help-27

[weak cove]

R being a regular region with.....

okay yes

and then instead of saying

[wicked turtle]

R being as in the problem statement

[weak cove]

Omega = S1 U S2

it should be

[wooden veldt]

But then it's interior wouldn't include the origin?

[wicked turtle]

why are you taking unions of things

[weak cove]

boundary omega = boundary S1 U boundary S2

[wicked turtle]

don't you want to remove stuff from R

[weak cove]

for the purpose of the integration

I want to split this into two things

because I found that

integrating across the boundary of a sphere of any radius

will be -4pi

[wicked turtle]

so S1 and S2 represent the "outer" and "inner" boundaries of Omega?

[weak cove]

yes they should

because divergence theorem will give us that the integral overall is 0

[wicked turtle]

yes

[weak cove]

but the integral = int over S1 + int over S2

and one of those will be -4pi

so the other I will conclude must be 4pi

is the goal

[weak cove]

because it is over the boundary of a sphere centered at the origin which in a previous part I showed is always (-4pi )

[wicked turtle]

the problem statement calls for -4pi

any concern about the fact that you are getting +4pi?

[weak cove]

I think I must've said something wrong

[wicked turtle]

maybe it's just a difference in convention for the normal vector's direction?

[weak cove]

Okay well earlier in the problem I proved without a doubt in my mind that

integrating over a spher

will give -4pi

so

I think I should not be getting pos 4pi..

[wicked turtle]

presumably that's the case when the sphere is the "outer" boundary of the region

in your case it's the "inner" boundary

[weak cove]

should be like this I think

and then over the delta B eps

is -4pi

and overall it is 0... so I will get 4pi this way aswell no?

wait no

I have it mixed up

the middle integral will be 0

the one on the right will be -4pi

so thus the one on the left must also be -4pi

[wicked turtle]

yea that sounds better

[weak cove]

Okay so I should say

let S be a regular region with piecewise smooth boundary whose interior contains the origin

then

[wicked turtle]

isn't this region already called R

[crisp niche]

I don't want to interfere for now, but you will have to use two balls, a large and a small Epsilon ball, which will tend to zero, this is a technical detail, but at this level you would have to write about it in my opinion, and besides, your task is the way to Poisson's equation, As far as I can see. In a sense, you are preparing to prove the derivation of this equation

[wicked turtle]

why introduce S

[weak cove]

I don't think we are going to do Poisson's equation

[weak cove]

right it is

good point

okay

so let R be the regular region yadda yadda

then the integral over the bounadry of R

[wicked turtle]

seems for the purpose of this problem it suffices to take any ball containing 0 that's in the interior of R

no need to shrink it to zero

[weak cove]

yes

yes

[weak cove]

will be equal to the integral over the boundary of R disjoint the B_eps + the integral over the boundary of the B_eps

right?

and then I'm basically done

I say by the divergence theorem and by the earlier part..

and get the result I want

that's worded much better I think

I have a confusion about this

what if R is some region right that doesn't contain the origin but say like

excludes a cube about the origin

then how could the boundary of R be represented by unioning with the boundary of a sphere about the origin?

surely that wont have the same boundary of the cube

[crisp niche]

okay, I'm not stubborn 🙂 but we are teaching, after the Gauss-Ostrogradzkei theorem and the proof of the Stokes theorem, the Poisson equation, where the Laplacian has exactly your value -4pi

[weak cove]

I think we might stop at the proof of the Stokes theorem :p but maybe you know more than me what is coming next XD

[crisp niche]

in some spare time i can show the proof

[weak cove]

this is like a HW problem so not very consequential

in terms of like what the next material is

[weak cove]

@wicked turtle Would you know about this?

[wicked turtle]

if R already excludes the origin then why do you need a sphere at the origin?

won't the integral just be zero?

[weak cove]

it will be -4pi

that's the point

I thought

wait no I see what you mean

this is a much better write up I think

[wicked turtle]

if your region doesn't include the origin then (in the physics lingo) the incoming "flux" equals the outgoing "flux" and so the surface integral is zero
if i'm remembering this stuff correctly anyway

it has been a while haha

[weak cove]

well it is 0 by the divergence theorem, because the divergence theorem applies when we don't include the origin, and it so happens that the divF for this function is 0

[wicked turtle]

yeah only the origin is a point of non-differentiability

[weak cove]

demonstrated here if you wish to see

why the laplacian is 0

kinda funky problem

these are my favorite once they are done though

I feel like a 10 year old who just built a new lego set I want to go show my parents and all my friends

[wicked turtle]

haha, i feel like the last time i saw this stuff i was actually 10 years old
not literally but it was a long time ago

[weak cove]

XD

[arctic field]

