elementary-number-theory

[hardy iron]

you're assuming the fractions aren't in simplified form yet

i think?

or even things like 6/3

15/5

etc.

[unkempt hedge]

oh shit

yes

That is a big mistake

[hardy iron]

lol dw i was convinced by it the first time i saw it

[unkempt hedge]

hmmm, what to do now

[hardy iron]

solve for k instead of m maybe?

2 - 13k = m(53 + 11k)

still looks ugly af

=tex /frac{2-53m}{11m+13}

[hardy iron]

=tex \frac{2-53m}{11m+13}

[unkempt hedge]

I know

[hardy iron]

oh wait

can't you solve it like the way you solved the first part?

[unkempt hedge]

I thought so too

[hardy iron]

oh i guess it's hard finding that first solution

oh

[unkempt hedge]

like if m = a/b then gcd(a,b) = b so ax+by = b

or something like that

but I don't know how I should do that

[hardy iron]

(2 -13k) = a(53 +11k) where a must be an integer?

idk how that would help

oh wtf... that's just a=m

i think you're more on the right track than me

[unkempt hedge]

@hardy iron You could rewrite m as -2 + (9k + 108)/(11k + 53)

[unkempt hedge]

@hardy iron I have been trying for a while now.
I found out that you could rewrite m as
11(m+1)(k+1) + 2(21m+k) = 13
which gives us another diophantine equation
(m+1)(k+1) = a
21m + k = b
11a + 2b = 13
a= 1 and b = 1 is a solution, so if we generalize we get
a = 1 + 2t and b = 1 - 11t

I: (m+1)(k+1) = 1 + 2t
II: 21m + k = 1 - 2t

II: k + 1 = 1 - 2t -21m + 1
substitute in "I"

I: (m+1)(1-2t-21m + 1) = 1+2t
now we have a solution for m given any integer value of t.

if you expand you will get a polynomial

21m^2 + m(19 + 11t) + (13t + 1) = 0

use quadratic formula, and then the possible values of t should be bounded between (some value which makes the square root negative) < t < (some value which makes the square root negative) because t is squared in under the square root, we get a polynomial.

(very unsure about this though)

Hmm, m is not forced to be an integer though 🤔 nvm, then

[sacred junco]

Has anybody thought of a solution for how to find prime numbers

[mossy folio]

Eratosthenes did.

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes @sacred junco

[sacred junco]

@mossy folio how about a formula to get you from one prime number to the next consecutive one

[mossy folio]

https://en.wikipedia.org/wiki/Formula_for_primes

No such formula which is efficiently computable is known.

@sacred junco

[north orbit]

a = (4d^2-30d)/(30-3d)
If a and d are positive integers, find all possible pairs of a and d

Could somebody help me with this please?

[drifting trench]

Long division probably?

[north orbit]

Ok I ended up with a remainder of 100/(30-3d)

First of all 0<d<10

What do I do from here? Do I find a value for d that makes 100/(30-3d) an integer?

Or what other ways are there to do this?

[unkempt hedge]

@north orbit
100/(30-3d) cannot be equal to an integer if d is an integer because we can rewrite the fraction as
100/ 3(10-d)
and we only get an integer if gcd(100, 3(10-d)) = 3(10-d)
and this cannot happen because
3 does not divide 100.

[sturdy dirge]

$$a=d\cdot\frac{4d-30}{30-3d}=d\left(\frac{-4}{3}+\frac{10}{30-3d}\right)$$.

[unkempt hedge]

@sturdy dirge 11(m+1)(k+1) + 2(21m+k) = 13 help me too😢

[sturdy dirge]

What is it ?

[unkempt hedge]

integer solutions

for m, k Z

[sturdy dirge]

$$11u+2v=13\iff (u, v)\in{(13+2w, -60-11w)|,\forall w\in\mathbb{Z}}$$\$$\left{\begin{matrix}21(m+1)+(k+1) &=-38-11w \ (m+1)(k+1) &=13+2w\end{matrix}\right.$$

[sturdy dirge]

@north orbit .

[unkempt hedge]

@sturdy dirge uhmm, so there are infinite solutions?

[sturdy dirge]

Maybe, multiply the last equation by 21, you will be able to express m, k.

[north orbit]

@unkempt hedge oh I might have screwed something up because I know there is supposed to be a solution

@sturdy dirge what does that tell us?

Your equation that you wrote above

[digital kettle]

hey

how do i solve

=tex x^2 + xy + y^2 = x^2 y^2

[digital kettle]

in the integers

?

the hint says use infinite descent

but i don't see how

<@&286206848099549185>

[drifting trench]

$$(x+y)^2=xy(xy+1)$$

[digital kettle]

um

so?

[drifting trench]

Idk

I was thinking

Wait.

Or $$x^2+y^2 +xy(1-xy)$$

[drifting trench]

=pup x^2 + xy + y^2 = x^2 y^2 where x and y are integers

[stoic basin]

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[drifting trench]

=pup x^2 + xy + y^2 = (xy)^2 where x and y are integers

[stoic basin]

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[drifting trench]

=wolf solutions of (x+y)^2=xy(xy+1) ; were x and y are integers

[stoic basin]

Do more with Wolfram|Alpha pro

Query made by

@drifting trench

[drifting trench]

Hmm, you could try to get the equation in either the form of x or y and solve it to get the results

That should work

[digital kettle]

how tho

[drifting trench]

$$\frac{1}{y^2}+ \frac{1}{xy}+ \frac{1}{x^2} =1$$

[drifting trench]

Nop

Rip, idk

[digital kettle]

wait

this might even work

because i solved an equation of the form 1/a + 1/b + 1/c = 1

and since that has no solns (iirc)

this must not have any solution either

now the only thing left is to prove that 1/a + 1/b + 1/c = 1 has no solns

[drifting trench]

That's wrong as 1,-1 are solutions

Unless the constraints are same a single proof doesn't apply over everything

[digital kettle]

-_-

[drifting trench]

$$\frac{1}{y}+ \frac{1}{x}+ \frac{y}{x^2} =y$$

[drifting trench]

$$\frac{1}{y}+ \frac{1}{x^2}(x+y) =y$$

[drifting trench]

Hmmm

Neh

[north orbit]

@drifting trench I guess the problem with long division was that the leading coefficients of the polynomials were 4 and 3

But if you factor out the 3 from the bottom like quantic did, you can divide but I don’t know what to do from there because of the factored 3

[sturdy dirge]

@north orbit You can multiple the equation by 3 then study the fraction or gcd(10, 10 - d).

[hollow basalt]

@digital kettle

Let u=x+y, v=xy. The equation becomes u^2=v(v+1). (* )

The RHS is only a perfect square if v=0 or v=-1, as otherwise sqrt(v(v+1)) lies strictly between |v| and |v+1|.

If v=0, then u=0 from (* ). Thus x=y=0.

If v=-1, then u=0 from (*) and we have x=1/x. This gives the solution pairs (1,-1) and (-1,1).

So the solutions are (0,0), (-1,1), (1,-1).

[ocean ravine]

[digital kettle]

okay...

[hollow basalt]

you sound hesitant, do you not follow some part of this proof?

