help-42

[lucid knot]

hows this for a comparison of joint and combined variation?

[calm coral]

@lucid knot Has your question been resolved?

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[remote mural]

can someone tell me how to do this step by step?

[calm coral]

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[white mountain]

Can someone please explain why they can assume f(a)=g(a)=0

[white mountain]

<@&286206848099549185>

[manic oriole]

its a weaker version of the proof

that doesnt actually work

[white mountain]

Wdym

[manic oriole]

basically it assumes that the functions are continuous

just cuz the lim x->a f(x) = 0 doesnt mean f(a) = 0

[white mountain]

But there is no limit there

It just states it

[manic oriole]

yeah

they are saying that lim x->a f(x) = 0 implies f(a) = 0

which is only true for continuous functions

the way you have to actually do the proof is define F and G such that F(x) = f(x) and G(x) = g(x) except at x = a when F(x) = 0 and G(x) = 0

and then use the cauchy mean value theorem on that

[white mountain]

Well it’s the only proof my book gives😭

Can you recommend somewhere I can learn a more robust one

[manic oriole]

https://proofwiki.org/wiki/L'Hôpital's_Rule

[manic oriole]

its the same basic idea

you just have to be careful cuz you dont actually know that f(a) = g(a) = 0

[white mountain]

This one doesn’t cite a limit for the f(a)=g(a) either

I know it’s referring to the weaker one

But I just don’t get how they are stating it

Even if you assume continuity how can you necessarily assume they are 0?

[manic oriole]

well the standard definition of continuity is that lim x->a f(x) = f(a)

we are working with the indeterminate form where the limits are zero

because 0 = lim x->a f(x) = f(a) we know f(a) = 0 if f is continuous

same thing for g

[white mountain]

Where was that stated

Nvm I think I see

Because otherwise it’s not indeterminate

[calm coral]

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[lucid knot]

i keep getting 1/ (5x)^3/2

but its not an option

[old falcon]

you are not fully simplified

1/x = x^(-1)

[onyx wadi]

u need to bring the (5x)^3/2 up

so that it isnt a decimal

[maiden plover]

btw

[onyx wadi]

do this to simplify further

@lucid knot

and ur gonna get ur answer

[lucid knot]

ohhh

ty!

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[gray briar]

(8x-1)/(3)=7-(4x+2)/(5) solve for x

[gray briar]

I didnt get the correct answer which is x=2 i was left with x=116/28 which simplified is x=58

where did I mess up to not get x=2

[kind kraken]

You put the product of the LCD in the numerator

[gray briar]

ohh that makes more sense know I got lost in my own math. Thank you for that .

[kind kraken]

Or actually.. having a hard time following your work, but.. looks like you corrected for it

[gray briar]

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[exotic falcon]

do not post images that contain your personal information

I can see your school, city, email address, etc.

all from this one image

you should delete it and take a better screenshot

[remote mural]

ok

[calm coral]

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[exotic falcon]

this has all of the same info...

also

[remote mural]

bruh

oops

[exotic falcon]

use this if you're incapable of getting a better SC

[remote mural]

yas

that one

[exotic falcon]

open a new channel

[remote mural]

oh ok

[exotic falcon]

@remote mural how old are you btw?

[remote mural]

14

wdym

.close

.close

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[thorny sage]

Disclaimer: Self-teaching calculus two years before I took the actual class so forgive me if this is the most bare-bones calculus question you can come across.

=====

Does anyone know exactly what purpose Sigma represents in an equation regarding area approximations on curved functions?

For example, I know how to work this problem. You would use the symbol Sigma to prepare one's equation for adding together the approximate areas of all the other, smaller rectangles in correspondence to the function. Still, it's never actually told me what Sigma itself represents. I've never had to do anything regarding it because the two values associated with Sigma I can easily define off to the side.

[thorny sage]

Funnily enough I got this one wrong only because I forgot to multiply my answer by delta x 💀

[calm coral]

@thorny sage Has your question been resolved?

[thorny sage]

<@&286206848099549185>

[drifting seal]

sigma is just the notation for summation

the same way the integral symbol is notation

sigma is just notation to represent summing up/adding up the rectangles

to give an area

and you obviously want infinitely many rectangles

[thorny sage]

Oh ok, so it's what I originally thought

[drifting seal]

to have an exact answer for area

it’s not a "value"

the sum has a clue

value*

but

it’s literally just telling you to add the inputs

[thorny sage]

Yeah

[drifting seal]

at each value of n

until the upper limit

[thorny sage]

Well thank you kind sir

[drifting seal]

you’re welcome good sir

[thorny sage]

Just wanted to know, since I'd rather understand the math than just input a bunch of values

[drifting seal]

mhm understandable

[thorny sage]

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[quartz nexus]

I enter a pizza store where they have 10 slices of pepperoni, 8 slices of margarita, and 5 slices of hawaii on display. In how many ways can I give it out to 10 people where each person can get anywhere from 0 to 23 slices of pizza?

I enter a pizza store where they have 10 slices of pepperoni, 8 slices of margarita, and 5 slices of hawaii on display. In how many ways can I give exactly one slice each out to 5 people?

[calm coral]

Please don't occupy multiple help channels.

[quartz nexus]

@sly geyser

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[rocky flower]

Is there an efficient way of calculating the sum the rounded down function of n/i given i ranges from 1 to n?