[weak cove]

oh god

don't tell me i'm wrong snow

[arctic field]

your answer to (a) lmfao

nah i won't

[weak cove]

what's wrong with ti

wait do tell me if I am though

[arctic field]

it's just written really funkily

[weak cove]

why

that's how I differentiate

I express my thoughts artisticly

[arctic field]

well i wouldve written it something like uh

,, \nabla g(\vb x) = -\f {\vb x} {\norm {\vb x}^3}

[weak cove]

well we can't all be you now can we

[arctic field]

[weak cove]

I did kinda actualy breeze through this one though

other than this funk up at the end

so u were kinda right about it

[arctic field]

anyway

the last part is looking better now

[wicked turtle]

seems kinda nasty to use x/|x|^3, gives the impression that it falls off as the cube of |x|
nicer would be (whatever notation you use for a unit vector in the direction of x) / |x|^2

[weak cove]

I'm scared for if there comes a day when I know more math than you guys and you can't help me anymore

I don't think it will come though because I basically only know what you all teach me

[arctic field]

lol this is how i write the inverse square law for gravity

[wicked turtle]

well it's not wrong haha

[arctic field]

this is still kinda sus

[weak cove]

damnit

[arctic field]

first of all

you need more brackets otherwise its really confusing to read

like

,, \del(R \setminus B_\eps(0)) \cup \del B_\eps(0)

[wicked turtle]

boundary(R \ B), not boundary R \ B, maybe?

[weak cove]

I was convinced

ah

ya that's fari

fair*

[arctic field]

but also the equation isnt right i dont think

like

,, \del R \ne \del (R \setminus B_\eps(0)) \cup \del B_\eps(0)

[wicked turtle]

yea that is dubious, i was looking only at the integrals which seem ok (if you use parentheses)

[weak cove]

you're right it isn't but the integrals line up

maybe I just don't need to include that part

[arctic field]

you want to phrase this a bit better

[wicked turtle]

you're introducing a new common boundary between the two regions, it's not part of the original del-R

[weak cove]

yes

so the surface integral will remain the same

not neccesarily having the same boundarys

but being the same region

[wicked turtle]

i would word it something like, let S = R \ B, then the boundary of S is the boundary of R union the boundary of B

[arctic field]

,, \int_{\del R} \pdv [g] n \dd A = \int_{\del R - \del B_\eps} \pdv [g] n \dd A + \int_{\del B_\eps} \pdv [g] n \dd A

[arctic field]

then appeal to the fact that like

[wicked turtle]

for the surface integral over the boundary of S, the normal vector of the boundary B portion points in the opposite direction to the normal vector in the B integral, so these cancel

[weak cove]

• being set minus?

[arctic field]

,, \del R - \del B_\eps = \del(R \setminus B_\eps)

[arctic field]

this is largely an orientation issue

[weak cove]

well it's because you're integrating over the same boundary twice

it's included in both dell (R\Beps) and dell(B eps)

and they have opposite orientation

and thus will cancel like Bungo said

[wicked turtle]

(side note, this delta g / delta n notation is odd, i don't recall seeing it before, it's the same as gradient of g dotted with n?)

[arctic field]

normal derivative

,, \pdv [g] n = \nabla g \cdot n

[weak cove]

yes

[arctic field]

ive seen this a few times

[weak cove]

I hate that too

[arctic field]

the integral over $\del R - \del B_\eps$ with the $-\del B_\eps$ is indicating that you're integrating over $\del B_\eps$ with the opposite orientation

[wicked turtle]

yea the rhs is what i'm used to, the lhs looks weird to me but i guess it couldn't mean anything else really

[weak cove]

I've never seen this

[arctic field]

essentially
[ \int_{\del R - \del B_\eps} \pdv [g] n \dd A = \int_{\del R} \pdv [g] n \dd A - \int_{\del B_\eps} \pdv [g] n \dd A ]

[wicked turtle]

i like that convention

[wicked turtle]

-(boundary) meaning the opposite orientation

[weak cove]

yah that makes snese

[wicked turtle]

generalizes the same notation that is common for line integrals

[arctic field]

then you can claim that
[ \int_{\del R - \del B_\eps} = \int_{\del (R \setminus B_\eps)} ]
because the origin is contained in the interior of $B_\eps$ but in the exterior of $R \setminus B_\eps$

[arctic field]

which is how you get the two different orientations

[wicked turtle]

👍

[weak cove]

👍

[arctic field]

👍

[wicked turtle]

three thumbs up, it can't possibly be wrong then

[weak cove]

This was very fruitful

well you both are welcome for my tutelage

If you need more help anytime soon just let me know

[wicked turtle]

gracias por ayudarnos

[weak cove]

I can't .close when this is the last message

it'd be like betraying my kind

[arctic field]

[wicked turtle]

.coose instead ig

[weak cove]

.coose

.close

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[weak cove]

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[wooden finch]

The students of a school decided to beautify the school on an annual day by fixing colourful flags on the straight
passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the
position of the middlemost flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books
where the flags were stored. She could carry only one flag at a time.
[4]
(i) How much distance did she cover in pacing 6 flags on either side of center point?
(ii) Represent above information in Arithmetic progression
(iii) How much distance did she cover in completing this job and returning to collect her books?
(iv) What is the maximum distance she travelled carrying a flag?

