help-49

[oak nymph]

Yes

Think why

[narrow edge]

cuz 5 linear terms are being multiplied

hey im actually understanding ts

:D

there are 5 ways for x^4 x y because there are 5 terms in the expression

shouldnt the same be applicable to x^3 x y

[oak nymph]

You mean x³y?

That term will not occyr in the expansion

[narrow edge]

sorry x^3 x y^2

[oak nymph]

Yes

[sudden yacht]

(Don't use x as multiplication symbol)
If you really really want, at least write * not x.

× is typically used in elementary schools when you have no numbers, especially no x appearing

[narrow edge]

why is x^4 * y wrong

[sudden yacht]

But here I strongly suggest to use either nothing (the best) or •

[narrow edge]

how to use those symbols

[oak nymph]

No you are correct there are 5 ways to get x^4*y

I meant the x for multiplying and x variable was confusing

[sudden yacht]

I think you have them on your keyboard (if you're using your mobile phone)

[narrow edge]

im using my pc

[oak nymph]

How many ways will be there to get x^3*y^2

[narrow edge]

by our previous logic shouldnt it be 5?

why is it 10

[sudden yacht]

Then you can copy-paste from the Internet or maybe you could try finding them with the combination "Win + ."

[oak nymph]

Because there are 10 ways to pick 2 things from 5 things. So there are 10 ways to pick two xs from the brackets and the rest have to be ys which makes x^3*y^2

[narrow edge]

are brackets fine?

[sudden yacht]

Very weird

I don't suggest you use them to denote multiplication

[narrow edge]

ohhh

and for the x^4 * y case its 5 choose 1

[oak nymph]

Yes

[lethal path]

yeah if you have to choose exactly 1 y, you must have the rest being xs, or 4 xs

[oak nymph]

What about the x^5*y^0 case

[narrow edge]

got it now

[sudden yacht]

From the best to the worst notation (in my opinion):

• x³y²
• x³ • y²
• x³ * y²

@oak nymph

(Now I won't interrupt you any more, sorry)

[narrow edge]

0! has 1 arrangement

[lethal path]

so it's pretty clear there are 5 ways to do that, since you can choose 1 y from any of the 5 brackets

[oak nymph]

Correct

[lethal path]

similarly x * y^4 is 5 choose 1

it's the ones in the middle that are tricky

[narrow edge]

ahh

this is beautiful

thanks guys

sorry for opening a new channel

.close

[midnight plank]

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[lethal path]

you can do:
XXOOO
XOXOO
XOOXO
XOOOX

OXXOO
OXOXO
OXOOX

OOXXO
OOXOX

OOOXX

so that gives you 4 + 3 + 2 + 1 ways for 5 choose 2 and 5 choose 3

[narrow edge]

XO?

[lethal path]

in general n choose 2 = 1 + 2 + 3 + ... + (n - 1)
= (n - 1)(n - 1 + 1)/2 = n(n + 1)/2

[midnight plank]

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[languid mica]

How do i find new limits of integration on Gaussian integral if its from 1/2 to 3/2

[languid mica]

When converting to polar

[sharp coral]

if the gaussian integral doesn't go to infinity then the polar conversion trick doesn't really help because then you have a square region

[languid mica]

How can i convert it to polar?

[sharp coral]

and the only way to solve it is either in terms of the error function or numerically

[languid mica]

Or i cant

[languid mica]

Why would it not work

[sharp coral]

you can convert it to polar, but polar only works well for circular regions, and doing the trick where you square it would give you a square region

[languid mica]

Cant i find what the rectangle 1x1 starting at 1/2 1/2 transofroms to in polar?

[sharp coral]

you certainly can, but it will not be nice to integrate

[languid mica]

Wdym

How do i find the new limits

[sharp coral]

well you would need to convert the lines x = 1/2, x = 3/2, y = 1/2, y = 3/2 into polar coordinates

[languid mica]

Meaning ?

Ill get a rectangle no?

[sharp coral]

well do you know the polar coordinate conversion formulas?

[languid mica]

As in with matrix ?

Like the 2x2 matrix

[sharp coral]

no like [ x = r\cos\theta \qquad y = r\sin\theta ]

[grand pond]

cloud

[sharp coral]

not the jacobian

[languid mica]

Yea ofc

But thats what i substitute

To have a jacobian

[sharp coral]

but you also substitute those to get your bounds

[languid mica]

Yes

Problem is its 2 variables

For each bound

So how do i find them

[sharp coral]

well you would end up with variable bounds for the inner integral

the usual way is for the inner integral to be with respect to r, then the bounds will be the bounds on r as a function of theta

are you familiar with double integrals with variable bounds?

[languid mica]

How would i calculate it then

If im stuck with cosθ and sinθ in bounds

[sharp coral]

because there is an outer integral with respect to theta

your integral ends up looking like
[ \int_{\alpha}^{\beta} \int_{g(\theta)}^{h(\theta)} f(r, \theta) r \odif{r,\theta} ]

[grand pond]

cloud

[languid mica]

Yea how will i integrate e^(1/^2cosθ....

What if i put first dθ

[sharp coral]

well that's the point where you realize that converting to polar wasn't very helpful in calculating this integral

[languid mica]

In the inner integral i first do θ

[languid mica]

So how do i calculate it

[sharp coral]

either in terms of the error function or numerically

[languid mica]

Like calculate it fully

Not approximation

[sharp coral]

there is no formula that will give it to you in terms of elementary functions

if you want a "precise" formula then there is no avoiding the error function

[languid mica]

Can i not do any trick to find it

Like some coordinate system conversation or anything

[sudden yacht]

No you can't.
Even if you insist lol

[languid mica]

Idrk what the error function is you keep referring to

[languid mica]

Is that proven or

[sharp coral]

the tricks for finding the gaussian integral only worked because the limits were from -infinity to infinity. you could do a similar trick for 0 to infinity, but for any other bounds you run out of luck

[sudden yacht]

Well then you really have no way to do

[languid mica]

Is there any other coordinate change or jdk

[sharp coral]

https://en.wikipedia.org/wiki/Error_function
https://mathworld.wolfram.com/Erf.html

[languid mica]

Like its proven that it cant be calculated in terms of elementary functions

[sharp coral]

you can do plenty of coordinate changes but none that would give an elementary formula for the integral

[languid mica]

How do you know

[languid mica]

What is this

[sudden yacht]

The error function...