[digital kettle]

oh lol

no

i was shady about the 'RHS is a perfect square' part

before thinking about it

-_-

[hollow basalt]

cool 😃 yeah the GM of two consecutive positive integers has to be strictly between them.

[digital kettle]

but the hint mentioned infinite descent

[hollow basalt]

There are often multiple ways to do such problems. I can't imagine a solution exists much shorter than this one though.

[digital kettle]

true

but curious 😅

nice soln tho

[hollow basalt]

thanks, yeah it's not obvious to me what they had in mind.

Often with these kinds of quadratic equations, you can use sum/product of roots to jump from one solution to the next and this is how you manufacture your infinite descent. This equation doesn't appear to just be a routine application of that though.

[drifting trench]

Why did I not think of this

Maybe because I wasn't really paying attention, but eh

[somber rampart]

viEtA jUmPiNg

[vast vessel]

kek

[ebon kettle]

Can anyone recommend me a book to study number theory at olympiad level? I'm a beginner so please keep that in mind when recommending me a book. Thanks!

[royal zodiac]

Rosen Number Theory

[ebon kettle]

Thanks I'll have a look at it!

[unkempt hedge]

How do you prove that m(m-4)(m+1) cannot be congruent to 1 mod 9?

I tried to do it through induction, but I am not too familiar with the method

[cinder hatch]

I wouldn't do this by induction

just doesn't strike me as good for that

[unkempt hedge]

why, it looks to me like the perfect place to use induction

[cinder hatch]

I just don't see how the induction step would work here at all

my though would be to figure out the possibilities mod 3

then go from there somehow

that shouldn't be too bad to compute directly

[unkempt hedge]

I tried a few numbers, and didn't get it to be congruent to 1 mod 9, which is why I would like to prove it through induction

Yeah, I could try mod 3

[somber rampart]

You can just brute force it lol

[cinder hatch]

you can always do that lol

[unkempt hedge]

but then I am coming at it with the assumption that it is untrue. (It might not be)

[somber rampart]

Calculate m(m-4)(m+1) for m from 0 to 8

[unkempt hedge]

Oh, how do you know that it repeats after 8?

[somber rampart]

Multiplication preserves modulus

[unkempt hedge]

I know it loops, but how do you know that it is exactly 8

and not 6

[cinder hatch]

you know it will definitely loop at 9

[somber rampart]

It's mod 9

[cinder hatch]

because 0 = 9 mod 9

[unkempt hedge]

oh

[somber rampart]

So it will definitely loop from 0 to 8

[unkempt hedge]

nice

[somber rampart]

Technically it loops at 1

[unkempt hedge]

makes sense

[somber rampart]

Since it's all 1 lol

But you can only use what you know

[unkempt hedge]

That would mean I should only use induction when you have a large modulus

[somber rampart]

Depends on the case tbh

[unkempt hedge]

I guess

[somber rampart]

Sometimes you can make clever arguments

[cinder hatch]

you need some sort of symmetry for the induction step to work out, that's not usually the case in my experience

[somber rampart]

But induction would probably work fine for this too

[cinder hatch]

for a problem like this

[somber rampart]

Nah you could do it

Let me sketch an induction priof

[unkempt hedge]

I was doing this problem, and I just had the assumption that there would be no solution and had a number theory approch. But how would you solve it normally?

[cinder hatch]

oh you would only approach that with number theory lol

[unkempt hedge]

But what if it had solutions?

[cinder hatch]

at least if you want integer solutions

[somber rampart]

Diophantine meme nice

You want to factor and work based on that

[unkempt hedge]

but LHS does not have any real roots

[somber rampart]

Lol

[unkempt hedge]

not sure what you mean by factor tbh

[somber rampart]

Then yeah easy meme

I didn't check roots

You can factor it out

As an equation

Also why do you need roots?

It doesn't need real roots

[unkempt hedge]

I don't know anything about imaginary numbers beside i = sqrt(-1)

[somber rampart]

Lol

[unkempt hedge]

@somber rampart What do you mean by factoring it out as an equation?

[somber rampart]

You're overthinking it

Literally polynomial factoring

You should have learned it in algebra 1 or 2

[unkempt hedge]

I know what polynomial factoring is

but how should you use it

[somber rampart]

So factor both sides

Then use divisibility memes

[unkempt hedge]

Sooo, you guess the roots?

These memes are to advanced for me to understand

[somber rampart]

Nah

Here's how you factor

Literally distributive

First add 3 to both sides

Sorry 2

Ultimately 1 + 2 = 3

27n^3 + 9n + 9n^2 + 3
3n(9n^2 + 3) +1(9n^2 + 3)

(3n+1)(9n^2 + 3)

3(3n+1)(3n^2 +1)

You know left hand side is divisible by 3

So m^3 + 5m + 2 is divisible by 3

And basically you keep making these kinds of arguments

Tired and on my phone so rip

[unkempt hedge]

yeah, this would be better. because you only need to check 3 cases

[somber rampart]

Uhhhhhhhh

Not really you still don't have quite that much info

[unkempt hedge]

because you have mod 3 instead of mod 9

[somber rampart]

Basically just you keep mining at the problem with these kinds of methods

Lolwot?

I'm so confused those are two separate problems

This is why I should not be awake at 3:00

[unkempt hedge]

It is like 10 in the morning here

LUL

[somber rampart]

Lol

Well anyway

These are two separate problems

Data mine the one with cubics by factoring on different offsets until something happens and using divisibility arguments and you'll be good

[unborn horizon]

hey guys

what does LCD(35,84)=7|14 mean

i know that the lcd of 35 and 84 is 7

[hardy iron]

never seen it like that lol

usually it's just 7

[fathom sierra]

well a|b means a divides b

still I don't understand its meaning here :/

[unborn horizon]

its used here, the problem is i do not know the english term for this

NZD -> LCD

[hardy iron]

what does it say under it?

[unborn horizon]

=> a solution exists

the first number(LCD) represents the number of solutions this equation has

[hardy iron]

hmm

from the 35x + 84y, i think it's saying that we can make it 7 (due to that formula i forgot the name of , basically gcd is the lowest possible sum)

and then multiply that by 2 to get a solution for x and y

(think it should be gcd actually)

lcd should be bigger than both numbers

[fathom sierra]

" the problem is i do not know the english term for this" : the whole equation is called a Diophantine equation

[unborn horizon]

after reading online a bit i found that the |14 is completely irrelevant

[remote tendon]

I assume it's just saying "the gcd is 7, which divides 14"

[hardy iron]

yeah that's what i thought too since you're looking to make sure you can find a sum multiple of 7

[unborn horizon]

It says that if the gdc divides c=> there are valid solutions

just found that rule

thanks a bunch guys

[remote tendon]

👍

[worthy saffron]

Is anyone interested in talking about the Collatz Conjecture?

[sacred junco]

I wrote a simple python script and gave it a number with 100,000 digits and it converged to 1 after 2402980 steps

[hollow basalt]

What about it?