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[restive river]

I don't really know what a sign chart is but I can help you solve it

Use the rational root theorem to try to find a root of the polynomial

[quaint citrus]

just find zeroes, put them on a number line, and plug in numbers in between the zeroes

thats how u use the sign chart i think

[open sluice]

Ohhh

Ok ok thank you

.close

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[void otter]

I've been stuck on this induction problem for a long time. Above is the best attempt I've made at it, but something seems to be going wrong. Did I do something wrong somewhere, or is there an easier method to solve it?

My method was basically assume true for n = k (eq 1), then i did n = k + 1 (eq 2). Then I divided eq 2 by 1 + x, so that eq 1 and eq 2 have the same left side. From there it was obvious that the first term of eq 2 is smaller than the first term of eq 1, so all that was left to prove was that the second term of eq 2 was smaller than the second term of eq 1, but i couldn't do that.

I figured that the reason is because i neglected the different between term 1 of eq 1 and term 1 of eq 2, so i calculated that difference and rewrote eq 2 with 3 terms, with the first term being equal to the first term in eq 1, and then tried to show that the second two terms of eq 2 were bigger than the second term of eq 1, but i had no luck. Did I do something wrong in my method? And is there an easier way of doing it?

[devout snow]

@void otter Has your question been resolved?

[devout snow]

@void otter Has your question been resolved?

[restive river]

me podrian ayudar a resolver este problema

[void otter]

<@&286206848099549185>

[devout snow]

@void otter Has your question been resolved?

[small jackal]

proof by induction??

[void otter]

yeah that's the method i was trying

[small jackal]

when n = 1

RHS = LHS

which is true

when n = k

lets see

1+ kcomb1*x + ... + x^k

[void otter]

i put my attempt here btw

and the explanation is under it

[small jackal]

oh ok

[void otter]

wait i accidentally sent the same screenshot 3 times

i'm dumb

that was my method/attempt

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[vital cradle]

guys, is absolute extrema or relative extrema a point or a number?

[vital cradle]

this is what book shows me the relative maximum, and I think it is a point right?

[woven radish]

woods

[vital cradle]

Thank you so much

another question is

if i have two value on "absolute minimum or absolute maximum" in an interval like [-1,22]

for example, i have (0,2) and (6,2)

(11,12) and (20,12)

so the abs minimum is (0,2) and abs maximum is (20,12) depends on the x right?

yeah, I think so

but I'm not sure how can I correctly find the abs extrema if I have a same value for y at two points in an interval, do I need to see the value of x and see which one is smaller or bigger in the interval?

yeah!!!

is it correct if i do like this

ok sure

thank you so much!!!

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[misty sun]

Is 6 correct for this question?

[supple knot]

it's not 6

[misty sun]

I thought so

[misty sun]

Is it because I came from the left instead of right? Does it have something to do with the solid dots?

[supple knot]

0+ means from the right

[misty sun]

Not ->, it is <-

[misty sun]

So that means it’s zero?

[supple knot]

this means, in words,
"The limit as x goes to zero from the right of the sum f(x) plus g(x)"

[misty sun]

So not the limit of both f(x) and g(x) at zero from the right summed?

[pseudo basin]

hold up

what did you get for $\lim_{x \to 0^+} f(x)$?

[woven radish]

Ann

[devout snow]

@misty sun Has your question been resolved?

[misty sun]

I got 3

[pseudo basin]

that's wrong tho

$\lim_{x\to 0^-} f(x)$ is 3, but $\lim_{x\to 0^+} f(x)$ isn't

[woven radish]

Ann

[devout snow]

@misty sun Has your question been resolved?

[misty sun]

So I believe the correct answer is zero

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[misty sun]

.reopen

[devout snow]

✅

[pseudo basin]

well the right limit of f is 0 yes

what about g

what's $\lim_{x \to 0^+} g(x)$?

[woven radish]

Ann

[devout snow]

@misty sun Has your question been resolved?

[misty sun]

I see it approaching 0 but having an actual value at 2, I’m not sure if it’s position between the two affect the value of the limit

[pseudo basin]

you are mixing things up.

[misty sun]

Thank you so much, I think it’s zero. I’m not sure what the correct answer is.

[pseudo basin]

no, lim[x->0+] g(x) isn't 0

g is continuous near 0, and g(0) is 2, not 0

the graph of g doesn't go near the origin as it would

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[lost laurel]

A bag contains 6 balls, two are drawn and are found to be black, the probability that the bag contains atleast 5 black balls is?

[lost laurel]

Now the bag obviously has atleast two black balls

but what do I do from here?

summation of probabilities should be 1?

or baye's theorm

I think bayes theorm, would that be right?

nvm, got it

.close

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[lost laurel]

what does this even mean?