You said you don't know what it is

[languid mica]

Oh

What if i do dθ first

[sudden yacht]

Try it and see what you get

[languid mica]

What θ bounds though

[sharp coral]

do the same thing you did to find the r bounds but solve for theta instead

[crystal plover]

It is known that no antiderivative of e^(-x²) can be written as finite combination of elementary simple functions

The error function erf(x) is basically defined to be its antiderivative (up to some constants)

You can keep trying, but it won't get you anywhere

[sudden yacht]

He won't like this fact lol

But this is a matter of fact 🤷‍♂️

[dawn dagger]

bruh he opened a help channel

[languid mica]

I opened before

[midnight plank]

@languid mica Has your question been resolved?

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[woeful turret]

can someone help me do this using binomial approximation

[fallow scarab]

do you mean taylor expansion?

[woeful turret]

sorry

yes

i mean ill send u a pic..im not really sure of the method name

i put theta = h+pi/4

so h--->0

and this is what i got

im kind of new to limits..so are we allowed to expand it like this?

[fallow scarab]

quite messy and hard to follow

[woeful turret]

i got the answer by simplifying
cos theta + sin theta as sqrt 2 ( cos (pi/4 - theta)

[woeful turret]

but im unsure why this method was not successful

[woeful turret]

^^

[dreamy lichen]

uh so you used the fact that sinx ~ x for x near 0?

It's quite hard to follow, but if you used that, which it seems like you did, then its incorrect

[woeful turret]

uh no i used sin x expansion

[dreamy lichen]

which is?

x - x^3/3!?

[woeful turret]

x-(x^3)/3!

yeah

[dreamy lichen]

this works pretty well, as long as x is near 0

unfortunately, theta + pi/4 isnt near 0 (as theta -> 0)

[woeful turret]

yeah

[woeful turret]

thats why i put theta as h+pi/4

so that h--->)

h-->0

oh

[dreamy lichen]

That doesnt solve the issue

[woeful turret]

wait i get ur point

[dreamy lichen]

the argument of sine is still around pi/4

[woeful turret]

i did not know it was only applicable for x near to 0

[dreamy lichen]

if x is far from 0, you would have to use the entire expansion

which is infinite

so that wouldnt be very helpful

[woeful turret]

thats why when i simplified it to this form it worked

[dreamy lichen]

or alternatively, you would have to make an expansion around x = pi/4

[woeful turret]

oh ok got it 👍

thank you

[dreamy lichen]

now when theta -> pi/4, theta - pi/4 -> 0

so the argument of cosine goes to 0

so the approximation should work fine

but it's also one of the well known limits

(1 - cosx) / x^2 -> 1/2 (as x -> 0) iirc

[woeful turret]

whats iirc

[dreamy lichen]

"if i remember correctly"

[woeful turret]

oh lmao

okay thank you

.close

[midnight plank]

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[desert siren]

General result is (1-cos(mx))/x^2 limit as x approaches 0 is m^2/2 :)

[midnight plank]

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[mint ravine]

Wait, are the cousins considered distinguishable or not?

[mint ravine]

I can't really tell by the quesiton

question*

[quartz hornet]

i would start by asuming humans are distinguishable yes

[desert siren]

Lolz

[mint ravine]

i mean i got 14 for that case

which is wrong

[desert siren]

14 is so less

Wait

[mint ravine]

wait i copied the question wrong

there are 3 rooms not 4

[desert siren]

i just worked and got 15

and then the question gets changed

[mint ravine]

sorry i have to go now i'll keep this channel open

[quartz hornet]

[4] 1
[3,1] 4
[2,2] 6
[2,1,1] 6
?

i liked your method , did i miss something brandon

[mint ravine]

how do you get 6 for 2,2?

you're overcounting

[quartz hornet]

oh right 3.

2,2
4C2/2

[desert siren]

Whats the formula

Ah

I was using groups

Which should be valid too

[quartz hornet]

s(4,1)+s(4,2)+s(4,3)?

[desert siren]

Officially zonked

[quartz hornet]

oh this is further generalised too tf

duly noted thanks

[mint ravine]

.close

[midnight plank]

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[sand flume]

So for the part where he starts -a≤x≤a he says -x≤a. So I understand he uses the given and just multiplies it by negative one. But what about if you looked at it like this. If x less than 0 |x|=-x. -x<x≤a. So -x<a but then we have two definitions

[velvet compass]

If x is less than zero then -x < x is false.

[sand flume]

Oh didn't think it through sorry. Thanks again

.close

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[bright shoal]

I have some doubt about Goldbach Conjecture

[lyric charm]

!da2a

[midnight plank]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[lyric charm]

ask away ig

[oak nymph]

why did you delete your question (same as this) in https://discord.com/channels/268882317391429632/1018703538517454919?

[lyric charm]

let him spit it out

[oak nymph]

he asked already, "can we solve it"

[lyric charm]

@bright shoal

[midnight plank]

@bright shoal Has your question been resolved?