[rocky flower]

if someone could help me latex this, it would be great

[modern peak]

hm

so like

hang on

$\sum_{i=1}^{n}\lfloor{\frac{n}{i}}\rfloor$

[potent lotus]

ren

[rocky flower]

yes

[modern peak]

basically, this

hm

like are you looking for the general formula?

[rocky flower]

i dont think there is a general formula to that

[modern peak]

then?

limit?

[rocky flower]

no, im looking for an algorithm to find the quickest way of calculating that

[modern peak]

so CS

yea no im out

[exotic falcon]

I think my ma'am @graceful dust might know about this

[modern peak]

LMAO

the squad rolling up fr

[graceful dust]

lmfao

[modern peak]

and why am i here is here too

nice

[blazing coyote]

so you want $\left[\frac{1}{n}\right]+\left[\frac{2}{n}\right]....\left[1\right]$?

[modern peak]

harvey, mike, jessica, and louis

[potent lotus]

Why am. I here

[modern peak]

no it's n/i

[blazing coyote]

oh

[modern peak]

not i/n

[blazing coyote]

right

[rocky flower]

yes, this

[modern peak]

and btw u put square bracs, it's floor

[blazing coyote]

$\left[n\right]+\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]....\left[1\right]$

[potent lotus]

Why am. I here

[blazing coyote]

that;s what I meant

[modern peak]

so basically he wants
f(n) + f(0.5n) ...

[graceful dust]

square braces are used for floor by some people

[modern peak]

i see

thinks

[blazing coyote]

hmm,w $\sum_{i=1}^{n}\lfloor{\frac{n}{i}}\rfloor$ (edited)

[potent lotus]

Why am. I here

[modern peak]

n is an integer, right?
prime factorization+mods maybe

[blazing coyote]

,w $\sum_{i=1}^{n}\lfloor{\frac{n}{i}}\rfloor$ (edited)

[modern peak]

dollar signs

+edited

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[modern peak]

sighs

WA doesn't do latex bro

,w sum from i = 1 to n of floor (n/i)

[potent lotus]

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[modern peak]

welp

[blazing coyote]

lol

[modern peak]

aint even a "more" button

;-;

[blazing coyote]

try using induction maybe

[modern peak]

,w algorithms for sum from i = 1 to n of floor (n/i)

no he wants an algorithm to solve this ASAP

THAT'S NOT EVEN WHAT I TYPED

[blazing coyote]

try for say i from 1 to 5

[modern peak]

BRO

THAT'S NOT EVEN WHAT I WROTE

[rocky flower]

huh

[winged smelt]

$\sum_{i=1}^{n}\lfloor{\frac{n}{i}}\rfloor$ (edited)

[potent lotus]

ColdTee

[modern peak]

he wants the fastest algorithm for this

[rocky flower]

yes this

[modern peak]

try using mods and prime factorization maybe?

[winged smelt]

Lemme think

[modern peak]

if n is an int

[winged smelt]

n is an integer?

[rocky flower]

yes

[modern peak]

hm

[blazing coyote]

ok, say n=a, until i=a ,it's 1

[modern peak]

?

explain

[blazing coyote]

say summation to n=10

[modern peak]

hm i see

wait wait wait

[modern peak]

explain again, idt i got this

[blazing coyote]

ooops, I was solving for its reciprocal

sorry

[modern peak]

xDX

@rocky flower what u want exactly

[blazing coyote]

hmm, say n=5, it becomes 5+2+1+1+1

right?

[graceful dust]

this might have a lot of dependence on the prime factors of n which is bad

[rocky flower]

i was wondering if there was a way of simplifying the expression for any given n

[modern peak]

yessir

[rocky flower]

yeah

[winged smelt]

The sum will depend on whether n is bigger or smaller than i

[modern peak]

what i was saying b4 ;-;

bruh

[winged smelt]

Lmao

[rocky flower]

oh .-.

[blazing coyote]

n=6 , 6+3+2+1+1+1

[modern peak]

if n > i then it's 1

but if n < i it's 0

but the thing is

sum's upper bound is n

so its never 0

omfg im an idiot

[blazing coyote]

,w summation from i=1 to 10 of floor(11/i)

[modern peak]

im a genuine idiot

one second

[potent lotus]

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[blazing coyote]

23,27,28

[rocky flower]

i was wondering for example, if i is a factor of n, we can calculate the exact values without need of the rounded down function, but i was wondering if there is a formula for when i is not a factor of n

[blazing coyote]

doesn't follow any immediate patter

oo

[modern peak]

$\sum_{i=1}^{n}\left[\frac{n}{i}\right]=\left[n\right]+\left[n\cdot\frac{1}{2}\right]+\left[n\cdot\frac{1}{3}\right]\ ...\ 1$

[potent lotus]

ren

[blazing coyote]

maybe use the defn of floor(x)

$x-{x}$

[modern peak]

no use related fracs i think

[potent lotus]

Why am. I here

[modern peak]

like how it varies

[graceful dust]

you can express that with mod or something

[modern peak]

this is hard to explain.

sighs

[rocky flower]

how would we do that, could u give an example?

[modern peak]

$\sum_{i=1}^{n}\left[\frac{n}{i}\right]=\left[n\right]+\left[n-\frac{n}{2}\right]+\left[n-\frac{2n}{3}\right]\ +\left[n-\frac{3n}{4}\right]...$

[potent lotus]

ren

[blazing coyote]

$\sum_{i=1}^n\frac{n}{i}-\left{\frac{n}{i}\right}$

[modern peak]

{} refers to?

