#Weird gamma / poly-gamma equation.

1 messages ยท Page 3 of 1

left sigil
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But bro

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We will get cooked

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Product of gammas in denominator ๐Ÿ˜ญ

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See

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We are getting cooked by products

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We don't know product of gammas

hallow geyser
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wait

left sigil
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Nor we can evaluate this double denominator product

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Ezily

hallow geyser
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1/gamma(x) has a product representation

left sigil
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What

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Yus

hallow geyser
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i remebered

left sigil
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Bro

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;-;

hallow geyser
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i forgor that weistras existed

left sigil
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Why did u not tell

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Then we go with gamma

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Bose

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Form

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For real

hallow geyser
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or the bernullie number idea

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look this is the representation

left sigil
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Okok

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Imma try bernoulli

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Show

hallow geyser
left sigil
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Bro zeta and Bernoulli have ;-;

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Aah

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Bro but this for every complex number right?

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Bro but we are cooked

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Bro

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How are we even gonna multiply

hallow geyser
left sigil
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Every gamma in denominator

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It will become product within product

hallow geyser
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yeah

left sigil
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We only have two options

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Bernoulli on me

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And gamma Bose on u

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Eta definition?

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If we use eta definition?

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Do u know product of etas?

hallow geyser
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eta dirichlet?

left sigil
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Yes

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(-1)^k/(k^s)

hallow geyser
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yeah but how is it usefull

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?

left sigil
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It's helpful

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If we can derive products of

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Etas

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Ezily

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Can we?

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Like eta 2 eta4 .... Eta(2n)

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Fr

hallow geyser
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i mean they have a strong relation

left sigil
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Yes

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Like

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zeta=eta/(1-2^(1-s(

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Something

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Like that

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Imma check fast

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Yea it's correct

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Now

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How do find product of etas?

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Any definition

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Or basic series

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@hallow geyser

hallow geyser
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the same way we been doing for the past hour

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i think

left sigil
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Bro but see

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We had used zeta in terms of bernoulli

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I don't much definitions for eta

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Aah

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Do u know

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If yes pls tel

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L

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We begin

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@hallow geyser

hallow geyser
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a yyea i know a def of eta

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in terms of zeta

left sigil
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Ok tell

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๐Ÿ’€

hallow geyser
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thats the problem

left sigil
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Bro it's good

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But we need product of etas

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For zetas

hallow geyser
left sigil
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Bro

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It's same as what i gave

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Zeta=eta/(1-2^(1-s)

hallow geyser
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there shoul be a parenthesis before the Zetsa

left sigil
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We need zeta

hallow geyser
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yeah

left sigil
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Any other definition

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Wait imma check quick

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Wait

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@hallow geyser

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U good at integrals

hallow geyser
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yeah

left sigil
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We have eta(s)=1/gamma(s) * integral from 0 to infinity x^(s-1)/e^x+1

hallow geyser
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yse

left sigil
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But same problem product of gammas

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Wait checking other

hallow geyser
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yes

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k man

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now i really gota go

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let's drop this for a while

left sigil
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Aah

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Okok imma research

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With the Bernoulli form

hallow geyser
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i will leave you some home work

left sigil
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Okok

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Okk

hallow geyser
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before you leavke this is the exercisse i want to leave to you

left sigil
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No bro don't leave

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Okok tell

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Btw tmrw I am bz

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Tell fast bro

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Gonna go to sleep

hallow geyser
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prove this

image.png
left sigil
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It's 11:01 here

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Bro

hallow geyser
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k

left sigil
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For real

hallow geyser
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but with integrals

left sigil
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Ok

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Imma use the Bose integral forn

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Form lol

hallow geyser
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k

left sigil
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Okok

hallow geyser
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youll have some fun its a math snack

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go sleep now

left sigil
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Okok

hallow geyser
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bye

left sigil
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But imma keep working

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On the Bernoulli form

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Cya

left sigil
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Ok

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@hallow geyser u there?

