how to do this

[rustic field]

use the laws of exponents

[marsh citrus]

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[valid plinth]

i was wondering if I could get some help on where to start. I've been stuck on this problem for a couple of days

[valid plinth]

this is what I've thought of so far
first the cup itself is always parallel to the tangent line at a point it's on, so that means that it is parallel to g'(x), and so would that imply that the slope itself of the water spilling = g'(x)

the larger |g'(x)| the larger the magnitude of the rate of change of the volume of the cup gets and so there is a positive correlation between the two. So then the crux of this question is to find the relation btwn g'(x) and the ROC of volume of the water.

the volume of water = h(w^2) and that can be written as 2w^3. But it's at this point i get stuck. I was thinking maybe since the width of the cup is parallel to g'(x) we can sub it into that but I don't know if that makes sense
i'm trying to think of it as a related rates problem and idk if that is the right approach and there are other concepts connected to it

[marsh citrus]

@valid plinth Has your question been resolved?

[indigo lantern]

Hello. I think might have overthought a bit. See if you catch this: the amount of water remained is determined by the biggest absolute value of slope (equates g'(x)).

[valid plinth]

ohh okokoay so thats what i was thinking right now

the largest the slope can be is 2

in terms of absolute value

[indigo lantern]

Right

[valid plinth]

but from there i thought that maybe

we find the angle the made between the graph with a slope 2 and the horizontal axis

and the angle of elevation could help

but i don't even know if that is right

i don't think that is what you're insinuating either with this

[indigo lantern]

It's a square cup remember?

the base is perpendicular to the side, so the minimum slope of the side is 1/2

So the diagonal of the cup is parallel to the x-axis

in that case

that's how you contain the rest of the water

[valid plinth]

I'm sorry I don't think I'm following, the side of the cup is 2x that of the base because the side is = 2w and the base is w^2

could I see a sketch of this then if it is possible

sorry for the rough sketh

[indigo lantern]

It has nothing to do with the volumn of water

the base is w and the height is 2w

right

[valid plinth]

okay

yea

[indigo lantern]

You agree this is the position that spills most of the water right?

[valid plinth]

yea i agree

[indigo lantern]

At after one spill no more can be spilled out unless there's a larger |g'(x)|

[valid plinth]

yea I agree with that part too

[indigo lantern]

Now draw a diagonal of the cup starting from the top of the triangle

[indigo lantern]

See that line? That's parallel to the x-axis.

Below that is the water you can contain

exactly half

[valid plinth]

ohh okay i see

and so this is exactly half of the square and thus half of the volume of the cup

[indigo lantern]

Yes

[valid plinth]

that makes sense

how can i come to this conclusion mathematically without drawing a picture

because now it seems obvious if i draw a picture

[indigo lantern]

First, the surface of the water is always parallel to the x-axis

this is commonsense so you don't really need to explain here

[valid plinth]

yes i see

[indigo lantern]

So the largest angle the cup is turned determines the water remained

[valid plinth]

at that point it would be when the slope of the tangent is the largest which is 2

[indigo lantern]

Yes

Which happens to be the "tangent" of your cup

h/w = 2

I got to go, see if you can phraze it better

[valid plinth]

oh okay thank you very much

.close

[marsh citrus]

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[opal vector]

how would I go about solving this?

[opal vector]

I started with this:

[elfin berry]

KXLI

[opal vector]

.close

[marsh citrus]

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[exotic parrot]

Let $E/F$ be a Galois extension with Galois group $G$. Let $f \in F[x]$ be a polynomial and let $a, b$ be roots of $f$ in $E$. Is there always a $\sigma \in G$ with $\sigma(a) = b$?

[elfin berry]

DavidL1450

[marsh citrus]

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[marsh citrus]

@humble pasture Has your question been resolved?

[humble pasture]

<@&286206848099549185>

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[tight meteor]

I need to find the z coordinate of the center of mass of a wire described by $\vb{r}(t) = (t\cos t, t\sin t, 2\sqrt{2}t^{3/2}/3)$ where $t \in [0, 1]$.