[modern peak]

also check this out

[blazing coyote]

fractional part function

[modern peak]

it's - not plus

[potent lotus]

Why am. I here

[blazing coyote]

my bad

[modern peak]

and put brackets around the function

[graceful dust]

example: floor(10/3) = 10/3 - (10 mod 3)/3 i think?

[modern peak]

otherwise it's (sum of n/i) - frac

yep

[blazing coyote]

the summ of fractional parts shouldn't be too hard IMO

[modern peak]

this is my thinking. idk why.

[blazing coyote]

1/2+2/3...

[graceful dust]

me thinks it will be as they depend on prime factorizations

[rocky flower]

so 3+1/3-1/3 = 3, hmmm that seems to work, could u explain why though?

[blazing coyote]

,w summation of n/(n+1) from 1 to n

oh boy

[graceful dust]

10 mod 3 is the remainder that 10 leaves when divided by 3. so 10/3 has (10 mod 3)/3 "extra" past its integer part

[potent lotus]

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[rocky flower]

ohhh yeah, i get it

[winged smelt]

,w summation of (1-i)/i from i=1 to n

[potent lotus]

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[modern peak]

ykw imma go

[winged smelt]

....

I got $n + n\sum_{i=1}^{n}\frac{1-i}{i}$

[graceful dust]

where did this problem come from/what do you need it for?

[potent lotus]

ColdTee

[rocky flower]

i made the problem myself, and i need it for an experiment

[blazing coyote]

the harmonic number don't have a closed form for theri summation if I remember right

[winged smelt]

I made a mistake nvm

[graceful dust]

it seems hard

[winged smelt]

$\left[\frac{n}{i}\right] = \left[\f1i\right] + \left[\f1i + \f1n\right] + \left[\f1i + \f 2n\right] + \ldots + \left[\f1i + \frac{n-1}{n}\right]$

[blazing coyote]

https://math.stackexchange.com/questions/486002/finding-the-summation-of-the-floor-of-the-series?noredirect=1

[winged smelt]

This works for $n \in N$

[potent lotus]

ColdTee

[graceful dust]

yep that confirms every single suspicion i had about this problem

[rocky flower]

thank you

[potent lotus]

ColdTee

[rocky flower]

yes, i knew there wouldnt be a formula for this, i was just wondering if there was a faster way of calculating the summation

[blazing coyote]

approximate it using an integral for sufficently large values?

say, n=100

this could be appoximated as 100ln(100) ti think

,w 100ln(100)

[potent lotus]

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[blazing coyote]

$\sum_{i=1}^{100}\lfloor{\frac{100}{i}}\rfloor$ (edited)

[winged smelt]

,w summation from i=1 to 100 of floor(100/i)

[potent lotus]

Why am. I here

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[blazing coyote]

n is too small

but close

[winged smelt]

Pretty much

[blazing coyote]

,w \sum_{i=1}^{1000}\lfloor{\frac{n}{i}}\rfloor

,w 1000ln(1000)

[potent lotus]

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[blazing coyote]

,w summation from i=1 to 1000 of floor(1000/i)

[potent lotus]

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[winged smelt]

,w summation from i=1 to 10000 of floor(10000/i)

[potent lotus]

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[graceful dust]

i'm curious...

[blazing coyote]

,w 10000ln(10000)

[graceful dust]

,w summation from i=1 to 10000 of (10000/i)-(1/2)

[potent lotus]

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[graceful dust]

why am i here your estimate seems to be missing something

,w 10000ln(10000) + (2*0.577215665-1)*10000

[potent lotus]

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[graceful dust]

pretty close

[rocky flower]

is this
a(n) = n * ( log(n) + 2*gamma - 1 ) + O(sqrt(n)) ?

[graceful dust]

yea

[rocky flower]

could u explain how it gives a good approximation

[graceful dust]

i have no idea and it probably involves analytic number theory way above my paygrade

i do happen to know an expert in this

[blazing coyote]

austin?

[graceful dust]

no definitely not

[blazing coyote]

look up rienman summation

[graceful dust]

@queen wave hi daddy explain why asymptotic formula here works pls
https://math.stackexchange.com/questions/486002/finding-the-summation-of-the-floor-of-the-series?noredirect=1

[blazing coyote]

that may provide some insights on why this works

[winged smelt]

I givw up

[graceful dust]

coldtee really out here trying to resolve open problems

[winged smelt]

[rocky flower]

no worries, thank you all so much for your help :))

.close

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[steel tide]

Ohh i solved it

Nevernind

.close

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[glacial raptor]

how can I find the probability in this

[glacial raptor]

so I can get the mew and sigma

[queen wave]

[topaz raft]

https://tenor.com/view/mewing-skeleton-nikita-gif-6132901180355580261

[queen wave]

The problem you are referring to can be rewriting using the divisor function. Namely, $$\sum_{n \le x} d(n)=\sum_{n \le x} \sum_{d|n}1= \sum_{d \le x} \left\lfloor x/d \right\rfloor$$
Now you can go ahead and express the floor function as x-{x} and get a rough estimate of $x \log x + O(x)$ but we can do better than this by taking a different method of estimating our sum via Dirichlet hyperbola method. With that method you get all these additional constants that come for free in the big O term in $x \log x + O(x)$

[potent lotus]

Deltoid

[glacial raptor]

bruh

.close

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[graceful dust]

thank you

[queen wave]

I love just waking up at 4 am and see this

[calm coral]

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[maiden crater]

hey guys need help on this one

[maiden crater]

I don't understand why number 42's answer is a negative

how did it get the negative

[blazing coyote]

arctan's principal branch is (-pi/2, pi/2)