[keen crane]

you calculate n(1 - |x-2n-1|) for n=1, n=2, .. and then choose the maximum. That is your f(x)

[lost laurel]

wait,what? is there any efficent way to do that?

I'm guessing I would find n by differentiating it assuming x to be. a constant , right?

[restive river]

the maximum value is when x = 2n+1

[lost laurel]

why?

[lost laurel]

then how am I supposed to integrate the function?

[restive river]

because the minimum value of an absolute value of something is 0

[lost laurel]

hmm, OK. Thanks

[restive river]

i dont know

[lost laurel]

<@&286206848099549185>

.close

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[eager nova]

Ok what is p(x)?

[hasty saffron]

,rotate ccw

[hasty saffron]

Use the fact that if a=b and a=c, then b=c

Hi

Yes

[hasty saffron]

That's what I mentioned here

it's given p(x)= 1/(x+1)
And it's also given that the same p(x)=1/4

What can that mean

$p(x)= \frac{1}{x+1}$ and $p(x)=\frac{1}{4}$
$$\implies \frac{1}{x+1}=\frac{1}{4}$$

[woven radish]

Lorentz

[hasty saffron]

Ig you can figure out the rest

[rustic yew]

yep u did it correctly

im struggling to understand what exactly did u do

did u perhaps squared on both sides?

it'll be okay

equate both the g(x)

then

u'll get

√(5x+1)=4

square on both sides

u'll get

5x+1=16

then solve

how old are you?

squaring on both sides means do whole LHS raised to power of 2
and whole RHS raised to power of 2

i did this because there was an under root on LHS

√(5x+1)

by squaring i removed it

oh

lemme break it down don't worry

[rustic yew]

LHS: left hand side, terms written on left side of the "equal to sign"(=)

so

in case of

5x-1=10

5x-1 will be LHS

RHS means right hand side, terms written on right side

[rustic yew]

therefore 10 is the RHS in this equation

equation refers to numbers that have equal to sign (=)

[rustic yew]

when a number has the power of two to it

for example 5²

some call this, 5 raised to the power of 2

some call this 5 square

@restive river

read up

tell me if u don't understand

i used the equation as an example

don't relate this to your question

any point which u do not understand?

we'll move to the next step only if everything's clear

okay so

are u aware of √

under root

@restive river

nope

the SQUARE of 4 is 16

cause the meaning of square is basically raised to the power of 2

so 4²= 4×4

which is 16

they're opposites, lemme explain with an example
can u tell me what the square of 2 will be?

2²

what will it be?

ye so 4?

ye

now

when you under root a number

for example when u under root 4

you get 2

under root of a number will be equal to a number who's square is equal to that number you just under rooted

the definition is complicated

but examples are simple

first understand under roots mate, they're extremely basic

we'll get to ur question

now uh

tell me what the value of 3² will be

@restive river

ye

now if someone asks you

what the value of under root 9 is

what are you gonna say

@restive river

YE

now tell me what under root 25 is

think about it and answer accordingly

cal?

no

think of a number who's square is 25

YES

you got it

it's okay

do u see a pattern?

if √4=2

then if i square √4

what should i get?

im squaring an under root number

√4=2, is correct
now if i square √4 itself

so {√4)²

u square an under root

try

if √4= 2
(√4)² must be

√4×√4

no

ye

which is 4

so when u square an under rooted number

the under root gets removed

(√9)²= 9

now can you tell me what square of under root 25 should be?

YESSIRR

you're getting a hold of it

when you see an equation such as 7x+2=4

you cannot square ONLY left hand side
or ONLY right hand side

lmao

if you wanna square an equation, u would need to add squaring to both the sides

add power of 2

in the equation

okay let's say

√(7x+2)=4

is your equation

what i CAN'T do is, i cannot just square only one side of the equation

BLUD

don't use calculator till u turn 18+

or u'll be dumb as shit

[rustic yew]

.

we know that squaring a number will remove the unde rroot

so

[rustic yew]

this euqation looks very complicated correct

?

we'll square on both sides YESSS

so we can remove that √ sign from one side (LHS in this case)

after squaring on both sides we'll get

7x+2=4²

@sul_silly#0000

u're with me?

under root

ye

same thing

by squaring on both sides

back to your question now

√(5x+1)=4

this is your question

doesn't look so complicated anymore, does it?

so uh

OH WAIT

i thought u were talking about ur question

old up

hold

x= 2

what'd u get?

squared on both sides

will get you

7x+2=4×4

both the sides of the equation

i squared 4 (right hand side of the equation)

and i squared √(7x+2)

(left hand side of the equation)

uh?

ye

square on both sides

raise both sides to the power of 2

ye

do that

we covered the fact that squaring removes the under root

remember?

because solving the under root will be complicated

if there's an under root and solving it is out of your league, u gotta square both sides

under roots are complicated to find out, √3's value will be smt like 1.78

[upbeat pawn]

sqrt = 1/2

[rustic yew]

yes

4²=16?

write it

and then solve it like you would solve any normal equation to find out the value of x

