help-49

[surreal moon]

what limit?

[last slate]

What was this question for?

[sage helm]

before beginning though we need to take care of the worst case scenario for $1/|b_n| \leq M \implies |b_n| \geq M' > 0$

[sage helm]

[ \lim \f{a_n}{b_n} = \f ab \quad b \neq 0 ]

[grand pond]

neon

[sage helm]

if both a_n and b_n are convergent sequences

[sage helm]

hope this cleared that up

[last slate]

Bro's cooking 🗣️🗣️

[grand pond]

neon

[surreal moon]

Oh. So you want to prove that, if $\lim a_n=a$ and $\lim b_n=b$, then $\lim \frac{a_n}{b_n} = \frac ab$?

[grand pond]

SWR

[sage helm]

yes

[surreal moon]

Okay. Let me read what you've done so far. You going hella fast rn'

[sage helm]

yeah sorry about that

im spewing my thoughts

before I write it down

[surreal moon]

Yeah. This is what you want to do.

Then you can use product of limits, which is (slightly) easier than quotient

[sage helm]

this is kind of worrying me though

I need an M'

[surreal moon]

Yeah. It's annoying

I don't remember the exact strat, but I have it saved somewhere. Let me go look and I can at least make sure you end up going in the right direction.

[sage helm]

Okay, thanks for that

[last slate]

What's the question 😭😭?? Congrats on solving lol

[surreal moon]

This is the question

[sage helm]

SWR wrote it up there properly

[surreal moon]

Okay yeah. It's a bit subtle how to go forward. Just go as far as you can and ping me if you need advice.

[last slate]

Ohhhhh thanks

[surreal moon]

You have it good here, yeah

[sage helm]

I did the only thing I could

[surreal moon]

Oh. You know what?

hmm

[sage helm]

Okay so for now let's just pick some M'

call it M'

[surreal moon]

there's an extra step you need here

[sage helm]

Oh?

[surreal moon]

limit composition

[sage helm]

Will I figure I need it if I keep going

[sage helm]

oh huh

I didn't prove that yet

[surreal moon]

Can you use continuous functions at all?

[sage helm]

nope

[surreal moon]

hmm

[sage helm]

whats a function

i only know what a sequence is

[surreal moon]

okay I need to cook too then.

I did this with the continuous function 1/x, and then limit composition. I thought it would translate nicely to sequence limits.

But yeah I need to think more a bit too

[sage helm]

I mean end of the day I need to do something like:
[ |b_n - b| \f{1}{M'|b|} = \epsilon ]

[grand pond]

neon

[sage helm]

so $|b_n - b| = M'|b|\epsilon$

[grand pond]

neon

[surreal moon]

But M' depends on n, right?

[sage helm]

Can't I just let that be my epsilon

[sage helm]

Yeah well it does

I need a sufficiently large M'

[surreal moon]

then it can't be your epsilon yet. You need a bound for M' independent of n

[sage helm]

yeah

[surreal moon]

I did it with 1/x by using a reverse triangle inequality, but I don't know how to translate that here

[sage helm]

I see

[surreal moon]

yeah I'll think on this too.

Luck to you in the meantime

[sage helm]

😭

good luck

@surreal moon couldn't we consider M' = |b|/k where k is an arbitrarily large positive number

[surreal moon]

yes, but you need to find k now

[sage helm]

hmm

is "arbitrarily large" not good enough a claim to fight in court

It should be fine except for if there's a random 0 in the line

[surreal moon]

no it is not

[sage helm]

hmm alright

@surreal moon what if we looked for a sufficiently large n

would that be good

[surreal moon]

that's what we are trying to do

[sage helm]

Then can I just not pick any |b|/k

after a large n >= N_1

[surreal moon]

What do you mean?

[sage helm]

okay lets just say k = 3 for now

for some value of n say N_1

|b - b_n| < |b|/3

is that alright to say

should be actually yeah

there isn't a problem

this is regular old epsilon-N

our epsilon is just |b|/3

[surreal moon]

hold on I'm having trouble understanding what you are trying to say. Lemme take a deeper look

[surreal moon]

What is k now?

[sage helm]

3

idk

whatever you like

I don't think it matters

[surreal moon]

no but like, what does k represent?

[sage helm]

uhh

just a positive real

Am I struggling and dragging this out more than usual for this proof

[surreal moon]

hard to say. Your proof is kind of fragmented so far, so it's hard for me to follow along. Would it be alright if you wrote out everything you had so far?

[cosmic mica]

i'm struggling to understand why you can't just use multiply or divide the limits since they are both convergent

[surreal moon]

(perhaps just a SS of where you are actually working on this)

[sage helm]

that's what we're proving

[surreal moon]

You need to prove that multiplying works through limits. That's the task

[sage helm]

Okay

[cosmic mica]

oh okay

[sage helm]

wait I think I got it

articulation in process

We aim to prove that given $\lim a_n = a$ and $\lim b_n = b \neq 0$, $\lim a_n/b_n = a/b$. If it is proven that $(b_n) \to b \implies (b_n^{-1}) \to b^{-1}$, then the rest of the proof follows from an established statement. Consider:
[ \left | \f{1}{b_n} - \f 1n \right | = \f{|b_n - b|}{|b||b_n|} ]
We wish to accomodate for the boundedness of $1/|b|_n$ requiring an $M'$ such that $|b_n| \geq M' > 0$. Consider for some $n \geq N_1 \in \set N$, $|b_n - b| < |b|/2$. With the triangle inequality, we can conclude that $|b_n| > |b|/2$. Next, for suitable $n \geq N_2 \in \set N$, we have:
[ |b_n - b| = \epsilon \cdot \f{|b|}{2} \cdot |b| ]
Piecing everything together, we have:
[ \left | \f{1}{b_n} - \f 1b \right | = \f{|b - b_n|}{|b|(|b|/2)} < \f{\epsilon |b|^2}{2} \cdot \f{1}{|b|(|b|/2)} = \epsilon ]

[grand pond]

neon

[sage helm]

NOTHING BROKE YESSSSSS

@surreal moon that look good to you?