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He went

left sigil
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@hallow geyser

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Hlo

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I found something

halcyon hearth
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found what

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@left sigil

left sigil
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A ray of light to find the product of even zetas

halcyon hearth
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im doing that right now

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nearly done

left sigil
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Just evaluate this limit

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And u get the product of even zetas

halcyon hearth
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i dont even need limits

left sigil
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Ok

halcyon hearth
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ill be back in 30 minutes

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show u my final product

left sigil
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Ok

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Ok

halcyon hearth
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nvm

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fucking failed

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whats ur idea

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@left sigil

left sigil
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;-;

halcyon hearth
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can u explain it for a bit i havent read the entire conversation

left sigil
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My idea is hypothetical

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Leave it

halcyon hearth
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ok

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say

left sigil
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It involves 3 core function

halcyon hearth
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i wanna see whether i can get ahything from it

left sigil
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Leave it

halcyon hearth
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ok go on

left sigil
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Leave it bro

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But yes see this

halcyon hearth
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u aint getting any progress if you aint tellin

left sigil
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See this

halcyon hearth
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?

left sigil
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I tried the basic method

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Not the op method

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Or not my method

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The basic method is

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P=product of zetas

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Take log both sides

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logP=sum of log zetas

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raise to e power

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P=e^lim n tends to infinity sum of log zetas

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See this

halcyon hearth
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any pictures

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can u write it down

left sigil
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,w summation from r=1 to infinity ln[zeta(2r)]

dusky sapphireBOT
left sigil
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So product is simply P=e^(0.599395)

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;-;

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I cannot find it manually

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I failed manually

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,w e^(0.599395)

dusky sapphireBOT
left sigil
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๐Ÿ’€

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@halcyon hearth

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The product is

halcyon hearth
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yes?

left sigil
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P=1.8102....

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Surprising fact

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,w zeta(2)xzeta(4)xzeta(6)xzeta(8)xzeta(10)xzeta(12)

dusky sapphireBOT
halcyon hearth
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wha

left sigil
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Did u understand

halcyon hearth
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yea but thats just

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ya i did

left sigil
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I feel it does stop growing

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Very fast

halcyon hearth
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yes cuz it approaches 1 very fast

left sigil
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Like see only 6 zetas and it is 1.8208

halcyon hearth
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because the bernoulli definition of the zeta function has a factorial in the denominator

left sigil
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Yes

halcyon hearth
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hence of course the denom grows very fast

left sigil
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;-;

halcyon hearth
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๐Ÿคฆ๐Ÿปโ€โ™‚๏ธ

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i thought you were doing more than this

left sigil
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But leave the Bernoulli definition

halcyon hearth
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no\

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im working on it

left sigil
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No

halcyon hearth
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found something online i might integrate it

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seriously

left sigil
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What

halcyon hearth
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trust

left sigil
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See what I send you

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Is this

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1.821...

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It is derived from the Bernoulli definition

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What I sent u above

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But it involves b(n^2+n) and the Barnes G function

halcyon hearth
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well what is it represented in terms of bernoulli definition

left sigil
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๐Ÿ’€

halcyon hearth
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can u give me one

left sigil
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Zeta

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Okok

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Zeta(2n)=B(2n).(2ฯ€)^2n/2.(2n)!

halcyon hearth
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well given the fact that the sum of all even bernoulli numbers is zeta(2)-1

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you can represent it better

left sigil
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Okok

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Hlo

halcyon hearth
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hi

left sigil
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Did it work

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Give me imma integrate

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What u found

halcyon hearth
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read on ur own

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takes a while

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good luck ๐Ÿ˜ธ

left sigil
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!filetype

glass parrotBOT
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Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

halcyon hearth
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you dont have to download dit

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its a link

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to the website

left sigil
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Got it

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Clear as hell

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Now

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Okok

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@halcyon hearth

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So what are u doing now

halcyon hearth
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trying to combine the polygamma definition of the bernoulli numbers we just found into the fact that the sum of bernoulli numbers is zeta(2)-1

left sigil
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Oof

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Don't u think we need the product?

halcyon hearth
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hence i will have a definition of the product in terms of zeta(2)-1

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you get me?