[elfin berry]

jewels!

jewels!

[tight meteor]

Is this integral correct?

the density is 1

[marsh citrus]

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[raw sand]

what the answer

[valid cape]

what have you tried?

[raw sand]

for e = 16

because the other side equals 16

[valid cape]

wrong

[raw sand]

for a = 28 the first diagram , second one b= 56

[tawdry rampart]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[raw sand]

third one idk

[tawdry rampart]

Can you label the diagram

Like A, B, C ....

[raw sand]

its already shown in the picture

topic is circle theorm

[tawdry rampart]

the angles are labelled

can you label the vertices

you will see why its important in a minute

@raw sand

ping me when you get back

<@&268886789983436800>

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[night lion]

area of parallelogram formed by body diagonals is twice the area of the original parallelogram
area of parallepiped formed by face diagonals is twice the area of the original parallelopied

[night lion]

does this follow in higher dimensions? just curious i have no idea

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[exotic echo]

I was trying to implement the letter "A" from the font OCR-A. I ran into the problem of not knowing how to find the distance from the zero line to the bottom line of the "A". I think I came up with a solution that works, but I am not entirely sure.

Basically, I wanted to find the length of a line that both intersects each sides of the isosceles and is parallel to the base line. I think the following method could work:

1. Start measuring from the point that both of the equal sides meet, then measure to the base in a perpendicular manner.
2. Take the height from the base to the desired "new base" and subtract that from the total height.
3. Take the current remaining height and times the original base length to get the "new base" length.

In the photo, the new base is colored in red. The drawing is a bit of a mess, so if you want me to further clarify, just let me know.

[exotic echo]

The OCR-A letter.

[marsh citrus]

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[exotic echo]

Welp. I have to go now.

.close

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[tired ore]

hello I need help proving that the number of non-increasing functions from N to N is equal to continuum

[tired ore]

firstly can I just say that it is <= continuum as there are continuum functions from N to N?

[tight kite]

wym by there are continuum functions from N to N?

[tired ore]

the number of functions from naturals to naturals is equal to continuum

[tight kite]

yeah, the cardinallity of a subset is always <= than the cardinallity of the set itself, you just need to consider if you need or not to include the proof for the cardinallity of the functions N to N

[tired ore]

ok great

this was proved in the lecture

so to prove >= I have to construct a function which is onto from the set of these functions to a set with cardinality continuum correct?

what about this: every function is uniquely defined by the infinite sequence f(1),f(2),..., define a function g(x) = 1 if f(x+1) < f(x) else 0

oh but infinite 1's isnt a value ever

ugh

[rugged ember]

What set do you consider cardinality of continuum?

[tired ore]

I tried set of all binary sequences

[rugged ember]

All functions?

[tired ore]

wait wym

[rugged ember]

I meant

But that was deducible

Nvm

[rugged ember]

I think that could work

Yes it should

I think i got this

[tired ore]

how

[rugged ember]

f(0)=0

f(n)=f(n-1)+1 if nth term in a string is 0

f(n)=f(n-1)+2 if nth term in a string is 1

[tired ore]

by string you mean binary string?

[rugged ember]

Yes

Binary sequence

[tired ore]

sorry im a little confused we need an onto function from binary strings to our functions or from functions to binary strings

cuz I thought we need the latter

[rugged ember]

Wait

You want to prove cardinality of this set is >= continuum?

[tired ore]

y

[tired ore]

<= here

[rugged ember]

So from the set of the cardinality continnuum

You need

Injection

[tired ore]

oh so this

works

why is this function an injection tho

[rugged ember]

If you have 2 different sequences

Take the smallest different term

Then f(n-1)=g(n-1)

Thus

f(n)=f(n-1)+1

And g(n)=g(n-1)+2

So f(n) and g(n) are different

[marsh citrus]

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[wheat pewter]

Am I right so far?

[marsh citrus]

@wheat pewter Has your question been resolved?

