help-19

[autumn flare]

if there are only two vectors

but it would not reach each and every place

am i wrong or right?

[forest sky]

well it is an a plane in 3d space

but yes

[autumn flare]

plan in 3d space...

hmm

its like if i flip the 2d plane and make it vertical

is that how it would look like?

In r3, two independent vectors also span a flat 2D surface —a plane — but that plane might be “tilted” or “vertical” inside the 3D space.

its a flat 2d surface + a 3d infintely long plane that is tilted or vertical based on the a and b value varied

[autumn flare]

thanks a lot for your effort

.close

[odd edge]

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[idle spear]

I'm continuing on an assignment for direct proofs and quite stuck on how to get n=(a+b)(a-b) into the form n=2k+1 for all k in Z.

The proposition is that every odd number can be expressed as a^2-b^2.

I have noticed that when b = a - 1, a^2-b^2 seems to iterate through odd numbers. The result of evaluating any particular a seems to be n = 2a - 1, but I'm not sure how to prove that algebraically for all integers.

[idle spear]

So as it stands now, the P side of the proof is basically at

Suppose n is odd.
n=2k+1.

And the Q side of the proof is at

Let a = k+1.
n = (2a - 1)(a - a + 1).
Let b = a - 1.
n = (a + b)(a - b).
Therefore n = a^2 - b^2 for every odd n.

I also have various fragments like solutions for the roots of the polynomial like a = sqrt(...), b = sqrt(...), and like a = (2k+1)/(a+b) + b

I'm a little unsure how to bridge the gap between intuition and a formal proof. "Let b = a - 1" feels like a deus ex machina or non-sequitur that maybe needs justification?

Is it sufficient to make a direct proof by pulling magic numbers from a hat and demonstrating that the proof is satisfied, or am I missing a tool (besides observation) to demonstrate where they come from?

[odd edge]

@idle spear Has your question been resolved?

[quasi falcon]

For a^2 - b^2, either a or b exclusively have to be odd. Based on the fact that:
odd . odd = odd
even . even = even

even + odd = odd
odd + odd = even
even + even = even

Order doesnt really matter that much

so, for the sake of ease, its pretty easy to prove that if:
a = 2m
b = 2n+1
then a^2 - b^2 is of the form 2k + 1

tho, the problem becomes proving that the the quotient of the difference of squares can produce all integer numbers k

[idle spear]

I think I have the beginnings of... something that spans the two ends of a direct proof, but I don't like how it gets there with some arbitrary let statements and injecting an a - a + 1 term:

Suppose n is odd.
n = 2k+1.
Let a = k+1.
n = 2a - 1.
Let b = a - 1.
n = (2a - 1)(a - a + 1) = (a + b)(a - b).
(a + b)(a - b) = a^2 - b^2
Therefore n = a^2 - b^2 for every odd n.

[quasi falcon]

It probably is correct, imma head off to sleep though, good luck 🥀

[idle spear]

No worries, thanks. I was also stuck on proving every odd integer satisfied the form.

Closing for now, as I'm also off to bed

.close

[odd edge]

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[static patrol]

Hello

[static patrol]

How did they get this new denominator in the numerator

[wooden python]

dear god... this typesetting's making it kinda hard to figure out wtf is meant

are you asking abt this

if you are: it looks like an application of l'hôpital.

[static patrol]

Ann!

I haven't seen you in a long long time

Man if you were close, I'd peck you on the forehead

You don't know me, but you helped me a lot two years ago on this discord server

[stable zealot]

anybody wants help?

[static patrol]

I want help

[wooden python]

uhhhhh yeah i dont remember you at all lmao

that was mildly weird.

2 years ago...

yeah sorry nothing is coming up

[bronze canyon]

ann has fans apparently

[wooden python]

well i had one "fan" who fucking stole my pfp and continues to do so (whom i witnessed in #chill...)

[wooden python]

i mean... read what i wrote yeah?

& respond to me

[static patrol]

Sorry

I went to take another photograph

They haven't applied Lopital to the denominator.

[wooden python]

yes they have actually

the denominator is h*log(STUFF)

the new denominator is log(STUFF) + h * (STUFF')/STUFF

and most importantly of all

[static patrol]

I missed the h.

got it, thanks

Take care of yourself, Ann.

Bye!

.close

[odd edge]

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[static patrol]

Yes, that is the big hint that I missed

😆

.close

[odd edge]

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[turbid warren]

I know this is false from use of venn diagram

[turbid warren]

but

how can I prove otherwise

[wooden python]

cook up some three sets for which there is in fact a difference

demonstrate it

[turbid warren]

oh so just find one counter example

?

ill give it a go ty

[wooden python]

use your venn diagram to figure out which regions definitely need to have sth in them

[turbid warren]

i think this works

if i let A = {1,2}

B = {2,3}

C = {3,4}

Im essentially seeing something in C that isnt in A right?

and so the LHS becomes {2}

RHS is {2,3,4}

which isnt true

[errant dew]

That works, you could also have A and B be empty sets and C be non-empty

For the “smallest” counter example

[turbid warren]

this was the aim right?