[lyric charm]

no question was ever asked

[bright shoal]

So Sorry

i am struggling with network issues

When i wrote the question in help channel 48

help channel did not get occupied by me

[lyric charm]

click ❌ on this

[bright shoal]

no like i have a doubt

[lyric charm]

well then repost the question here

[bright shoal]

Does anyone have any info on goldbach conjecture

i need to ensure before asking the actual doubt

[lyric charm]

click ❌ on bot's prompt unless you want the channel to close

[lyric charm]

no you need to ask the doubt now

[bright shoal]

So like is there any exact formula to calculate the number of goldbach pairs for even number

I tried looking up but like could not find any valid source

Mind looking on this?

This is like even number v/s goldbach pairs for each upto 100000 number

s

[pine wave]

Probably computed by computer program

[bright shoal]

i mean number of goldbach pairs

[bright shoal]

Yes but like i was just interested in asking do we have any formula to calculate the number

of pairs

Sorry guys if like this is not suitable for here but like i asked this in Python and WoC

[pine wave]

Probably doesnt get any easier than "sieve out primes and check"

[bright shoal]

AND No one had knowledge

I mean the servers

[oak nymph]

if we did, it would probably not be hard to decide then if no. of pairs is >0 for all even numbers or =0 for some even numbers thus solving the conjecture

[bright shoal]

Yes

Yes

i am researching on this

[pine wave]

Granted some fast convolution algorithm can help but... thats about all the computational speedup you have

[carmine sigil]

Yes, there is, but it involves calculating it by simply enumerating the possible pairs, and checking if both numbers are prime

[bright shoal]

I have been working and learning prime checking algos goldbach conj primality test etc since few days

[carmine sigil]

oh I was scrolled up lol

[bright shoal]

So like i thought maybe working on a formula would lead in the direction of solving it

Then i came here to like

cross verify that are their any approximations

and there were as Omnipotent told

[bright shoal]

Convolution algorithm?

Could you describe some more i could apply that

cause current algo is slow and like sieve takes up space

lets say for n>10^12

[pine wave]

Sorry cant help for that

Sieve + convolution can only take you to 10^9

[bright shoal]

No like describe some about how to define a convolution algorithm

I am writing this down one sec

[pine wave]

Look for FFT (fast fourier transform)

[bright shoal]

scipy.fft ?

[carmine sigil]

@bright shoal do you understand what a convolution is?

if not, you might need some additional background education before you can hope to tackle and open problem 😉

[midnight plank]

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[pine wave]

You kinda want exact answers so you need to work in a finite field

[bright shoal]

rough idea like somewhat combining two functions

[pine wave]

.reopen

[midnight plank]

✅

[bright shoal]

to like produce another right?

Ok i am writing this down too

[carmine sigil]

it's a specific method for doing that, but yeah that's the gist.

[bright shoal]

So like fft s would help in this case

[pine wave]

basically "count number of pairs summing up to x" if you need answers across different x can be processed together by running something like a fourier transform

[bright shoal]

i checked fft and i could implement them using scipy

but how to define a convolution for this

[pine wave]

Its a discrete convolution

[bright shoal]

yes

wait

No like we need to count sum of prime numbers summing to x right?

[pine wave]

Another way is encoding the entire list inside a big integer and doing multiplication

But of course number of digits increases with how big your n is

[bright shoal]

oh so you mean like first calculate all by running a sum and sieve and then use some fourier transform or so

[bright shoal]

the sieve?

[pine wave]

So you would be working with squaring roughly a 10^10 digit number when doing up to n=10^9

[bright shoal]

oof

So like implementing a convolution algorithm would speed up ?

Man sorry but i am liking getting a little overwhelmed here

Sorry

[pine wave]

No better if the multiplication already uses convolution

[carmine sigil]

well, here's a question: what would checking a whole bunch of numbers accomplish in this case?

[bright shoal]

I am thinking

[carmine sigil]

if the goal is to prove the goldbach conjecture, then verifying it up to 10^20 or 10^100 or 10^10^10^100 squintillion isn't sufficient.

[bright shoal]

the goal is to analyse more data

to like derive an EXACT formula

A question

[pine wave]

sounds like xy problem

!xt

!xy

[midnight plank]

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

[pine wave]

Oh wait not exactly

[bright shoal]

Lets say one derives a exact formula to like find the number of goldbach pairs

how we will verify it that it works for all n

like back to the whiteboard typa thing

until its a rigorous proof

[pine wave]

Is it going to be easy looking at 10^9 numbers and picking out a formula

[pine wave]

You already can see this and you cant imagine a formula which does this

What will more data give you?

[bright shoal]

Look i am just asking if on rigorous analysis lets say you define a formula or function which fits even 10^4 numbers and fails for 10^7 ?

i just mean how would it be verified that formula is true for all n until its derived from a rigorous proof?

right?

Like omnipotent said the case is same with the goldbach conjecture

[bright shoal]

Man my original question was like is there a formula but when you told about convolution algorithm i got eager sorry

i was eager to like speed up

and run for more numbers faster

Could you help me like develop and understand how would this help in working out number of pairs for more numbers

If it is pointless then sorry for taking time of you all and thanks for the help to all the people who came here

[pine wave]

You run against output limits rather easily when you want to break past 10^9

To produce m bits of output takes O(m) time at least

[bright shoal]

Right

Man i want to like understand what was about the convolution algorithm you were telling

sieve can be implemented how to implement that

lets say we only want to test upto 10^9

Man i am not interested in asking ai or something thats i would be grateful if you could help

😔 🥲

[pine wave]

Task: calculate 110110111 * 110110111 and see how the digits of the result match with number of ways to make 6 to 20 with 2 primes

[bright shoal]

through sieve we would get O(Nlog(logN)) to find prime list

lets say N is 10^9 here

thats done

[bright shoal]

ok lemme check

6 to 20 with 2 primes?