[surreal moon]

okay lemme read

I got a solution too, so if yours doesn't work, I can help

[sage helm]

Oh wait I forgor a statement

N = max{N_1, N_2}

[surreal moon]

I assume you mean -1/b here?

[sage helm]

oh yeah

[surreal moon]

So you want to find an $M'$ (which depends only on $\epsilon$ I assume?) that serves as an upper bound for $|b_n|$ for sufficiently large $n$?

[grand pond]

SWR

[sage helm]

I mean my M' here is |b|/2 it doesn't depend on epsilon

[surreal moon]

oh ok

[sage helm]

the real jump was going from |b_n - b| to |b_n|

because that satisfies the condition on M'

[surreal moon]

So wait, you're saying that $M'=\frac{|b|}{2}$?

[grand pond]

SWR

[sage helm]

yeah

it's also entirely arbitrary I think

oh wait this should be <

[surreal moon]

And you're saying that $|b_n|\ge\frac{|b|}{2}$ for sufficiently large $n$?

[sage helm]

also you can modify this bit to whatever we want to accomodate the M'

[grand pond]

SWR

[sage helm]

yeah

[surreal moon]

oh I see

yeah I guess that will need to be true

[sage helm]

if we did |b|/3 then this bit would become 2|b|/3 I guess

[sage helm]

and here we have 2|b|/3 instead of |b|/2

so it still all cancels back to epsilon

is it satisfactory?

[surreal moon]

yeah it looks good

I'd say you did it

[sage helm]

phew

[surreal moon]

and it was cleaner than mine

[sage helm]

it only took what

an hour

[surreal moon]

yeah they can

I had a proof a while ago that took a whole weekend

[sage helm]

¯_(ツ)_/¯

[surreal moon]

[sage helm]

I'll get on your level soon

@twilit field in case you were wondering you're not alone in the whole proof taking a lot of time department

Alright then, thanks again SWR, I'll head to bed now

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[grave locust]

Hello

i think the only statement that makes sense is B and C

but its not right

[surreal moon]

!show

[midnight plank]

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

[grave locust]

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[meager ore]

$\phi$ is the Number-Theoretic Euler's function

[grand pond]

Halex

[midnight plank]

@meager ore Has your question been resolved?

[midnight plank]

@meager ore Has your question been resolved?

[junior flower]

is this question is poorly written. i believe the a_i's are supposed to be relatively prime to n

not just any phi(n) residues

[meager ore]

what do you mean with "not just any phi(n) residues"?

[junior flower]

like... take n = 4. then phi(n) = 2 but letting a_1 = 0 and a_2 = 1 makes a_1 + a_2 not 0 mod n

you cannot just take any 2 residues

they should be distinct, and the ones relatively prime to 4 (so, 1 and 3)

so i'm just pointing out the problem is unclear with that but that's probably what was intended

but anyway for the problem, maybe you can think about pairing up the a_i's so that they add up to 0 (mod n)

you may want to look at example (say, n = 20) to see what's going on

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[twilit field]

So I'm trying to prove gcd(a,b) can be expressed as ax+by

[twilit field]

where x an y are integers

so my book started off with the claim the $S= {au+bv|au+bv>0; u,v \text { are integers }}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

and from this it claims d=ax+by

[runic hamlet]

thats not a claim, thats a set

and thats jumping a couple of points

[twilit field]

my bad

I know, I'm skipping some points to get to my question

[zealous schooner]

Ah good ol’ bezout’s identity

[twilit field]

what now

my question is how can we claim d=ax+by

[runic hamlet]

well you skipped over so much

[zealous schooner]

I’m fairly certain this comes directly from the division algorithm

[runic hamlet]

we dont know what your book covered until then

[twilit field]

ok, sorry

[twilit field]

got it

thanks

[zealous schooner]

Just write the division algorithm backwards

[twilit field]

so $x=aq+d$

[grand pond]

ƒ(Why am. I here)=I don't Know

[runic hamlet]

dont mix up division algorithm and euclidean algorithm

[pearl idol]

(When you say "division algorithm", that's the same thing as the Euclidean algorithm, right?)

[runic hamlet]

the book here uses wop and not euclidean algorithm

[zealous schooner]

ah, my bad

[runic hamlet]

the book calls the result that division with remainder exists the division algorithm

[zealous schooner]

oh okay

[zealous schooner]

sorry, I meant the Euclidian algorithm

[twilit field]

hmm, ok

thanks

[runic hamlet]

you dont need the euclidean algorithm for the proof

[twilit field]

I'll send my books proof

just a minute

[zealous schooner]

hmm? there's another proof?

[runic hamlet]

yeah consider the set from above and use wop. or wait a min until they send the proof

[zealous schooner]

wop?

well ordering principle?