[turbid warren]

oh i see

oh thats neat icl

just thought abt it

[errant dew]

The simpler the counter example, the easier the verification is

[turbid warren]

for this one, can we just say intersection is associatve?

wait no, i know its true so there would have to be some sort of rigour involved

alright what am i even looking at here

wait lemme just thinka abt it then come back to it

[odd edge]

@turbid warren Has your question been resolved?

[odd edge]

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[warm bridge]

just need somone to check if this is right

[odd edge]

@warm bridge Has your question been resolved?

[warm bridge]

no and im gonna die if i dont get an answer in the next 5 min

<@&286206848099549185>

[wicked olive]

Are you alive?

[warm bridge]

its been 4 min so yes

now i am

[wicked olive]

All good

Its correct if im not wrong

And im not wrong

Good job mate

[opaque hearth]

that problem looks cool

[wicked olive]

Yeah

[errant dew]

If the helpee is dead do we close the help channel?

[warm bridge]

NO its not cool its making me rip my hair out

i hate physics

yes i came back to life just to say that

[turbid warren]

this is mechanics, dw, nothing w phys

[warm bridge]

also thanks guys

[turbid warren]

well no theory anway

[warm bridge]

well uhh this is from a book called univeristy physics

and im taking a physics course

its included

[turbid warren]

yea looks close to A level mechanics

[turbid warren]

oh nice nice

gl

[warm bridge]

thank you

[odd edge]

@warm bridge Has your question been resolved?

[opaque hearth]

ah i used to use that

my university uses matter and interactions

really unique approach... like you start off with vectors then newton's laws, momenum, universal gravitation

[odd edge]

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[warm bridge]

.reopen

[odd edge]

✅ Original question: #help-19 message

[warm bridge]

yes we did this

but our prof is so

insufferable

like

bad teaching style (at least for me and my mates) and keeps bragging about his phd and how physics is just better than math

💀

skull emohi

[odd edge]

@warm bridge Has your question been resolved?

[odd edge]

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[random helm]

hello, i am working on this problem, but i feel like something has gone wrong/my understanding of how triple integrals work isnt correct, mainly because in the final point that i am up to, i am getting an integration that will give 0, after cos changes to sin, limits of 2 pi and 0 go in and we get 0 for both of them

this is the content that was shown just before this example, so i am pretty sure that this is the way in which the question is meant to be solved

[odd edge]

@random helm Has your question been resolved?

[random helm]

<@&286206848099549185>

[nocturne jetty]

What is the issu

[random helm]

i would just like someone to help me understand where i have gone wrong

[nocturne jetty]

Which one of all is the exercise

[random helm]

the first image i sent is the question

[nocturne jetty]

this?

[random helm]

yes

[nocturne jetty]

okey Okay, let me read it

[nocturne jetty]

why do you add those numbers and where do you get them from

[random helm]

which numbers?

[nocturne jetty]

0.056...

[random helm]

ah yes

so you see the integral with "M"

that represents the missing parts

so the 2 vertical columns that are missing (from what would otherwise be make the object a complete cuboid instead of an I shaped object), either side of the central "web", they make up the 0.056

and the other integral with "C" represents the circular sections of the object that are also missing

[clever fjord]

ρ♗☾♄☋_➊➋➌➎

[random helm]

@nocturne jetty the part I am using polar coordinates for is circular, it is the missing circular sections

and I believe that the intended way to complete the question is with integration since that is what was taught just before this example

[nocturne jetty]

Okay, let me do it with integrals

[random helm]

thanks

I have to go for tea now @nocturne jetty will be back in around 20-30 minutes, so i wont be able to respond to anything until then

[nocturne jetty]

I need to know where you get those constant numbers from

[nocturne jetty]

okey

[random helm]

which constant numbers?

[nocturne jetty]

385042,5
0.056...

Where do those numbers come from?

[random helm]

385042.5 is from taking out the constants of g and sigma and 5 out of the integration and multiplying them together

i took 5 out because the object is just 5 identical parts stuck together, each containing 1 hole and having length of x=0.865

ill explain 0.056 after

i gtg

ah wait no @nocturne jetty i realised just now that splitting it up into 5 sections isnt possible

because they are not identical

we are calculating the gravitational torque, so the contribution from each of the 5 sections is not equal

I will try it again after tea

[odd edge]

@random helm Has your question been resolved?

[random helm]

@nocturne jetty <@&286206848099549185> Here I have attempted it again with a different approach that you suggested, but i am still unsure about how to go about the triple integral of the holes?

[random helm]

<@&286206848099549185>

[random helm]

<@&286206848099549185> is anyone able to help me with this please?