left sigil
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Yea gotchu

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Got it

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What was the polygamma definition

halcyon hearth
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@left sigil

left sigil
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Okok

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,w 1-ln(2)

dusky sapphireBOT
halcyon hearth
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now what the fuck is that

left sigil
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,w summation from r=1 to inf. (zeta(r)-1)

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Bruh

halcyon hearth
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lllooooooooool

left sigil
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Lol

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M

halcyon hearth
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try riemanzeta

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or riemann_zeta

left sigil
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No it does it

halcyon hearth
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k

left sigil
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Idk why it's not doing

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Like see

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,w zeta(4)

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Sum of all zetas -1 is simply 1-eta(1) lol

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Sum of all etas-1 is simply -eta(1)

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Do u understand

halcyon hearth
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are you using the dirichlet eta

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but thats regularization only

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dont think it applies to positive even zetas

left sigil
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No

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Sum of all [zetas-1] is 1-eta(1)

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And sum of all [etas-1] is simply -eta(1)

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Lol

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Let me try sum of all gamm-1

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Lol

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Yo

halcyon hearth
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,w Sum[i*pi/2+ln(digamma(2n-1,1)*4n/(2pi)^2n),{n,1,infinity}]

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what

left sigil
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Wait let me try

halcyon hearth
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,w Sum[ipi/2+ln(digamma(2n-1,1)4n/(2pi)^2n),{n,1,infinity}]

dusky sapphireBOT
halcyon hearth
#

this thing i shella dumb

left sigil
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,w summation from r=1 to inf. ln(digamma(2r-1,1)4r/(2ฯ€)^(2n)])

halcyon hearth
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,w Sum[i*Divide[pi,2]+ln(40)digamma(40)2n-1(44)1(41)*4Divide[n,Power[(40)2pi(41),2n]](41),{n,1,infinity}]

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bruh

dusky sapphireBOT
halcyon hearth
#

its sitalinning my text

left sigil
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Mf

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@halcyon hearth

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Oof

halcyon hearth
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,w Limit[ln(digamma(2n-1,1)/(2*pi)^2n),n->infinity]

dusky sapphireBOT
halcyon hearth
#

FUCK

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FUKC

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FUCK

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FUCK

left sigil
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Lol

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Why are u doing

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This when u can simply do this

halcyon hearth
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do what

left sigil
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,w summation from r=1 to infinity ln[zeta(2r)]

dusky sapphireBOT
halcyon hearth
#

bro thats not it trust

left sigil
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so Product of all zetas is

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e^(0.599395)

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,w e^(0.599395)

dusky sapphireBOT
halcyon hearth
#

yes im tryna do the math just leave me alone pls

left sigil
#

Okok

halcyon hearth
#

IT DOESNT

left sigil
#

What

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What was I saying bro

halcyon hearth
#

IT DOES NOT DIVERGE

left sigil
#

U sure?

halcyon hearth
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THIS DOESNT EVEN DIVERGE

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SO WHY WOULD THE LOG OF IT DIVERGE

left sigil
#

Ik

halcyon hearth
left sigil
#

Ask the calculator

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Yes i

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I am not saying this

halcyon hearth
#

just use norlund sum

left sigil
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I am saying why u playing with Bernoulli

halcyon hearth
#

i am 1000000000% sure it does not diverge using norlund sum

halcyon hearth
#

do you not listen

left sigil
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I listened

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But see

halcyon hearth
left sigil
#

,ln(0.1)

halcyon hearth
#

wtf does that fucking do

left sigil
#

,w ln(0.001)

dusky sapphireBOT
left sigil
#

Understand

halcyon hearth
#

listen bro that doesnt do shit

left sigil
#

We have -ves bro

halcyon hearth
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we're using norlund sum

