help-4

[glass kelp]

fucking hell

[inner mirage]

Interesting

[glass kelp]

happened to me once

in a competition

[inner mirage]

Yeah thats annoying

Ok then thanks for the help k and riemann

[glass kelp]

it happens becuz u didnt use alpha * gamma

in ur derivation

[glass kelp]

ABSOLUTELY

[inner mirage]

How much did it cost you?

[glass kelp]

like 4 points

and there were only 25 questions 😂

[inner mirage]

.close

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[inner mirage]

A similar thing happened to me on an exam, it was infuriating because I think it cost me the perfect score

[vale dock]

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[merry geode]

Ello

[merry geode]

What’s the difference between exponential and linear

Like here

[glass kelp]

exponential => a^x
linear => ax

[merry geode]

oh

So this is exponential

Linear is like R = 4(t)

[glass kelp]

linear is like

[merry geode]

exponential is R = 4^t

[glass kelp]

if i pick up a rose, i will increase by 4, by 4, by everyday

4,8,12,..

[glass kelp]

yes

[merry geode]

man ur the rose

🌹

thank u

[teal thicket]

Hi

[glass kelp]

hi crystal

[merry geode]

Hey

R increases everytime t increases

Exponential relation

Mhm?

[glass kelp]

both linear and exponential where a is positive increase as t increases

so the argument isnt the clearest

[merry geode]

But our question is exponential

[teal thicket]

Oh no I don't have any idea about English maths. I study in my language. I've to learn English

[glass kelp]

the nature in which they increase is different for each

[merry geode]

But still increase

[glass kelp]

are there other choices

other than a and b

[merry geode]

Mhm

It’s either A or D

Lemme show u the rest

[glass kelp]

aight

[glass kelp]

now this

is much a better argument

[merry geode]

What does factor mean tho

U said that’s linear

4,8,12,16

[glass kelp]

factor as in multiplied factor

[merry geode]

Aha

[glass kelp]

4, 4 * 4 = 16, 4 * 4 * 4 = 64, ...

[merry geode]

Mhm

Thank u

🫡

💗

.rate @glass kelp ♾️/1

[glacial kelp]

any jee aspirants here

[glass kelp]

a lot

but not me

[merry geode]

if someone messes with professor k ima mess with him

[glacial kelp]

cool ig

[merry geode]

Them*

k

What does

Percent mean

In math

like

15% of 120

Is it like

15/100*120

Or how exactly

[glass kelp]

per cent is from 'per centum' meaning 'out of 100' in whatever language

[merry geode]

Mhm

[glass kelp]

so its like

[merry geode]

of means *?

[glass kelp]

if the thing is divided into 100 pieces

u take x% from it

say i want 10% of 1000 bananas

[merry geode]

10/100

[glass kelp]

yup

so

it is equivalent to saying

[merry geode]

mhm

[glass kelp]

10/100 * 1000 = 100

[merry geode]

love u k

[glass kelp]

its like saying if 1000 bananas is divided into 100 groups (each containing 10 bananas). if i pick up 10 groups, i will therefore have 100 bananas

[glass kelp]

read the graph

[merry geode]

how to calculate kinetic energy

hmm

[inner mirage]

You don't need to calculate

Just look at the graph

[merry geode]

Ohhh

Tysm

Nuclear interaction

Anyways thank u

.forceclose

.close

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[whole plank]

why is this saying its wrong

[glass kelp]

x^5/2

[merry crystal]

What does the power rule say?

[whole plank]

ohhhh

nvm

im legit so dumb

.close

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[little pier]

Is 3!^sqrt((((phi!)!)!)) irrational?

[ebon glade]

probably. who knows

doubt anyone ever thought about that number

[little pier]

I called this my number

[sleek nebula]

by ! you mean the gamma function of phi, the golden ratio?

[little pier]

Surprisingly my number ± 0.1 is rational

[sleek nebula]

gl proving irrationality or rationality

you know

e + π hasnt been proved irrational yet iirc

[ebon glade]

we dont even know that pi^pi^pi^pi isnt an integer

[little pier]

Is it the gamma function?

[ebon glade]

you cant just do a factorial of a non natural number

[little pier]

I was thinking, since it has phi in it, it has to be irrational

[sleek nebula]

dosent necessary work like that

[sleek nebula]

are you familiar with the gamma function

(an extension of the factorial function)

[stark wedge]

what do you mean by this

[little pier]

• 0.1 is 172913923/24839361 and - 0.1 is 410991793/60786029

[stark wedge]

so you're saying there exists a rational less than 0.1 away from your number in either direction

do i understand your claim correctly? y/n

[ebon glade]

are these just numbers your calculator spit out?

that doesnt mean shit

calculators have limited precision

[little pier]

yes

[ebon glade]

all numbers in a calculator are rational numbers

[little pier]

I used Wolfram Alpha

[stark wedge]

ok then in that case

the same claim is true of literally any real number.

[little pier]

you can take the factorial of a non-natural number?

[ebon glade]

you need the gamma function

thats one possible way to generalize the factorial to non natural numbers

anyway, I dont know what you did to WA so that it spit out a rational number for you, but you probably messed up somewhere

[cloud coral]

just checking, you mean this

$3!^{\sqrt{((\varphi!)!)!}}$

[rocky lotus]

Element118

[little pier]

yes

[cloud coral]

well, firstly we can simplify 3! to 6, that's obvious

[full lake]

hello

I am a mathmatics noob

[little pier]

yeah

[full lake]

who can give a book list about maths

me

[cloud coral]

so the question is whether there are integers $a,b$ where

$\sqrt{((\varphi!)!)!}=\log_6(a/b)$

[rocky lotus]

Element118

[little pier]

Wolfram Alpha says it is unknown.

[cloud coral]

pulling away the square root is probably pretty hard

[little pier]

how?