[maiden crater]

yes im aware that's the restriction for tan

why couldn't it just be pi/2

[blazing coyote]

wdym

[maiden crater]

i meant positive pi/4

[blazing coyote]

what;s tan(3pi/4) equal to

[maiden crater]

it's -1

[blazing coyote]

ok

and what's tan(pi/4)

[maiden crater]

wait hold on

why is it negative one

[blazing coyote]

it;s not

that's why arctan(3pi/4) isn't pi/4

[maiden crater]

i meant for this, why is it negative one

[fleet verge]

It is not

[blazing coyote]

it is

[fleet verge]

,w tan^(-1)(3×π/4)

[maiden crater]

whattt lmaooo

[potent lotus]

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[maiden crater]

im doubting everything now

[fleet verge]

Oh sorry u said tan

My bad reading

[maiden crater]

where did the negative come from tho

[blazing coyote]

,w tan(3pi/4)

[potent lotus]

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[blazing coyote]

try deriving it using a unit circle

[fleet verge]

It is in the second quadrant

It is negative

[maiden crater]

ohh okay

[maiden crater]

how bout for number 36 then

4π/3 is in the third quadrant, why is it not negative

[fleet verge]

From where u get 3

I read 2

Maybe it is a 3 i cant read

Ok it is a 3 yes

But cos^-(1)(cos(3pi4)) is in the second quadrant

So it has to be positive

[maiden crater]

i dont get it WASHSHAHSHSAHAS

isnt sin the only trig function thats supposed to be positive on the second quadrant

[fleet verge]

Ok let see

Cos(3pi/4) how much is it

Now we want to find the angle whose cos is -1/2, but arcsin gives the value from 0 to pi for negative values of cos. In the unit circle the -1/2 is pi/4 and 3pi/4

But because we are interested in 0, pi range

We get the second cuadrant

Pi/4

[calm coral]

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[calm coral]

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[remote mural]

how do I determine where the turning points of f(x) is coordinate wise?

I know that since the f'(x)=0 at 1 and -1 that will be the x coordinate of the turning point in the graph of f(x)

but how am I able to determine the y-coordinate of the respective x coordinate?

[manic grove]

can we assume that f'(0) = -1?

[remote mural]

I read this earlier thinking it was insignificant to the question but is this what you're looking for?

[manic grove]

did you get the f'(x)=0 at 1 and -1 also from the exercise or did you just look at the graph

[remote mural]

I just looked at the graph of f'(x)

[manic grove]

if f'(0)=-1 we know that f'(x) = x^2-1 so we can get f(x) by antiderivative

[remote mural]

mm okay I see where you're getting that point

but how does that help us?

[manic grove]

well if you know f(x) you can sketch it

[remote mural]

okay

alright I think we're on the same boat then

[manic grove]

or well it says rough sketch so my idea is maybe a bit over the top

[remote mural]

yeah I personally haven't dealt with antiderivatives before lol but I'm open to learning

aka integrals

[manic grove]

i could explain both ways your choice

[remote mural]

I mean if you'd like to explain it the antiderivative way go ahead, but that hasn't been covered yet

[manic grove]

ok sure

so the goal of the integral is to find the function that, if you take the derivative, results again in the function you want to find the integral of

[remote mural]

basically just reversing the derivative right

[manic grove]

so e.g. f(x) = x^2 and g'(x) = x^2 whats g(x)

[remote mural]

okok gotcha

[manic grove]

yeah you could look at it that way

so what would g(x) be in that case f(x) = x^2 and g'(x) = x^2?

[remote mural]

what are we trying to do here?

[manic grove]

find g(x)

so that if you take the derivative you get x^2

[remote mural]

hm

so if you take the derivative of g(x) you'll get x^2

[manic grove]

yeah

[remote mural]

that would mean

g(x) = 1/3 x^3

?

[manic grove]

yup nice

but note that you could add something e.g. 1/3 x^3 + 5 and it would result in the same derivative

[remote mural]

yeah cause thats a constant

[manic grove]

yeah

there is an infinite number of integrals for a function

[remote mural]

definetly

[manic grove]

so to generalize that you just add a +C at the end always

so like g(x) = 1/3 x^3 + C

[remote mural]

mhm yea makes sense

[manic grove]

but thats just for looking nice

so now you know the basics how to get the integral of a function

so now back to your question if you look at the graph we get f'(-1) = 0 = f'(1) and f'(0) = -1

what does that tell us about the function f'(x)?

[remote mural]

so visibly the slope is horizontal at -1 and 1

wait a minute

nvm

[manic grove]

you talk about f(x) not f'(x)

[remote mural]

oh right

[manic grove]

the goal currently is to get a function for f'(x)

[remote mural]

right

[manic grove]

so what function does the picture remind you of

[remote mural]

a parabola

[manic grove]

yep

and whats the basic function of a parabola?

[remote mural]

wdym by this?

oh

x^2?

[manic grove]

yep

but the normal parabola starts at 0,0 the one in the picture doesnt

[remote mural]

x^2-1 then

[manic grove]

yea

so now we got f'(x) = x^2-1
whats f(x) then

[remote mural]

$\tfrac13 x^3 -x + c$

[potent lotus]

Safiya

[manic grove]

yeah nice

so to be easier just set the c=0 and ignore it
and then the goal would be to sketch that function

[remote mural]

I see why this is important

we're basically given c then right

[manic grove]

ah yeah didnt remember that info but yeah

[remote mural]

$f(x)=\tfrac13 x^3 -x + 2$

[potent lotus]

Safiya

[manic grove]

yup

so because of the info you could find the one definite integral
thats often the case in exercise if i remember correctly

so how would you sketch f(x) ?