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REDAD

left sigil
#

See

halcyon hearth
#

READ

#

THE

#

LINK

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READ THE LINK

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READ THE LINK

left sigil
#

Ik it

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Ik

halcyon hearth
#

you obviously havent

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you obviously havent bro

left sigil
#

But it is not saying bout sum of log(B2n)

halcyon hearth
#

logically

left sigil
#

Understand

halcyon hearth
#

if the sum of B2n is convergent

left sigil
#

And stop shouting

halcyon hearth
#

so is the sum of log B2n

halcyon hearth
left sigil
#

Bernoulli numbers are very less

halcyon hearth
#

if you read the link

left sigil
#

Like 1/30

halcyon hearth
#

if you read the link

#

if you read the link

left sigil
#

,w ln(1/30)

halcyon hearth
#

if

#

uyou

dusky sapphireBOT
halcyon hearth
#

read

#

the

#

link

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if you read the link

left sigil
#

Ok ok

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Wait we check

halcyon hearth
#

listen

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if wolfram alpha gets sum b2n wrong

left sigil
#

I read the link till the point he reached the sum

halcyon hearth
#

then why should i get sum log b2n right

left sigil
#

Is zeta(2)-1

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,w summation from r=1 to infinity B(2r)

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Lol

halcyon hearth
#

see?

#

see?

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this bot is wrong af

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no idea how to use norlund sum

dusky sapphireBOT
halcyon hearth
#

bro has no bloody clue

left sigil
#

Wait wait

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Is he even

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Considering B(2r)

halcyon hearth
#

THATS MY POINT

left sigil
#

As 2rth Bernoulli number

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I don't think so

halcyon hearth
#

yes he is

left sigil
#

No

halcyon hearth
#

its in the bracket

left sigil
#

Id think

halcyon hearth
#

why would it be otherwise

left sigil
#

No

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It's not

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,w compute 2nd Bernoulli number

dusky sapphireBOT
left sigil
#

See

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Bernoulli numbers are no functions

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We represent their number is sub script

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In*

halcyon hearth
#

alr what now

#

huh

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HUH

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HUH

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??

left sigil
#

Now idk

halcyon hearth
#

.

left sigil
#

Wait

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,w compute summation of all even Bernoulli numbers

dusky sapphireBOT
left sigil
#

Lol

halcyon hearth
#

i give up

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@hallow geyser would really appreciate it if you could read our convo and provide us with some info

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tysmmmm

left sigil
#

The sum is a simple value

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But to reach to it

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Is not simple

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It's 1.8210

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Lol

halcyon hearth
left sigil
#

See if

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U differentiate the function x/(e^x-1) n times and put lim x tends to 0

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U get the nth Bernoulli

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Number

halcyon hearth
#

tf

left sigil
#

@halcyon hearth

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Yes

halcyon hearth
#

ok

left sigil
#

That is lim x tends to 0 nth derivative fo x/(e^x-1)

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Is nth Bernoulli number

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U can find 2,3

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Using this

halcyon hearth
#

i dont get

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what ur tryna say

left sigil
#

That is B2 B3 and so on

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Using this

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Wait

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This means

halcyon hearth
#

why dont u demonstrate

left sigil
#

Pro tip

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As sum of all bernoulli is

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Zeta(2)-1

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Sum of lim x tends to 0 all derivatives of x/(e^x-1) is also zeta(2)-1

halcyon hearth
#

why

left sigil
#

Cz nth Bernoulli number is

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I told above

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Bye

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Cya

halcyon hearth
#

can u write it down

left sigil
#

Later

halcyon hearth
#

cant identify your jumbled up mess

left sigil
#

Going

halcyon hearth
#

ok

#

i give up anyways

hallow geyser
#

hello

#

seem you reached a dead end

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@halcyon hearth you might wanna see this

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it seem that there is no discovered closed form for the product

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By elementary estimates, the constant lies in the open interval (Pi/6, exp(3/4)). - Bernd C. Kellner, May 18 2024

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my take on this is that one may get a closed form of the product using inequalities

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Decimal expansion of zeta(2)zeta(4)...zeta(2k)...
If u(k) denotes the number of Abelian groups with group order k , then Product_{k>=1} zeta(2*k) = Sum_{k>=1} u(k)/k^2.