[odd edge]

@random helm Has your question been resolved?

[fallow pond]

I've arrived.

[random helm]

integrand is gsigmax

g times sigma times x

[fallow pond]

It's not sigma?

Isn't that alpha?

Ohh nevermind.

Read it wrong.

[fallow pond]

What is g?

[random helm]

9.81

gravity

and sigma is given in question aswell

[fallow pond]

Isn't that just a constant.

So can't you take g out of the integral.

[random helm]

yeah i could take it out but its ok

my main issue is with the circles

you see the 5 circles in the original question?

i want to integrate this integrand over these regions

[fallow pond]

Do you know polar coordinate integration.

[mystic saffron]

Why don't you integrate the simple planar regions and then subtract the repeated circular regions?

[fallow pond]

It's usually more effective for circles.

[random helm]

wouldnt that mean doing 5 seperate integrals?

but maybe thats simply the only way

[random helm]

i think that is what i have done if i am not mistaken

[olive needle]

That's what is implied here.

[fallow pond]

Right but you simplify it a whole lot more right, the limits are now only the rdrdtheta.

[mystic saffron]

hmm, that should've worked in that case

[fallow pond]

I don't think polar works here either way because we're in 3d.

Isn't cylindrical the way to go?

[random helm]

ah possibly

[mystic saffron]

cylindrical should work you just have to account for that x

[random helm]

ok let me remind myself of cylindrical and i will give it a go

thanks

but @fallow pond can i ask why wouldnt polar work here

[fallow pond]

Polar Coordinates are only 2 dimensional right.

[random helm]

oh ok

[fallow pond]

Yeah.

This is a tough problem nonetheless tbh.

[random helm]

so you couldnt integrate for example wrt r then wrt theta then wrt z?

[fallow pond]

Very application driven.

[random helm]

with z being in the direction of the height of cylinders

[fallow pond]

Yeah this is cylindrical.

rdr, dtheta, dz

[random helm]

oh right

so does the z even change?

[fallow pond]

The limits for z should be the same as height is given.

I believe.

[random helm]

yes i seem to remember z=z for the conversion

[fallow pond]

Oh I see.

[random helm]

i mean that would make sense i think

[fallow pond]

Try it and see what happens.

[random helm]

yes

thank you very much @fallow pond

[fallow pond]

Good luck!

[random helm]

@fallow pond

there will be 5 integrations for the 5 holes

and these 5 integrations should in theory produce 5 different results

with the hole furthest away from x=0 having the highest result since it produces the greatest torque

but with this integration with these limits it is going to produce a result of 0

because this integration calculates the torque produced

but these limits imply the circle is centred at x=0

and therefore doesnt produce any torque

[fallow pond]

I see.

[cinder tundra]

Dang calc 4

Hot

[odd edge]

@random helm Has your question been resolved?

[odd edge]

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[pastel pasture]

hi

[pastel pasture]

i need some help in learning mathematical induction... i know the base case and the inductive step, inductive hypothesis my struggles is with the algebra and proving the answer

<@&286206848099549185>

[stoic cloud]

!15min

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[jagged maple]

have you proved yet the cases for n=1 ?

[waxen talon]

induction got 3 steps

[pastel pasture]

yes that isn't the hard part,
for example in 1) it would just replace every occurence of n with 1

[waxen talon]

1. proving the base case

2. assume the formula is correct for some $k \in \mathbb N$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

3. prove it's correct for $k + 1$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

this step troubles you the most right?

[pastel pasture]

well

yes

normally step 2 is after step 3

[waxen talon]

for this step, this should not trouble you

[pastel pasture]

but yes step 2 is the struggle

[waxen talon]

replace every occurence of $n$ by $k$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

as in n = k + 1

[waxen talon]

it's like you do in step 1, but you replace $k$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

every occurence of n is replaced with k

[waxen talon]

this one is important

[waxen talon]

now if you want to prove for this

[pastel pasture]

yes thats also pretty straightforward

[waxen talon]

for example: 2 (i)

so if you replace $n$ by $k$, what do you get?

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

or that's called substitute

[pastel pasture]

3k = 3k(k + 1)/2

[waxen talon]

and that is $3 + 6 + 9 + ... + 3k$ right?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

yes

[waxen talon]

now if instead, you substitute $k + 1$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

the expression should look like this:

$3 + 6 + 9 + ... + 3k + 3(k + 1)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

do you see something?

[pastel pasture]

yes ok

the first 3k i put

isn't there anymore?

[waxen talon]

no no

[waxen talon]

let me add in a parentheses so it's easier to visualize

$(3 + 6 + 9 + ... + 3k) + 3(k + 1)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

do you see it now?

[pastel pasture]

yes one 3k is seperated from the other 3k?

[waxen talon]

...