[stark wedge]

hello and welcome to the server

if you want to just chat, go to #discussion or #chill

#book-recommendations also exists

[full lake]

OK thanks

[cloud coral]

one would have to deal with a product of logarithms, there's not many identities about that

[little pier]

oh

[cloud coral]

and then getting all the way into phi would be another issue too

factorials are pretty much only known to be nice on integers

maybe factorial of a rational has a closed form

[little pier]

$79445709/81494272 + 1/10 = {\sqrt{((\varphi!)!)!}}?$

[rocky lotus]

F

[ebon glade]

unlikely

[cloud coral]

probably easy to prove that is wrong with sufficiently precise calculations

[little pier]

Wolfram Alpha said that ${\sqrt{((\varphi!)!)!}} - 0.1$ isn't irrational

[rocky lotus]

F

[little pier]

it could be a coincidence that it happen.

[cloud coral]

the way it tries to detect rational numbers is probably by continued fraction

[ebon glade]

I dont know what you messed up, but WA does not say that

did you compute the first part first, then take the result of that, copy paste it in and subtract 0.1?

[little pier]

oh it isn't lol

[turbid valve]

The Gamma function turning irrational numbers into rational ones would be such a breakthrough

idk what for but still

[ebon glade]

well given that its continuous, very likely happens from time to time

[little pier]

the second alternate form is scary

[ebon glade]

but for specific irrationals probably not

[turbid valve]

It's only rewriting the gamma form as a factorial, using the definition of phi = 1/2 (1 + sqrt(5) )

[little pier]

ok

[turbid valve]

Ugly though it is, that is precisely what's being computed

[little pier]

a natural number to the power of an unknown can be irrational.

[turbid valve]

...yes

2 to the power 1/2 is the obvious example

[little pier]

also a natural number to the power of an irrational is irrational.

like the Gelfond-Schneider Constant.

[cloud coral]

sometimes it's rational

[little pier]

example?

[cloud coral]

$3^{\log_32}$

[rocky lotus]

Element118

[little pier]

hmm.

[cloud coral]

anyway if you want, I can probably show you what you need to learn in order to prove that this is false

[little pier]

ok.

[cloud coral]

so firstly, you would probably need to know some programming language

Python should be enough if you already have that

assuming you have that, the next step is learning how to compute gamma of an input to an arbitrary precision

[little pier]

I used Python before. But I don't remember how to make a file on IDLE.

um....

oh, got it

[little pier]

so what do I do?

[cloud coral]

kinda the difficult research step

https://math.stackexchange.com/questions/19236/algorithm-to-compute-gamma-function

here's a list of options, you'll have to read through to see which meets your needs

which likely, you would want to find an implementation that can hit thousands or millions of decimal digits

and you'll have to implement that yourself in the programming language of your choice

also you need an algorithm to calculate phi, Babylonian method for square roots should be enough

[little pier]

that's C++

[cloud coral]

Python is probably a good idea because you can use the big integer or decimal library

C++ should be also fine if you are familiar enough with it

[ebon glade]

WA computes to arbitrary precision. just click on "more digits" until it doesnt fit anymore

shouldnt take more than once or twice

[little pier]

time to boot up Unity

[cloud coral]

remember you want arbitrary precision

so, not floating point

[little pier]

no floating points so no mes

[cloud coral]

and so far, with all the steps I outlined here, you should be able to formulate a pretty convincing demonstration of two numbers being different

the fine print to make it a proof that the two numbers are different is to do the numerical analysis to bound your errors for all the computation

[little pier]

how am I going to write the code for it.

[cloud coral]

with lots of effort

it's not going to be easy

but it's not impossible

[little pier]

I'm looking at this.

[cloud coral]

if you prefer proofreading code, you can get an LLM to help you write

but you'll be stuck proofreading code so, take your pick what you want to do

[little pier]

I don't want to write the code because it will take a long time.

but I don't want to use an LLM to code it for me because of errors

[cloud coral]

yeah, it might not be a problem worth it to tackle with limited time

[little pier]

well I sometimes I have java homework because I am taking summer classes (yes I do it online)

maybe I should close this?

.close

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[halcyon plank]

Help

<@&286206848099549185>

[jovial edge]

!15m

[vale dock]

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

[halcyon plank]

Ok

I'm not well acquainted with this server sorry

[golden moss]

there the pin msg by bot

[halcyon plank]

Can you help me solve the problem?

[ashen prawn]

aha

fellow indian

[lyric bloom]

So they are asking q for the largest r?

[stark wedge]

yes

[halcyon plank]

yes

yes

[stark wedge]

you can bruteforce this with some code but i don't think that's the intended solution

[lyric bloom]

that's solvable for an ioqm problem 😭

It's a maths olympiad question, I don't think they allow computers in there

[marsh forge]

r = 100p + q and then some divisibility/gcd stuff

[tidal swift]

you need to rewrite 100p + q into a form that involves (p + q), i believe

[lyric bloom]

Yeah thought the same

[tidal swift]

actually how far are you in, OP?

[lyric bloom]

p+q can either be 33 or 99 I think

[tidal swift]

but first how much does OP know

[lyric bloom]

Ye

[tidal swift]

!status

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[halcyon plank]

I dont know where to begin

[tidal swift]

write r in terms of p and q

[halcyon plank]

ive tried using chatgpt

its giving wrong answers

and i also tried to solve it myself but couldn't

[tidal swift]

!nogpt

[vale dock]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[lyric bloom]

What is the answer it's giving?

[tidal swift]

that's why i'm giving you steps to try

[halcyon plank]

I can directly tell you answer because Ik what the answer is

[lyric bloom]

No no what gpt is telling

Ik the answer

[halcyon plank]

the answer is 13 but chatgpt is telling 60

[lyric bloom]

Ohk

13 or 12?

[halcyon plank]

13

[lyric bloom]

Ohkk

Got it

[tidal swift]

then do you know why it's 13?

not much point knowing the answer but not how to get there

[halcyon plank]

no its from a book

the direct answer

[tidal swift]

^

this is step #1

[halcyon plank]

try to solve it on a page paper

and send photo cuz i cant generate latex here

[lyric bloom]

Huh?