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This constant C is connected with the product of values of the Dedekind eta function on the upper imaginary axis. The product runs over the primes, where i is the imaginary unit: 1/C = Product_{prime p} (p^(1/12) * eta(i * log(p) / Pi)). - Bernd C. Kellner, May 18 2024

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this is the best i found

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and i sugest that if you are not interested in spending too muchh time in this to drop it and move on

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@left sigil

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i will be going now sincei have work to get done

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bye guys

left sigil
#

Yo

#

@hallow geyser

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@hallow geyser Come aa

hallow geyser
#

can't rn

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im doing homework

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:/

left sigil
#

Imma help

#

@hallow geyser

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What is the homework

hallow geyser
#

wut

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its an exam

left sigil
#

Oh ok

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Nvm

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Cya

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I understand

hallow geyser
#

bye

left sigil
#

Bye

left sigil
#

@hallow geyser

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Yo you free?

hallow geyser
#

yeah

#

@left sigil sup

#

xsecdrtvbnbgvftdcrsexrdctfvbgyhui'

left sigil
#

Yo

#

Want to ball

#

,w zeta(phi)

dusky sapphireBOT
left sigil
#

Bruh

hallow geyser
#

bruh

left sigil
#

Our mission is today

#

Yo find zeta(phi) by hand

hallow geyser
#

have you done the proof?

left sigil
#

That one

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Oh no

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I forgor

hallow geyser
#

do it

left sigil
#

Skull

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๐Ÿ’€

hallow geyser
#

very good excerssice

left sigil
#

Ok i

#

@hallow geyser

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Now can we

#

Bro went offline

hallow geyser
#

wut

left sigil
#

Now can we discuss

hallow geyser
#

wut

left sigil
#

Can u tell me zeta definition

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Which has operation range for every rational number

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1

hallow geyser
#

les than 1 of bigger?

left sigil
#

Bigger

hallow geyser
#

this one

left sigil
#

Is this valid for every rational number ?

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I don't think this is general definition right?

hallow geyser
#

yeah

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every real bigger than 1

left sigil
#

Sure?

hallow geyser
#

yeah

left sigil
#

Ok

left sigil
#

,w summation from r=2 to 5 1/[(r)^(phi)-1]

dusky sapphireBOT
halcyon hearth
#

im dropping this

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too much

left sigil
#

Lol

#

@halcyon hearth

#

What

left sigil
#

@halcyon hearth what are u dropping

halcyon hearth
#

yes

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you dont need me anyways

#

im only 15

#

ill learn more about different stuff then ill take this on (and probably fail

#

)

left sigil
#

Everyone fails

#

But those who accept that they failed

#

Become better

#

@halcyon hearth

left sigil
#

@halcyon hearth yo

#

Come

hallow geyser
#

@halcyon hearth it's better to have a more positive mindset

#

try to focus on what you can learn and how you can grow from the challenges you propose to yourself.

#

Every step you take provides you with new perspectives and knowledge.

hallow geyser
left sigil
#

Oh

#

@hallow geyser

#

@halcyon hearth

halcyon hearth
#

tbf my dream career isnt even maths

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not my best thing

#

its sorta just a side hobby i do

left sigil
#

Okok

#

Bro

#

Chill

#

U are just 15

#

U have years to find your dream career

#

We just discuss here casually

#

@halcyon hearth

left sigil
#

@hallow geyser

hallow geyser
#

sup

left sigil
#

Yo

#

Wanna do something

#

Let me make myself clear

#

Here for a personal project imma here

#

U tell the topic

hallow geyser
#

k

left sigil
#

We ball

hallow geyser
#

hmmm

left sigil
#

Enough of me discussing

#

We providing free service fr

#

Ur topic ur choice

hallow geyser
#

look

left sigil
#

Yea

hallow geyser
#

this might be a very interesting exercice

left sigil
#

Ok

hallow geyser
left sigil
#

Okok

#

I see

hallow geyser
#

so i think for a series representation here

left sigil
#

Ok

#

General representation?

hallow geyser
#

i guess

#

for the reiman zeta function

left sigil
#

See

#

We can write

#

Gamma(1+z)=zgamma(z) write?