[still temple]

Your graph of g(x) is wrong

[wheat pewter]

Oh-

How comes and how can I fix

[still temple]

Trying to plot the graphing by plotting the points is a good approach

Try it again,it isn't (0,-4)

Rather than converting it into a quadratic and using the formula for finding it's roots making a big mess and a huge chance of going wrong

[still temple]

Why did you miss out 2.d?

@wheat pewter ?

[wheat pewter]

I didn't quite understand it

[still temple]

Which one?

[wheat pewter]

2d

I think I need a walk through on that mess I did

[still temple]

It's the same as 2.c

But reverse and they haven't mentioned a relation ,but it's already given in the table

:p it's not the same

[still temple]

Nice

[marsh citrus]

@wheat pewter Has your question been resolved?

[wheat pewter]

Is it 8?

[still temple]

Yep

[wheat pewter]

Raeee

Uhm

I need help with the rest😭

The one I did the mess with and the uhm other incompleted ones

[marsh citrus]

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[worn field]

i got using snells formula that a) is 30 degrees

[worn field]

but how do i do b

[marsh citrus]

@worn field Has your question been resolved?

[marsh citrus]

@worn field Has your question been resolved?

[marsh citrus]

@worn field Has your question been resolved?

[languid anvil]

<@&286206848099549185>

[languid anvil]

.

[elfin berry]

| ~ вєℓℓα ❄⛄

LaTeX source sent via direct message.

[queen sundial]

Hang on

There

Hope that helps

[marsh citrus]

@worn field Has your question been resolved?

[worn field]

could you maybe show me a diagram of the

angle of indiedcen at the second interface

i dont quite get that part

[marsh citrus]

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[cosmic raptor]

how to find C?

[marsh citrus]

@cosmic raptor Has your question been resolved?

[red nimbus]

You can consider x'(t)

look for which gamma you get x'(t) > 0 or < 0 for all t

[cosmic raptor]

I tried doing it

positive whjen 1/2 * sech^2(6*t) which was there, i tried putting it as an interval but still wrong

[red nimbus]

show

[red nimbus]

hahahaha

i meant to show your work

[cosmic raptor]

oh hang on

hahahha

[red nimbus]

[ x(t) = 4t-\gamma\tanh(6t) ]
[ x'(t) = 4-6\gamma\sech^2(6t) ]

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

So consider first x'(t) > 0

,, 4 \geq 6\gamma\sech^2(6t) \Leftrightarrow \gamma \leq \frac{2}{3\sech^2(6t)}

[cosmic raptor]

ahh i mislooked the question then

i just tried solving it again, apprently incorrect

wait, wouldn't it be 2 / 3*sech^2(6t) ?

[red nimbus]

oh yea

[cosmic raptor]

apparently, still incorrect 😦

[red nimbus]

ya hold on

The max value of sech²(6t) is 1

so the least value for gamma would be 2/3

gamma < 2/3

[cosmic raptor]

ohh yeah that's true

[red nimbus]

Now the other case

[cosmic raptor]

i just typed it in and it partially worked

so now i'm curious, why is it <=

and not just <

[red nimbus]

oh yea

actually <=

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

i considered strict monotony

\textbf{2. Case} : $x' \leq 0$
[ \gamma \geq \frac{2}{3\sech^2(6t)} ]

[elfin berry]

𝔸dωn𝓲²s

[cosmic raptor]

but like why is it <= thought?

let's say it is equal two 0 and function is neither increasing nor decreasing

wouldn't there be 2 values of x with the same y?

[red nimbus]

no

e.g. x^3

it would have saddle points

[cosmic raptor]

wait, that's for multivar calc, we're not there yet hahaha

[red nimbus]

no

[cosmic raptor]

so with x^3

so function is increasing when x>0

when x=0, it's not moving at all so 0 and so on would be strictly increasing

makes sense

but when do we know when to use < vs <= ?

[red nimbus]

here <= is sufficient

we just gotta make sure the derivative doesn't cross the x-axis

so either above or below x-axis

[cosmic raptor]

sorry, i'm confused