[pastel pasture]

so like they are on different sides

[waxen talon]

it's this expression

and we know what value it is

[pastel pasture]

ohhh ok yea i got it

[waxen talon]

so that's where you substitute

[pastel pasture]

the algebraic part how does that come in

as in the inductive step

[waxen talon]

substitute first

[waxen talon]

what do you mean?

[waxen talon]

so basically, this is $P(k)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

and this is $P(k+1) = P(k) + 3(k + 1)$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok

[waxen talon]

so try substituting the value of $P(k)$ in

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

3 + 6 + 9 ...+ 3k substituted into = P (K) + 3(k + 1)

[waxen talon]

yes

and $P(k) = \frac{3k(k + 1)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

see theres where my problem lies where do i start

how do i substitute this

do i expand

[waxen talon]

$P(k + 1) = \frac{3k(k + 1)}{2} + 3(k + 1)$

[pastel pasture]

3k multiple k and multiply 1

[waxen talon]

it's something like this

[pastel pasture]

3k^2 + 3k/2 + 3k (k + 1)

[waxen talon]

don't expand yet

bring the second term with the denominator 2 first

[pastel pasture]

k+1/2

[waxen talon]

$3k(k + 1) = \frac{6k(k + 1)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

jesus

[waxen talon]

that's no problem

[pastel pasture]

ok

where did 6 come from

did u add it

did you put everything over two?

[waxen talon]

does your expression look like: $P(k + 1) = \frac{3k(k + 1)}{2} + \frac{6(k + 1)}{2}$?

[waxen talon]

3 x 2 = 6

[waxen talon]

yes

[pastel pasture]

Ohhhhh

u put both expressions together

[waxen talon]

lel

[waxen talon]

yes

[pastel pasture]

6k ( k + 1) /2 ok

[waxen talon]

wait actually

remove the $k$

[clever fjord]

1 divided by 0 equals Infinity

1 divided by 0 equals Infinity

[waxen talon]

my bad

that should be your $P(k + 1)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

but the idea is the same, bring the same denominator

[pastel pasture]

could it still be 6 since it has a k or would it be different?

[waxen talon]

it's still 6

basically timesing by $\frac{2}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

mmm ok i wouldve thought it wouldnt work

6k(k+1)/2

[waxen talon]

...

okay

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

your expressions should look like this

[pastel pasture]

no it doesnt so we have do SIMPLIFY

[waxen talon]

and do some fraction additions with factorization

[waxen talon]

our target is to bring to something like this

$P(k + 1) = \frac{3(k + 1)(k + 2)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok

[waxen talon]

because we want $P(n)$ to be true for $n = k+1$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

so from here

[waxen talon]

we want to turn it into something like this

[pastel pasture]

so ok can we do like terms here?

[waxen talon]

do factorization

[waxen talon]

it's technically like terms lel

what do you get?

[pastel pasture]

so 3k(k+1)(k+1)/2 + 3

i may of

been so off

ngl i have no idea how to factorize mb

[waxen talon]

💀

waait wait

did your $P(k + 1) = \frac{3k(k + 1) + 6(k + 1)}{2}$?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

no

it didnt

[waxen talon]

isn't that just adding with the same denominator?

show me your working

[pastel pasture]

see um

i just been kinda

winging the working and trying to get the steps cuz ngl the test is tomorrow

[waxen talon]

dang

[pastel pasture]

i get everything except this topic

and the algebra the factorization

[waxen talon]

but you gotta show your working so i can help fixing for you

[pastel pasture]

idk what is required im trying to figure out my methodology

ok hold on

3k^2 + 1 + 6k + 6/2

[waxen talon]

don't expand yet

[waxen talon]

there's a common factor here

[pastel pasture]

k + 1

[waxen talon]

so factorize

[pastel pasture]

3k (k+1) + 6 / 2

WAIT

[waxen talon]

boom

mistake found

[pastel pasture]

3K (K + 1) 3 / 2

[waxen talon]

$(k + 1)(3k + 6)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

add a denominator too

[pastel pasture]

so (k + 1) (3k + 6) / 2

wait... its cooking..

[waxen talon]

yes

[pastel pasture]

there may be hope

[waxen talon]

3k + 6 got a common factor

[pastel pasture]

oh mb

3

[waxen talon]

COOK IT

YES

FACTOR 3 OUT

[pastel pasture]

so k + 2

[waxen talon]

so what's your result

[pastel pasture]

(k + 1) (k + 2)

[waxen talon]

3(k + 1)(k + 2)/2

you cooked

🔥

[pastel pasture]

Woooo!!

wait

[waxen talon]

do you want to know a second way

that you don't need to do any work?

[pastel pasture]

where did the 3 come from

[waxen talon]

3k + 6 = 3(k + 2)

not k + 2

[pastel pasture]

oh ok got you

[waxen talon]

?

[pastel pasture]

yes if theres a 2nd way that is a bit easier for e

me

[waxen talon]

remember this?