[pine wave]

Essentially only the even numbers

[bright shoal]

like every number of pairs for each number from 6 to 20 right

[pine wave]

Since the digits of 110110111 encode 19 17 13 11 7 5 3

Actually it works all the way up to 22

,calc 110110111*110110111

[grand pond]

Result:

1.2124236544432e+16

[bright shoal]

12,12,42,36,54,44,32,321

yes

[pine wave]

6: 3+3
8: 3+5 5+3
10: 3+7 5+5 7+3
12: 5+7 7+5
14: 3+11 7+7 11+3

[bright shoal]

right

[pine wave]

So essentially something similar can calculate number of ways

Though you may need to pad the digits further apart if it gets to more than 10/100/1000/... ways

[bright shoal]

man one doubt

this like calculated number of ways to make 6 to 22 with 2 primes

[pine wave]

Yeah

[bright shoal]

oh

Bro sorry

but i am like confused how this relates to the number sorry again

i am like slow

Sorry again

[pine wave]

Ones digit: ways to make 6
Tens digit: ways to make 8
Hundreds digit: ways to make 10
... all the way until 22 (24 is also correct but 26 is missing 23+3 and is hence wrong)

[bright shoal]

no way

what is this sorcery

no way

Man i cant still believe it

No way

like you wanna say

encoding and multiplying two large numbers gives me no of pairs

beginning from ones digit

no way

[pine wave]

$110110111=\sum_{3\leq p\leq19,\text{prime}}10^{(p-3)/2}$

[bright shoal]

What kind of magic math is this

no way

no way

[grand pond]

Element118

[pine wave]

$110110111^2=\sum_{3\leq p,q\leq19,\text{prime}}10^{(p+q-6)/2}$

[grand pond]

Element118

[bright shoal]

Man this is sorcery indeed where do we learn such math

[pine wave]

By living in my head

[bright shoal]

Man

you are too genius please stay

i now definitely know with your help imma gonna build the algo

please

[pine wave]

idk if the algo is going to achieve anything

[bright shoal]

but like this is like incoding all number of pairs how

[pine wave]

i have better projects for myself

[bright shoal]

🥲

man you single handedly calculated 6 to 22 number of ways to prime

using like multiplication

which can be made faster using ffts

🥲

man your knowledge is helping in a good way

🥲

So like how do we encode for big N

lets say 100

[pine wave]

lemme show you for up to 100

[bright shoal]

thanks man

just teach me this i can implement fft for faster multiplication myself thanks

[oak nymph]

i think most programming languages already implement it

[carmine sigil]

Well, I wouldn't say most. You generally have to use a bignum library to get fft based multiplication

[bright shoal]

Yes

[pine wave]

$100000001000001000100000101000100000101000001000100010100010000010100000100010100010100010101=\sum_{3\leq p\leq97,\text{prime}}100^{(p-3)/2}$

[grand pond]

Element118

[bright shoal]

so like fft would do multiplication with time complexity O(nlogn)

right

what

Man such big numbers is this possible to achive 10^6 with this?

i mean like pairs for prime numbers

[carmine sigil]

https://en.wikipedia.org/wiki/Schönhage–Strassen_algorithm

[pine wave]

100000001000001000100000101000100000101000001000100010100010000010100000100010100010100010101 if i didnt make a mistake

[bright shoal]

Yes i will implement that

[carmine sigil]

but like, don't

use a library

where it's already implemented

[bright shoal]

I am like wanting to work as much as i can this summer on primes

without ai

i feel scipy has it l

[quiet hinge]

the libraries are made without ai

hopefully

[bright shoal]

Omnipotent do you have any knowledge

[bright shoal]

yes these are like libraries built by people

these are for years

Python is approx 33 or 34 years old man

Thats why people say everything is there in python

Omnipotent how is Element118 doing this trickery

is there any math i can learn?

[pine wave]

I mean i finished my undergrad in math

[bright shoal]

lemme check the digits of this number

[pine wave]

In general use the summation to calculate the number

[bright shoal]

93 digits

oof

[pine wave]

Its gonna go up to about 10^10 digits for going up to 10^9

[bright shoal]

Man this is straight up sorcery

oh

like 10^10 digits

oof

[oak nymph]

if we want to start encoding primes from some q, we can make the exponents in the summation (p-q)/2

[bright shoal]

and like fft has nlogn where n is number of digits right?

[oak nymph]

does that help in saving computation?

[bright shoal]

fft would save

[carmine sigil]

I showed you the page with the algorithm, you just need to read and understand if you want to learn how to multiply. There's also, like, the Russian Peasant's algorithm for multiplying numbers. element 118 could have been using that as well.

[bright shoal]

no like how is encoding these primes

and like the sum?

[pine wave]

Russian peasant is still quadratic time tho

[bright shoal]

Thanks again

Sure i am gonna read

[pine wave]

@bright shoal

Check that you can derive this

[bright shoal]

Which

oh that one

so we need to sum

oh

so like we just need to sum for the prime numbers we want?

and we get the number

like the bounds

right?

[pine wave]

Hard to get bounds this way

[bright shoal]

But lets say we do for 3<=p,q<=10^9

so we need 10^9 prime number right

oh shi

we need all 10^9 numbers

by sieve

[pine wave]

Yeah you still need to sieve

[bright shoal]

O(nlog(logn))

ok

and like lets say we get the number

how much complexity does the sum have

[pine wave]

But after you sieve you construct the big number, square it then read the digits

[bright shoal]

this sum has what complexity

no i mean at last we like sum

thats why i am saying

[pine wave]

You may need to work in the computers native base to speed up

[bright shoal]

like numbers would be summed up

[pine wave]

Which is likely a power of 2

[bright shoal]

I am like tryna estimate

this like how much tc this all has

[pine wave]

So instead of 100^... you might want to do 1024^...