[sacred junco]

For the collatz conjecture

[worthy saffron]

I looked into the problem for awhile, but there's stuff I don't think people usually talk about. One thing I found interesting is I found 3n+3 and 3n+7 inside of 3n+21

@sacred junco was that the largest total stopping time number you found? 2,402,980 steps is...huge...

[sacred junco]

My computer screen started to blink because it was running out of RAM doing calculations on 100,000 digits but I’m sure I can push it further

[worthy saffron]

I'm really curious what the trajectory looks like plotted in an excel chart...

[sacred junco]

I scrawled some notes...

[worthy saffron]

Did you find a number that takes exactly 1,000,000 steps to reach 1?

[sacred junco]

No, but I’m sure you can run a regression and see what is close

[somber rampart]

Collatz meme

You can't do a regression

[fathom sierra]

well 2^1000000 does griffon

[somber rampart]

Because it doesn't follow any pattern

[sacred junco]

That is true lol

[somber rampart]

So the regression will be wildly inaccurate

[sacred junco]

Just run a regression on the collection of regressions

[somber rampart]

[worthy saffron]

I would simply ask a python script to write down what number it used to get 1,000,000 steps in the first place... although I doubt my computer would handle that computation very well.

[somber rampart]

Lol

Why are we talking about collatz though

It's intractable

[worthy saffron]

I'm obsessed with it and I have the dream to prove it in my lifetime, 2 years down hopefully more to go!

[somber rampart]

Lol

[celest plinth]

yay obsession

[somber rampart]

So do you have any background in analytic number theory and such things?

Probably want to start with that

[worthy saffron]

No, that's either a plus or a con. I'm still taking calculus classes...

[celest plinth]

o that reminds me, I saw a cool question on advmath

[somber rampart]

Lol definitely a con to proving it in your lifetime

Self-study some number theory

Some => a lot

[celest plinth]

prove every positive rational can be expressed (p+1)/(q+1) for some primes p,q

[worthy saffron]

Maybe not. 80 years later, elite math nerds couldn't solve it.

[celest plinth]

I am quite bamboozled

[somber rampart]

Sure but you have to have some background to approach the problem front

[sacred junco]

Yeah my script is actually set up as a game for people who don’t know what the Collatz Conjecture is can try to find the “magic number” but it obviously proves difficult for them lol

[celest plinth]

we don't know enough about interactions between addition and multiplication

too spooky

[worthy saffron]

I doubt any of my work now has any value to it, but once I do learn the necessary math to work with it I may have some interesting ideas

[remote tendon]

@celest plinth so that comes down to showing that for any rational a/b there is an n such that an-1 and bn-1 are both prime

that seems v difficult

[celest plinth]

wait

[somber rampart]

Addition and multiplication are spooky

[celest plinth]

^ too spooky

[worthy saffron]

This sequence is Collatz-related but I have never seen it before in a paper: 1,1,2,1,1,2,1,1,3,1,1,2,1,1,2,1,1,3,1,1,2,1,1,2,1,1,4,1,1,.....

[somber rampart]

The natural numbers are spooky and unintuitive

Cmv

[remote tendon]

my favourite sequence:
1,1,2,1,1,2,2,2,3,1,1,2,1,1,2,2,2,3,2,1,1,2,1,1,2,2,2,3,1,1,2,1,1,2,2,2,3,2,2,2,3,2,...

[celest plinth]

nice

[worthy saffron]

I need a minute to plot that in excel, I'll be back...

[remote tendon]

my sequence is defined by looking at the sequence so far.
f.ex.
the first few terms are 1,1,2,1,1,2,2,2,3,1,1,...
this can be written as 1,1,2,1,1,2,2,2,3 + (1)^2
so the next number is 2

1,1,2,1,1,2,2,2,3,1,1,2,1,1,2,2,2,3,
is
(1,1,2,1,1,2,2,2,3)^2 so the next number is 2

1,1,2,1,1,2,2,2,3 is just 1,1,2,1,1,2,2,2 + (3)^1 so next number is 1

Puzzle: does the number 4 ever appear? 5?

[west glade]

I just saw that one today. was that on reddit?

[remote tendon]

oh, maybe!
i haven't seen it there - where did you?

[west glade]

not sure

something something large counterexamples?

[remote tendon]

i'll have a look 🙂

[west glade]

I could be remembering it wrong

[worthy saffron]

What a nice pattern...

[west glade]

https://www.reddit.com/r/math/comments/92g9um/what_famous_theorems_in_number_theory_have_been/

top comment, stackexchange link, scroll down a bit

[remote tendon]

thanks! :D
for the graph did you just copy oeis or did you program it?

[worthy saffron]

I copied it from OEIS

[remote tendon]

ok cool

[worthy saffron]

I don't want to rely on my processing speed to figure out how the pattern works before uploading the graph.

[remote tendon]

yeah i was thinking and couldn't think of a good way to program it

[worthy saffron]

Thanks to whoever decided on the OEIS it was a good idea to paste a "simple chart" that was all in one line...

I pasted the whole thing in word just to take out 1-99 and the blank spaces before dumping it into excel.

...wait this sequence is based on words?

OEIS: "Gijswijt's sequence: a(1) = 1; for n>1, a(n) = largest integer k such that the word a(1)a(2)...a(n-1) is of the form xy^k for words x and y (where y has positive length), i.e., the maximal number of repeating blocks at the end of the sequence so far."

[remote tendon]

"words" as in "sequences of symbols"

[worthy saffron]

I'm still having trouble wrapping my head around this... how do you calculate the first 3 terms?

[west glade]

1

11 because you can repeat the 1 one time

112 because you can repeat the 1 2 times

1121 because you can repeat the 2 one time

11211

112112

and then 112 112 2 because you have '112' two times

[worthy saffron]

so then 122 122 2 1, because it's that one time?

[remote tendon]

and then 112 11 22 2 as you have "2" twice

[west glade]

https://math.stackexchange.com/a/1881963

here with a little bit of colours

[remote tendon]

and then 112 11 222 3 as you have "2" three times

and then 112 112 223 1 as you have "3" once

basically you work out the highest possible number of repeats at the end of the string

[worthy saffron]

oh ok, that makes more sense, thank you

A051064 in the OEIS is based on dividing 3n exactly using the 3-adic metric... How does this sequence work?

[remote tendon]

ok
so the 3-adic valuation of a natural number n is just the number of threes in its factoriation

so f.ex the 3-adic valuation of 54 is 3 because

54 = 2*3*3*3

and that sequence is a(n) = 3-adic valuation of 3n

[west glade]

is there a bot which prints for you the start of the sequence? so that you don't have to look it up?

[remote tendon]

i don't know 🤷🏼

[worthy saffron]

I thought p-adic theory was way more complicated than that...

[remote tendon]

it does get harder and weirder

but the "p-adic valuation" number of p in the factorisation is the very first thing

[worthy saffron]

okay, got it. I think it's interesting I came across that same sequence but in a completely different way. I would start with any 5 (mod 6) number and after multiplying by 2 and applying (n-1)/3, checked to see if the resulting number was still a 5 (mod 6) number or not.

So 5 = 1, 11 = 1, 17 = 2, and so on

[remote tendon]

ooh interesting
are those definitely the same?

is that "number of iterations until it's no longer of the form 6n+5"?