[pastel pasture]

i need like some steps or some type of methodolgy my brain works like that

yes

[waxen talon]

so for 2 (i), factor 3 out and copy what they do 💀

[pastel pasture]

ok fair enough idk if that'll be on the test

[waxen talon]

some methods huh?

okay

[pastel pasture]

like after the inductive step like some visual cues like if i see this then do that

like a flowchart

of some sort

[waxen talon]

do 2 (ii)

[pastel pasture]

ok

so lets start w the base case

[waxen talon]

if you see a series of sum like exercise 2

[pastel pasture]

im writing as i go along

[waxen talon]

for example 2i

the sum can always be written in the form of $P(n) = P(n-1) + \text{something}$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

okay

[pastel pasture]

ohh ok got it

[waxen talon]

step 1 and 2 is easy enough

also

[pastel pasture]

is it (3(1) - 2) = n ( 3(1) -1)/2

[waxen talon]

set the function like they do

[waxen talon]

you left out an n lol

[pastel pasture]

mb

and then n = k

[waxen talon]

cool

and prove it's correct for n = k+1

[pastel pasture]

3k - 2 = k(3k -1) /2

[waxen talon]

yes

[pastel pasture]

where do i put the k + 1

[waxen talon]

$3k - 2 = \frac{k(3k - 1)}{2}$

[waxen talon]

wait actually no

[waxen talon]

take a look at the expression

the left hand side does not only have $3k - 2$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

they have more expressions

[pastel pasture]

1 + 4 + 7 + ... + 3k-2 = k(3k-1)/2

[waxen talon]

now that's correct

[waxen talon]

let $P(n)$ be $1 + 4 + 7 + ... + 3n-2$

[pastel pasture]

ok

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

now to the gritty

[waxen talon]

so $P(k)$ is this

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

you remember this?

[pastel pasture]

i am TRYING to prove P(K) = 3n - 2

[waxen talon]

no no no no no no n o no no

[pastel pasture]

oh

[waxen talon]

you need to prove $P(k + 1) = \frac{(k + 1)(3(k + 1) - 1)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok

prove the LHS = RHS

[waxen talon]

or simplify it: $P(k + 1) = \frac{(k + 1)(3k + 2)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

oh you actually know the notations

[pastel pasture]

Yea man i pay attention

ish

[waxen talon]

remember this

$P(k + 1) = P(k) + (3k - 2)$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok got it

[waxen talon]

what's $P(k)$?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok

[waxen talon]

you just substituted $k$ in

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

3k - 2?

[waxen talon]

this is $P(k)$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

1 + 4 + 7 + ... + 3k-2 = k(3k-1)/2

[waxen talon]

k(3k-1)/2

[pastel pasture]

so the full expression ?

okk ok

[waxen talon]

this one is P(k)

[pastel pasture]

k(3k -1) / 2

got it

so stick with the RHS

[waxen talon]

so that you can substitute in here

[waxen talon]

yeah basically

the LHS is too complicated

so you stick with the RHS so it's easier to work with

the LHS is to turn into P(k)

[pastel pasture]

k ( 3k - 1) / 2 + (3k -2) / 2

[waxen talon]

try bringing to the same denominator

[pastel pasture]

Oh im seeing a trend nowww

ok

[waxen talon]

wait wait

[pastel pasture]

well shit

[waxen talon]

you multiply by 2/2, not 1/2 XD

k ( 3k - 1) / 2 + 2(3k -2) / 2

[pastel pasture]

oh mb

yea

ur right

[waxen talon]

okay

add the fractions

[pastel pasture]

could u do the magic thingy

so i can see how that looks

[waxen talon]

$\frac{k(3k - 1)}{2} + \frac{2(3k - 2)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

can i just ask where the 2 came from sorry

2(3k -2)

[waxen talon]

uh

multiply by $\frac{2}{2}$?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok so u can simplify it easier

so u can do that?

[waxen talon]

ye

[pastel pasture]

u can just add a two

[waxen talon]

that's basically multiplying by 1

[pastel pasture]

fair

[waxen talon]

notice that 2/2 = 1

so i multiplied with a 1

[pastel pasture]

k(3k - 1)(3k -2) / 2

[waxen talon]

wait wait

[pastel pasture]

did i mess u

[waxen talon]

too quick

[pastel pasture]

ok

[waxen talon]

step by step

should be $\frac{k(3k - 1) + 2(3k - 2)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

and try to work around with this

[pastel pasture]

so from there

2k (3k -1) (3k -2)?

idk from this step do i use my brackets and multiply

[waxen talon]

wait wait

notice how 3k - 2 can be seperated into 3k - 1 - 1

[pastel pasture]

yea

[waxen talon]

so you can seperate $2(3k - 2) = 2(3k - 1 - 1) = 2(3k - 1) - 2$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

pulling the -1 out of the bracket

[pastel pasture]

my brain

ok

[waxen talon]

visualize this

$a(b + c) = ab + ac$ right?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

correct

[waxen talon]

that's expanding

so here, $a$ is $2$, $b$ is $3k - 1$ and $c$ is $-1$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

2 ( 3k - 1) 2 ( 3k -2)?