[tidal swift]

!nosols

[vale dock]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lyric bloom]

Yeah we can try to take you there

Can't solve for you

[tidal swift]

and

!noans

[vale dock]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tidal swift]

well i mean noans is kinda pointless here because you have the answers, but you're looking for the steps and i would like you to work it out yourself with prompts from us

and btw, you can write latex with texit

[halcyon plank]

i need help

atleast tell me how do i approach the problem

[tidal swift]

but i told you how to?

[lyric bloom]

How did you approach the problem at first?

[tidal swift]

yeah maybe let's hear your initial approach

[halcyon plank]

what i did is fro mknown given i found out that r=100p+q

and from there

I used modular arithmetic

[lyric bloom]

Can you specify what you did?

[halcyon plank]

and took 100p+q/p=q and tried to input values from 99< and then i didnt know what to do further

yea thats what i told i couldn't find what to do further

[tidal swift]

then you probably should have said you got r = 100p + q

[lyric bloom]

Ye true

But seems like the right approach

[tidal swift]

but here p + q are separate

you need to rewrite this so that you get (p+q)

[halcyon plank]

yea i said the same

[tidal swift]

i didn't see any hint of you saying you got there

cuz you replied to me saying you used gpt

but anyway

[tidal swift]

^

[halcyon plank]

.

[tidal swift]

ye i asked that half an hour ago sir hahah

[halcyon plank]

im telling you my approach

[tidal swift]

nvm, forget it

[tidal swift]

this is your next step here

[lyric bloom]

100p + q = r and r is divisible by p+q

How can you write r

[halcyon plank]

so i write r-99=p+q

[tidal swift]

don't think that form is exactly right

[lyric bloom]

Other way around

and 99p

[halcyon plank]

Idk its an ioqm problem

[tidal swift]

i'll give you a hint

[lyric bloom]

Yes we can see

[halcyon plank]

so according to it seems to be right

[tidal swift]

you need a 100q to show up somehow

so that you can factor 100p + 100q

you need to get a 100q from the 1q that you have

[halcyon plank]

add 99q both sides?

that wouldn't work4

[tidal swift]

r + 99q = 100p + q + 99q
r + 99q = 100p + 100q

now isolate r

[halcyon plank]

How do i do that

im not good at mathematical terms'

[tidal swift]

one step literally

move the 99q over

[halcyon plank]

but wouldn't that create a loophole

[tidal swift]

no

[lyric bloom]

Are you in 9th or 10th rn? Just asking

[tidal swift]

move the 99q over, but don't simplify the 99q with the 100q

[halcyon plank]

I'm in 11th preparing for ioqm

[lyric bloom]

Ah ok

I thought prodigy kid

[halcyon plank]

then i factorize?

[tidal swift]

correct

[lyric bloom]

Yes

[halcyon plank]

Nah no prodigy

[lyric bloom]

Yes

[halcyon plank]

Oh wait

what do i get after factorization

[tidal swift]

what do you think you will get?

[halcyon plank]

because i dont thhink that will change any thing

[tidal swift]

write down the current form of r first

[halcyon plank]

ok wait don't go

[tidal swift]

you can assume anything you want after writing it down

[halcyon plank]

100(p+q)-99q

[tidal swift]

there we go

now, you said this won't change anything

i'll tell you what this changes

r = 100(p+q) - 99q

we know p + q must divide r

p+q obviously divides 100(p+q)

agreed?

[halcyon plank]

yes

[tidal swift]

so for p+q to divide r

p+q must now divide 99q

[halcyon plank]

oh yes

[tidal swift]

so we know p+q must be a factor of 99

but before we consider which factors

let's consider the range of values of p+q

or rather, p and q separately

p and q are 2-digit numbers

so they are at least 10 and at most 99

[lost marlin]

??? I don't get this

[halcyon plank]

yes

so it will have maximum value 99+98

as their gcd is 1

[tidal swift]

p+q does not share any factors with q

because p and q are coprime

so for p+q to divide 99q, p+q must divide 99 if it cannot divide q

[lost marlin]

Oh i forgot to see that

Thanks

[tidal swift]

aight back to the issue at hand

[halcyon plank]

.

[tidal swift]

we know that the minimum value of p+q must be 11+12 = 23

[halcyon plank]

yes

[tidal swift]

p and q cannot be 10 because neither ends in 0

and if p = 11, q cannot be 11 because of relative coprimeness

so what are the factors of 99 that are greater than 23?

[halcyon plank]

33

[tidal swift]

and?

[halcyon plank]

wait lemme think

its 99

itself

[tidal swift]

good

so we need to consider p+q = 33 and p+q = 99

but here's the thing

if p+q is 33

we said earlier that the minimum p or q is 11

so the maximum p can be is 33 - 11 = 22

[halcyon plank]

yes

[tidal swift]

compare this to p+q = 99

the maximum p can be is 99 - 11 = 88

since p forms the first two digits of r, larger p = larger r

so we focus on p+q = 99 because that will give us way larger p to work with

[halcyon plank]

yes

[tidal swift]

so from here it's kind of enumaration/bruteforcing

we start from p = 88 and go down by 1

then check every possible p

let's start with p = 88

[halcyon plank]

we try to substitute values/

[tidal swift]

is p = 88 possible?

[halcyon plank]

?

wait

no

[tidal swift]

you're right that p = 88 is not possible

why?

[halcyon plank]

because if we divide 8811 by 99 it doesn't give us a natural number

as the question clearly states that p+q divides r

[tidal swift]

but 8811/99 = 89

right answer, wrong reason

[halcyon plank]

oh

wait

because 88 and 11 are not coprime

[tidal swift]

good!

so p = 88 is invalid

moving on

is p = 87 ok?

[halcyon plank]

wait

it is

[tidal swift]

why?

[halcyon plank]

because 87 and 12 are coprime and they 8712 is divisible by 99

[tidal swift]

87 and 12 are coprime?

are you sure?

[halcyon plank]

wait

no

sorry

i made a mistake

[inland coral]

whats going on

[halcyon plank]

its not coprime

[tidal swift]

good

[halcyon plank]

i rushed

[tidal swift]

so we can conclude p = 87 is out

agreed?