[worthy saffron]

yeah

I like to think of it as the "backwards" version of 1,2,1,3,1,2,1,4,1,2,1,3,1,... in terms of the Collatz Conjecture anyway.

[remote tendon]

yeah :D

[worthy saffron]

I want to see if I can prove it...

[remote tendon]

yeah it's not immediately obvious that the two sequences are the same.
i'll have a go too :D

ok I have shown it 😀
let me know if you'd like me to share it
(there's a bit of complicated notation but I'd be happy to explain any of it)

[worthy saffron]

I'm stuck on [(12n+10)-1]/3 = cuberoot(3n/2)... If that's even right...

[remote tendon]

i should probably share mine

[worthy saffron]

yeah

[remote tendon]

ok:

basically, every time you do the reverse-Collatz, you reduce the number of threes in the factorisation of n by one

[worthy saffron]

Man you made that look easy... what the heck was I trying to do...

[remote tendon]

i've had a lot of recent practice with this area of maths haha

do you understand the argument?

[worthy saffron]

I think my understanding of calculus and function notation are failing me... For a minute there I thought you did a derivative...

[remote tendon]

I accidentally wrote f^(k) a couple time actually haha

and had to correct it to f^k

[worthy saffron]

Is this gone over in modern algebra...? Or am I better off finding a book somewhere and starting from there?

[remote tendon]

hm

[glad nacelle]

I think an algebra class probably won't go over it

Mine didn't at least

[remote tendon]

like most of the ideas above are basic number theory

the only weird thing is p-adic valuation

i didn't look at that at all until i didn't a course covering (among other things) metric spaces

[glad nacelle]

Did they introduce Q_p as the completion of Q with that metric?

[worthy saffron]

I'm starting to wonder if what work I did on the Collatz problem is either super obvious or is traces of higher level math in the disguise of made up notation and subtraction...

For example, I know that it is possible to figure out how to find numbers that start their trajectories through defining a specific behavior... For example, if I wanted to find numbers that after applying 3n+1 only divide by 2 once, there's a formula for that: 3+4n. I actually just finished writing a java program that allows for me to do this...

[celest plinth]

(Both)

[worthy saffron]

I'm sorry in advance for the made-up notation. "nß(x)" basically means how many times you can divide by n, x number of times. For example, 2ß(2) means that applying 3n+1 to n will result in an even number that divides by 2 twice. And then it will do the exact same thing a second time.

[worthy saffron]

I colored in parts of the graph to show how the formula applied to the trajectory of 3883.

[wind spear]

hi i'm new to the channel. Is anybody willing to point out what are the relations between the Collatz conjecture and prime numbers or maybe specific cases regarding uneven numbers, if there are any? Possibly link some articles? I'm not a mathematician nor a student! I'm asking on behalf of someone else.

[real peak]

Hi guys, I'm trying to find a space efficient way to get all primes below a certain number. I'm not allowed to use any additional arrays (other than the output array), so all Sieves are out.

can you see any way to optimize this? can I use multiples or sqrt somehow to iterate fewer times?

void genPrimes(int num) {
    primes3.push_back(2);
    primes3.push_back(3);

    for (int candidate = 5; candidate < num; candidate += 2) {
        bool bIsPrime = true;

        for (int prime : primes3) {
            iterationsCustom++;

            if (candidate % prime == 0) {
                bIsPrime = false;
                break;
            }
        }

        if (bIsPrime)
            primes3.push_back(candidate);
    }
}```

like in the outer loop, could I iterate up to the Sqrt(num) or somehting? (already tried taht, doesnt work) - but something similar?

i just cant see it! Please help!

[worthy saffron]

@wind spear From what I did in my own time, I have not found any connections between prime numbers and the Collatz Conjecture. However, several attempts have been made to connect the two. I can't say how successful those attempts were.

[snow basin]

I m looking for a proof of a result related to miller rabin's test. It states that if a numer n is composite then there are at least 3/4*(n-1) classes a in (Zn)* such that a ensures that n is composite.

(ie the necessary condition given by miller rabin tests for n to be prime fails)

fails when you test with a*

[lyric adder]

@real peak you can check all divisors up to the sqrt of your number

i think this is one of the most efficient prime checking methods

[sturdy dirge]

@real peak If the square root doesn't help you then you can't do better than that. (Sieves and sorting maybe)

[sturdy dirge]

@real peak Try this: ```haskell
#include <iostream>
#include <vector>

void genPrimes(int num, std::vector<int>& primes3) {
primes3.push_back(2);
primes3.push_back(3);

for (int candidate = 5; candidate < num; candidate += 2) {
    bool bIsPrime = true;

    for (int prime : primes3) {
        //iterationsCustom++;
        if (prime * prime > candidate)
            break;

        if (candidate % prime == 0) {
            bIsPrime = false;
            break;
        }
    }

    if (bIsPrime)
        primes3.push_back(candidate);
}

}

int main()
{
std::vector<int> prime;
int n = 1 << 20;
genPrimes(n, prime);
std::cout << prime[prime.size() - 1] << std::endl;
return 0;
}

[real peak]

@sturdy dirge omg that's magical!

how did you do that?

I dont understand why it works...

this is almost as fast as sieve of eratosthenes but with ZERO additional space!

I've never seen this implementation anywhere, why doesnt anybody know about this?!

[fringe knot]

@sturdy dirge you could probably make it a bit more efficient by only checking +- 1 mod 6

checks like 1/3 fewer numbers

[sturdy dirge]

I start to see the difference only for $$n > 2^{23}$$.

[wind spear]

https://www.reddit.com/r/math/comments/93fs43/about_the_collatz_conjecture/

[lyric adder]

you really should only check for divisors up to the sqrt

@sturdy dirge

[sturdy dirge]

That's in the code.

[ebon kettle]

https://kheavan.files.wordpress.com/2010/06/paul-zeitz-author-the-art-and-craft-of-problem-solving-2edwiley20060471789011.pdf

in page 67 of the pdf I just sent, one part says: The solution: strengthen the hypothesis from the start. Let us replace 3n with 3n + 1.

the middle part to be more specific.

Why do you replace 3n with 3n + 1? I don't understand how this is allowed or how this proves the original question (bottom of page 66 in the pdf)

[hardy iron]

what's the actual page number?

pg50?

didn't read the whole thing but...

=tex \frac{1}{\sqrt{3n+1}} \leq \frac{1}{\sqrt{3n}}

$$ \frac{1}{\sqrt{3n+1}} \leq \frac{1}{\sqrt{3n}} $$

bot's dead but u get the idea

(\leq means less than or equal to)

[ebon kettle]

sorry for the late response

the actual page number is 50 yes

I don't understand the notation that you used .-.

could you just tell me why the author uses 3n + 1 in words? @hardy iron

[hardy iron]

1/sqrt(3n+1) is less than or equal to 1/sqrt(3n)

related example:

show -3 < 0

well if i know -3 < -1 then i know that -3 < 0

since -3 < -1 < 0

[ebon kettle]

oh yeah

[hardy iron]

(that's the kind of logic they used)

[ebon kettle]

ohh but what exaclty was the -3, -1 and 0 in that case?