#

Right*

#

Or is it only for Re(z)>1

hallow geyser
#

for all non negatie integers]

left sigil
#

Ok

#

So leave I cannot get it to gamma

#

See I was thinking to use the Bose integral definition

hallow geyser
#

hmmm

left sigil
#

Ok we try simply

#

We integrate 1 and diff zeta

hallow geyser
#

gona be a masive problem

left sigil
#

No see

#

I got

#

(ฯ€^3)/3-integral from 0 to 2ฯ€ zeta'(1+e^it).e^it.t.dt

#

Is the eta zeta definition for complex ?

#

Numbers?

hallow geyser
#

yes

left sigil
#

I think its for s>0 right

#

It's for complex numbers

#

Lol

#

Okok

#

Wait idea

#

Do u know the zeta functional equation

hallow geyser
#

yeah

left sigil
#

Zeta(s)=2^(ฯ€)^s-1 sin(ns/2) zeta(1-s)gamma(1-s)

#

Ya

#

Ook

#

Now what if we use

#

s=-e^(itheta)

hallow geyser
#

yeah

left sigil
#

I mean we can get zeta(1+e^it) in terms of zeta(-e^it)

hallow geyser
#

i think we can't

left sigil
#

See

#

Another idea

#

U know kings rule

#

?

hallow geyser
#

no

left sigil
#

If u apply it it becomes zeta(1+e^-it)

#

kings rule cz I am in india currently

#

It's said like it

#

let me xplain

#

f(b+a-x)=f(x) rule

#

Do u know?

hallow geyser
#

no

left sigil
#

Bro

#

int from a to b f(b+a-x)= int from a to b f(x)

#

U not know this?

hallow geyser
#

nah

left sigil
#

,w State kings rule of integration

#

No

#

,w kings rule

hallow geyser
#

bro

#

i justhad a revelation

left sigil
#

Oof

#

Ok

hallow geyser
#

hear me ou

#

t

left sigil
#

Ok

hallow geyser
left sigil
#

Oh

#

You gonna do complex analysis

#

Ok

hallow geyser
#

you know that that e^ix its the complex that lie down that circumference?

left sigil
#

Yes

hallow geyser
left sigil
#

Ik this

#

But we need zeta of this

#

And it's integral

#

๐Ÿ’€โ˜ ๏ธ

hallow geyser
left sigil
#

Ya ok

hallow geyser
left sigil
#

What

hallow geyser
#

then you know complex analysis theorem?

left sigil
#

What

#

What

#

e^itheta=z it's good

#

Then u directly wrote

#

idtheta=dz

hallow geyser
#

crap g

left sigil
#

Okok

#

Wait

#

Ok hear me out

hallow geyser
left sigil
#

don't u think there is another way too

hallow geyser
#

k nows its good

left sigil
#

Ok

#

Actually

#

yea ok

#

U can find this generally also

hallow geyser
#

now

left sigil
#

Yea

hallow geyser
#

be prapared\

left sigil
#

This only I was saying about

#

Okok

hallow geyser
#

gamma was a closed smooth curve in the complex plane

left sigil
#

Ok

hallow geyser
#

then we can aply the following theorem

left sigil
#

Ok

hallow geyser
left sigil
#

Yes ok i see

hallow geyser
#

sowe gota find the res of the zeta(1+z)/z

left sigil
#

Fr

hallow geyser
#

i think there is one in 0

#

but idk if there are any other

left sigil
#

Imma expressing them as series

#

And summing

#

Wait imma tell u something

#

See this

hallow geyser
#

i see

left sigil
#

Now if u apply this

#

see bro

#

We treat our original integral as I

hallow geyser
#

aaaaaaa this thing

left sigil
#

And now

#

We apply this propert

#

So our integral remains i

#

So we get

#

2I=int from 0 to 2ฯ€ zeta(e^it)+zeta(e^-it)

#

Fr

hallow geyser
#

true

left sigil
#

Now do u know

#

cosx=(e^ix+e^-ix)/2

#

We reaching the point with this one fr ๐Ÿ”ฅ๐Ÿ”ฅ๐Ÿ—ฃ๏ธ

hallow geyser
#

yeah

left sigil
#

Now imma just thinking this only

#

Should I use this

#

Or I should simply express zeta(1+e^it) and zeta(1+e^-it)