[bright shoal]

Man you all have so much knowledge why dont write a paper about this

[pine wave]

because its essentially useless for this problem

[bright shoal]

Man i am shocked

[bright shoal]

why?

like its making finding pairs faster right

its useful

[pine wave]

Calculating lots of values brings you no closer to solving

[bright shoal]

;-;

i know you are right but this solves my current problem of computing

bigger number pairs

ok so like

Sieve

to find bounds

then we sum to get number

what is tc of sum ?

fft has O(nlogn) for n bits right?

[pine wave]

You may have to optimize that part

Summing up - best to see how the numbers work to see what base to use instead of power of 10

I only did power of 10 for human understanding

[midnight plank]

@bright shoal Has your question been resolved?

[bright shoal]

@pine wave Sorry i went

actually i need to go

i will come

hopefully this does not reset due to timeout

[pine wave]

It kinda does

But at least the messages stay on this server

If you need to dm me, you may, but i may not be available at all times

[midnight plank]

@bright shoal Has your question been resolved?

[analog canyon]

hiiii i'm new here

[sudden yacht]

!occupied

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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).

[analog canyon]

oh sorry

[sudden yacht]

Don't worry

[midnight plank]

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[inland patio]

I'm reading on Wikipedia that polynomials in one variable over a field F are zero on finite sets or on all of F. The latter case occurs iff it is the zero polynomial, correct?

[lyric charm]

yes

[inland patio]

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[chilly cobalt]

Let ABC be a triangle with orthocenter H. The tangent lines from A to the circle with diameter BC touch this circle at P and Q. Prove that H,P,Q are collinear

[chilly cobalt]

no idea😓

i know that PQ is polar of A wrt to the circle but how do i show that H is on the polar??

[midnight plank]

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[waxen knot]

For what values ​​of the parameter m the equation |x − 1| + |x + 2| = m has
maximum number of solutions?

[topaz epoch]

Consider the x-values where the moduli change sign

Then just do case bashing

[midnight plank]

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[sand flume]

How can I properly format this proof

[runic hamlet]

you should be more concerned with the fact that what you are trying to prove is false

[blissful pier]

Counterexample: x=1

[dreamy lichen]

Seems like you intended to write |x| = 0 <=> x = 0

[sand flume]

Yes but if you plug that in the definition you get 1

[sand flume]

How so?

[sand flume]

Yes

[blissful pier]

but |1|=1, however 1≠0

[dreamy lichen]

You wrote |x| = x <=> x = 0 though

not |x| = 0

[sand flume]

Oh oops

Ok sorry for that but then what is the right way to format this because it looks ugly

[sage helm]

Generally when you're doing proofs of an iff (<=>) statement, you'd just break it in two parts, and there's really no need for the bullet point or the "Work" subheading

[dreamy lichen]

You could either go for a one liner
|x| = 0 iff ...... iff x = 0
or split it into the two directions as neon suggested

[sand flume]

So I need to show that if |x| = 0 then x=0 and the other way around right?

But what about the actual formating itself with latex

[sage helm]

If. Lorem ipsum dolor...
<par> Only if. Lorem ipsum dolor...

¯_(ツ)_/¯

[runic hamlet]

you can also do

\begin{enumerate}
  \item[$\implies$] this direction
  \item[$\impliedby$] the other direction
\end{enumerate}```

[sand flume]

Also I just want to make sure it would be necessary to define the absolute value definition correct?

[runic hamlet]

at some point, sure

but not in the proof

[sand flume]

Why is that because don't we need to refer to it when evaluating |x|

[runic hamlet]

sure but that doesnt mean we have to copy it down from whereever else we defined it

\begin{enumerate}
  \item[$\implies$] Let $\abs{x}=0$. We consider two cases. Let first $x\geq 0$. From the definition of the absolute value it follows that $0=\abs{x}=x$, i.e. $x=0$. Let now $x<0$. Then from the definition of the absolute value it follows that $0=\abs{x}=-x$, hence $x=0$. In both cases we see that $x=0$.
  \item[$\impliedby$] Let $x=0$. Then in particular $x\geq 0$ and from the definition of the absolute value it follows that $\abs{x}=x=0$.
\end{enumerate}```

for example something like that

but proof writing and formatting is somewhat subjective and other people will write and format it differently

[sand flume]

Ok thanks so much

.close

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[sage helm]

also if you think the arrows are too long, you can use \Rightarrow and \Leftarrow instead

[runic hamlet]

you can also put them in () or in "" or something else

[sand flume]

Ok thank you

[midnight plank]

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[sullen scaffold]

Hello,

Bayesian inference: Where does it comes the ' in the integral?

[sullen scaffold]

Or am I overcomplicating stuffs and it's simply the theta?

[subtle blaze]

It’s just another symbol

It’s being integrated away

It’s just the dummy symbol for integration

[sullen scaffold]

Aaah right the dummy symol.

I did overcomplicating stuffs. Thank you.

[midnight plank]

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[upbeat timber]

,, \int_2^n \frac{\ln(x)}{x^3} \dd{x}

[grand pond]

criRata

[upbeat timber]

is ibp the only way to do this

like its really long bruh

[lyric charm]

not a lot else you can do

[upbeat timber]

how do i verify if my work is correct?

[lyric charm]

look over it painfully

[dire tundra]

hi anyone?

[upbeat timber]

no

this is my channel , go read those introduction channels

rules and other shit

[dire tundra]

i need to know what are the prerequisites of abstract algebra

[upbeat timber]

bruh

or ask in #math-discussion

if its a question related to a hormework or smth , open a channel

[zealous schooner]

Do you know the tabular method of IBP? Might prove to be a bit easier

[dire tundra]

bro its fine i'll ask them

[upbeat timber]

hmm , i just use u ,du

and then integrate by parts

should i learn it ?