[waxen talon]

wait a moment

[pastel pasture]

i need 5 minutes ngl my brain is going in 360s

do you mind

[waxen talon]

alr

[pastel pasture]

ok im back

[pastel pasture]

needed an orange talk to me

[waxen talon]

remember our target is to turn into this

[pastel pasture]

we are here

[waxen talon]

but we want to turn into this

so i think we should find a way to get the 3k + 2 factor

[pastel pasture]

ok u mentioned 3k - 2 is the same as 3k -1 -1

[waxen talon]

,w simplify k(3k - 1) + 2(3k - 2)

[clever fjord]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[waxen talon]

wait

i think i calculated wrong or smth

[pastel pasture]

ok phew

[waxen talon]

$1 + 4 + 7 + ... + 3(k + 1) - 2$ that should be your expression

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

of $P(k + 1)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

or $P(k + 1) = P(k) + 3(k + 1) - 2$

yeah i see where im wrong

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

it should be like this

[waxen talon]

not this

[pastel pasture]

so k(3k -1) / 2 + 3(k+1) -2?

[waxen talon]

yeah

remember

$P(k + 1) = 1 + 4 + 7 + ... + 3k - 2 + (3(k + 1) - 2)$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

the next term is $3(k + 1) - 2$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

NOT $3k - 2$

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

ok lets simplify

[waxen talon]

rewrite the expression

$\frac{k(3k - 1)}{2} + \frac{2(3(k + 1) - 2)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

this should be it

and if you are asking why is there a 2, you should understand lol

[pastel pasture]

so k(3k -1) / 2 + 2 (3(k+1) -2/2?

[waxen talon]

i just did the magic thing for you lol

[waxen talon]

for the right term, the parentheses looks yucky

[pastel pasture]

so this is the step what do i do here

add them?

[waxen talon]

yeah

[pastel pasture]

k (3k - 1) + 2(3(k +1) -2)/2

well im wrong

[waxen talon]

now for $k(3k - 1) + 2(3(k + 1) - 2)$

[pastel pasture]

hold on

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

no no

you're correct

you see the $2(3(k + 1) - 2)$?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

(3k + 1) -2)?

[waxen talon]

that looks kinda yucky 🤮

[waxen talon]

yes

[waxen talon]

$3(k + 1) - 2$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

not $(3k + 1) - 2$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

so anyways $3(k + 1) - 2$ can be simplified

can you do it?

[clever fjord]

1 divided by 0 equals Infinity

[pastel pasture]

(k + 1)

[waxen talon]

3k + 1

3(k + 1) is 3k + 3, not k + 3

[pastel pasture]

yes for sure

[waxen talon]

so what does this look like?

[pastel pasture]

k(3k -1) + (3k + 1) / 2

[waxen talon]

where's the 2

[pastel pasture]

good?

[waxen talon]

k(3k -1) + 2(3k + 1) / 2

...

[pastel pasture]

Ohhh the -2 goes to front...

https://tenor.com/view/231-gif-25095509

[waxen talon]

whatever

[pastel pasture]

now we simplify?

[waxen talon]

let me do the magic thing

$\frac{k(3k - 1) + 2(3k + 1)}{2}$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

this is what we want right?

notice that we want to have a $3k + 2$ factor in the fraction

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

and there are $3k - 1$ and $3k + 1$

[pastel pasture]

.

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

so what do we do with them

[pastel pasture]

seperate them

(3k - 1) (3k + 1)

or maybe just add them

[waxen talon]

yes

[pastel pasture]

this gives u 3k + 2?

i am a bit overworked but u helped a lot i will take this info and in this channel and go from there im kinda sleepy thank you so much

ima find my methodology

[waxen talon]

$3k - 1 = 3k + 2 - 3$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

do you see it now

[waxen talon]

same for $3k + 1$

[clever fjord]

1 divided by 0 equals Infinity

[waxen talon]

try it

[waxen talon]

and just like what i told you here

[odd edge]

@pastel pasture Has your question been resolved?

[odd edge]

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[latent hemlock]

Find the average number of people who picked any particular number. I am trying to understand what this means for a programming problem.

[latent hemlock]

There are 10000 people who picked a four digit number from 0000 to 9999

Does it mean I need to find the unique numbers then divide the people by that?

[chilly ridge]

is it not just 1...

[latent hemlock]

thats why im confused lol

I think it means to find the number of unique entries and then use that to see how many people on average picked a number

[odd edge]

@latent hemlock Has your question been resolved?

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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[plain coral]

P is a randomly chosen dot that can be inside or outside of the rectangle ABCD

how can i prove it ?