#

As seriss

#

Infinite series*

hallow geyser
#

wait

left sigil
#

Fr

#

Imma show your

#

See this

#

Wait

hallow geyser
#

no i was going to see if the series expansion was valid to all negative integres

left sigil
#

Fr

#

We gonna ball with this one

#

Now I use the general definition

#

For real

#

See

#

General definition

#

Valid for

#

Real>1

hallow geyser
left sigil
#

We have

#

Re>1

#

For all values in the interval

#

No wait

#

I feel imma wrong

#

I feel need to break

hallow geyser
#

so you cant use it since in 3pi/4 it goes to negative 1

left sigil
#

No prob

#

We gonna break

#

At each ฯ€/2 intevral

#

Let me compute the sum first

#

Na this not working bro

#

It getting complicated for intervals

#

Aah

#

@hallow geyser proceed with the complex analysis

hallow geyser
#

we just gota fin the Residues of the function

left sigil
#

We were at gamma is a smooth closed curve

#

Over the

hallow geyser
left sigil
#

,w residue zeta(1+z)/z

dusky sapphireBOT
left sigil
#

Lol

hallow geyser
#

WAAAAAAAAAAAAAAAAT?

#

!!!!

left sigil
#

Ik tricks

#

Fr

hallow geyser
#

Fire AF

left sigil
#

We have residue as gamma

hallow geyser
#

then we recall

#

mascheroni constant

left sigil
#

Yes

#

Gamma Euler macheroni constant whatever

#

Okok u proceed now

hallow geyser
#

so since we only have 1 residue

#

its 1 term

left sigil
#

Ok

#

2ฯ€igamma?

hallow geyser
#

yes i think

left sigil
#

Lol

hallow geyser
#

ah but then

left sigil
#

For real

hallow geyser
#

we have that i there

left sigil
#

So we get

#

2ฯ€Y

#

Lol

hallow geyser
#

so its 2pigamma

left sigil
#

Lol goated fr

hallow geyser
#

epic

#

yeah

#

lemme check tho

left sigil
#

Ok

hallow geyser
#

man

#

we got it right

#

les gooooo

left sigil
#

Lol

#

We reaching it to the top with this one

#

@hallow geyser Let's go next topic

hallow geyser
#

fr ๐Ÿ—ฃ๏ธ ๐Ÿ”ฅ

left sigil
#

Lol bro a exotic sum

hallow geyser
left sigil
#

No no

#

Today I don't have anything specific that much

#

Will discuss later

hallow geyser
#

k

left sigil
#

Your turn

hallow geyser
#

hmmmmmmmm

left sigil
#

Ok u type imma get water

hallow geyser
#

k

left sigil
#

We getting hydrated with this one ๐Ÿ˜€

#

Ok imma here

#

Lesss gooooo

hallow geyser
#

there are two thing i wanted to discuss

left sigil
#

Ok

hallow geyser
#

one related with actuall math

left sigil
#

Ok

hallow geyser
#

and the other one is more filosofical related

left sigil
#

Ok

#

We here for both

hallow geyser
#

i only go time like for halve of any of the topics :/

left sigil
#

What?

hallow geyser
#

its getting mad late in my country

left sigil
#

You wanna say

#

You got time for only half of the topics

#

Ok we be fast

hallow geyser
#

kk

#

so the filosofical math it goes like this

left sigil
#

Ok

hallow geyser
#

you remember goodel shit?