[halcyon plank]

yes

[tidal swift]

p = 86?

[halcyon plank]

it is

[tidal swift]

triple check

what is q if p = 86

[halcyon plank]

13

[tidal swift]

we know 13 is prime

[halcyon plank]

yes

[tidal swift]

so we just need to check if 86 is a multiple of 13

is it?

[halcyon plank]

no

[tidal swift]

and does 8613 divide p+q?

[halcyon plank]

99 divides 8613

[tidal swift]

good

there you have it

the last two digits of the largest possible r is 13

[halcyon plank]

so 13 is the answer

THANKS ALOT SEIA

[tidal swift]

nps

glad to help

[halcyon plank]

I WAS STUCK ON THIS PROBLEM SO LONG

[tidal swift]

yeah sometimes you just need a fresh pair of eyes

happens to the best of us

[halcyon plank]

THANKS FOR CONTRIBUTING TIME AND TOLERATINGME

!close

[tidal swift]

.close

[halcyon plank]

thanks once again

.close

[tidal swift]

nps

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[vestal merlin]

sorry

anyways

im pretty certain there's a way to use cauchy schwarz here

but idk how

[autumn lantern]

express f(1) in terms of f’

and express f’ in terms of f’’ to get f(1) in terms of f’’

and use cauchy schwarz

[vestal merlin]

$f(1)=\int_{0}^{1}f'(x),dx$

[rocky lotus]

TargetVN

[autumn lantern]

yep

[vestal merlin]

"express f' in terms of f" "

so, ibp?

[autumn lantern]

no use the same method

since f’(0)=0

[vestal merlin]

ah

wait how do u do it

[autumn lantern]

with an integral

idk how to express myself without giving the answer lol

like use FTC

[vestal merlin]

im a bit confused here :L

[autumn lantern]

$f(x)=f(0)+\int_0^x f’(x)dx$ right

[rocky lotus]

tm

[vestal merlin]

is it simply just f'(x) replaced as the integral from 0 to x of f"(t)

oh

[autumn lantern]

yes $f’(t)=f’(0)+\int_0^t f’’(x)dx$

[rocky lotus]

tm
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[autumn lantern]

and f’(0)=0 so we only have the integral

so now rewrite f(1)

[vestal merlin]

so do i do it like this

$f(1)=\int_{0}^{1}\left ( \int_{0}^{x} f''(t),dt\right ),dx$

[rocky lotus]

TargetVN

[autumn lantern]

yes exactly, but we would like a single integral to apply cauchy schwarz

[vestal merlin]

👀 i need more clues

this is still going nowhere

[autumn lantern]

rewrite f(1) as a double integral over a triangular region

[vestal merlin]

huh??

[autumn lantern]

the idea is that to use cauchy schwarz we would like to get $f(1)=\int_0^1 g(t) f’’(t)dt$

[vestal merlin]

the thing is that that's calculus 2, whereas the problem is from calculus 1

[rocky lotus]

tm
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[autumn lantern]

ah

hm let me try something else

[rocky lotus]

tm
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[vestal merlin]

i mean, i can understand why since i learned the basic of calculus 2 in advance

though, is that allowed in calc 1?

[autumn lantern]

idk the programs sorry

but i don’t think lol

I also thought about the Lagrangian but I don’t think it’s in calc 1

I can't see any other method than swapping the order of integration and using cauchy schwarz

[vestal merlin]

apparently i realized we had always been stupid

[vestal merlin]

just ibp this LOL

$f(1)=\int_{0}^{1}f'(x),dx=\left [ f'(x)(x-1) \right ]{0}^{1}-\int{0}^{1}(x-1)f''(x),dx$

[rocky lotus]

TargetVN

[vestal merlin]

1st one is 0

and it would return exactly like what u said

[autumn lantern]

wait

why x-1?

It’s been too long since I’ve had an ipp...

[vestal merlin]

f'(1)(1 - 1) = 0

that's the idea

[autumn lantern]

yeah but where did it come from lol

[vestal merlin]

it's just there to cancel out the uv

[autumn lantern]

u took u’=1 and v=f’(x) right

[vestal merlin]

u=f'(x) and dv=dx

v=x+C and i choose C = -1 specifically for that purpose

[autumn lantern]

i might be wrong but the formule give me $f’(1)-\int_0^1 xf’’(x)dx$

[rocky lotus]

tm
Compile Error! Click the reaction for more information.
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[autumn lantern]

or i’m fckng dumb idk

[vestal merlin]

because u chose v=x

[autumn lantern]

no i integrate 1

[vestal merlin]

yes

integral of 1 is x + C

not just x

[autumn lantern]

hm smart

i never did ipp like that with this constant lol

[autumn lantern]

okay and now u apply cauchy schwarz

[vestal merlin]

it's also the first time i have to do something like that

:L

[vestal merlin]

alright

[autumn lantern]

I’d never have thought of integrating v’ with a non-zero constant

[vestal merlin]

$f(1)=\int_{0}^{1}(1-x)f''(x),dx\leq \left ( \int_{0}^{1} (1-x)^2,dx\right )\left ( \int_{0}^{1}[f''(x)]^2,dx \right )$

[rocky lotus]

TargetVN

[vestal merlin]

got it

answer is 2

[autumn lantern]

yes

thx vro you taught me something new 😭

[vestal merlin]

bro u also taught me something new 😭

[autumn lantern]

so we’re fair 😌

[vestal merlin]

ye

thx bro

.close

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[glass kelp]

cauchy schwarz for integral doesnt have square on the lhs??

nvm

[autumn lantern]

yes but i think he did it with his head

[glass kelp]

ye i noticed 1/3 * 12 = 4 => 2

thx

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[light pumice]

What is 2^2^-1? I thought it would be (2^2)^-1, or 0.25, but Google calculator seems to think it's 2^(2^-1), or root 2.