[odd edge]

@plain coral Has your question been resolved?

[plain coral]

<@&286206848099549185>

[outer hinge]

hint is it has something to do with the fact ABCD is a rectangle

[odd edge]

@plain coral Has your question been resolved?

[plain coral]

<@&286206848099549185>

[nocturne jetty]

What is the problem?

[nocturne jetty]

.

[plain coral]

i want the proof of the equation

[nocturne jetty]

Let me check it out.

[tall ivy]

Maybe try congruency?

[nocturne jetty]

“Hey, I was checking that equality you posted:
𝑃A^2 +𝑃C^2 =𝑃B^2 +𝑃A^2
, and in general it doesn't hold. It only works if the diagonals of the quadrilateral are perpendicular. If they aren't,

[plain coral]

i saw it in a book

and

PB^2 + PD^2 *

i also got the resault that it would be ez to proof if they were perpendicular

but the book says it works for every rectangle

[nocturne jetty]

Yes, but the diagonals have to be perpendicular to each other.

[nocturne jetty]

If they are perpendicular everything is perfect

[plain coral]

so u mean the book is wrong here ?

[nocturne jetty]

Since it is a rectangle there is actually a 90 degree angle so they would be perpendicular

[inner swift]

what

[nocturne jetty]

So you make some sense.

[inner swift]

@plain coral try dropping perpendiculars

from P onto the sides / their projections

assume some lengths, you should see stuff cancel when you pythagoras and add

[nocturne jetty]

That's what I said

[inner swift]

no

what

you first refuted the question by saying that it holds only if diagonals of quadrilaterals are perpendiculae to each other

that does not equal dropping perpendicular

[nocturne jetty]

Well, I'll let you check it out then.

[inner swift]

@plain coral any progress?

[plain coral]

U mean something like this?

Is AE^2 = BF^2 ?

And the same for CF and DE ?

If AE^2 = BF^2 and CF^2 = DE^2 then it's solved

@inner swift

[wary fox]

Ok so im guessing youre done then?

[plain coral]

yea ig

[wary fox]

Well you can close the channel if you dont have any other questions

[plain coral]

.close

[odd edge]

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[light hawk]

did i do this correctly:

[weary pelican]

By u' and v' you mean the gradient?

[light hawk]

yes

[weary pelican]

Ok the quotient rule is valid for gradients so that's fine

[light hawk]

do you know if the final answer okay? I didn't think id be able to solve for w because there are so many vectors and matrix division isn't defined

[odd edge]

@light hawk Has your question been resolved?

[light hawk]

<@&286206848099549185>

[odd edge]

@light hawk Has your question been resolved?

[light hawk]

<@&286206848099549185>

[odd edge]

@light hawk Has your question been resolved?

[light hawk]

<@&286206848099549185>

[amber veldt]

my recommendation is

to write

$$\langle w^\top C w \rangle = \langle w, w \rangle_C$$
$$w^\top w = \langle w, w \rangle$$

and calculate the quotient rule , relying on the produt rule, without ever writing down the letter "C". once you get the answer, you can plug C back in

[clever fjord]

gfauxpas

[amber veldt]

I think it will be easier that way

C is a constant matrix I am assuming

if C varies with time its a bit more complex

but not much

[light hawk]

so does this mean the method i used is wrong?

[amber veldt]

idk, but it's too long for me to want to check

and annoying to read with all the symbols

[light hawk]

ok. im trying to understand, does this mean set u = <w, w>c,
v = <w, w>?

[amber veldt]

no, it's just another notaztion for multiplying

I just would rather not deal with "C" as a variable at all, it makes it easier to see your work

[amber veldt]

it's like how in calc i we sometimes introduced variables to make the chain rule cleaner, remember?

D_x [ sin(log(x+1))] = D_x [ g(u(v))] = dg/du du/dv dv/dx

also I see I made a typo

I meant

$$w^\top C w = \langle w, Cw \rangle$$

[clever fjord]

gfauxpas

[amber veldt]

ah wait i think i see what you did

you replaced $C$ with $\frac{C+C^\top}{2}$?

[clever fjord]

gfauxpas

[light hawk]

.close

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[static locust]

so i have k(x) -|-2x + 1|/2 + 5

[static locust]

what i did here i did

|-2x +1|/2 = 5

how do i get rid of the 2

[earnest radish]

Can anyone help me with number 10

[quasi sparrow]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[quasi sparrow]

use definition of absolute value function

,tex .abs def

[clever fjord]

riemann

[static locust]

x >/ 0

because its absolute value

[quasi sparrow]

what are you saying exactly with
x > / 0

/ is often used to represent division

[static locust]

i found how all ihad to do was put -5 on the other side of the equality

.close

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[umbral epoch]

why is this wrong

[umbral epoch]

here is my work

<@&286206848099549185>

wait

nvm lol

silly billy me i plugged in wrong pts

.close

[odd edge]

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[umbral epoch]

.reopen

[odd edge]

✅ Original question: #help-19 message

[umbral epoch]

what did i do wrong

i did Duf = fxcos + fysin

[mystic saffron]

do yk what the goal is?