left sigil
#

Goodel

#

So

hallow geyser
#

so there are matha statements that can't be proven true or false

left sigil
#

Yes fr

hallow geyser
#

that there are statements that are unsolvable and all that thing

#

so

left sigil
#

I feel cz they are both true and false w.r.t to the observation or reference

#

Like 1+1/2+1/3+1/4+.... Seems finite from a pre-uni one's prospect

#

But is infinite from uni one's prospect

hallow geyser
#

fr

left sigil
#

Just a philosophical example

#

It's like this

hallow geyser
#

then, imagine that there are probles that can't be proven using math

left sigil
#

There are finite no of atoms in this universe

#

But it's volume is infinite ๐Ÿ’€

hallow geyser
#

:/

left sigil
hallow geyser
#

and there is no way of prooving that are unsolvable

left sigil
#

Fr

hallow geyser
#

is;t like a shiti carboard game

left sigil
#

Like for real

#

Take example of riemann hypothesis

hallow geyser
#

imagine being stuck in a problem that can't be solved

#

yeah

left sigil
#

We have checked it for billions of values

#

That for re(z)=1/2

hallow geyser
#

but noo one

left sigil
#

It's a not trivial zero

#

But we can't agree it's true

#

Even if it's true

hallow geyser
#

fr

left sigil
#

So it's neither false nor true simultaneously

hallow geyser
#

bro

left sigil
#

Once a philosopher said

hallow geyser
#

hmm?

left sigil
#

A problem is a problem until it's proved

#

He also said

hallow geyser
#

facts

left sigil
#

Every problem is neither true or false

#

Till it's a problem

hallow geyser
#

imagine saying you proved a problem using numerical proof for some fintie number a

left sigil
#

Yes ok

hallow geyser
#

and then in number a+1 it dosn't work

#

that was something you said last time

left sigil
#

For real

hallow geyser
#

that of computational limitations

#

fo irational numbers

left sigil
#

For real

#

I got it know

hallow geyser
#

that one can say 2.0000000000000000000000000000000000

left sigil
#

Imma say it again

#

For real

#

Let me give an example

#

Assume our computational limit is

#

10^8

hallow geyser
#

k

left sigil
#

,w zeta(242)

dusky sapphireBOT
left sigil
#

Now

#

One with the limit of 10^8 will says it's 1

#

At that time

#

But as computational limit increases one says it's not 1

#

It's irrational

hallow geyser
#

true

left sigil
#

The statement which made sense 20 years ago

#

Does not make sense now

hallow geyser
#

and imagine that the other way around

left sigil
#

Fr

hallow geyser
#

lets say 1.123425675734215658696875 its irrational

left sigil
#

For real

hallow geyser
#

but then you compute a little and see that its just

left sigil
#

We can say its irrational cz we don't have the way to express in terms of ratio

#

Who knows when we advance

hallow geyser
#

1.123425675734215658696875777000000000000000000000

left sigil
#

Yes for real

hallow geyser
#

like its wierd to be studying this and then realizing that

left sigil
#

This is the only reason I said ฯ€+e is irrational

#

They have infinite digits

hallow geyser
left sigil
#

So even if their digits add and become something like

hallow geyser
left sigil
#

A.00000000000000

#

You cannot say it's rational

hallow geyser
#

true

left sigil
#

Cz like u checked till 10^(101)

#

But did u check till 10^(1010)

#

No

hallow geyser
#

facts

left sigil
#

The only reason why we cannot prove Riemann hypothesis numerically

#

Cz u cannot compute infinity numerically

hallow geyser
#

the only reliable proofs are that of logic ones, that fall from theorems

left sigil
#

Yea maybe we can say ฯ€+e tends to a

#

But is not

#

A

hallow geyser
#

true

left sigil
#

The only reason I say

#

ฯ€^ฯ€^ฯ€

#

Can tend to rational

hallow geyser
left sigil
#

But is not

#

Rational

hallow geyser
#

then pi+e=6

left sigil
#

Lmao ๐Ÿคฃ

#

I once heard

#

That

hallow geyser
#

hmm?

left sigil
#

From someone

#

That we are a simple civilization

#

Because of how simple we take things

hallow geyser
#

huh

left sigil
#

He wrote that

#

The small assumptions and neglects we make

#

If we start considering it

#

We may change

#

Like see

hallow geyser
#

k

#

i see

left sigil
#

He said

#

That

#

If for once

#

researchers

#

Lol

#

Don't focus on problem solving

#

But focus how accurately they are solving it

#

We can be better

hallow geyser
left sigil
#

He said the small assumptions we make

#

Make a big difference in reality

#

He said if we can narrow down the difference

hallow geyser
#

man

left sigil
#

We can predict nature

hallow geyser
#

that guy

left sigil
#

Bro he is my friend

#

He is goated