[hardy iron]

original is -3<0

the new thing they added in is the -1

[ebon kettle]

OHHH

Thanks so much XD

[hardy iron]

np

[ebon kettle]

I get it now 😄

[ebon kettle]

wait @hardy iron sorry for pinging you again but how do you solve 2.3.13? It reads: Prove that there is no smallest real number ( at the bottom of page 50 in the problems and exercises section)

[hardy iron]

i'm thinking by saying that there is a smallest real number and then find a contradiction

Let n be arbitrary.
Suppose n is the smallest real number.

[insert contradiction here]

Thus, n is not the smallest number.

Therefore, since n is arbitrary, there is no smallest real number.

[ebon kettle]

ohh

i'll try that, Thanks again!

[hardy iron]

yw

[placid lintel]

https://en.wikipedia.org/wiki/Cooperative_principle

[rocky spear]

I have difficulty understanding complex number anyone explain

[shrewd marsh]

What exactly are you having trouble with?

[stray gale]

To understand an imaginary number it helps to physically look at a complex plane. The Real axis is just your good old number line from 1st grade

The Imaginary axis is just an extension

When you read numbers, 3 would mean to count three on your number line right? How about -3? Think of it as -1 times 3. Count 3, then flip 180° to the west axis about the origin. Seems like a silly way to get to the answer, right? Well, it can help you understand "I" a little better I think.

What would 3i be? It's 3 times i. Count to three, then rotate 90° to the north axis. This is what i does.

What about -3i? Well, count to 3 on the east axis as usual, then the negative tells you to flip 180°, and the i a further 90°, landing you on the southern axis.

You probably once learned this set of rules:
1 = 1
i = i
i² = -1
i³ = -i
i⁴ = 1

Think about the "rotations" I told you about, and it should make sense now

If you multiply 3 by i twice, or 3i², doesn't it make sense that it would be -3? You're turning 90° twice, which is just like saying a 180° turn, which is why it turns into a negative :)

Not sure if this helps your understanding or not, you're welcome to ask questions @rocky spear

As for operations, they can all be represented visually in the complex plane above. You can see how those numbers cannot be described in just one term, because they're not on an axis. They're self-explanatory. If it says 1+2i, just go over 1 on the + axis and up 2 on the i axis, like reading coordinate pairs. This is easier than thinking about it the rotation way, and you should do it like this in most cases. But it helps to understand the "rotations" when you're new because that explains what's really happening. Add i, you slide up the i axis. Multiply by i, turn 90° about the origin. Get it?

I would recommend this source to get started
https://www.mathsisfun.com/numbers/complex-numbers.html

Best wishes!

[rocky spear]

@stray gale thank u ana so much it really helps

[stray gale]

@rocky spear let me know if you need help getting started but I hope that site opens enough doors for you to explore on your own :)

[blissful venture]

They are technically numbers

👀

[flat rock]

Isnt number theory about integers?

👀

[shrewd marsh]

[vast vessel]

\👀 wut about complex numbres with real part 1/2?

[stray gale]

Wait so if i is the square root of -1,

Wouldn't the fourth root of -1 be like,

sqrt(2)+sqrt(2)i?

[serene ferry]

‘The fourth square root’

[fringe knot]

yes @stray gale

although there are more solutions to x^4 = -1

so idk which one would count as the 4th root

[ember cove]

@stray gale the principal fourth root (the fourth root with smallest argument in [0,2pi)) of -1 is (1+i)/sqrt(2)

[trim matrix]

Hey guys does anyone have a good resource on the Second Principle of Finite Induction.

I'm still kind of confused as to how the First and Second differ in execution

maybe theres an example that shows the First and Second Principles being used to prove the same question kind of side by side

[ember cove]

basically you use the first principle when you only need to know about the immediately previous state in order to prove the proposition in the next state

but sometimes you need to know that the proposition holds in the last two states, or the last three, or even all of the previous states, to prove that the proposition is true in the next state

rough example: suppose you want to prove an exact formula for the Fibonacci number F(n).
You know that F(n) = F(n-1) + F(n-2), so to prove the formula for F(n) it's not enough just to assume that the formula is true for F(n-1), you also have to assume it's true for F(n-2).

[trim matrix]

ok so it kind of expands on the first principle

[ember cove]

So you can't use the first principle in that case. You need to use something stronger. The second principle, in which you would assume that the formula holds for all previous Fibonacci numbers, works though.

yeah

ultimately they imply the same thing, that the proposition is true for the whole sequence, but they get there in different ways

[trim matrix]

ok cool

[unkempt hedge]

If n is an integer, n/2 is a square number and n/3 is a cubic number. What is the lowest possible value for n?

[blissful venture]

2 * a * a = 3 * b * b * b

=> 2 divides b, and 3 divides a

=> 9 divides b, and 8 divides a

[unkempt hedge]

🤔

[blissful venture]

Aye, I'll go get a paper

=> 273 divides a

I think I'm approaching this wrong

Ok, it might be simpler to do this,

write n = 2^x * 3^y

n/2 = 2^(x-1) * 3^y

x-1 is even, y is even

similarly

you get x is divisible by 3, and y-1 is divisible by 3

smallest numbers to fit this gives 2^3 * 3^4

[drifting trench]

n should be divisible by 6

[blissful venture]

I hope I didn't mess up somewhere

[drifting trench]

I don't think you did

[blissful venture]

Aye, seems to fit the answer

bonus to use CRT would be to add n/5 is a fifth root 👀

[drifting trench]

648....Hmm

It works, yeah

Hmmm

[unkempt hedge]

@blissful venture What do you mean by x-1 is even?

[blissful venture]

n = 2^x * 3^y

[unkempt hedge]

yes and then you say that x-1 is even because n/2 = 2^(x-1) * 3^y

[blissful venture]

yes

because n/2 is a perfect square

[unkempt hedge]

oh, it has to be even because it is a square

and then you say y-1 needs to be divisible 3 and x needs to be divisible by 3

Ok, thank you

[blissful venture]

🍻

[ebon kettle]

Hi I was just wondering what this is called in english? I learned it recently in a korean high school math book

P(n,k) and S(n,k)

its a partition (something in number theory and combinatorics apparently) but what is P(n,k) and S(n,k) called?

is P(n,k) something like partition of real numbers

and S(n,k) partition of a domain?

[mossy folio]

Look up the Korean wikipedia page and switch the language to English. Works every time.

[ebon kettle]

wait but do u know what im talking about?

like for example P(4,2) = 2 because
4 = 1+3
4 = 2+2

so n is the sum of all the digits

and k is the number of digits that you use to create that value

[mossy folio]

I have no idea.

Maybe this helps: https://en.m.wikipedia.org/wiki/Partition_(number_theory)

[graceful notch]

yes partition

[dense adder]

Help

[elder bobcat]

no

jk

[drifting mantle]

I think you chased him away -_-

[pearl storm]

Hey

I have a question, and it just popped up in my mind when doing something online so I don't know how it will look like in hindsight

If I have 1 dollar, and you have 100, we can say that you have 100 times as many dollars as many

but what if I have no dollars ?