[rustic yew]

might get too complicated for him that way

[upbeat pawn]

im afraid he would square both side if its cube root

[rustic yew]

ye now u have
7x+2=16

[rustic yew]

i doubt he'll get cube rooted questions anytime soon though

yes

keep the x terms on one side

switch 2, it'll be easier

always prioritise to keep the alphabetical/variable term on one side

yes

[upbeat pawn]

W

[rustic yew]

dubs

firstly, try the question given in ur book

question number 10

YESSIRRR

[upbeat pawn]

lmaoo bluds prob underage

[rustic yew]

he's 15

or she

ion know

[upbeat pawn]

loll

[rustic yew]

WHAT THE

u said u were 15

wyd on this app 😭🙏

[upbeat pawn]

its weird if he still stumble on basic algebra at 15

[rustic yew]

do NOT lie about ur age to a person trying to explain u maths

[rustic yew]

i thought that too tbh, i was worried about him

yes that is correct

you did it !!

congratulations

it's all you mate

i didn't do anything

[upbeat pawn]

u still got like 4-5 yrs chill

[rustic yew]

ah i see

[rustic yew]

let him cook now

[upbeat pawn]

aight

cook smtg good

[rustic yew]

he gonna cook smt good 🗣️

[upbeat pawn]

igcse?

igcse at 12?

tf

nah

[rustic yew]

what is IGCSE

why do i not know what it is

[upbeat pawn]

cambridge syllabus

exam

[rustic yew]

kinda like jee or gaokao?

[upbeat pawn]

nah gaokao way harder

isnt jee like uni stuff

15 taking igcse kindda early tho

[rustic yew]

u give it right after high school, same syllabus as gaokao

like at 18

[upbeat pawn]

igcse is like the final final final exam yk

[rustic yew]

very tough tho, and huge competition

im preparing for jee

[upbeat pawn]

mhm

jee is?

i heard people complaining its hard

[rustic yew]

joint entrance exam for indian colleges related to engineering

[upbeat pawn]

oh

[rustic yew]

syllabus is high school maths, physics and chemistry

[upbeat pawn]

so ur still high school?

[rustic yew]

ye

[upbeat pawn]

glgl to u guys

[rustic yew]

im 17

[rustic yew]

thanks man

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[devout snow]

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[restive river]

appointment procedure

[devout snow]

Please don't occupy multiple help channels.

[restive river]

ok but i need help

i dont know how to do the appointment procedure thing i really need help 😰

<@&286206848099549185>

[plucky maple]

Ask ur question

[restive river]

i dont understand how appointment procedure works

[cedar hare]

[restive river]

idk i used google translator

[median tiger]

helloo

any doubts?

[restive river]

is this math related?

and also send the original problem (if it is math related), even if it is in your language

[restive river]

yes

ok show the original problem

ill show the example thats on my book

okay

you are trying to solve a system of linear equations

thats the sentence for it, for your information

anyways what do you not understand

everything

okay i guess i will try to build up on what you know

[median tiger]

shall i help?

[restive river]

,align
3x+2y &= 40 \ y - 2x &= 6

[restive river]

if you know german, that would be great

[median tiger]

unfortunately not

[restive river]

its ok

[wraith patio]

what is the appointment procedure

was ist die deutsche version davon

[restive river]

its a system of linear equations

given the picture

but i will leave you to it

[wraith patio]

no continue i can't stay here for long

i'm just wondering where she got that word from and hoped i could clarify

[restive river]

oh okay yeah

that would be awesome

[restive river]

@restive river

maybe say it to them in german and clarify your problem. Timo would relay the info back to us in English

[wraith patio]

ok it's probably Einsetzungsverfahren

[restive river]

yes

[wraith patio]

she doesn't understand the process of isolating a variable to get an expression for it in terms of the other variable

[restive river]

yeah okay

thats fine

[wraith patio]

and then plugging that expression into the other equation so that it only depends on one variable now

[restive river]

gotcha

thanks Timo!

[wraith patio]

[restive river]

anyways @restive river let's have a closer look

[wraith patio]

ping me again if there's still confusion about something maybe i'm still here by then

[restive river]

so i will describe to you the issue and what we are trying to do: You are trying to find a suitable (x,y) that satisifes both the first equation and the second

[restive river]

alrighty

[restive river]

yes

[dry geyser]

You can ping me as well I can help translate too

[restive river]

but like

[dry geyser]

If Timo isn’t there

[restive river]

this start

(1) bla bla
(2) bla bla

okay so what exactly of the process do you not understand

i want to know

why does it changes position and why do we only calculate (2)

alright i guess you understand the why part so i will jump straight to the how

[restive river]

lets walk through it together

ok

you have y - 2x = 6

ok

,align
y-2x &= 6 \
y &= {???}

[restive river]

would you be able to tell me what the

??? are meant to be

try not to look at ur book's solution

uhh

idk

okay so let's see what we are trying to do

do you notice how the -2x disappears on the left?