[zealous schooner]

It's basically the same but using the tabular method can be easier to write

https://www.youtube.com/watch?v=2I-_SV8cwsw

[upbeat timber]

alr thx

.close

[midnight plank]

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[dusty portal]

You can get an easier IBP with the sub u=ln(x)

[upbeat timber]

.reopen

[midnight plank]

✅

[dusty portal]

(or just a faster one)

[upbeat timber]

x =e^u

,, \int \frac{u}{e^{2u}} \dd{u}

[grand pond]

criRata

[upbeat timber]

is this where i ibp @dusty portal ?

[dusty portal]

Yes

[old ravine]

proof writing

parsing logical statements

[dusty portal]

$\int ue^{-2u}du$

[old ravine]

set theory

[grand pond]

;(

[dusty portal]

!occupied

[midnight plank]

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).

[upbeat timber]

fair

.close

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[midnight plank]

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[sand flume]

How can I prove |. |x|-|y|. |≤|x-y|

[exotic stratus]

!showyourwork

[midnight plank]

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

[sand flume]

Ok so I have attempted it in two ways

@exotic stratus

I then tried using the triangle inequality but that was also a dead end

[exotic stratus]

What domain are you working in?

i.e. are x and y real or complex numbers?

[sand flume]

Real

[exotic stratus]

If you go case by case this will be lengthy but that is the simplest way

[sand flume]

Yes I tried that but I must have made a mistake which I cannot seem to find since it gets equal not less than or equal

Quick question you state that case work is the longest what is a more efficient way then?

[exotic stratus]

Yes, you aren't considering the differnt outputs when |x| > |y|, |x| = |y|, |x| < |y|

[exotic stratus]

Prolly something with triangle inequality as you also guessed, though I don't see it immediately, I think there prolly does exist one.

[sand flume]

Yes I saw this on stack exchange: https://math.stackexchange.com/questions/127372/prove-that-x-y-le-x-y

But it wasn't clear how they achieved the final results

[sand flume]

Why do we have to consider these? So if I consider these do I need to look at the different values of x and y. Or is this more important than that. And if so why

[midnight plank]

@sand flume Has your question been resolved?

[dreamy lichen]

ill have to go soon so i prob wont be able to explain it, but this is wrong

if x < 0, then |x| should be replaced with -x

you replaced it with |-x|

it's not true that ||x| - |y|| = |x-y| in general

it's true for some specific cases (x, y > 0 or x, y < 0) but not in general

[sand flume]

.close

[midnight plank]

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[midnight plank]

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[safe onyx]

https://www.desmos.com/3d/gyvt8pb7mg

I tried graphing it

but like the equation for their intersection only gives it in x and y

$6x^{2}+4y^{2}+5\left(x^{2}+y^{2}\right)^{2}=30$

[grand pond]

smeagol

[safe onyx]

but I'm not quite sure how to get the tangent lines

I think z is constant when they intersect

since it's like a slice of the elipsoid

but this makes me doubt that now

[fierce arrow]

z isn't constant when they intersect

[safe onyx]

since I see purple inside of the red and blue on the outsdie

[fierce arrow]

z = x2 + y2

[safe onyx]

but I am not sure what strategy to use to get the tangent line

[desert siren]

Implicit diff for the tangent line from this, I think

[safe onyx]

F(x,y) = ...

so

[fierce arrow]

Do the gradient of both and then cross product

[safe onyx]

the cross product gives a line perpendicular to both

so it would be tangent to both?

[fierce arrow]

The cross product of the gradients at that point will give you a directional vector of the tangent of the tangent line to the curve of intersection

[safe onyx]

I got <48,21,-10>

as the cross product

wait i messed up

<-2,2,-1> gradient at (-1,1,2) for the first eq

[fierce arrow]

I got <48,52,8>

[safe onyx]

<-12,8,20> for the gradient of the sec

[fierce arrow]

gradients: -2 2 -1 and -12 8 20

[safe onyx]

I messed up with the first eq gradient's x comp

[safe onyx]

I got this now yeah

so now that I have the direction and point I am all done?

[fierce arrow]

Yeah

[safe onyx]

just x= 48t-1 etc

[fierce arrow]

just do x = xo + at

yeah

[safe onyx]

thank you

.close

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[dusty portal]

googoogaga

[dusty portal]

Mb bro

anyways this is just killing me

Note: $\kappa(a, r)={z\in\bC:|z-a|=r}$\
$I=\int_0^{2\pi}\frac{\frac12+\frac12\cos(4\theta)}{1-2p\cos(\theta)+p^2}d\theta}=\frac12I_2+\frac12\int_0^{2\pi}\frac{\cos(4\theta)}{1-2p\cos(\theta)+p^2}d\theta=\frac12I_2+\frac12I_3$, where $I_2=\int_0^{2\pi}\frac{1}{1-2p\cos(\theta)+p^2}d\theta$ and $I_3=\int_0^{2\pi}\frac{\cos(4\theta)}{1-2p\cos(\theta)+p^2}d\theta$

[grand pond]

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

[dusty portal]

I'm slow at typing today

Jesus

So I know that $I_2=\int_{\kappa(0, 1)}\frac1{1-p(z+z^{-1})+p^2}\frac{dz}{iz}$ with the sub $z=e^{i\theta}$

Nah wrong

[grand pond]

;(

[autumn canopy]

I've seen something very similar; try to re-writing it as [\frac{A}{2i(1 - qz)} - \frac{B}{2i(1 -q \overline z)}] where $z \coloneqq e^{ix} = \cos(x) + i \sin(x)$, in the case of what I've seen, namely [\frac{q \sin(x)}{1 - 2q \cos(x) + q^2},] that was the way to go and rewrite this into a series

[grand pond]

Kepe

[dusty portal]

Oh

I don't think I need series here

[autumn canopy]

This is really funny because it was on the last homework sheet at my uni

[dusty portal]

💀

[dusty portal]

Trying to find a fourier transform or something?