How many more do you have ?

Can we simply state this in a finite quantity ?

I have a feeling that the answer should be infinity

[olive raptor]

Just say, "Imbroke, and you're not"

I'm broke*

[pearl storm]

LOL

[final field]

@pearl storm thats similar to asking whats something divided by 0

it cant be infinity because infinity * 0 is still 0

[pine geyser]

NOOOMBA FEORY

[pine geyser]

if that's true

what happens if u multiply both sides by 0, does the equation still hold?

100=/=infinity*0

[somber rampart]

@sacred junco a/x as x goes to 0 is not infinity

It's +/- infinity

[somber rampart]

?

it's like undefined

the limit does not exist

the absolute value tends to infinity

it doesn't even tend to infinity though

you can say 1/0+ is infinity

and 1/0- is -infinity

and split 0 into {0+, 0-}

[final field]

assume 1/0 is infinity

1 = infinity * 0

anything times 0 is 0

1 = 0

infinity doesn't work

and plus-minus infinity doesn't help

do you see the problem when you try to trivialize division by zero?

[somber rampart]

infinity * 0 ≠ 0

"is infinity'" can mean a lot of things

[final field]

are we assume some magic infinity that can make nothing into something

[somber rampart]

that's how infinity can do things sometimes

[final field]

ok then

1/0 = k

k is our magic infinity

[somber rampart]

so k * 0 ≠ 0

[final field]

2/0 = 2k

whats 0/0

[somber rampart]

undefined

1/0 is defined, 0/0 isn't ^^

look into wheels

[final field]

mhm

thats the problem

[somber rampart]

that's not too much of a problem

[final field]

division by 0 can never be entirely consistent

[somber rampart]

I mean you could make it complex infinity

it can be though

[final field]

do it

even if you pull out your fancy riemann spheres there will always be a flaw somewhere

even if its a small problem its still a problem, faye

[somber rampart]

take the {0} field

0/0 = 0

0 = 1

0 + 0 = 0

0 * 0 = 0

[final field]

what

0 = 1 what

[somber rampart]

this is not the real numbers it's the field of {0}

[final field]

ok heres the thing

if you bend the rules you can make anything happen

[somber rampart]

some axiom lists say a field cannot have 0 = 1 but tbh it's just because you have to leave it out of usual theorems about fields

Yeah that's the point

[final field]

lets have normal rules

[somber rampart]

@sacred junco was too broad thinking, you're too narrow thinking

but the bended rules are important because you get 0 maps

and 0 maps are useful in things like homology

[sacred junco]

define normal rules @final field

[vast vessel]

@final field said normal rules, as you've shown, prove that 1/0 cannot exist

[final field]

1/0 = n
1 = n * 0
1 = 0
false

and n * 0 will always be 0 because thats how real numbers work

[sacred junco]

who says 1/0 is a real number

i sin't a real number and we have no problem with that

[somber rampart]

^

sqrt(ab) = sqrt(a)sqrt(b) holds normally

but then we get
1 = sqrt(1) = sqrt((-1)(-1)) = sqrt(-1) sqrt(-1) = i * i = -1

oops

broke maffs

when you want to do things outside of the normal scope you have to broaden your definitions

sometimes it makes things trivial

like the {0} field

[sacred junco]

How many of u guys are preparing for the imo ??

[torn moat]

how to number theory

also is this hannnel dead

[digital wigeon]

can someone explain the number 5 to me.

im mostly good with the other numbers, but 5 always confuses me

[sacred junco]

understandable

I try to think of the numbers less than 5 as complex numbers so that I'm not so lost

since it's easier to do stuff like i * i=-1 instead of 2 * 2=4 or like i+(-i) = 0 instead of 2+3 = 0 mod 5

[sacred junco]

5 is hard

Not all of us are born with the ability to think of 5 as what it really is

But for us who can, we know that 5 is 5

[torn moat]

lol

[sacred junco]

https://abstrusegoose.com/strips/NUM63R5.png

fun problem to try to generalize this

[lofty ridge]

12345678910

e = mc cube

[digital wigeon]

🕓 🔺

[flat rock]

3^2+4^2=5^2

👀

[digital wigeon]

0^2 = 0

[sacred junco]

good start @flat rock

[flat rock]

No integer solution for 7 consecutive squares

The first 4 adding up to the last three

:/

[misty pebble]

21² + 22² + 23² + 24² = 25² + 26² + 27²

[sacred junco]

nice

I guess it's no fun for me, I already worked out the answer

[flat rock]

Alternating triangulars

The smallest terms are 3, 10, 21, 36...

I guess the next one will be 55

[lofty ridge]

answer is 1

[sacred junco]

there's an explicit formula for the smallest term, given the number of terms

[flat rock]

Yeah

[sacred junco]

so like let's call 3^2+4^2=5^2 the n=1 term and so on

what's your formula for the smallest number

[flat rock]

0 term is 0?

[digital wigeon]

0^2 = 0

[flat rock]

Ok

[lofty ridge]

wrong

1

[flat rock]

Gimme a sec

[sacred junco]

or rather, show your derivation

that's more interesting than trying to brute force curve fit or something lol

[lofty ridge]

use calculato

[flat rock]

2n^2+n

Easy as that

[misty pebble]

Muh hexagons

[sacred junco]

lol insightful answer matt

[flat rock]

I used a thing that i found a while ago in a book

I think it was gardner

[digital wigeon]

it is better to find the middle term for some hot steamy symmetry

[sacred junco]

^

that's how I solved it

[flat rock]

Its called finite difference method

Or something like that

It works to find the discrete function for a sequence of integers

[sacred junco]

so you'd need to have cases to use to find it then?

[flat rock]

Just a sequence of terms of a polynomial function

[sacred junco]

show your working

[misty pebble]

Eh, isn't that still just fitting a curve to it?

[sacred junco]

yeah^

[flat rock]

Where x=1, 2, 3...

Yes

It is

[lofty ridge]

45678910

[sacred junco]

so your solution just assumes that there's always a solution

[lofty ridge]

wrong

its answer

[flat rock]

It looked like triangular numbers

[misty pebble]

Dangerous route to go there :x

[lofty ridge]

numbers arnt triangles

[flat rock]

Because 3, 10, 21, 36...

So i just assumed it was like that

[sacred junco]

lol

=tex \sum_{k=1}^N (n-k)^2 + n^2 = \sum_{k=1}^N (n+k)^2

[wild zinc]

"looks like" :^)

[sacred junco]

this is how I solved it

[misty pebble]

Place n points on the edge of a circle and join them, what's the maximal number of sections in the circle?

[sacred junco]

=tex n^2 = \sum_{k=1}^N 4 nk

[misty pebble]

First few look like 1, 2, 4, 8, 16, ... :^)

[sacred junco]

hahaha good one

[lofty ridge]

17

18

19

20

[sacred junco]

what's the sequence?

1, 1, 2, ...?