[runic hamlet]

yes

[zealous schooner]

oh I think I understand how

||considering the minimum element and proving it's the GCD?||

[runic hamlet]

yes

[zealous schooner]

nice

[runic hamlet]

ok, so now what exactly is your question

d is in S. by definition of S that means d=ax+by for some integers x,y

[twilit field]

that was my question, d is the divisor, right, how can we claim it will be of the form ax+by

[runic hamlet]

at that point we havent said anything about d being a divisor

[twilit field]

ah, then it's fine

thanks

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[viral dagger]

BGD=BCD-BGC-CGD

no?

i mean they are both the same ig

alr do you know what the length of the side of the square is

can you simplify it

uh

i meant like 32=16×2 and take the 16 out

ye

alr, do you know how to find area BCD?

yes

the next easiest is finding CGD

if we let CD be the base, do you know the height of the triangle?

which is?

would you say the distance from the base to G to be half of BC?

since G is the midpoint of F to C and the height of F to the base is the same as BC, so G would be half of BC

if you manage to get the area of CGD, you can do the same process for BGC

what

do you understand that if GC is half of FC, basically the height from F to CD is half of the G to CD

i mean you can prove this with simmilar triangles ig

sait your asking q2 right

blue and yellow are simmilar

so like, the height of blue/hyp of blue=height of yellow/hyp of yellow

hyp of yellow is 2 hyp of blue

so height of yellow would mean 2 times height of blue

huh

uh

try bc and imagine the midpoint, now try moving point b horizontally, if you think about it, the height of the midpoint woulf still be the same just that the horizontal distance changes

why would they match up? BD is the hypotenuse

heres me graphing it

the one on tbe black line the midpoint and the one on the red line is the B when you move it, and 0,0 is C

you can see that the midpoint still sits at the black line

hopefully you can unferstand my bad teaching 🙏

do you get it? like hoe the height is half of BC

i dont understand what your asking

yea i get that but what are you asking

are you askibg why they are diffrent?

wdym meet up??

btw what grade of math is this?

what

what grade are you

oh oj

have you been g9 for a while or just got to g9?

ooo same

can i ask what you mean with "meet up"?

huh

like why if its half of the hypotenuse it would also be half of the veritcal thing

[viral dagger]

.

[astral talon]

i think this might help

ignore the left side of the figure

wait let me draw it for you

[viral dagger]

half of the hypotenuse means if one of them is at (0,0), the midpoint of the hypotenuse would be (x/2,y/2) if the hypotenuse is (x,y)

im not really sure how to explain it

[astral talon]

what shape is only this part

yes good

now what is line FC

1 sec lemme show u more clearly

what is FC of the rectangle

exactly

now if you get the midpoint of the diagonal of the rectangle....

it's the midpoint of the rectangle!

and that is point G

therefore the verticles are equal

oh no it's completely fine

umm do you know properties of a rectangle

"diagonals bisect eachother"

the point where both diagonals meet

what bisection means is that it divides the line into 2 equal parts

think of the centre of a plus sign

those 2 lines bisect eachother

meaning the centre of the line

yeppers

that meeting point is not only the centre of the rectangle

but also the centre of each diagonal

lemme show u what i mean

AX = XC

BX = XD

i don't understand what you mean by proof

give me a second here i'll give you something a little more concrete

the strongest proof is just construction

if you go and plot a rectangle on a graph and draw its diagonals, you'll see that the intersection point divides the line into 2 equal parts

what do you want

an algebraic proof?

what proof do you want

with similarity or something?

i mean

[astral talon]

keep this figure in mind

we know it's a rectangle right

AB = CD, AD = BC

so we can prove congruency

use side angle angle

AXB is congruent to CXD

AB = CD opposite sides of a rectangle

and ABX = CDX by alternate angles, AB || CD

BAX = DCX

after proving congruency you can assume AX = CX

BX = DX

do the same for triangles AXD and CXB

and voila

[viral dagger]

ye

huh

uhh did you understand how the height of CGD is half of BC now?

alright, so whats the height of the triangle?

then whats the area

well BC to G is half of BF, and BF is half of AB

and you know AB

use the same logic as how you get height of CDG as half of BC, instead of half of BC, its half of BF

alr so whats BF?

so whats the height of the yriangle then

[midnight plank]

@last slate Has your question been resolved?

[viral dagger]

BCG

[viral dagger]

yes

yhen whats thr area

[viral dagger]

now use this

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[twilit field]

if $a|b$ and $c|d$ then show $ac|bd$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $b=ka$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$d=k'c$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so so $db =(k k ' ac$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $db = Kac$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

thus $ac|db$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is that right

[junior flower]

yes

[twilit field]

thanks !

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[twilit field]

Prove $8|5^{2n}+7$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

using induction

for $ n \geq 1$

Now I can solve this using the binomial theorm

For induction, these are my thoughts

yeah, not too sure of how I'd use induction here

becomes rather messy

no?

wait

I think I know a theorm that may help

[jagged saffron]

you dont need any theorems here

just set up your induction as usual and use your inductive step to prove the n+1 case

[twilit field]

so $8|25 \cdot 5^{2n}+7$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is ehat I have to prove

maybe I express 25 as 28-3

[jagged saffron]

using the fact that 8 | 5^2n + 7

its easier imo if you unpack what "divides" means

and use equations

[twilit field]

am I allowed to do that in NT

[jagged saffron]

use the definition of divides? yes

[twilit field]

ok so $\frac{5^{2n}+7}{8} =k$

[grand pond]

ƒ(Why am. I here)=I don't Know

[jagged saffron]

okay but you should steer clear of division whenever possible

[twilit field]

ok, so I can write this as

$25\frac{5^{2n}+1}{8} - \frac{18}{8}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

ok, that doesn't seem to work

[jagged saffron]

you never need to divide anything to do this

[twilit field]

oh

$5^{2n}+7=8k$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I use this?