[glass kelp]

$2^{(2^{-1})}$ or ${(2^2)^{-1}}$

[somber sequoia]

calculators generally read exponentials right to left instead of the usual left to right

[rocky lotus]

k

[somber sequoia]

so 2^2^-1 = 2^1/2 = sqrt(2)

[glass kelp]

it depends on what u're trying to say

[glass kelp]

if its this, ye

[light pumice]

I'm working on a calculator project, so I'm wondering which I should implement

[somber sequoia]

for exponents reading right to left is standard

[light pumice]

Alrighty then

Ty

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[sleek nebula]

Good luck on your calculator project!

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[marsh forge]

can someone explains what this answer means😭

[marsh forge]

i suck at reading math notation

(i dont think i have to translate it since its pretty universal)

[hybrid hound]

do you have the problem to go with this answer?

[marsh forge]

i do yeahh but i just dont understand the notation

gimme a sec

[hybrid hound]

do you know what the notation means?

[marsh forge]

Let $\mathbb{N}_0$ be the set of nonnegative integers. Find all functions $f:\mathbb{N}_0\rightarrow\mathbb{N}_0$ such that $$\ f(m+f(n)) = f(f(m)) + f(n)$$

[rocky lotus]

Copter

[marsh forge]

i know what each symbol means yeah

just not like "the big picture"

[hybrid hound]

if you could, do you mind translating the stuff, because there's a lot that's just written out and not notation

[marsh forge]

sure

[hybrid hound]

we just need the parts that aren't in english, the notation is fine

[marsh forge]

f(i) = ... for i = 0,1,..., a -1 where a in N, and n_0,...,n(a-1) in N0, and f(a+n) = a + f(n) for all n in N

therefore f(ap + q) = a(p + n_a) where q in {0,1,..,a-1) and p in N0

[vale dock]

@marsh forge Has your question been resolved?

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[whole ibex]

hello, here is aproblem. i don't need the solution just to know how to find it.

[whole ibex]

absolute values and what they do to inequalities conceptually is confusing me

[jovial edge]

$|x|<a \Rightarrow -a<x<a$

[rocky lotus]

Alexis_Fx

[jovial edge]

a >=0 ofc

[marsh forge]

think of |x| <= a as the set of all points with distance to the origin less than a

and |x| > a as the converse

[whole ibex]

hmm

i will think about this give me a moment to ponder

distance to origin

seems useful

way of thinkingabout it

[marsh forge]

you should get x is ||from -a to a||

[marsh forge]

and this is just everywhere else, so you obviously get x > a, and x< -a

[whole ibex]

oh my god i got it

you are a gentleperson and a scholar

i was breaking into two inequalities when turning it into one is easier for this situation

for me anyway

[marsh forge]

well is your question answered?

[whole ibex]

yes ty!

.close

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[whole ibex]

inequality issues again 😭

[whole ibex]

same situation as before, but the -5 outside the absolute value symbol makes me think adifferent emthod is necesary

[thorn grail]

Just add +5 to both sides and continue from there?

[whole ibex]

bruh

why didnt i think of that

ty

[thorn grail]

Cheers

[whole ibex]

literally the same as an equation

lmao

.close

ty

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[slim fossil]

\text{Given a polynomial} $P(x) = a_1 x^n + a_2 x^{n-1} + \ldots + a_n$. \text{Prove that if} ${a_n}$ \text{is a positive, real valued, non-decreasing sequence then} $P(x)$ monotonically increases on $\mathhbb{R}$

[rocky lotus]

AvidDavid
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[slim fossil]

I'm just conjecturing this

[stark wedge]

i think your numbering is a bit off

[slim fossil]

kinda want to explore more about this and get it on a high-school research competition

[stark wedge]

also you dont need those \text commands

if you're indexing the coefficients starting from $a_1$ like this then the last coefficient will be $a_{n+1}$

[rocky lotus]

Ann

[slim fossil]

can u pls see if it's alr been done before

[slim fossil]

oh yeah

[stark wedge]

so maybe you wanted $P(x) = a_0x^n + a_1 x^{n-1} + \dots + a_n$ ?

[rocky lotus]

Ann

[slim fossil]

yep

sry lol

[stark wedge]

you're given that 0 < a_0 < a_1 < ... < a_n then and you want to show P(x) is a monotone increasing function

[slim fossil]

not strictly less thatn

it just needs to be non decreasing

[stark wedge]

0 < a_0 ≤ a_1 ≤ ... ≤ a_n

[slim fossil]

if they're all constant then it's a geometric series and that's monotone increasing

[slim fossil]

yeah

[slim fossil]

for odd n btw

[stark wedge]

oh so we also know n is odd, ok

well the obvious idea is to look at P'(x) and attempt to show it's positive

[slim fossil]

the coefficients of P'(x) would be a decreasing sequence and the largest value will dominate

[stark wedge]

wait what

[slim fossil]

there'll also be a constant at the far end

[stark wedge]

since when do we know the coeffs of P'(x) are decreasing

[slim fossil]

oh sry i was working w the case that a0 = a1 = ... = const

and nvm i found a counterexample

hell

[frozen ledge]

yea i mean this is obviously false

it’s very easy to construct p satisfying your conditions but having a turning point

[slim fossil]

yea i just constructed one

it works with a lot of stuff so i still have some hope of refining the conditions

[frozen ledge]

it’s pretty clear when this would happen no? you just need to avoid turning points

[slim fossil]

yeah im trying to describe that rigourously

seems like we need to make P''(x) > 0 too?

[frozen ledge]

no

you just need the roots of p’ to be of even multiplicity

and the leading coefficient of x^n to be > 0

your example of the coefficients being in some sequence is just refining it even more

which is aside from the point of what you’re trying to get at

[slim fossil]

backtracking this condition would be tough...

[frozen ledge]

then don’t write the polynomial in standard form just write it in factored form

[slim fossil]

but having the factored form of P(x) doesnt yield the factored form of P'(x) easily no?

[frozen ledge]

sure

i was more so saying the derivative but honestly i’m failing to see what’s interesting about this

why are you unsatisfied with "the roots of p’ need to be of even multiplicity"

[craggy mist]

Won't it be "increasing" in place of non-decreasing ?