[lofty ridge]

12345

[flat rock]

3

5

8

👀

[sacred junco]

lol

[flat rock]

Or anything else

[wild zinc]

6,24,120,720,5040,40320,362880,3628800

[sacred junco]

closer but

actually I was evaluating n!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

at n=0, 1, and 2

[flat rock]

👀

[lofty ridge]

[sacred junco]

kek

[lofty ridge]

solve this problem

[misty pebble]

Okay how about this one:

[lofty ridge]

123456789

whats next

[misty pebble]

Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, ....?

[sacred junco]

lol

[misty pebble]

:^)

[sacred junco]

I know this integral

[misty pebble]

Aaaa, spoilers

[flat rock]

Not rational

👀

[misty pebble]

Eh

1/2 then

[lofty ridge]

123456789

123456789

123456789

123456789

😮

[flat rock]

Arcsin (1+0n)

[lofty ridge]

[flat rock]

👀

[lofty ridge]

guys lets play minecraft

[misty pebble]

Close, but the next one was actually 1/2 - 6879714958723010531/935615849440640907310521750000

[sacred junco]

oof

[lofty ridge]

[misty pebble]

Google "surprising sinc sums" or something :^)

[flat rock]

There are infinitely many functions that intersect a finite set of points

[sacred junco]

https://en.wikipedia.org/wiki/Borwein_integral

[misty pebble]

Ah that was the name, couldn't think of it

[flat rock]

I just took what seemed a nice solution

[sacred junco]

I think the problem you posted earlier with the circle was from polya

[digital wigeon]

number of different differentiable structures on R^n (starting at n = 1): 1, 1, 1, ∞, 1 ,1, 1, 1,...

[sacred junco]

if I were god that's what I'd do

[flat rock]

My formula worked for n=5

👀

[sacred junco]

nice just gotta check it empirically for all n>5 and you're done

[flat rock]

Sad

Well i guess ill prove it

=tex n^2 =4n~ \sum_{k=1}^N k

This is what you get when you simplify the thing

4n is a constant

So

[flat rock]

Divide by n

=tex n = 4~\sum_{k=1}^N k

[flat rock]

Then you expand the thing and should get it

But its the middle number this time

Now cubes ;)

3^3+4^3+5^3=6^3

Ew cubic equations

[sacred junco]

nice

yeah I was somewhere else I would like to look into the cubic stuff

I guess is this the natural way to go about it, instead of n+1 on the left and n on the right it's n+2 on the left and n on the right?

[lofty ridge]

no

n for noobs

[flat rock]

@sacred junco yes

[sacred junco]

hello

is it possible to solve this

=tex \sum _{n=0}^{\left\lfloor \frac{\log (x)}{\log (10)}\right\rfloor } 10^{-5 n} \left((x \bmod 10^{n+1})-(x \bmod 10^n)\right)^5 = x : { x \in Z }

[stoic basin]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[vast vessel]

@sacred junco didn't I already explain that one

[sacred junco]

yeah check the posting time tho @vast vessel

[vast vessel]

exactly

🔨

[torn moat]

can anyone teach me like basic euler toitent stuff like if i wanted to find the last 3 digits of 3^185 lets say how would i do that

i remember looking it up and figuring out but i completely forgot

[wild zinc]

3^185 mod 1000,phi (1000) =400 so I'm not sure euler's totient theorem is going to help you much here

[torn moat]

heh

[wild zinc]

you could try exponentiation by squaring but without a calculator it's still going to be a pain

[mossy folio]

Just type it into a Python interpreter or WolframAlpha:
18511095575145533338447403836109581505353226740782032803932380476024584374357028238763043

[torn moat]

kek

what if i did

2^29340923480238

do u have enough bits

to store that

xd

[wild zinc]

mod 1000? sure

[mossy folio]

Python has arbitrary precision integer handling.

[wild zinc]

2^238 mod 1000

[torn moat]

oof python

ik ik but what if i cant use python

:^)

[mossy folio]

You use C++.

[torn moat]

i mean like if its during a math competition lol

[wild zinc]

2^21 = 2097152, 128152104856, 124152128, 872152, 544

I think the last three digits are 544

nope wolfram says 944

[torn moat]

lol

[wild zinc]

close enough

in a competition you'd almost certainly get paper :^)

[torn moat]

lol

i need help on most of the questions tbh

i keep like

understanding

and then forgetting later

how to do them

[sacred junco]

In a long competition you could just try to find the period

I mean, whatever is necessary amirite

[torn moat]

i find the period ... i find it every month

[mossy folio]

well, this was unexpected

[torn moat]

:^)

[fringe girder]

Why is it that some numbers are hard to check primality for?

I was reading about the fibonacci primes and I took the largest one and added 1000 to it then tried checking whether that number was prime in mathematica. I left it running for an hour and it couldn't compute it

I tried all the odd numbers near it and they instantly gave results

[vast vessel]

wdym u took the largest and added 1000

[wild zinc]

some numbers have prime factors that are easy to find (small or p-1 or p+1 is very smooth)

Which makes it easier to say that they aren't prime

[fringe girder]

@vast vessel I took the largest know fibinacci prime (n = 3341167) then tried seeing iif that number + 1000 (n = 3342167)

[sturdy dirge]

For which $$n\in\mathbb{N}, m = 2n+1, \sum\limits_{k=0}^{m-1}\genfrac{(}{)}{}{}{k}{m}\binom{m}{k}=0$$ ?

[sacred junco]

Like 3?

[sturdy dirge]

Is it the only ?

[sacred junco]

The only what?

[sturdy dirge]

Is that the only integer n that verifies the equation ?

[sacred junco]

m=2n+1, m in Z?

[sturdy dirge]

$$m\in\mathbb{N}$$.

[sacred junco]

What's the E mean?

[sturdy dirge]

E or $$\in$$ (it means 'in') ?

[sacred junco]

Shouldn't those be quotation marks, not apostrophes?

[sturdy dirge]

Maybe.

[sacred junco]

Are you sure?

[sturdy dirge]

The question is more interessing that my notation. Can you answer it ?

[sacred junco]

Answear?

[torn moat]

m-

[sacred junco]

this is a fun problem @sturdy dirge

I'm assuming this is a fraction not a legendre symbol?

=tex (1+x)^m = \sum_{k=0}^m x^k \binom{m}{k}

[sacred junco]

I'd start here

then take the derivative wrt x

=tex m (1+x)^{m-1} = \sum_{k=0}^m k x^{k-1} \binom{m}{k}

[sacred junco]

divide both sides by m and then plug in x=1

=tex 2^{m-1} = \sum_{k=0}^m \left( \frac{k}{m} \right) \binom{m}{k}

[sacred junco]

then I'd subtract out the k=m term from both sides

=tex 2^{m-1} - 1 = \sum_{k=0}^{m-1} \left( \frac{k}{m} \right) \binom{m}{k}

[sacred junco]

so now we want that to equal 0, so

=tex 2^{m-1} -1 = 0

[sacred junco]

that means m=1

which, since m=2n+1 means that n=0

[sturdy dirge]

@sacred junco Sorry, but it is a Jacobi symbol .

[sacred junco]

haha rip

[sturdy dirge]

I partially solved, is there a solution for n even ?