[autumn canopy]

Maybe it could be rewritten in a similar fashion

[dusty portal]

Or just a conversion

[autumn canopy]

Just conversion

[dusty portal]

I think I have the gist of it

Like, I "solved" it in my notebook but my working is very convoluted

I know how to get to the answer, I'm just not being precise

[dusty portal]

For convienience I'm just going to call $\kappa(0, 1)$ $\gamma$

[grand pond]

;(

;(

;(

[dusty portal]

(quadratic explaining the simple pole part)

and thennnnnn

Ugh, since 0<p<1 we have that 1/p is outside K(0, 1) but p is inside

So we only need the residue at z=p

[grand pond]

;(

;(

[dusty portal]

So that is what I was missing

Now I guess we need to calculate I_3 with the same approach

[grand pond]

;(

[dusty portal]

Ok I see it now

Uhmm

[grand pond]

;(

;(

[dusty portal]

YIPEE

I am NOT TeXing all of this into Overleaf

[patent snow]

Use LaTeXOCR

[dusty portal]

@gray widget Does it look good?

Fucking hell

<@&286206848099549185>

I gotta eat people

brr

GOOGOOGAGA

I NEED DA VERIFICATION

[midnight plank]

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[spark nebula]

please help me understand what is going on here

[floral depot]

<@&286206848099549185>

[floral depot]

Ik whats going on

math

jk

<@&286206848099549185>

[velvet compass]

Any info about how g(x) and f(x) are related?

[fallow scarab]

Show the full context

[velvet compass]

The rest if pretty simple

[spark nebula]

that's the full context

question 2

let me translate

[fallow scarab]

and which line confuses you first

[spark nebula]

second

let u = x + 1 , x: 0 -> 2pi that means that u : 1 -> 2pi + 1

but what they wrote is u : -1 -> 2pi - 1

that's the translation

[velvet compass]

This is why they shifted the bounds in the 2nd line, so that after the u-sub they become the same as beforehand

[spark nebula]

yeah but the bounds seems wrong

how it's u = x + 1

[velvet compass]

u = x + 1
for the 1st integral:
when x = 2pi - 1 --> u = 2pi
when x = -1 ---> u = 0
(works)
2nd:
u = x - 1
when x = 2pi + 1 ---> u = 2pi
when x = 1 ---> u = 0
(works)

[spark nebula]

alright alright let me try to understand so let's say that you trying to integrate x from a to b and then from a - 1 to b-1 you will not get the same result meaning the changing the bounds without a changing the variable does affect the integral's result

i don't know how it worked here

and i really want to understand so i can use it later

but

hold on

does that have any relation with the 2pi period

[midnight plank]

@spark nebula Has your question been resolved?

[velvet compass]

Yes

[grand pond]

Modus

[velvet compass]

In fact you can change the bounds in any way as long as they difference is 2pi

wait

I wrote it wrong

[spark nebula]

okay but again it works only for periodic funcs

[velvet compass]

Yes

[spark nebula]

Okay thank you so much

[velvet compass]

https://math.stackexchange.com/questions/4271615/when-computing-fourier-series-how-do-you-know-when-you-can-change-the-bounds-of

I recommend you to read this article

The guy literally had the same problem as yours

[midnight plank]

@spark nebula Has your question been resolved?

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[tidal turret]

f^-1 denotes the preimage

[midnight plank]

@tidal turret Has your question been resolved?

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@tidal turret Has your question been resolved?

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@tidal turret Has your question been resolved?

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[hybrid crane]

,w solve x - 3y = 1, -2x + y + az = 2, y + bz = b - 3

[grand pond]

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[midnight plank]

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[azure oracle]

yea can u translate it into english

[lyric charm]

Let B = ... be a basis of R^3 and f : R^3 -> R^3 be the linear map whose matrix wrt B for input and E for output is [see image].

Find a ∈ R such that f(1,2,-1) = (1,3,2) and for this value of a find all v ∈ R^3 such that f(v) = (1,3,2).

[safe yew]

thanks ann

[lyric charm]

E stands for the canonical/standard basis, yes? @tidal turret

[safe yew]

I was about to use my basic french to try and translate it

[tidal turret]

yes!!

i got some ideas

[lyric charm]

i'm riding on that and latin lmfao

[tidal turret]

i can find a such that the first condition holds

the problem is when i try to figure out the preimage of (1,3,2)

,w rref {{3,2,1,1},{1,2,1,3},{2,2,1,2}}

[grand pond]

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[tidal turret]

@lyric charm do you happen to see the mistake?

the answer is different, though, a = 3 is correct

[lyric charm]

uhh sorry im a little busy with other stuff and dont feel like reviewing your arithmetic and shit

[midnight plank]

@tidal turret Has your question been resolved?

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[tidal turret]

.reopen

[midnight plank]

✅

@tidal turret Has your question been resolved?

[hybrid crane]

@tidal turret $$ \beta = 2 - \frac{\gamma}{2} $$

[grand pond]

StrangeQuarkAL

[tidal turret]

yes xD

not (2-2gamma)/2

[tidal turret]

good catch, i found it a bit ago, fixed it, ty

[hybrid crane]

no problem

[hybrid crane]

isn't the (alpha, beta, gamma) you get from solving for the preimage expressed in B?

[tidal turret]

yes (x)B is equivalent to (f^-1(1,3,2))B

is a line of coordinates

[hybrid crane]

yeah that's what I got as well

[tidal turret]

what?