[remote mural]

I would use a table of values

[manic grove]

yup

[remote mural]

hmm okay

this doesn't seem too bad

are you sure there's a different way of doing it?

[manic grove]

yeah if you didnt have integrals until now how would you get the function of f(x) then

[remote mural]

I probably would've done something similar because it seems intuitive

[manic grove]

you could try to find a few fixpoints like -1,0,1 because you know the value or know how the function behaves at that point
and then you also know its a x^3 function because the derivative is a parabola
so try fitting that function somehow on those fixpoints

because it says you just need a rough sketch i guess that was the goal

[remote mural]

so you'd have it as y=ax^3+bx^2+cx+d

but d=2

and then find the original fxn based off that?

[manic grove]

yeah kinda

[remote mural]

hmm interesting

okay well, thank you

[manic grove]

yup no problem 🙂

e.g. like this

just focus on 0 where we know the value and -1,1 because there is the slope 0 and then between -1,1 the slope needs to be negative because f' is negative there and outside the slope is all positive

and rising

@remote mural

[remote mural]

yeah that makes sense

[calm coral]

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[bronze adder]

$\dv{y}{x} = \frac{y(x - y\ln y)}{x(x\ln x - y)}$

[potent lotus]

NEON

[bronze adder]

Don't know where to begin

tried writing it in terms of differentials and nothing seems to work

[blazing coyote]

sure it has a soln?

,w $\dv{y}{x} = \frac{y(x - y\ln y)}{x(x\ln x - y)}

[bronze adder]

i wouldnt be here if it didnt 😭

.close

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[blazing coyote]

really sillly question, but why isn't the energy stored in a region with no electric field not necessarily zero?

[vivid vine]

ask this question in a phyaics server

[blazing coyote]

ok

.close

[calm coral]

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[remote mural]

.reopen

[calm coral]

✅

[remote mural]

@blazing coyote to give you a good analogy

your question can be asked as
"If some land is flat or horizontal, does it imply that its height above sea level has to be zero?"

[blazing coyote]

no

but $\frac{1}{2}\epsilon_0E^2$ is the average energy density

[potent lotus]

Why am. I here

[blazing coyote]

if E=0, wouldn't that mean energy density is zero

[remote mural]

actually, no

because what you have is infact an integral

,, \412 \eps_0 \int E^2 \dd \tau

[remote mural]

E = 0 just implies that the integral is equal to a constant

[blazing coyote]

ah

[restive perch]

you can associate the energy with a potential.
your question is essentially does the potential need to be zero if electric field is zero, which is clearly false.

[bronze adder]

whats tau

[restive perch]

This is half of the energy.

[restive perch]

Time

[blazing coyote]

I thought the energy was stored in the field

[remote mural]

volume element

not time

its dV

[restive perch]

yeah yes,

[bronze adder]

yeah it should be volume

[restive perch]

it's energy density, my bad

(hence volume)

[remote mural]

this is an eternal debate between physicists for some reason

is energy stored in the field or in the charge?

[bronze adder]

field obvs 🙄

[remote mural]

if its in the charge you hvae $\ds \412\rho V$ as your energy instead

[remote mural]

i mean both are correct but like you really shouldnt care

[blazing coyote]

hmm, so to find the energy stored inside a sphere for instance, I first find the energy required to construct it, and then remove the energy outside the sphere, right?

[bronze adder]

self energy stuff

[blazing coyote]

yeah

,w integrate 1/r^2 from r to infty

[potent lotus]

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[blazing coyote]

IMO, I'm getting 0

[bronze adder]

what?

[blazing coyote]

nvm

made a mistake, let me retry

hmm, so considering the energy required to construct the system first

let the surface charge density be $\sigma$

[potent lotus]

Why am. I here

[blazing coyote]

so the radius of a ring element varies with the polar angle, say $\theta$ as $Rsin(\theta)$

[potent lotus]

Why am. I here

[blazing coyote]

so the area varies as $2\pi R^2 (\sin(\theta)) d\theta$

[potent lotus]

Why am. I here

[blazing coyote]

nah. this won't work

I'm trying to find the energy stored inside a sphere of charge Q

[bronze adder]

$U = \frac 12 \epsilon_0 \int_0^R \left ( \frac{1}{4\pi \epsilon_0}Q \frac rR \right )^2 4\pi r \dd{r}$

just that?

[potent lotus]

NEON

[blazing coyote]

wait, how did you get that

E_in=0

ah

potential

you integrated that?

thanks

[bronze adder]

potential?

electric field inside a solid sphere is $\frac{kQr}{R^3}$

[potent lotus]

NEON

[bronze adder]

oh you mean the derivation of this

sure

[blazing coyote]

This is a hollow sphere

I know the derivation no need of that

It's a hollow conducting sphere

[bronze adder]

uh

pretty sure it's solid

I mean the field inside a hollow sphere is obviously 0 because of gauss's law

[blazing coyote]

Yeah, that was my question, so the energy stored inside a solid sphere is zero, right?

Thanks !

[bronze adder]

no it's not

[bronze adder]

it's this

that's meant to be an R^3 in the denominator btw

[blazing coyote]

I meant hollow sphere sorry

[bronze adder]

0 yes

[blazing coyote]

Thanks

. close

.close

[calm coral]

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[unborn stone]

How come in the final equation it’s not sec(x)tan(x) times tan(x)sec(x)tan^2(x)?