[torn moat]

,

[bronze breach]

I managed to turn a piecewise function into a non-piecewise function :D

These two functions are equivalent at all natural numbers.

Can all piecewise functions be collapsed this way?

[sacred junco]

yes

if you have something that's true for particular congruences, it boils down to writing a fourier series essentially

I guess the 0^0 part is how you can get the stepwise aspect of n<=3 in there

=tex g_0(x) = \frac{1+(-1)^x}{2} \ g_1(x) = \frac{1-(-1)^x}{2}

[sacred junco]

so in the even odd case it's pretty easy to decompose the 0 mod 2 and 1 mod 2 parts apart like this

it's probably better to write these as e^{i 2pi / 2} to see the generalization clearer

=tex h_0(x) = \frac{1 +e^{\frac{i 2\pi }{3}x} + e^{\frac{i 2\pi }{3}2x}}{3}

[sacred junco]

0 mod 3 version

you can differentiate as a trick to generate the next two

etc etc

[bronze breach]

I've never seen that trick before, very interesting. I was taking advantage of how remainder of n/k can be rewritten as x - kfloor(x/k)

Do you know where I could find more information on using fourier series this way?

[sacred junco]

idk I came up with this myself

I'm sure I didn't invent it first though or anything

I can explain it better maybe type something up for you if you like

show some examples or something

I wouldn't mind figuring out how to do this a bit more generally with some inequalities and stuff too

[bronze breach]

Sure, that sounds interesting

[sacred junco]

I think you could even do this piecewise function as an infinite series if you try to fit it to a polynomial with a vandermonde matrix or something too

I dunno it's been a while since I last talked to you I forget what sorta stuff you know lol

actually what do you mean when you write this anyways

=tex 0^{0 \lfloor \frac{x}{3} \rfloor}

[sacred junco]

I was kind of assuming it's like an indicator function

[bronze breach]

"Zero to the power of (zero to the power of the floor of x/3)"

[sacred junco]

oh the 0 is to the power as well

[bronze breach]

I needed something that was 0 at [0, 2] but 1 at (2, infinity)

That worked nicely

[sacred junco]

I see the way I've come to this same sort of thing is to put the fourier series in the exponent of 0 or something like this

I haven't through about this kinda thing in a while

[bronze breach]

Ah shoot I mistyped the piecewise function. It's supposed to be n <= 2, not 3

[sacred junco]

let me think here

=tex \left\lfloor \frac{1}{n} \right\rfloor

[sacred junco]

this is 1 at n=1 and 0 for n>1

I don't know if this can be doctored up into something a bit nicer or if it's just about as ugly as 0^0^stuff

just a thought maybe

[bronze breach]

Hm

I wanted the function defined at 0

But actually that may have been a mistake haha

This is a solution to a sequence, and typically those start at 1, correct?

Oh boy. I might as well rewrite it so that f(1) represents the first term of the sequence instead of f(0). Not sure what I was thinking

[sacred junco]

really doesn't matter

if you want to shift it, that's not really anything to worry about redefine it as g(x) = f(x-1)

or vice versa

=tex \left\lfloor \frac{1}{1+n} \right\rfloor

[sacred junco]

this is 1 at n=0 and 0 for n>0

I really only ever worry about whether the first term should have index 0 or 1 if I end up with some kind of n+1 or n-1 in my formula that I want to get rid of

I think generally a lot of number theory stuff just ends up starting at 0 because 0, 1, 2, 3, 4 is a nice way to think mod 5 instead of 1,2,3,4,5

[bronze breach]

=tex \left\lfloor \frac{3}{n+1} \right\rfloor

[sacred junco]

actually here's something quite ugly but workable

[bronze breach]

That works for 1 and 2, just not 3

[sacred junco]

=tex J(n):= \left\lfloor \frac{1}{(1+n)^2} \right\rfloor

[sacred junco]

oh ok yours might be fine then

this is more like just a single spot

J(0)=1 and J(n)=0 otherwise

hmm maybe not actually I am distracted but

I just imagined once you have one single point, you can add and subtract a function arbitrarily many times

[bronze breach]

=tex \left\lfloor \frac{5}{n+2} \right\rfloor

[sacred junco]

pretend J(n) =1 at n=0 and J(n)=0 otherwise, then you can make A(n) =[J(n-1)+J(n-2)] and B(n) = 1-A(n) to make a stepwise

[bronze breach]

How could you translate that to "be 1 from [a, b] and 0 elsewhere?"

Also, I wonder if $$J(k) := \left\lfloor \frac{2k-1}{n+k-1} \right\rfloor$$ works for [0, k]

[bronze breach]

I believe it does

Now I'm wondering what the compliment would be.

In the case of 0^0^floor(x/4), that's 0 from 1 to 3 but 1 elsewhere.

Meanwhile 0^floor(x/4) is inverted: it's 1 from 1 to 3 but 0 elsewhere

Oh I suppose just flip it: $$\left\lfloor \frac{n+k-1}{2k-1} \right\rfloor$$

[sacred junco]

I guess this is over all doable and a fun exercise

I think it kind of obfuscates the writing of the function

even though piecewise does kinda suck in its own way

[bronze breach]

Oh it absolutely does obfusctate things. I was just curious if this was possible.

A generalized method of transforming a piecewise function would still be exciting to see haha.

Also, what would you recommend for making the piecewise function not suck?

[sacred junco]

oh nothing, it's just the least of two evils I think haha

I think it's not a bad idea to just use a step function though

maybe define like

=tex H(n) := \left{ \begin{matrix} 0 & n<0 \ 1 & n\ge 0 \end{matrix} \right.

[sacred junco]

then by shifting and subtracting you can make anything

=tex f(n) = g(n) H(n-3) + h(n)(1-H(n-3))

[sacred junco]

let me see did I do that right

it should be f(n)=g(n) when n<3 and f(n)=h(n) when n>=3

actually I did it backwards

oh well, that's probably the cleanest way if I had to do it

at the end of the day floor functions are just some other arbitrarily defined function

using them to make a step function is just hacky

[bronze breach]

I found this definition of floor especially fun:

https://media.discordapp.net/attachments/153337086466981888/481591707142455297/unknown.png

[sacred junco]

hahaha I've done similar things in the past

like arccos(cos(x)) to make a triangle wave

[bronze breach]

Really? I'd love to see

I have a sort of kink for seeing ways of defining "arbitrarily defined functions" using less arbitrarily defined functions

My gateway drug was the realization that sqrt(x^2) was synonymous with abs(x), at least for real numbers

[sacred junco]

https://www.desmos.com/calculator/xljljml3h6

well I guess some of these are using different functions but arcsin(sin x) is in there, same thing basically as arccos(cos x) not much to that I think

but it can be fun to fool around with stuff

[bronze breach]

It really is

The sequence the function was a solution for went 1, 1, 1, 2, 2, 3, 2, 4, 3, 5...

Essentially, you have 3 counters A, B, and C. You increment A, then B, then C, then back to B, then A, then B, then C, etc. Each turn you keep track of what each counter is at.

[sacred junco]

oh I see you're zig zagging back and forth

or no I don't quite get it actually