yes

okay so we must have eliminated it from that side

+2x

yes

great

yay

so we add 2x both sides

[restive river]

so what is ??? now

12?

no!

oh

how did u arrive at that conclusion?

uhh

2•6

idk

why multiply? didnt we just say we are adding +2x to both sides instead?

y-0=6
2x-y =?

idk

.close

[devout snow]

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[restive river]

okay bye lol

[devout snow]

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[real pewter]

help

[real pewter]

ignore my answer but how would you do this?

the integral isnt something i can solve within whats expected in this course

derivative of xF(x) = xF'(x) + F(x)

[grand siren]

i think they might be expecting you to just write F(x) in the answer

[real pewter]

as in just F(x)? or an evaluated expression for F(x), as in the solution to the integral

[grand siren]

just F(x)

[real pewter]

how would that work?

js not sure how you got that

[feral agate]

here, you can calculate F'(x) using FTC

type out the rest as it is

[zenith vortex]

because cos(t)/t doesn't have an elementary anti derivative

[feral agate]

since you can't simplify the F(x) term in any way

[real pewter]

yeah i got that

[real pewter]

ohh okay i get what you mean

thanks

i thought you meant js F(x) on its own

and nothing else

its correct btw

[grand siren]

...oh
yeah that would be completely wrong lol sorry for the ambiguity

[real pewter]

no worries

.close

[devout snow]

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[restive river]

waits

hello

<@&286206848099549185>

..

[brittle tree]

which question

[restive river]

all of it :3

besides 2

[brittle tree]

for the first one you can use the property QV = 2/3 QU, PV = 2/3 PT and RV = 2/3 RS

for something like this

sane for all from 3 - 13

[restive river]

wait

for cj do i just divide cd by 2

[brittle tree]

first question/

??

[brittle tree]

you see we know that DC = (2/3) DJ right?

and DC + CJ = DJ

so we can derive that CJ = (1/3)DJ

get it?

you there?

[restive river]

yes

[brittle tree]

did you understand?

[restive river]

no

[brittle tree]

Did you get this part?

[restive river]

no...

[brittle tree]

well there is this property of centroid that it divides the median in a ration of 2:1

[brittle tree]

??

[restive river]

what

[brittle tree]

just a sec

Try this

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[devout snow]

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[stoic wasp]

can someone help me understand this process

I understand the first line, and then the second line factoring out the -1/2, and the substitution

then the integralis split up into two

I get how cos(2x +y) is integrated into 1/2 sin(2x+y)

I'm confused why cos(y) gets the cos(y) part integrated out and left with 1, but I can see how the choice makes sense, and how it becomes -2picos(y)

but I don't understand why 1/2 sin(2x+y) has become zero

[trail eagle]

This is integrated with respect to x, and when you plug in you'll have sin(something) - sin(something + 2pi*n), which is 0 since sin is 2pi- periodic

[trail eagle]

And this happens because cos(y) is independent of x, so when integrating with respect to x it's just a constant, so it's taken out of the integral.

[stoic wasp]

so you'll have 1/2 sin(o+y) - 1/2sin(2pi +y) but they both equal zero?

[trail eagle]

No, they'll both be equal to eachother, so they cancel out.

[stoic wasp]

because it's the same point on the graph due to the periodic nature?

[trail eagle]

Yes.

[stoic wasp]

Ohhh okay, that makes sense

[stoic wasp]

This makes sense too

[trail eagle]

In general, sin(x)=sin(x+ 2pi*n) for any integer n

[stoic wasp]

What would the n be in the context of the question?

would it just be 2?

Oh wait I get it,

.close

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[untold nova]

I did
a x c = -b
a x (a x c) = a x -b

Then applied vector triple product on LHS

(a.c)a - (a.a)c = i+j
4a - |a|²c = i+j

[violet wind]

Vector triple product?

[untold nova]

But I end up with 9/2

[untold nova]

3 vectors multiplying

[violet wind]

The triple product I'm familiar with is a • (b×c)

[untold nova]

that's vector dot product

Or scalar product

[violet wind]

well a•b is dot product yeah

[untold nova]

That gives you a determinant

Vector triple product has 1 in cross multiply

This

[thin fern]

You can also approach this by seeing that c has to be in the form (x+4, x, y) based on the fact that a dotted with c equals 4

And then compute a crossed with c and set that equal to -b

[violet wind]

hmm ok

I'll look closer at the problem then

[untold nova]

I would like to use triple product here tho, cuz then there are questions where i can't decode that

I did use a similar method before when getting the right answer now I don't recall how I did it 😃

[untold nova]

Like, I get ur point but I don't get why my method isn't working

[thin fern]

ah

[untold nova]

Most I could think was that I got -b x a wrong?