[slim fossil]

constant sequence works

[slim fossil]

im just trying to look for interesting things lol

[frozen ledge]

i doubt there’s a nicer answer in terms of coefficients

anything involving roots with n degree polynomials is just not going to go great

[slim fossil]

yeah, makes sense

[craggy mist]

x²+4x+4 = 0

[frozen ledge]

n is odd

and its already been disproven

[craggy mist]

Oh, i see.

[slim fossil]

@frozen ledge

[frozen ledge]

yes

[slim fossil]

if the coefficients form a concave sequence

wait

im being stupid arent i

[frozen ledge]

you’re saying this is sufficient?

[slim fossil]

Yeah

Im working on this

[frozen ledge]

what makes you say that

🤔

[slim fossil]

Cuz turning points are caused by changes in concavity

[frozen ledge]

that would be a point of inflection but what does that have to do with the coefficients?

[slim fossil]

That’s what im trying to figure out

[frozen ledge]

try making a counterexample

choose a_1 to be anything

[slim fossil]

alr lemme try some

[frozen ledge]

for simplicity just take it to be 1

then you also need to take a_2 satisfying your conditions but of nondecreasing

it’s simplest just to consider n = 3 btw when you’re looking

the derivative is a nice quadratic and you can use the discriminant

[slim fossil]

havent seen any counterexamples yet

lemme try

[vale dock]

@slim fossil Has your question been resolved?

[slim fossil]

So I’ve tested with a lot of stuff

And I found that all powers of x work

Nested logarithms work

Polynomials fail

If the nth coefficient is the nth prime then it also works

constant works

[slim fossil]

nth fibonacci also works

Tried with lots of sequences from oeis, as long as there are no “sharp” points it works

So both convex and concave

Even the nth harmonic number, if it’s the coefficient of x^n+1

<@&286206848099549185> any clues pls

[slim summit]

what is your question

[slim fossil]

I’m seeing a pattern about odd degree polynomials, about its coefficients and the polynomial’s monotonicity

I thought it works for every non increasing sequence for its coefficients but it isnt

Ive been trying with diff things to try to come up w a condition

[woeful frost]

i have a solution

me + you = yes

(joke)

[slim summit]

What is your pattern

[slim fossil]

if it the coefficients make up a sequence that grows nicely then the polynomial is monotonically increasing

im struggling to rigourously state the "grows nicely" part

[slim summit]

we can't invent things I think there is no link with the growth of a polynomial function and the growth of this one on the other hand we could look at a condition of the roots of the polynomials in itself there is necessarily a link you will tell me because the coefficients of your polynomial function and your roots are linked but the formulas are horrible and irregular so we can't get much out of them

[slim fossil]

all the sequences i tried have polynomial/exponential/factorial growth and/or linear combinations of em

well not all polynomials work but im guessing something

if i define the coefficient $a_k = f(k)$ then if $f(k) >0$ and $f'(k)$ > 0 then shit works

[rocky lotus]

AvidDavid

[slim fossil]

im trying to look into this

[slim summit]

Proof

[slim fossil]

im trying

[vale dock]

@slim fossil Has your question been resolved?

[sharp whale]

for n=3, we first assume P(x) = a₀ x³ + a₁ x² + a₂ x + a₃ is increasing without any extra conditions, then find out the minimum requirements for the coefficients

P is increasing iff its derivative P' > 0
so 3 a₀ x² + 2 a₁ x + a₂ > 0

a quadratic is non-negative iff its leading coefficient > 0 and it does not have two distinct roots
or that ax² + bx + c > 0 for all x iff a > 0 and b² - 4ac < 0

using this, 3 a₀ x² + 2 a₁ x + a₂ > 0 shows that
3 a₀ > 0 and (2a₁)² - 4(3a₀)(a₂) < 0
so a₀ > 0 and a₁² - 3 a₀ a₂ < 0
so a₀ > 0 and a₁² < 3 a₀ a₂

so P(x) = a₀ x³ + a₁ x² + a₂ x is increasing iff a₀ > 0 and a₁² < 3 a₀ a₂

the requirement a₁² < 3 a₀ a₂ can be rewritten to be easier to understand

since a₁² > 0 and a₀ > 0, we know a₂ > 0
but notice we ignore the sign of a₁ due to the requirements squaring it away
so a₀ x³ + a₁ x² + a₂ x is just as increasing as a₀ x³ - a₁ x² + a₂ x

we can then consider something nice which works if a₁ is not 0
if a₁ = 0, then a₁² < 3 a₀ a₂ is already true, so a₀ x³ + a₂ x is increasing iff a₀ > 0 and a₂ > 0
if a₁ ≠ 0,
a₁² < 3 a₀ a₂
|a₁²| < |3 a₀ a₂|
|a₁|² < 3 a₀ a₂
|a₁| < 3 a₀ a₂ / |a₁|
|a₁| / a₀ < 3 a₂ / |a₁|
now this can be easier read as:
(the ratio from a₀ to |a₁|) can only be as large as 3 * (the ratio from |a₁| to a₂)

so for a₀ x³ + a₁ x² + a₂ x + a₃, all you check is

• that a₀ > 0 and a₂ > 0
• if a₁ ≠ 0, that |a₁| / a₀ < 3 a₂ / |a₁|

[slim fossil]

i think i got a proof

using Descartes' sign rule

[slim fossil]

nah i give up this is beyond me

[slim summit]

As I told you, I don't think we can find anything satisfactory. It could even be false. We'll spend hours trying to find a counter-example.

[slim fossil]

well, it was worth working on. i thought i was onto smth for a while. thanks for the guidance

hope you have a great day

.close

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[slim summit]

U can try this if u want challenge Let ( P \in \mathbb{R}[X] ) be a non-constant polynomial such that ( P(x) \geq 0 ) for all real numbers ( x ). Show that there exist polynomials ( A, B \in \mathbb{R}[X] ) such that
[
P = A^2 + B^2.
]

[vale dock]

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[rocky lotus]

Clint

[slim fossil]

do you also mean nonnegative?