[jagged saffron]

yep, and now youre trying to show that 5^2(n+1) + 7 is a multiple of 8 using that

[twilit field]

so $7=8k-5^{2n}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so that gives me

,, \frac{5^{2n}(24)+8k}{8}

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is indeed divisible by 8

got it

Thanks a lot!

[jagged saffron]

nice

[twilit field]

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[sage helm]

is this fine?

[frail carbon]

right idea, although you've not said what N is in your proof

[sage helm]

uh

it's just some sufficiently large natural I guess?

[frail carbon]

specifically the one such that |xn|<ε^2 yeah

you should write it down though

[sage helm]

Okay

"Choose N such that..."

Would adding that fix it?

[frail carbon]

yeah

[sage helm]

Do you mind if I ping you in a bit regarding (b)?

[frail carbon]

go ahead

[soft token]

Which book is this from?

[sage helm]

Abbott

[sage helm]

Kinda hit a roadblock

The problem is that the quantity on the right is greater than epsilon

[soft token]

Full title? Understanding Analysis?

[sage helm]

That's the one

[frail carbon]

So you've noticed essentially that [|\sqrt{x_n}-\sqrt x|=\frac{|x_n-x|}{|\sqrt{x_n}+\sqrt{x}|}]

[hard umbra]

might need to split cases

[grand pond]

Edward II

[sage helm]

Oh and that is lesser than replacing x_n with M

[sage helm]

split cases how?

[hard umbra]

to ensure the denominator doesn't die

[frail carbon]

The denominator can't die

[sage helm]

why would it die

[frail carbon]

We've assumed everything is >0

[sage helm]

[hard umbra]

have you though

neon hasn't written that

[frail carbon]

oh that's a good point

[sage helm]

x_n >= 0

and x is whatever

[hard umbra]

you need to split cases based on whether x = 0

[sage helm]

Hmm

[hard umbra]

if it is, you've already done the proof

if it isn't, you can ensure the denom doesn't die

[sage helm]

Right I can declare x isn't 0 here

[hard umbra]

yes

[sage helm]

I mean alright but that doesn't adress my point of the inequality

[hard umbra]

well you see

[sage helm]

Do I restructure around this

[hard umbra]

if x > 0

you can bound x_n from below as well

x_n > ε > 0

[sage helm]

Hmm true

[hard umbra]

that's enough to save you

[sage helm]

Okay let me write it in mathematician

[hard umbra]

ok you physicist

<@&268886789983436800> 963077882035466310

[hard umbra]

uh

are you sure

that last inequality is false

[sage helm]

oh

bruh

wait how is this supposed to help me then

It just kicks the can down the road

[hard umbra]

you could've just start with sqrt(epsilon') + sqrt(x)

instead of M

[sage helm]

Okay that makes sense

@hard umbra sorry it took a while

The only thing I'm confused is if I need to do the N1 N2 thing

for epsilon' and the main proof statement

[hard umbra]

technically yes

[sage helm]

And then choose max{N1, N2}

[hard umbra]

you'd need to justify why epsilon' exists

[sage helm]

Alright, beyond that it's all good?

[hard umbra]

yes

[sage helm]

was the justification in the () not enough

[hard umbra]

well i mean thats leading up to why

but to get into the weeds, you'd have to unpack the definition of convergence

[sage helm]

ughughugh

problem for future me

[hard umbra]

👍

[sage helm]

Thanks though :)

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[twilit field]

Prove that for any integer a, one of the integers, a,a+2,a+4 is divisible by 3

[twilit field]

so I first assumed a is of the form 3n

in that case it's trivially divisible by 3

then in the from 3n+1

in which case 3n+3 will be divisible by 3

and lastly 3n+2

in which case 3n+6 is divisible by3

would this be enough

or is it lacking rigour

[sage helm]

looks good

write it properly

[zealous schooner]

Just stipulate that every integer is of one of the forms 3n, 3n+1 or 3n+2

Rest is good

[twilit field]

don't I have to prove that in. a NT class?

I mean it makes sense

but do I have to prove that first?

[zealous schooner]

There’s a theorem that says that

[zealous schooner]

I mean, you can

[zealous schooner]

I think*

[twilit field]

hmm

Let me think of how I'd prove it

I guess I could prove that a number n, and it's successor are always co prime

But that doesn't help much

[jagged saffron]

what happens if you add 1 to 3n+2

[twilit field]

you get 3n+3

which is also divisible by 3

[zealous schooner]

Yeah a proof by induction is what I was thinking

[twilit field]

hmm

yeah, that works

but wherever possible I try to avoid induction

[humble torrent]

i like how the sort of thinking youd use when using mod 3 is sort of already intuitive to @twilit field now, without learning it yet

[twilit field]

Thanks!

[zealous schooner]

Yeah, modular arithmetic is gonna come really naturally to you

[twilit field]

I hope so, I need a perfect score in my UG

[zealous schooner]

You’re forming the base ideas of it naturally

[twilit field]

thanks

[humble torrent]

oh and youre doing all this before even starting, yea it will go smoothly for you!