[hybrid crane]

for f(v) = (1,3,2)

[tidal turret]

show your work

because we differ in the last coordinate,

[hybrid crane]

one sec

oh nvm, it copied it wrong, it's -1

[tidal turret]

good

that's a nice way of approaching it

[hybrid crane]

thanks

[tidal turret]

but i think there is something

[hybrid crane]

what else

[tidal turret]

this should be the answer

[tidal turret]

i don't get the same line of solutions prof got, where is the mistake? 🤔

please don't leave, there must be a logical explanation

[hybrid crane]

i'm looking don't worry

[tidal turret]

we found (v)B correctly

[tidal turret]

if you check in geogebra this two (x)B lines are the same line

[hybrid crane]

oh

[tidal turret]

so we found (v)B correctly, but fucked up when finding all the v

there must be a logical explanation

[hybrid crane]

no, we found it correctly

(5, 0, -1) + t (-1, 1, 0)
and (3, 0, -1) + s(1, -1, 0) describe the same line

wait no

yeah no, my bad

[tidal turret]

we made a computational mistake somewhere

[hybrid crane]

ohhhhhh

[tidal turret]

🤔

[hybrid crane]

it's because of what we chose as our variable

[tidal turret]

i did both possibilities

[tidal turret]

i did (x,y,z)=(-1,y,4-2y) and (x,y,z)=(-1,2-z/2,z)

[hybrid crane]

your teacher copied the answer wrong

[tidal turret]

this is how i got the lines earlier, which both are correct representation of (v)B

[hybrid crane]

should be (5, -2, -1) not (5, 0, -1)
notice when they summed the vectors, they got (-L + 5, L - 2, -1)

so what we got before should still be correct

$$ (3, 0, -1) + t(1, -1, 0) $$
$$ (5, -2, -1) + s(-1, 1, 0) $$

are the same line

[grand pond]

StrangeQuarkAL

[tidal turret]

bro my teacher is stupid

[tidal turret]

you are a genius mate!!

[hybrid crane]

everyone makes mistakes :/

[hybrid crane]

ha idk about that but thanks

[tidal turret]

sure, but you shouldnt do them where other people will use your solution to correct themswlves

because i have been stuck in this one for 24h

[hybrid crane]

true true, it definitely becomes a hassle for everyone else

[tidal turret]

i appreciate it and sorry for wasting your time

[hybrid crane]

you're welcome and it's all good! Maths is what summer break is for

[tidal turret]

you are a genius

ty for your time !!

.solved

[midnight plank]

Channel closed

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[cursive swan]

no

[desert siren]

[midnight plank]

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[nimble kraken]

Hullo

[nimble kraken]

Should my answer give the eigenvector solution in terms of basis for W?

Like always?

[prime hornet]

there's nothing wrong with that

[nimble kraken]

for example here

[runic hamlet]

what you shouldnt do is give the eigenvectors as for example (0,0,1)

cause that is not an element of W

[nimble kraken]

Got it

thank you for confirming that's a good catch i should have thought of

.close

[midnight plank]

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[woeful turret]

is there a way to simplify or do i just notice that denominator is bigger than numerator

[carmine sigil]

@woeful turret what have you tried so far?

[silent zodiac]

Just use standard limit properties

[fallow scarab]

latter if you can prove it

[carmine sigil]

You can divide the numerator and denominator by 5^n and handle it term by term

[silent zodiac]

I think answer should be quite obvious

[carmine sigil]

(notice that 5^-n / 5^-n = 1, so this is an allowed operation)

[woeful turret]

by that u mean handle each term in the numerator?

[silent zodiac]

The thing is

[patent snow]

There's a hidden 9^n in the denominator

[carmine sigil]

and denominator

[silent zodiac]

9 is there in denominator

And 5 is there in numerator

And 9>5

[carmine sigil]

I see that now

[silent zodiac]

so 9^n>>>>>5^n

[carmine sigil]

So instead you'll want to use 9^-n / 9^-n

[silent zodiac]

I think u can tell its over

[silent zodiac]

Dont need it

Just neglect everything smaller

[carmine sigil]

do need it if you want to prove it

[patent snow]

The best solution is to factor out the numerator by the "largest growth term", and same for the denominator

[woeful turret]

can one person help

[patent snow]

So factor out the 5^n above and the 9^n below

[silent zodiac]

5^n/9^n

So

Some big number by a BIGGGGER number

[nova yoke]

proof by BIGGGG

[grim vector]

Also can be seen as q^n where |q| < 1 so it converges to 0

[woeful turret]

after this?

[silent zodiac]

Do u know

9^n is wayyyyyyy bigger

Than 5^n

When n approaches infinity

[patent snow]

It's (5/9)^n factor some things that converge to a real number

So by product you should get the answer

[silent zodiac]

0

[woeful turret]

oh ok so that is the only term we need to worry about? if that term was >1 then it would not converge?

[silent zodiac]

5/9 is less than 1

so its power

[patent snow]

It'd diverge to +oo (or -oo depending on signs)

[silent zodiac]

Approaches to 0

[patent snow]

This factoring out method is quite neat in general

[woeful turret]

ohh alr 👍 thank you

.close

[midnight plank]

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[midnight plank]

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[nimble kraken]

dim(U+W) = dim(U)+ dim(W) - dim(U interesct W)

dim(U+V) = P_3 = 4

dim(U) = 3

dim(W) = 1+ dim(U intersect W)

Hence D is correct

I stumbled looking at E. I felt like D would have be correct on exam, but I was unsure about why E was for sure wrong

can someone explain?

[prime hornet]

imagine if dim(W) = 4

the only 4-dimensional subspace of P_3 is P_3 itself

so W = P_3

then the intersection of U and W is just U

[nimble kraken]

Oh

ohhh

then its 4

and not <= 3

[prime hornet]

uhh, hold on

[nimble kraken]

did i oopsie

[prime hornet]

okay yeah. the formula is gonna read
dim(U + W) = dim(U) + dim(W) - dim(U intersect W) => 4 = 3 + dim(W) - 4

if you assume that W is dim 4

then you rearrange and get that dim(W) = 5

that's bigger than P_3

contradiction