[surreal patio]

u mean this?

the plus being a multiplication sign? if so then
(1+tanx)(secxtanx) = secxtanx + secxtan^2x

[unborn stone]

Oh I thought u we’re supposed to distribute tan to both sec and tan

So after the plus sign secxtanx times tan^2x

[surreal patio]

secxtanx is basically 1 term so u first multiply with 1 and then with tanx and get

[unborn stone]

Ok

.close

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[manic cape]

The problem:
Given a convex pentagon ABCDE with AB + CD = BC + DE. It is known that there is a circle centered on AE which forms points of tangency with the sides AB, BC, CD and DE P, Q, R, and S, respectively. Prove that PS and AE are parallel.

So, there is already a given solution (shown below), but I don't understand one part.

Let AP = x, P B = y, QC = z, RD = t. From tangency we get BQ = y, CR = z, DS = t, so AB = x + y, BC = y + z, CD = z + t, DE = t + SE. Thus AB + CD = BC + DE gives x + y + z + t = y + z + t + SE, SE = x = AP. Together with OP = OS (radii) and ∠AP O = ∠ESO = 90◦ from the tangent

Why are the angles APO and ESO = 90? What kind of rule is this?

[calm coral]

@manic cape Has your question been resolved?

[calm coral]

@manic cape Has your question been resolved?

[manic cape]

nope

[calm coral]

@manic cape Has your question been resolved?

[limber swift]

Put the diagram

[calm coral]

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[blazing coyote]

!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

[blazing coyote]

also, is this a test?

[lime valve]

make this into cosa - cosb form

[tall hemlock]

,rotate

[blazing coyote]

no need

[lime valve]

and then it'll be easier

[tall hemlock]

its just a small problem i think

[blazing coyote]

sin(x) when integrated from 0 to 2pi is 0

the entire integral is 0

[tall hemlock]

yes i tried that but its wrong

[blazing coyote]

ah

it;s not odd

my bad

product to sum formulae then

[tall hemlock]

i did IBP twice

[wraith robin]

Use the product sum formulae

[tall hemlock]

but the problem is that the initial integral cancels out

[lime valve]

first convert it..

[wraith robin]

No point doing IBP, use sin(x)sin(y) = 1/2(cos(y-x)-cos(y+x))

[lime valve]

to cos a - cosb

then integrate it

it is easier

[wraith robin]

Yep

[tall hemlock]

oh its because we only learnt IBP

we only did the beginning of integration so i only know IBP and u-sub

[lime valve]

yes but this is like product to sum formula and its taught before calculus.. well as far as I know

[tall hemlock]

yes but it should still work

if i do double IBP

[lime valve]

yeah.. it isn't like one method would work and the other would give a wrong answer lol

[tall hemlock]

yeah but my double IBP cancels the initial integral out which makes me think the answer is 0 but its not

[faint ether]

First reduce the integral

2times

Then do kings rule

@tall hemlock

you will get πcos1 ig

[calm coral]

@tall hemlock Has your question been resolved?

[calm coral]

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[fierce cradle]

could someone help me with problem 2?

[kind wagon]

images are not working discord issue

[glad plover]

what have you tried

[calm coral]

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[calm coral]

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[fierce cradle]

.reopen

[calm coral]

✅

[fierce cradle]

Working on it right now, do you underatand part 1?

[calm coral]

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[calm coral]

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[opal zodiac]

chat whats this

[exotic falcon]

Math

What is your doubt?

@opal zodiac

[opal zodiac]

i need to know how to do that

@exotic falcon

[exotic falcon]

What have you tried

[opal zodiac]

i thought it was the y intercept

how do i do it

@exotic falcon

plz

i need the answer

rn

its almost due

[remote mural]

do you know integration

[opal zodiac]

no

plz answer

i need it

its due in 11 mins

PLZ

[calm coral]

@opal zodiac Has your question been resolved?

[opal zodiac]

no

😭

plz help

[calm coral]

@opal zodiac Has your question been resolved?

[calm coral]

@opal zodiac Has your question been resolved?

[calm coral]

@opal zodiac Has your question been resolved?

[calm coral]

@opal zodiac Has your question been resolved?

[fossil spruce]

Hi

Did you figure it out?

In case you want to know, if you look at the graph carefully you will look that at x= -2 , y= -2 so f(-2) = -2 then just plug the -2 and solve for b

[remote mural]

guys

i have 5000+ messages

from this server

[fossil spruce]

Then we get 2×(-2)² + b×(-2) + 10 = -2

<=> 18 - 2b = -2 <=> -2b = -20 <=> b = 10

[fossil spruce]

κέχοδα

[remote mural]

the hell does that mean

[remote mural]

ok sure

[fossil spruce]

Ἓν οἶδα ὅτι οὐδὲν οἶδα

[calm coral]

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[buoyant grove]

Can someone explain why there's also $h > 4$? Shouldn't it only be $0 < h < 1$ since we can split these two by the purple line?

[potent lotus]

skyglow

[calm coral]

@buoyant grove Has your question been resolved?

[calm coral]

@buoyant grove Has your question been resolved?

[buoyant grove]

<@&286206848099549185> This is for SVM.

[calm coral]

@buoyant grove Has your question been resolved?