[humble torrent]

oh you need a perfect score? I honestly never think its that necessary unless you maybe plan on doing a PhD or something

[runic hamlet]

you have the division algorithm. use that to prove this claim. no need to use induction explicitly. (tho it wouldnt be bad to practice it of course)

[zealous schooner]

yeah division algorithm also works

[twilit field]

i'm not studying at a great college,it's good, but that; about it, so i need that for my msc

[twilit field]

hmm, thanks

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[shadow sparrow]

How to factorise x³-6x²+11x-6

[visual tiger]

!status

[midnight plank]

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

[sweet wing]

try finding one root first

[shadow sparrow]

1

[sweet wing]

indeed

so now divide the expression by x-1

[shadow sparrow]

I want to learn how to factorise these type of question

[sweet wing]

$\frac{x^3-6x^2+11x-6}{x-1}$

[grand pond]

Flappie

[shadow sparrow]

Ok

[zealous schooner]

not necessary to do long division here, you can do it by using sum and product of roots

[shadow sparrow]

How sir pls teach me

[zealous schooner]

the original equation has roots whose sum equals 6

[shadow sparrow]

How do we know that can you pls tell from basic

I mean about that roots

What is that

[grim vector]

you try lil numbers like 1, 2, 3 or -1, -2, -3 and check if it's equal to zero

[shadow sparrow]

Zero of a polynomial

Ok

[visual tiger]

if you have a polynomial that goes like x^n.... + k

you're not forced to check through all "small" numbers to find the root

[visual tiger]

the rational root theorem tells you that in those cases, only divisors of k can be integer roots

so +/- 1, +/- k, etc...

[shadow sparrow]

Ok

[visual tiger]

example : x^3 - 3x^2 - 9x - 5

which integer roots can we hope to check for?

[shadow sparrow]

5

[visual tiger]

ok so yes

+/- 1

and +/- 5

are the only possible integer roots

[visual tiger]

in our case

[shadow sparrow]

Yes

[visual tiger]

we have all divisors of 6, plus or minus

so 1,2,3,6 or -1,...

we found 1 so we're good to go

then we can factor

[shadow sparrow]

How to further

Do

[visual tiger]

well

[visual tiger]

factor this by (x-1)

and then you're left with a quadratic

[shadow sparrow]

In this case it should be 6??

[visual tiger]

-> quadratic formula or whatever you prefer

[visual tiger]

?

[shadow sparrow]

That root thing??

[visual tiger]

I said the integer roots can only be 1,2,3,6 or -1,-2,-3,-6

we never said

all roots are in here

[grim vector]

x^3 -6x^2+11x-6 = (x-1)(ax^2 +bx + c) you can do like this too

[visual tiger]

and we found that "1" is a root

[visual tiger]

so we can factor exactly like this

and then you're left to factor ax^2+bx+c

[shadow sparrow]

Like x is a polynomial and the value of x is 1

[zealous schooner]

again, it's not necessary to factor like this

factoring using Viete's relations is quite faster

[shadow sparrow]

What I am reading things like viete and quadrictic equation is going over my head

Pls tell for a class 8 boy

[zealous schooner]

you probably do know what viete's relations are, just not by the name

if you have a quadratic $ax^2+bx+c=0$ can you tell me the sum of its roots?

[grand pond]

kheerii

[shadow sparrow]

1???

I didn't really understand what is (sum of roots)

Pls explain one more time

??

Sir??

. Close

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[drowsy thunder]

When doing integration by parts how do we decide which term will be “u” and “dv”

[drowsy thunder]

Do we just pick at random

[night egret]

just think about which one is easier to integrate and which one is easier to differentiate

there's also an acronym you can use to help you but it's not always correct

[zealous schooner]

Either LIATE or ILATE

[drowsy thunder]

?

What is the acronym

[zealous schooner]

L: logarithmic
I: inverse trigonometric functions
A: algebraic (polynomials)
T: trigonometric
E: exponential

These tell you in which order you should choose your u

L and I are interchangeable

Note that this isn’t a hard and fast rule, it’s just a guide

[drowsy thunder]

Ok thx

[zealous schooner]

You will find instances where this isn’t the best approach

[zealous schooner]

This is a general rule

[drowsy thunder]

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[twilit field]

prove that the difference if two consecutive cubes is never divisible by 2

[pearl idol]

what have you tried so far

[twilit field]

so I assumed the integers to be a, a+1

so then I have $(a+1)^3-a^3$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is

$(1)((a+1)^2+a^2+a)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

oh

right

got it

sorry

and thanks

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[pearl idol]

If you want a simpler argument: n^3 is always the same parity (odd or even) as n. So since n+1 and n are opposite parities, (n+1)^3 and n^3 will also be opposite parities. Therefore, their difference is odd because even - odd or odd - even are always odd.

(I left some details out about how you prove the claims I made, but that's the gist)

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[uneven sandal]

Prove the Lax-Milgram lemma:
Let \( a : X \times X \to \mathbb{C} \) be a bounded coercive sesquilinear form on the Hilbert space \( X \). Then there exists a bijective linear bounded operator \( A : X \to X \) such that for all \( x, y \in X \),
\[ a(y, Ax) = (y, x)_X. \]
Hint: According to the Riesz representation theorem, there exists a unique element \( Ax \) for each \( x \) satisfying \( a(y, x) = (y, Ax)_X \) for all \( y \in X \). Show sequentially: (i) \( N(A) = \{0\} \), (ii) the image space \( R(A) \) is closed, and (iii) \( R(A) = X \).