[slim summit]

No P>0 but A and B not necessarily

[merry crystal]

can you use the fact that R[X] is a ring over + and *?

to establish closure and existence

[slim summit]

Idk Wdym but go

[merry crystal]

oh we would be going backwards

[slim summit]

I don't know, try and tell me if you need any clues, I'll just give you some. ping me.

Here is a hint: we can start by showing that there exists a polynomial ( C \in \mathbb{C}[X] ) such that ( P = \overline{C} C ).

[rocky lotus]

Clint

[slim fossil]

using the FTA we can factor it into a bunch of linear factors (with even multiplicity because P is nonnegative, hence square) and quadratic factors

then for the quadratic factors $x^2 + px + q$ substitute $a = sqrt{\Delta}$ and $b = x+ \frac{b}{2}$ they become $a^2 + b^2$

[rocky lotus]

AvidDavid

[slim fossil]

repeatedly applying $(a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2 $and we're done w the quadratics

let

let b = 0 and apply this to the linear factors repeatedly and we're done

[slim fossil]

i didn't use complex conjugates but ig it's implied in the substitution step?

[slim summit]

Yep great idea

[slim summit]

What is c and d ?

[slim fossil]

other quadratics

[slim summit]

I don't see how you conclude on that.

[slim fossil]

like one we can write it as (a^2 + b^2) the other we write it (c^2 + d^2)

[slim summit]

Okay

[slim fossil]

then it's of this form, which is again sum of 2 squares: (a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2

mistyped the sign there

[slim summit]

mmh ok clever

just one fact not to forget is to prove that the dominant coefficient is positive

[slim fossil]

ur complex conjugation thingy is encoded in there somewhere right

[slim fossil]

oh yeah, that follows from the fact that P is nonnegative

[slim summit]

Yep ur décomposition is clever but it’s exactly what we do with the conjugate

[slim fossil]

when i did it i just factored out the coefficient c then work w the monic linears and quadratics for simplification's sake

[slim summit]

It's not easy to prove the same for multiplicity, I don't know your level but these are still non-trivial things.

we generally use analysis results for this if two functions are equivalent in the neighborhood of a point then the sign of the two functions in this neighborhood is the same

Use this result on +- inf for an and on a root for the multiplicity

[slim fossil]

odd multiplicity = change signs. P stays nonnegative => must only have even multiplicity roots

that was my intuition

[slim summit]

Yep idea not a proof

[slim fossil]

i mean it's sufficient as a proof isnt it

[slim summit]

No

[slim fossil]

if there exists a root with odd multiplicity then P will change signs at that root

but P stays nonnegative (no sign changes)

therefore no such roots

[slim summit]

Why

[slim fossil]

💀 my teacher said so

ig we could go deeper and prove that

but idk lol

[slim summit]

but you're right, eh, but you have to see why you do not have to accumulate knowledge of results without knowing how prove it. There are two ways to prove it, one as you stated: Let ( P(x) \in \mathbb{R}[X] ) be a real polynomial.
Suppose that ( a \in \mathbb{R} ) is a root of ( P ) with odd multiplicity.
Then we can write:

[
P(x) = (x - a)^{2k + 1} Q(x)
]

where ( k \in \mathbb{N} ) and ( Q(a) \neq 0 ).

We now study the sign of ( P(x) ) near ( a ):

• When ( x < a ), the factor ( (x - a)^{2k + 1} ) is strictly negative.
• When ( x > a ), the factor ( (x - a)^{2k + 1} ) is strictly positive.
• The function ( Q(x) ) remains either strictly positive or strictly negative in a neighborhood of ( a ), since it is continuous and nonzero at ( a ).

Therefore:

• If ( Q(a) > 0 ), the sign of ( P(x) ) is governed by ( (x - a)^{2k + 1} ), and thus ( P ) changes sign at ( a ).
• If ( Q(a) < 0 ), the conclusion is the same, but with opposite signs.

[
\Rightarrow P(x) \text{ changes sign at } a.
]

This contradicts the assumption that ( P(x) \geq 0 ) for all ( x \in \mathbb{R} ).
Hence, all real roots of ( P ) must have even multiplicity.

[rocky lotus]

Clint

[slim fossil]

oooh

cool

[slim summit]

Let ( a ) be a root of ( P ) with multiplicity ( m ).
We know that:

[
P(x) \sim \lambda (x - a)^m \quad \text{as } x \to a,
]

for some ( \lambda \in \mathbb{R}^* ).
If ( m ) is odd, then ( (x - a)^m ) changes sign when ( x ) crosses ( a ), and thus so does ( P(x) ).
This contradicts the assumption that ( P(x) \geq 0 ) for all ( x ) near ( a ).
Therefore, all roots of ( P ) must have even multiplicity.

[rocky lotus]

Clint

[slim fossil]

isn't this the same as the one above

just stated a bit more compactly

[slim summit]

No, here there is something hidden, it's what I said above, two continuous functions have the same sign in a neighborhood when they are equivalent in this one.

the first is an algebraic proof using the factorization theorem and the second using analysis

[slim fossil]

well i haven't learnt analysis yet. intuitively they both conveys sort of the same idea

[slim summit]

Yep

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[sick fossil]

Translation : Suppose f(x) is continuous on its domains and has the following table of variations (on the right).

Examine the function of g(x) = f(-x^2+3x) and draw its table of variations.

(yes, it is a multiple choice question but i want to know i should go about doing this generally with excerises that gives higher degree nested functions and/or higher derivatives.)