[untold compass]

it's annoying since it's off the grid but if you keep going there's another line

https://imgur.com/tjYuzN2

[buoyant grove]

So we would have two lines. For the second line, the blue at (0,0), circled in pink, would be an outlier for the linear SVM?
https://imgur.com/a/9cYZ774
Is that essentially what I'm missing?

[untold compass]

oh there'd be no outlier, swap where you're writing the o's and +'s

[buoyant grove]

OH, because the red dot we're looking at is (h,1) and can be shifted.

[untold compass]

the intuition for visualizing this is that if you connect the blue ones with a line (or like convex hull in general), you want to draw lines that don't cut through that line

[buoyant grove]

Yea, I think I was just stuck on trying to draw a second line with (h,1) at where it's currently drawn. Not where it could potentially be.

[buoyant grove]

Thanks!

[untold compass]

makes sense

[buoyant grove]

.close

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[remote mural]

on d rn

apparently iii) is x=2?

need someone to explain that to me

that isn't a local minimum or maximum

[swift laurel]

it's a graph of the derivative of f

a local maximum or minimum corresponds to a point where the derivative is 0

[remote mural]

oh right

thanks

.close

[calm coral]

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[opal zodiac]

yes

it was 10

but idk how

[calm coral]

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[dry glacier]

Need help for discretization. (I will apply explicit FTCS scheme) In the explicit FTCS (Forward Time Centered Space) method, the time derivative and the spatial derivative terms of the Navier-Stokes equation are discretized separately. For the temporal derivative, a forward difference approximation is employed Meanwhile, the spatial derivative term is discretized using central differences.

[dry glacier]

,tex \rho\frac{\partial u}{\partial t}=-\frac{\partial p}{\partial x}+\mu\frac1r\frac\partial{\partial r}\Big(r\frac{\partial u}{\partial r}\Big)

[potent lotus]

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

[dry glacier]

How do I discretisize this?

[calm coral]

@dry glacier Has your question been resolved?

[calm coral]

@dry glacier Has your question been resolved?

[calm coral]

Channel closed

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[calm coral]

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[shell spruce]

Question 48

[honest quiver]

This is going to be the line that perpendicularly bisects line FF’

So you need to find line FF’, find an equation for lines perpendicular to that, and then find an equation for the specific line that passes through the midpoint of FF’

[shell spruce]

How do I find the fr line

Fr

Ff line

[honest quiver]

Well what’s the slope of that line

[shell spruce]

0?

[honest quiver]

Oh wait that’s a vertical line

So what would be the slope of a line perpendicular to it

[shell spruce]

Um

[honest quiver]

What kind of line is perpendicular to a vertical line

[shell spruce]

Also o

[honest quiver]

That one would be 0

[shell spruce]

0 slope

Infinite?

[honest quiver]

Line FF’ has undefined slope

The perpendicular line has 0 slope

You follow that?

[shell spruce]

Yes.

[honest quiver]

Ok

Now what’s the midpoint of FF’

[shell spruce]

So the midpoint formula with the foci points?

[honest quiver]

Yes

[shell spruce]

(0,0)

[honest quiver]

So what’s a line with slope 0 that passes through (0,0)

[shell spruce]

The FF line?

[honest quiver]

No, the FF’ line is a vertical line with undefined slope

[shell spruce]

Sorry I’m like new to hyperbolas 😭

[honest quiver]

We want its perpendicular bisector

[shell spruce]

So

Undefined slope?

[honest quiver]

Ok draw this on graph paper

It will make sense

FF’ has undefined slope as it is vertical

It’s perpendicular bisector has 0 slope as it is horizontal

FF’ goes from (0,-17) to (0,17) and has a midpoint at 0

Now what line with slope 0 passes through (0,0)?

Think y=mx+b

[shell spruce]

Vertical line

[honest quiver]

A vertical line does not have slope 0

A vertical line has undefined slope

You cannot make a vertical line with slope intercept form

A horizontal line has 0 slope because it is of the form y=b

[shell spruce]

So a horizontal line?

[honest quiver]

Yes but what is the equation of a horizontal line passing through (0,0)

Don’t overthink it

[shell spruce]

Isn’t that the only type of line that can have slope 0

[honest quiver]

Yes

Slope 0 means horizontal by definition essentially

Because it means y does not change with x

It stays the same for all x

If we have a line of the form y=b, and it passes through (0,0), what is b?

[shell spruce]

0

[honest quiver]

So what is the equation of the line?

[shell spruce]

Y= o

0

[honest quiver]

Yes

That’s the answer

y=0

[shell spruce]

But what is the equation

[honest quiver]

That is the equation

[shell spruce]

O

[honest quiver]

Like I said don’t overthink it

It’s just a line

[shell spruce]

But what about k

[honest quiver]

Oh

You raise a good point my friend

[shell spruce]

LMAO

[honest quiver]

That was how to find a line equidistant from the foci

Anyway

[shell spruce]

yeah…

[honest quiver]

Onto the actual problem

Let's consider both points in isolation

What represents all points 30 units away from F?

And similarly, what represents all points 30 units away from F'?

[shell spruce]

Ummmmmm the center???

Or vertex?

[honest quiver]

no....