[grand pond]

tobi

[uneven sandal]

I already managed to proof that the kernel of A is trivial.

N denoted the kernel and R denotes the Imagine of A (I dont know why they use this notation in functional analysis its confusing to me).

I Would need help with the 2nd step

[pearl idol]

You probably want to post this in #advanced-analysis instead of in this help channel

[pearl idol]

(N = null space and R = range, probably)

[uneven sandal]

okay thanks

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[wanton trail]

Show that: (α,β,γ are random vectors) I am not sure what the first step is or how should I approach similar questions. you can use a,b,c

[nova yoke]

do you already know this inequality if there were only 2 vectors instead of 3?

[wanton trail]

yes

[nova yoke]

ok, try applying it twice

|a+b+c| <= |a+b| + |c| ...

[wanton trail]

what do you do after wont you just get |a+b| + |c| <= |a| + |b| + |c|?

[nova yoke]

yep pretty much

just two quick steps

can do the same thing with 4 vectors or 5, etc

[wanton trail]

i still dont understand

|a+b| <= |a| + |b|

|a+b| + |c| <= |a| + |b| + |c|

i can only get to here what do i need to do after

isnt |a+b+c| different from |a+b| + |c|? i mean they equal different things right?

[pearl idol]

Yes, they are different

but how do they relate?

Hint: a+b+c = (a+b)+c

[wanton trail]

|a+b+c| <= |a+b|+|c|?

oh i got it now

|a+b+c| = |(a+b)+c| <= |a+b| + |c| right?

.close

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[mystic elk]

I'm trying to show that the definition of the covariant derivative of a tensor density implies that the covariant derivative of the Levi-Civita tensor density vanishes.

This definition is given:
$\nabla_c T^{a...}{b...} = \text{usual terms if $T^{a...}{b...}$ were a tensor} - W\Gamma^{d}{dc}T^{a...}{b...}$

For the Levi-Civita tensor $\epsilon^{abcd}$, I have:
$\nabla_e \epsilon^{abcd} = \partial_e \epsilon^{abcd} + \Gamma^{a}{fe}\epsilon^{fbcd} + \Gamma^{b}{fe}\epsilon^{afcd} + \Gamma^{c}{fe}\epsilon^{abfd} + \Gamma^{d}{fe}\epsilon^{abcf} - \Gamma^{f}{fe}\epsilon^{abcd} $
The middle terms cancel out due to the antisymmetric properties of the LC tensor density, so that I have:
$\nabla_e \epsilon^{abcd} = \partial_e \epsilon^{abcd} - \Gamma^{f}{fe}\epsilon^{abcd} $
The LC tensor density is just a generalisation of the Dirac delta function which vanishes under partial derivatives, leaving:
$\nabla_e \epsilon^{abcd} = - \Gamma^{f}_{fe}\epsilon^{abcd} $
But I'm not sure how to argue that the last term must be zero. I think one could argue that it's to do with the connection being symmetric and the LC tensor density being antisymmetric, but they don't have the same rank, so I'm not sure if that follows, exactly.

[grand pond]

momoji

[mystic elk]

Oh, sorry about the poor formatting. I can try to make it a little clearer.

Objective: I'm trying to show that the definition of the covariant derivative of a tensor density implies that the covariant derivative of the Levi-Civita tensor density vanishes.\newline
The definition given:\newline\newline
$\nablac T^{a...}{b...} = \text{usual terms if $T^{a...}{b...}$ were a tensor} - W\Gamma^{d}{dc}T^{a...}{b...}$
\newline\newline
For the Levi-Civita tensor $\epsilon^{abcd}$, I have: \newline
$\nabla_e \epsilon^{abcd} = \partial_e \epsilon^{abcd} + \Gamma^{a}{fe}\epsilon^{fbcd} + \Gamma^{b}{fe}\epsilon^{afcd} + \Gamma^{c}{fe}\epsilon^{abfd} + \Gamma^{d}{fe}\epsilon^{abcf} - \Gamma^{f}{fe}\epsilon^{abcd} $
\newline\newline
The middle terms cancel out due to the antisymmetric properties of the LC tensor density, so that I have:
$\nabla_e \epsilon^{abcd} = \partial_e \epsilon^{abcd} - \Gamma^{f}{fe}\epsilon^{abcd} $
\newline\newline
The LC tensor density is just a generalisation of the Dirac delta function which vanishes under partial derivatives, leaving:\newline
$\nabla_e \epsilon^{abcd} = - \Gamma^{f}_{fe}\epsilon^{abcd} $
\newline\newline
But I'm not sure how to argue that the last term must be zero. I think one could argue that it's to do with the connection being symmetric and the LC tensor density being antisymmetric, but they don't have the same rank, so I'm not sure if that follows, exactly.

[grand pond]

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

[mystic elk]

(Hopefully that's readable aha)

Oh, some additional notes, I suppose. There's a derivation of this online that uses the metric compatibility of the LC tensor to get to the end result. But my textbook (D'Inverno) seems pretty explicit about using the definition given.

[midnight plank]

@mystic elk Has your question been resolved?

[midnight plank]

@mystic elk Has your question been resolved?

[midnight plank]

@mystic elk Has your question been resolved?

[midnight plank]

@mystic elk Has your question been resolved?