But I tried changing a few stuff and still couldn't get the answer

[violet wind]

I get -i+j-2k

[untold nova]

Yeah I did that too

[violet wind]

for b×a

[untold nova]

Still I got something like 11/2

[violet wind]

so i-j+2k for negative ig

[untold nova]

Tho why did u take the -ve away from b

[violet wind]

forgor

[untold nova]

Yup still got 11/2

[violet wind]

the error isn't just due to losing information somewhere right

I would expect doing a× to both sides to lose information generally but can't be sure

[untold nova]

Wdym by losing info?

[empty flame]

from first eq you can get $|\vec{a}$x$\vec{c}|=|\vec{b}|$

[woven radish]

calculus is fun

[empty flame]

where θ is the angle between a and c

this gives |a||c|sinθ=|b|

from 2nd one you get |a||c|cosθ=4

now square both sides of each eq and both to each other

[violet wind]

$x \mapsto a\cross x$ isn't injective

[woven radish]

thewizardofOU

[untold nova]

Oh

[violet wind]

there's infinitely many inputs for any possible output

[untold nova]

Yeah it's not but like

[untold nova]

Ik that but the thing is, what else can you do

Even if I try to put the vector product that u said, I won't get an answer

You can try using it

[violet wind]

I wasn't recommending using that

Actually it would have the same problem

[empty flame]

this gives $|\vec{a}|^2|\vec{c}|^2cos^2\theta+|\vec{a}|^2|\vec{c}|^2sin^2\theta=16+|\vec{b}|^2$

[violet wind]

I just was confused because when I heard triple product, that's always what was meant

[untold nova]

Yeah, so like we just have 2 choices and dot product won't work

[untold nova]

No I got that, I was just explaining why cross product was the way to go

[upper schooner]

($\mathbf{a}$ for bold btw)

[woven radish]

@upper schooner

[violet wind]

The fact that dot producting something with both sides doesn't work doesn't mean cross producting something with both sides has to work

Maybe it does, idk

[untold nova]

Yes that's true, but again what else can we do then

[woven radish]

calculus is fun

[empty flame]

tysm

[untold nova]

And I did similar questions with the same exact method idk why this isn't working

[violet wind]

well we know a×c and we know a

[untold nova]

Can we get theta?

[empty flame]

no need

[untold nova]

Oh yeah

Didn't realise that

[empty flame]

notice that you can take $|\vec{a}|^2|\vec{c}|^2$ as a common factor

[woven radish]

calculus is fun

[empty flame]

you have vecs a and b you can get their norms and then you can get |c|^2

[untold nova]

Yeah got that

[empty flame]

👍

[untold nova]

Thankyou

[empty flame]

np i didnt do anything

[violet wind]

if we plug in the components of c and do some linear algebra we can find the line this forces c to be on

[untold nova]

How will we plug the components of c?

[violet wind]

like, make up variable names for them

[untold nova]

I suppose we can

Tho it will get complicated

[violet wind]

You end up with 3 equations and 3 variables in the end

Which sounds nice but one of the equations should be redundant unfortunately

So instead we'll pull the 3rd equation from the fact that a•c = 4

[untold nova]

Fair

[devout snow]

@untold nova Has your question been resolved?

[untold nova]

.close

[devout snow]

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[bright juniper]

Need help with this

[devout snow]

@bright juniper Has your question been resolved?

[bright juniper]

<@&286206848099549185>

.close

[devout snow]

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[olive raptor]

for this q, I’ve started by assuming that the expression on the left is strictly lesser than 2

then have evolved it into (m^2 + n^2)/nm

but have no idea how to prove such

[glad shuttle]

Well if ur gonna use proof by contradiction u gotta assume that the opposite of what ur trying to prove is true

[olive raptor]

ah on I think I have it

actually no

[glad shuttle]

So in that case I suppose u assume there exists positive integers m and n such that that equations is less than 2

[olive raptor]

I have that down,

[glad shuttle]

I feel like this has to do something with odd and even numbers

[olive raptor]

ah ok, I’ll try that approach

[glad shuttle]

But maybe not cuz that might not really be proof by contradiction

[olive raptor]

if that reaches a conclusion that is false from my assumption, then wouldn’t it still be pbc?

although when using even number for m and n, wouldn’t it still be the same expression I’ve written down?

[glad shuttle]

Well an even number is just a number that can be written down as 2x an integer

An odd number is a number that can be written down as 2x an integer + 1

[olive raptor]

I’ve hit another wall, I’ve sort of got a fraction the same as before

[glad shuttle]

Hmmm

Maybe the thing u got was onto something

Like before

[olive raptor]

idk, I’ll give it some rest thanks for the help

.close

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[winged jasper]

Hello,

I feel like this is probably too much to ask, but I'm writing my first proofs while taking discrete math, and I'm hoping someone would be able to look one over and let me know if I'm on the right track. It's in written english mostly and is fairly long (just over 2 pages), so if there are any takers I've hosted a pdf copy on tiiny.site (an image hosting webservice) and can provide the link, or send the pdf itself if they'd prefer.