[sick fossil]

What I did is finding g'(x) zeroes like so :

If f'(x)=0 means that x is either -4 or 0, then that means g'(x)=f'(-x^2+3x)=0 has to mean -x^2+3x has to be equals to either -4 or 0. This turns out to be x=-1, x=4 or x=0

What I am having trouble with is how should i go about telling if g'(x) is larger or smaller than 0 in the domain from (-infinity;1)

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[slim fossil]

1 trò khá là mẹo ông có thể làm là giải phương trình hàm để tìm f(x)

sau đó ông lắp f(-x^2 +3x) vào ra g(x)

lúc này có g(x) rồi thì tính đạo hàm của nó đơn giản

giả sử f(x) là đa thức thì mọi thứ đơn giản th

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[small marsh]

This questionMathematical Olympiad National Round of this year of my country. So the question is,

There are infinitely many positive integers called a, b, c, d & e such that ab+4, bc+1, cd+49, de+25, ea+1 are perfect square numbers. Now,
a) Find such 3 positive integer solutions of a, b, c, d & e.
b) Prove that, there are infinitely many positive integers, such that ab+4, bc+1, cd+49, de+25 & ea+1 are perfect square numbers.

On the exam itself, I found a single solution for a, b, c, d & e. But since then, I tried to find more through trial and error but failed.

[small marsh]

<@&286206848099549185>

[cloud coral]

hmm, oh i see something

difference of squares hmm

think about that

from there it should be trivial

[small marsh]

Btw, does the proof require finding patterns between solutions? Cause taht's the only thing I could think of while finding the proof.

[cloud coral]

i guess not necessarily

[small marsh]

So there are other ways?

[cloud coral]

the difference of squares thing

so you understand what relation there is between a and b

[small marsh]

So we have to use Pell's Equation?

[cloud coral]

factor $x^2-y^2$

[rocky lotus]

Element118

[small marsh]

I got to this :

(k-√4)(k-√4) = ab
(k-√1)(k+√1) = bc
(k-√49)(k+√49) = cd
(k-√25)(k+√25) = de
(k-√1)(k+√1) = ea

[cloud coral]

wouldn't it be nice if the two products corresponded

[small marsh]

Ye

[cloud coral]

what if they did

[small marsh]

For ab + 4 = k^2
Then a could be = k-2 and b could be = k+2
Or a could be = 1 and b could be (k-2)(k+2)

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[midnight pier]

Is the solution ok

@unborn apex

@stark wedge

Look is this right

?

[jovial edge]

!noping

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[midnight pier]

Ohk

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[slate basin]

Set of all bijective function composition over a finite set what is the identity of this group?

[merry crystal]

the identity map?

[slate basin]

@merry crystal the waht

[merry crystal]

id(x) = x

[slate basin]

Ooooo

Great tbanks

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[spark gull]

There are four sockets on a bookshelf, each numbered 1, 2, 3, 4, and eight different books. An (a name) divided all of the eight books into the four sockets such that each socket has at least one book, and the books are arranged vertically in a horizontal row with the spines facing out on each shelf. When done, An's two methods of arranging are similar if they satisfy the following conditions at the same time:

• For each socket, the number of books in that socket in the first method is the same as that in the second method.
• For each socket, the order from left to right of all the books is the same in both methods.

Call T the number of ways, different from each other (way 1 != way 2 != … idk the word in English lol), that An can arrange the books. What is the value of T/600?

[spark gull]

Can someone please help guide me through this, i'm so lost T.T

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[spark gull]

<@&286206848099549185>

[jovial edge]

Why is this so similar to the one in the Vietnam national exam

[spark gull]

cause it is lol

i just don't get how they figured out the way to do it, thought I would ask here for help

[clever sentinel]

do you have link to original question?

[jovial edge]

Okay, so from the given condition we know the order in each "socket" does matter

That means we will need to use permutations

[spark gull]

ok

[spark gull]

here if u can read vietnamese

[jovial edge]

Let's do a smaller problem, we have 6 books a,b,c,d,e,f

How many ways can we arrange these 6 books into 4 "sockets"

[spark gull]

6p4

[jovial edge]

You know what, screw this

I'm gonna speak Vietnamese

[spark gull]

=))))))

[jovial edge]

Bây h nha

Có 4 ngăn đúng k

Tức là có 3 vách

H tìm số cách đặt 3 vách đó vào khoảng trống giữa cách quyển sách

V có bao nhiêu khoảng trống trong 8 quyển sách ở đầu bài

[spark gull]

7

[clever sentinel]

thanks

[spark gull]

tức là 7p3 =210

sao nữa bro T.T

[jovial edge]

Chọn 3 khoảng trống trong 7 khoảng trống có bn cách?

[spark gull]

7c3

[jovial edge]

Oke, h xử lí vụ hoán vị này

Vì ta tính cả hoán vị trong cả từng khoang chứa sách

Nên ta chỉ cần hoán vị cả 8 quyển

[spark gull]

hmm

là 7c3 x 8!

à à à ok

tức là hoán vị 8 quyển sách

sau đó đặt các vách vào giữa chúng thôi

nhanh gọn v

[jovial edge]

Um

Yeah bài này dễ mà

Nhân tiện hỏi phát, bn lớp mấy đó

[spark gull]

mình ms test đề này 🙂

ý là 12 á

[jovial edge]

Oh

Okay

[spark gull]

bn lớp mấy v

[jovial edge]

Năm nay lên 12

[spark gull]

=)) ok

[spark gull]

xin típ tư duy mấy bài xs được không :V

ý là ní nhìn rùi lm sao biết cách lm á

chắc lúc thi đến bài n là áp lực r nên nhìn cái đề hết muốn lm bỏ luôn T.T

[jovial edge]

Bài này là vận dụng bài toán chia kẹo euler vs hoán vị thôi

Làm nhiều là biết

[spark gull]

hm ok nha

cảm ơn chỉ bài nhé

[jovial edge]

Nếu xong rồi thì đóng kênh này nhé:)

[spark gull]

ýe ýe

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[jovial edge]

Cái bài tổ số tam giác sài chia hết

[spark gull]

à bài đó t ngồi đếm =))

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[spark gull]

hông mấy chỉ cách đó luôn ik

[jovial edge]

Nahh, chỉ lâu lắm, đêm rồi

Có j sáng dậy mở kênh

[spark gull]

à okok

ngủ ngon

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[maiden panther]

How do I do this🙏

[glass kelp]

hello "susgusinabus"

[maiden panther]

Ahh yes

What I was hoping for

I did most of the work for you

By the way t is tanx/2