[shell spruce]

Or line

[honest quiver]

if i ask you to draw a curve with all the points some distance away from a certain point, what would you draw

[shell spruce]

A circle ⭕️

[honest quiver]

Indeed

[shell spruce]

o I thought it was a hyperbola

[honest quiver]

so what would be a circle of all the points 30 units away from F

oh wait

[shell spruce]

Parabola

[honest quiver]

ok ok ok

i finally read the question properly

[shell spruce]

Yup

[honest quiver]

you want an equation for the distance of a point to each one

[shell spruce]

I think the answer is in this format

But not sure

[honest quiver]

you should get there eventually

but first find what is the expression for the distance from F to a point (x,y)

And a similar expression for F'

[shell spruce]

So

Just a random point

[honest quiver]

yes

[shell spruce]

Like the origin

17

[honest quiver]

no no no

as an expression of x,y

So you could plug in x,y for any point and get the distance to F

[shell spruce]

🧠 an eclipse formula

[honest quiver]

?

[shell spruce]

Since the distance is always the same ?

[honest quiver]

no forget that for now

think just simply in terms of the distance from (x,y) to F

use pythagorean theorem

[shell spruce]

oh yeah

Do we solve for a and b?

[honest quiver]

no

do you know distance formula

[shell spruce]

Yup

[honest quiver]

use that

[shell spruce]

So the distance between the F and F’

[honest quiver]

no, the distance between point (x,y) and F

Leaving x and y as variables

[shell spruce]

Ok

[honest quiver]

yes

now what about for F'

[honest quiver]

yes

now we know that |d1 - d2| = 30 right?

as that is what the question is asking

[shell spruce]

So we subtract the two distances?

[honest quiver]

yes

and set them equal to 30

square both sides to eliminate the absolute value and it's all algebra from there

[shell spruce]

Do make d 30 for both

?

[honest quiver]

you need to plug them in to this

d1 and d2 are the two distance expressions you derived

[shell spruce]

Is this set correctly

[honest quiver]

there should be a minus sign instead of a plus sign between the two distances

but otherwise yes

this equation literally states the premises of the question, now you just have to simplify it

[shell spruce]

Idk how to simplify this 💀 it seems wrong

[honest quiver]

i'm not sure, there might be an easier way of doing this, but this is just taking the question and putting it in mathematical form

i am trying to simplify it myself, you have to start by squaring both sides

[shell spruce]

Should I check on Photomath this seems like it’s going to be a really long answer

[honest quiver]

i have it simplifying down into a regular conic section

[shell spruce]

?

[honest quiver]

you just have to do a lot of annoying algebra to get rid of the square roots

why is it taking a derivative...

that is not at all what you need

[shell spruce]

O I see

[honest quiver]

i put it in desmos and it is a hyperbola

it just needs simplification

[shell spruce]

Doesn’t everything cross out

[honest quiver]

you just have to do the algebra

square both sides and do all the multiplication and whatnot

[shell spruce]

Is it y=900\68

[honest quiver]

it should be a hyperbola not a line

[shell spruce]

I can’t

[honest quiver]

the algebra is...not fun

there may be a formula or something

[shell spruce]

@honest quiver tysm for you help

[honest quiver]

sorry i couldn't help much :/

[shell spruce]

I got to go but that was fun and it help me make a lot of progress

No you did

[honest quiver]

alright

[shell spruce]

👍ty

[honest quiver]

you're welcome

[shell spruce]

Now

[honest quiver]

[shell spruce]

How do I close this

[honest quiver]

.close

[shell spruce]

.close

[calm coral]

Channel closed

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Use .reopen if this was a mistake.

[calm coral]

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[willow cove]

Hey! I have an exact different equation here that I'm trying to solve but I'm getting stuck when it comes to integration constants (...yes, really.)

I need to find a solution for this:

$2 \frac{x}{y} dx + \left(1 - \frac{y^2}{x^2}\right) dy = 0$

[potent lotus]

Lucas

[willow cove]

Where x and y are non-zero. My step by step got me this far:

$2 \frac{x}{y} dx = - \left(1 - \frac{x^2}{y^2}\right) dy$

$\text{Find } \phi(x, y) \text{ s.t. } \frac{\partial \phi}{\partial x} = \phi_x = 2\frac{x}{y}, \ \frac{\partial \phi}{\partial y} = \phi_y = 1 - \frac{x^2}{y^2}$

$\phi = \int \phi_x dx = \int 2 \frac{x}{y} dx = \int \phi_y dy = \int 1 - \frac{x^2}{y^2} dy$

[potent lotus]

Lucas

[willow cove]

$\phi = \frac{x^2}{y} + g(y) = y + \frac{x^2}{y} + h(x)$

[potent lotus]

Lucas

[willow cove]

$g(y) = y + h(x)$

And then I got stuck. The expected answer is $\phi(x, y) = y + \frac{x^2}{y} + K$ where K is a constant. What I don't understand, however, is how on earth g(y) and h(x) became K. Can anyone enlighten me please?

From my understanding, both g(y) and h(x) are functions, not necessarily constants

let me reopen this with proper formatting. god damn.

.close

[calm coral]

Channel closed

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[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[sterile wharf]

I have no clue where to start from 😐

[quaint sphinx]

would rewrite 8 in terms of (sqrt2) to the power of something

[sterile wharf]

8^2?

[quaint sphinx]

like

(sqrt2)^something = 8

[sterile wharf]

oh

2^3

[quaint sphinx]

do u know what "sqrt" means

[sterile wharf]

square?

[quaint sphinx]

is shorthand for "square root"

[sterile wharf]

Oh ☠️

[quaint sphinx]

valid tbh

[sterile wharf]

umm no clue

[quaint sphinx]

well do u know what (sqrt2)^2 is

[sterile wharf]

2?

[quaint sphinx]

ye

by definition the square of sqrt2 is 2

[sterile wharf]

so m = 2?