[pearl idol]

(yup, that's right!)

[pearl idol]

wait what does the W mean in your definition of covariant derivative?

also you might get better help in #diff-geo-diff-top than in this help channel

[mystic elk]

It's the weight of the tensor density. In the case of the Levi-Civita tensor density with all contravariant indices, it's +1

Ahh okay, I thought maybe I should've posted it there haha.

[midnight plank]

@mystic elk Has your question been resolved?

[last slate]

.close

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[twilit field]

Prove that $1 \cdot x = x \forall x \in F^n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so I thought of starting my adding the additive -1(x) to both sides

$1(x)-1(x)=x-1(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $1(x-x)=x-1(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so 0= x-(1)x

so $x=(1)x$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

is this right?

[runic hamlet]

how about actually just doing the multiplication?

I dont know if your axioms allow you to factor the 1 out like you did already

[twilit field]

I have to prove that (1)x=x though

[runic hamlet]

well just compute what (1)x is

you know how multiplication works

[twilit field]

1(x) is just x

[runic hamlet]

thats what you are supposed to show

use the definition of scalar multiplication in F^n to compute 1*x

[fleet moss]

like x in F^n means you have some sense of x consisting of n scalars

[twilit field]

$1(x_1,x_2,x_3....x_n)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is $(x_1,x_2....x_n)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[fleet moss]

right

[runic hamlet]

dont skip steps

[twilit field]

$(1x_1,1x_2....1x_n)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[runic hamlet]

and now this is equal to (x1,...xn) why?

[twilit field]

1 is the multiplicative identity

[fleet moss]

in?

[twilit field]

oh right

over $\mathbb{F}^n$?

[grand pond]

ƒ(Why am. I here)=I don't Know

[runic hamlet]

well thats what you are supposed to show

[fleet moss]

here 1 is the multiplicative identity in F, which is why you can safely say... ?

[twilit field]

in that case I'll probably use the fact that $1(a)'s additive inverse is -1(a)

so 1(a)-1(a)=a-(1)a

so a=1(a)

[fleet moss]

what is a

its important to define and quantify your variables

[twilit field]

I meant x

sorry

[fleet moss]

x in F^n?

[twilit field]

I have to go now( have a class), can I close this for now?

[twilit field]

yes

[fleet moss]

but you already tried that at the beginning

[visual tiger]

you're making it more complicated than it should be

[visual tiger]

you just come back to this

[fleet moss]

i mean yeah you can always close it. but heres what i was getting at: since 1 is the multiplicative identity in F, and since $x_1,\dots,x_n$ are all in F, you can safely say $1\cdot x_i = x_i$

[visual tiger]

and knowing 1 is multiplicative identity in F

[grand pond]

esca (@ with reply)

[twilit field]

Sorry

. close

[fleet moss]

why sorry

[twilit field]

For closing it midway

Forgot that I had a class now

.close

[fleet moss]

oh no worries lol. good luck

[midnight plank]

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[midnight plank]

@bleak pier Has your question been resolved?

[full marlin]

write it as a piecewise function and find the conditions for differentiability at x=0

[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

Piecewise how?

[carmine sigil]

@bleak pier piecewise means write it using cases:

[
f(x) = \begin{cases}
a_0 + a_1 x + a_2 x^2 + a_3 x^3, & \text{if } x \ge 0 \
a_0 + a_1 (-x) + a_2 (-x)^2 + a_3 (-x)^3, & \text{otherwise}
\end{cases}
]

[grand pond]

OmnipotentEntity

[bleak pier]

I see and what should we do next?

Should I put x=0?

[carmine sigil]

@bleak pier the idea is if f(x) is differentiable at x=0 then lim h->0 (f(h) - f(0))/h exists.

In order for this limit to exist the limit h->0+ and limit h->0- must both exist and be equal

An easier way to express this is if we say f+(x) = a0 + a1 x + a2 x^2 + a3 x^3, and f-(x) is the version of the function with x replaced by (-x), then we want f+'(0) to be equal to f-'(0)

How can we select our ai to make this true?

[bleak pier]

a1 will be zero? And x^3..... odd terms

???

@carmine sigil

[carmine sigil]

So what is f+'(x) and f-'(x)?

Let's take this step by step

@bleak pier ^

[bleak pier]

For x>=0 f'(x)=a1+2a2x+3a3x^2+....
f'(x-)=-a1+2a2x-3a3x^2

a1=0 and negative terms should be 0?

2(3a3x^2+5a5x^4+.....)

[carmine sigil]

Well

We only care about the value of f+'(0) and f-'(0)

We only need that these are equal to each other at the one point, other points do not matter

@bleak pier

So

f+'(0) = a1 + 2a2 (0) + 3a3 (0)^2 = a1
f-'(0) = -a1 + 2a2 (0) - 3a3 (0)^2 = -a1

So all we need is a1 = -a1 or a1 = 0

[bleak pier]

Yes sir

It will be continuous rest all the points

X>0 and x<0

[carmine sigil]

Oh right!

You also need that f+(0) = f-(0)

But that is fairly trivial, because it reduces to a tautology. a0 = a0

But yes, the only point we need to be concerned about is the point x = 0. Because it's just a normal polynomial at every other point.