Thanks

[proud bridge]

about what

[winged jasper]

i was asked to write a proof for this:
The sum of the squares of 4 consecutive integers is an even integer.

[proud bridge]

understand

where it is

[winged jasper]

https://amaranth-may-36.tiiny.site/

that's the link to the web service hosting pdf

[proud bridge]

let me check it thanks

[winged jasper]

thank you

no rush I'll be here at the computer

[proud bridge]

@winged jasper You said somewhere that it is odd because it does not have "2" common factors. Actually, I think this is wrong.

It was 4x²+4x+1

[winged jasper]

i'll look at it again, but I meant that 4x²+4x+1 does not have a common factor of 2

[proud bridge]

Yes but

3x²+4x+1

also does not have

Bur when x=1

It is even

I think you should start from the fact that 4x²+4x is even

And when you add 1

To any even number

[winged jasper]

ok it's no longer

[proud bridge]

It is odd

Yes

exactly

[winged jasper]

right, that seems like a good clarification

[proud bridge]

I'm not finished yet

[winged jasper]

just avoid the factor thing

trying to change the wording to fix the inaccuracy you pointed out, something like

Suppose a is an odd integer, such that a = 2k + 1 for some integer k.
b = (a + 1) = (2k + 2). In b, the value being added to the even number 2k is a multiple of 2, so b is even.

(I think) that wording is actually saying what I'm trying to convey

by making it clear that 2k is an even number, and then the value added on top being even/odd

[proud bridge]

So it's like well

Or it can be expressed as 2(k+1)

[winged jasper]

because it has a common factor of 2

which circles back

oh sorry wrong reply

ignore that lol

[proud bridge]

ok np

[winged jasper]

but the 2k + 2 factors to 2(k+1)

[proud bridge]

We already guarantee that in such a case there is a even number

But it is not healthy to say that this is not even because they do not have 2 factors in common

[proud bridge]

I mentioned it with reference to the previous example

Like 4x²+4x+1

[winged jasper]

yeah

[proud bridge]

It's not like this for this

It's always odd

But what I'm trying to say is that there are many expressions that do not have 2 factors in common

[winged jasper]

ok maybe it's just the way i'm phrasing it, or I'm completely misunderstanding. All the numbers share the factor of 2, not that they have 2 factors in common

I'm a bit confused at how the quoted expression could ever come up in the context of this proof. As far as I know, an it is just odd integers that are being described as x = 2k + 1 for some integer k. In the context of an odd integer being squared, the expression would never be 3x^2 + 4x + 1, because 3 is not 2 squared, and any odd number fits into the definition 2k + 1 for some int k.

[proud bridge]

Yes

There is no such expression when squared

[winged jasper]

but I still shouldn't phrase it that way?

[proud bridge]

But now we are free from this incident

Actually, we will try to say that the square of an odd number is still odd

So it doesn't really matter how you do it

if the coefficient of all x terms is even

[winged jasper]

right

great, I appreciate your help Muhammet

.close

[devout snow]

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[light zinc]

hello

[light zinc]

does anyone know how to solve this cuz i cant find any tutorial in yt with this problem
the topic is about electfic charges coloumbs law in physics

[near stone]

there's a physics server listed in #old-network

[restive river]

Hmm

I can try to help

[light zinc]

ok thx

[restive river]

Let me solve it once first

[restive river]

its just vector addition

pretty much lol

only two things you should keep in mind are 1- like charges repel 2- opposite charges attract

and you can use that to determine ur charge directions

[light zinc]

but how can i start Fc on a if c is goin up

[restive river]

Now u can just put the values in Resultant of vector formula

[restive river]

We wont have to see that c is going up , we only have to calculate the force exerted on c by a and b

Because it said , "at the third charge"

[light zinc]

oh ok thx I thought therrs a trick to it

[restive river]

We just calculate the force exerted on it by all other charges

[light zinc]

ty im gonna close this now

.close

[devout snow]

Channel closed

Closed by @light zinc

Use .reopen if this was a mistake.

[restive river]

Ok

[devout snow]

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[light zinc]

wait can i ask this

[restive river]

?

[light zinc]

how did fq3 on q1 have right arrow on top if they dont attract each other

[restive river]

You mean Fq3q2 or Fq3q1

Wait

[restive river]

Do u mean the red arrow?

Or

[restive river]

Ignore all the other arrows they are not forces , only green and Red arrow is force

The green arrow

Green arrow , shows force of a on c , as both of them are positive they repel

Red arrow shows force of b on c , as both of them are opposite charges , so they attract eachother

What you have to do is

You have to fix the charge which was asked

Then think of forces on it by other charges

If u wanna ask more then .reopen or itll be gone

Any second now

[restive river]

Ignore all the other arrows they are not forces , only green and Red arrow is force

R→ is the Resultant force from green + red

[light zinc]

wait not thay

[restive river]

?? Which

[restive river]

I meant to write F(on q3 due to q2)

I added the arrow on top of F because

It shows vector form of force

And not magnitude