[umbral epoch]

of?

[mystic saffron]

the problem 😭

[umbral epoch]

yes???

id hope so

[mystic saffron]

i would too 😭

[umbral epoch]

not even 2nd grade reading level right there

do you know how to get the answer?

[mystic saffron]

ofc

well first u want to find the directional derivative

[umbral epoch]

dont pmo

[mystic saffron]

u want help tight

right

[umbral epoch]

i do want help from someone who knows what they're doing

[mystic saffron]

ik what im doing

[umbral epoch]

okay

continue

[mystic saffron]

😭

ok first u want to find the directional derivative

[umbral epoch]

[mystic saffron]

so df/dx

and df/dy

[umbral epoch]

i did that

[mystic saffron]

what would df/dx be

[umbral epoch]

i got fx = 2x

[mystic saffron]

yes

[umbral epoch]

and fy = 14y

[mystic saffron]

great

so now u sub 2x in for for i hat

and 14y in for j hat

[umbral epoch]

i did that also

[mystic saffron]

yes

[umbral epoch]

and then i plugged in theta

[mystic saffron]

yes

u did that incorrectly

because cos (pi/6) isnt 1/2

its root 3/2

[umbral epoch]

oh shiz

i was thinking pi/3

damn

[mystic saffron]

so itd be x root 3

[umbral epoch]

society

[mystic saffron]

and the y term woiuld be 7y

so itd be

xroot3 + 7y

[umbral epoch]

sorry for being an ass 😔 i lowk thought you were boutta troll me

[mystic saffron]

yea idk what the need to be rude was

[umbral epoch]

neither do i

im sorry

[mystic saffron]

anything else?

@umbral epoch

[umbral epoch]

maybe

[mystic saffron]

alr alr

tag me if u need help alr

[umbral epoch]

i need to see the next q

okay

thank you

[mystic saffron]

got u

[umbral epoch]

for this one, how do i know which direction is increases the most rapidly?

[mystic saffron]

well what are u thinking rn

[umbral epoch]

i really dont know

i was thinking plug in points but that doesnt make sense

[mystic saffron]

well we're looking for a direction right

[umbral epoch]

yeah

[mystic saffron]

so why not find and check both

df/dx

[umbral epoch]

i got fx and fy

[mystic saffron]

and df/dy

yes

[umbral epoch]

wym check?

plug in P?

[mystic saffron]

well nto check

plug in poiin ts

plug in the point

into dfdx

and dfdy

youll get

(-10,-11) im pretty sure

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yeah

[mystic saffron]

so then u want to find the magnitude

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oki

[mystic saffron]

tell me what u get alr

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sqrt(221)

[mystic saffron]

yup

thats the answer

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okay i see

how tho?

[mystic saffron]

u get it?

oh alr

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no

[mystic saffron]

well the direction will always be the gradient vector right

so gradient gives direction

what do u think mag of the gradient gives us

@umbral epoch

[umbral epoch]

well im inclined to say direction of most rapid increase

[mystic saffron]

thats what the gradient vector gives us

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i dont understand it conceptually but i will remember

[mystic saffron]

alr so gradient gives us direction of maximum increase

magnitude gives u the rate of that maximum increase

does that make any sense?

[umbral epoch]

ohhh

okay yeah

that makes sense

thank you

[mystic saffron]

does it actually 😭

yk what the gradient represents right

or yk what a vector is right

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bro yes ik what a vector is

[mystic saffron]

yea what qualities does a vector possess

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[mystic saffron]

direction and what else

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i know what a vector is

[mystic saffron]

might as well answer then

any other problems?

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what is the point of the question evidently ik what vectors are

[mystic saffron]

well the problem was asking for a vector

stemming from the point (-3,4)

on the curve f(x,y)

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why is this incorrect

i got fx = y(1/1+(y/x)^2)

and fy = (1/x)(1/1+(y/x)^2)

@mystic saffron

[mystic saffron]

one sec

ok so fx

is

(-y)/(x^2 + y^2)

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how

[mystic saffron]

yk what the derivative of arctan is...

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1/1+x^2

is it not

[mystic saffron]

yea

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okay

so then where is the x^2 +y^2 coming from

and -y

[mystic saffron]

alr alr i might be wrogn

but i got

(root 2)/14

as my final answere

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how did you get fx tho

[mystic saffron]

check if i got it correct first

ill explain if it was correct

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well my numbers randomize each time

so it wont be that

[mystic saffron]

wdym randomize each time

oh u get a new problem

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the question is at p(-6,6) now

[mystic saffron]

alr alr

wanna send it to me?

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its all the same but at (-6,6)

f(x,y) is the same