[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

Tq very much

.close

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[twilit field]

if a and b are odd integers prove that $8|a^2-b^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[zealous schooner]

hey there

[zealous schooner]

!status

[midnight plank]

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

[twilit field]

$\frac{(a-b)(a+b)}{8}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

a= 2k+1

$b=2l+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

the thing is I'm only able to factor out a 4 here

what am I doing wrong

[zealous schooner]

what do you get after the substitutions

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

[grand pond]

kheerii

[twilit field]

yeah

[zealous schooner]

consider the parities of both the numbers on top

in particular consider their difference

[twilit field]

No idea what that means

parity

what's that

[hollow oyster]

is it odd or even

[twilit field]

ah

[zealous schooner]

odd-even

[twilit field]

one is even one is odd

oh

right

[zealous schooner]

yes, since their difference is odd

[twilit field]

hmm, thanks

.close

[midnight plank]

Channel closed

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[midnight plank]

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[twilit field]

I'm not sure I understand how $F^{\infty}$ is a vector space

[grand pond]

ƒ(Why am. I here)=I don't Know

[surreal moon]

What are the vector space axioms that you feel would not apply?

[twilit field]

I didn't undestand the part where they said it's the set of all sequences of elements of F

[surreal moon]

Every infinite ordered list whose elements are all elements of F

[twilit field]

hmm

ok so it;s somewhat like a power set

[runic hamlet]

uhm

depends on what you mean by "somewhat"

[twilit field]

as in a power set ia. set of sets

[runic hamlet]

I mean you believe without problem that {(x1,...,xn): x_i in F} is a set, no?

[twilit field]

yes

[runic hamlet]

its not a set of sets

well and now instead of the objects being (x1,...,xn), they are (x1,x2,....)

[twilit field]

I don't follow

oh

the second set is non-finite ?

[runic hamlet]

its not a set

its a sequence

[twilit field]

you're talking about F

right

[runic hamlet]

what

[twilit field]

I'm confused

doesn't the picture say $F^{\infty}$ is the set of all sequences ...

[grand pond]

ƒ(Why am. I here)=I don't Know

[runic hamlet]

yes

[runic hamlet]

but these two arent sets

[twilit field]

ok, yes, that makes sense

thanks

.close

[midnight plank]

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[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

yes?

[runic hamlet]

yes what?

[twilit field]

@dreamy lichen was typing

[dreamy lichen]

Nothing

That was like 3 mins ago i think

[twilit field]

.close

[midnight plank]

Channel closed

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[dreamy lichen]

Sorry

[midnight plank]

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[livid bramble]

any hints?

[livid bramble]

trying to find a non brute force way

[grim vector]

Informatic science or no ?

[livid bramble]

its from euler project

similar to leetcode but mathy

[grim vector]

Im not good in computer science, but i can give you a website where they give some helps if you want

https://stackoverflow.com/questions/51447940/finding-sum-of-consecutive-primes

[livid bramble]

thanks

.close

[midnight plank]

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[midnight plank]

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[white loom]

would this be the integral?

first step I did was y^3 = 81y

[fresh sparrow]

what is this y for

between the pi and the brackets

also i think you forgot ||to square it||

[white loom]

good point

that is not supposed to be there

well for +-9 right

but do i make the bound x = -9, 9

or 0,9

[fresh sparrow]

that's your hint

[white loom]

yeaa

that dont help me

ahh ok yeah

quandrat 1

would 0 to 9

@fresh sparrow

<@&286206848099549185>

is this integral right?

its saying my answer isnt right tho

[fresh sparrow]

no

(y^3 - 81y)^2

[midnight plank]

@white loom Has your question been resolved?

[white loom]

still doesnt

[midnight plank]

Channel closed

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[white loom]

.reopen

[midnight plank]

✅

[white loom]

<@&286206848099549185>

[midnight plank]

@white loom Has your question been resolved?

[midnight plank]

@white loom Has your question been resolved?

[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[vast remnant]

help me pls

[midnight plank]

@vast remnant Has your question been resolved?

[vast remnant]

<@&286206848099549185>

[midnight plank]

@vast remnant Has your question been resolved?

[high orchid]

better to go through the congruent cases again dude

[midnight plank]

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[wooden cape]

can someone guide me on how to do A

[wooden cape]

my previous answer for A was 24310 but its wrong

[tribal tartan]

do you know what the difference between permutation and combination is?

the difference between A and B is exactly that

[midnight plank]

@wooden cape Has your question been resolved?

[wooden cape]

I found the answer

but thank you

[midnight plank]

@wooden cape Has your question been resolved?

[midnight plank]

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[steel cairn]

My ans is c^11/2 d8

[night egret]

should be 2(c^6)*(d^7)

[steel cairn]

Oh sirry i add not minus

Why is it 2(c^6)

Shouldnt jt be c^6 / 2

[slender walrus]

how are you getting division by 2

[steel cairn]

Wait no

I meant

[slender walrus]

yeh, so like i asked, how are you getting division by 2
why is 2 in the denominator

[steel cairn]

When i asked a guy

For this proglem

He said ans was

[slender walrus]

yes, that would be correct for that problem

based on the values initially present

$\frac{8}{32} = \frac14$

[grand pond]

ℝαμΩℕωⅤ

[slender walrus]

$\frac{14}{7}$ is NOT $\frac12$

[grand pond]

ℝαμΩℕωⅤ

[midnight plank]

@steel cairn Has your question been resolved?

[steel cairn]

Oh okay, tysm

.close

[midnight plank]

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[midnight plank]

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[twilit field]

if $d|n$, then prove $2^d-1|2^n-1$